Nimbers

Stream: combinatorial-games

[Violeta Hernández]

@Django Peeters Hi! We have our own channel now.

[Violeta Hernández]

Here's a list of some pending work on nimbers:

• Prove that nimbers are algebraically complete. There's an incomplete argument I co-authored somewhere that I could dig up and resurrect.
• Prove that nimbers below w^w^w are algebraically complete. This should basically boil down to formalizing Lenstra's paper.
• Prove that *(2^x + y) = *(2^x) + *y for y < 2^x. This should be a straightforward inductive argument.
• Characterize nimber multiplication on natural numbers. And as a bonus, define some sort of NatNimber structure we can use to eval expressions with finite nimbers. (Division might be a bit troublesome...)

[Violeta Hernández]

Point 3 seems simple enough, I'll write it down.

[Violeta Hernández]

Just found the branch where I had progress on proving that they were algebraically complete:
https://github.com/leanprover-community/mathlib4/blob/701ad05f447002ee685df2218e1c6c3ca903e7e1/Mathlib/SetTheory/Nimber/Algebraic.lean

[Violeta Hernández]

I think it's simpler (pun intended) to first prove the first simplicity theorem: if ∗x is a group and y < x, then ∗(x + y) = ∗x + ∗y.

[Violeta Hernández]

Then the induction only needs to show that ∗(2 ^ x) is a group for any x.

[Violeta Hernández]

Is there a way to make Ordinal.toNimber x pretty-print as ∗x?

[Violeta Hernández]

Well, I just managed to prove the result on powers of 2. It was not in fact a straightforward inductive argument! Or rather, to give a nicer proof, we needed a bit of preliminary theory.
https://github.com/vihdzp/combinatorial-games/pull/153

[Django Peeters]

Violeta Hernández said:

Is there a way to make Ordinal.toNimber x pretty-print as ∗x?

I think you need prefix " ∗ " => Ordinal.toNimber.

[Violeta Hernández]

I have that, but it still pretty-prints as Ordinal.toNimber.

[Django Peeters]

Maybe it needs a higher precedence?

[Violeta Hernández]

The notation itself works, it just doesn't show up in the infoview
image.png

[Django Peeters]

Any progress?

[Django Peeters]

On another note, I have a paper proof of exercise 12 in Lenstra's "nim multiplication":
$\prod_{i < 2^n}{\ast a + \ast i} = \sum_{\ast j + \ast (n-j)}{(\ast a)^{2^j}}$

[Violeta Hernández]

Django Peeters said:

Any progress?

I went to sleep, I'm going to start work on the second simplest extension theorem now!

[Aaron Liu]

Violeta Hernández said:

The notation itself works, it just doesn't show up in the infoview
image.png

You can write a delab :) Maybe notation3 can do something for you?

[Violeta Hernández]

I don't know how to write delabs :frown:

[Violeta Hernández]

Django Peeters said:

On another note, I have a paper proof of exercise 12 in Lenstra's "nim multiplication":
$\prod_{i < 2^n}{\ast a + \ast i} = \sum_{\ast j + \ast (n-j)}{(\ast a)^{2^j}}$

Nice! I don't think that follows from the simplest extension theorems directly, how do you prove that?

[Django Peeters]

Induction, of course :clown:.
Jokes aside, the exercise itself says "let a,n be naturals" but I think a can be any ordinal. I fidled around with some ideas and the hardest part is controlling the sum.

[Django Peeters]

Btw, am I seeing this wrong, or is my math formatting not working?

[Violeta Hernández]

It doesn't work on mobile, but it works on desktop.

[Violeta Hernández]

@Django Peeters If you have the time, I'd appreciate if you could review my PR on (the first part of) the simplest extension theorem
https://github.com/vihdzp/combinatorial-games/pull/153

[Django Peeters]

I have time today. The multiplication in w^w^w using Python (actually Sage) is coming along too :)

[Django Peeters]

I think this will be my first review.

[Django Peeters]

My review is done.

[Violeta Hernández]

Thanks for the review! And sorry @Tristan Figueroa-Reid for the overeager merge.

[Tristan Figueroa-Reid]

All good! I didn't assign myself as a reviewer on that PR beforehand regardless :+1:

[Violeta Hernández]

I've got part 2 ready at #161, it's quite a smaller PR

[Violeta Hernández]

Currently working on part 3... Dreading having to work with nimber polynomials.

[Violeta Hernández]

Or ordinal polynomials, for that matter!

[Violeta Hernández]

Lean doesn't even let you talk about Ordinal[X], since it requires that polynomials be defined over semirings, which have commutative addition... which Ordinal doesn't.

[Violeta Hernández]

The limitation makes sense. How would Lean know whether to multiply c * x ^ n or x ^ n * c? And in which order to add up the terms?

[Violeta Hernández]

I think what I'll do is just write down the sum explicitly. Is there a definition for ordered sums indexed by naturals in Lean?

[Tristan Figueroa-Reid]

Not sure of anything more specific than List.sum.

[Violeta Hernández]

I think I'll have to use that. There's docs#Finset.sum, but that requires commutativity for obvious reasons.

[Aaron Liu]

docs#Finset.noncommSum

[Aaron Liu]

It only requires the elements of the finset to commute, instead of requiring everything to commute

[Violeta Hernández]

Isn't what I want either. Ordinals only satisfy ω ^ a + ω ^ b = ω ^ b + ω ^ a when a = b.

[Violeta Hernández]

I do need to take the sum in a very specific order

[Violeta Hernández]

So I think I'll just use List.sum

[Aaron Liu]

Oh wait yeah I get it now

[Django Peeters]

Django Peeters said:

I have time today. The multiplication in w^w^w using Python (actually Sage) is coming along too :)

Done. I've tested until excess(107) and it seems there are no bugs. Regardless of my initial plan, the only thing I've used from Sage is some prime-related methods and arbitrary-precision integers (already in python). This may be doable to convert to C or C++.

[Django Peeters]

nim_multiplication.py
Here it is in Python. I've tried for the next prime not cached, but it looks like that would take ~9 days :dizzy: .
PS: I forgot to mark the second argument of power(?,?) for needing arbitrary precision.
Most of this code is a translation from scala of Aaron Siegel's NimFieldCalculator and related files.

[Junyan Xu]

Violeta Hernández said:

Lean doesn't even let you talk about Ordinal[X], since it requires that polynomials be defined over semirings, which have commutative addition... which Ordinal doesn't.

Are you sure you don't want to use Nimber[X] instead?

[Violeta Hernández]

Junyan Xu said:

Violeta Hernández said:

Lean doesn't even let you talk about Ordinal[X], since it requires that polynomials be defined over semirings, which have commutative addition... which Ordinal doesn't.

Are you sure you don't want to use Nimber[X] instead?

I think the better question is whether I don't want to use docs#Ordinal.CNF instead.

[Violeta Hernández]

I've just remembered that exists.

[Violeta Hernández]

I kind of dread working with the cantor normal form. I've tried to refactor that file multiple times but it always ends up stuck in the queue.

[Violeta Hernández]

My idea was to redefine it as an docs#AList, since that more naturally expresses the idea that the CNF is a function from exponents to coefficients, rather than just a list of pairs. I think Mario dissented that it was odd that the entries now had type Σ _, Ordinal instead of just Ordinal × Ordinal. And the PR fell through.

[Violeta Hernández]

Another nice thing about using an AList is that we can then use docs#AList.lookupFinsupp, and express the CNF as a finitely supported function Ordinal →₀ Ordinal.

[Violeta Hernández]

What do you think about these ideas? Should I re-open the PR?

[Violeta Hernández]

If you think this is a good idea, I can put my stuff on the combinatorial-games repo and work on top of it, while I slowly PR it into Mathlib.

[Violeta Hernández]

(I think that's been our modus operandi so far)

[Django Peeters]

Django Peeters said:

Django Peeters said:

I have time today. The multiplication in w^w^w using Python (actually Sage) is coming along too :)

Done. I've tested until excess(107) and it seems there are no bugs. Regardless of my initial plan, the only thing I've used from Sage is some prime-related methods and arbitrary-precision integers (already in python). This may be doable to convert to C or C++.

C++ file is about 40% done. All headers are written down and now I have to implement ImpartialTermAlgebra and some common functions.

[Violeta Hernández]

Violeta Hernández said:

If you think this is a good idea, I can put my stuff on the combinatorial-games repo and work on top of it, while I slowly PR it into Mathlib.

Thought about this a bit more. I think we can keep CNF with its current type and use AList only as an auxiliary definition to define the finsupp variant.

[Violeta Hernández]

Django Peeters said:

C++ file is about 40% done. All headers are written down and now I have to implement ImpartialTermAlgebra and some common functions.

I'm interested in how this algorithm works, I'll check out your Python code a bit later today.

[Django Peeters]

Now ImpartialTermAlgebra is done. The next stretch is implementing the interface between this representation and the ordinals below w^w^w

[Django Peeters]

I think I'm 70% through. 1 big function left to implement and after that it's time to test&debug.

[Django Peeters]

The Python version has some bugs where it adds 2 ordinals in the wrong order (aka the Cantor sum is absorbing some terms which should actually be added in the reverse order)...

[Django Peeters]

But only the interface between the algebra and the ordinals I think (e.g. alpha).

[Django Peeters]

I've looked at the code and alpha may be the only function with an error in the Python version.

[Django Peeters]

Nope, the Python version (hence also the C++ version) has more errors. excess and qSet seem to fail both on some primes so I think I translated ImpartialTermAlgebra wrong from Scala...

[Django Peeters]

It should be fixed now. I'm gonna send the corrected Python version again and put my debugged C++ version on github later.

[Django Peeters]

nim_multiplication.py is bug-free I hope.

[Django Peeters]

Mostly translation bugs. Now time for github.

[Django Peeters]

It's on github. So we essentially want something like this, but in Lean and appropriate API, right? And hopefully a program that's efficient enough for most calculations.

[Django Peeters]

I'm gonna try and make it faster for now.

[Django Peeters]

I made it faster.

[Django Peeters]

I still have some optimization ideas left, but tomorrow I'll try to do some formalization in Lean. Maybe characterize finite nimber multiplication.

[Django Peeters]

Violeta Hernández said:

Here's a list of some pending work on nimbers:

• Prove that nimbers are algebraically complete. There's an incomplete argument I co-authored somewhere that I could dig up and resurrect.
• Prove that nimbers below w^w^w are algebraically complete. This should basically boil down to formalizing Lenstra's paper.
• Prove that *(2^x + y) = *(2^x) + *y for y < 2^x. This should be a straightforward inductive argument.
• Characterize nimber multiplication on natural numbers. And as a bonus, define some sort of NatNimber structure we can use to eval expressions with finite nimbers. (Division might be a bit troublesome...)

I don't know how to do the NatNumber though.

[Aaron Liu]

how does division work

[Django Peeters]

You could do it the naive way, but there's also a recursive version which is already formalized in the repository.

[Violeta Hernández]

Aaron Liu said:

how does division work

I don't know of an algorithm to find nim inverses of natural numbers. There's a sequence on OEIS but it isn't very helpful either. AFAIK the best way is to just use Nat.find.

[Violeta Hernández]

Django Peeters said:

You could do it the naive way, but there's also a recursive version which is already formalized in the repository.

This definition characterizes the inverse as the least nimber not in some infinite set, so I don't think it can be used for computation.

[Violeta Hernández]

https://oeis.org/A051917

[Violeta Hernández]

In fact this sequence only goes up to 255. Surely it's not too difficult to extend it to the square?

[Violeta Hernández]

I just came across this
https://docs.rs/nimber/0.1.1/src/nimber/finite.rs.html

[Violeta Hernández]

It gives algorithms for nim sums, products, inverses, and even roots of quadratic polyonmials for finite nimbers.

[Violeta Hernández]

This would be really cool to formalize in Lean!

[Violeta Hernández]

It seems like the general idea for nimber computation with naturals is to split $n = a 2^{2^t} + b$, where $a, b < 2^{2^t}$. Since Fermat two powers work really nicely under all the arithmetical operations, we're able to recurse on $t$.

[Violeta Hernández]

To multiply two nimbers, you can use the formula:

[Violeta Hernández]

(deleted)

[Violeta Hernández]

image.png

[Violeta Hernández]

This reduces the calculation to either nimber multiplications with a smaller value of t, or nimber multiplications $x2^{2^t}$ where $x < 2^{2^t}$, which are equal to the usual multiplication on natural numbers.

[Violeta Hernández]

In fact, plugging in $c=2^{2^t-1}$ and $d=0$, you get this

[Violeta Hernández]

370964201478553600.png

[Violeta Hernández]

Which implies you only need a subroutine to multiply $x < 2^{2^t}$ by $2^{2^t-1}$

[Violeta Hernández]

Now, the problem. How do we actually show 2^2^n is a ring?

[Aaron Liu]

show it's a docs#Subring of the nimbers

[Violeta Hernández]

No I mean how do we prove that 2^2^n IsRing

[Violeta Hernández]

I'm sure there has to be some simple proof involving the simplicity theorems

[Violeta Hernández]

Actually, I presume you have to simultaneously prove IsRing (2^2^n) and (2^2^n)^2 = 2^(2^n-1) + 2^2^n

[Violeta Hernández]

Those two claims, alongside the multiplication formula I gave, suffice to prove IsRing (2^2^(n+1))

[Violeta Hernández]

So really the question is how to prove that identity about (2^2^n)^2.

[Aaron Liu]

follow your nose

[Violeta Hernández]

Let me get my bourbon

[Violeta Hernández]

Oh oops notebook I meant notebook

[Aaron Liu]

what are the excludents for (2^2^n)^2?

[Violeta Hernández]

(a + b) * 2^2^n + a * b for a, b < 2^2^n

[Violeta Hernández]

These are not pretty

[Violeta Hernández]

In fact to even prove nimbers < 2^2^n are in there, you have to prove 2^2^n is quadratically closed

[Aaron Liu]

oh it can't be too hard

[Violeta Hernández]

Oh it totally isnt

[Violeta Hernández]

In fact, every IsField x is quadratically closed

[Violeta Hernández]

Since the square function is injective on char 2 fields

[Aaron Liu]

oh yeah right

[Aaron Liu]

it looks like you also need a certain amount of independence

[Aaron Liu]

linear independence of characters

[Violeta Hernández]

Hm yeah actually I'm starting to doubt what I said

[Violeta Hernández]

There's no way any field of char 2 is quadratically closed

[Violeta Hernández]

Every finite field of char 2 should be

[Aaron Liu]

since injective implies surjective

[Violeta Hernández]

So nimbers below 2^2^n are excludents for (2^2^n)^2

[Aaron Liu]

now try ones between 2^2^n and 2^2^n * 2

[Violeta Hernández]

Those are of the form 2^2^n + a * (a + 1)

[Violeta Hernández]

For a < 2^2^n

[Violeta Hernández]

...what's the quadratic formula in a char 2 field?

[Aaron Liu]

no dividing by two

[Aaron Liu]

the quadratic formula won't serve you here

[Violeta Hernández]

Damn

[Aaron Liu]

why is that the first thing you thought of

[Violeta Hernández]

Well, we want to find the first k such that a^2 + a + k has no root below 2^2^n

[Aaron Liu]

so far I know that a * (a + 1) on 2^2^n is not identically zero

[Violeta Hernández]

Wait I have an idea

[Aaron Liu]

oh I'm listening

[Violeta Hernández]

What if we substitute b = 2^(2^n - 1) + a in there?

[Aaron Liu]

you're the one with the notepad

[Violeta Hernández]

I mean the idea is that if a * (a + 1) is not surjective, it should take exactly half of all values

[Violeta Hernández]

Though maybe I'm thinking too much about fields that aren't char 2

[Violeta Hernández]

What is (2^(2^n - 1))^2 equal to?

[Aaron Liu]

maybe we can set up something explicit with docs#GaloisField

[Violeta Hernández]

No idea how

[Aaron Liu]

maybe with some of those simplest extension theorems

[Aaron Liu]

do we know that powers of two are closed under addition?

[Violeta Hernández]

We do

[Violeta Hernández]

https://vihdzp.github.io/combinatorial-games/CombinatorialGames/Nimber/Simplicity.html#Nimber.IsGroup.two_opow

[Aaron Liu]

do we know that nimbers are an integral domain

[Violeta Hernández]

We know that they're a field

[Aaron Liu]

ok that works

[Violeta Hernández]

Hmmm

[Violeta Hernández]

(k + a) * (k + a + 1) = k * (k + 1) + a * (a + 1)

[Violeta Hernández]

That's a very peculiar equation

[Violeta Hernández]

It implies the image of a * (a + 1) is closed under addition

[Aaron Liu]

it implies fun a => a * (a + 1) is an AddMonoidHom

[Aaron Liu]

what happens when you change 1 to some other nimber

[Violeta Hernández]

(a + b) * (a + b + c) = a^2 + (a + b) * c + b^2

[Violeta Hernández]

Not quite as nice

[Aaron Liu]

= a * (a + c) + b * (b + c) I don't know what you're talking about this is plenty nice

[Violeta Hernández]

Oh you're right lol

[Violeta Hernández]

Whoa

[Aaron Liu]

wait it's the sum of squaring (which is an add monoid hom in char 2) and multiplying by c so obviously it's an add monoid hom

[Violeta Hernández]

Oh yeah

[Violeta Hernández]

Makes sense

[Aaron Liu]

so so far we have

• fun a => a * (a + c) add monoid hom
• fun a => a * (a + c) not identically zero

[Violeta Hernández]

Wait I just realized something which makes me feel dumb

[Violeta Hernández]

a * (a + 1) = (a + 1) * (a + 1 + 1)

[Violeta Hernández]

So if we have a root r of a * (a + 1) + k, the other root is r + 1

[Aaron Liu]

this is the truth

[Junyan Xu]

You're discovering some Artin--Schreier theory

For IsField (2^2^n) have you read the proof of theorem 49 on ONAG

[Violeta Hernández]

Oh I probably should have started there

[Aaron Liu]

oh it was already done 100 years ago

[Aaron Liu]

Junyan Xu said:

For IsField (2^2^n) have you read the proof of theorem 49 on ONAG

I knew about this but discovering is so much fun

[Junyan Xu]

There's also something in VIII.4 of Siegel's CGT
possibly a more organized approach to simplest extension theorems

[Violeta Hernández]

Oh so what you prove inductively is that 2^2^n is a field, and that x^2 + x takes on the values less than 2^(2^n - 1)

[Violeta Hernández]

Junyan Xu said:

There's also something in VIII.4 of Siegel's CGT
possibly a more organized approach to simplest extension theorems

That's the reference I've used to formalize them, Conway's argument actually has an error in it!

[Violeta Hernández]

https://mathoverflow.net/questions/477566/clarification-on-proof-of-the-algebraic-completeness-of-nimbers

[Violeta Hernández]

Violeta Hernández said:

Oh so what you prove inductively is that 2^2^n is a field, and that x^2 + x takes on the values less than 2^(2^n - 1)

Oh and actually you need the very last simplicity theorem to do this

[Violeta Hernández]

Since what you do is show that (2^2^n)^2 + 2^2^n = 2^(2^n - 1) by using that x^2 + x + 2^(2^n - 1) is the lexicographically earliest polynomial without a root in 2^2^n

[Violeta Hernández]

I wasn't expecting nimber arithmetic on naturals to be this difficult to pin down

[Aaron Liu]

it's just a finite field

[Violeta Hernández]

I mean yeah

[Violeta Hernández]

But as to actually proving that

[Aaron Liu]

I wonder if there are char-p analogues for p ≠ 2

[Violeta Hernández]

https://arxiv.org/abs/1108.0962

[Violeta Hernández]

Haven't read this thoroughly, but it seems like the construction is quite different in a few regards

[Violeta Hernández]

ON the topic of the last simplicity theorem, I've been having trouble even just writing down a lemma that I need for it
image.png

[Violeta Hernández]

I'm thinking, perhaps we can make a new predicate IsNthDegClosed n x for nimbers that contain roots of all non-constant polynomials with coefficients < n, and redefine IsAlgClosed x as ∀ n, IsNthDegClosed n x

[Violeta Hernández]

(better name suggestions welcome)

[Violeta Hernández]

Then we could state results such as IsNthDegClosed 2 (∗ω)

[Violeta Hernández]

Yeah, this IsNthDegClosed predicate was a great idea

[Violeta Hernández]

There's a lot of convenience results to be proven

[Violeta Hernández]

Such that an IsNthDegClosed 1 ring is a field

[Violeta Hernández]

Or that to prove IsNthDegClosed n x, for a field x, you just need to check monic polynomials

[Violeta Hernández]

https://github.com/vihdzp/combinatorial-games/pull/190

[Violeta Hernández]

So we can now state lemma 4.4.c' in Siegel as

theorem IsNthDegreeClosed.foldr_mul_add_eq_of_lt {n : ℕ} {x : Nimber} (h : IsNthDegreeClosed n x)
    (l : List Nimber) (hl : ∀ y ∈ l, y < x) (hln : l.length ≤ n + 1) :
    l.zipIdx.foldr (fun a b ↦ a.1 * x ^ a.2 + b) 0 =
    of (l.zipIdx.foldr (fun a b ↦ a.1.val * x.val ^ a.2 + b) 0) := by
  sorry

[Violeta Hernández]

Actually, it's probably idea to wrap "evaluate a nimber polynomial with ordinal operations" as its own function; then we can do

/-- Evaluate a nimber polynomial using ordinal arithmetic. -/
def ordinalEval (p : Nimber[X]) (x : Nimber) : Nimber :=
  of <| (List.range (p.natDegree + 1)).foldr (fun k a ↦ a + (p.coeff k).val * x.val ^ k) 0

theorem IsNthDegreeClosed.eval_eq_of_lt {n : ℕ} {x : Nimber} (h : IsNthDegreeClosed n x)
    {p : Nimber[X]} (hpn : p.degree ≤ n) (hp : ∀ k, p.coeff k < x) :
    p.eval x = ordinalEval p x := by
  sorry

[Violeta Hernández]

Made good progress on this tonight

[Violeta Hernández]

I want to say that we'll know that nimbers are algebraically closed before next week

[Violeta Hernández]

Just found this really weird lemma which generalizes a lot of the arithmetic from both theorems 4.4b' and 4.4c'
image.png

[Violeta Hernández]

Really proud of myself, finding this saved me like an hour

[Aaron Liu]

congrats

[Violeta Hernández]

my next class is in 2 hours

[Violeta Hernández]

i'll see if i can finish 4.4c' proper in that timespan

[Django Peeters]

This one is not explicitely mentioned in CGT:20250813_165846.jpg

[Violeta Hernández]

You mean Theorem 44 from ONAG?

[Violeta Hernández]

That one is outright wrong

[Django Peeters]

Yea, this is supposed to be a corrected version

[Django Peeters]

Maybe from Lenstra

[Violeta Hernández]

Wait, what's the difference between the two statements?

[Django Peeters]

Let me find my copy of ONAG

[Violeta Hernández]

image.png
Here's a screencap for you

[Django Peeters]

Thx

[Django Peeters]

Iirc the proof in ONAG is wrong according to the note on p.63

[Django Peeters]

What's up with WithTop.lean failing to build?

[Violeta Hernández]

Damn it, I probably forgot to lake exe mk_all

[Violeta Hernández]

oh i see the mistake

[Violeta Hernández]

it's a small fix, I'll push it directly to master

[Violeta Hernández]

should work now

[Violeta Hernández]

i was nearly able to prove that the induction hypothesis implies x^n = of (x.val ^ n)

[Violeta Hernández]

i'll continue later

[Django Peeters]

I'm gonna start on the w^w^w point and name it WWW.lean for now.

[Violeta Hernández]

image.png
Holy I actually did it

[Violeta Hernández]

this is like 90% of the way to the last part of the simplicity theorem

[Violeta Hernández]

I think I genuinely need a break though, will continue tomorrow

[Django Peeters]

I got this far lol
image.png

[Violeta Hernández]

Surely there's some simp lemma stating nth Prime 0 = 2?

[Violeta Hernández]

There's no way that isn't in Mathlib

[Violeta Hernández]

Btw for alpha and kappa you're missing a few of

[Tristan Figueroa-Reid]

Violeta Hernández said:

Surely there's some simp lemma stating nth Prime 0 = 2?

Surprisingly there isn't!

[Violeta Hernández]

Huh

[Violeta Hernández]

docs#Nat.nth_prime_zero_eq_two

[Tristan Figueroa-Reid]

??

[Violeta Hernández]

There it is

[Tristan Figueroa-Reid]

https://loogle.lean-lang.org/?q=Nat.nth%2C+Prime turned up empty for me - have I been using loogle wrong this entire time?

[Tristan Figueroa-Reid]

nth_prime_zero_eq_two isn't new either. That's really strange..

[Violeta Hernández]

Nat.Prime and Prime are two different predicates

[Django Peeters]

Violeta Hernández said:

Btw for alpha and kappa you're missing a few of

Then why isn't it complaining?

[Violeta Hernández]

Nimbers are definitionally equal to ordinals

[Violeta Hernández]

I'm not sure if you've worked with other type aliases before, like docs#OrderDual

[Violeta Hernández]

The order dual of a type is definitionally equal to the type, what changes are the instances on the type

[Django Peeters]

No, I've never worked with type aliases before

[Violeta Hernández]

So for instance you have of x = x and of y = y, but if you try to write down of x + of y = x + y you'll run into issues, because these are two different additions

[Django Peeters]

I don't see why that would be a problem here

[Aaron Liu]

It can stop simp from working

[Violeta Hernández]

If you start abusing def-eqs, you might run into difficult to debug instance problems, and simp might not be able to solve some goals

[Aaron Liu]

If you do defeq abuse

[Aaron Liu]

Also tactics like abel and ring

[Django Peeters]

So this is actually us abusing Lean in a way?

[Violeta Hernández]

Not really, this is just a nuanced pattern to be aware of

[Violeta Hernández]

We could prohibit def-eq abuse altogether by tagging Nimber as irreducible

[Violeta Hernández]

But there are some situations where making use of def-eq is fine and ends up simplifying proofs

[Violeta Hernández]

So I've chosen not to prohibit it

[Django Peeters]

Hmm

[Django Peeters]

Okay

[Django Peeters]

Where did you learn this

[Violeta Hernández]

Just by working with OrderDual really

[Aaron Liu]

I have no idea where I learned this

[Aaron Liu]

I just learned it

[Violeta Hernández]

You can often prove order-swapped versions of theorems using one-liners with OrderDual def-eq abuse

[Violeta Hernández]

But you have to be careful to keep the public API abuse free, lest you run into the aforementioned problems

[Violeta Hernández]

Similar remarks go for docs#toLex

[Aaron Liu]

Is (dual a < dual b) = (b < a) and other similar defeqs really defeq abuse?

[Violeta Hernández]

Def-eq abuse is a spectrum

[Violeta Hernández]

You can do weirder things like IsLUB (a := a^od) = IsGLB (a := a) or IsSuccLimit (a := a^od) = IsPredLimit (a := a)

[Violeta Hernández]

I guess the idea is that as long as the "abuse" is contained within proofs, it's fine

[Django Peeters]

Django Peeters said:

I got this far lol
image.png

Any other remarks on these definitions?

[Violeta Hernández]

alpha and kappa shouldn't take those nth Prime k = _ arguments

[Violeta Hernández]

You can define them in terms of k, rather than p

[Aaron Liu]

especially not implicitly, since they're impossible to infer

[Django Peeters]

But what if I want the definition to align with it's literature counterpart.

[Aaron Liu]

just don't

[Violeta Hernández]

Lean defines a / 0 = 0

[Django Peeters]

That's different

[Violeta Hernández]

Obviously we try to align things to literature, but there are other considerations

[Aaron Liu]

you're taking q along with a proof saying q = ...

[Aaron Liu]

The only time I have ever done this was for performance reasons

[Violeta Hernández]

I think it will make our life much easier if we define alpha k for all k, rather than just primes

[Django Peeters]

Interesting, I'll continue tomorrow

[Violeta Hernández]

Would it be a design antipattern to define a LinearOrder (Nimber[X]) instance?

[Violeta Hernández]

I'm from the Rust school of thought "don't add upstream instances on upstream types" but I imagine Lean has other considerations

[Violeta Hernández]

Surely no one will ever PR a Preorder (R[X]) instance that just breaks all my code, right?

[Tristan Figueroa-Reid]

My impression of Lean is that they avoid the orphan rule since newtyping is much more costly than it is in Rust. (afaik this is borrowed from haskell and the haskell community is still split on the orphan rule)

I would say go for making a linear order on Nimber polynomials? As long as there isn't some canonical ordering on generall polynomials.

[Violeta Hernández]

Having a CompleteLinearOrder instance is definitely much nicer than to work with the "raw" order relation.

[Violeta Hernández]

Specifically, this was my idea:

/-- Returns the lexicographically earliest polynomial, all of whose coefficients are less than `x`,
without any roots less than `x`. If none exists, returns `⊤`.

This function takes values on `WithTop (Nimber[X])`, which is a well-ordered complete lattice (the
order on `Nimber[X]` is the lexicographic order). -/
noncomputable def simplestIrreducible (x : Nimber) : WithTop (Nimber[X]) :=
  sInf (WithTop.some '' {p | 0 < p.degree ∧ (∀ k, p.coeff k < x) ∧ ∀ r < x, ¬ p.IsRoot r})

[Violeta Hernández]

As it turns out, most of the results I was proving on IsNthDegreeClosed can be re-stated to take a p < simplestIrreducible x assumption, rather than p.degree ≤ n

[Violeta Hernández]

Of course, the former implies the latter on an n-th degree closed field

[Violeta Hernández]

(The design of the proof is still rapidly changing, I think I won't be PR-ing anything until I have the full proof)

[Django Peeters]

Do we already have if Nimbers *a and *b are fields, a < b, then b = a^c for some Ordinal c?

[Django Peeters]

( Mod our dummy statement that *0 is a field)

[Violeta Hernández]

Django Peeters said:

Do we already have if Nimbers *a and *b are fields, a < b, then b = a^c for some Ordinal c?

We don't!

[Violeta Hernández]

How do you prove that?

[Django Peeters]

I don't know, but I think a bit of field theory and then relating it to the order type with some basis-argument.

[Django Peeters]

And iirc both Lenstra and Aaron Siegel give this as an exercise without proof.

[Django Peeters]

So no peeking

[Violeta Hernández]

I'll take a look at that later

[Django Peeters]

Where should we put this theorem?

[Violeta Hernández]

Realistically we should probably split the simplicity file at some point

[Violeta Hernández]

But I think we should prove first and organize later!

[Django Peeters]

Works for me!

[Violeta Hernández]

Violeta Hernández said:

Realistically we should probably split the simplicity file at some point

My work branch has 1300 lines of code in here...

[Violeta Hernández]

My idea is to split it as:

• Simplicity.Basic: the first 3 parts of the simplicity theorem
• Simplicity.Polynomial: various auxiliary definitions and theorems about nimber polynomials
• Simplicity.Algebraic: the last part of the simplicity theorem + the proof nimbers are algebraically complete

[Violeta Hernández]

Also you have to start correcting me when I call it the simplicity theorem instead of the simplest extension theorem, haha

[Violeta Hernández]

Or the name might actually stick!

[Django Peeters]

I also saw "algebraically complete", which should be "algebraically closed".

[Violeta Hernández]

Really? I've been working on this for this long and you haven't corrected me on that? :rofl:

[Violeta Hernández]

TIL I've been calling these the wrong thing my whole life

[Aaron Liu]

Django Peeters said:

I don't know, but I think a bit of field theory and then relating it to the order type with some basis-argument.

do we know what the multiples of an element look like

[Django Peeters]

Violeta Hernández said:

Really? I've been working on this for this long and you haven't corrected me on that? :rofl:

I just noticed it yesterday lol

[Django Peeters]

Aaron Liu said:

Django Peeters said:

I don't know, but I think a bit of field theory and then relating it to the order type with some basis-argument.

do we know what the multiples of an element look like

No, but my gut tells me to use the simplest extension theorems.

[Django Peeters]

Or yes, kinda

[Django Peeters]

If we take an a-basis {b_i | i in I} for b, then the elements in b look like a_1*b_1 +...+a_n*b_n.

[Django Peeters]

As Nimbers

[Aaron Liu]

Django Peeters said:

If we take an a-basis {b_i | i in I} for b, then the elements in b look like a_1*b_1 +...+a_n*b_n.

The definition of a basis?

[Django Peeters]

What

[Violeta Hernández]

aesop is incredible
image.png

[Django Peeters]

If it works it ain't broken, lol

[Violeta Hernández]

Oh I'm definitely not complaining!

[Aaron Liu]

I just hope it's fast

[Violeta Hernández]

Didn't need extra heartbeats or anything

[Django Peeters]

Polynomial.coeff_eq_zero_of_degree_lt is causing name collisions. It's already in mathlib.

[Aaron Liu]

If it's the same statement then you can just get rid of it

[Django Peeters]

I'll see what I can do

[Violeta Hernández]

Django Peeters said:

Polynomial.coeff_eq_zero_of_degree_lt is causing name collisions. It's already in mathlib.

I've left a lot of auxiliary theorems all over the repository, the idea is for them to slowly make their way into Mathlib

[Violeta Hernández]

If they're already there you can just remove them!

[Violeta Hernández]

Why is ω quadratically closed?

[Aaron Liu]

since every quadratic polynomial with coefficients in ω has a root

[Violeta Hernández]

As far as I can tell, ONAG only really proves that it contains roots for x^2 + x + n

[Aaron Liu]

it's probably equivalent

[Violeta Hernández]

Oh it's yet another thing Siegel outright explains better
image.png

[Aaron Liu]

oh no we're in char 2

[Violeta Hernández]

Here $f_\alpha = x^2 + x + \alpha$

[Aaron Liu]

oh I remember this argument

[Django Peeters]

Is there a way to not make your VSCode look like a battle field

[Aaron Liu]

Django Peeters said:

Is there a way to not make your VSCode look like a battle field

what is battling

[Tristan Figueroa-Reid]

(if this is about the infoview) clearing hypotheses you don't need while writing proofs helps somewhat, but it's a double-edged sword.

[Violeta Hernández]

Once we implement NatNimber, we should implement a function to solve quadratic equations

[Violeta Hernández]

The Rust repo I sent earlier has an algorithm for that, though I haven't quite combed through it

[Aaron Liu]

Violeta Hernández said:

Once we implement NatNimber, we should implement a function to solve quadratic equations

like a function or a tactic

[Violeta Hernández]

Computable function

[Violeta Hernández]

But using it to make a norm_num impl would be a great goal as well!

[Aaron Liu]

I hope norm_num will work for this

[Aaron Liu]

oh we can do division too

[Aaron Liu]

and addition and multiplication and subtraction

[Violeta Hernández]

yep, repo contains an algorithm for that as well

[Violeta Hernández]

(and square roots)

[Tristan Figueroa-Reid]

(I haven't used CGSuite much, but perhaps CGSuite has some impls for this as well? I'm not inclined to trust the correctness of a random rust library)

[Aaron Liu]

Tristan Figueroa-Reid said:

(I haven't used CGSuite much, but perhaps CGSuite has some impls for this as well? I'm not inclined to trust the correctness of a random rust library)

the fun part is proving correctness

[Violeta Hernández]

does CGSuite have browsable docs?

[Django Peeters]

Do you use attributes for this to work with norm_num?

[Aaron Liu]

you write a norm_num extension

[Tristan Figueroa-Reid]

Violeta Hernández said:

does CGSuite have browsable docs?

Don't think so (nothing on the cgsuite site at least), but it's just Scala/Java so you can auto-generate some locally.

[Aaron Liu]

examples can be found in file#Tactic/NormNum/Basic

[Tristan Figueroa-Reid]

Tristan Figueroa-Reid said:

Violeta Hernández said:

does CGSuite have browsable docs?

Don't think so (nothing on the cgsuite site at least), but it's just Scala/Java so you can auto-generate some locally.

(oh! Misinterpeted this - if you open the CGSuite app there are docs locally)

[Django Peeters]

Aaron Liu said:

examples can be found in file#Tactic/NormNum/Basic

Thank you. This is all Chinese for me.

[Violeta Hernández]

Aaron Liu said:

examples can be found in file#Tactic/NormNum/Basic

This seems to depend mostly on the docs#Mathlib.Meta.NormNum.IsNat structure. Is that a problem? ∗n really is not the same as ↑n.

[Django Peeters]

How fast can Lean algorithms be compared to Rust or C?

[Tristan Figueroa-Reid]

I'm unaware of any lean4 benchmark, but I don't think it particularly matters? Lean4 does produce native artifacts (does anyone know if lean4 uses LLVM as its backend?)

[Violeta Hernández]

There's obviously some tradeoff between correctness and performance

[Violeta Hernández]

But if you're using native_decide or #eval then Lean isn't half bad

[Aaron Liu]

Violeta Hernández said:

Aaron Liu said:

examples can be found in file#Tactic/NormNum/Basic

This seems to depend mostly on the docs#Mathlib.Meta.NormNum.IsNat structure. Is that a problem? ∗n really is not the same as ↑n.

oh no

[Violeta Hernández]

Though, maybe that just means we have to search for patterns e.g. ∗_ + ∗_ rather than just _ + _

[Violeta Hernández]

Idk, I can't make out most of what that file is doing

[Violeta Hernández]

We already have a proof that nimber natural addition is the same as XOR, if anyone wants to give writing this tactic a try

[Aaron Liu]

Violeta Hernández said:

Though, maybe that just means we have to search for patterns e.g. ∗_ + ∗_ rather than just _ + _

that also doesn't work

[Django Peeters]

No thx, it'd take me like a whole year

[Aaron Liu]

you need to output one of the following

• nat
• int
• nnrat
• rat
• True
• False

[Tristan Figueroa-Reid]

oh

[Aaron Liu]

there are no other options at the moment

[Violeta Hernández]

That sucks

[Violeta Hernández]

Wait, but that's not the full story, is it?

[Aaron Liu]

so you can't output *n

[Violeta Hernández]

norm_num afaik works with Ordinal which is clearly not one of those types

[Violeta Hernández]

(with finite ordinals, that is)

[Aaron Liu]

well it outputs nat via natcast

[Aaron Liu]

and of course by int I really mean intcast of an int literal, and nnrat cast and rat cast

[Aaron Liu]

but True and False are literally just docs#True and docs#False

[Aaron Liu]

no casting going on there

[Violeta Hernández]

Perhaps we can create a wrapper? Maybe to prove an equality on Nimber we first cast it to ℕ in terms of new functions nim_add, nim_mul, etc. Then we write norm_num extensions for those functions, then we cast back to Nimber.

[Aaron Liu]

that could work

[Tristan Figueroa-Reid]

Though for what benefit? It seems nicer to extend norm_num in the first place (or create a more abstract base and specialize norm_num to its specific Nat/Int/Rat/... types) unless we need a norm_num-like tactic for something.

[Violeta Hernández]

Well, I don't really know what's possible here

[Violeta Hernández]

I'm trying to figure out what the limits of norm num are

[Violeta Hernández]

Perhaps that's worth opening a thread elsewhere

[Django Peeters]

Would it be possible to do something similar for arithmetic in w^w^w?

[Violeta Hernández]

We'd have to use docs#NONote for that

[Violeta Hernández]

but in theory we could

[Violeta Hernández]

well, not the norm_num thing

[Violeta Hernández]

rather the "computable arithmetic" thing

[Violeta Hernández]

That file is a mess. I did a complete rewrite of it about a year ago, but it's 1k lines of code so I've been hesitant on PRing it

[Violeta Hernández]

Violeta Hernández said:

norm_num afaik works with Ordinal which is clearly not one of those types

Hm, I think this only really works because arithmetic on finite ordinals is exactly the same as arithmetic on natural numbers

[Violeta Hernández]

In fact, it doesn't even seem to work for multiplication... only addition

[Aaron Liu]

someone should write a norm_num extension

[Aaron Liu]

for multiplication

[Aaron Liu]

on finite ordinals

[Aaron Liu]

Violeta Hernández said:

In fact, it doesn't even seem to work for multiplication... only addition

probably because the extension assumes some sort of ring, and ordinal addition is noncommutative

[Aaron Liu]

for addition you only need docs#AddMonoidWithOne which doesn't assume commutative addition

[Violeta Hernández]

I just found this: https://github.com/leanprover-community/mathlib4/blob/ed5dc595cc78f827821728092d34c935aa261d9c/Mathlib/Tactic/NormNum/LegendreSymbol.lean

[Violeta Hernández]

Potentially a good template for what we'll be doing

[Violeta Hernández]

Aaron Liu said:

for multiplication

oh my god, literally three days ago
https://github.com/leanprover-community/mathlib4/blob/ed5dc595cc78f827821728092d34c935aa261d9c/Mathlib/Tactic/NormNum/Ordinal.lean#L35

[Violeta Hernández]

oh nvm, the change 3 days ago was just fixing a typo

[Violeta Hernández]

still, literally 3 weeks ago

[Violeta Hernández]

From reading these examples I think I've gotten a better idea of how norm_cast works

[Violeta Hernández]

And yeah, the design means you actually just can't return something that isn't NatCast.

[Violeta Hernández]

Although. Perhaps we can sidestep this by defining an internal type Nimber' inheriting the operations of Nimber but with natCast n re-defined as ∗n.

[Violeta Hernández]

Then the norm cast extension would return .isNat for this Nimber' type, except that whatever it proves would be def-eq to the actual claim on Nimber.

[Violeta Hernández]

i.e. this type would be entirely internal to the norm_num extension

[Violeta Hernández]

I might be saying nonsense though

[Tristan Figueroa-Reid]

No, that's a good design pattern if you want to explicitly avoid modifying norm_num.

[Aaron Liu]

Violeta Hernández said:

From reading these examples I think I've gotten a better idea of how norm_cast works

norm_num

[Violeta Hernández]

that

[Violeta Hernández]

hate that we have two very different tactics with very similar names

[Violeta Hernández]

Well, in any case, we can't create this tactic if we haven't even defined the operations!

[Violeta Hernández]

I am beyond exhausted at the moment, but I think I've proven basically everything required for the final simplest extension theorem

[Violeta Hernández]

Will be out of town tomorrow but by Sunday that should be done

[Violeta Hernández]

(really I'd finish this tonight if I wasn't running on like 4 hours of sleep)

[Tristan Figueroa-Reid]

(is the proof attempt on a branch?)

[Violeta Hernández]

vi.simplicity_5

[Violeta Hernández]

(i'm not good at names)

[Aaron Liu]

Violeta Hernández said:

(really I'd finish this tonight if I wasn't running on like 4 hours of sleep)

That's not good

[Aaron Liu]

Violeta Hernández said:

(i'm not good at names)

It's not a bad name

[Tristan Figueroa-Reid]

(If anyone else also wants to try to fill in the sorrys, I'll be happy to help golf as well.)

[Violeta Hernández]

I have a very clear idea of how to fill in the last two sorries

[Violeta Hernández]

What I do need help with is moving all the random auxiliary results I proved to Mathlib!

[Aaron Liu]

I am also good at filling in sorrys, even when I only have a vague idea of a proof

[Aaron Liu]

But of course it's a lot easier to follow an existing proof

[Violeta Hernández]

If you really want to work with proofs, I think you could start proving that IsField (∗2^2^n)

[Aaron Liu]

However I dislike trying to figure out the best way to state something

[Violeta Hernández]

Actually you know what I've got nothing going on in the next two hours

[Violeta Hernández]

I'll grab some coffee and see what I can do in that timeframe

[Aaron Liu]

You could try catching up on some sleep instead

[Violeta Hernández]

I wish

[Violeta Hernández]

I have to go to the capital in two hours :frown:

[Aaron Liu]

Violeta Hernández said:

If you really want to work with proofs, I think you could start proving that IsField (∗2^2^n)

Do you have a statement I can use

[Aaron Liu]

A statement in Lean

[Violeta Hernández]

proof_wanted IsField.two_two_pow (n : ℕ) : IsField (∗(2 ^ 2 ^ n))

[Aaron Liu]

Great themst works

[Violeta Hernández]

of course the very first thing you need to do is prove

proof_wanted IsRing.toIsField_of_finite {n : ℕ} (h : IsRing (∗n)) : IsField (∗n)

[Violeta Hernández]

Which is the injectivity → surjectivity argument

[Aaron Liu]

That's trivial

[Violeta Hernández]

yes, but we haven't written it down yet

[Aaron Liu]

ok

[Violeta Hernández]

Another lemma we should prove is

proof_wanted IsRing.X_sq_add_X_lt_simplestIrreducible {n : ℕ} (h : IsRing (∗n)) :
    X^2 + X < simplestIrreducible (∗n)

[Violeta Hernández]

where simplestIrreducible is the definition on my branch

[Violeta Hernández]

X^2 ≤ _ follows from IsField.X_sq_le_simplestIrreducible which I just pushed

[Violeta Hernández]

X^2 + C a ≠ _ is because these fields have all square roots

[Violeta Hernández]

X^2 + X ≠ _ is because of coeff_simplestIrreducible_zero_ne

[Aaron Liu]

which branch do I go to if I want to start proving

[Aaron Liu]

for some reason master isn't building

[Tristan Figueroa-Reid]

vi.simplicity_5

[Violeta Hernández]

Master isn't building? Again?

[Violeta Hernández]

So much for CI

[Tristan Figueroa-Reid]

I'll see if I can reproduce that - if it isn't building locally, it's a CI issue since CI isn't reproducing it.

[Violeta Hernández]

will re-run CI on master to see if it says anything

[Tristan Figueroa-Reid]

(it shouldn't - I would be much more concerned if it failed)

[Tristan Figueroa-Reid]

Aaron Liu said:

for some reason master isn't building

It compiles fine for me locally - what was the error that you got?

[Violeta Hernández]

Btw I know one of you will complain about this but

[Violeta Hernández]

The reason I made simplestIrreducible take values in WithTop rather than just assigning the dummy value 0 is so that statements of the form "for all p < simplestIrreducible" hold for all polynomials in the alg closed case, rather than just being vacuously true

[Violeta Hernández]

Also, simplestIrreducible is uncomfortably long for a name

[Violeta Hernández]

Suggestions welcome

[Aaron Liu]

Well I have finite subring -> subfield now

import Mathlib

def Subring.toSubfieldOfFinite {α : Type u} [DivisionRing α] (s : Subring α)
    (hs : Set.Finite (SetLike.coe s)) : Subfield α where
  toSubring := s
  inv_mem' := by
    intro x hx
    -- import Mathlib.RingTheory.LittleWedderburn
    have : Finite s := hs.to_subtype
    have hsi : IsField s := Finite.isDomain_to_isField s
    -- for this one you have to assume `Field α` instead of `DivisionRing α`
    -- import Mathlib.RingTheory.IntegralDomain
    -- have : IsField s := Finite.isField_of_domain
    obtain rfl | hx0 := eq_or_ne x 0
    · simp
    let ux : s := ⟨x, hx⟩
    have hux0 : ux ≠ 0 := ne_of_apply_ne Subtype.val hx0
    obtain ⟨y, hy⟩ := hsi.mul_inv_cancel hux0
    rw [Subtype.ext_iff, Subring.coe_mul, Subring.coe_one] at hy
    have hxi : x⁻¹ = y := inv_eq_of_mul_eq_one_right hy
    rw [hxi]
    exact y.prop

[Aaron Liu]

I think I can reduce imports so that you don't have to import little wedderburn

[Violeta Hernández]

The idea is that we should write this in terms of IsRing and IsField

[Aaron Liu]

def Subring.toSubfieldOfFinite {α : Type u} [DivisionRing α] (s : Subring α)
    (hs : Set.Finite (SetLike.coe s)) : Subfield α where
  toSubring := s
  inv_mem' := by
    intro x hx
    have : Finite s := hs.to_subtype
    have : Fintype s := .ofFinite s
    classical
    let hsi : DivisionRing s := Fintype.divisionRingOfIsDomain s
    obtain rfl | hx0 := eq_or_ne x 0
    · simp
    let ux : s := ⟨x, hx⟩
    have hux0 : ux ≠ 0 := ne_of_apply_ne Subtype.val hx0
    have hxi : x⁻¹ = (ux⁻¹ : s) :=
      inv_eq_of_mul_eq_one_right (congrArg Subtype.val (mul_inv_cancel₀ hux0))
    rw [hxi]
    exact (ux⁻¹).prop

[Aaron Liu]

Violeta Hernández said:

The idea is that we should write this in terms of IsRing and IsField

The idea is that this is a theorem that should go in mathlib

[Aaron Liu]

and we can translate between IsRing and Subring and IsField and Subfield excluding the edge cases of zero and one

[Violeta Hernández]

It's a bit painful to work with the subtypes

[Violeta Hernández]

But if it makes the subtypeless result easier, then go for it

[Violeta Hernández]

Btw this stuff about natural numbers should go in another file imo

[Aaron Liu]

how do I show that finite ordinals are finite

[Violeta Hernández]

docs#Ordinal.lt_omega0 ?

[Aaron Liu]

ok let me clarify

[Aaron Liu]

how do I show that Set.Finite (Set.Iio (Nat.cast n : Ordinal))

[Aaron Liu]

I guess I'll just provide a surjection from Fin n

[Aaron Liu]

that was so much more work than I thought it was going to be

theorem IsField.two_two_pow (n : ℕ) : IsField (∗(2 ^ 2 ^ n)) := by
  let sk : Set Nimber := Set.Iio (∗(2 ^ 2 ^ n))
  have rsk : IsRing (∗(2 ^ 2 ^ n)) := IsRing.two_two_pow n
  have u1 : 1 ∈ sk := by
    rw [← Nimber.of_one]
    apply Set.mem_Iio.2
    apply of.lt_iff_lt.2
    rw [← Nat.cast_one, ← @Nat.cast_ofNat Ordinal (nat_lit 2),
      ← Ordinal.opow_natCast, ← Ordinal.natCast_opow, Nat.cast_lt]
    simp
  let rk : Subring Nimber := {
    carrier := sk
    zero_mem' := (zero_lt_one' Nimber).trans u1
    one_mem' := u1
    neg_mem' := id
    add_mem' := @IsGroup.add_lt _ rsk.toIsGroup
    mul_mem' := @IsRing.mul_lt _ rsk
  }
  let uf (i : Fin (2 ^ 2 ^ n)) : sk := by
    refine ⟨of (Nat.cast i.val), of.lt_iff_lt.2 ?_⟩
    conv_rhs => rw [← Nat.cast_ofNat, ← Ordinal.opow_natCast, ← Ordinal.natCast_opow]
    exact Nat.cast_lt.2 i.isLt
  have fk : Finite sk := by
    apply Finite.of_surjective uf
    intro x
    obtain ⟨k, hk⟩ : ∃ n : Nat, val x = n := Ordinal.lt_omega0.1 ((of.lt_iff_lt.1 x.prop).trans
      (Ordinal.lt_omega0.2 ⟨2 ^ 2 ^ n, by simp [← Ordinal.opow_natCast]⟩))
    have uk : k < 2 ^ 2 ^ n := by
      refine Nat.cast_lt.1 (hk.symm.trans_lt ?_)
      rw [Nimber.val_lt_iff, Ordinal.natCast_opow, Ordinal.opow_natCast, Nat.cast_ofNat]
      exact x.prop
    exact ⟨⟨k, uk⟩, Subtype.ext (val_eq_iff.1 hk).symm⟩
  let uk : Subfield Nimber := rk.toSubfieldOfFinite (Set.toFinite sk)
  exact ⟨rsk, fun x _ => uk.inv_mem⟩

[Aaron Liu]

now to prove IsRing.two_two_pow

[Violeta Hernández]

...is this taking IsRing (2^2^n) as an assumption?

[Aaron Liu]

Violeta Hernández said:

...is this taking IsRing (2^2^n) as an assumption?

yes

[Violeta Hernández]

Really you have to prove both that it's a ring and a field in the induction

[Aaron Liu]

well fine

[Violeta Hernández]

So you're better off proving IsRing (*n) -> IsField (*n)

[Aaron Liu]

we can generalize it to 1 < n → IsRing n → IsField n

[Violeta Hernández]

Works for all n

[Violeta Hernández]

Just do 0 and 1 separately

[Aaron Liu]

I guess we can also prove the case where n ≤ 1

[Violeta Hernández]

It seems unreasonable that 1 in Iio (*2^2^n) can't be proven by just simp

[Violeta Hernández]

You could add 1 < *x iff 1 < x as a simp lemma

[Violeta Hernández]

Alongside variants for other order relations and for 0

[Violeta Hernández]

And for removing val instead, if you want to go the extra mile

[Aaron Liu]

Well I changed it to be IsRing → IsField

-- universe monomorphic
theorem IsField.of_isRing_natCast (n : ℕ) (hn : IsRing.{u} (∗n)) : IsField.{u} (∗n) := by
  obtain hn | h1n := le_or_gt n 1
  · interval_cases n <;> simp
  let sk : Set Nimber := Set.Iio (∗n)
  have u1 : 1 ∈ sk := by simp [sk, ← Nimber.of_one, h1n]
  let rk : Subring Nimber := {
    carrier := sk
    zero_mem' := (zero_lt_one' Nimber).trans u1
    one_mem' := u1
    neg_mem' := id
    add_mem' := @IsGroup.add_lt _ hn.toIsGroup
    mul_mem' := @IsRing.mul_lt _ hn
  }
  let uf (i : Fin n) : sk :=
    ⟨of (Nat.cast i.val), of.lt_iff_lt.2 (Nat.cast_lt.2 i.isLt)⟩
  have fk : Finite sk := by
    apply Finite.of_surjective uf
    intro x
    obtain ⟨k, hk⟩ : ∃ n : Nat, val x = n := Ordinal.lt_omega0.1
      ((of.lt_iff_lt.1 x.prop).trans (Ordinal.nat_lt_omega0 n))
    have uk : k < n := Nat.cast_lt.1 (hk.symm.trans_lt (Nimber.val_lt_iff.2 x.prop))
    exact ⟨⟨k, uk⟩, Subtype.ext (val_eq_iff.1 hk).symm⟩
  let uk : Subfield Nimber := rk.toSubfieldOfFinite (Set.toFinite sk)
  exact ⟨hn, fun x _ => uk.inv_mem⟩

[Aaron Liu]

should I make it polymorphic

[Violeta Hernández]

Wdym

[Violeta Hernández]

Oh I see what you mean

[Tristan Figueroa-Reid]

universe polymorphic

[Violeta Hernández]

Nah

[Violeta Hernández]

We haven't really had a need to move things between universes

[Violeta Hernández]

We haven't even defined IGame.lift

[Violeta Hernández]

Btw perhaps you can prove Finite (Iio (*n)) by proving it's equal to (of o natCast) '' Iio n?

[Violeta Hernández]

This feels like the sort of goal that should be an aesop one liner

[Violeta Hernández]

In fact that might be one of the random "todo upstream" lemmas I left in the repo

[Violeta Hernández]

Check the nim file

[Violeta Hernández]

In fact, are we sure it's not easier to omit the subtypes and simply use docs#mul_right_injective

[Aaron Liu]

@Violeta Hernández if you have an easier proof then go for it!

[Violeta Hernández]

Currently on a bus

[Django Peeters]

I want to make a Field instance from IsField.

[Tristan Figueroa-Reid]

https://loogle.lean-lang.org/?q=IsField%2C+%E2%8A%A2+Field+_ ?

[Django Peeters]

I mean Nimber.IsField.

[Django Peeters]

Tristan Figueroa-Reid said:

https://loogle.lean-lang.org/?q=IsField%2C+%E2%8A%A2+Field+_ ?

Did you find this using loogle?

[Django Peeters]

I'm kinda stuck. Does this look anywhere close to good?
image.png

[Aaron Liu]

No that does not look good

[Aaron Liu]

You should make a subfield and then use docs#Subfield.toField

[Aaron Liu]

you should also not take hypotheses implicitly

[Aaron Liu]

they will never be found by unification

[Aaron Liu]

h' and h'' should be explicit arguments

[Django Peeters]

Why doesn't unification find them?

[Aaron Liu]

nothing to unify with

[Django Peeters]

I need to lookup what unification is

[Violeta Hernández]

Django Peeters said:

Why doesn't unification find them?

Unification finds things that are already in the expression. For instance if f : (n : Nat) -> (hn : n ≠ 0) -> Nat, you can generally write f _ hn, and n will get inferred from whatever hn is.

You usually can't unify proofs, since save for rare circumstances, your expression won't contain them built in.

[Violeta Hernández]

Django Peeters said:

I mean Nimber.IsField.

Use IsField.toSubfield, there's a type coercion on Subfield and a Field instance on that.

[Violeta Hernández]

But also, subtypes can be more annoying to work with than you expect. Are you sure you need to use them?

[Aaron Liu]

Violeta Hernández said:

But also, subtypes can be more annoying to work with than you expect. Are you sure you need to use them?

you seem to have some terrible experience with subtypes in the past

[Violeta Hernández]

Well it's more of a terrible experience with subtypes in the present

[Violeta Hernández]

Having to carry around 1 < x assumptions just because polynomials require a semiring to even be defined, having to work on Subfield because Lean won't let me declare a Field instance on Subring without creating diamonds

[Violeta Hernández]

And having to constantly carry around assumptions of type a < x

[Violeta Hernández]

It's much easier to state results in terms of Nimber, and use the subtypes as auxiliary constructions to get there

[Aaron Liu]

you still have to know a < x even without subtypes?

[Aaron Liu]

I guess it's the partial operations

[Violeta Hernández]

Yeah, but I don't have to carry these terms around in my expressions

[Violeta Hernández]

There's a reason we like to separate proofs and data

[Violeta Hernández]

And the subtype pattern sort of breaks that

[Aaron Liu]

sure I guess

[Violeta Hernández]

By the way, how much progress did you make yesterday?

[Violeta Hernández]

I can work on this for like an hour or so rn

[Aaron Liu]

I showed that the finite rings are fields

[Django Peeters]

Violeta Hernández said:

But also, subtypes can be more annoying to work with than you expect. Are you sure you need to use them?

I need some way to access the machinery from fieldtheory to prove b = a^c, so yes, I think I need to use them.

[Django Peeters]

Violeta Hernández said:

Django Peeters said:

I mean Nimber.IsField.

Use IsField.toSubfield, there's a type coercion on Subfield and a Field instance on that.

I can't find IsField.toSubfield in Mathlib, nor in the repo's master. Is it on one of the non-master-branches?

[Violeta Hernández]

It's in the vi.simplicity_5 branch I keep alluding to

[Violeta Hernández]

@Aaron Liu This is slightly simpler than your proof I think

theorem IsRing.toIsField_of_finite {n : ℕ} (h : IsRing.{u} (∗n)) : IsField.{u} (∗n) where
  inv_lt' x hx₀ hx := by
    obtain hn₁ | hn₁ := le_or_gt n 1
    · interval_cases n <;> simp_all
    have := (finite_Iio_of_natCast n).to_subtype
    let f (y : Iio (∗n)) : Iio (∗n) := ⟨x * y, h.mul_lt hx y.2⟩
    have hf : Function.Injective f := fun _ ↦ by aesop
    obtain ⟨⟨y, hy' : _ < ∗n⟩, hy⟩ := Finite.surjective_of_injective hf
      ⟨1, (Nat.one_lt_cast (α := Ordinal)).2 hn₁⟩
    convert hy'
    apply inv_eq_of_mul_eq_one_right
    simpa [f] using hy
  __ := h

[Aaron Liu]

good job

[Violeta Hernández]

The lemma finite_Iio_of_natCast is one I just added to the OrdinalAlias file

[Aaron Liu]

so you're hiding part of the proof in a new proof

[Violeta Hernández]

Well, if you want to be cynical about it

[Violeta Hernández]

Actually yeah, I think your proof might be shorter once you factor that out

[Violeta Hernández]

Yeah, once you factor that out the entire proof is just

theorem IsRing.toIsField_of_finite {n : ℕ} (h : IsRing.{u} (∗n)) : IsField.{u} (∗n) where
  inv_lt' x _ hx := by
    obtain hn₁ | hn₁ := le_or_gt n 1
    · interval_cases n <;> simp_all
    · exact ((h.toSubring ((Nat.one_lt_cast (α := Ordinal)).2 hn₁)).toSubfield_of_finite
        (finite_Iio_of_natCast n)).inv_mem hx
  __ := h

[Violeta Hernández]

Maybe subtypes aren't so bad

[Aaron Liu]

Violeta Hernández said:

Maybe subtypes aren't so bad

yeah they really aren't bad

[Aaron Liu]

people just don't like them

[Django Peeters]

Apparently mathlib doesn't have nth Prime n > 1 yet.

[Violeta Hernández]

We have docs#Nat.add_two_le_nth_prime

[Django Peeters]

How did you find this?

[Aaron Liu]

loogle probably

[Violeta Hernández]

Just searching around in the docs

[Aaron Liu]

or maybe docs

[Django Peeters]

Hmm, my search skills are also lacking :(

[Aaron Liu]

@loogle Nat.nth, Nat.Prime

[loogle]

:search: Nat.nth_prime_four_eq_eleven, Nat.nth_prime_one_eq_three, and 6 more

[Django Peeters]

Okay, I proved some lexicographic ordering.
image.png

[Django Peeters]

What's up with all these proof_wanted's instead of using sorry's?

[Violeta Hernández]

They're basically the same

[Violeta Hernández]

The difference is that you can't use a result marked as proof_wanted, while you can use a result proved with sorry

[Violeta Hernández]

Mathlib allows the former, but not the latter

[Violeta Hernández]

Django Peeters said:

Okay, I proved some lexicographic ordering.
image.png

Would it make sense to define Nimber.kappa_q as a function ℕ ⊗ₗ ℕ → Nimber, and prove it strictly monotone?

[Django Peeters]

Would it make more sense than taking the two inputs one by one?

[Django Peeters]

I don't know.

[Django Peeters]

I'll have to formalize more to see if this is the case.

[Violeta Hernández]

Django Peeters said:

Would it make more sense than taking the two inputs one by one?

Likely not, I just thought to ask

[Django Peeters]

It's linear in both arguments, right?

[Violeta Hernández]

Linear?

[Violeta Hernández]

You mean monotone?

[Django Peeters]

The tensor product

[Violeta Hernández]

Oh no I didn't mean the tensor product

[Violeta Hernández]

I meant the product of types (with the lexicographic ordering)

[Django Peeters]

My phone rendered ℕ ⊗ₗ ℕ.

[Violeta Hernández]

Yeah, that's how you write that in Lean

[Aaron Liu]

you wrote it with the tensor symbol

[Violeta Hernández]

oh wait you're right

[Aaron Liu]

ℕ ×ₗ ℕ

[Violeta Hernández]

I meant that yeah

[Django Peeters]

No, my phone rendered ℕ ⊗□ ℕ.

[Aaron Liu]

mine too

[Django Peeters]

Wut

[Violeta Hernández]

Well to be more explicit I mean Lex (ℕ × ℕ)

[Django Peeters]

Why is it a box, that's so confusing.

[Aaron Liu]

but from context clues I could tell that the box was supposed to be a subscript l and the tensor was supposed to be a product

[Violeta Hernández]

I've got time (and energy) to work on this again

[Violeta Hernández]

Just proved that simplestIrreducible is monic in a field

[Aaron Liu]

I'll see if I can show that 2 ^ 2 ^ n is a ring for finite n

[Violeta Hernández]

reminder that the way you prove that is by simultaneously proving simplestIrreducible (∗(2^2^n)) = X^2 + X + (∗(2^(2^n - 1)))

[Aaron Liu]

I don't remember how to do this

[Aaron Liu]

do you know of a good place to read about it

[Aaron Liu]

or is it simple enough that you can just explain it

[Aaron Liu]

over Zulip

[Violeta Hernández]

ONAG gives a good enough explanation but it's also simple enough for me to do it

[Violeta Hernández]

IsField.X_sq_lt_simplestIrreducible proves all linear polynomials have roots

[Aaron Liu]

alright I'll see what ONAG has to say about this

[Violeta Hernández]

Then you prove polynomials X^2 + a have roots by that argument about the square function being surjective and thus injective

[Violeta Hernández]

For X^2 + X + a it's some algebra I don't remember and yeah you should just check ONAG

[Aaron Liu]

can I import some field theory

[Violeta Hernández]

Yeah sure

[Violeta Hernández]

This is presumably going on another file anyways

[Violeta Hernández]

Do we actually have "all char 2 finite fields have all square roots" in Mathlib?

[Aaron Liu]

probably

[Aaron Liu]

I'll take a look

[Aaron Liu]

is it docs#PerfectRing.ofFiniteOfIsReduced

[Aaron Liu]

mathlib generality makes everything slightly hard to find

[Violeta Hernández]

Perfect ring?

[Aaron Liu]

you can read the definition

[Violeta Hernández]

Ohh that x^p is bijective

[Violeta Hernández]

Ok so IsField.isRoot_simplestIrreducible is literally the only sorry remaining in my branch

[Violeta Hernández]

I have a paper pseudo-Lean proof over here somewhere

[Violeta Hernández]

Shouldn't take more than like two hours

[Violeta Hernández]

(clueless)

[Aaron Liu]

just make sure to sleep

[Violeta Hernández]

I slept pretty well last night, dw

[Violeta Hernández]

Had some personal life situations going on which have now been resolved

[Aaron Liu]

great

[Aaron Liu]

keep it up

[Violeta Hernández]

btw did no one figure out a better name for simplestIrreducible

[Violeta Hernández]

that's 19 characters long I don't like it

[Aaron Liu]

ONAG is omitting a lot and it's a bit tricky to piece it all together

[Violeta Hernández]

perhaps Siegel does it better

[Aaron Liu]

perhaps

[Aaron Liu]

but I would have to get out another book

[Violeta Hernández]

let me get the screencap

[Violeta Hernández]

image.png

[Violeta Hernández]

Wait I'm struggling

[Violeta Hernández]

How do I prove this
image.png

[Aaron Liu]

you might have to explain why you think this is true

[Violeta Hernández]

Well the reason I thought it was true is because x.simplestIrreducible doesn't split

[Violeta Hernández]

But clearly I'm missing something

[Violeta Hernández]

Because this is an inequality of values, not of polynomials

[Violeta Hernández]

Siegel doesn't really explain this either? Or maybe I'm not understanding him
image.png

[Violeta Hernández]

...I think I see the missing step, though

[Aaron Liu]

I really don't

[Violeta Hernández]

∏ i, (x + f i) = eval (∏ i, (X + C (f i))) which is a polynomial of degree n, which we know is equal to oeval, which we know is less of (val x ^ n) = x ^ n

[Violeta Hernández]

Like, you have to run the argument backwards

[Aaron Liu]

great, go do that then

[Violeta Hernández]

You don't have to tell me!

[Violeta Hernández]

Thanks for the rubberducking

[Violeta Hernández]

wait no that's nonsense

[Aaron Liu]

oh no

[Violeta Hernández]

We don't want to prove that this is less than x ^ n, we want to prove that it's unequal to f(x)

[Violeta Hernández]

Translating to the Siegel notation

[Violeta Hernández]

I don't see why $f'(\gamma) \ne f(\gamma)$

[Aaron Liu]

I still don't know what $f$ is

[Violeta Hernández]

The simplest irreducible polynomial

[Violeta Hernández]

And $f'$ is some other polynomial that splits and all of whose roots are less than $\gamma$

[Violeta Hernández]

Obviously $f' \ne f$ since $f'$ splits and $f$ doesn't

[Violeta Hernández]

But $f'(\gamma) \ne f(\gamma)$ isn't a consequence of that

[Aaron Liu]

what are they polynomials in

[Aaron Liu]

polynomials in $\gamma$?

[Violeta Hernández]

Yep

[Violeta Hernández]

In the Siegel treatment, at least

[Aaron Liu]

and $\gamma$ is a field

[Aaron Liu]

ok thinking time

[Violeta Hernández]

I'm pretty sure the other part of the proof is correct

[Violeta Hernández]

"Every $g'(\gamma)$ is an excludent for $\gamma^{\boxed{n}}$"

[Violeta Hernández]

My only gripe is "$g(\gamma)$ itself is not"

[Aaron Liu]

wait what's n

[Violeta Hernández]

The degree of $f$

[Aaron Liu]

everything coming before the simplest irreducible has to split?

[Aaron Liu]

wait what's the definition of "simplest irreducible"

[Aaron Liu]

it can't literally be that since the polynomial X is obviously irreducible

[Aaron Liu]

along with all the other linear polynomials

[Violeta Hernández]

Aaron Liu said:

wait what's the definition of "simplest irreducible"

Lexicographically first polynomial in $\gamma$ of non-zero degree without roots less than $\gamma$

[Violeta Hernández]

Again, the name is bad

[Violeta Hernández]

Please give it a better one

[Aaron Liu]

I think that means everything before it splits

[Violeta Hernández]

Yep

[Violeta Hernández]

That's not what's in contention

[Aaron Liu]

so it's the first polynomial which doesn't split

[Aaron Liu]

maybe that could be inspiration for a better name

[Violeta Hernández]

And as a corollary, it has no roots less than $\gamma$

[Violeta Hernández]

Is there a word for "polynomial that doesn't split"?

[Violeta Hernández]

what if we just name it something generic and ambiguous like "characteristic polynomial"

[Aaron Liu]

ok I think I proved the thing

[Aaron Liu]

so we need to show that $f'(\gamma) + f(\gamma) \ne 0$

[Violeta Hernández]

Yep

[Aaron Liu]

but $f'$ is monic and $f$ is monic and they have the same degree so $f' + f$ has degree less than $n$

[Aaron Liu]

so it splits

[Violeta Hernández]

Oh!

[Violeta Hernández]

That's it!

[Violeta Hernández]

So really the fact that $f'$ splits was a red herring

[Violeta Hernández]

Well, not quite

[Violeta Hernández]

You do need to use that to prove $f' \ne f$

[Violeta Hernández]

Alright! Just proved this part of the argument in Lean.

[Violeta Hernández]

Other part should not be much more complicated.

[Violeta Hernández]

I'm mostly done, though there's a bit of a hefty subgoal I realize I have to prove, namely:

theorem forall_lt_oeval_iff {x : Nimber} (hx₁ : 1 < x) {P : Ordinal → Prop}
    {p : Nimber[X]} (hpk : ∀ k, p.coeff k < x) :
    (∀ y < oeval x p, P y) ↔ ∀ q < p, (∀ k, q.coeff k < x) → P (oeval x q) := sorry

[Aaron Liu]

have fun

[Violeta Hernández]

anything involving the CNF is not fun

[Violeta Hernández]

Hi I'm back

[Violeta Hernández]

Good news: the proof of the last simplest extension theorem is pretty much done

[Violeta Hernández]

Bad news: by "pretty much done" I mean "barring some really annoying lemmas about oeval I haven't proven"

[Aaron Liu]

did you have fun

[Violeta Hernández]

No :frown:

[Violeta Hernández]

I have like an hour left to work on this before I have to take another bus

[Violeta Hernández]

I'll try to leave all the sorries contained to oeval

[Violeta Hernández]

So maybe one of you can try to fill them in for me

[Aaron Liu]

I'm stuck trying to figure out how to show that IsRing 2

[Django Peeters]

aesop?

[Violeta Hernández]

You are trying to prove IsRing (∗2), right? And not IsRing 2 = IsRing 0

[Violeta Hernández]

Presumably we're missing a lemma x < ∗2 ↔ x = 0 ∨ x = 1

[Aaron Liu]

that should probably be automatically generated

[Violeta Hernández]

I'd prove it by showing ∗2 = succ (∗1), which should be def-eq to the statement on Ordinal

[Violeta Hernández]

We already have the lemma x ≤ 1 ↔ x = 0 ∨ x = 1

[Violeta Hernández]

Aaron Liu said:

that should probably be automatically generated

I wish there was a way to make interval_cases work with stuff like this

[Aaron Liu]

unfortunately it's not extensible

[Violeta Hernández]

Yeah, simple enough

@[simp] theorem succ_one : succ 1 = ∗2 := Ordinal.succ_one

theorem lt_two_iff {x : Nimber} : x < ∗2 ↔ x = 0 ∨ x = 1 := by
  rw [← succ_one, lt_succ_iff, le_one_iff]

[Violeta Hernández]

Arguably you don't even need the latter lemma, it's easy enough to simply rewrite ∗2 = succ 1 when it comes up

[Violeta Hernández]

Violeta Hernández said:

I'll try to leave all the sorries contained to oeval

I think I achieved this

[Aaron Liu]

great

[Aaron Liu]

how do I show ∗2 * ∗2 = ∗3

[Violeta Hernández]

Brute casework

[Violeta Hernández]

• This can't be 0 because ∗2 ≠ 0 (and nimbers are a domain)
• This can't be 1 because ∗2 ≠ 1 (and squaring is injective)
• This can't be 2 because ∗2 ≠ ∗1 (and nimbers are a domain)
• This is less or equal to ∗3, since ∗a * ∗2 + ∗2 * ∗b + ∗a * ∗b ≠ ∗3 for a, b ≤ 1.

[Violeta Hernández]

To be even more explicit,

• ∗a * ∗2 + ∗2 * ∗b + ∗a * ∗b = 0 when a = b = 0
• ∗a * ∗2 + ∗2 * ∗b + ∗a * ∗b = 1 when a = 0, b = 1 or viceversa
• ∗a * ∗2 + ∗2 * ∗b + ∗a * ∗b = ∗2 when a = b = 1

It might be faster to prove ∗2 * ∗2 = ∗3 using these subgoals.

[Aaron Liu]

I can figure out a case bash

[Aaron Liu]

I was just thinking surely there must be an easier way

[Violeta Hernández]

Hm no I don't think so

[Violeta Hernández]

Obviously once we've proven correctness of NatNimber we could just use that and decidability

[Violeta Hernández]

But that'd be putting the cart before the horse

[Aaron Liu]

Violeta Hernández said:

image.png

ok I still don't get it

[Violeta Hernández]

What's the thing you're confused about?

[Aaron Liu]

how do you get the next field

[Violeta Hernández]

Oh

[Violeta Hernández]

Ok so, suppose ∗(2^2^n) is a field

[Violeta Hernández]

Then you can write any element < ∗(2^2^(n+1)) in the form ∗(a * 2^2^n + b), for a, b < 2^2^n

[Aaron Liu]

why

[Violeta Hernández]

This is just a lemma about the natural numbers

[Aaron Liu]

makes sense

[Violeta Hernández]

A number less than x^2 can be written as a * x + b for a, b < x

[Aaron Liu]

something something docs#finProdFinEquiv

[Violeta Hernández]

By IsGroup.mul_add_eq_of_lt and IsRing.mul_eq_of_lt, you can show ∗(a * 2^2^n + b) = ∗a * ∗(2^2^n) + ∗b

[Violeta Hernández]

Which means you can multiply two nimbers less than ∗(2^2^(n+1)) by just expanding this entire thing out

[Aaron Liu]

great that works

[Violeta Hernández]

And you can show that what you get is less than 2^2^(n+1)

[Aaron Liu]

believable

[Violeta Hernández]

That last part involves a lot of calculations and a few more instances of IsGroup.mul_add_eq_of_lt and IsRing.mul_eq_of_lt so you might want to do it on paper first

[Violeta Hernández]

Violeta Hernández said:

I'm mostly done, though there's a bit of a hefty subgoal I realize I have to prove, namely:

theorem forall_lt_oeval_iff {x : Nimber} (hx₁ : 1 < x) {P : Ordinal → Prop}
    {p : Nimber[X]} (hpk : ∀ k, p.coeff k < x) :
    (∀ y < oeval x p, P y) ↔ ∀ q < p, (∀ k, q.coeff k < x) → P (oeval x q) := sorry

This is literally the last sorry I've got

[Violeta Hernández]

(Well, that and some other sublemmas I think I'll need to show this)

[Violeta Hernández]

I realize I can rename simplestIrreducible to leastIrreducible now that I've define the ordering on Nimber[X] proper

[Violeta Hernández]

The name still has issues but at least it's shorter

[Django Peeters]

How about leastNotSplit?

[Violeta Hernández]

hmmmmm

[Violeta Hernández]

Yeah that's even better actually

[Aaron Liu]

Let us use, as sometimes customary, $\varepsilon_{\alpha}$ for the $\alpha$ th solution of $\alpha = [\omega^{\alpha}]$ (counting from $\alpha = 0$), then $\zeta_{\alpha}$ for the $\alpha$ th solution of $\alpha = \varepsilon_{\alpha}$, $\eta_{\alpha}$ for the $\alpha$ th solution of $\alpha = \zeta_{\alpha}$, and so on through the (transfinite) Greek Alphabet. We shall use the symbol $\boxed{\alpha}$ to denote the $\alpha$ th letter of this Alphabet. Then we can state:

I love this guy

[Django Peeters]

Lol, Conway is legendary

[Violeta Hernández]

Ah yes, the Veblen function $\boxed{\alpha}_{\beta}$, just as I remember it

[Aaron Liu]

this is immediately not enough as in the next paragraph I see

(b) the least $\alpha$ with $\boxed{\alpha}_0 = \alpha$.

[Violeta Hernández]

I think that's a weird way of referring to the ordinal $\varphi(1, 0, 0)$

[Aaron Liu]

yes that's how I interpreted it

[Aaron Liu]

I think it has a name

[Aaron Liu]

https://en.wikipedia.org/wiki/Feferman%E2%80%93Sch%C3%BCtte_ordinal

[Violeta Hernández]

Oh yeah!

[Violeta Hernández]

We actually have that in Mathlib

[Violeta Hernández]

docs#Ordinal.gamma

[Violeta Hernández]

(this ordinal is gamma 0)

[Violeta Hernández]

Ok so I'm actually kind of stuck with these last theorems

[Aaron Liu]

oh no

[Violeta Hernández]

These are the last sorries
image.png

[Aaron Liu]

try induction

[Violeta Hernández]

I know it's induction!

[Violeta Hernández]

I'm just struggling to set it up

[Aaron Liu]

you can set it up later

[Aaron Liu]

just focus on proving the theorem for now

[Violeta Hernández]

What I mean is, I think there's other auxiliary results that I need to prove first

[Aaron Liu]

usually what I do is I just do it and then if I find out I need an auxiliary result in the course of the proof I recursively start proving the auxiliary result

[Violeta Hernández]

Let me try and write the pen-and-paper argument for these theorems.

[Violeta Hernández]

oeval_lt_of_lt: the hypothesis p < q implies there's some n such that all higher coefficients of p and q match. So we can remove then to get p' and q', and it suffices to prove oeval x p' < oeval x q'. This is true because oeval x p' = ∗(x.val ^ n * (p.coeff n).val + (oeval x p.eraseLead).val) < ∗(x.val ^ n * (p.coeff n + 1)) ≤ ∗(x.val ^ n * (q.coeff n).val) ≤ oeval x q'.

[Violeta Hernández]

So the main issue is that whole p' and q' thing

[Aaron Liu]

Have you started the proof yet?

[Aaron Liu]

Just do induction

[Violeta Hernández]

...on n?

[Aaron Liu]

On n, or on p, or q, or even on h

[Aaron Liu]

Well not n actually

[Violeta Hernández]

You know actually I think the other sorry might be much easier

[Violeta Hernández]

I'll start there

[Aaron Liu]

if it's too hard I can just do them for you

[Violeta Hernández]

That one should just be induction on p.degree

[Violeta Hernández]

Aaron Liu said:

if it's too hard I can just do them for you

Ok that's a bit condescending, but I could use the help

[Aaron Liu]

Violeta Hernández said:

Aaron Liu said:

if it's too hard I can just do them for you

Ok that's a bit condescending, but I could use the help

oh sorry I just want to help

[Violeta Hernández]

I've pushed to the simplicity_5 branch I've been using

[Django Peeters]

How much did you formalize today?

[Violeta Hernández]

Well, this is all that's missing for the full simplest extension theorem

[Violeta Hernández]

Violeta Hernández said:

You know actually I think the other sorry might be much easier

Oh yeah it totally was

[Violeta Hernández]

So we have a single sorry now

[Aaron Liu]

I am working on it

[Violeta Hernández]

I'll leave you to it!

[Aaron Liu]

after seeing the definition of oeval I suddenly understand why everything is so difficult

[Violeta Hernández]

You're more than welcome to come up with something better

[Aaron Liu]

oh I probably won't change it

[Violeta Hernández]

The idea is that you're supposed to be working using oeval_C_mul_X_pow_add

[Aaron Liu]

makes sense

[Django Peeters]

I'm gonna switch back to C++ tomorrow. Maybe we'll know alpha(263) in the near future.

[Aaron Liu]

anticipation

[Aaron Liu]

@Violeta Hernández I did it!

theorem oeval_lt_oeval {x : Nimber} {p q : Nimber[X]} (h : p < q)
    (hpk : ∀ k, p.coeff k < x) (hqk : ∀ k, q.coeff k < x) : oeval x p < oeval x q := by
  rw [Nimber.Lex.lt_def] at h
  obtain ⟨n, hnl, hnr⟩ := h
  have hx : 0 < x := (zero_le (p.coeff 0)).trans_lt (hpk 0)
  induction hk : p.natDegree - n using Nat.caseStrongRecOn generalizing p q with
  | zero =>
    rw [Nat.sub_eq_zero_iff_le] at hk
    apply (mul_coeff_le_oeval x q n).trans_lt'
    have hpn : (erase n p).degree < n := by
      apply lt_of_le_of_ne ((p.degree_erase_le n).trans
        (degree_le_natDegree.trans (WithBot.coe_mono hk)))
      intro hdd
      exact coeff_ne_zero_of_eq_degree hdd (erase_same p n)
    rw [← p.monomial_add_erase n, ← C_mul_X_pow_eq_monomial]
    rw [oeval_C_mul_X_pow_add hpn, of.lt_iff_lt]
    have hpe (k : Nat) : (p.erase n).coeff k < x := by
      rw [p.coeff_erase]
      split <;> simp [hpk, hx]
    apply (add_lt_add_left (val_lt_iff.2 (oeval_lt_opow hpe hpn)) _).trans_le
    rw [← mul_add_one]
    apply mul_le_mul_left'
    rwa [add_one_le_iff, val.lt_iff_lt]
  | ind k ih =>
    have hp0 : p ≠ 0 := by rintro rfl; simp at hk
    have hq0 : q ≠ 0 := by rintro rfl; simp at hnr
    have hqd : n ≤ q.natDegree := by
      by_contra! hn
      apply coeff_eq_zero_of_natDegree_lt at hn
      simp [hn] at hnr
    have hpqd : p.natDegree ≤ q.natDegree := by
      rw [Polynomial.natDegree_le_iff_coeff_eq_zero]
      intro N hN
      exact (hnl N (hqd.trans_lt hN)).trans (coeff_eq_zero_of_natDegree_lt hN)
    replace hqd : n < q.natDegree := by
      apply lt_of_le_of_ne hqd
      rintro rfl
      simp [hpqd] at hk
    replace hpqd : p.natDegree = q.natDegree := by
      apply Polynomial.natDegree_eq_of_le_of_coeff_ne_zero hpqd
      rw [hnl _ hqd]
      simp [hq0]
    rw [← p.eraseLead_add_C_mul_X_pow, add_comm p.eraseLead,
      ← q.eraseLead_add_C_mul_X_pow, add_comm q.eraseLead]
    rw [oeval_C_mul_X_pow_add ((degree_eraseLead_lt hp0).trans_le degree_le_natDegree),
      oeval_C_mul_X_pow_add ((degree_eraseLead_lt hq0).trans_le degree_le_natDegree)]
    rw [hpqd, ← coeff_natDegree, hpqd, (hnl _ hqd), coeff_natDegree, of.lt_iff_lt,
      add_lt_add_iff_left, val.lt_iff_lt]
    by_cases hpe : p.eraseLead = 0
    · rw [hpe, oeval_zero]
      rw [← eraseLead_coeff_of_ne n (hqd.trans_eq hpqd.symm).ne, hpe, coeff_zero,
        ← eraseLead_coeff_of_ne n hqd.ne] at hnr
      have hqe : q.eraseLead ≠ 0 := fun h => by simp [h] at hnr
      apply (opow_natDegree_le_oeval x hqe).trans_lt'
      rw [← val_lt_iff, val_zero, ← Ordinal.opow_natCast]
      apply Ordinal.opow_pos
      rwa [← of_lt_iff, of_zero]
    apply ih (p.eraseLead.natDegree - n)
    · rw [Nat.sub_le_iff_le_add, ← Nat.lt_succ_iff,
        ← Nat.succ_add, ← hk, Nat.sub_add_cancel (hqd.le.trans_eq hpqd.symm)]
      exact p.eraseLead_natDegree_lt_or_eraseLead_eq_zero.resolve_right hpe
    · intro u
      rw [eraseLead_coeff]
      split <;> simp [hx, hpk]
    · intro u
      rw [eraseLead_coeff]
      split <;> simp [hx, hqk]
    · intro u hu
      rw [eraseLead_coeff, eraseLead_coeff, hpqd]
      split
      · rfl
      · exact hnl u hu
    · rwa [eraseLead_coeff, eraseLead_coeff, hpqd, if_neg hqd.ne, if_neg hqd.ne]
    · rfl

[Aaron Liu]

I did in fact end up using induction

[Violeta Hernández]

Awesome!

[Violeta Hernández]

I'll take a look once I get home

[Violeta Hernández]

I'm thinking about how to PR all of this

[Violeta Hernández]

Since it's like a thousand lines of code

[Violeta Hernández]

I feel like there's an argument for separately PRing

• lemmas on polynomials
• the order relation on polynomials
• oeval
• leastNotSplit

[Violeta Hernández]

...of course there's the question, who is even going to review this?

[Violeta Hernández]

@Aaron Liu @Django Peeters would you be up to the task?

[Django Peeters]

Uuuuuhm I can give it a try

[Django Peeters]

Have you ever used a VPS before?

[Django Peeters]

A virtual private server, I'm gonna try it for alpha(263) cuz it might take 3 days.

[Aaron Liu]

How do you know it won't take forever

[Django Peeters]

Theoretically, because Lenstra proved so, and practically, debugging and optimization :melting_face:

[Violeta Hernández]

Just to summarize some DMs between me and Aaron: we should leave it as a TODO to more generally define the function (Ordinal →ₒ Ordinal) → Ordinalthat evaluates a CNF; then oeval can be a special case of it.

[Violeta Hernández]

Opened #196 which simply splits our existing Simplicity file into three.

[Violeta Hernández]

(I'll be slowly moving stuff from my work branch into master, but while that happens, I'll get working on proving the nimbers are alg closed!)

[Django Peeters]

Do I need to pay attention to something specific when reviewing?

[Violeta Hernández]

Well, for a general PR, it's mostly about making sure the API makes sense, and the proofs are as simple as can be.

[Violeta Hernández]

I'm inclined to say this file splitting one is uncontroversial (save for the folder rename, which I agree with), so I'll go ahead and merge it.

[Violeta Hernández]

(well, if Junyan Xu doesn't have anything more to say)

[Violeta Hernández]

To prove that the nimbers are algebraically closed, I'll introduce a definition algClosure x, e.g. the least nimber containing all sums, products, and roots of polynomials with coefficients less than x. The idea is that if we suppose all non-constant polynomials < p have roots but p doesn't, and if m is larger than the coefficients of p, then algClosure m is a field with leastNotSplit (algClosure m) = p, so p.eval (algClosure m) = 0, a contradiction.

[Violeta Hernández]

Now the question is, is it worth adding the other closures? Namely groupClosure, ringClosure, and fieldClosure? Or should I just have this last one?

[Django Peeters]

I haven't seen the literature use those closures...

[Violeta Hernández]

groupClosure x in particular would be a bit silly, it's just the next power of 2

[Django Peeters]

Is there a characterization of nimbers which are rings? Or is that already too complicated?

[Violeta Hernández]

If there is, then it's not known.

[Aaron Liu]

Well the finite rings are 2^2^n

[Django Peeters]

For now, I don't think we need those other closures.

[Django Peeters]

Btw, here are some speed gains:
http://www.neverendingbooks.org/aaron-siegel-on-transfinite-number-hacking/
now becomes
image.png

[Violeta Hernández]

That's awesome! You should probably communicate this to Siegel.

[Django Peeters]

Is he on this Zulip chat?

[Violeta Hernández]

No, I don't think so.

[Violeta Hernández]

Perhaps not the most professional avenue, but he's in the Gamesters Discord server

[Violeta Hernández]

I can DM a link to that

[Django Peeters]

I'm gonna go with e-mail, that seems appropriate.

[Violeta Hernández]

Yeah, that tracks

[Violeta Hernández]

I couldn't find his e-mail though

[Django Peeters]

Me neither :/

[Violeta Hernández]

I did it.
image.png

[Aaron Liu]

congrulatis

[Violeta Hernández]

Aaron Liu said:

this seems like just a regular argument, which is not by contradiction

Btw it was a proof by contradiction. If you have a polynomial without roots, you can build some huge nimber such that it's a root of that, contradiction. But if you know that it has a root, you know nothing about what it is!

[Violeta Hernández]

Btw, a fun fact I haven't seen explicitly mentioned in the literature. Check this out:
image.png

[Aaron Liu]

it seems that just by taking the algebraic closure you can get an algebraically closed nimber above any other

[Violeta Hernández]

Indeed

[Violeta Hernández]

I mean, it's slightly nontrivial to show the algebraic closure is a set, rather than a class

[Violeta Hernández]

And in any case, it might be an absolutely huge set!

[Aaron Liu]

well the algebraic closure of a set-sized field is set-sized

[Violeta Hernández]

Hmm fair enough

[Violeta Hernández]

Well, I guess what's slightly nontrivial is to show that these algebraic closures look like a lower set of nimbers

[Aaron Liu]

Wikipedia says

The algebraic closure of a field K has the same cardinality as K if K is infinite, and is countably infinite if K is finite.

[Violeta Hernández]

Actually, on that note

[Django Peeters]

Do you think Lenstra's gonna be right about the next member after w^w^w?

[Violeta Hernández]

Can we prove the following?

example : algClosure (∗ (ω ^ ω ^ ω + 1)) ≤ ∗ω₁ := sorry

[Violeta Hernández]

Django Peeters said:

Do you think Lenstra's gonna be right about the next member after w^w^w?

What's his conjecture?

[Django Peeters]

Complicated

[Violeta Hernández]

Γ₀? SVO? LVO? Something else entirely?

[Aaron Liu]

why are there so many Veblen ordinals

[Violeta Hernández]

There's only two of them

[Aaron Liu]

that's a lot

[Violeta Hernández]

SVO = φ(1@ω)
LVO = smallest fixed point of φ(1@x)

[Violeta Hernández]

where a@b is notation for Finsupp.single b a

[Aaron Liu]

this is why I don't like infinitary functions

[Violeta Hernández]

It's not an infinitary function, it's a function (Ordinal →₀ Ordinal) → Ordinal

[Aaron Liu]

so (Fin n →₀ α) → β is an n-ary function

[Violeta Hernández]

Ok fine fair enough

[Django Peeters]

He had to define a new function like Veblen's infinitary one.

[Aaron Liu]

nothing is "new"

[Django Peeters]

I think at least \boxed{alpha}_0

[Violeta Hernández]

I had the idea of formalizing the transfinite Veblen function in Lean at one point, but then I realized I had enough trouble getting docs#Ordinal.veblen accepted to begin with

[Violeta Hernández]

Django Peeters said:

I think at least \boxed{alpha}_0

So, Γ₀?

[Violeta Hernández]

Is that a proven theorem?

[Django Peeters]

I don't think so

[Violeta Hernández]

Violeta Hernández said:

Can we prove the following?

example : algClosure (∗ (ω ^ ω ^ ω + 1)) ≤ ∗ω₁ := sorry

What about this?

[Violeta Hernández]

Do we have any upper bounds whatsoever?

[Aaron Liu]

Violeta Hernández said:

Well, I guess what's slightly nontrivial is to show that these algebraic closures look like a lower set of nimbers

Wikipedia also says

if M is any algebraically closed field containing K, then the elements of M that are algebraic over K form an algebraic closure of K.

[Aaron Liu]

and also

algebraic closure of a field K is unique up to an isomorphism that fixes every member of K

[Violeta Hernández]

Aaron Liu said:

Violeta Hernández said:

Well, I guess what's slightly nontrivial is to show that these algebraic closures look like a lower set of nimbers

Wikipedia also says

if M is any algebraically closed field containing K, then the elements of M that are algebraic over K form an algebraic closure of K.

Yes, but why would that have to be a lower set?

[Aaron Liu]

great question

[Violeta Hernández]

In fact, it's slightly wrong to say that algClosed x is the algebraic closure of x

[Violeta Hernández]

Rather, it's the smallest algebraically closed nimber that's at least x

[Violeta Hernández]

There could be some smaller, random subset that's also algebraically closed

[Django Peeters]

image.png

[Violeta Hernández]

I think that's just the SVO

[Violeta Hernández]

I have a really nice definition of infinitary Veblen somewhere

[Violeta Hernández]

One that doesn't require all that splitting between successor and limit cases

[Violeta Hernández]

Let me look it up

[Violeta Hernández]

(by which I mean, ask my friend who I know remembers it)

[Aaron Liu]

that looks like SVO

[Violeta Hernández]

btw, I've always found it fascinating that for all we know, "the x-th algebraically closed nimber" could be the fastest-growing known function ω₁ → ω₁

[Violeta Hernández]

Probably isn't, but y'know

[Django Peeters]

Like busy beaver? But then on transfinite ordinal

[Aaron Liu]

0, 1, w^w^w, ???

[Aaron Liu]

Violeta Hernández said:

Aaron Liu said:

Violeta Hernández said:

Well, I guess what's slightly nontrivial is to show that these algebraic closures look like a lower set of nimbers

Wikipedia also says

if M is any algebraically closed field containing K, then the elements of M that are algebraic over K form an algebraic closure of K.

Yes, but why would that have to be a lower set?

I found

theorem eq_oeval_of_lt_oeval {x y : Nimber} {p : Nimber[X]} (hx₀ : x ≠ 0)
    (hpk : ∀ k, p.coeff k < x) (h : y < oeval x p) :
    ∃ q : Nimber[X], q < p ∧ (∀ k, q.coeff k < x) ∧ oeval x q = y

in one of the files

[Aaron Liu]

this is taking the closure of Iio x, by the way

[Aaron Liu]

if you just take the closure of {x} you probably get something different

[Aaron Liu]

oh that didn't say what I thought it said

[Violeta Hernández]

Opened #198 which proves some more stuff on groups/rings/fields, and #199 which IMO is an uncontroversial rename.

[Django Peeters]

Violeta Hernández said:

Can we prove the following?

example : algClosure (∗ (ω ^ ω ^ ω + 1)) ≤ ∗ω₁ := sorry

My gut tells me yes, but besides cardinality arguments, I don't have much to go on.

[Violeta Hernández]

Well the thing is, we don't a priori know that a random polynomial like x ^ 2 + ω ^ ω ^ ω + 1 doesn't have a root with a much larger cardinality

[Violeta Hernández]

(or do we?)

[Aaron Liu]

you know how many roots it has

[Tristan Figueroa-Reid]

Django Peeters said:

A virtual private server, I'm gonna try it for alpha(263) cuz it might take 3 days.

(I have access to a currently unused compute machine - if you want to run your program there, feel free to DM me about it :+1:)

[Violeta Hernández]

Yeah, but I don't know that one of its roots isn't aleph_37

[Violeta Hernández]

If you look at my construction of algClosure, you'll notice that it's defined as the iterated supremum of a set algClosureSet

[Violeta Hernández]

We can prove algClosureSet x is countable when x is, but the supremum might be any huge ordinal of countable cofinality for all we know.

[Violeta Hernández]

And the proof that nimbers are algebraically closed doesn't help much either, as again, it's non-constructive!

[Aaron Liu]

can we cps-transform it to be constructive

[Violeta Hernández]

No idea what that means

[Aaron Liu]

magic

[Violeta Hernández]

I'm inclined to ask about this in MathOverflow

[Violeta Hernández]

You know, on the topic of the next algebraically closed nimber

[Violeta Hernández]

Do we have any lower bounds better than w^w^w?

[Violeta Hernández]

I think Siegel claims e_1 without proof

[Aaron Liu]

Violeta Hernández said:

Do we have any lower bounds better than w^w^w?

w^w^w+1

[Violeta Hernández]

Any nontrivial lower bounds

[Violeta Hernández]

i.e. something larger than 2^(w^w + 1)

[Aaron Liu]

ONAG claims that the first cubic extension "probably" happens at ε₀

[Aaron Liu]

or is that ε₁ I really can't tell

[Django Peeters]

Violeta Hernández said:

I think Siegel claims e_1 without proof

Lenstra proved that lower bound.

[Django Peeters]

e_1 is the quadratic closure of e_0

[Violeta Hernández]

That's yet another formalization goal!

[Aaron Liu]

so many goals

[Violeta Hernández]

That's the fun thing about mathematics

[Django Peeters]

:shaking_face:

[Aaron Liu]

Violeta Hernández said:

If you look at my construction of algClosure, you'll notice that it's defined as the iterated supremum of a set algClosureSet

can't find this construction :(

[Violeta Hernández]

@Aaron Liu check branch vi.alg_closed

[Violeta Hernández]

Also, I have a proof that w_a is algebraically closed for every a ≠ 0.

[Violeta Hernández]

Not in Lean obv, but we could work on that

[Violeta Hernández]

Let me explain

[Aaron Liu]

what properties of w_a does it use

[Violeta Hernández]

Just a sec I literally just figured this out in the shower

[Violeta Hernández]

Ok so, w_a is obviously a group, since w_a = 2^(w_a)

[Violeta Hernández]

We can prove quite easily that it's a ring, by proving x, y < w_a -> x * y < w_a inductively. x * y is the mex of a set with cardinality x.card * y.card < aleph_a * aleph_a = aleph_a, all whose elements are less than w_a (by induction), so x * y < w _a.

[Violeta Hernández]

We can prove w_a is a field (for a ≠ 0) by using induction, and the inductive characterization. If x < w_a, x^-1 is the mex of a set of cardinality max(aleph_0, x.card) < w_a of nimbers less than w_a.

[Violeta Hernández]

Now, to prove that it's algebraically closed. You can use essentially the same algClosed argument, but this time, only including roots of polynomials that are less than w_a.

[Violeta Hernández]

i.e. in rootSet, you say r \in rootSet x when it's the root of a polynomial with coefficients < x, and also, r < w_a.

[Violeta Hernández]

So as before, you can prove algClosed x is a field, but now, x < w_a.

[Violeta Hernández]

Prove by induction that all polynomials in w_a have roots less than w_a. If one p doesn't but all the previous ones do, you can take x < w_a greater than all coefficients, and you have leastNotSplit (algClosed x) = p, which means algClosed x < w_a is a root, a contradiction.

[Violeta Hernández]

Actually, small caveat. I think this doesn't work for arbitrary w_a, only for w_(a + 1)

[Django Peeters]

Interesting

[Violeta Hernández]

Since you need to use cof (w_(a + 1)) = aleph_(a + 1)

[Aaron Liu]

docs#Cardinal.IsRegular

[Violeta Hernández]

Yep

[Aaron Liu]

so does it work for successor cardinals or regular cardinals

[Violeta Hernández]

Regular cardinals, I think

[Django Peeters]

Surely, not every algebraically closed field would be of this form, right? Besides the first 3.

[Violeta Hernández]

Or rather x.ord for x regular

[Violeta Hernández]

Django Peeters said:

Surely, not every algebraically closed field would be of this form, right? Besides the first 3.

You can prove that as a corollary. As I mentioned, algClosure' a x < w_a for x < w_a, where algClosure' a restricts to roots below w_a. But if you know IsAlgClosed (w_a), then algClosure' a x = algClosure x.

[Violeta Hernández]

Which means there's actually countably many countable ordinals that are algebraically closed!

[Violeta Hernández]

I hope I'm not saying nonsense I really do have to emphasize this is a shower thought

[Aaron Liu]

the way you know it's not nonsense is by putting it in Lean

[Violeta Hernández]

We should start by proving that regular cardinals are fields

[Aaron Liu]

that's how I know my proofs aren't nonsense

[Violeta Hernández]

I suspect we will eventually run into issues surrounding cofinality. That file is a mess, and I haven't been able to refactor it.

[Violeta Hernández]

Maybe I'll do that first

[Violeta Hernández]

Btw, w_a is a group for any a

[Violeta Hernández]

You don't need to assume regularity

[Violeta Hernández]

So a conjecture that would make sense is that the same is true for rings/fields/alg closed nimbers

[Violeta Hernández]

No actually, that should be true too! Assuming that what I said in the regular case is true.

[Violeta Hernández]

e.g. w_w is a ring, because if x, y < w_w, you can find n with x < w_n and y < w_n, meaning x * y < w_n < w_w

[Aaron Liu]

every cardinal is a limit of regular cardinals?

[Violeta Hernández]

So yeah, in reality, all infinite initial ordinals (save for potentially w) are groups, rings, fields, and algebraically closed

[Violeta Hernández]

Again, if you believe what I said earlier.

[Aaron Liu]

well of course since every limit cardinal is a limit of successor cardinals

[Violeta Hernández]

There's an important question to be asked here

[Violeta Hernández]

Is this a novel result?

[Aaron Liu]

well it'd be good to try to formalize either way

[Violeta Hernández]

well I ask because if it isn't then I have much more confidence that it's true

[Violeta Hernández]

(though if it is, I should probably look into publishing it somewhere, assuming it's correct)

[Violeta Hernández]

I think I'll start by just writing this down on a notebook

[Violeta Hernández]

My head is an unreliable sea of monsters

[Django Peeters]

Lol, gn

[Violeta Hernández]

Oh yeah, everything up to IsField (∗ω_ x) is kind of trivial

[Violeta Hernández]

It follows from the fact that for s : Set Ordinal.{u} and x : Cardinal.{u}, #s < lift.{u + 1} x implies sInf (sᶜ) < x.ord

[Violeta Hernández]

I might have PR'd that to Mathlib already

[Aaron Liu]

Cardinals and Ordinals are tricky

[Violeta Hernández]

well i'm sort of the expert when it comes to working in them in Lean

[Violeta Hernández]

In the case of mutiplication, if x, y < ω_ a, the set of excludents for x * y has cardinality at most x.card * y.card < ℵ_ a

[Violeta Hernández]

For inverses, the excludents are invSet x, which have cardinality at most #(List x.toType) ≤ max x.card ℵ₀ < ℵ_ a

[Violeta Hernández]

So yeah, I'm pretty confident in the claim a ≠ 0 → IsField (∗ω_ a)

[Violeta Hernández]

As for being algebraically complete, it probably serves to write the argument with a bit more detail.

[Violeta Hernández]

(I'll let you guys speak first)

[Aaron Liu]

I have a bunch of stuff about transfinitely iterating monotone inflationary functions on Set X that I didn't end up doing anything with

[Aaron Liu]

I proved some theorems like that this eventually stabilizes to a closure operator

[Violeta Hernández]

That does seem potentially useful for some of this.

[Violeta Hernández]

I'll continue.

[Violeta Hernández]

Define rootSet' a x (for x < ω_ a) as the set of all roots less than ω_ a of all polynomials with coefficients less than x. By a straightforward cardinality argument, if a ≠ 0, then #(rootSet' a x) ≤ #(ℕ × (ℕ → Iio x)) ≤ max ℵ₀ x.card < ℵ_ a (here omitting lifts for clarity).

[Violeta Hernández]

Define algClosureSet' a x = Iio x + Iio x ∪ Iio x * Iio x ∪ rootSet' a x, which is (by what we've established) a subset of Iio (ω_ a) with cardinality less than ℵ_ a.

[Violeta Hernández]

Define algClosure' a x as ⨆ n : ℕ, (fun y ↦ sSup (succ '' algClosureSet' a y))^[n] x. That is, you take algClosureSet' a x, take its least strict upper bound, and repeat this ω₀ many times. This will give you something at most ω_ a.

[Violeta Hernández]

It's straightforward to see that IsField (algClosure' a x), and that any root less than ω_ a of any polynomial with coefficients less than algClosure' a x is less than algClosure' a x.

[Junyan Xu]

This is probably an easier way to show that any uncountable initial ordinal x is a group, ring, field, and algebraically closed: e.g. to show it's a group it suffices to show every ordinal y < x there is some group z such that y < z < x. If there is no such z, then every such z is the sum of two smaller ordinals; by induction every such z is in the additive closure of [0,y], whose cardinality doesn't exceed max(#[0,y], aleph0), so #[0,x)=#(y,x)≤max(#[0,y], aleph0), contradicting that y is an uncountable initial ordinal. For the algebraically closed case you might want to use an expression like algebraicClosure (Subfield.closure _) Nimber.

[Violeta Hernández]

Hmm wait, let me think about how that would look like for the algebraically closed case

[Violeta Hernández]

To prove x is algebraically closed, it suffices to prove that for every y < x there's a y < z < x that is algebraically closed.

[Violeta Hernández]

Suppose that's not the case. We already know (or rather it's easy to show) there's a cofinal subset of fields between y and x , and each element in them has to be a root of a polynomial using smaller coefficients.

[Violeta Hernández]

...but does it follow that they have to be in the algebraic closure of Iio y?

[Aaron Liu]

why do you like contradiction so much

[Violeta Hernández]

I'm just following Junyan's argument

[Violeta Hernández]

I see how it works for groups, but I don't see how it generalizes for rings/fields/algebraically closed fields

[Aaron Liu]

ok suppose you have x where cof x ≥ ω₀ and a polynomial with coefficients all smaller than x, then it's a polynomial with coefficients all smaller than some y < x so there exists z ∈ Ioo y x algebraically closed so the polynomial has a root

[Aaron Liu]

does that work?

[Violeta Hernández]

so there exists z ∈ Ioo y x algebraically closed

That's what we're trying to prove!

[Aaron Liu]

it's by assumption

[Aaron Liu]

Violeta Hernández said:

To prove x is algebraically closed, it suffices to prove that for every y < x there's a y < z < x that is algebraically closed.

[Aaron Liu]

so we're assuming that for every y < x there's a y < z < x that is algebraically closed.

[Violeta Hernández]

No, we're trying to prove that

[Aaron Liu]

oh I must have misunderstood

[Violeta Hernández]

Well, let me continue writing down my proof

[Violeta Hernández]

We can then see if this other argument works

[Aaron Liu]

wait so what's x

[Violeta Hernández]

The initial ordinal

[Violeta Hernández]

Violeta Hernández said:

Define algClosure' a x as ⨆ n : ℕ, (fun y ↦ sSup (succ '' algClosureSet' a y))^[n] x. That is, you take algClosureSet' a x, take its least strict upper bound, and repeat this ω₀ many times. This will give you something at most ω_ a.

Let's suppose ω_ a is regular, so that we get something strictly smaller than ω_ a. The general case follows from the fact every cardinal is the limit of regular cardinals (or a regular cardinal itself)

[Violeta Hernández]

Violeta Hernández said:

It's straightforward to see that IsField (algClosure' a x), and that any root less than ω_ a of any polynomial with coefficients less than algClosure' a x is less than algClosure' a x.

So, we now prove that ω_ a is algebraically closed by contradication. Suppose it isn't, let p be the (lexicographically) least polynomial of positive degree in ω_ a without a root in ω_ a. Let x < ω_ a be larger than all coefficients of x. Then leastNotSplit (algClosure' a x) = p, meaning p has the root algClosure' a x, which is less than ω_ a. And we're done.

[Junyan Xu]

If z is not an algebraically closed field, then either it's the sum of two earlier ordinals, or it's the product, or it's an inverse of an earlier ordinal, or it's a root of a polynomial with earlier coefficients; in all cases it's contained in algebraicClosure (Subfield.closure (Iio z)) Nimber.

[Violeta Hernández]

Oh! So you're using all the simplest extension theorems at once!

[Violeta Hernández]

That's really elegant.

[Violeta Hernández]

Yeah ok that's way better

[Violeta Hernández]

So you can inductively prove the entire interval Iio (ω_ a) is contained in the algebraic closure, which is a contradiction by a straightforward cardinality argument.

[Violeta Hernández]

Awesome!

[Violeta Hernández]

And we don't have to use all of these extra definitions

[Violeta Hernández]

This is way simpler than I was expecting

[Junyan Xu]

To do induction we need to show algebraicClosure (Subfield.closure ·) Nimber is a closure operator to get idempotency/transitivity. I think we have some similar code in mathlib ... Nevermind, I think it's easier to prove inductively every y < z < x is contained in algebraicClosure (Subfield.closure (Iic y)) Nimber using that it's an algebraically closed field. There's e.g. Polynomial.Splits.mem_subfield_of_isRoot.

[Violeta Hernández]

Wait. This might be even simpler.

[Violeta Hernández]

Can't you prove that any subgroup of the nimbers is a lower set?

[Aaron Liu]

no

[Aaron Liu]

{0, 2} is an AddSubgroup

[Violeta Hernández]

Oh, right :frown:

[Violeta Hernández]

Well, you can at least prove the interval up to the first missing element is a group

[Violeta Hernández]

Similarly for rings, fields, and algebraically closed fields

[Violeta Hernández]

Which I guess is another way to think of what we're doing

[Violeta Hernández]

If x < w_ a, we can take the algebraic closure of Iio x and take the smallest y not in it, then x ≤ y < w_ a is an algebraically closed field.

[Violeta Hernández]

These are completely equivalent approaches, though arguably this result of "the least nimber not in a sub-X is a sub-X" is a worthwhile lemma to PR in its own right.

[Violeta Hernández]

Wait. So not every subgroup/etc. is a lower set. But it is true that the group/ring/field/algebraic closure of a lower set is a lower set, right?

[Violeta Hernández]

By the argument I gave of "the smallest nimber not in it is a subgroup/etc."

[Aaron Liu]

which closure are you using

[Violeta Hernández]

Any of the four

[Aaron Liu]

I will think a bit

[Aaron Liu]

yeas that sounds correct

[Violeta Hernández]

That's a really cool result

[Aaron Liu]

wait that proves that the algebraic closure of w^w^w is countable

[Violeta Hernández]

Indeed

[Aaron Liu]

the algebraic closure of any countable ordinal is countable

[Violeta Hernández]

In fact, it proves there's uncountably many countable ordinals that are algebraically closed

[Aaron Liu]

Violeta Hernández said:

Which means there's actually countably many countable ordinals that are algebraically closed!

[Aaron Liu]

so are there countably many or uncountably many?

[Aaron Liu]

I sure hope it's not both

[Violeta Hernández]

w_1 ordinals below w_1

[Aaron Liu]

so uncountably many

[Violeta Hernández]

Yep

[Junyan Xu]

Violeta Hernández said:

Wait. So not every subgroup/etc. is a lower set. But it is true that the group/ring/field/algebraic closure of a lower set is a lower set, right?

Great observation! Here's a simple proof in terms of general closure operators: every lower set in Ordinal can be written as [0,y) for some y. Given y there's a minimal ordinal x not contained in the closure of [0,y) (here we assume that the closure is set-sized), and we can show that [0,x) is precisely the closure of [0,y), assuming a "simplest extension theorem" for the closure operator.

By the choice of x we already know that [0,x) is contained in the closure of [0,y). For the other containment, it suffices to show [0,x) is closed; suppose it's not, then the simplest extension theorem says that x is contained in the closure of [0,x), which is contained in the closure of the closure of [0,y), which equals the closure of [0,y) by idempotency, and this contradicts the choice of x.

[Violeta Hernández]

Do we have definitions in Lean to talk about e.g. the smallest subgroup, or smallest subring, containing a given set?

[Yuyang Zhao]

docs#AddSubgroup.closure docs#Subring.closure

[Violeta Hernández]

Opened #200, which defines IsField.embed, i.e. the map from Nimber[X] to (h.toSubfield _)[X].

[Violeta Hernández]

Yuyang Zhao said:

docs#AddSubgroup.closure docs#Subring.closure

Why isn't there a simp lemma turning s.carrier into ↑s?

[Violeta Hernández]

Tried proving that the AddSubgroup.closure of a lower set is a lower set, but ran into two issues. One is the lack of docs#IsLowerSet.eq_univ_or_Iio, which is just a matter of bumping Mathlib, I think. But the other is proving that Small.{u} s → Small.{u} (AddSubgroup.closure s).

[Violeta Hernández]

Presumably this is just a cardinality argument. Do we have anything like that in Mathlib?

[Violeta Hernández]

More specifically, the sets AddSubgroup.closure s, Subring.closure s, Subfield.closure s, and algebraicClosure s K should all be bounded in cardinality by max (#s) (ℵ₀), which proves they're small through docs#Cardinal.small_iff_lift_mk_lt_univ.

[Violeta Hernández]

I think I proved some results about the cardinality of algebraic closures at some point, but now I can't find them, which presumably means they got reworked into something better

[Violeta Hernández]

We have docs#Subfield.cardinalMk_closure_le_max which is great, but it begs the question of why we don't have the other ones.

[Violeta Hernández]

Currently reviving some old Mathlib PRs on the Cantor normal form. Hopefully we're able to use its API as a basis for oeval soon.

[Violeta Hernández]

#28586 defines the CNF as a finitely supported function.

[Junyan Xu]

For algebraic closure there is Algebra.IsAlgebraic.cardinalMk_le_max. This should be enough when combined with docs#Subfield.cardinalMk_closure_le_max. If you want the results for e.g. Subring.closure, you can combine Algebra.adjoin_eq_ring_closure, FreeAlgebra.adjoin_range_ι and FreeAlgebra.cardinalMk_le_max. FreeAlgebra is a "universal object" that the docstring of docs#Subfield.cardinalMk_closure_le_max refers to. For AddSubgroup.closure there is FreeAddGroup.range_lift_eq_closure and Cardinal.mk_freeAddGroup; for AddSubmonoid.closure there is FreeAddMonoid.mrange_lift and FreeAddMonoid is just List.

[Junyan Xu]

Django Peeters said:

I'm gonna switch back to C++ tomorrow. Maybe we'll know alpha(263) in the near future.

image.png
263 was beaten, 283 is the next boss
https://discord.com/channels/842810949706055690/895374513020207175/1332371437817692243

[Tristan Figueroa-Reid]

( @Django Peeters if you do want to parallelize it, this same machine has a weak single-core performance, but it has 128 cores.)

[Django Peeters]

We're going for alpha(263) in 1 day. I'm not sure about alpha(283) but it might take 10 days.

[Django Peeters]

Tristan Figueroa-Reid said:

( Django Peeters if you do want to parallelize it, this same machine has a weak single-core performance, but it has 128 cores.)

What's there to parallelize?

[Django Peeters]

Oh for different excesses

[Django Peeters]

That might be possible yes.

[Django Peeters]

It already uses 2 threads, 1 for calculating and another for logging. I also noticed the big exponents had a lot of repetition in them that I could leverage.

[Junyan Xu]

it's not written in cgsuite, it's all heavily optimized scala code. Rewriting it in C or Rust would probably give a speedup but not an earth-shattering one, I'd be very surprised if it were more than 2x.

from Aaron Siegel

[Django Peeters]

Django Peeters said:

Btw, here are some speed gains:
http://www.neverendingbooks.org/aaron-siegel-on-transfinite-number-hacking/
now becomes
image.png

I haven't run the scala code, only translated from it. And I'm not sure which code gave these time results.

[Junyan Xu]

Are the results in http://www.neverendingbooks.org/aaron-siegel-on-transfinite-number-hacking/ generated by the Scala code?

[Django Peeters]

You'd have to ask Aaron.

[Junyan Xu]

I think probably yes, but maybe many optimizations have been made since 2012.

[Junyan Xu]

He isn't expecting much gain from porting to C, but there should still be gains.

[Tristan Figueroa-Reid]

I would rarely ever trust anyone with guessing performance this close to the system. @Django Peeters's program (assuming Siegel's 5.2 day time for calculating alpha(263)) seem to be nearly 3x as fast, at least based on the current output.

[Django Peeters]

While this is on 1 core and 5.2 days for Scala on 128 cores?

[Tristan Figueroa-Reid]

(In fairness, while both of these are single-core, we also don't know the clock speed of the respective CPUs. This one is 2.9GHz - did Siegel ever say anything about system specs?)

[Django Peeters]

If he did, I'm not aware of any.

[Junyan Xu]

it's a 3 GHz Intel Xeon. There are faster CPUs out there. Memory speed is probably a bottleneck too.

[Junyan Xu]

I can see that Tristan and Violeta both replied in the #cgsuite channel of the Gamester Discord server where the p=263 result was announced, but Tristan only before the announcement. (Actually Aaron Siegel redid the computation prompted by questions from Violeta.) Maybe it's just been too long that you've both forgotten about it ... (tbh I also forgot about the exact prime)

[Tristan Figueroa-Reid]

I remembered that Violeta asked about this earlier, but I forgot that it resulted in alpha(263) being calculated.

[Violeta Hernández]

Probably should have checked...

[Violeta Hernández]

Yeah, I did ask about this, but I forgot the exact values that were calculated

[Violeta Hernández]

Yeah, I did ask about this, but I forgot the exact values that were calculated

[Violeta Hernández]

We should really make an OEIS sequence for this

[Aaron Liu]

They aren't really integers

[Django Peeters]

But the excesses are

[Django Peeters]

And the q-sets are more easily deduced from them

[Aaron Liu]

I have no idea what that means

[Django Peeters]

alpha(p) = kappa(f(p)) + excess(p) where f(p) is the least h>0 such that p|(2^h-1).

[Aaron Liu]

What's a kappa

[Django Peeters]

kappa(h) is the least nimber which has a degree divisible by h

[Django Peeters]

Over the prime field

[Aaron Liu]

f(p) seems like it can get pretty big (long division)

[Django Peeters]

Yes

[Aaron Liu]

I think f(p) < p but I'm not too sure

[Violeta Hernández]

What's the paper where these values are defined again?

[Django Peeters]

f(p) | (p-1)

[Django Peeters]

It's in Lenstra's "On the algebraic closure of 2"

[Violeta Hernández]

@Aaron Liu have you managed to prove Fermat powers are fields?

[Violeta Hernández]

On an unrelated note, I just computed this:
b051917.txt

[Violeta Hernández]

Look at the first few terms after 256

[Violeta Hernández]

Is there a reason $n^{-1}$ and $(n + 1)^{-1}$ seem to be numerically close to each other?

[Aaron Liu]

Violeta Hernández said:

Aaron Liu have you managed to prove Fermat powers are fields?

Working on it

[Aaron Liu]

Violeta Hernández said:

Is there a reason $n^{-1}$ and $(n + 1)^{-1}$ seem to be numerically close to each other?

Have you tried induction

[Violeta Hernández]

Well I'm not making any concrete mathematical claim

[Violeta Hernández]

Just a numerical observation

[Violeta Hernández]

Also, just uploaded b-files up to 65536 for the OEIS sequences on nim squares and nim square roots as well

[Violeta Hernández]

Made them with the aforementioned Rust library; obviously that's not super reassuring in terms of proving correctness, but the output matched all existing entries, and it passed some double-checking

[Aaron Liu]

Violeta Hernández said:

Aaron Liu have you managed to prove Fermat powers are fields?

which branch should I do this on

[Violeta Hernández]

You should probably work on top of the alg_closed branch

[Violeta Hernández]

Which reminds me I should continue moving stuff from there to master

[Violeta Hernández]

Oh yeah, #200 hasn't been reviewed yet

[Aaron Liu]

ok I'll stash everything and unstash them on the other branch

[Violeta Hernández]

I can start work on the NatNimber stuff, e.g. doing the algorithms

[Violeta Hernández]

(proving correctness will require your work)

[Violeta Hernández]

Oh I just found the compact definition of transfinite Veblen

[Aaron Liu]

what is it

[Aaron Liu]

Some fixpoint probably

[Violeta Hernández]

Define the degree of a finitely supported function Ord -> Ord as the smallest non-zero index (so the opposite of how you'd do it for a polynomial). Let f be a finitely supported function Ord -> Ord of degree n ≠ 0. Then x => phi(f + x@0) enumerates the common fixed points of x => phi(g, x@m) for g < f, deg(g) ≥ deg(f), and m < n.

[Violeta Hernández]

"compact"

[Violeta Hernández]

It's nicer to give examples

[Violeta Hernández]

e.g. phi(1,2,0,4) is the 4th fixed point of x => phi(1,1,x,0) (and a bunch of other functions whose fixed points are a superset of this)

[Violeta Hernández]

SVO = phi(1@w) is the smallest fixed point of x => phi(x@n) for n < w

[Violeta Hernández]

And then of course the LVO is the smallest fixed point of x => phi(1@x)

[Violeta Hernández]

nvm

[Aaron Liu]

?????

[Violeta Hernández]

No yeah I was missing that condition

[Violeta Hernández]

deg(g) ≥ deg(f) and g < f is too weak

[Aaron Liu]

This causes me great confusion

[Violeta Hernández]

Otherwise you get nonsense such as phi(1, 1, 0) being a fixed point of x => phi(LVO, x)

[Violeta Hernández]

(or replace LVO by your favorite huge cardinal)

[Aaron Liu]

Is it not?

[Violeta Hernández]

Of course not

[Aaron Liu]

oh I guess that wouldn't be countable

[Violeta Hernández]

You do have that phi(1, 1, 0) is a fixed point of x => phi(1, 0, x)

[Violeta Hernández]

And I think you can prove it's actually a fixed point of x => phi(y, x) for y < phi(1, 1, 0)

[Violeta Hernández]

But this latter point isn't part of the definition

[Violeta Hernández]

So yeah, if n = deg(f) ≠ 0, then x => phi(f + x@0) enumerates the fixed points of x => phi(g, x@m), where m < n, but where also g is f with the leading coefficient decreased

[Violeta Hernández]

This ensures that you're only taking fixed points of a set of functions, rather than potentially a class

[Aaron Liu]

sure

[Violeta Hernández]

Or I guess another way to fix my original definition is to require g < f pointwise, and not lexicographically

[Violeta Hernández]

That should be equivalent I think?

[Violeta Hernández]

I feel like this is one of those topics in mathematics that's really difficult to work with not because the math is difficult but because the notation is godawful

[Violeta Hernández]

I need to do a write-up on this and the LVO someday

[Aaron Liu]

so do you have any better notation

[Django Peeters]

There's BMS

[Django Peeters]

Taranovsky's C notation

[Django Peeters]

Y-sequence

[Django Peeters]

A lot of notations from the googology community...

[Django Peeters]

Idk if they're better

[Violeta Hernández]

Ah yes, the googology community

[Violeta Hernández]

I'm well acquainted with them

[Django Peeters]

Lol

[Aaron Liu]

I don't get it

[Django Peeters]

Buchholz's psi is a good one I think, from an actual mathematician

[Django Peeters]

Taranovsky's C is also from a n actual mathematician I think

[Violeta Hernández]

Aaron Liu said:

so do you have any better notation

Well I just introduced that degree definition to make the definition nicer

[Violeta Hernández]

The a@b notation isn't mine, but I've seen a few texts try to define the w-argument Veblen function without it and end up a horrible mess

[Violeta Hernández]

(when really the w-argument version is but a special case of the transfinite argument version)

[Aaron Liu]

everything is a special case

[Violeta Hernández]

Django Peeters said:

Buchholz's psi is a good one I think, from an actual mathematician

Oh I'm not talking about ordinal notations, but rather notation for ordinals!

[Violeta Hernández]

At some point I tried formalizing Buchholz's psi in Lean, but that sort of fell through.

[Violeta Hernández]

Aaron Liu said:

so do you have any better notation

Oh and also a lot of sources try to define successor and limit cases separately, which really isn't necessary

[Violeta Hernández]

Which is the main reason I'm of the view that "most things about ordinals only need recursion, and not limit recursion"

[Django Peeters]

Violeta Hernández said:

Django Peeters said:

Buchholz's psi is a good one I think, from an actual mathematician

Oh I'm not talking about ordinal notations, but rather notation for ordinals!

What's the difference?

[Violeta Hernández]

Ordinal notations are well-ordered syntax trees with effective arithmetic modeling that of the ordinals

[Violeta Hernández]

I'm just talking about how to write them down

[Django Peeters]

Violeta Hernández said:

Ordinal notations are well-ordered syntax trees with effective arithmetic modeling that of the ordinals

Like ordinal collapsing functions?

[Violeta Hernández]

Ordinal collapsing functions are that, functions

[Violeta Hernández]

But they often give rise to an ordinal notation

[Violeta Hernández]

Violeta Hernández said:

Ordinal notations are well-ordered syntax trees with effective arithmetic modeling that of the ordinals

In simpler words, "ordinals as strings"

[Violeta Hernández]

So for instance, every ordinal has a Cantor normal form

[Violeta Hernández]

But there's also a Cantor normal form ordinal notation, containing strings such as "1" and "ω^2 * 3" and "ω^(ω^(ω^6 * 7 + ω^2 + 4) + ω * 15) + 37"

[Violeta Hernández]

It contains a unique representation for every ordinal below ε₀, and there are algorithms to perform all basic ordinal operations on them

[Django Peeters]

So as a corollary, notation for ordinals only gives you a countable subset of ordinals.

[Violeta Hernández]

Yep

[Violeta Hernández]

So of course the holy grail is to find the "simplest" ordinal notation giving you the largest subset of countable ordinals

[Violeta Hernández]

Buchholz's ψ is pretty good for what it is

[Django Peeters]

Is this related to the first Church-Kleene ordinal?

[Violeta Hernández]

The LVO is Ψ₀(Ω^Ω^Ω), the limit of the notation is Ψ₀(ε_(Ω+1))

[Aaron Liu]

What even is the church kleene ordinal

[Violeta Hernández]

Django Peeters said:

Is this related to the first Church-Kleene ordinal?

Yes, in the sense that an ordinal notation with effective ordering cannot get to or exceed said ordinal

[Violeta Hernández]

Aaron Liu said:

What even is the church kleene ordinal

The smallest ordinal such that there is no computable well-ordering of (a subset of) the naturals with said order type

[Django Peeters]

Violeta Hernández said:

So of course the holy grail is to find the "simplest" ordinal notation giving you the largest subset of countable ordinals

I'm surprised it exists!

[Violeta Hernández]

Well, we don't really know if it exists! There are ordinal notations going beyond Buchholz, but they're increasingly convoluted, except for a particular one I believe that no one understands?

[Aaron Liu]

what does that mean

[Violeta Hernández]

I don't know, that's beyond my scope

[Aaron Liu]

Violeta Hernández said:

The LVO is Ψ₀(Ω^Ω^Ω), the limit of the notation is Ψ₀(ε_(Ω+1))

Ψ₀? 0??? Is there a Ψ₁ too?

[Violeta Hernández]

Yes, but Ψ₁ returns ordinals larger than ω₁

[Violeta Hernández]

In fact I'm just remembering that the extended Buchholz ordinal notation exists

[Aaron Liu]

who even needs all these ordinals

[Violeta Hernández]

The idea is that you have functions Ψᵥ giving ordinals of cardinality ℵᵥ (except when v = 0, where you get countable ordinals)

[Aaron Liu]

so it's another ordinal index

[Aaron Liu]

all the numbers are ordinals

[Django Peeters]

Violeta Hernández said:

Well, we don't really know if it exists! There are ordinal notations going beyond Buchholz, but they're increasingly convoluted, except for a particular one I believe that no one understands?

Which one?

[Violeta Hernández]

The limit of that notation is the limit of Ψ₀(0), Ψ₀(Ψ₀(0)), Ψ₀(Ψ_(Ψ₀(0))), Ψ₀(Ψ_Ψ_(Ψ₀(0))), ...

[Violeta Hernández]

really big ordinal

[Aaron Liu]

Violeta Hernández said:

Aaron Liu said:

What even is the church kleene ordinal

The smallest ordinal such that there is no computable well-ordering of (a subset of) the naturals with said order type

a computable subset or any subset

[Violeta Hernández]

Aaron Liu said:

Violeta Hernández said:

Aaron Liu said:

What even is the church kleene ordinal

The smallest ordinal such that there is no computable well-ordering of (a subset of) the naturals with said order type

a computable subset or any subset

computable subset

[Violeta Hernández]

Really the subset clause is just there so that the answer isn't 0 or some finite natural

[Violeta Hernández]

I think it's equivalent to say "the smallest infinite ordinal such that there's no computable well-ordering of the naturals with said order type"

[Violeta Hernández]

Aaron Liu said:

who even needs all these ordinals

Well depending on how long the next algebraically closed subfield of nimbers takes to arrive, perhaps us

[Django Peeters]

Who knows, maybe nimbers give a unified way of going to better and better ordinal notations

[Violeta Hernández]

We can dream

[Violeta Hernández]

I think every single possible answer for "what the next algebraically closed subfield is" would be really interesting

[Aaron Liu]

can't we just compute it?

[Violeta Hernández]

Do you have a computer that can perform operations with arbitrary countable ordinals

[Aaron Liu]

of course not

[Aaron Liu]

that would solve the halting problem

[Violeta Hernández]

Violeta Hernández said:

I think every single possible answer for "what the next algebraically closed subfield is" would be really interesting

I think it'd be even more interesting if the first ω algebraically closed fields had simple representations in some known or unknown ordinal notation

[Violeta Hernández]

Like, maybe the next ordinal is Ψ₀(Ω^Ω^Ω), and the next one is Ψ₀(Ω^Ω^Ω^Ω), and so on

[Django Peeters]

I wonder at what point the arithmetic isn't going to be effectively computable

[Violeta Hernández]

Is it still effectively computable past ω^ω^ω?

[Aaron Liu]

yes

[Django Peeters]

Lenstra thinks yes in e_0

[Violeta Hernández]

I really need to read Lenstra's paper

[Violeta Hernández]

(deleted)

[Violeta Hernández]

Wait, something I'm not clear about

[Violeta Hernández]

Are the values alpha and kappa originally defined by Conway, or by Lenstra?

[Django Peeters]

Alpha by conway, and kappa by Lenstra because he needed them for effective computing.

[Django Peeters]

I think.

[Django Peeters]

Conway already encounterd kappa(prime powers) but he didn't give em a name.

[Violeta Hernández]

hmm, I'm struggling a bit.
image.png

[Violeta Hernández]

"every power of 2 has a unique expression as a decreasing product"
Why?

[Aaron Liu]

what's a [power]

[Violeta Hernández]

They're using the conway notation where [a + b] means ordinal addition and [a] + [b] means nimber addition

[Aaron Liu]

try taking the log

[Django Peeters]

Violeta Hernández said:

"every power of 2 has a unique expression as a decreasing product"
Why?

I think they mean: for 2^n we take the binary expansion of n (so 2^n_k + ... + 2^n_0) where n_k>...>n_0, and then put that in 2^•

[Aaron Liu]

oh they're also considering infinite powers

[Violeta Hernández]

Oh wait nevermind I see why that's true

[Violeta Hernández]

The terms κ_q^m(q) for q = p^n give you the term ω^k where p is the k-th prime

[Violeta Hernández]

(the k + 1 thing is just zero-indexing shenanigans)

[Aaron Liu]

why not zero indexing

[Violeta Hernández]

Well I'm zero-indexing but I think Lenstra isn't

[Aaron Liu]

oh no

[Violeta Hernández]

Now, what's going on here?
image.png

[Violeta Hernández]

Does Lenstra prove this? Or is this something Conway said somewhere else

[Django Peeters]

Well, 1st is for index 0 I guess

[Django Peeters]

Violeta Hernández said:

Does Lenstra prove this? Or is this something Conway said somewhere else

I don't think he does.

[Aaron Liu]

Violeta Hernández said:

Now, what's going on here?
image.png

math

[Aaron Liu]

try using the simplest extension theorems

[Aaron Liu]

and induction too

[Violeta Hernández]

Actually I feel like I should be reading theorem 49 in ONAG first

[Violeta Hernández]

I swear Conway's writing has some of the highest "Lean LOC to text" ratio I've seen

[Violeta Hernández]

I don't think ONAG contains that theorem about multiplication of kappas either

[Violeta Hernández]

time to try Siegel

[Aaron Liu]

Violeta Hernández said:

I swear Conway's writing has some of the highest "Lean LOC to text" ratio I've seen

what about the theory of finiteness

[Violeta Hernández]

wdym?

[Aaron Liu]

stuff like Nonempty (Fin n ≃ Fin m) ↔ n = m

[Violeta Hernández]

Well that stuff isn't trivial to prove in pen-and-paper math either

[Violeta Hernández]

You just usually take it for granted, which is a different problem

[Aaron Liu]

but this stuff also isn't trivial to prove

[Aaron Liu]

Conway just skips a bunch

[Violeta Hernández]

Hm yeah that's a better assessment of the situation

[Violeta Hernández]

He does in fact skip a bunch

[Violeta Hernández]

(surreal multiplication PTSD)

[Violeta Hernández]

Oh unrelated, but I just found this
image.png

[Aaron Liu]

oh I saw that too

[Violeta Hernández]

So it's not only that every finite ring is a field, but every ring below the first transcendental is a field

[Violeta Hernández]

...why?

[Aaron Liu]

did you read the thing

[Violeta Hernández]

Yeah but

[Violeta Hernández]

Why does every element of an algebraic field extension have finite order?

[Aaron Liu]

check the field theory

[Violeta Hernández]

well yeah that's why I'm asking

[Violeta Hernández]

holy moly deepseek is cracked

[Aaron Liu]

did it tell you the answer

[Violeta Hernández]

It did

[Violeta Hernández]

If $\alpha$ is an element in the algebraic extension, then $[\mathbb F_2(\alpha):\mathbb F_2]$ is finite, meaning $\mathbb F_2(\alpha)$ is a finite field, meaning the multiplicative order of $\alpha$ is finite

[Violeta Hernández]

So it of course has an inverse

[Aaron Liu]

since the algebraic closure is galois?

[Aaron Liu]

I'm a bit rusty on field theory

[Violeta Hernández]

Wait what step are you asking about

[Aaron Liu]

the part where you conclude the index is finite

[Violeta Hernández]

The degree of $[\mathbf F_2(\alpha) : \mathbf F_2]$ is the degree of the minimal polynomial of $\alpha$ over the algebraic extension

[Aaron Liu]

oh yeah of course

[Aaron Liu]

that makes sense

[Violeta Hernández]

I think this results generalizes to "if every element of a field has finite multiplicative order, so does every element in an algebraic extension"

[Violeta Hernández]

(is there a name/any known properties of such a field?)

[Aaron Liu]

for field where every element has finite multiplicative order?

[Violeta Hernández]

Yep

[Violeta Hernández]

This random Math SE user claims these fields are exactly the subfields of $\overline{\mathbf F_p}$
image.png

[Aaron Liu]

AI keeps trying to convince me that it's only finite fields

[Violeta Hernández]

yeah AI is very hit or miss

[Aaron Liu]

https://math.stackexchange.com/q/4193256

[Aaron Liu]

look what I found

[Violeta Hernández]

Whoa

[Violeta Hernández]

You know, the other day I was going to ask whether we knew what the roots of unity in the nimbers are

[Aaron Liu]

now we know

[Violeta Hernández]

But I guess the answer is literally just "every single nimber below ω^ω^ω"

[Violeta Hernández]

which is awesome

[Violeta Hernández]

Oh and awesome, by this same argument every subfield below ω^ω^ω is perfect

[Django Peeters]

Niiice

[Violeta Hernández]

Wait, why is w^3 a field?

[Violeta Hernández]

I get that the next field has to be at least w^3, but why is that a field already?

[Violeta Hernández]

Ohhhhh of course because w^3 = 2

[Violeta Hernández]

I'm thinking about that result "if x ≤ y are fields, then y is an ordinal power of x"

[Violeta Hernández]

This would follow from proving it when y is the next largest field after x

[Violeta Hernández]

Is it always true y must be at most x^w?

[Violeta Hernández]

It feels true which means it absolutely isn't

[Aaron Liu]

try it out some

[Violeta Hernández]

I conjecture it's true below w^w^w under the argument "all rings are fields"

[Violeta Hernández]

Ok you know what

[Violeta Hernández]

I just realized Siegel gives exercises

[Violeta Hernández]

I'm going to work through them

[Aaron Liu]

wait there's exercises???

[Violeta Hernández]

image.png

[Violeta Hernández]

Exercise 4.2 is the theorem I was mentioning

[Aaron Liu]

goodluck

[Violeta Hernández]

Wait.

[Violeta Hernández]

How do we know $\alpha_k$ exists?

[Aaron Liu]

wait what's $\alpha_k$ again?

[Violeta Hernández]

If $p_k$ is the k-th odd prime ($p_0=3$) then $\alpha_{p_k}$ is the least nimber with no $p$-th root in $\omega^{\omega^k}$

[Violeta Hernández]

so e.g. $\alpha_3=2$

[Aaron Liu]

so you're asking why $\omega^{\omega^k}$ does not have all $p_k$ th roots

[Violeta Hernández]

yep

[Aaron Liu]

try induction?

[Aaron Liu]

so all these fields are made by adjoining a root of unity to the previous field

[Aaron Liu]

or is it multiple roots of unity?

[Violeta Hernández]

I think it's $\omega^{\omega^k p_k^{i+1}}$ adds a root of unity to $\omega^{\omega^k p_k^i}$

[Aaron Liu]

and you start with $\omega^{\omega^0}$ which has all power of two roots but not all of any other root

[Violeta Hernández]

I imagine that's what we'd have to prove first

[Violeta Hernández]

Probably just an argument about divisors of 2^2^n - 1

[Violeta Hernández]

Actually, $\alpha_7 = \omega + 1$

[Violeta Hernández]

Does that mean $\omega$ has all seventh powers?

[Aaron Liu]

Oh

[Violeta Hernández]

(this isn't necessarily true, but still)

[Violeta Hernández]

(could be)

[Violeta Hernández]

Well actually, can a finite field of characteristic k have all p-th roots for p ≠ k?

[Aaron Liu]

the finite field $\mathbb{F}_2$ has all roots

[Violeta Hernández]

True

[Violeta Hernández]

Since $2^3 = 3^3 = 1$ no other field $\mathbb F_{2^{2^n}}$ can have all cube roots

[Violeta Hernández]

In fact, if $p$ divides some $2^{2^n}-1$ then $\mathbb F_{2^{2^n}}$ can't have all $p$-th roots and neither can any larger finite field

[Aaron Liu]

what about their limit colimit

[Violeta Hernández]

Those can't either

[Violeta Hernández]

no actually idk

[Aaron Liu]

Fin n doesn't have all successors but their colimit ℕ does

[Violeta Hernández]

Hmm

[Violeta Hernández]

Well we can think about the argument for why w doesn't have all cube roots

[Aaron Liu]

apparently 2 doesn't have a cube root

[Violeta Hernández]

You take 2 which has order 3, a cube root would have order 3 * 3 = 9, but that doesn't exist in any of the finite fields

[Violeta Hernández]

Because 9 never divides 2^2^n - 1

[Aaron Liu]

so we need a number which never divides 2^2^n-1

[Violeta Hernández]

Actually, what primes can divide that?

[Violeta Hernández]

Mersenne primes can

[Violeta Hernández]

Idk how helpful that is

[Aaron Liu]

but we care about not dividing

[Aaron Liu]

why does 9 never divide 2^2^n-1

[Aaron Liu]

well since 2 has period 6 in 9

[Violeta Hernández]

ord_2(9) = 6 and 6 |\ 2^n

[Aaron Liu]

wait don't we have an algorithm that calculates $\alpha_k$?

[Aaron Liu]

why not just prove the algorithm correct and have that they exist as a consequence

[Violeta Hernández]

Yeah but surely proving that it exists is easier than to prove the algorithm correct?

[Violeta Hernández]

Surely...

[Aaron Liu]

Violeta Hernández said:

Does that mean $\omega$ has all seventh powers?

I found this in ONAG

It follows that every finite number has a finite seventh root

[Aaron Liu]

I'll cut out this paragraph and put it here

Now the ordinals below the first transcendental are algebraic over previous ones, and so by induction algebraic over the field 2 whose only elements are 0 and 1. It follows that any finite number of such ordinals generate a finite field. Each of these ordinals $\Delta$ which is itself a field defines an algebraic extension of itself. Since these extensions are taken in order of degree where possible, the first extensions will be quadratic, and then when the field is quadratically closed we shall take cubic extensions, then quintic ones, etc. [Since the Galois group of every finite field is abelian, the quadratically closed field remains quadratically closed after taking cubic extensions, etc.]
Moreover, the quadratic extensions will all be by equations of the form $x^2 + x = \alpha$, since the only lexicographically earlier quadratics are $x^2 = \alpha$, and every element of a finite field of characteristic 2 already has a square root in that field. The cubic extensions will be by cube roots, however, since the equation $x^3 = \alpha x^2 + \beta x + \gamma$ defines an extension of the field generated by $\alpha, \beta, \gamma$ to a larger finite field, and any finite field extension of degree 3 (and characteristic 2) can be made by a cube root, corresponding to a lexicographically earlier equation. Similar comments apply to the later extensions by fifth roots, seventh roots, etc.

[Violeta Hernández]

Oh awesome the OEIS guys got back to me

[Violeta Hernández]

Told me 65k terms was way too much

[Violeta Hernández]

...even though I've seen sequences with up to 100k terms

[Aaron Liu]

So if 2^2^4 is too much then how about 2^2^3

[Violeta Hernández]

All the sequences I was expanding the b-files of already had 256 terms

[Violeta Hernández]

But that feels like too little

[Aaron Liu]

Sucks there's no integers between 3 and 4

[Violeta Hernández]

I'm wondering if they would accept the argument of "the next field of nimbers after 256 is 65536, so it's a natural point to stop the sequence"

[Violeta Hernández]

Ugh, I hate bureaucracy

[Violeta Hernández]

Actually question

[Violeta Hernández]

Why is x^5 + 4 the simplest non-split polynomial in ω^ω?

[Violeta Hernández]

I mean, I can believe it's cubically closed, but why can't it be a quartic?

[Violeta Hernández]

Oh wait I just saw the paragraph you posted

[Violeta Hernández]

"since the Galois group of every finite field is abelian, the quadratically closed field remains quadratically closed after taking cubic extensions"
Ok I really need to brush up on Galois theory

[Violeta Hernández]

Ohh wait the galois group of a finite field means, over its prime field

[Violeta Hernández]

So just the automorphism group

[Violeta Hernández]

Which is cyclic, hence abelian

[Violeta Hernández]

after reading about galois theory for two hours i still don't understand this statement

[Django Peeters]

Violeta Hernández said:

I mean, I can believe it's cubically closed, but why can't it be a quartic?

Cuz every polynomial in x with degree m can be written as a multivariate polynomial in p_1(x),...,p_k(x) with p_1,...,p_k having degrees that cover every unique prime divisor of m?

[Django Peeters]

Nah, not multivariate

[Django Peeters]

That doesn't feel right

[Django Peeters]

Maybe it's just p_1(p_2(....p_k(x)...))

[Violeta Hernández]

I think the reasoning is that any extension of degree > 2 has to be by a n-th root

[Violeta Hernández]

And if every element of the field has a square root then it of course has a fourth root too

[Django Peeters]

This really needs to be translated to Lean.

[Violeta Hernández]

If I can even figure it out!

[Violeta Hernández]

This paragraph in particular seems to hold the answer to some of my questions, but I can't quite decipher it
image.png

[Violeta Hernández]

By quadratic extension, we mean an extension of degree 2, right?

[Django Peeters]

Yes

[Violeta Hernández]

There's two things I don't get

• How do you deduce $\mathcal F = \mathcal E(\omega)$?
• Why does the fact that any quadratic extension of $\mathcal P_2(\omega)$ is isomorphic to a subfield of $\mathcal P_{\omega^3}$ prove it quadratically closed?

[Django Peeters]

For $\mathcal{F} = \mathcal{E}(\omega)$, I think you take the fixfield some subgroup of order 2.

[Django Peeters]

To find $\mathcal{E}$.

[Django Peeters]

Cuz $\mathcal{F}$ has Galoisgroup $C_6$ over $\mathcal{P}_2$.

[Violeta Hernández]

Order 2, or index 2?

[Django Peeters]

I think order 2, because then you adjoin $\omega$ which triples it to $C_6$.

[Django Peeters]

If $\mathcal{P}_{\omega^3}$ isn't quadratically closed, there is some quadratic extension which adjoins $a \geq \omega^3$. Then we look at $\mathcal{P}_2(\omega, a)$ I think.

[Django Peeters]

This must contain a quadratic extension of $\mathcal{P}_2(\omega)$.

[Violeta Hernández]

Django Peeters said:

If $\mathcal{P}_{\omega^3}$ isn't quadratically closed, there is some quadratic extension which adjoins $a \geq \omega^3$. Then we look at $\mathcal{P}_2(\omega, a)$ I think.

But there is, isn't it? e.g adjoining a root of $x^2 - \omega_1$

[Aaron Liu]

I don't think that's an algebraic extension of $\omega^3$

[Violeta Hernández]

Isn't every extension by adjoining a root of a polynomial algebraic?

[Aaron Liu]

which polynomial with coefficients in $\omega^3$ is this new root a root of

[Aaron Liu]

Violeta Hernández said:

Isn't every extension by adjoining a root of a polynomial algebraic?

I can adjoin a root of the polynomial $x - \pi$ to get $\mathbb{Q}(\pi)$ but that's not an algebraic extension of $\mathbb{Q}$ since $\pi$ isn't algebraic over $\mathbb{Q}$.

[Violeta Hernández]

Ah, of course

[Violeta Hernández]

$F(\alpha)$ is algebraic when $\alpha$ is a root of a polynomial in $F[X]$, rather

[Violeta Hernández]

So you consider that extension. Then $\mathcal P_2(a, \omega)$ is an abelian extension of degree 6, hence has Galois group $C_6$. There's a subgroup with degree 2, which means there's a subextension $\mathcal E$ of degree 2, which has to be a subfield of $\mathcal P_\omega$.

[Violeta Hernández]

Do we know that $\mathcal E = \mathcal P_2(a)$ specifically?

[Aaron Liu]

the only other choice is $\mathcal P_2(\omega)$

[Aaron Liu]

which has the wrong degree

[Violeta Hernández]

Can't $a$ be, say, a root of $x^2 - \omega$?

[Violeta Hernández]

That is an algebraic extension of $\mathcal P_2(\omega)$

[Aaron Liu]

ok I lost you

[Aaron Liu]

doesn't help that I barely know any galois theory

[Violeta Hernández]

I took an entire course on Galois theory like two years ago, but the stuff about the fundamental theorem I didn't remember all that well

[Aaron Liu]

oh I guess it could be like $\mathcal P_2(a^3)$ or something

[Django Peeters]

I got Galois theory together with more group and field theory.

[Django Peeters]

Do we have other literature to look for an argument?

[Violeta Hernández]

I haven't found another source going over this proof besides Conway and Siegel

[Django Peeters]

Violeta Hernández said:

Can't $a$ be, say, a root of $x^2 - \omega$?

It can.

[Django Peeters]

Notebook time

[Violeta Hernández]

Actually that would be equivalent to adjoining 2

[Violeta Hernández]

Since $\sqrt\omega = 3*\omega^2$

[Violeta Hernández]

So yeah, $\mathcal E = \mathcal P_2(2)$

[Django Peeters]

$\sqrt{\omega} = 2\omega^2$

[Violeta Hernández]

Ok yeah I get it now, since $\mathcal E$ is a quadratic extension of $\mathcal P_2$ it must be a subset of $\mathcal P_\omega$, and since $\omega$ has a cubic minimal polynomial over $\mathcal P_\omega$, the extension $\mathcal E(\omega)$ must be of degree 6, hence the entire thing

[Violeta Hernández]

Django Peeters said:

$\sqrt{\omega} = 2\omega^2$

No, isn't $(3\omega^2)^2 = 2*3*\omega = \omega$?

[Aaron Liu]

no I don't think so

[Aaron Liu]

I'm getting out my multiplication table

[Violeta Hernández]

Violeta Hernández said:

This paragraph in particular seems to hold the answer to some of my questions, but I can't quite decipher it
image.png

I think I was a bit confused about the use of "isomorphic" here, but since we're talking about subfields of an algebraically closed field I think you can just say "equal"

[Violeta Hernández]

Aaron Liu said:

I'm getting out my multiplication table

Oh no wait I forgot to square the 3

[Aaron Liu]

Django Peeters said:

$\sqrt{\omega} = 2\omega^2$

I think this one is correct

[Violeta Hernández]

Out here doing research level math and I still can't square monomials

[Django Peeters]

Lol, it's normal. Everyone does that once in a while.

[Aaron Liu]

Violeta Hernández said:

Violeta Hernández said:

This paragraph in particular seems to hold the answer to some of my questions, but I can't quite decipher it
image.png

I think I was a bit confused about the use of "isomorphic" here, but since we're talking about subfields of an algebraically closed field I think you can just say "equal"

You can have disequal subfields which are isomorphic

[Violeta Hernández]

Yeah, but I think in this case Siegel did mean equality

[Django Peeters]

Django Peeters said:

If $\mathcal{P}_{\omega^3}$ isn't quadratically closed, there is some quadratic extension which adjoins $a \geq \omega^3$. Then we look at $\mathcal{P}_2(\omega, a)$ I think.

How would you construct the quadratic extension of \mathcal{P}_2(\omega) from this? Because a might have a bigger degree over that subfield.

[Violeta Hernández]

Yeah, I'm not sure

[Violeta Hernández]

I think the general argument is this: if you have a quadratic extension of a finite degree subfield of $\mathcal F_{\omega^3}$, then the Galois group has a subgroup of degree 2, which must give a quadratic extension of $\mathcal F_2$, hence a subfield of $\mathcal F_\omega$, so then the entire thing is a subfield of $\mathcal F_{\omega^3}$ (?)

[Violeta Hernández]

There has to be some general theorem I'm missing

[Aaron Liu]

let's walk through it step by step

[Violeta Hernández]

"A cubic extension of a quadratically closed field with abelian Galois group is quadratically closed"

[Violeta Hernández]

Is that true in general?

[Django Peeters]

Idk, never seen that theorem

[Aaron Liu]

AI is telling me that adjoining the cube root of 2 to the quadratic closure of the rationals is a counterexample

[Violeta Hernández]

That's not quadratically closed

[Aaron Liu]

which one

[Violeta Hernández]

Oh wait the quadratic closure of the rationals

[Violeta Hernández]

Isn't the Galois group of that infinite

[Violeta Hernández]

No idea how it looks

[Aaron Liu]

what do you mean "the abelian group"

[Aaron Liu]

there are lots of abelian groups

[Aaron Liu]

Violeta Hernández said:

Isn't the Galois group of that infinite

Galois group of what over what?

[Violeta Hernández]

Over the prime field?

[Aaron Liu]

oh sure yeah that's infinite I think

[Violeta Hernández]

Maybe you need the original field to be perfect or something like that

[Violeta Hernández]

Do you think this is worth asking on Math SE

[Aaron Liu]

I'm sure we can figure it out ourselves

[Violeta Hernández]

No yeah I'm kind of stuck

[Violeta Hernández]

I feel like the puzzle pieces are there

[Violeta Hernández]

Every element below w^w^w is algebraic, so every subfield generated by a finite subset is also algebraic and of finite degree

[Violeta Hernández]

If you have extensions of a finite field, you can swap the order

[Violeta Hernández]

But w isn't a finite field

[Violeta Hernández]

Maybe I just need a break from this

[Aaron Liu]

Violeta Hernández said:

I feel like the puzzle pieces are there

Did you try aesop, simp_all, and grind?

[Violeta Hernández]

I'm far from writing any of this down in Lean

[Violeta Hernández]

I think I'll start with the NatNimber stuff instead

[Violeta Hernández]

Opened #204 defining nim addition and multiplication on natural numbers (though the latter hasn't been proven correct yet, #eval certainly suggests that it is!)

[Violeta Hernández]

Oh, #28586 just got approved thanks to Bhavik's kind help

[Violeta Hernández]

So yeah, we really ought to add CNF.eval, and transfer the results about Nimber.oeval there.

[Violeta Hernández]

Then we can prove CNF.eval (CNF.coeff b x) = x

[Aaron Liu]

so we're doing CNF as Ordinal →₀ Ordinal?

[Violeta Hernández]

Well, the list version is still there

[Violeta Hernández]

I just presume that we won't really need it much

[Aaron Liu]

can we redefine the list version in terms of the finsupp?

[Aaron Liu]

I feel like lists are just so much harder to work with

[Aaron Liu]

I tried proving an "obvious" theorem with lists and had to give up

[Violeta Hernández]

It's probably possible to define the finsupp version without going through lists but I don't really think it's bad to have both versions

[Violeta Hernández]

For instance, the result "the CNF evaluates to the original ordinal" is actually nicer to state in terms of lists

[Aaron Liu]

what's the statement?

[Violeta Hernández]

docs#Ordinal.CNF.foldr

[Aaron Liu]

the way I would do it is I would define an evaluation function as the inverse of CNF, and then prove a bunch of stuff about how it works

[Violeta Hernández]

Well yeah, as I just mentioned we should in fact do that

[Violeta Hernández]

CNF.eval is my proposed name

[Aaron Liu]

I still don't really think lists are necessary at all

[Violeta Hernández]

They are at least necessary to define CNF.eval

[Aaron Liu]

no they aren't

[Aaron Liu]

you don't need them

[Violeta Hernández]

I don't think we have any mechanisms for multivariate noncommutative addition besides List.sum

[Aaron Liu]

docs#FreeAddMonoid

[Violeta Hernández]

That's just lists with a wrapper

[Violeta Hernández]

But also how would that help here?

[Aaron Liu]

lets you define noncommutative sums with docs#FreeAddMonoid.lift

[Aaron Liu]

I really thought that they're have somewhere that the algebras of the FreeAddMonoid monad are the add monoids but couldn't find it

[Aaron Liu]

Violeta Hernández said:

That's just lists with a wrapper

implementation detail

[Aaron Liu]

Multiset is also using List but I don't think of it that way

[Violeta Hernández]

Interesting

[Violeta Hernández]

Well if you want to try that out, go ahead

[Violeta Hernández]

My idea for defining CNF.eval was simply summing b ^ f x * x over the sorted support of f

[Violeta Hernández]

I think I'll take a few days break from this. I have some unfinished business relating to Mathlib and ordinals.

[Violeta Hernández]

(also, I'd like to redo the oeval stuff using the CNF proper)

[Violeta Hernández]

Aaron Liu said:

lets you define noncommutative sums with docs#FreeAddMonoid.lift

I don't really get the connection here

[Violeta Hernández]

I tried defining eval through List.sum and honestly the results have been much better than expected

[Violeta Hernández]

docs#Finsupp.induction_on_max does a lot of the heavy lifting

[Violeta Hernández]

And I think the abstraction of working with raw finsupps instead of juggling degrees and polynomial multiplication really makes things easier

[Django Peeters]

Violeta Hernández said:

I think the reasoning is that any extension of degree > 2 has to be by a n-th root

I'm gonna try this in my notebook.

[Violeta Hernández]

Haven't come back to this. Tbh this whole ordinal refactoring stuff is hurting my brain, might need to actually take a day or two off from Lean.

[Violeta Hernández]

Django Peeters said:

Violeta Hernández said:

I think the reasoning is that any extension of degree > 2 has to be by a n-th root

I'm gonna try this in my notebook.

Any success?

[Django Peeters]

Well, I got to the finite field extension that Conway talks about (all roots of $x^{3n}-x$ over all roots of $x^n-x$). But not much after that.

[Django Peeters]

Apart from n=2, I need to show x^3 isn't surjective on the smaller field, I think.

[Violeta Hernández]

I got this answer regarding $P_{\omega^3}$ the other day
image.png

[Violeta Hernández]

I think I might be somewhat mindblocked on this though, I can't really understand the argument

[Aaron Liu]

It's okay I don't understand it either

[Django Peeters]

Lol

[Aaron Liu]

which field is being considered here

[Violeta Hernández]

idk

[Violeta Hernández]

So, I'm at a bit of a crossroads. I'm currently working on adding the colexicographic ordering on finsupps to mathlib (i.e. the order on Lex (αᵒᵈ →₀ β). This is the relevant ordering for both nimber polynomials and Cantor normal forms (used in oeval). In particular, there is some material relating to docs#Ordinal.CNF.coeff I want to state in terms of this Colex order, which would (hopefully) largely trivialize the material on oeval within our own repo.

[Violeta Hernández]

I do expect this to take a while. #29058 is the initial PR where I define the Colex type alias (and use it in lieu of docs#Finset.Colex). I still need to make a followup PR defining the colex ordering on pi types, and a followup doing the same for DFinsupp, and a followup doing the same for Finsupp.

[Violeta Hernández]

On the other hand, there are things we could be working on right now that depend on the full simplest extension theorem. The easier one would be continuing the work on NatNimber, the harder one would be to figure out arithmetic before ω ^ ω ^ ω!

[Violeta Hernández]

So basically, the two proposals are:

• Wait for all this material on Colex/the CNF to be added to Mathlib, reprove the stuff on oeval by using it
• Move what we have now (in the alg_closed branch) to master, refactor it at some later point

[Violeta Hernández]

What do you think?

[Django Peeters]

I would like to refactor at a later point and work on NatNimber now. Then we can bundle forces later to work on $\omega^{\omega^\omega}$ and fill in the gaps.

[Violeta Hernández]

I agree.

[Violeta Hernández]

We shouldn't block our work for months for what's realistically at most like two hundred lines of code less

[Violeta Hernández]

I'll PR the stuff on oeval tomorrow.

[Violeta Hernández]

On an unrelated note, I felt like it'd be fun trying to define the multivariate Veblen function in Lean.

[Violeta Hernández]

If I'm not mistaken, this should be correct:

import Mathlib.Data.Finset.Max
import Mathlib.Data.Finsupp.Single
import Mathlib.SetTheory.Ordinal.Veblen

namespace Ordinal

noncomputable def veblen₀ (f : Ordinal →₀ Ordinal) : Ordinal :=
  let g := f.erase 0
  if hg : g = 0 then ω ^ f 0 else
    let e := g.support.min' (Finsupp.support_nonempty_iff.2 hg)
    derivFamily
      (fun (x : Set.Iio (f e) × Set.Iio e) o ↦ veblen₀ ((g.update e x.1).update x.2 o)) (f 0)
termination_by f -- with colex order
decreasing_by sorry

end Ordinal

[Django Peeters]

Cool! How would you use this?

[Violeta Hernández]

If nothing else we'd be able to state Lenstra's conjecture about the next transcendental using this

[Violeta Hernández]

Again assuming that I'm correct,

noncomputable def LVO : Ordinal :=
  nfp (fun a ↦ veblen₀ (.single a 0)) 0

[Django Peeters]

Maybe we can do something similar to the simplest extension theorems. First the next group, then ring, then field, then perfect closure, quadratic, ...

[Violeta Hernández]

I imagine the way to build the next transcendental looks like finding some sequence of nth closed fields, for increasing prime numbers n, then taking the supremum

[Violeta Hernández]

Though I'm not clear on which of the theorems that hold below w^w^w also hold beyond it

[Violeta Hernández]

e.g. every quartic closed field below w^w^w is in fact quintic closed. Is that true in general?

[Django Peeters]

Well, the biggest difference is now we're talking about trancendentals that aren't the first one.

[Django Peeters]

So less using finite fields I think.

[Violeta Hernández]

Yeah, I'd imagine any results that depended on finite field machinery will not generalize nicely.

[Violeta Hernández]

(though I think we'd need to first prove the relevant results below ω ^ ω ^ ω to get full clarity on which results that encompasses)

[Django Peeters]

Violeta Hernández said:

Django Peeters said:

Violeta Hernández said:

I think the reasoning is that any extension of degree > 2 has to be by a n-th root

I'm gonna try this in my notebook.

Any success?

Now in the other way. This isn't true at all when the base field has 2 or 8 elements and the extension degree 3. I'm missing something.

[Django Peeters]

It works when the base field has a 4-power of elements.

[Violeta Hernández]

Violeta Hernández said:

I think I might be somewhat mindblocked on this though, I can't really understand the argument

I've decided to come back to this.

[Violeta Hernández]

I'm still trying to figure out what the general form of the theorem we want is. Is the following true always? Adjoining a cube root to a quadratically closed field yields a quadratically closed field.

[Violeta Hernández]

In other words, what about $P_{\omega}$ is relevant to the fact that $P_{\omega^3}=P_{\omega}[\sqrt[3]{\ast 2}]$ is quadratically closed, besides the fact that it's itself quadratically closed?

[Violeta Hernández]

I think I am missing one other condition: every element of $P_\omega$ is algebraic over the prime field.

[Violeta Hernández]

Wait a second I think I've spent so much time thinking about this that I actually just get what Siegel said about this

[Aaron Liu]

explain

[Violeta Hernández]

So, proving that a field $P_x$ is quadratically closed is equivalent to proving that any quadratic extension is isomorphic to a subfield of $P_x$, right?

[Violeta Hernández]

For one direction, a quadratic extension of a quadratically closed field is isomorphic to itself

[Violeta Hernández]

Wait I don't think the other direction is true?

[Violeta Hernández]

Unless you further specify that the isomorphism fixes $P_x$

[Violeta Hernández]

Well, at the very least I understand one part of Siegel's argument

[Violeta Hernández]

If $\mathcal F$ is a quadratic extension of $P_2(\omega)$, then it's isomorphic to $\mathcal E(\omega)$ for some $\mathcal E$ isomorphic to a subfield of $P_\omega$

[Violeta Hernández]

This is pretty easy to show, actually

[Violeta Hernández]

That whole argument about "the Galois group is abelian" can be replaced by some lemmas on finite fields

[Violeta Hernández]

$[\mathcal F : P_2] = [\mathcal F : P_2(\omega)][P_2(\omega) : P_2] = 6$, which means $\mathcal F$ is isomorphic to $\mathbb F_{64}$.

[Violeta Hernández]

It's a known fact that $m | n$ iff $\mathbb F_n$ has a subfield isomorphic to $\mathbb F_m$.

[Violeta Hernández]

So we can get to $\mathbb F_{64}$ the other way around, finding an intermediate field isomorphic to $\mathbb F_4$.

[Violeta Hernández]

i.e. a quadratic extension of $P_2$

[Violeta Hernández]

i.e. a field isomorphic to a subfield of $P_\omega$

[Violeta Hernández]

Violeta Hernández said:

If $\mathcal F$ is a quadratic extension of $P_2(\omega)$, then it's isomorphic to $\mathcal E(\omega)$ for some $\mathcal E$ isomorphic to a subfield of $P_\omega$

Actually, we can skip the double isomorphism: if $\mathcal F$ is a quadratic extension of $P_2(\omega)$, then it's isomorphic to $\mathcal E(\omega)$ for some subfield $\mathcal E$ of $P_\omega$. In fact, $\mathcal E = P_4$ works.

[Violeta Hernández]

So $\mathcal F$ is isomorphic to a subfield of $P_{\omega^3}$ (in fact $P_4(\omega)$).

[Violeta Hernández]

Now the last thing I don't understand is, why does that mean $P_{\omega^3}$ is quadratically closed? It feels like it shouldn't be more than a single step to finish.

[Aaron Liu]

I will understand this once it goes into Lean

[Violeta Hernández]

It's a bit hard to put something into Lean if you don't understand it

[Aaron Liu]

I don't understand why things being isomorphic means other things are quadratically closed

[Violeta Hernández]

Yeah that's the step I'm not seeing either

[Violeta Hernández]

Though I do have an idea.

[Violeta Hernández]

Take a quadratic polynomial in $P_{\omega^3}$. We want to show there's a root in $P_{\omega^3}$.

[Aaron Liu]

yup

[Aaron Liu]

wlog monic

[Violeta Hernández]

The coefficients of this polynomial have the form $\omega^k a + b$ for $k \le 2$, $a, b < \omega$.

[Violeta Hernández]

Let $\mathcal F$ be the subfield generated by these $a, b$. It's a finite field.

[Aaron Liu]

makes sense

[Violeta Hernández]

So the root of this polynomial exists in some quadratic extension of $\mathcal F(\omega)$

[Aaron Liu]

wait what

[Violeta Hernández]

The polynomial has coefficients in $\mathcal F(\omega)$ so adding a root of the polynomial means taking a quadratic extension of that

[Aaron Liu]

oh that makes sense

[Violeta Hernández]

Call the quadratic extension $\mathcal K$

[Violeta Hernández]

So we have a tower $\mathcal K/\mathcal F(\omega)/\mathcal F$

[Aaron Liu]

so many letters

[Violeta Hernández]

And I claim that you can instead find some $\alpha$ such that $\mathcal K / \mathcal F(\alpha) / \mathcal F$ and such that $[\mathcal F(\alpha) : \mathcal F]=2$

[Aaron Liu]

interesting claim

[Violeta Hernández]

This is again just lemmas on finite fields

[Aaron Liu]

something about galois groups I think?

[Violeta Hernández]

"The Galois group is Abelian"

[Violeta Hernández]

But I think you don't need that

[Aaron Liu]

we can come back to the claim after we use it to prove the lemma

[Violeta Hernández]

If $k | m | n$ are prime powers then there's a field tower $\mathbb F_n / \mathbb F_m / \mathbb F_k$

[Aaron Liu]

that sounds correct

[Violeta Hernández]

Anyways, since $\mathcal F$ is a subfield of $P_\omega$, and $\mathcal F(\alpha)$ is a quadratic extension, $\mathcal F(\alpha)$ will be isomorphic to a subfield of $P_\omega$

[Aaron Liu]

$P_\omega$?

[Violeta Hernández]

yep

[Aaron Liu]

is that supposed to be the nimbers $< \omega$

[Violeta Hernández]

yes

[Aaron Liu]

ok that makes sense

[Violeta Hernández]

So $\mathcal K = \mathcal F(\alpha, \omega)$ is isomorphic to $\mathcal E(\omega)$ for some subfield $\mathcal E$ of $P_\omega$

[Aaron Liu]

that sounds reasonable

[Violeta Hernández]

I feel like that should imply that $P_{\omega^3}$ is quadratically closed but I am missing the last step

[Aaron Liu]

I feel like I'm missing the point of all this

[Aaron Liu]

you do some field magic and then

[Violeta Hernández]

The general idea is that a quadratic extension of a cubic extension should roughly be the same as a cubic extension of a quadratic extension, which should land you back in your field

[Aaron Liu]

why is $\mathcal K = \mathcal F (\alpha, \omega)$

[Violeta Hernández]

Not entirely sure

[Violeta Hernández]

Let's file that one under "Galois shenanigans" and come back to it later

[Violeta Hernández]

Actually, is $\alpha$ the root that we wanted to find?

[Aaron Liu]

what happens if it is?

[Aaron Liu]

do we like pull it across the isomorphisms or something

[Violeta Hernández]

Wait I got it!

[Aaron Liu]

yay

[Aaron Liu]

tell me tell me

[Violeta Hernández]

Ok so

[Violeta Hernández]

Violeta Hernández said:

The polynomial has coefficients in $\mathcal F(\omega)$ so adding a root of the polynomial means taking a quadratic extension of that

Let $\alpha$ be a root of that

[Aaron Liu]

$\alpha$ is in the extension?

[Violeta Hernández]

Yep

[Violeta Hernández]

So we have our rhombus of four fields

[Violeta Hernández]

$\mathcal F(\alpha, \omega)$ containing $\mathcal F(\alpha)$, $\mathcal F(\omega)$, and $\mathcal F$ at the bottom

[Violeta Hernández]

So, $\mathcal F(\alpha)$ is isomorphic to a subfield of $P_\omega$, meaning $\mathcal F(\alpha, \omega)$ is isomorphic to a subfield of $P_{\omega^3}$, but crucially, the isomorphism preserves both $\mathcal F$ and $\omega$, i.e. it preserves the coefficients of the polynomial

[Violeta Hernández]

Meaning that whatever $\alpha$ gets mapped to is the root of the polynomial we wanted

[Aaron Liu]

uhm ok

[Aaron Liu]

yeah sure

[Aaron Liu]

why is $\mathcal F (\alpha)$ isomorphic to a subfield of $P_\omega$

[Violeta Hernández]

Because $\mathcal F$ is a subfield of $P_\omega$, and $\mathcal F(\alpha)$ is a quadratic extension, and $P_\omega$ is quadratically closed

[Aaron Liu]

why is $\mathcal F (\alpha)$ a quadratic extension

[Violeta Hernández]

Ok I think it's worth thinking about the Galois shenanigans now

[Violeta Hernández]

We know that $[\mathcal F(\alpha,\omega):\mathcal F(\omega)]=2$

[Violeta Hernández]

The claim is that $[\mathcal F(\alpha):\mathcal F]=2$

[Aaron Liu]

yes that sounds correct

[Violeta Hernández]

Yeah idk

[Violeta Hernández]

Can we use the fundamental theorem of Galois theory?

[Violeta Hernández]

What's the automorphism group of $\mathcal F(\alpha,\omega)/\mathcal F$? Do we know for a fact it's just $C_2\times C_3$?

[Violeta Hernández]

Actually yes we do! Every finite extension of a finite field is Abelian

[Violeta Hernández]

Actually wait a second

[Aaron Liu]

ok so my polynomial is $x ^ 2 - \omega$ which has a double root $\omega ^ 2 2$ of order 18 9 and $\mathcal F = \mathbb{F}_2$ having two elements and $\mathcal F (\alpha)$ has at least 18 9

[Aaron Liu]

that seems to contradict what you said but I'm not sure

[Violeta Hernández]

That does in fact seem to contradict what I said

[Aaron Liu]

sometimes you just need to put a little effort into finding counterexamples

[Violeta Hernández]

Actually isn't the order 9

[Violeta Hernández]

I mean that still contradicts what I said

[Aaron Liu]

is it 9?

[Aaron Liu]

oh it is 9

[Violeta Hernández]

In fact this is interesting because $\omega^22 = \omega^5$ meaning $\mathcal F(\alpha,\omega)=\mathcal F(\omega) = \mathcal F(\alpha)$

[Aaron Liu]

you do get the subfield $\mathcal F (2)$

[Aaron Liu]

which is quadratic

[Violeta Hernández]

I think what happens here is that if your quadratic polynomial already has a root in $\mathcal F(\omega)$ then $\mathcal F(\alpha,\omega)$ is no longer a quadratic extension

[Violeta Hernández]

Of course, if that happens, then you already have the root that you wanted

[Violeta Hernández]

And in fact... isn't the entire point that this always happens?

[Violeta Hernández]

So really this is moreso a proof by contradiction

[Violeta Hernández]

...actually I don't think this always happens but I don't have a counterexample at hand

[Violeta Hernández]

In either case, the lemma we're missing is the following. If $\mathcal F$ is a finite field, and $\alpha,\beta$ are elements in some algebraic completion such that $[\mathcal F(\alpha):\mathcal F]$ and $[\mathcal F(\alpha,\beta):\mathcal F(\alpha)]$ are coprime (distinct primes?), then $[\mathcal F(\alpha,\beta):\mathcal F(\alpha)]=[\mathcal F(\beta):\mathcal F]$.

[Violeta Hernández]

No, I don't think that's true. What if $\alpha$ is a cube root of $x$, and $\beta$ is a sixth root of $x$?

[Aaron Liu]

I found an irreducible polynomial

[Aaron Liu]

$X ^ 2 + X + \omega + 2$ does not factor in $P_2 (\omega)$

[Violeta Hernández]

$2$ isn't in $P_2(\omega)$ though

[Violeta Hernández]

oh no wait it is

[Aaron Liu]

it's $\omega ^ 3$

[Aaron Liu]

how can we compute its root in $P_{\omega^3}$

[Violeta Hernández]

First we turn it into a polynomial of the form $X^2 + a$

[Violeta Hernández]

Actually maybe we don't need to do that

[Violeta Hernández]

Can't we just write $X = \omega^2 a + \omega b + c$, expand everything, match like terms

[Aaron Liu]

that sounds tedious

[Violeta Hernández]

It's slightly easier when you remember $(a + b)^2 = a^2 + b^2$

[Aaron Liu]

$\omega ^ 2 (a + b) + \omega (a + b) + c ^ 2 + c + \omega + 2 = 0$

[Aaron Liu]

is that correct

[Aaron Liu]

no I missed a factor

[Aaron Liu]

$\omega ^ 2 (a + b ^ 2) + \omega (2a ^ 2 + b + 1) + c ^ 2 + c + 2 = 0$

[Violeta Hernández]

Yep that's exactly what I got

[Aaron Liu]

this looks slightly tricky

[Aaron Liu]

how do we solve all of these simultaneously

[Violeta Hernández]

$c = 4$

[Violeta Hernández]

We have $a = b^2$ so we really only want to solve $2b^4 + b + 1=0$

[Violeta Hernández]

$b = 2$

[Violeta Hernández]

$a = 3$

[Aaron Liu]

does this method work to solve every equation of type $X ^ 2 + X + \omega^2 a + \omega b + c = 0$

[Violeta Hernández]

My method was trial and error lol

[Aaron Liu]

I mean assuming you can solve quadratics in $P_\omega$

[Violeta Hernández]

I don't know

[Violeta Hernández]

We kinda ended up with a quartic polynomial in there

[Aaron Liu]

whoops

[Aaron Liu]

should I find another irreducible

[Violeta Hernández]

Why did we want to find irreducibles?

[Aaron Liu]

because the reducibles have roots obviously

[Violeta Hernández]

Ah yes of course

[Violeta Hernández]

Hm I want to test something first

[Violeta Hernández]

Let $\alpha = \omega^23 + \omega2 + 4$

[Violeta Hernández]

Is $P_4(\alpha)$ a quadratic extension?

[Aaron Liu]

just a sec I have to tell my CAS to pass to a field extension

[Aaron Liu]

this is the field with 4096 elements?

[Violeta Hernández]

yikes

[Violeta Hernández]

So it's a sixth degree extension

[Aaron Liu]

well $P_{16} (\omega)$ has 4096 elements

[Aaron Liu]

I haven't checked $P_{16} (\alpha)$ yet

[Violeta Hernández]

Isn't $\mathcal F = P_4$ though

[Aaron Liu]

wait no it was supposed to be $P_4$

[Aaron Liu]

so I want the minimal polynomial of $\omega ^ 2 3 + \omega 2 + 4$ over $P_4$

[Aaron Liu]

I'll just test all the polynomials lexographically

[Violeta Hernández]

I'm starting to think that my argument might just not work in this form

[Violeta Hernández]

We know that $\mathcal F(\alpha,\omega)$ has degree 6, so we know there's some intermediate field between that and $\mathcal F$ with degree 2

[Violeta Hernández]

We don't know if it's $\mathcal F(\alpha)$

[Violeta Hernández]

Call this intermediate field $\mathcal E$

[Violeta Hernández]

We know $\omega \notin \mathcal E$ since otherwise $\mathcal F(\omega)\subseteq \mathcal E$ would have degree at least 3

[Violeta Hernández]

Therefore $[\mathcal E(\omega):\mathcal E]>1$

[Violeta Hernández]

Which means $\mathcal E(\omega) = \mathcal F(\alpha,\omega)$

[Violeta Hernández]

i.e. $\alpha \in \mathcal E(\omega)$

[Violeta Hernández]

Since $\mathcal E$ is a quadratic extension of $\mathcal F\subseteq P_\omega$, it's isomorphic to a subfield of $P_\omega$

[Violeta Hernández]

So $\mathcal E(\omega)$ is isomorphic to a subfield of $P_{\omega^3}$

[Violeta Hernández]

And again, the isomorphism preserves both $\mathcal F$ and $\omega$

[Violeta Hernández]

Meaning that it preserves our polynomial

[Violeta Hernández]

Meaning that this isomorphism sends $\alpha$ to a root of the polynomial

[Violeta Hernández]

QED

[Violeta Hernández]

please tell me this works

[Aaron Liu]

can you use it to find some roots

[Violeta Hernández]

I don't know if we can find $\mathcal E$ constructively

[Aaron Liu]

it's a finite field

[Aaron Liu]

worst case test all the subfields

[Violeta Hernández]

Thing is, we know $\alpha \in\mathcal E(\omega)$, but the proof doesn't tell you how to represent $\alpha$ in terms of elements of $\mathcal E$ or $\omega$

[Aaron Liu]

Violeta Hernández said:

Is $P_4(\alpha)$ a quadratic extension?

I found the minimal polynomial

[Violeta Hernández]

Which is?

[Aaron Liu]

$\alpha$ is a root of $X ^ 3 + 4 X ^ 2 + 4 X + 7$ and this polynomial is irreducible over $P_{16}$ (wait did you want me to find the minpoly over $P_{16}$ or $P_4$?)

[Aaron Liu]

$\alpha$ is a root of $X ^ 6 + X ^ 5 + 3 X ^ 4 + X ^ 3 + X ^ 2 + 3 X + 3$ and this polynomial is irreducible over $P_4$

[Aaron Liu]

it became a lot faster to test after I found out the minpoly over a field extension is an irreducible factor of the minpoly over the prime field

[Violeta Hernández]

Yep, so basically $\mathcal E\ne\mathcal F(\alpha)$ in general

[Aaron Liu]

Violeta Hernández said:

Thing is, we know $\alpha \in\mathcal E(\omega)$, but the proof doesn't tell you how to represent $\alpha$ in terms of elements of $\mathcal E$ or $\omega$

we want "degree 2 intermediate field" which is a decidable predicate on subsets of $\mathcal F (\alpha, \omega)$ so check them all

[Aaron Liu]

and if $\alpha \in \mathcal E (\omega)$ then check all the elements

[Aaron Liu]

there's only finitely many after all

[Aaron Liu]

this is the power of Fintype + DecidableEq

[Aaron Liu]

so do you think you could do this in Lean now

[Violeta Hernández]

yikes

[Violeta Hernández]

Can I do this in Lean? Not yet

[Violeta Hernández]

I need to first figure out the general form of this argument

[Aaron Liu]

what's the general statement?

[Violeta Hernández]

Well we haven't even finished the very next sentence

[Aaron Liu]

what's the next sentence

[Violeta Hernández]

image.png
Why is that the case?

[Violeta Hernández]

Don't tell me I need to expand $(\omega^2a+\omega b+c)^3$ to figure this out

[Aaron Liu]

I have an argument that feels too simple to be correct

[Violeta Hernández]

Go ahead

[Aaron Liu]

So let $\alpha < \omega$ then consider the field $\mathcal F$ generated by $2$ and $\alpha$ and assume it doesn't have a cube root of $\alpha$ then adjoing a cube root of $\alpha$ gives a degree 3 extension and $\mathcal F (\omega)$ is a degree 3 extension so they're isomorphic

[Aaron Liu]

I think all the degree 3 extensions are isomorphic

[Aaron Liu]

oh and the isomorphism has to fix $\mathcal F$

[Aaron Liu]

does this work at all???

[Violeta Hernández]

Let me think

[Violeta Hernández]

Yeah it feels correct

[Aaron Liu]

oh that's great

[Violeta Hernández]

Though actually wait

[Violeta Hernández]

Why do we know there's an automorphism fixing $\mathcal F$?

[Violeta Hernández]

Any two splitting fields of the same field by the same polynomial are isomorphic, but do we further know that some isomorphism fixes the base field?

[Aaron Liu]

uhm all the automorphism of the base field are powers of frobenius

[Aaron Liu]

so you can patch it up by composing with an appropriate power of frobenius

[Violeta Hernández]

Actually wait our splitting fields are not by the same polynomial so this doesn't matter

[Aaron Liu]

they have the same finite cardinality

[Violeta Hernández]

Rather we have two different extensions $\mathcal F(\alpha)$ and $\mathcal F(\beta)$ of the same degree, and the question is whether some isomorphism between both preserves $\mathcal F$

[Aaron Liu]

They have the same finite cardinality so an isomorphism exists which will act on the base field by a power of frobenius

[Aaron Liu]

compose with the inverse power of frobenius on either side

[Violeta Hernández]

This feels correct yeah

[Aaron Liu]

wow that was a lot easier than expected

[Violeta Hernández]

Well, there's also the other direction

[Violeta Hernández]

Why does $\omega$ not have a cube root in $P_{\omega^3}$?

[Violeta Hernández]

Put another way, why does adding a cube root of 2 preclude from adding a ninth root as well?

[Violeta Hernández]

Ok ok ok ok I think I Know why this is

[Violeta Hernández]

Had a bit of help from someone way more intelligent than me

[Violeta Hernández]

Suppose $\omega$ had a cube root $\omega^2a + \omega b + c$ for $a,b,c<\omega$. Take the finite field $\mathcal F\subseteq P_{\omega}$ generated by $2,a,b,c$. Thus, the hypothesis is that $\omega$ has a cube root in $\mathcal F(\omega)$.

[Violeta Hernández]

Let $x$ be a primitive root of $\mathcal F(\omega)$. Let $k=\frac{|F|^3-1}{|F|-1}$. Then $x^k$ is a primitive root of $\mathcal F$. Write $x^{nk} = 2$. Since $2$ is not a cube root in $\mathcal F$, we have $3\nmid n$. Since $3$ divides $k$ exactly once (by Lifting the Exponent), $2$ can't have any ninth roots, i.e. $\omega$ can't have any cube roots.

[Violeta Hernández]

Awesome! We have a pen-and-paper proof that $P_{\omega^3}$ is a quadratically closed field and $[\omega^3]^3=\omega$ and it only took like two weeks

[Violeta Hernández]

So the very first thing I think we should do is figure out the general forms of the lemmas we made use of

[Violeta Hernández]

Theorem 1: Suppose $\mathcal F$ is algebraic over its prime field and is $n$-th degree closed. Let $p > n$ be prime, and consider an algebraic extension $\mathcal F(\alpha)$ of degree $p$. Then $\mathcal F(\alpha)$ is $n$-th degree closed.

[Violeta Hernández]

I think that's the general form of that first result

[Violeta Hernández]

...we're still missing things.

[Violeta Hernández]

Why is an $n$-th degree closed field also $m$-th degree closed for $m$ less than the next prime $p$ after $n$?

[Violeta Hernández]

Page 447 seems relevant
image.png

[Violeta Hernández]

So actually I have no idea if e.g. every quartic closed field is quintic closed

[Violeta Hernández]

I guess the argument doesn't extend past $\omega^{\omega^\omega}$

[Violeta Hernández]

Which admittedly seems to be a common theme here

[Violeta Hernández]

Violeta Hernández said:

Theorem 1: Suppose $\mathcal F$ is algebraic over its prime field and is $n$-th degree closed. Let $p > n$ be prime, and consider an algebraic extension $\mathcal F(\alpha)$ of degree $p$. Then $\mathcal F(\alpha)$ is $n$-th degree closed.

Proof: Let $q_1\in\mathcal F[X]$ be the minimal polynomial of $\alpha$. Let $q_2\in\mathcal F(\alpha)[X]$ be a polynomial of degree $\le n$, and let $\beta$ be a root in the algebraic completion. Let $d = [\mathcal G(\alpha,\beta) : \mathcal G(\alpha)] \le n$. Let $\mathcal G\subseteq\mathcal F$ be the finite subfield generated by the coefficients of $q_1$ and $q_2$.

The field extension $G(\alpha,\beta)$ has degree $dp$ over $G$. We can find a subfield $\mathcal E$ with degree $d$ over $G$. We can't have $\alpha\in \mathcal E$, as otherwise $G(\alpha)\subseteq \mathcal E$ would have degree at least $p > d$. Thus $\mathcal E(\alpha) = \mathcal G(\alpha,\beta).$

Since $\mathcal F$ is $d$-th degree closed, $\mathcal E$ is isomorphic to a subfield of $\mathcal F$, and thus $\mathcal E(\alpha)$ is isomorphic to a subfield of $\mathcal F(\alpha)$. Moreover, there's an isomorphism which fixes $\mathcal G$ and $\alpha$. Therefore, the map of $\beta$ under this isomorphism must be a root of $q_2$.

[Violeta Hernández]

Actually, there's an auxiliary lemma we can get out of this.

Theorem 0: Suppose $\mathcal F$ is algebraic over its prime field, and $\mathcal F(\alpha)$ is a finite extension. There exists a finite subfield $\mathcal G\subseteq\mathcal F$ such that for any subfield $\mathcal G\subseteq \mathcal H$, we have $[\mathcal H(\alpha):\mathcal H] = [\mathcal F(\alpha) : \mathcal F]$.

[Violeta Hernández]

This is just the "take the subfield generated by the minimal polynomial's coefficients" argument

[Violeta Hernández]

But I imagine we will be using this more than once

[Violeta Hernández]

Oh btw I asked Siegel and he mentioned a much better way to show ω doesn't have a cube root below [ω³]

[Violeta Hernández]

2^(3*2^n) - 1 can't be divisible by 27

[Violeta Hernández]

I wonder if this is the sort of argument we need to establish the existence of α_p.

[Kevin Buzzard]

I'm not following this thread closely so sorry if this is not relevant, but the way the Galois theory of finite fields works is that if $k$ is a finite field then up to $k$-isomorphism there's a unique extension of degree $n$ for any positive integer $n$, which is Galois, meaning that if $\overline{k}$ is an algebraic closure of $k$ then the Galois group of $\overline{k}/k$ is $\widehat{\Z}$, the profinite completion of $\Z$. Closed subgroups of this group thus correspond to subfields $k\subseteq L\subseteq\overline{k}$. Now $\widehat{\Z}=\prod_p\Z_p$ and if one fixes a prime $p$ and looks at the projection onto $\prod_{q\ne p}\Z_q$ then this has kernel $\Z_p$ which is then the absolute Galois group of the corresponding fixed field $k_p$, an infinite field whose only finite extensions are Galois with group cyclic of $p$-power order. Such a field, it seems to me, will be $n$ th degree closed iff ... wait. What does $n$ th degree closed actually even mean? All polynomials of degree $n$ have a root, or all polynomials of degree $n$ factor into linear factors? The two things are the same for $n=2$ but differ in general. I'm suddenly unclear about whether what I'm writing is relevant.

[Aaron Liu]

Kevin Buzzard said:

... wait. What does $$n$$th degree closed actually even mean? All polynomials of degree $n$ have a root, or all polynomials of degree $n$ factor into linear factors? The two things are the same for $n=2$ but differ in general. I'm suddenly unclear about whether what I'm writing is relevant.

It's "all polynomials of degree ≤ n have a root" I think

[Kevin Buzzard]

(deleted)

[Aaron Liu]

do you have a better suggestion?

[Kevin Buzzard]

Oh degree at most $n$?

[Aaron Liu]

yes

[Kevin Buzzard]

So that's the same as "all polynomials of degree at most n split into linear factors"?

[Aaron Liu]

I think that's equivalent

[Aaron Liu]

since you can repeatedly factor out the root

[Kevin Buzzard]

So my field $k_p$ is $n$ th degree closed iff $n<p$.

[Kevin Buzzard]

There are no irreducible polynomials of degree $p+1$ over this field but it's not $p+1$ th degree closed because you do $X$ times an irreducible poly of degree $p$.

[Aaron Liu]

The Nimbers are an algebraically closed Field of characteristic 2 whose elements are the ordinals. We can learn about this Field by studying the sets Iio x. In particular, if Iio x is closed under the field operations (it is a subfield) and is not algebraically closed, then x is the root of the lexographically earliest polynomial with coefficients in Iio x which does not have a root in Iio x, and if this polynomial has degree n then for m < n and o < x and p < x ^ m we have x ^ m * o + p (where addition and multiplication and exponentiation are the nimber operations) is equal to [x ^ m * o + p] (where addition and multiplication and exponentiation are the ordinal operations)

[Violeta Hernández]

I think we can actually characterize the intermediate subfields below the first transcendental quite well

[Aaron Liu]

but can we prove the characterization is the problem

[Violeta Hernández]

Actually wait I'm not so sure

[Aaron Liu]

uh oh

[Violeta Hernández]

Nimbers below ω are just the direct limit of fields GF(2^2^n)

[Violeta Hernández]

Simple enough

[Violeta Hernández]

Is it accurate to say nimbers below ω^ω are a direct limit of fields GF(2^2^n) and also GF(2^3^n)?

[Violeta Hernández]

Or rather GF(2^(2^m * 3^n))?

[Aaron Liu]

what's your diagram

[Violeta Hernández]

GF(2^a) → GF(2^b) through the unique embedding when a | b

[Aaron Liu]

you take all the quadratic extensions and you take all the cubic extensions right

[Violeta Hernández]

and then the quintic extensions and so on

[Violeta Hernández]

Actually on this note

[Violeta Hernández]

Is it correct to say the algebraic completion of 2 is the direct limit of literally every single finite field of characteristic 2?

[Aaron Liu]

probably

[Aaron Liu]

well every nonzero element of the algebraic closure is a root of unity and generates a finite field

[Aaron Liu]

and every element of a finite field is algebraic over the prime field

[Violeta Hernández]

Actually, conjecture

[Violeta Hernández]

Each of these intermediate fields of nimbers looks like the direct limit of fields GF(2^(2^a_1 * 3^a_2 * ... * p^a_n)) where all the exponents vary over the integers, save for potentially the last one which might be bounded

[Aaron Liu]

which one is bounded

[Aaron Liu]

oh I get what you mean

[Aaron Liu]

yeah this seems true enough

[Aaron Liu]

hold on this is some category theory

[Aaron Liu]

even though I don't usually think of field theory as category theory since galois theory exists

[Violeta Hernández]

I mean I don't think this characterization will help much

[Violeta Hernández]

But it at least means we can state our intermediate results without alluding to the nimbers

[Aaron Liu]

this colimit seems like it's obviously nth degree closed

[Aaron Liu]

very obviously

[Aaron Liu]

for n < p

[Aaron Liu]

and it's probably unique up to non-canonical isomorphism with certain properties

[Aaron Liu]

the category of fields has all filtered colimits so I can start doing some category theory

[Aaron Liu]

to get a filtered colimit we need some squares to commute

[Aaron Liu]

specifically, embedding GF a into GF a*b into GF a*b*c should be the same as embedding GF a into GF a*c into GF a*c*b

[Aaron Liu]

is this true?

[Violeta Hernández]

of course it is

[Violeta Hernández]

Wait no I don't know

[Violeta Hernández]

GF(a*b) has a unique subfield of size GF(a) but there's as many embeddings as there's automorphisms of said field

[Violeta Hernández]

In fact that sort of calls into question how we're defining the direct limit to begin with

[Aaron Liu]

wait are these inclusions unique

[Aaron Liu]

well no since we have nontrivial automorphisms

[Aaron Liu]

well it should be possible to make them all commute

[Aaron Liu]

just choose embeddings $\text{GF}_{2^n} \hookrightarrow \overline{\mathbb{F}_2}$ for every $n$ and use the inclusions

[Aaron Liu]

ok I have a question

[Aaron Liu]

Suppose I have numbers $a, b, c, d \in \mathbb{N}^+$ such that $a = \gcd(b, c)$, and $d = \operatorname{lcm}(b, c)$. Suppose I have field homomorphisms $f : \text{GF}_{2^a} \hookrightarrow \text{GF}_{2^b}$ and $g : \text{GF}_{2^a} \hookrightarrow \text{GF}_{2^c}$. Is does there always exist $\text{inl} : \text{GF}_{2^b} \hookrightarrow \text{GF}_{2^d}$ and $\text{inr} : \text{GF}_{2^b} \hookrightarrow \text{GF}_{2^d}$ making the square a docs#CategoryTheory.BicartesianSq

[Aaron Liu]

I was able to get AI to spit out a reasonable-looking proof

[Violeta Hernández]

Unrelated but I think life would be easier if we wrote $\textrm{GF}_p(\alpha)$ instead of $\textrm{GF}(p^{\alpha})$

[Aaron Liu]

we can do that no one's stopping us

[Aaron Liu]

look what I found https://math.stackexchange.com/a/2627324

[Aaron Liu]

isn't this what we're looking for

[Aaron Liu]

I think it's saying essentially the same thing but this seems like much more understandable

[Violeta Hernández]

Aaron Liu said:

look what I found https://math.stackexchange.com/a/2627324

Ok so, how do we formalize this in Lean?

[Violeta Hernández]

Do we need/should we have a typeclass for "every polynomial of degree n splits"?

[Violeta Hernández]

How do we write down the condition "every algebraic extension is abelian"?

[Violeta Hernández]

Actually, how do we know that e.g. every algebraic extension of the quadratic closure of $\mathbb F_2$ is abelian?

[Violeta Hernández]

Presumably it's the same argument of passing to finite fields, but I'm unsure on the details

[Violeta Hernández]

Oh on an unrelated note, I finally remembered to do this: https://github.com/vihdzp/combinatorial-games/pull/231

[Kevin Buzzard]

Violeta Hernández said:

Is it correct to say the algebraic completion of 2 is the direct limit of literally every single finite field of characteristic 2?

What do you mean by "the algebraic completion"? You've said this phrase several times. Do you mean "an algebraic closure"? There are set-theoretic difficulties with the phrase "direct limit of literally every single finite field of characteristic 2" and there are also issues with the fact that usually in a direct limit you have precisely one map from X_i to X_j if i<=j, so your statement needs to be made more precise, but if you believe in universes then yes one algebraic closure of F_2 (this is not a thing which is unique up to unique isomorphism so I'm reluctant to use "the" and would rather use "a") is a certain filtered colimit of all finite fields of characteristic 2. I'm reluctant to call it a direct limit because of all the automorphisms.

You can also describe the filtered colimit of all these F_{2^2^n} fields; the absolute Galois group of F_2 is Z-hat which has a quotient isomorphic to Z_2, the 2-adic integers, and within a fixed alg closure the subfield corresponding to this quotient is the colimt of the F_{2^2^n}, a field extension of F_2 with Galois group Z_2.

[Violeta Hernández]

I mean algebraic closure, yes. That's a slip-up I've been corrected on multiple times...

Yeah, I realized a bit later that maps $\textrm{GF}(m) \to \textrm{GF}(n)$ are not generally unique. But as Aaron said, the fact that $\overline{\mathbb F_2}$ is a direct limit of a particular directed system that does include every finite field of characteristic 2, means that you can unambiguously talk about direct limits of any subset of these fields.

[Violeta Hernández]

Well, "every finite field of characteristic 2" really means "one field of characteristic 2 for every possible cardinality", which provides the skeleton of the direct system so to speak.

[Kevin Buzzard]

Aaron Liu said:

Suppose I have numbers $a, b, c, d \in \mathbb{N}^+$ such that $a = \gcd(b, c)$, and $d = \operatorname{lcm}(b, c)$. Suppose I have field homomorphisms $f : \text{GF}_{2^a} \hookrightarrow \text{GF}_{2^b}$ and $g : \text{GF}_{2^a} \hookrightarrow \text{GF}_{2^c}$. Is does there always exist $\text{inl} : \text{GF}_{2^b} \hookrightarrow \text{GF}_{2^d}$ and $\text{inr} : \text{GF}_{2^b} \hookrightarrow \text{GF}_{2^d}$ making the square a docs#CategoryTheory.BicartesianSq

I don't see why you need any of these gcd/lcm assumptions. they can just be replaced with divisibility assumptions to ensure the maps exist. If you have fields $K,L,M$ and injections $K\to L$ and $K\to M$ making $M$ and $L$ algebraic over $K$ then take an algebraic closure $\overline{M}$ of $M$; this is an algebraic closure of $K$, so there's a $K$-algebra map $L\to\overline{M}$ and now just let $N$ be the smallest subfield of $\overline{M}$ containing $M$ and the image of $L$. My mental model is $K\subseteq M\subseteq \overline{M}$ and then $L$ is this random other thing but you can just squish $L$ into $\overline{M}$ $K$-linearly and now everything becomes an inclusion.

[Kevin Buzzard]

Violeta Hernández said:

Actually, how do we know that e.g. every algebraic extension of the quadratic closure of $\mathbb F_2$ is abelian?

Galois theory! Any subquotient of an abelian group is abelian.

[Violeta Hernández]

Just to make sure, the abelian group in question here is $\widehat{\mathbb Z}$, right?

[Kevin Buzzard]

Right. The fundamental theorem of (infinite) Galois theory goes like this. If $L/K$ is an algebraic, normal and separable extension (a.k.a. a "Galois extension") then the $K$-algebra automorphisms of $L$ (aka the Galois group of $L$ over $K$) is a group $G$ (group law is composition) which is naturally endowed with a topology called the Krull topology, making it into a profinite group. $G$ is finite iff $L$ is a finite-dimensional $K$-vector space, in which case the topology is discrete (a.k.a. "not there"), and this is the case taught to undergraduates (without any mention of topology). You're interested in the infinite case. The theorem in the infinite case is that there's a canonical bijection between the closed subgroups $H$ of $G$ and the intermediate fields $K\subseteq M\subseteq L$; the maps in both directions are "the first thing you'd think of", there is nothing deep in the maps; the hard part is proving that they're inverse bijections. The maps are: given the subgroup, look at the subfield of things fixed pointwise by all elements of the subgroup, and given the subfield, look at the group elements which fix the subfield pointwise.

If $G$ is abelian (the only case you're interested in, by the looks of it) then both $L/M$ and $M/K$ are also Galois (this would not be true in general for $G$ nonabelian because some subgroups are not normal so $M/K$ might not be normal), and the thing which you want to be true is true; the Galois group of $L/M$ is algebraically and topologically $H$, and the Galois group of $M/K$ is algebraically and topologically $G/H$, via the obvious maps (again the maps are obvious, it's checking that everything is a bijection which is hard).

Galois theory of finite fields; if $k$ is a finite field and $\overline{k}$ is an algebraic closure of $k$ then the Galois group of $\overline{k}/k$ is algebraically and topologically $\widehat{\Z}$, and the usual isomorphism sends $1\in\widehat{\Z}$ to $x\mapsto x^n$ where $n=|k|$.

If $K$ is a field of characteristic $p$ and is algebraic over $\mathbb{F}_p$ (which seems to be the only case you're interested in) then every algebraic extension of $K$ is automatically normal and separable, so you can forget these adjectives and just think always about algebraic extensions. So your question written carefully, if I've understood it correctly, is a special case of "if $\overline{\mathbb{F}}_2$ is a fixed algebraic closure of $\mathbb{F}_2$ and if $\mathbb{F}_2\subseteq K\subseteq L\subseteq \overline{\mathbb{F}_2}$ are intermediate fields then do I know that $L/K$ is Galois with abelian Galois group?" and the answer is "yes, because general field theory tells you it's algebraic, normal (because $\widehat{\Z}$ is abelian as are all its subgroups and quotients) and separable, and then the fundamental theorem Galois theory applied twice tells you that the Galois group is a quotient of a subgroup of $\widehat{\Z}$"

[Violeta Hernández]

Thank you! This is all really helpful.

[Violeta Hernández]

(Wasn't expecting having to get this deep into field theory to formalize funny multiplication on ordinals)

[Kevin Buzzard]

(I just edited a bit, fixed typos, added a couple of comments etc, let me know if there's anything unclear).

Usually people think of $\widehat{\Z}$ as the projective limit of $\Z/n\Z$ for $n\geq1$ but in your case with base $\mathbb{F}_2$ it seems also helpful to think of it as $\prod_p\Z_p$, product over all primes, because then quotients like $\Z_2$ correspond to extensions which are just the union of all subfields of size $2^{2^n}$ and $\Z_3$ corresponds to the extension which is the union of all the subfields of size $2^{3^n}$ etc.

[Aaron Liu]

I learned some Galois theory today

[Violeta Hernández]

Wait, question

[Aaron Liu]

yes?

[Violeta Hernández]

Suppose x is a nimber that is a field that isn't algebraically closed

[Violeta Hernández]

Does that necessarily mean the next ring is also a field?

[Aaron Liu]

yes

[Aaron Liu]

since every nonzero element is a root of unity

[Violeta Hernández]

That's true if x < w^w^w

[Violeta Hernández]

But is it true in general?

[Aaron Liu]

oh that's not what I thought you asked for

[Aaron Liu]

I must have misinterpreted

[Violeta Hernández]

Not all rings are fields

[Violeta Hernández]

...I think, I don't have a concrete counterexample

[Violeta Hernández]

The question is rather, if you have a field, do you keep getting fields until the next transcendental?

[Aaron Liu]

I have ONAG open here

[Violeta Hernández]

I would imagine yes, since the simplest extension theorems tell you that x is the root of some irreducible polynomial in the field

[Violeta Hernández]

Which means the next ring is an algebraic extension

[Violeta Hernández]

And an algebraic extension of a field is also a field, right?

[Aaron Liu]

ONAG claims that $\omega^{\omega^\omega}$ is the first trancendental, and $\omega^{\omega^{\omega+1}}$ is the next ring after that (which is not a field), and the next field is $\omega^{\omega^{\omega^\omega}}$

[Violeta Hernández]

So what I'm claiming is that every single ring from $\omega^{\omega^{\omega^\omega}}$ all the way to the next transcendental is also a field

[Violeta Hernández]

Pretty crazy if true

[Aaron Liu]

so since $\omega^{\omega^{\omega^\omega}}$ is a field which is not algebraically closed, we have that it must be the root of some irreducible polynomial (that polynomial happens to be $X^2 - \omega^{\omega^\omega}$), so the next ring must be generated by this element, so you have a field adjoining the root of an irreducible polynomial, which must be a field

[Aaron Liu]

is that the argument?

[Violeta Hernández]

I think so yes

[Violeta Hernández]

It sounds too good to be true

[Aaron Liu]

the argument is very simple

[Aaron Liu]

try checking it into Lean

[Violeta Hernández]

We can write it in Lean like this

[Aaron Liu]

the only part I'm not sure about is the one where the new root has to generate the extension

[Violeta Hernández]

IsField x and (simplestIrreducible x).degree = n implies IsField (of (x.val ^ n))

[Aaron Liu]

but the new root generating the extension seems like the "simplest" thing to do

[Aaron Liu]

yeah it does generate it

[Aaron Liu]

that's one of the simplest extension theorems or a corollary or something

[Violeta Hernández]

Well, if the simplest irreducible polynomial has degree n then the first n - 1 powers of x are the same as the ordinal powers

[Violeta Hernández]

IsRing (of (x.val ^ n)) should be pretty easy

[Violeta Hernández]

What I don't know is how to write this argument in Lean

[Violeta Hernández]

The field part

[Aaron Liu]

looks like docs#AdjoinRoot.instField

[Violeta Hernández]

Ah yeah that's definitely it

[Aaron Liu]

you can pull the field-ness along docs#IsAdjoinRoot.adjoinRootAlgEquiv

[Aaron Liu]

using docs#MulEquiv.isField

[Violeta Hernández]

Nice

[Violeta Hernández]

So we don't even need to use that argument of "finite rings are fields"

[Violeta Hernández]

to prove that 2^2^n is a field

[Aaron Liu]

oh yeah I guess that's true

[Aaron Liu]

we just need "finite fields aren't algebraically closed"

[Violeta Hernández]

Well we don't even need that

[Violeta Hernández]

The inductive argument proves both that 2^2^n is a field and that x^2 + x + 2^(2^n - 1) is the simplest irreducible polynomial for it

[Violeta Hernández]

Now, question

[Violeta Hernández]

Can we use this to show problem 3?

[Aaron Liu]

what's problem 3

[Violeta Hernández]

That if x ≤ y are fields, then y is an ordinal power of x

[Violeta Hernández]

We can show that this is true below the first transcendental

[Violeta Hernández]

but why do we know that e.g. the next field after w^w^w is a power of it?

[Aaron Liu]

well what do we have so far?

[Violeta Hernández]

If y is the next largest field after x, and x is not algebraically closed, then y is an ordinal power of x

[Violeta Hernández]

The supremum of fields is a field

[Violeta Hernández]

So that's enough to prove the result below w^w^w

[Aaron Liu]

I guess think of all the things you need to adjoin to make it a field

[Violeta Hernández]

If x is algebraically closed then the next ring is x^w

[Aaron Liu]

yes

[Violeta Hernández]

...actually, is there a chance this result is true for rings as well?

[Aaron Liu]

this ring is isomorphic to the polynomials w^w^w[X]

[Aaron Liu]

Violeta Hernández said:

...actually, is there a chance this result is true for rings as well?

who knows do you want to try it

[Violeta Hernández]

w^w^(w + 1) is a ring that isn't a field so it is the inverse of the first element without an inverse which is w

[Aaron Liu]

correct

[Violeta Hernández]

So the next ring is... Laurent series over w^w^w?

[Violeta Hernández]

Does that make any sense?

[Aaron Liu]

not series

[Aaron Liu]

only finitely many terms

[Aaron Liu]

and this ring is w^w^(w+2)

[Aaron Liu]

and w^w^(w+2) is the inverse of w^w^w+1

[Violeta Hernández]

huh

[Aaron Liu]

since all the earlier things have an inverse already

[Aaron Liu]

is that true?

[Violeta Hernández]

It seems true

[Aaron Liu]

did I screw up the arithmetic

[Aaron Liu]

maybe that was supposed to be w^(w^(w+1)*2)

[Violeta Hernández]

no I think it was correct

[Violeta Hernández]

(w^w^x)^w = w^(w^x * w) = w^w^(x + 1)

[Aaron Liu]

these power towers are really big

[Violeta Hernández]

Wait until we get to e0

[Aaron Liu]

ok so w^w^(w+1) is the inverse of the w^w^w

[Violeta Hernández]

Yeah

[Violeta Hernández]

Question: will the rings until the next field all just be the previous one to the w power?

[Aaron Liu]

so w^(w^(w+1)+w^w) is the square of that

[Aaron Liu]

and in general w^(w^(w+1)+w^w*n) is the -n-th power of w^w^w, for finite n

[Aaron Liu]

and then taking all the powers together, w^(w^(w+1)+w^w*w) is the next ring

[Violeta Hernández]

Aaron Liu said:

so w^(w^(w+1)+w^w) is the square of that

wait why?

[Aaron Liu]

w^w^(w+1) * w^w^(w+1) is the smallest product without a solution

[Violeta Hernández]

without a solution?

[Aaron Liu]

the smallest product not less than w^(w^(w+1)+w^w)

[Aaron Liu]

lexographically

[Aaron Liu]

by simplest extension theorem (multiplication), w^w^(w+1) * w^w^(w+1) = w^(w^(w+1)+w^w)

[Violeta Hernández]

wait a sec I'm confused

[Violeta Hernández]

I know the simplest extension theorem I just don't know where w^(w^(w+1)+w^w) is coming from

[Aaron Liu]

oh

[Aaron Liu]

let me check the simplest extension theorems

[Aaron Liu]

I think it comes from w^w^(w+1)*w^w^w

[Violeta Hernández]

Ah yes

[Violeta Hernández]

in a ring multiplying by a smaller number is the same in ordinal and nimber arithmetic

[Aaron Liu]

yeah something like that probably

[Violeta Hernández]

IsRing.mul_eq_of_lt

[Aaron Liu]

yeah that one I'm looking at it right now

[Aaron Liu]

well, I'm looking at the primed version

[Violeta Hernández]

I wish we could have docs# be customized for channels like this

[Aaron Liu]

anyways so we have the next ring is w^(w^(w+1)*2) which is generated by the transcendental element and its inverse

[Aaron Liu]

then by the same argument, w^(w^(w+1)*2+1) is its square

[Aaron Liu]

and the next ring is w^(w^(w+1)*3)

[Aaron Liu]

which is then the inverse of w^w^w+2

[Aaron Liu]

probably make this argument universally quantified

[Violeta Hernández]

Aaron Liu said:

w^w^(w+1) * w^w^(w+1) is the smallest product without a solution

Wait, aren't you misapplying that theorem?

[Aaron Liu]

am I?

[Violeta Hernández]

"z is the lexicographically earliest product of things below z" only applies when z is a group but not a ring

[Violeta Hernández]

But here we do have a ring

[Violeta Hernández]

So what gives?

[Aaron Liu]

it's not a ring though

[Aaron Liu]

since w^w^(w+1) does not have a square

[Violeta Hernández]

w^w^(w+1) is a ring

[Violeta Hernández]

Or what group are you applying it with?

[Aaron Liu]

applying it with w^(w^(w+1)+w^w)

[Aaron Liu]

in which w^w^(w+1) does not have a square, so w^(w^(w+1)+w^w) must be the square of w^w^(w+1)

[Violeta Hernández]

Ok I think I see the argument

[Violeta Hernández]

lemme think about it a bit though

[Violeta Hernández]

Wait

[Violeta Hernández]

Suppose x is a ring but not a field

[Violeta Hernández]

Does that mean x * x is always the next group after x?

[Violeta Hernández]

No absolutely not

[Violeta Hernández]

oh wait lol my counterexample was 2*2 = 3 but that is a field

[Violeta Hernández]

yeah no I'm completely stuck on why w^w^(w+1) * w^w^(w+1) = w^(w^(w+1)+w^w). How do you know its product is at least w^(w^(w+1)+w^w)? How do you know all lexicographically smaller products are less than that?

[Violeta Hernández]

Is this a general instance of a theorem of the form "if x is a ring but not a field then x*x equals y"?

[Aaron Liu]

observe that (w^w^w)^w*w^w is a group
observe that (w^w^w)^w is a ring
observe that w^w^w is a group
observe that (w^w^w)^w is the inverse of w^w^w

let x, y < (w^w^w)^w, then x * y < (w^w^w)^w since it is a ring, and this is less than (w^w^w)^w
let x = (w^w^w)^w and let y < (w^w^w)^w, then express y as a "polynomial" sum of (w^w^w)^n*a_n+...+a_0 for a_i < w^w^w, then x * y = (w^w^w)^(n-1)a_n + ... + xa_0, since w^w^w is an n-th degree closed field and x is the inverse of w^w^w, and rearranging gives x*a_0 + (polynomial) = (w^w^w)^w*a_0 + (polynomial) < (w^w^w)^w*a_0 + (w^w^w)^w < (w^w^w)^w*(a_0+1) < (w^w^w)^w*w^w

[Aaron Liu]

is that clear

[Aaron Liu]

basically if you think of w^w^w as the base field

[Aaron Liu]

then (w^w^w)^w is the polynomials X^n*a_n + X^(n-1)*a_{n-1} + ... + a_0

[Aaron Liu]

and (w^w^w)^w*w^w is the extended polynomials X^n*a_n + ... + a_0 + X^(-1)*a_{-1} where we have negative one terms

[Violeta Hernández]

one sec

[Violeta Hernández]

So if y < (w^w^w)^w then (w^w^w)^w * y < (w^w^w)^w * w^w^w

[Violeta Hernández]

ahh yes

[Violeta Hernández]

More generally if x is algebraically closed then x^w * x^w = x^(w+1) right?

[Aaron Liu]

maybe

[Aaron Liu]

try proving it

[Violeta Hernández]

It should be literally the same argument

[Violeta Hernández]

x^w is a ring, x^(w+1) is a group

[Violeta Hernández]

If y, z < x^w then y * z < x^w

[Aaron Liu]

yeah I believe you

[Violeta Hernández]

If y < x^w then x^w * y = a + x^w * z for a < x^w and z < x

[Aaron Liu]

it's the same argument

[Violeta Hernández]

Actually just one missing step

[Violeta Hernández]

Why is x^w * x^w not smaller than x^(w+1)?

[Aaron Liu]

things smaller then x^(w+1) are those polynomials with the extra X^-1 term

[Aaron Liu]

X^-2 is not one of those

[Violeta Hernández]

Convinces me

[Violeta Hernández]

So the next rings after an algebraically closed x are x^w, then x^(w+1)

[Violeta Hernández]

and then?

[Aaron Liu]

x^(w+1) still isn't a ring

[Violeta Hernández]

hmm

[Aaron Liu]

since x^w doesn't have a cube

[Violeta Hernández]

x^w is the inverse of x, x^(w+1) is the inverse of x^2

[Violeta Hernández]

Do you also have x^(w+2) be the inverse of x^3?

[Violeta Hernández]

Wait, before we continue

[Violeta Hernández]

Wait, before we continue

[Violeta Hernández]

Is there a chance these results are published somewhere?

[Violeta Hernández]

Conway is completely capable of saying "w^w^w^w is the next field" and refusing to elaborate but perhaps Lenstra discusses this?

[Aaron Liu]

I wouldn't know

[Aaron Liu]

but so far I'm reading straight out of ONAG

[Violeta Hernández]

There's a grand total of four sources I know for results about nimbers:

• Conway, On Numbers and Games
• Siegel, Combinatorial Games
• Lenstra, On the Algebraic Closure of Two
• Lenstra, Nim Multiplication

[Violeta Hernández]

I know for a fact none of the first three have anything beyond ω^ω^ω, will look at the last

[Violeta Hernández]

Oh wow that last paper is a gold mine

[Violeta Hernández]

Comes with a lot of exercises i.e. formalization targets

[Violeta Hernández]

There's a fifth simplest extension theorem?!

[Violeta Hernández]

image.png
Excuse me?

[Violeta Hernández]

Oh nevermind, this is just a trivial consequence of the fourth one

[Violeta Hernández]

A field being non-perfect in characteristic 2 just means some element is not a square

[Aaron Liu]

remind me what a perfect field is

[Aaron Liu]

and why it's important

[Violeta Hernández]

In this case the simplest irreducible polynomial is just x² + β

[Aaron Liu]

oh ok

[Violeta Hernández]

Found something which seems relevant

[Violeta Hernández]

image.png
image.png

[Violeta Hernández]

t ⚬⚬ n is typewriter speak for t ^ n

[Aaron Liu]

typewriter?

[Violeta Hernández]

well you can see the font

[Violeta Hernández]

this feels pre-LaTeX

[Aaron Liu]

This all looks pretty similar to the stuff in ONAG

[Violeta Hernández]

In fact Lenstra proves below that if x is algebraically closed, then the next perfect field (is this just the next quadratic field?) is the next epsilon number after x

[Violeta Hernández]

Yeah I think this paper contains the state of the art regarding nim arithmetic above ω^ω^ω;

[Aaron Liu]

that's interesting since ONAG claims that the first perfect field after w^w^w is w^w^(w^w+1)

[Violeta Hernández]

Presumably Lenstra means quadratic closure, then

[Aaron Liu]

yeah that makes sense

[Violeta Hernández]

You know after opening ONAG myself things make a lot more sense

[Violeta Hernández]

I didn't remember that whole page about the nimbers below ω^ω^ω^ω

[Violeta Hernández]

So basically what we want to prove is this
image.png

[Violeta Hernández]

Which presumably works for any t that's algebraically closed, α < t and n : ℕ

[Violeta Hernández]

Which then, presumably, implies the next field after an algebraically closed t is always t ^ t

[Violeta Hernández]

It's probably worth nothing that the equation t ^ t = ω ^ t doesn't hold in general

[Violeta Hernández]

At the very least, it doesn't hold for any epsilon numbers t, since t ^ t > t while ω ^ t = t

[Violeta Hernández]

So it won't hold for even the next algebraically closed nimber

[Violeta Hernández]

It is true that t ^ t = ω ^ t for t = ω ^^ n and n : ℕ+, funnily enough

[Violeta Hernández]

Actually, wait. Does this analysis imply that the next rings after an algebraically closed t are precisely t ^ (ω * x) for x < t?

[Violeta Hernández]

And doesn't this imply that problem 2 (I checked Siegel it's actually problem 4.2) holds for rings?

[Violeta Hernández]

IsRing x → IsRing y → x ≤ y → ∃ z : Ordinal, x.val ^ z = y.val

[Violeta Hernández]

It absolutely does not and i should learn to think before I sspeak

[Violeta Hernández]

obviously t ^ (ω * 3) is not a power of t ^ (ω * 2)

[Violeta Hernández]

But the result that t ^ t is the next field should imply the field version

[Violeta Hernández]

Also, this gives us a pretty good characterization of the rings that aren't fields

[Violeta Hernández]

They're precisely of the form t ^ x for 1 < x < t and t algebraically closed

[Violeta Hernández]

Again, assuming I understand correctly

[Violeta Hernández]

You know, we technically don't have the full first/second simplest extension theorems in our repo

[Violeta Hernández]

I proved that a nimber that isn't a group is the sum of two smaller ones, and a nimber that is a group but not a ring is a product of two smaller ones

[Violeta Hernández]

The proof uses that you can take the lexicographically smallest pair whose sum/product isn't in the group/ring

[Violeta Hernández]

But the statement technically doesn't say that

[Violeta Hernández]

That's a quickfix though

[Violeta Hernández]

Oh nevermind I remembered why I didn't add them

[Violeta Hernández]

They're direct consequences of add_le_of_forall_ne and mul_le_of_forall_ne

[Violeta Hernández]

Ok I think I understand how things work until t^(w2)

[Violeta Hernández]

t^(w+n) = t^-(n+1) is a relatively straightforward induction

[Violeta Hernández]

true for n = 0 by the SET

[Violeta Hernández]

For n + 1, you show t^(w+n+1) = t^w * t^(w+n) by an antisymmetry argument

[Violeta Hernández]

But what about t^(w2)? Why is that the inverse of t+1?

[Violeta Hernández]

Or put another way, why does X+1 have no inverse in R[X,X^-1]?

[Violeta Hernández]

Maybe the idea is to work within the field of rational functions R(X)

[Violeta Hernández]

Actually, here's a super informal argument for why [t^(w2)] = (t-1)^-1

[Violeta Hernández]

In the ring of formal power series P_t[[t]], the inverse of t-1 is 1+t+t^2+... which is not in the subring generated by P_t, t, and t^-1

[Violeta Hernández]

If I'm not mistaken what we want to prove is that [t^(w(1+a)+n)] is the P_t-submodule of P_t(t) generated by t, (t+b)^-m for b < a and natural m, as well as (t+a)^-(m+1) for m < n

[Violeta Hernández]

In fact, let's call this submodule S_(a, n)

[Violeta Hernández]

I think what we want to show is that (t+a)^-(n+1) is not in S_(a, n)

[Violeta Hernández]

Which is now (mostly) a claim about fields of fractions rather than about nimbers

[Violeta Hernández]

Maybe we can start with an easier claim

[Violeta Hernández]

Let F be a field, consider the transcendental extension F(X). Is the subset (X+a)^-1 for a in F linearly independent?

[Violeta Hernández]

Here is a proof so informal it might as well be a joke

[Violeta Hernández]

Suppose the sum a_i(X+b_i)^-1 is 0 with a_i ≠ 0

[Violeta Hernández]

Evaluate at X = -b_i

[Violeta Hernández]

Then infinity = 0

[Violeta Hernández]

lmao

[Violeta Hernández]

I mean, the proof isn't completely informal

[Violeta Hernández]

If you take F = the real numbers then you can formalize this by using limits

[Violeta Hernández]

The function a_i(X+b_i)^-1 has an asymptote at -b_i which means it can't be constantly 0 even on the subset where it's well-defined

[Violeta Hernández]

What would be the way to write down in Lean "F with an infinite element"? Would it be WithTop F even if F wasn't ordered?

[Violeta Hernández]

Does there exist an evaluation map FractionRing F[X] -> WithTop F with enough algebraic properties for this argument to work?

[Violeta Hernández]

You know, sometimes you really just have to Google things: https://math.stackexchange.com/questions/3430357/linearly-independent-set-in-field-of-rational-functions

[Violeta Hernández]

Oh, of course. You can just multiply out the denominators.

[Violeta Hernández]

In fact, this same technique works to prove the stronger claim that $X^n$ and $(X + a)^{-(n+1)}$, for $a \in \mathbb F$ and $n\in\mathbb N$, are all linearly independent

[Violeta Hernández]

Take a linear combination that equals 0. For every term $(X+a)^{-(n+1)}$ with a non-zero coefficient and $n$ maximal, multiply out by $(X+a)^{n+1}$. You now have an equality between polynomials. Evaluate at $X=-a$. You'll get that the coefficients of $(X+a)^{-(n+1)}$ had to all be zero.

[Violeta Hernández]

This all seems highly Lean-able

[Violeta Hernández]

...wait, something I hadn't paid attention to

[Violeta Hernández]

For $a < t$, why is $[t^\omega a] = [t^\omega]a$?

[Django Peeters]

I have no clue

[Violeta Hernández]

oh nevermind this is just IsRing.mul_eq_of_lt again

[Violeta Hernández]

To be more precise, it's this: https://github.com/vihdzp/combinatorial-games/pull/233

[Violeta Hernández]

Which in particular implies that $[ba]=ba$ for $t \le b$ a ring and $a < t$ with $t$ algebraically closed

[Violeta Hernández]

I think the more important question is, why is $[t^{\omega+1}a]=[t^{\omega+1}]a$?

[Violeta Hernández]

Wait, I think I know

[Violeta Hernández]

It's IsGroup.mul_eq_of_lt with $x = [t^{\omega+1}]$, $y=t$, $z=a$, $w=[t^\omega]$

[Violeta Hernández]

To prove $\forall a<x,\forall b<w, a b < x$ you basically expand everything out as polynomials, and use that $[t^\omega][t^{n+1}]=[t^n]$

[Violeta Hernández]

Following this same line of reasoning you should be able to prove $[t^ba]=[t^b]a$ for $a,b<t$

[Violeta Hernández]

Which is pretty insane, since it means you're essentially constructing a $P_t$-basis for $P_t(t)$

[Violeta Hernández]

I want to try and make a Lean proof for all this but the fourth simplest extension theorem hasn't landed in master yet :frown:

[Violeta Hernández]

(which is mostly my fault, don't get me wrong)

[Django Peeters]

I'm still stuck on the same Lean file as 10 days ago lol.

[Violeta Hernández]

Well, to be completely fair, we've only mostly figured out nim arithmetic below ω^ω^ω

[Violeta Hernández]

Mostly I've figured out that it depends on a whole lot of Galois theory

[Django Peeters]

And some Galois theory we need isn't in Mathlib I guess.

[Violeta Hernández]

I don't actually know, or rather I haven't checked

[Violeta Hernández]

In fact, I think the main problem here is that I don't know enough Galois theory to know the fullest generality of the lemmas we want to use.

[Violeta Hernández]

For instance, part of the argument needed to show $[\omega^3]^3=\omega$ is that the cubic extension $P_{\omega}(\omega)$ contains cube roots for every element in $P_{\omega}$. There's presumably some general theorem of the form "if $F$ is a field such that (...), and $F(\alpha)$ is an algebraic extension of prime degree $p$, then all elements in $F$ have $p$-th roots in $F(\alpha)$". But I don't know how to fill in that parenthesis.

[Violeta Hernández]

Maybe it's the same condition as that other theorem? "every algebraic extension is abelian"

[Django Peeters]

I think the parenthesis will contain "every prime below p divides $|F|-1$"

[Violeta Hernández]

$F$ here might be an infinite field!

[Django Peeters]

Hmm

[Django Peeters]

What about "F has $q$-th roots for every prime $q < p$"

[Violeta Hernández]

I think the result holds for all finite $F$. Suppose $x\in F$ has no $p$-th root in $F(\alpha)$. Adjoin a $p$-th root $\beta$ to $F$. Since $F(\alpha)$ and $F(\beta)$ have the same cardinality, they're isomorphic, and by repeatedly composing with Frobenius we can find an isomorphism $\varphi:F(\beta)\to F(\alpha)$ which preserves $F$. Then $\varphi(\beta)$ will be an $p$-th root of $x$.

[Django Peeters]

Wait, no.

[Django Peeters]

I'm confusing this theorem for something else.

[Violeta Hernández]

The question here is how to correctly generalize this for non-finite fields.

[Violeta Hernández]

If $F$ is algebraic over its prime field, then I believe the theorem remains true, since you can just take a finite subfield containing all relevant values.

[Violeta Hernández]

That is enough for what we want to do, but I do still wonder if this can be generalized further

[Aaron Liu]

good morning

[Aaron Liu]

I woke up

[Violeta Hernández]

hi hello

[Violeta Hernández]

I've been busy hehe

[Aaron Liu]

I can tell

[Aaron Liu]

why do we need to prove this in generality

[Violeta Hernández]

well we don't need to, but we probably should

[Aaron Liu]

alright

[Violeta Hernández]

especially if we plan on figuring anything out past ω^ω^ω

[Aaron Liu]

well it only work because finite fields are nice

[Violeta Hernández]

most likely yeah

[Violeta Hernández]

but maybe a few of the theorems do generalize?

[Aaron Liu]

which ones

[Violeta Hernández]

well that's what I want to figure out

[Aaron Liu]

what theorems are you looking at

[Violeta Hernández]

Violeta Hernández said:

For instance, part of the argument needed to show $[\omega^3]^3=\omega$ is that the cubic extension $P_{\omega}(\omega)$ contains cube roots for every element in $P_{\omega}$. There's presumably some general theorem of the form "if $F$ is a field such that (...), and $F(\alpha)$ is an algebraic extension of prime degree $p$, then all elements in $F$ have $p$-th roots in $F(\alpha)$". But I don't know how to fill in that parenthesis.

this one, currently

[Aaron Liu]

I think all you need is algebraic over the prime subfield

[Violeta Hernández]

That works yes

[Violeta Hernández]

My question is whether there's any generalization of that

[Aaron Liu]

and also positive characteristic of course

[Aaron Liu]

try the perfect fields

[Aaron Liu]

maybe it generalizes to those

[Aaron Liu]

for example, I don't think $\mathbb{F}_2(X)(Y)(Y^{-p})$ has a pth root of $X$, so clearly it doesn't generalize that much

[Violeta Hernández]

Fair enough. I suspect we won't be able to find any cool conditions like that "every algebraic extension is abelian" condition from earlier

[Violeta Hernández]

(Is that condition even particularly good? Are there any fields that satisfy that that aren't just subfields of an algebraically closed field?)

[Aaron Liu]

Violeta Hernández said:

Are there any fields that satisfy that that aren't just subfields of an algebraically closed field?)

all field are subfields of an algebraically closed field

[Violeta Hernández]

sorry, an algebraic closure of the prime field

[Aaron Liu]

the complex numbers

[Violeta Hernández]

lol

[Aaron Liu]

not algebraic over the prime field, and no proper algebraic extensions

[Aaron Liu]

galois groups all trivial, so all abelian

[Violeta Hernández]

fair enough

[Violeta Hernández]

Well, there's another theorem that we need

[Violeta Hernández]

Wait before that

[Violeta Hernández]

Does that result even need $p$ to be prime?

[Violeta Hernández]

The "adjoining a p-th root adjoins all p-th roots" result

[Aaron Liu]

Now $E$ is abelian over $K$, and any finite abelian group of order divisible by $p$ has an index $p$ subgroup

look like this part

[Violeta Hernández]

Yeah! This result doesn't need $p$ to be prime

[Violeta Hernández]

The proof I gave works just the same

[Aaron Liu]

wow really

[Violeta Hernández]

And I'm pretty sure you can prove this using primitive roots as well

[Violeta Hernández]

I'm talking about the "every element in $F$ has an $n$-th root in $F(\alpha)$" theorem just to be completely clear

[Kevin Buzzard]

Violeta Hernández said:

Let F be a field, consider the transcendental extension F(X). Is the subset (X+a)^-1 for a in F linearly independent?

This is called the theory of partial fraction decomposition and I definitely had a student do it once but I don't know if it made it into mathlib

[Aaron Liu]

oh I remember doing partial fractions in calculus class

[Violeta Hernández]

You know I'm ashamed it took me this long to see that connection

[Aaron Liu]

it said so in ONAG

[Violeta Hernández]

I must have skipped that sentence :sob:

[Aaron Liu]

based on what I remember from calculus, is it more generally true that inverses of irreducible polynomials in F(X) are F-linearly independent?

[Aaron Liu]

I guess you could have associated elements

[Aaron Liu]

what if you only take one polynomial from each associated equivalence class

[Violeta Hernández]

Presumably the same proof technique works? Take a linear combination, multiply by all the denominators, then substitute a root of one of the irreducible polynomials (in some splitting field).

[Violeta Hernández]

What if you also consider reciprocals of powers of irreducible polynomials?

[Violeta Hernández]

wait yeah we're totally just reinventing Calculus II aren't we

[Aaron Liu]

what's Calculus II

[Aaron Liu]

they made a sequel?

[Violeta Hernández]

I mean, when you do partial fraction decomposition, the terms you end up with are precisely of the form $a/(X+b)^n$

[Aaron Liu]

when I'm working over the real numbers sometimes I get degree two irreducible polynomials

[Violeta Hernández]

Well, partial fraction decomposition over the complex numbers

[Violeta Hernández]

And the reason is that these polynomials (plus $X$) form a basis for the $\mathbb C$-vector space of rational functions $\mathbb C(X)$

[Violeta Hernández]

Which is literally the result that we're making use of here

[Violeta Hernández]

Just for $\mathbb C=\overline{\mathbb F_2}$

[Aaron Liu]

but $\mathbb{C} \ne \overline{\mathbb{F}_2}$

[Violeta Hernández]

For $\overline{\mathbb F_2}$ rather than $\mathbb C$

[Violeta Hernández]

Is what I meant to say

[Aaron Liu]

the proof for $\mathbb{C}$ I just came up with involves Cauchy's integral theorem, which won't generalize to $\overline{\mathbb{F}_2}$

[Aaron Liu]

is there a way to do it without complex analysis

[Violeta Hernández]

Can't you just do the thing where you multiply out denominators in a linear combination for 0, then substitute values to show that every coefficient is 0?

[Aaron Liu]

no idea

[Aaron Liu]

you have powers now right

[Violeta Hernández]

Yeah, but I think it all still works

[Violeta Hernández]

Just multiply by the largest power of each term that appears

[Aaron Liu]

and how does the substituting work

[Aaron Liu]

I think you get some linear equations

[Aaron Liu]

oh you get linear equations

[Aaron Liu]

no but I don't know if it's full rank so I can't use those theorems

[Violeta Hernández]

Suppose $\sum_{i<m,j<n}\frac{a_{i,j}}{p_i^j}=0$, multiply by $\prod_{i<m}p_i^n$ to get something horrible

[Violeta Hernández]

Evaluate at a root of $p_i$ in a splitting field, and you should get $a_{i,n}=0$

[Aaron Liu]

yes that should work

[Aaron Liu]

and then you can induction step

[Violeta Hernández]

I don't think you need induction

[Violeta Hernández]

Well, as I wrote it down you probably do

[Aaron Liu]

do you have a different way

[Violeta Hernández]

Make $n$ depend on $i$

[Violeta Hernández]

Suppose $\sum_{i<m,j\le n_i}\frac{a_{i,j}}{p_i^j}=0$, suppose that $a_{i,n_i}\ne 0$. Multiply by $\prod_{i<m}p_i^{n_i}$. Do the substitution again and find that $a_{i,n_i}= 0$, a contradiction.

[Aaron Liu]

that's induction in disguise

[Violeta Hernández]

Isn't everything

[Aaron Liu]

1+1=2 is not induction

[Violeta Hernández]

Wait until you find out how addition is defined

[Aaron Liu]

uhm

[Django Peeters]

Lol

[Aaron Liu]

docs#Classical.em is not induction

[Violeta Hernández]

You say that, and yet
image.png

[Aaron Liu]

anyways how do we show this spans

[Violeta Hernández]

Isn't that what partial fraction decomposition does

[Aaron Liu]

I mean, the negative powers of irreducibles + nonnegative powers of X

[Aaron Liu]

Violeta Hernández said:

Isn't that what partial fraction decomposition does

how do you know it always works

[Aaron Liu]

I think we do need that these span, in order to conclude that adjoining them all to our ring gives a field

[Violeta Hernández]

Actually, I'm not sure if it's true that these always span

[Violeta Hernández]

It should be true in the case of an algebraically closed base field, which we do have

[Violeta Hernández]

But in general, I'm not sure

[Violeta Hernández]

For instance, how do you do partial fraction decomposition for $(X(X^2+1))^{-1}$ over $\mathbb R$?

[Aaron Liu]

it decomposes into hold on let me calculate

[Violeta Hernández]

You can't write it as a linear combination of $X^{-1}$ and $(X^2+1)^{-1}$

[Aaron Liu]

why not

[Aaron Liu]

oh

[Aaron Liu]

well of course by linear combination I meant R[X]-linear combination

[Violeta Hernández]

...is that what people generally mean by partial fraction decompositions?

[Aaron Liu]

yes?

[Aaron Liu]

that's what I learned in calculus

[Violeta Hernández]

Yeah sorry it's been a while

[Aaron Liu]

the numerators of your fractions are polynomials with degree less than the denominator

[Violeta Hernández]

Does the linear independence result I gave still hold?

[Violeta Hernández]

Oh yeah it totally should

[Aaron Liu]

well no since X * X^-1 - 1 = 0

[Violeta Hernández]

ok nevermind

[Aaron Liu]

how about this

[Violeta Hernández]

Well, our reciprocals of irreducible polynomials should totally span as an R[X]-module

[Violeta Hernández]

Something something polynomials are a PID

[Violeta Hernández]

but in the case of nimbers we want the stronger result that they span as an R-module for R algebraically closed

[Aaron Liu]

we can do R-linear combinations but instead of just having negative powers of irreducible polynomials we add in a power of X times a negative power of an irreducible such the the denominator has larger degree than the power we took in the numerator

[Aaron Liu]

oh but we need to special-case the negative powers of X

[Violeta Hernández]

...are we sure someone hasn't done this in Mathlib yet?

[Aaron Liu]

@loogle Polynomial, RatFunc, _⁻¹

[loogle]

:search: RatFunc.inv_def, RatFunc.ofFractionRing_inv, and 4 more

[Aaron Liu]

maybe this works in more generality?

[Violeta Hernández]

Partial fraction decomposition? In more generality than rational functions?

[Aaron Liu]

or maybe it's just not there

[Violeta Hernández]

more likely

[Aaron Liu]

@loogle Polynomial, RatFunc, _ / _

[loogle]

Failure! Third party responded with 502

[Aaron Liu]

maybe I'll try again later

[Violeta Hernández]

Kevin Buzzard said:

Violeta Hernández said:

Let F be a field, consider the transcendental extension F(X). Is the subset (X+a)^-1 for a in F linearly independent?

This is called the theory of partial fraction decomposition and I definitely had a student do it once but I don't know if it made it into mathlib

Apparently the answer is "no". Does the code exist anywhere public?

[Violeta Hernández]

As a brief aside, I find it hilarious that we're nominally trying to do "game theory"

[Violeta Hernández]

This is so far removed from games it's honestly kind of astounding

[Aaron Liu]

this is Field theory

[Violeta Hernández]

this is a silly stone game we're overanalyzing

[Aaron Liu]

nim?

[Aaron Liu]

did that have multiplication?

[Violeta Hernández]

Nimber multiplication matches the multiplication of games applies to nim

[Aaron Liu]

what's the operation to multiply two piles

[Aaron Liu]

in the game

[Violeta Hernández]

IGame.mul'

[Violeta Hernández]

I mean, to be completely frank, game multiplication is basically nonsense if you view it as an operation on games

[Violeta Hernández]

It just so happens to have really nice algebraic properties when you restrict it to either the impartial games or the surreal games

[Aaron Liu]

we already solved transfinite nim with the powers of two decomposition

[Aaron Liu]

this is something else now

[Violeta Hernández]

We already solved transfinite nim: first player wins if the pile is non-empty, second player wins otherwise

[Aaron Liu]

that's one pile nim

[Aaron Liu]

very boring

[Violeta Hernández]

At a very large glance all we're currently doing is finding which one-pile nim games correspond to other impartial games built out of sums and products of nim piles

[Aaron Liu]

that is not

[Aaron Liu]

no

[Aaron Liu]

there's no such thing as a "product of nim piles"

[Aaron Liu]

the multiplication is happening in the nimbers

[Violeta Hernández]

• nim a + nim b is the game where you can move to nim c + nim b or nim a + nim d for c < a and d < b
• nim a * nim b is the game where you can move to nim c * nim b + nim a * nim d + nim c * nim d for c < a and d < b

[Aaron Liu]

what physical... never mind

[Violeta Hernández]

I mean, there's certainly ways you could represent that physically

[Violeta Hernández]

I don't claim they're particularly good

[Violeta Hernández]

Nim rectangles perhaps?

[Aaron Liu]

it has to be composable with itself

[Aaron Liu]

and with addition

[Violeta Hernández]

You turn a rectangle into 3 smaller rectangles, you can now move on any of the rectangles

[Violeta Hernández]

Unironically this would make a pretty fun game if you ignore the exponential amount of counters you'll probably need to play it

[Aaron Liu]

does it play nice with addition

[Aaron Liu]

what if I want 6+(6+2)*4

[Aaron Liu]

(I will let you go first in this game)

[Violeta Hernández]

Well now you're complicating my setup

[Violeta Hernández]

You can go first, I'll be the bigger woman

[Aaron Liu]

Violeta Hernández said:

Well now you're complicating my setup

but do you need like weird rectangles to handle this

[Violeta Hernández]

The nim rectangles game only really showcases nim a * nim b

[Violeta Hernández]

If you want to visualize a product of two arbitrary games then you're going to have a much harder time

[Django Peeters]

Django Peeters said:

We're going for alpha(263) in 1 day. I'm not sure about alpha(283) but it might take 10 days.

On an unrelated note, this has now been reduced to 2211 seconds for alpha(263).

[Violeta Hernández]

Violeta Hernández said:

If you want to visualize a product of two arbitrary games then you're going to have a much harder time

You can do it. The entire idea behind LGame multiplication is that a game x * y can be seen as a multiset of signed pairs of states in x and y.

[Violeta Hernández]

Should you do it? Yeah, I don't think so.

[Django Peeters]

image.png
What about this? It's what I need to continue working on $\omega^{\omega^\omega}$...

[Aaron Liu]

Django Peeters said:

image.png
What about this? It's what I need to continue working on $\omega^{\omega^\omega}$...

which one

[Django Peeters]

Where Conway says every cubic extension is by a cuberoot.

[Aaron Liu]

we just need to prove that the previous cube root has no cube root

[Django Peeters]

How does that work?

[Aaron Liu]

let's start at the base case

[Aaron Liu]

how to show that $\omega$ has no cube root in $P_{\omega^3}$

[Violeta Hernández]

Django Peeters said:

Where Conway says every cubic extension is by a cuberoot.

Isn't that essentially what I was saying like 50 messages above?

[Violeta Hernández]

Violeta Hernández said:

I think the result holds for all finite $F$. Suppose $x\in F$ has no $p$-th root in $F(\alpha)$. Adjoin a $p$-th root $\beta$ to $F$. Since $F(\alpha)$ and $F(\beta)$ have the same cardinality, they're isomorphic, and by repeatedly composing with Frobenius we can find an isomorphism $\varphi:F(\beta)\to F(\alpha)$ which preserves $F$. Then $\varphi(\beta)$ will be an $p$-th root of $x$.

here

[Django Peeters]

Not exactly.

[Aaron Liu]

not the same

[Violeta Hernández]

Hmm, you're right, it's not exactly that

[Django Peeters]

Even when restricting to finite fields of characteristic 2, it doesn't always hold because $x \mapsto x^3$ can be bijective.

[Aaron Liu]

this is "a cubic extension adds all the cube roots"

[Aaron Liu]

oh I guess maybe that's the same

[Violeta Hernández]

ok now I'm confused

[Violeta Hernández]

is it or is it not

[Aaron Liu]

it's not

[Aaron Liu]

we need a specific cube root

[Violeta Hernández]

Ok, check this out

[Aaron Liu]

in particular, we need that it didn't already exist

[Violeta Hernández]

Suppose $F$ is a finite field of characteristic $2$, take an algebraic extension $G$ of odd degree $n$. Let $k = \frac{|G|-1}{|F|-1}$, let $\xi$ be a primitive root of $G$. Then $n \mid k$, meaning $G=F(\xi^{k/n})$ is generated by an $n$-th root.

[Aaron Liu]

I think that works

[Django Peeters]

$n \mid k$ doesn't always hold in this context

[Violeta Hernández]

Yeah, I'm realizing you probably need $n$ to be prime

[Django Peeters]

It doesn't always work when $n = 3$. Then you need that $|F|$ is a 4-power.

[Aaron Liu]

wait why is this supposed to work

[Violeta Hernández]

wow ok I'm completely misremembering Fermat's little theorem apparently

[Aaron Liu]

$X^{|F|}-X$ splits into distinct linear factors in a finite field $F$

[Violeta Hernández]

If $n$ is prime and $n \mid |F|-1$ then $n \mid k$ by LTE

[Violeta Hernández]

Is that anything

[Aaron Liu]

what's LTE

[Violeta Hernández]

Lifting the Exponent

[Aaron Liu]

sorry I'm not very familiar with that theorem

[Violeta Hernández]

If $p$ is an odd prime and $p \mid a - b$ then $v_p(a^k - b^k) = v_p(a - b) + v_p(k)$

[Aaron Liu]

great

[Aaron Liu]

how does that apply here

[Violeta Hernández]

it says that in that very special case $n\mid k$

[Aaron Liu]

oh I get it now

[Aaron Liu]

set $a := |F|$ and $b := 1$ and $k := p$

[Violeta Hernández]

$3\mid 2^{2^n}-1$ for $n \ne 0$

[Violeta Hernández]

so my proof does work in that case I think?

[Violeta Hernández]

A cubic extension of a finite char 2 field is by a cube root

[Aaron Liu]

yes we knew that

[Violeta Hernández]

we did?

[Aaron Liu]

doesn't it follow from that theorem

[Aaron Liu]

uhm

[Aaron Liu]

well I guess with torsion...

[Violeta Hernández]

ok now I'm really confused

[Violeta Hernández]

We know that a degree $n$ extension of a finite field has all $n$-th roots for the base field unless I also fakesolved that

[Aaron Liu]

so in the finite field $\mathbb{F}_8$ every element has a cube root

[Django Peeters]

Violeta Hernández said:

$3\mid 2^{2^n}-1$ for $n \ne 0$

Yes, that's true

[Violeta Hernández]

If the base field didn't already have all $n$-th roots, does that imply that the extension is by an $n$-th root?

[Aaron Liu]

which extension

[Aaron Liu]

what

[Aaron Liu]

I got lost

[Violeta Hernández]

$[G:F]=n$ means $F \subseteq G^n$

[Aaron Liu]

what's $G^n$

[Violeta Hernández]

pointwise power

[Aaron Liu]

that's not a field

[Violeta Hernández]

It's a subset

[Django Peeters]

Violeta Hernández said:

If the base field didn't already have all $n$-th roots, does that imply that the extension is by an $n$-th root?

(For finite fields) I think yes.

[Aaron Liu]

but in the finite case they have the same cardinality?

[Aaron Liu]

I'm confused

[Aaron Liu]

where are $F$ and $G$ coming from

[Violeta Hernández]

They absolutely don't have the same cardinality

[Aaron Liu]

are they arbitrary field? arbitrary finite fields?

[Violeta Hernández]

finite fields

[Django Peeters]

Except maybe trouble with going from "has no roots" to "is irreducible"...

[Aaron Liu]

then I'm assuming $F$ is an extension of $G$ since you're talking about $[F:G]$

[Violeta Hernández]

yes

[Violeta Hernández]

I'm just rephrasing a result I proved earlier

[Aaron Liu]

and if $[F:G] = n$ then $|F| = |G|^n$

[Violeta Hernández]

oh oops

[Violeta Hernández]

I wrote the extension backwards

[Violeta Hernández]

I meant $[G:F]=n$

[Aaron Liu]

so the result is that if $[G:F]=n$ then there exists a ring homomorphism $F \hookrightarrow G^n$

[Violeta Hernández]

$G^n$ is just the pointwise power here

[Violeta Hernández]

the image of $x \mapsto x^n$ on the set $G$

[Aaron Liu]

oh

[Aaron Liu]

I thought you meant the cartesian product of $n$ copies of $G$ which inherits its ring structure pointwise

[Aaron Liu]

so $F$ is the subset of $G$ fixed by $x \mapsto x^{|F|}$

[Violeta Hernández]

That is true

[Violeta Hernández]

No, not quite

[Violeta Hernández]

It's the subset fixed by $x \mapsto x^{\frac{|G|-1}{|F|-1}}$

[Aaron Liu]

:confusedface:

[Violeta Hernández]

wait no you're right

[Django Peeters]

Lol

[Aaron Liu]

:relievedface:

[Violeta Hernández]

Violeta Hernández said:

It's the subset fixed by $x \mapsto x^{\frac{|G|-1}{|F|-1}}$

It's the range of this map, not the fixed subset

[Aaron Liu]

that I can believe

[Violeta Hernández]

So what exactly is the claim we're trying to prove again?

[Aaron Liu]

uhm

[Violeta Hernández]

I just found this in Siegel
image.png

[Violeta Hernández]

Presumably this is what we want

[Aaron Liu]

trying to prove that $\omega^{3^k}$ is the cube of its [cube]

[Violeta Hernández]

Wait, isn't the claim I just screenshotted false?

[Violeta Hernández]

How do you get $\mathbb F_{512}$ as an extension of $\mathbb F_8$ by a cube root?

[Aaron Liu]

you don't

[Aaron Liu]

every element already has a cube root

[Violeta Hernández]

yeah so that's false then

[Violeta Hernández]

Miscellaneous question that's been tripping me up

[Violeta Hernández]

If you have a field, and adjoin an nth root that isn't already in the field, is the extension always degree n?

[Aaron Liu]

if $n$ is prime

[Django Peeters]

Violeta Hernández said:

Miscellaneous question that's been tripping me up

Been tripping me up too :(

[Violeta Hernández]

yeah

[Aaron Liu]

but not if it's not prime

[Violeta Hernández]

So now I'm really starting to question that whole thing I said earlier

[Aaron Liu]

what did you say earlier

[Violeta Hernández]

$F\subseteq G^n$

[Violeta Hernández]

Actually that should still be true

[Violeta Hernández]

You can induct on the prime factorization of n

[Violeta Hernández]

You know I've definitely been awake for too long

[Aaron Liu]

did you not sleep

[Violeta Hernández]

My theorems aren't theoreming right now

[Aaron Liu]

I can help

[Violeta Hernández]

I'm taking a forced break

[Aaron Liu]

have fun

[Kevin Buzzard]

Violeta Hernández said:

If you have a field, and adjoin an nth root that isn't already in the field, is the extension always degree n?

The short answer is no but the long answer is that this is not even a well-defined concept. In general if $k$ is a field and $a\in k$ then $X^n-a$ can factor "randomly" over $k$ as a product of irreducible factors of different degrees, so "adjoining an $n$-th root of $a$" is not even a well-defined concept in the sense that you can even get extensions of different degrees which are both claiming to be the ill-defined notion $k(\sqrt[n]{a})$.

[Kevin Buzzard]

The situation is much better when $k$ contains $n$ $n$-th roots of 1, in which case the relevant phrase is "Kummer theory". In particular in this case "adjoining an $n$-th root of $a$" can be made rigorous and the answer is that the extension is always Galois with cyclic Galois group of order dividing $n$ (so if $n$ is prime then indeed you're correct).

[Aaron Liu]

Kevin Buzzard said:

Violeta Hernández said:

If you have a field, and adjoin an nth root that isn't already in the field, is the extension always degree n?

The short answer is no but the long answer is that this is not even a well-defined concept. In general if $k$ is a field and $a\in k$ then $X^n-a$ can factor "randomly" over $k$ as a product of irreducible factors of different degrees, so "adjoining an $n$-th root of $a$" is not even a well-defined concept in the sense that you can even get extensions of different degrees which are both claiming to be the ill-defined notion $k(\sqrt[n]{a})$.

I interpreted this to mean "I have a nontrivial field extension K/F which is generated by an nth root of an element in F (there exists u in K such that u^n in F and IntermediateField.adjoin F {u} = \top), do I know this is a degree n extension?" and I thought a bit and concluded "if n is prime" but is that correct?

[Aaron Liu]

oh it's not correct I just thought of a counterexample

[Kevin Buzzard]

$\mathbb{Q}(\omega)$ is a counterexample if $\omega:=\frac{-1+\sqrt{-3}}{2}$.

[Kevin Buzzard]

If you don't have $n$ $n$-th roots of 1 in your base field then adding $n$-th roots can be very counterintuitive (in fact far more counterintuitive than many of the people who just blithely write $k(\sqrt[n]{a})$ on sites like MO realise)

[Aaron Liu]

what does having roots of unity do for you

[Kevin Buzzard]

It stops $X^n-a$ from factoring into factors of different degrees, for a start. Example: $X^3-1$ factors over $\mathbb{Q}$ (which doesn't have three cube roots of 1) into a linear and irreducible quadratic, which is the example above.

[Violeta Hernández]

Well that explains why I kept tripping up with this concept!

[Violeta Hernández]

That does call into question most of my proofs until this point... I might have to reprove some things.

[Violeta Hernández]

Also, if adjoining an n-th root isn't a well-defined concept in general, does it even make sense to claim that an extension is "by an n-th root"?

[Kevin Buzzard]

Yes, the definition you gave above is fine although as Aaron pointed out it can behave oddly if there aren't enough roots of unity in the ground field

[Django Peeters]

PastedText.txt
New alpha's just dropped!

[Violeta Hernández]

Kevin Buzzard said:

Yes, the definition you gave above is fine although as Aaron pointed out it can behave oddly if there aren't enough roots of unity in the ground field

Is it at least true that if an extension is by a p-th root for prime p, the extension is of degree p?

[Aaron Liu]

Violeta Hernández said:

Kevin Buzzard said:

Yes, the definition you gave above is fine although as Aaron pointed out it can behave oddly if there aren't enough roots of unity in the ground field

Is it at least true that if an extension is by a p-th root for prime p, the extension is of degree p?

I extend the reals by a primitive cube root of unity to get the complex numbers (degree 2 extension)

[Violeta Hernández]

damn it I keep forgetting about that

[Violeta Hernández]

Isn't every finite extension of a finite field separable, or something like that?

[Kevin Buzzard]

Yes, and furthermore every finite extension of an algebraic extension of a finite field (the situation you sometimes seem to be in) is also separable, and hence simple, but this doesn't help. Basically if $X^n-1$ doesn't factor into linear factors then there's very little structure in how $X^n-a$ may factor (and in particular the factors can have different degrees). If $X^n-1$ factors into distinct linear factors then $X^n-a$ is guaranteed to factor into lots of factors of the same degree (which will clearly be a divisor of $n$) and all of the extensions corresponding to the irreducible factors will be abstractly isomorphic, meaning that $k(\sqrt[n]{a})$ is well-defined up to (non-unique) isomorphism in this case (and is meaningless otherwise).

[Django Peeters]

Django Peeters said:

PastedText.txt
New alpha's just dropped!

It's now at alpha(359) and I expect it'll take 26 days.

[Django Peeters]

Tristan is running it in the background.

[Django Peeters]

Violeta Hernández said:

I just found this in Siegel
image.png

I think I know how to prove this.

[Violeta Hernández]

How?

[Django Peeters]

Not the whole thing yet, but how part of it works.

[Aaron Liu]

which part

[Django Peeters]

When a finite field can be extended by a pth root.

[Aaron Liu]

probably when it doesn't have all pth roots

[Aaron Liu]

I got lost what do we have so far on w^w^w and what do we still need

[Django Peeters]

The lexicographic ordering on kappa_q...

[Django Peeters]

And that's as far as I got

[Django Peeters]

Django Peeters said:

Django Peeters said:

PastedText.txt
New alpha's just dropped!

It's now at alpha(359) and I expect it'll take 26 days.

Alpha(359) is done!

[Aaron Liu]

not 26 days

[Django Peeters]

Right you are!

[Django Peeters]

I made some more changes to speed the program up.

[Violeta Hernández]

Is the code open-source?

[Django Peeters]

Yes, I put it here about a month ago.

[Django Peeters]

Now alpha(463) is done!

[Violeta Hernández]

Nice!

[Violeta Hernández]

Is this list of new values anywhere?

[Django Peeters]

<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
    p           q_set excess                alpha t(sec)
    3             [2]      0                    2      0
    5             [4]      0                    4      0
    7             [3]      1                w + 1      0
   11             [5]      1              w^w + 1      0
   13          [4, 3]      0                w + 4      0
   17             [8]      0                   16      0
   19             [9]      4              w^3 + 4      0
   23            [11]      1          w^(w^3) + 1      0
   29          [4, 7]      0          w^(w^2) + 4      0
   31             [5]      1              w^w + 1      0
   37          [4, 9]      0              w^3 + 4      0
   41             [5]      1              w^w + 1      0
   43             [7]      1          w^(w^2) + 1      0
   47            [23]      1          w^(w^7) + 1      0
   53            [13]      1          w^(w^4) + 1      0
   59            [29]      1          w^(w^8) + 1      0
   61          [3, 5]      0              w^w + w      0
   67         [3, 11]      0          w^(w^3) + w      0
   71          [5, 7]      0        w^(w^2) + w^w      0
   73             [9]      1              w^3 + 1      0
   79            [13]      1          w^(w^4) + 1      0
   83            [41]      1         w^(w^11) + 1      0
   89            [11]      1          w^(w^3) + 1      0
   97         [16, 3]      0              w + 256      0
  101            [25]      1          w^(w*5) + 1      0
  103         [3, 17]      0          w^(w^5) + w      0
  107            [53]      1         w^(w^14) + 1      1
  109          [4, 9]      0              w^3 + 4      0
  113          [4, 7]      0          w^(w^2) + 4      0
  127             [7]      1          w^(w^2) + 1      0
  131         [5, 13]      0        w^(w^4) + w^w      0
  137            [17]      1          w^(w^5) + 1      0
  139         [3, 23]      0          w^(w^7) + w      4
  149            [37]      1         w^(w^10) + 1      0
  151          [3, 5]      0              w^w + w      0
  157            [13]      1          w^(w^4) + 1      0
  163            [81]      4             w^27 + 4      0
  167            [83]      1         w^(w^21) + 1     57
  173         [4, 43]      0         w^(w^12) + 4      0
  179            [89]      1         w^(w^22) + 1      6
  181          [9, 5]      0            w^w + w^3      0
  191         [5, 19]      0        w^(w^6) + w^w      1
  193         [32, 3]      0            w + 65536      0
  197         [4, 49]      0        w^(w^2*7) + 4      0
  199         [9, 11]      0        w^(w^3) + w^3      0
  211          [5, 7]      0        w^(w^2) + w^w      0
  223            [37]      1         w^(w^10) + 1      0
  227           [113]      1         w^(w^28) + 1      2
  229            [19]      1          w^(w^6) + 1      0
  233            [29]      1          w^(w^8) + 1      0
  239         [7, 17]      0    w^(w^5) + w^(w^2)      0
  241          [8, 3]      0               w + 16      0
  251            [25]      1          w^(w*5) + 1      0
  257            [16]      0                  256      0
  263           [131]      1         w^(w^30) + 1    131
  269            [67]      1         w^(w^17) + 1     28
  271         [27, 5]      0            w^w + w^9      0
  277            [23]      1          w^(w^7) + 1      1
  281          [5, 7]      0        w^(w^2) + w^w      0
  283            [47]      1         w^(w^13) + 1    894
  293         [4, 73]      0         w^(w^19) + 4      0
  307         [3, 17]      0          w^(w^5) + w      0
  311            [31]      1          w^(w^9) + 1      0
  313            [13]      1          w^(w^4) + 1      0
  317            [79]      1         w^(w^20) + 1      3
  331          [3, 5]      0              w^w + w      0
  337             [7]      1          w^(w^2) + 1      0
  347           [173]      1         w^(w^38) + 1   6255
  349            [29]      1          w^(w^8) + 1      0
  353         [8, 11]      0         w^(w^3) + 16      0
  359           [179]      1         w^(w^39) + 1 177379
  367            [61]      1         w^(w^16) + 1      0
  373         [3, 31]      0          w^(w^9) + w      0
  379         [27, 7]      0        w^(w^2) + w^9      0
  383           [191]      1         w^(w^41) + 1   5547
  389            [97]      1         w^(w^23) + 1      1
  397            [11]      1          w^(w^3) + 1      0
  401         [8, 25]      0         w^(w*5) + 16      0
  409         [3, 17]      0          w^(w^5) + w      0
  419        [11, 19]      0    w^(w^6) + w^(w^3)     19
  421          [5, 7]      0        w^(w^2) + w^w      1
  431            [43]      1         w^(w^12) + 1      0
  433          [8, 9]      0             w^3 + 16      0
  439            [73]      1         w^(w^19) + 1      0
  443        [13, 17]      0    w^(w^5) + w^(w^4)      0
  449         [32, 7]      0      w^(w^2) + 65536      0
  457            [19]      1          w^(w^6) + 1      0
  461            [23]      1          w^(w^7) + 1      1
  463         [7, 11]      0    w^(w^3) + w^(w^2)      0

[Django Peeters]

If we get to alpha(547), we'll be able to do nimber arithmetic below $\omega^{\omega^{100}}$.

[Violeta Hernández]

Hi! I'm back on the nimber grind

[Violeta Hernández]

This thread has gotten way too big for its own good. I can't even remember the main results we need to prove to characterize arithmetic below w^w^w, nor which of them we have proofs for.

[Aaron Liu]

I don't know what needs proving and what needs putting in lean and what's done

[Aaron Liu]

had there been like a list of all the things to do

[Violeta Hernández]

I don't believe so. We should make one.

[Violeta Hernández]

I'd like to mention that one of my PRs porting the proof that nimbers are alg closed has been unreviewed for three weeks. Might be worth starting there.

[Violeta Hernández]

CGT#231

[Django Peeters]

I'm also about to pivot back to Lean and theorem proving. The C++ sidetrack of making tables has almost come to an end. If the current table needs to be extended to the next prime, I estimate a wait time of ~4 years with the current setup. But the table can have holes so I'll try to push it a little further and let Tristan run this last iteration in the background as long as we want.

[Django Peeters]

Do you know how we could get this on the OEIS?

[Violeta Hernández]

I know little about OEIS submissions (though I've made one or two). maybe check the website for instructions? I don't remember it being hard as much as it was tedious and bureaucratic.

[Violeta Hernández]

I think there were basically two questions we didn't have answers to:

• why is [w^3] quadratically closed
• why does it have cube roots below [w]

[Violeta Hernández]

I think the former could be answered using a theorem from Math SE

[Violeta Hernández]

I don't remember how to answer the second one

[Violeta Hernández]

I recall I tried to find a general theorem for this but failed

[Aaron Liu]

Violeta Hernández said:

I think there were basically two questions we didn't have answers to:

• why is [w^3] quadratically closed
• why does it have cube roots below [w]

didn't we answer this already

[Violeta Hernández]

the second one?

[Aaron Liu]

which one

[Aaron Liu]

isn't it both?

[Aaron Liu]

don't we have the answer to both of them

[Aaron Liu]

or did we not answer these

[Aaron Liu]

I thought we answered both

[Violeta Hernández]

The first one we did answer

[Violeta Hernández]

it's on that one Math SE post

[Aaron Liu]

yeah ok

[Aaron Liu]

what about the other one I think we did the other one too

[Violeta Hernández]

How

[Aaron Liu]

just look at it

[Aaron Liu]

it's so true

[Aaron Liu]

if you want I can try throwing together a proof

[Django Peeters]

Proof by obviousness lol

[Violeta Hernández]

please do

[Aaron Liu]

oh

[Aaron Liu]

uhh

[Aaron Liu]

let me think a bit

[Django Peeters]

I was joking

[Aaron Liu]

Yeah I did this before

[Aaron Liu]

You can refer to past me for the proof

Aaron Liu said:

So let $\alpha < \omega$ then consider the field $\mathcal F$ generated by $2$ and $\alpha$ and assume it doesn't have a cube root of $\alpha$ then adjoing a cube root of $\alpha$ gives a degree 3 extension and $\mathcal F (\omega)$ is a degree 3 extension so they're isomorphic

[Aaron Liu]

no that's the square roots I think

[Aaron Liu]

no it's not that definitely says cube root right there

[Aaron Liu]

ok I think I understand my own proof again

[Django Peeters]

I remember you had a way to contact A. Siegel, right?

[Aaron Liu]

Kevin Buzzard said:

Violeta Hernández said:

If you have a field, and adjoin an nth root that isn't already in the field, is the extension always degree n?

The short answer is no but the long answer is that this is not even a well-defined concept. In general if $k$ is a field and $a\in k$ then $X^n-a$ can factor "randomly" over $k$ as a product of irreducible factors of different degrees, so "adjoining an $n$-th root of $a$" is not even a well-defined concept in the sense that you can even get extensions of different degrees which are both claiming to be the ill-defined notion $k(\sqrt[n]{a})$.

ok apparently I can't just say "adjoin a cube root" there's a bit more to check here

[Aaron Liu]

but I think in this case since $2$ is a cube root of unity and $\alpha$ doesn't have a cube root already then "adjoin a cube root" is well-defined and a degree 3 extension

[Violeta Hernández]

Well, [ω³] doesn't only have cube roots below [ω]. Isn't it more generally true that any cubic polynomial in [ω] splits in [ω³]?

[Violeta Hernández]

So presumably instead of writing "adjoin a cube root" you can write "take a cubic extension"

[Aaron Liu]

yeah I guess

[Violeta Hernández]

In fact, since [ω³] is quadratically closed, a cubic polynomial splits iff it's not irreducible

[Aaron Liu]

makes sense

[Violeta Hernández]

No wait

[Aaron Liu]

what's wrong

[Violeta Hernández]

Take a polynomial $p$ in $\mathcal P_{\omega}$. We want to show that it splits in $\mathcal P_{\omega^3}$. So what we do is take the field $\mathcal F$ generated by $2$ and the coefficients of $p$, right?

[Aaron Liu]

uhm

[Aaron Liu]

mayve

[Violeta Hernández]

And then what we want to show is that the splitting field of $p$ over $\mathcal F$ is a cubic extension, right? So that we can then show it isomorphic to $\mathcal F(\omega)$

[Aaron Liu]

that would suffice

[Violeta Hernández]

But why is that the case?

[Aaron Liu]

why what the case

[Violeta Hernández]

A splitting field for a cubic polynomial can have up to degree six

[Aaron Liu]

this is news to me

[Aaron Liu]

why is that the case

[Violeta Hernández]

From Math SE
image.png

[Aaron Liu]

interesting

[Aaron Liu]

I had no idea

[Violeta Hernández]

I imagine that what stops us from getting this worst case scenario is that once we have one root, we have the other two because of some field being quadratically closed

[Aaron Liu]

I guess 3! = 6

[Aaron Liu]

Violeta Hernández said:

I imagine that what stops us from getting this worst case scenario is that once we have one root, we have the other two because of some field being quadratically closed

in my original proof which only talked about cube roots this never happens because we have all the cube roots of unity

[Violeta Hernández]

You know, I'm actually not even sure if the stronger claim is true

[Aaron Liu]

and using the cube roots of unity once you have one root you can obtain the other two cube roots

[Aaron Liu]

Violeta Hernández said:

You know, I'm actually not even sure if the stronger claim is true

what makes you say this

[Aaron Liu]

ok what if you don't take the splitting field what if you just adjoin one root

[Aaron Liu]

and we can get the other two roots later

[Violeta Hernández]

Please remind me of something

[Violeta Hernández]

If $p$ is irreducible over $\mathcal F$, then adjoining a root of $p$ always gives a degree $\deg(p)$ extension, yes?

[Aaron Liu]

definitely

[Violeta Hernández]

And we know that if $p$ is a cubic polynomial in $\mathcal P_\omega$ without roots, it must be irreducible

[Violeta Hernández]

Since $P_\omega$ is quadratically closed

[Aaron Liu]

yes

[Violeta Hernández]

well actually just because $3 < 2 + 2$

[Aaron Liu]

yes you can't factor into multiple nonlinear factors without having at least two quadratic factors

[Violeta Hernández]

Ok so, let $\alpha$ be a root of $p$. The field $\mathcal F(\alpha)$ is a cubic extension of $\mathcal F$. The field $\mathcal F(\omega)$ is also a cubic extension of $\mathcal F$. Since they're both finite fields they're isomorphic, and there exists an isomorphism fixing $\mathcal F$. Whatever $\alpha$ gets sent to by this isomorphism is less than $\omega^3$, and is a root of the polynomial.

[Violeta Hernández]

Ok wow that was easy

[Aaron Liu]

yeah I know right

[Violeta Hernández]

How easy is it to make a blueprint?

[Violeta Hernández]

as in, the Lean webpages collecting informal proofs

[Violeta Hernández]

that would be very useful here

[Aaron Liu]

https://github.com/pitmonticone/LeanProject looks like it comes with a blueprint

[Violeta Hernández]

I'll check this out as soon as I'm on my PC again

[Violeta Hernández]

So, I believe that we completely understand nimbers below w^w

[Aaron Liu]

that would be great

[Violeta Hernández]

And we know why w^w is cubically closed

[Violeta Hernández]

but why is it quartically closed?

[Violeta Hernández]

that makes no sense to me

[Aaron Liu]

maybe because 4 is composite

[Violeta Hernández]

well clearly it has something to do with that

[Violeta Hernández]

because w^(w^2) is not only quintically but sextically closed, etc.

[Aaron Liu]

I'll see what it says in the ONAG book

[Violeta Hernández]

Maybe the argument is that a quartic extension of a finite field is the same as a quadratic extension of a quadratic extension which we do have?

[Aaron Liu]

ok I looked at the ONAG book and was not enlightened

[Aaron Liu]

maybe we can look at the absolute galois group

[Aaron Liu]

ok I'm looking at the absolute galois group now

[Aaron Liu]

Kevin said that algebraic extensions of $\mathbb{F}_2$ correspond to closed subgroups of $\widehat{\mathbb{Z}}$

[Aaron Liu]

we start by quotienting out by the subgroup corresponding to $\omega^\omega$ since that's the field we're interested in

[Aaron Liu]

or no not quotienting out

[Aaron Liu]

the opposite

[Aaron Liu]

since we want the galois group of the $\overline{\mathbb{F}_2}$ over $P_{\omega^\omega}$ not the other way around

[Aaron Liu]

well not the other way around but the

[Aaron Liu]

ok $\widehat{\mathbb{Z}}$ is the product of $\prod_p \mathbb{Z}_p$

[Aaron Liu]

and we have the quadratic closure whose galois group looks like some limit of $\mathbb{Z} / 2^n \mathbb{Z}$

[Violeta Hernández]

I asked my other mathematician friendsScreenshot_2025-10-02-18-02-14-166_com.discord-edit.jpg

[Aaron Liu]

Violeta Hernández said:

I asked my other mathematician friendsScreenshot_2025-10-02-18-02-14-166_com.discord-edit.jpg

do you understand the reply (I don't understand, can you explain it to me)

[Aaron Liu]

where was I

[Aaron Liu]

ah yes the galois group of $P_\omega$ seems to be $\mathbb{Z}_2$ the 2-adic integers

[Aaron Liu]

does $P_\omega$ have any nontrivial degree 4 extensions?

[Violeta Hernández]

Aaron Liu said:

Violeta Hernández said:

I asked my other mathematician friendsScreenshot_2025-10-02-18-02-14-166_com.discord-edit.jpg

do you understand the reply (I don't understand, can you explain it to me)

not really :frown:

[Violeta Hernández]

I'm not even sure if this argument works

[Kevin Buzzard]

Just catching up.

If your field does not have characteristic 3 and if it has three distinct cube roots of 1 in it then you can happily "adjoin a cube root of x" for any x in the field which doesn't have a cube root in the field; this is a well-defined degree 3 extension and it's the splitting field of X^3-x. The chaos only occurs if you don't have three distinct cube roots of 1, then pathological things can happen.

If you look at the closed subgroup of $\prod_p\Z_p$ which is $\prod_{p>2}\Z_p$ then by FTG this corresponds to an algebraic infinite degree extension $k$ of $\mathbb{F}_2$ whose Galois group over $\mathbb{F}_2$ is $\Z_2$ and such that the Galois group of $\overline{\mathbb{F}}_2$ over $k$ is $\prod_{p>2}\Z_p$.

[Violeta Hernández]

Oh actually, I do get why all quartic polynomials in $\mathcal P_{\omega^\omega}$ have roots, and it's basically the same argument as before.

[Violeta Hernández]

Take a quartic polynomial $p$ in $\mathcal P_{\omega^\omega}$. Suppose it doesn't have roots; since $\mathcal P_{\omega^\omega}$ has roots for all polynomials of degree $3$ or lower, $p$ must be irreducible. Take the field $\mathcal F$ generated by its coefficients, adjoin a root $\alpha$ to create a degree $4$ extension. By finite field properties, you can find a quadratic extension $\mathcal F\le\mathcal F(\beta)\le\mathcal F(\alpha)$. Since $\mathcal P_{\omega^\omega}$ is quadratically closed, $\beta$ and also $\alpha$ must be in $\mathcal P_{\omega^\omega}$.

[Violeta Hernández]

tl;dr, a quartic extension of a finite field is just a quadratic extension of a quadratic extension

[Violeta Hernández]

So I think there really is only one thing we don't understand: why do the $\alpha_p$ exist?

[Violeta Hernández]

e.g. why does $\mathcal P_{\omega^\omega}$ not have all fifth roots?

[Violeta Hernández]

Bonus question, the fact that $[\omega^\omega]^5=4$ implies that $[2]$ and $[3]$ have fifth roots below $[\omega^\omega]$. What are they?

[Violeta Hernández]

I think the Galois group of $\mathcal P_{\omega^\omega}$ should be $\mathbb Z_2 \times \mathbb Z_3$? Is there some way to just look at that and conclude that there exists a quintic extension?

[Violeta Hernández]

Actually, why is the Galois group of the quadratic closure of $\mathbb F_2$ equal to $\mathbb Z_2$ again?

[Violeta Hernández]

Does the Galois map preserve limits?

[Violeta Hernández]

ok ok so thanks to my friend again I think I know why $\alpha_p$ i.e. the smallest number without a $p$-th root in $\kappa_p$ exists

[Violeta Hernández]

so, the idea is that since $p-1$ has prime factors smaller than $p$, we should be able to find a subfield of order $2^{p-1}$, whose multiplicative subgroup has an order divisible by $p$.

[Violeta Hernández]

Which means that there exists some element of this subfield which does not have a $p$-th root within the subfield

[Violeta Hernández]

But any algebraic extensions that land within $\kappa_p$ must be of a degree whose prime factors are less than $p$, which means there aren't any $p$-th roots added for these original elements

[Violeta Hernández]

i.e. these values don't have a $p$-th root in $\kappa_p$ either

[Violeta Hernández]

To be completely honest I'm not 100% on the details of this but, it all seems legit

[Violeta Hernández]

I'm admittedly a bit worried that some of the results we proved via the "take a finite subfield, use finite field theory" method might have more direct proofs using Galois theory

[Aaron Liu]

I think I understand now why the degree of the next extension is always prime

[Violeta Hernández]

Isn't that just the same argument I gave? If the degree was composite you could pass to finite fields, find an intermediate extension, and then both steps of the extension would already exist in your field

[Aaron Liu]

no but like it's intuitive for me now

[Aaron Liu]

since the galois group of the extension is smooth or something

[Violeta Hernández]

wdym?

[Aaron Liu]

like its cardinality

[Violeta Hernández]

Yeah

[Violeta Hernández]

I'm actually not sure if that is sufficient, though

[Aaron Liu]

well all n-smooth numbers are p-smooth for p prime less than n

[Violeta Hernández]

Say $G$ is a group with 3-smooth order, and $H$ is a subgroup of index 4. Does there always exist an intermediate subgroup with index 2?

[Aaron Liu]

no the point is we've already "completed" the p-smooth stuff

[Violeta Hernández]

I think that's a valid way to think of it

[Violeta Hernández]

Not sure what the exact formal statement for that is

[Aaron Liu]

something something absolute galois group

[Django Peeters]

What is a 4-smooth Galois group?

[Aaron Liu]

The cardinality is a 4-smooth number

[Violeta Hernández]

hmmm wait

[Violeta Hernández]

why is $\omega^{\omega 5}$ a fifth root of $\omega^\omega$?

[Violeta Hernández]

or rather, why does $\omega^\omega$ have no fifth roots in $\mathcal P_{\omega^{\omega5}}$?

[Violeta Hernández]

I think the idea is that the order of $\omega^\omega$ must be a 3-smooth number times 25, but the order of everything in $\mathcal P_{\omega^{\omega5}}$ is a 3-smooth number times at most 5?

[Violeta Hernández]

Why is any of this true

[Aaron Liu]

Violeta Hernández said:

I think the idea is that the order of $\omega^\omega$ must be a 3-smooth number times 25, but the order of everything in $\mathcal P_{\omega^{\omega5}}$ is a 3-smooth number times at most 5?

No that doesn't sound quite right

[Aaron Liu]

what was the argument for why $\omega$ has no cube root

[Violeta Hernández]

A cube root would need to have order 27 but that's not possible below $\omega^3$

[Aaron Liu]

Why not

[Violeta Hernández]

image.png

[Violeta Hernández]

I'm trying to figure out how this argument generalizes

[Aaron Liu]

Violeta Hernández said:

Bonus question, the fact that $[\omega^\omega]^5=4$ implies that $[2]$ and $[3]$ have fifth roots below $[\omega^\omega]$. What are they?

My CAS tells me we have $[4]^5=[2]$ and $[6]^5=[3]$

[Violeta Hernández]

In fact you also have $[3]^5=[2]$ and $[2]^5=[3]$

[Aaron Liu]

well that works too I guess

[Aaron Liu]

I have five fifth roots here actually

[Violeta Hernández]

ooh what are they?

[Aaron Liu]

4,5,11,3,9 are all fifth roots of 2

[Violeta Hernández]

nice!

[Violeta Hernández]

Violeta Hernández said:

I'm trying to figure out how this argument generalizes

I think our original idea worked better

[Aaron Liu]

and then 6,2,7,12,15 are all fifth roots of 3

[Violeta Hernández]

If $\omega$ has a cube root below $\omega^3$, take a finite field containing the cube root and $\omega$, then this is a field with order $2^{2^k \times 3}$, so by a divisibility argument it can't contain an element of order $27$, i.e. a cube root of $\omega$.

[Aaron Liu]

what's the argument for $\omega^\omega$ which is order 75

[Aaron Liu]

and how are we supposed to calculate this order in general

[Violeta Hernández]

hmm great question

[Violeta Hernández]

We probably shouldn't be relying on divisibility arguments like that

[Violeta Hernández]

Question

[Violeta Hernández]

Does every finite subfield of $\mathcal P_{\omega^\omega}$ have order $2^{2^{k_1}\times 3^{k_2}}$?

[Violeta Hernández]

Aaron Liu said:

what's the argument for $\omega^\omega$ which is order 75

I think it should go like this. Suppose $\omega^\omega$ has a fifth root below $\omega^{\omega\times 5}$, take a finite field containing the fifth root and $\omega$, then this is a field with order $2^{2^m\times 3^n\times 5}$. Since $5$ is not a Wieferich prime (please tell me there is a way to avoid this step), the largest power of $5$ that can divide this is $25$. However, a fifth root of $\omega^\omega$ would have order $25\textrm{ord}(\alpha_5)$, and $\alpha_5$ must have order divisible by $5$ because (?), so this is impossible.

[Aaron Liu]

Violeta Hernández said:

Does every finite subfield of $\mathcal P_{\omega^\omega}$ have order $2^{2^{k_1}\times 3^{k_2}}$?

yes?

[Aaron Liu]

Violeta Hernández said:

Since $5$ is not a Wieferich prime (please tell me there is a way to avoid this step)

You mean A001220 right? We've already brought galois theory in I don't want to rope in number theory too

[Kevin Buzzard]

Violeta Hernández said:

Say $G$ is a group with 3-smooth order, and $H$ is a subgroup of index 4. Does there always exist an intermediate subgroup with index 2?

Not at all (e.g. S_3 in S_4). But it's certainly true in the abelian case (and 3-smoothness is irrelevant), and I thought that this was what you cared about.

[Aaron Liu]

Yes all the galois groups are abelian (for now)

[Aaron Liu]

Does a (galois) field tower $E/K/F$ give you an exact sequence $0 \to \text{Gal}(E/K) \to \text{Gal}(E/F) \to \text{Gal}(K/F) \to 0$

[Aaron Liu]

is that the fundamental theorem of galois theory

[Violeta Hernández]

Aaron Liu said:

Violeta Hernández said:

Since $5$ is not a Wieferich prime (please tell me there is a way to avoid this step)

You mean A001220 right? We've already brought galois theory in I don't want to rope in number theory too

I imagine you can avoid using this by just adjusting your exponent

[Violeta Hernández]

Although it would be hilarious if something broke at alpha_1093 and no one had noticed

[Django Peeters]

That's not actually that far away.

[Violeta Hernández]

I realize that our problem can be turned into a much simpler problem on finite fields

[Aaron Liu]

I think I figured it out

[Violeta Hernández]

Suppose F(a) is a degree p extension of F. Can a be a pth power?

[Violeta Hernández]

This should be simple to disprove by primitive roots

[Violeta Hernández]

Aaron Liu said:

I think I figured it out

oh go ahead

[Aaron Liu]

wait what were we trying to prove again

[Violeta Hernández]

that w doesn't have a cube root below w^3, etc.

[Aaron Liu]

oh of course

[Aaron Liu]

and the quadratic closure of $2$ is $\omega$

[Aaron Liu]

so any quadratic extension of $2$ is isomorphic to a subfield of $\omega$

[Django Peeters]

I didn't notice before, but we now have found the smallest prime p which has a q-set of size 3.

[Aaron Liu]

what does that mean

[Django Peeters]

It means that nimbers behave more interesting than I thought before.

[Aaron Liu]

sure

[Aaron Liu]

ohno

[Aaron Liu]

there is a prime between $n$ and $n^2$

[Django Peeters]

There sure is

[Aaron Liu]

oh I fixed it

[Aaron Liu]

hahaha I think I did it

[Aaron Liu]

for real this time

[Aaron Liu]

oh I assumed the coefficient is nonzero I guess I got to check what happens when it's zero

[Aaron Liu]

oh this just doesn't work

[Aaron Liu]

ok I think I got it for real for real this time

[Aaron Liu]

and another MSE answer https://math.stackexchange.com/a/584336

[Aaron Liu]

Let $\alpha$ be a cube root of $\omega$, so that $\alpha ^ 3 = \omega$, and let $\mathcal{F} = P_\omega$ be the quadratic closure of $\mathbb{F}_2$, then $\alpha ^ 9 \in \mathcal{F}$ but $\alpha ^ 3 \notin \mathcal{F}$ and the polynomial $X^3-1$ splits over $\mathcal{F}$ and $9$ is not divisible by $2$, so the minimal polynomial of $\alpha$ over $\mathcal{F}$ is $X^9-\alpha^9$ which is degree $9$ so $\alpha$ cannot lie in $\mathcal{F}(\omega)$ which is a degree $3$ extension of $\mathcal{F}$.

[Aaron Liu]

does that work

[Aaron Liu]

I think this argument generalizes

[Violeta Hernández]

Violeta Hernández said:

Suppose F(a) is a degree p extension of (a finite field) F. Can a be a pth power?

Let $|F| = k$, let $\xi$ be a primitive root in $\mathcal F(\alpha)$. We can write $\alpha = \xi^{mn/p}$ for $m = \frac{k^p-1}{k-1}$, and $p\nmid n$. If $p\mid k-1$, then LTE says $p^2\nmid m$, meaning $\alpha$ is not a $p$-th root.

[Violeta Hernández]

Small problem. If $p\nmid k - 1$, then it's totally possible that $p^2\mid k^p-1$. Wiefrich primes are an example when $k=2$.

[Aaron Liu]

uhoh

[Aaron Liu]

Violeta Hernández said:

We can write $\alpha = \xi^{mn/p}$ for $m = \frac{k^p-1}{k-1}$, and $p\nmid n$.

why

[Violeta Hernández]

The nonzero elements in $\mathcal F$ are of the form $\xi^{mn}$, and $\alpha$ is a $p$-th root of one of them

[Aaron Liu]

oh I think I get it now

[Aaron Liu]

no I don't get it

[Violeta Hernández]

the values $\xi^{mn}$ for varying $n$ are all the roots of $x^{k-1}=1$

[Violeta Hernández]

i.e. the elements of $\mathcal F$

[Aaron Liu]

where does $m = \frac{k^p-1}{k-1}$ come from

[Violeta Hernández]

$m=\frac{|\mathcal F(\alpha)|}{|\mathcal F|}$

[Aaron Liu]

oh I get it now

[Aaron Liu]

this only works for finite fields

[Aaron Liu]

we have a very infinite field

[Violeta Hernández]

ooopsie I forgot to say that

[Violeta Hernández]

well, but I do believe the result we want follows from the finite case

[Violeta Hernández]

by the usual trick of "take a field generated by everything you care about"

[Aaron Liu]

have you reviewed my proof yet

[Aaron Liu]

I finally got one after two false positives

[Violeta Hernández]

oh yeah that would be a good idea

[Violeta Hernández]

one sec

[Violeta Hernández]

Aaron Liu said:

Let $\alpha$ be a cube root of $\omega$, so that $\alpha ^ 3 = \omega$, and let $\mathcal{F} = P_\omega$ be the quadratic closure of $\mathbb{F}_2$, then $\alpha ^ 9 \in \mathcal{F}$ but $\alpha ^ 3 \notin \mathcal{F}$ and the polynomial $X^3-1$ splits over $\mathcal{F}$ and $9$ is not divisible by $2$, so the minimal polynomial of $\alpha$ over $\mathcal{F}$ is $X^9-\alpha^9$ which is degree $9$ so $\alpha$ cannot lie in $\mathcal{F}(\omega)$ which is a degree $3$ extension of $\mathcal{F}$.

why can't the minimal polynomial be smaller?

[Aaron Liu]

Definition. Let $L$ be a field and let $\alpha$ be an element of a field extending $L$. The radical degree of $\alpha$ over $L$ is the smallest $p \ge 1$ such that $\alpha^p \in L$ (or $+\infty$ if no such $p$ exists).

Then we have:
Lemma. If the radical degree $d$ of $\alpha$ is finite and not divisible by the characteristic of $L$, and if $L$ contains all $d$-roots of unity, then the minimal polynomial of $\alpha$ over $L$ is exactly $X^d-\alpha^d$. In particular, the radical degree coincides with the usual degree.

[Aaron Liu]

wait I forgot to prove that $\mathcal{F}$ contains all the $9$ th roots of unity (it doesn't)

[Violeta Hernández]

Violeta Hernández said:

Small problem. If $p\nmid k - 1$, then it's totally possible that $p^2\mid k^p-1$. Wiefrich primes are an example when $k=2$.

Wait. Surely we can just make $\mathcal F$ big enough so that $p\mid k-1$?

[Aaron Liu]

oh it just occurred to me that that's always possible

[Violeta Hernández]

yep, just make $k=2^{x}$ for $p-1\mid x$

[Violeta Hernández]

take as many lower-degree extensions as needed

[Aaron Liu]

no but we still have to remain a subfield of the big field

[Violeta Hernández]

Yes, and that field contains all extensions of degree less than $p$

[Aaron Liu]

does that work?

[Aaron Liu]

I don't know

[Violeta Hernández]

lemme give the example with $\omega$

[Violeta Hernández]

actually that example is kind of dumb, let me give the example with $p=7$ instead

[Violeta Hernández]

Suppose that $\kappa_7=\omega^{\omega^2}$ did have a seventh root in $\mathcal P_{\kappa_7^7}$. You can write this seventh root as $p(\kappa_7)$ for $p\in\mathcal P_{\kappa_7}$ with $\deg(p)<7$. Take $\mathcal F_0\subseteq \mathcal P_{\kappa_7}$ the field generated by the coefficients of $p$, and let $\mathcal F\subseteq \mathcal P_{\kappa_7}$ be some arbitrary sixth degree extension, so that $7\mid |\mathcal F|-1$. Then, by the theorem I proved above, $\kappa_7$ can't be a seventh root in $\mathcal F(\kappa_7)$, a contradiction.

[Aaron Liu]

interesting

[Violeta Hernández]

Actually, I think you might need to add $[\kappa_7]^7 \in \mathcal P_{\kappa_7}$ to $\mathcal F_0$, so that $\mathcal F(\kappa_7)$ has degree $7$ and not something larger

[Violeta Hernández]

I think the general theorem looks something like this: let p be prime, and F be a field algebraic over its prime field with no extensions of degree < p. If F(a) is an extension of degree p, then a has no pth root in F(a).

[Violeta Hernández]

This isn't much more general than what we already had but it helps abstract away from nimbers

[Aaron Liu]

oh no I have another proof

[Aaron Liu]

I read ONAG again and I think it enlightened me on this problem

[Violeta Hernández]

Go ahead

[Aaron Liu]

so let $p_n$ be the $n$ th odd prime, so $p_0=3$, $p_1=5$ etc. and let $\alpha_{p_n} = [\omega^{\omega^n}]^{p_n}$

[Aaron Liu]

have I done correctly so far

[Violeta Hernández]

sure

[Violeta Hernández]

that looks correct

[Aaron Liu]

alright so by induction or whatever all the finite subfields of $[\omega^{\omega^n}]$ have order $2^{2^k}$ where $k$ is $p_n-1$-smooth

[Violeta Hernández]

I'm still not really sure why this is true

[Aaron Liu]

it's because of the degrees of the extensions

[Violeta Hernández]

actually I think I see the induction argument

[Aaron Liu]

anyways since $[\omega^{\omega^n}]^{p_n}=\alpha_{p_n}$ we have that $\alpha_{p_n}$ has no $p_n$ th root in $[\omega^{\omega^n}]$

[Violeta Hernández]

e.g. a finite subfield of $\omega^3$ is contained in a cubic extension of a finite subfield of $\omega$ hence 3-smooth, etc.

[Violeta Hernández]

Aaron Liu said:

anyways since $[\omega^{\omega^n}]^{p_n}=\alpha_{p_n}$ we have that $\alpha_{p_n}$ has no $p_n$ th root in $[\omega^{\omega^n}]$

That's basically how $\alpha_p$ is defined yeah

[Aaron Liu]

and then... uhmmm...

[Aaron Liu]

I'll be back after a brief intermission

[Violeta Hernández]

sure

[Aaron Liu]

gotta stay hydrated :water_drop:

[Aaron Liu]

anyways fix $n$ and let $\alpha_{p_n}$ be a primitive $u$-th root of unity

[Aaron Liu]

obviously $p_n \nmid u$

[Violeta Hernández]

isn't $\alpha_3 = 2$ a third root of unity?

[Aaron Liu]

oh no was it the other way around

[Aaron Liu]

obviously $p_n \mid u$

[Aaron Liu]

disregard my earlier statement

[Violeta Hernández]

i'm not actually sure why this happens in general

[Aaron Liu]

consider the multiplicative subgroup generated by $\alpha_{p_n}$, which is cyclic, therefore divisible by any number coprime to its order

[Violeta Hernández]

yep

[Aaron Liu]

where was I

[Violeta Hernández]

wait but then what

[Aaron Liu]

suppose for sake of contradiction that $x < [\omega^{\omega^n p_n}]$ but $x^{p_n}=[\omega^{\omega^n}]$

[Violeta Hernández]

i still don't know why $p_n\mid u$ but sure

[Aaron Liu]

Violeta Hernández said:

i still don't know why $p_n\mid u$ but sure

since $\alpha_{p_n}$ is not divisible (rootable) by $p_n$, it follows that the multiplicative order of $\alpha_{p_n}$ is not coprime to $p_n$

[Violeta Hernández]

ah! ok

[Violeta Hernández]

continue

[Aaron Liu]

consider $\mathcal{F}$ the field generated by $x$ which has $2^{2^{kp_n}}$ many elements, where $k$ is $p_n-1$-smooth

[Violeta Hernández]

ok

[Aaron Liu]

and $x$ is a primitive $up_n^2$ root of unity

[Violeta Hernández]

yes

[Aaron Liu]

and no subfield of $[\omega^{\omega^n}]$ contains any $up_n$ root of unity

[Violeta Hernández]

why?

[Aaron Liu]

well they do contain $1$ which is a root of unity

[Aaron Liu]

I meant that they don't contains any primitive $up_n$ roots of unity

[Violeta Hernández]

why?

[Aaron Liu]

well because $\alpha_{p_n}$ is not a $p_n$ th power

[Aaron Liu]

if you have a primitive $up_n$ root of unity you can express any $u$ root of unity as a $p_n$ th power of a power of that

[Violeta Hernández]

ah

[Aaron Liu]

like if you have a primitive $441$ root of unity then any $63$ root of unity is a $7$ th power

[Violeta Hernández]

ok

[Aaron Liu]

for any $p_n-1$-smooth number $g$ there exists a subfield of $[\omega^{\omega^n}]$ which has order $2^{2^g}$

[Violeta Hernández]

yep

[Aaron Liu]

so then $up_n \nmid 2^{2^g}-1$

[Violeta Hernández]

for any such g?

[Aaron Liu]

yes, since otherwise the field with order $2^{2^g}$ would contains a primitive $up_n$ root of unity

[Violeta Hernández]

yep

[Aaron Liu]

so then I was hoping we could do some divisibility magic to conclude that $up_n^2 \nmid 2^{2^{gp_n}}-1$

[Violeta Hernández]

yeah I think that sort of sidesteps the part of my proof where I need to take a (p-1)-th degree extension

[Violeta Hernández]

The rest of your proof is just LTE

[Violeta Hernández]

same as mine

[Aaron Liu]

yay

[Violeta Hernández]

specifically, you want to show that the highest power of $p_n$ dividing $2^{2^g p_n}$ is at most that of $u$ plus 1

[Violeta Hernández]

since $p\mid 2^{2^g}-1$, you can use LTE to prove this

[Violeta Hernández]

actually idk if you know $p\mid 2^{2^g}-1$ but you can definitely make it so

[Aaron Liu]

when you set $g = k$ at the end you do know

[Violeta Hernández]

yep

[Violeta Hernández]

I think your proof is very similar to what I did

[Aaron Liu]

I had a look at ONAG and inferred the missing pieces

[Violeta Hernández]

ok, so I think we finally have all the pieces

[Violeta Hernández]

I'm struggling to figure out how to set up a blueprint for Lean

[Violeta Hernández]

but I can at least write down the theorems we need and their proofs on Overleaf

[Violeta Hernández]

question

[Violeta Hernández]

how much of what we've proved for fields above w also helps us below w?

[Aaron Liu]

well we aren't adjoining roots

[Aaron Liu]

well we are

[Violeta Hernández]

true

[Violeta Hernández]

not square roots

[Aaron Liu]

I have some stale half-finished work characterizing the multiplication

[Violeta Hernández]

there is literally a multiplication algorithm for finite nimbers in the repo already

[Violeta Hernández]

we just haven't proven correctness

[Aaron Liu]

yes my not-done-yet work says basically that $[2^{n+1}]$ is a root of $X^2+X+[2^n]$

[Violeta Hernández]

Aaron Liu said:

yes my not-done-yet work says basically that $[2^{n+1}]$ is a root of $X^2+X+[2^n]$

yep

[Violeta Hernández]

yeah i think nimbers below w need to be done separately

[Violeta Hernández]

anyways

[Violeta Hernández]

I'll make the Overleaf

[Violeta Hernández]

give me like two hours

[Aaron Liu]

Aaron Liu said:

yes my not-done-yet work says basically that $[2^{n+1}]$ is a root of $X^2+X+[2^n]$

that should be $[2^{2^{n+1}}]$ is a root of $X^2+X+[2^{2^n}]$

[Violeta Hernández]

yeah that

[Violeta Hernández]

actually

[Violeta Hernández]

does every quadratic over $[2^{2^n}]$ split over $[2^{2^{n+1}}]$?

[Aaron Liu]

yes that sounds correct

[Violeta Hernández]

it's still the same argument, right? any two extensions of a finite field with the same degree are isomorphic

[Aaron Liu]

yeah that

[Violeta Hernández]

so that gives a pretty succint proof that $\mathcal P_{\omega}$ is quadratically closed

[Aaron Liu]

oh yeah

[Violeta Hernández]

and it means we can reuse some of the work past $\omega$

[Aaron Liu]

perhaps

[Violeta Hernández]

e.g. this theorem

[Violeta Hernández]

Hmm

[Violeta Hernández]

Why did we know that every algebraic extension of a subfield of $\overline{\mathbb F}_2$ was an abelian extension?

[Aaron Liu]

I'm sure I can find it by scrolling up

[Aaron Liu]

Kevin Buzzard said:

Violeta Hernández said:

Actually, how do we know that e.g. every algebraic extension of the quadratic closure of $\mathbb F_2$ is abelian?

Galois theory! Any subquotient of an abelian group is abelian.

here we are

[Violeta Hernández]

Well, that answers half of my question

[Violeta Hernández]

Why do we know that the extension is Galois to begin with?

[Aaron Liu]

it's normal and separable

[Violeta Hernández]

Every algebraic extension of a finite field is separable

[Aaron Liu]

so we want normal

[Violeta Hernández]

I can believe Wikipedia on that one (and surely we have this in Lean already?)

[Aaron Liu]

Violeta Hernández said:

I can believe Wikipedia on that one (and surely we have this in Lean already?)

docs#PerfectField.ofFinite

[Violeta Hernández]

That plus docs#Algebra.IsAlgebraic.isSeparable_of_perfectField

[Aaron Liu]

we can also just use galois theory I guess

[Violeta Hernández]

Violeta Hernández said:

Why did we know that every algebraic extension of a subfield of $\overline{\mathbb F}_2$ was an abelian extension?

Ok so this probably isn't true in general BUT, if our subfield has the property of "all polynomials of lower degree than p split", then the extension should be normal hence abelian

[Aaron Liu]

every (closed) subgroup of $\widehat{\mathbb{Z}}$ is normal since the group is abelian

[Aaron Liu]

Violeta Hernández said:

Violeta Hernández said:

Why did we know that every algebraic extension of a subfield of $\overline{\mathbb F}_2$ was an abelian extension?

Ok so this probably isn't true in general BUT, if our subfield has the property of "all polynomials of lower degree than p split", then the extension should be normal hence abelian

nonono it's totally true probably

[Aaron Liu]

how do we know the absolute galois group is abelian

[Aaron Liu]

oh it's procyclic

[Aaron Liu]

all the finite extensions are cyclic galois group

[Violeta Hernández]

yeah, the Galois group of $\overline{\mathbb F}_2$ is $\widehat{\mathbb Z}$

[Aaron Liu]

yeah but is that in Lean

[Aaron Liu]

I couldn't find it in mathlib but maybe I didn't look hard enough

[Aaron Liu]

or maybe it was stated in a way I didn't expect

[Aaron Liu]

like with category theory or something

[Violeta Hernández]

we can ask later

[Violeta Hernández]

Violeta Hernández said:

yeah, the Galois group of $\overline{\mathbb F}_2$ is $\widehat{\mathbb Z}$

How does my claim follow from this?

[Aaron Liu]

$\widehat{\mathbb{Z}}$ is an abelian group

[Aaron Liu]

so every (closed) subgroup is a normal subgroup

[Aaron Liu]

so every intermediate field is a normal wait did I get the orientation right

[Violeta Hernández]

wait, normal extensions are the same as normal subgroups?

[Aaron Liu]

let me check

[Aaron Liu]

yeah I think I got the orientation right

[Aaron Liu]

so every intermediate field is a normal extension maybe?

[Violeta Hernández]

that seems too good to be true

[Aaron Liu]

it seems very reasonable

[Violeta Hernández]

A finite normal extension is the same as a splitting field

[Violeta Hernández]

The result seems too strong

[Violeta Hernández]

Though of course in our particular case, where we're considering extensions of the form $\mathcal P_{\omega^{\omega^k p_k^{n+1}}} / \mathcal P_{\omega^{\omega^k p_k^{n}}}$, we do in fact know that they are splitting fields

[Aaron Liu]

Violeta Hernández said:

wait, normal extensions are the same as normal subgroups?

I guess think about why they're both called "normal"

[Violeta Hernández]

The words "normal subgroup" are suspiciously absent from the Wikipedia article for "normal extension"

[Aaron Liu]

it's because they're invariant when you move them around in the big thing

[Violeta Hernández]

Is that also true in the infinite case?

[Aaron Liu]

infinite normal subgroups or infinite normal extensions

[Violeta Hernández]

infinite galois groups

[Aaron Liu]

well I just found the (not following the naming convention) docs#InfiniteGalois.normal_iff_isGalois

[Aaron Liu]

oh I guess InfiniteGalois is a namespace yeah this name is fine

[Aaron Liu]

some of the other names in that file are not fine though

[Violeta Hernández]

[Field k] [Field K] this makes me unreasonably angry

[Aaron Liu]

well you have $K/k$

[Aaron Liu]

this is fine

[Violeta Hernández]

am I blind or do k and K look nearly the same in the monospace font

[Aaron Liu]

I don't know what your font looks like

[Violeta Hernández]

image.png

[Aaron Liu]

oh it's the same font

[Violeta Hernández]

anyways. so this means that every single intermediate field between $\mathbb F_2$ and $\overline{\mathbb F}_2$ is Galois over $\mathbb F_2$?

[Aaron Liu]

well yeah

[Violeta Hernández]

whoa

[Aaron Liu]

actually for $K$ algebraic over $\mathbb{F}_2$, every single intermediate field between $K$ and $\mathbb{F}_2$ is also galois over $K$

[Aaron Liu]

this is basically the same argument

[Violeta Hernández]

not only galois, but abelian

[Aaron Liu]

yes yes of course

[Violeta Hernández]

Aaron Liu said:

this is basically the same argument

Actually, what are the details here? I get that $\overline{\mathbb F}_2/K$ must be Galois, but what is its Galois group? Is it a subgroup of the profinite integers? A quotient? Something else?

[Aaron Liu]

it's the subgroup corresponding to the elements of the galois group which fix $K$ pointwise

[Violeta Hernández]

Ah, of course

[Violeta Hernández]

so it's definitely abelian

[Violeta Hernández]

so then the argument applies

[Violeta Hernández]

It feels ludicrously strong that any algebraic extension is Galois when the absolute Galois group is abelian

[Aaron Liu]

I was kind of taking it for granted all this time

[Aaron Liu]

obviously all the algebraic extensions of a field algebraic over a finite field will be galois

[Violeta Hernández]

yeah, I don't really think that's obvious

[Aaron Liu]

well because they're all roots of unity right

[Violeta Hernández]

but if we can prove that in Lean then it doesn't matter what I think

[Violeta Hernández]

Aaron Liu said:

well because they're all roots of unity right

ohhh actually that's an interesting way to think about it

[Aaron Liu]

idk maybe it's not actually that obvious

[Violeta Hernández]

well, it makes sense that a field extension made out of distinct roots of unity would be separable

[Violeta Hernández]

normality seems way less obvious

[Violeta Hernández]

and I wonder if using it can somehow make our previous proofs simpler

[Aaron Liu]

it's a perfect field all the extensions are separable

[Violeta Hernández]

It still trips me up that a perfect field and a perfect ring are two different things and we have both of them

[Aaron Liu]

all the finite extensions at least

[Aaron Liu]

Violeta Hernández said:

It still trips me up that a perfect field and a perfect ring are two different things and we have both of them

docs#PerfectField.toPerfectRing docs#PerfectRing.toPerfectField

[Violeta Hernández]

sometimes names in mathematics do make sense

[Violeta Hernández]

What exactly is ExpChar though

[Aaron Liu]

read the definition

[Violeta Hernández]

is it literally just the characteristic but when it should give 0 it gives 1 instead

[Aaron Liu]

only prime

[Aaron Liu]

primes and one instead of zero

[Violeta Hernández]

what's so exponential about that

[Aaron Liu]

like a field, which can be prime or zero

[Violeta Hernández]

i thought it was going to be the period of the x^n map or something

[Aaron Liu]

that doesn't work

[Violeta Hernández]

${\mathbb F}_p$ has exponential characteristic $p-1$

[Aaron Liu]

no

[Violeta Hernández]

assuming the definition that I thought those words would have

[Aaron Liu]

oh

[Aaron Liu]

what about docs#PerfectRing

[Violeta Hernández]

hm

[Violeta Hernández]

every char 0 ring is perfect?

[Aaron Liu]

every char 0 field is perfect

[Aaron Liu]

so for rings too

[Violeta Hernández]

interesting

[Violeta Hernández]

i still don't know why it's called the "exponential" characteristic but I guess I can accept it

[Violeta Hernández]

or is the exponential in question the Frobenius map

[Violeta Hernández]

that would make a lot of sense actually

[Aaron Liu]

I don't know

[Aaron Liu]

it's not like I came up with this name

[Aaron Liu]

but it does give you that frobenius with the exponential characteristic is always a ring hom

[Violeta Hernández]

nice

[Violeta Hernández]

wait wait wait

[Violeta Hernández]

so every subfield of $\overline{\mathbb F}_2$ is a perfect ring?

[Aaron Liu]

yes?

[Violeta Hernández]

that feels super false somehow

[Violeta Hernández]

no no actually

[Violeta Hernández]

I get why it is

[Aaron Liu]

it feels super true to me

[Violeta Hernández]

it's obvious on finite fields

[Violeta Hernández]

and you can always pass to finite fields

[Aaron Liu]

yeah that

[Violeta Hernández]

that's awesome

[Violeta Hernández]

I feel like knowing all of this would have been a good starting point haha

[Aaron Liu]

I thought you knew this already

[Violeta Hernández]

you overestimate my familiarity with Galois theory

[Aaron Liu]

I'm also not familiar with field theory but I knew this

[Violeta Hernández]

yeah but you're just built different

[Violeta Hernández]

Question, is this still true when $\mathcal G_1,\mathcal G_2$ are algebraic, isomorphic, possibly infinite extensions?
image.png

[Aaron Liu]

I'll try thinking

[Violeta Hernández]

to rephrase, if $\mathcal G$ is an algebraic extension of a finite field $\mathcal F$, is there a unique subfield isomorphic to $\mathcal F$?

[Aaron Liu]

Yes that's correct probably

[Aaron Liu]

It's the fixed field of iterated frobenius

[Violeta Hernández]

Ah! Of course

[Violeta Hernández]

the set of roots of that polynomial

[Violeta Hernández]

btw I'm still making the Overleaf

[Violeta Hernández]

I fell asleep yesterday sorry hehe

[Django Peeters]

I like that you made progress on the alpha stuff

[Violeta Hernández]

Yeah! I think we now know all the results we

Aaron Liu said:

It's the fixed field of iterated frobenius

So in fact something even stronger is true, right? If $\mathcal G$ is any field, and $\mathcal F$ is a finite field, then there is at most one subfield of $\mathcal G$ that is isomorphic to $\mathcal F$.

[Violeta Hernández]

Because it would have to be the full set of roots of $x^{|\mathcal F|}=x$

[Violeta Hernández]

Another question

[Violeta Hernández]

Suppose $\mathcal F/\mathcal G$ is a degree $n$ extension. Let $\mathcal G'\subseteq\mathcal G$ be a subfield. Does there always exist a degree $n$ extension $\mathcal F'/\mathcal G'$ with $\mathcal F'\subseteq\mathcal F$?

[Violeta Hernández]

My idea is that we can choose a basis $\{\alpha_1,\ldots,\alpha_n\}$ for $\mathcal F/\mathcal G$, and then $\mathcal G'(\alpha_1,\ldots,\alpha_n)$ should be the desired extension.

[Violeta Hernández]

(this works, right?)

[Django Peeters]

Not sure

[Aaron Liu]

Aaron Liu said:

consider $\mathcal{F}$ the field generated by $x$ which has $2^{2^{kp_n}}$ many elements, where $k$ is $p_n-1$-smooth

I have been informed that there are too many 2s in the exponent tower

[Aaron Liu]

The proof still works right?

[Aaron Liu]

Violeta Hernández said:

Suppose $\mathcal F/\mathcal G$ is a degree $n$ extension. Let $\mathcal G'\subseteq\mathcal G$ be a subfield. Does there always exist a degree $n$ extension $\mathcal F'/\mathcal G'$ with $\mathcal F'\subseteq\mathcal F$?

Just an arbitrary extension of arbitrary fields?

[Violeta Hernández]

Aaron Liu said:

The proof still works right?

yeah probably

[Violeta Hernández]

Aaron Liu said:

Violeta Hernández said:

Suppose $\mathcal F/\mathcal G$ is a degree $n$ extension. Let $\mathcal G'\subseteq\mathcal G$ be a subfield. Does there always exist a degree $n$ extension $\mathcal F'/\mathcal G'$ with $\mathcal F'\subseteq\mathcal F$?

Just an arbitrary extension of arbitrary fields?

that's the idea yes

[Aaron Liu]

Yeah this seems false

[Violeta Hernández]

does my proof not work?

[Aaron Liu]

I don't see your proof

[Violeta Hernández]

Violeta Hernández said:

My idea is that we can choose a basis $\{\alpha_1,\ldots,\alpha_n\}$ for $\mathcal F/\mathcal G$, and then $\mathcal G'(\alpha_1,\ldots,\alpha_n)$ should be the desired extension.

this

[Aaron Liu]

But A5 has a subgroup of index 30 but no subgroup of order 30

[Violeta Hernández]

...how does that relate at all?

[Aaron Liu]

I guess it's the other way around

[Aaron Liu]

A5 has a subgroup of order 2 but no subgroup of index 2

[Aaron Liu]

maybe?

[Aaron Liu]

I can't tell anymore

[Aaron Liu]

Let me try again

[Aaron Liu]

Violeta Hernández said:

Violeta Hernández said:

My idea is that we can choose a basis $\{\alpha_1,\ldots,\alpha_n\}$ for $\mathcal F/\mathcal G$, and then $\mathcal G'(\alpha_1,\ldots,\alpha_n)$ should be the desired extension.

this

Why is that of the required degree

[Aaron Liu]

Is degrees of successive extensions multiplicative

[Aaron Liu]

ok I think I got it

[Aaron Liu]

There exists a galois extension $\mathcal{F}/\mathcal{G}'$ which has galois group $A_5$

[Aaron Liu]

Any subgroup of order 2 corresponds to an intermediate field $\mathcal{G}$ such that $\mathcal{F}/\mathcal{G}$ has degree 2

[Aaron Liu]

Then let $\mathcal{F}'\subseteq\mathcal{F}$ and let $\mathcal{F}'/\mathcal{G}'$ have degree 2

[Aaron Liu]

Then $\mathcal{F}/\mathcal{F}'$ has degree 30

[Aaron Liu]

And its corresponding subgroup of $A_5$ has order 30

[Aaron Liu]

But there are no order 30 subgroups of $A_5$

[Violeta Hernández]

Aaron Liu said:

Is degrees of successive extensions multiplicative

this is like the very first thing i learned about galois theory

[Violeta Hernández]

the tower lemma

[Violeta Hernández]

Aaron Liu said:

There exists a galois extension $\mathcal{F}/\mathcal{G}'$ which has galois group $A_5$

why? Can every group be a Galois group?

[Aaron Liu]

Violeta Hernández said:

Aaron Liu said:

There exists a galois extension $\mathcal{F}/\mathcal{G}'$ which has galois group $A_5$

why? Can every group be a Galois group?

Yes

[Aaron Liu]

Every profinite group at least

[Aaron Liu]

Not sure if the extension has to be galois though

[Aaron Liu]

But in this case I have an explicit construction that gives you a galois extension

[Violeta Hernández]

hm actually I see why my proof doesn't work

[Violeta Hernández]

The product $\alpha_i \alpha_j$ could be an $\mathcal F$-linear combination of the basis, but not a $\mathcal G$-linear combination

[Violeta Hernández]

Well, I need your help again

[Aaron Liu]

Let $K$ be your favorite field and let $F=K(x_1,x_2,x_3,x_4,x_5)$, let $L$ be the intermediate field generated by the symmetric polynomials, then $F/L$ has galois group $S_5$ and you take the fixed field of the normal subgroup $A_5$ which is then a normal extension

[Violeta Hernández]

Well, I believe that my original claim was false

[Violeta Hernández]

But then I need help patching up a proof

[Violeta Hernández]

When we argued that every cubic polynomial $p$ over $\mathcal P_{\omega}$ splits over $\mathcal P_{\omega^3}$, the idea was that you could take a finite field $\mathcal F$ containing the coefficients of $p$, and that the splitting field of $p$ over $\mathcal F$ (i.e. the field adjoining a root) would be isomorphic to some other cubic extension of $\mathcal F$, which would be contained in $\mathcal P_{\omega^3}$

[Aaron Liu]

sounds good

[Violeta Hernández]

In this specific case, we can simply choose $\mathcal F(\omega)$ as our other cubic extension

[Aaron Liu]

yes

[Violeta Hernández]

I wanted to see what the generalization of this argument would be

[Aaron Liu]

isn't it the same

[Violeta Hernández]

If $\mathcal F$ is algebraic over its finite prime field, and $\mathcal G$ is a degree $n$ extension, does that mean every degree $n$ polynomial over $\mathcal F$ splits over $\mathcal G$?

[Violeta Hernández]

That sounds too good to be true

[Aaron Liu]

That sounds very false

[Aaron Liu]

Actually I don't have a counterexample so maybe it's true

[Violeta Hernández]

It's true if $\mathcal F$ is finite, for basically the same reason

[Aaron Liu]

it's true for finite $\mathcal{F}$

[Aaron Liu]

can we take the subfield generated by the coefficients of the polynomials

[Violeta Hernández]

If that subfield has a degree $n$ extension contained in $\mathcal G$, then the argument should work

[Aaron Liu]

$\mathcal{G}$ is a splitting field

[Violeta Hernández]

my idea was that these extensions would always exist but you just showed that they don't

[Violeta Hernández]

But maybe my original idea can be adapted?

[Aaron Liu]

But $\mathcal{G}$ is a splitting field

[Violeta Hernández]

It's a splitting field for $\mathcal F$, yes

[Aaron Liu]

Take the subfield generated by the coefficients of the polynomial defining $\mathcal{G}$ and the degree $n$ polynomial you want to show splits

[Aaron Liu]

or does that not work

[Violeta Hernández]

I don't know, does it?

[Aaron Liu]

It seems like it should work

[Violeta Hernández]

hm

[Violeta Hernández]

So, $\mathcal G$ is a splitting field for some irreducible degree $p$ polynomial in $\mathcal F$, right?

[Violeta Hernández]

because it's a normal extension

[Aaron Liu]

Violeta Hernández said:

my idea was that these extensions would always exist but you just showed that they don't

But the galois group is abelian

[Violeta Hernández]

Does that somehow fix the problem?

[Aaron Liu]

oh it maybe doesn't

[Aaron Liu]

Violeta Hernández said:

So, $\mathcal G$ is a splitting field for some irreducible degree $p$ polynomial in $\mathcal F$, right?

Where did $p$ come from

[Violeta Hernández]

n

[Aaron Liu]

Are $p$ and $n$ related?

[Aaron Liu]

Is $p$ prime?

[Violeta Hernández]

i meant $n$

[Aaron Liu]

Oh ok

[Violeta Hernández]

actually, is that even true?

[Aaron Liu]

Not sure

[Violeta Hernández]

A finite normal extension is a splitting field, but is it a splitting field for a single polynomial?

[Aaron Liu]

If it's for multiple polynomials

[Aaron Liu]

Then just take the product

[Violeta Hernández]

ah yes of course

[Violeta Hernández]

followup question: must this polynomial be irreducible?

[Violeta Hernández]

presumably the answer depends on whether $n$ is prime or not

[Aaron Liu]

anyways my hope is that the splitting field of the polynomial over the subfield is also degree $n$

[Violeta Hernández]

oh, I see what you're going for

[Violeta Hernández]

hmm

[Violeta Hernández]

let me ask a different question

[Violeta Hernández]

Suppose $\mathcal G$ is a degree $n$ extension of $\mathcal F$. Does there necessarily exist some element in $\mathcal G$ whose minimal polynomial over $\mathcal F$ has degree $n$?

[Aaron Liu]

no

[Aaron Liu]

consider $\mathbb{Q}(\sqrt{2},\sqrt{3})$

[Aaron Liu]

maybe

[Violeta Hernández]

isn't the minimal polynomial of $\sqrt2+\sqrt3$ of degree $4$?

[Aaron Liu]

I just realized that that doesn't work

[Aaron Liu]

ahh

[Aaron Liu]

so embarrassing

[Violeta Hernández]

Well, if my claim were true, then we could complete that other proof

[Aaron Liu]

it should be true when $n$ is prime

[Violeta Hernández]

True

[Violeta Hernández]

In fact, if $n$ were prime, wouldn't every element not in $\mathcal F$ satisfy that property?

[Aaron Liu]

yeah

[Violeta Hernández]

let $x\in\mathcal G\setminus\mathcal F$, the value $[\mathcal F(x) : \mathcal F]$ must divide $n$ but it can't be $1$ so it's $n$

[Violeta Hernández]

Violeta Hernández said:

If $\mathcal F$ is algebraic over its finite prime field, and $\mathcal G$ is a degree $n$ extension, does that mean every degree $n$ polynomial over $\mathcal F$ splits over $\mathcal G$?

So then this is true when $n$ is prime

[Violeta Hernández]

Take $p$ a degree $n$ polynomial, take $q$ the minimal polynomial of some random element in $x \in \mathcal G\setminus\mathcal F$, take $\mathcal F_0$ the subfield generated by the coefficients of both. Then $\mathcal F_0(x)\subseteq \mathcal G$ is a degree $n$ extension of $\mathcal F_0$, and the rest of the argument follows as before

[Aaron Liu]

Try maybe $(\mathbb{F}_2(x,y))(\sqrt{x},\sqrt{y})$ is that a counterexample

[Violeta Hernández]

that seems like it could work

[Aaron Liu]

Let $\mathcal{F}=\mathbb{F}_2(x,y)$, then $\mathcal{G}=\mathcal{F}(\sqrt{x},\sqrt{y})$ is a degree four extension but all the elements in $\mathcal{G}$ are a square root of an element in $\mathcal{F}$

[Violeta Hernández]

interesting

[Violeta Hernández]

I'm wondering if something like this could be used to disprove my original claim

[Aaron Liu]

what claim

[Kevin Buzzard]

Violeta Hernández said:

Suppose $\mathcal G$ is a degree $n$ extension of $\mathcal F$. Does there necessarily exist some element in $\mathcal G$ whose minimal polynomial over $\mathcal F$ has degree $n$?

You're asking if every finite extension is simple. This is not true in general (Aaron posted the standard counterexample) but every finite separable extension is simple (as are some non-separable ones, but not Aaron's one).

[Aaron Liu]

it's a standard counterexample?

[Kevin Buzzard]

Yes, it's the first (counter)example that everyone lecturing Galois theory gives after proving that every finite separable extension is simple.

[Aaron Liu]

wow amazing

[Kevin Buzzard]

except usually I use $\mathbb{F}_p(x,y)$ and then adjoin $x^{1/p}$ and $y^{1/p}$.

[Kevin Buzzard]

This is the simplest non-simple extension

[Aaron Liu]

I specialized to $p=2$ for concreteness

[Kevin Buzzard]

yes, I assumed that a prime exists in my lectures

[Aaron Liu]

and it was easier to verify since I can compute stuff

[Violeta Hernández]

What is this, some kind of simplest extension theorem? :rofl:

[Violeta Hernández]

No but seriously, thanks for that word. That is exactly what I was asking about.

[Aaron Liu]

simple extension

[Aaron Liu]

not simplest

[Violeta Hernández]

Kevin Buzzard said:

Yes, it's the first (counter)example that everyone lecturing Galois theory gives after proving that every finite separable extension is simple.

Ah! But I do have a finite separable extension in my case!

[Aaron Liu]

well of course

[Aaron Liu]

it's perfect

[Violeta Hernández]

I agree, this is perfect

[Violeta Hernández]

Violeta Hernández said:

If $\mathcal F$ is algebraic over its finite prime field, and $\mathcal G$ is a degree $n$ extension, does that mean every degree $n$ polynomial over $\mathcal F$ splits over $\mathcal G$?

Slight correction here, I need the polynomial to be irreducible

[Violeta Hernández]

But I think this should all work

[Violeta Hernández]

This is a really cool result

[Violeta Hernández]

I'm glad that all of these results we've found about nimbers cleanly generalize to the bare language of field theory

[Violeta Hernández]

Is "every finite separable extension is simple" in Mathlib?

[Aaron Liu]

is it docs#Field.exists_primitive_element

[Violeta Hernández]

that is a special case of that theorem

[Aaron Liu]

I see a finite separable extension

[Violeta Hernández]

nevermind then

[Violeta Hernández]

my fault for assuming a theorem statement from a theorem name

[Aaron Liu]

what did you think it was

[Violeta Hernández]

"every finite field has a primitive root"

[Aaron Liu]

oh that's probably docs#instIsCyclicUnitsOfFinite

[Aaron Liu]

"an integral domain with a finite amount of units has cyclic units"

[Violeta Hernández]

whoa

[Violeta Hernández]

that's a pretty cool generalization

[Violeta Hernández]

ok I need to keep working on the Overleaf

[Violeta Hernández]

I can show you the first page, in case you have any notes/comments/corrections
image.png

[Django Peeters]

Lenstra eventually defines kappa_n for all natural n>0.

[Django Peeters]

That's why my draft contains 2 versions. We need to prove they're consistent.

[Violeta Hernández]

Ah, so it's $\alpha_p$ but $\kappa_n$?

[Violeta Hernández]

Or how are they defined?

[Django Peeters]

I guess you could define alpha_n as kappa_n^n

[Django Peeters]

So at least the junk values have some meaning

[Violeta Hernández]

Yeah, I was imagining that in our actual formalization, we'd probably write $\alpha_0=\ast\omega$ instead of $\alpha_3=\ast\omega$

[Violeta Hernández]

So that we don't need to carry around proofs of primality

[Violeta Hernández]

Django Peeters said:

I guess you could define alpha_n as kappa_n^n

I think that makes more sense. We can then have "$\alpha_n$ is the least nimber without a $p_{n+1}$-th root below $\kappa_n$" as a theorem.

[Django Peeters]

Did we already define 'algClosure' for nimbers in Lean?

[Violeta Hernández]

I think I might have defined that in some branch. But it's not on the main branch yet.

[Django Peeters]

I like that

[Aaron Liu]

Violeta Hernández said:

I can show you the first page, in case you have any notes/comments/corrections
image.png

I see

We can generalize this theorem to requiring only the assumptions of Theorem 2 for $\mathcal{F}$.

What does this mean

[Violeta Hernández]

Gentle reminder that CGT#231 has been open for a month

[Violeta Hernández]

Aaron Liu said:

We can generalize this theorem to requiring only the assumptions of Theorem 2 for $\mathcal{F}$.

What does this mean

It means that instead of $\mathcal F$ being an algebraic extension of $\mathbb F_q$, we only (?) need that every algebraic extension of $\mathcal F$ is abelian

[Aaron Liu]

can you write that instead

[Violeta Hernández]

sure

[Violeta Hernández]

I do wonder, is that hypothesis equivalent to saying "the absolute Galois group of $\mathcal F$ is abelian"?

[Aaron Liu]

if every algebraic extension is abelian then in particular the algebraic closure would be

[Violeta Hernández]

yeah, but the separable closure of $\mathcal F$ isn't necessarily the same as its algebraic closure

[Aaron Liu]

hmm what if it's not perfect

[Violeta Hernández]

unless $\mathcal F$ is perfect

[Violeta Hernández]

yeah

[Violeta Hernández]

I guess you could have $\mathcal F$ be algebraically closed, in which case the assumption holds vacuously

[Violeta Hernández]

Did I write this correctly?
image.png

[Aaron Liu]

sounds legit

[Django Peeters]

Looks correct

[Aaron Liu]

slightly bothered about how isomorphisms use the $\to$ arrow instead of the $\cong$ arrow or the $\simeq$ arrow

[Violeta Hernández]

I am bothered about that too

[Violeta Hernández]

blame all those 1Xth century mathematicians for being born before category theory was invented

[Aaron Liu]

you're the one writing this you can fix it

[Violeta Hernández]

you are so correct

[Violeta Hernández]

off topic but, learning category theory, I've really gotten the impression that the notation Lean uses makes way more sense for it than just trying to shoehorn set theoretic notations

[Violeta Hernández]

I have writing stuff like "$f:X\to Y$ is an isomorphism"

[Aaron Liu]

that's because it's not set theory

[Violeta Hernández]

true! for as much as my textbooks insist that morphisms aren't just functions you'd expect that they'd adapt the notation correspondingly

[Aaron Liu]

what notation do you have in mind

[Violeta Hernández]

Well, another thing that also annoys me is how people also write functors as if they were just a function

[Violeta Hernández]

They're two separate functions bundled together!

[Aaron Liu]

it's a morphism

[Violeta Hernández]

hmm

[Violeta Hernández]

you know what, fair enough

[Violeta Hernández]

I do feel that writing $F(X)$ and $F(f)$ for two separate function applications can get pretty confusing though

[Aaron Liu]

maybe but you have the big letters and the small letters

[Violeta Hernández]

Sure, but then people start getting compositional and writing things like $F(\textrm{Hom}(f, Y))$ and it becomes harder to tell at a glance if it's a morphism or an object

[Aaron Liu]

oh no

[Violeta Hernández]

Anyways! We shouldn't be clogging the channel

[Aaron Liu]

but the thing with functors is you can map anything, not just objects and morphisms

[Aaron Liu]

like you can map commutative squares too for example

[Aaron Liu]

so we don't want to invent too many notations

[Aaron Liu]

just use the same one for all of them

[Violeta Hernández]

fiiiiine

[Violeta Hernández]

I guess it's similar to how you write $f(X)$ for both application and images

[Violeta Hernández]

Ok one last hot take

[Violeta Hernández]

It's weird that we write $f:X \to Y$ yet insist on writing $\textrm{Hom}(X, Y)$ for the entire set of morphisms

[Violeta Hernández]

Why not just write $X \to Y$?

[Aaron Liu]

why not indeed

[Violeta Hernández]

Back to nimbers

[Violeta Hernández]

I just opened Lenstra's paper

[Violeta Hernández]

I hadn't noticed that $\kappa_n$ was indeed defined for every prime power $n$

[Violeta Hernández]

Specifically, for $q=p_k^n$, $\kappa_q=[2^{\omega^k p_k^{n-1}}]$

[Aaron Liu]

what does that mean

[Violeta Hernández]

maybe this makes it clearer:
image.png

[Aaron Liu]

oh?

[Aaron Liu]

oh

[Violeta Hernández]

It might be better to, in our Lean code, write this as kappa p n

[Django Peeters]

It's a nice definition, what can I say

[Violeta Hernández]

Following docs#GaloisField

[Violeta Hernández]

Unrelated, but what's going on here: docs#instInhabitedGaloisFieldOfNatNat

[Aaron Liu]

404

[Violeta Hernández]

oh sorry apparently I was in some random outdated mirror of the docs?

[Aaron Liu]

how

[Violeta Hernández]

Great question

[Violeta Hernández]

the thing I linked to was an Inhabited (GaloisField 2 1) instance setting the default value to 37

[Violeta Hernández]

I think I know why it was a 404 lol

[Django Peeters]

Would it be a nice challenge to calculate alpha(719) using all this theory, as opposed to numbercrunching?

[Aaron Liu]

Violeta Hernández said:

the thing I linked to was an Inhabited (GaloisField 2 1) instance setting the default value to 37

I think I know who wrote that

[Violeta Hernández]

Well, none of what we've proven over the past month really gives a way to calculate $\alpha_n$

[Aaron Liu]

what's the algorithm we're using

[Aaron Liu]

and can we prove it correct

[Violeta Hernández]

I gave an existence proof

[Violeta Hernández]

I haven't tried to read through the algorithm

[Django Peeters]

I meant after proving all the necessary stuff from Lenstra's papers.

[Violeta Hernández]

Well, it'd be great if we could somehow find a faster way to calculate these values

[Violeta Hernández]

But atm I don't really know enough to judge whether that's possible

[Django Peeters]

Yes, totally. Until now, Lenstra's lower bound for these values seems to always be sharp. Maybe we can try proving that in the far future.

[Violeta Hernández]

I'd love for our formalization work to at least bring us a bit closer to all these unsolved conjectures

[Django Peeters]

Btw, you can find the table here.

[Violeta Hernández]

I have that message pinned :slight_smile:

[Violeta Hernández]

regarding kappa, I think I have an even better idea:

def kappa (k n : ℕ) : Nimber :=
  ∗(2 ^ (ω ^ k * (Nat.nth Nat.Prime k) ^ n))

[Violeta Hernández]

That way, the sequence of fields goes kappa 0 0, kappa 0 1, ..., kappa 1 0, kappa 1 1, ...

[Aaron Liu]

yeah I thought of that too

[Violeta Hernández]

as for alpha, I guess we can do this?

def alpha (k : ℕ) : Nimber :=
  (kappa k 0) ^ (Nat.nth Nat.Prime k)

[Aaron Liu]

sure thing

[Aaron Liu]

if you can get all the proofs to work

[Violeta Hernández]

this gives us the value alpha 0 = 3 which I don't think means anything, but other than that alpha enumerates precisely the values we want

[Violeta Hernández]

2, 4, ω + 1, etc.

[Aaron Liu]

Violeta Hernández said:

this gives us the value alpha 0 = 3 which I don't think means anything, but other than that alpha enumerates precisely the values we want

this looks like you're saying kappa 0 0 ^ 0 = 3

[Violeta Hernández]

oh oops that's because I wrote it wrong

[Aaron Liu]

oh yeah that would do it

[Violeta Hernández]

fixed

[Aaron Liu]

now we have 2 ^ 2 = 3 yeah that checks out

[Django Peeters]

And if, for some reason we need kappa_n^n, we can always introduce other names.

[Violeta Hernández]

Oh I doubt we'll need it! I just wrote it down wrong.

[Violeta Hernández]

Question

[Violeta Hernández]

Does every finite abelian group of composite order have a nontrivial subgroup?

[Aaron Liu]

yes???

[Violeta Hernández]

is this just Sylow

[Aaron Liu]

structure theorem

[Aaron Liu]

every finitely generated abelian group is a direct sum of copies of $\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z}$

[Violeta Hernández]

Ah yeah that works

[Aaron Liu]

alternatively, pick any non-identity element and look at the subgroup it generates

[Aaron Liu]

if it's the whole group then you have a cyclic group of composite order

[Aaron Liu]

if it's not the whole group then you have a nontrivial subgroup

[Violeta Hernández]

Though now I'm curious if this is also true for non-abelian groups

[Violeta Hernández]

Aaron Liu said:

if it's the whole group then you have a cyclic group of composite order

oh yeah that's easy

[Violeta Hernández]

nice

[Aaron Liu]

group theory

[Violeta Hernández]

Did I write this down properly?
image.png

[Aaron Liu]

makes sense

[Kevin Buzzard]

I don't know what assumptions you have on your \F but in general if you adjoin one root of a polynomial you don't get a Galois extension (it won't in general be normal). You're OK if F is algebraic over a finite field though.

[Aaron Liu]

yeah that's the assumption

[Violeta Hernández]

Those are exactly the assumptions on F, hehe

[Aaron Liu]

Violeta Hernández said:

I can show you the first page, in case you have any notes/comments/corrections
image.png

see it says at the top "In the following, unless otherwise specified, let $\mathcal{F}$ be an algebraic extension of a finite prime field $\mathbb{F}_q$"

[Aaron Liu]

not sure why you need the finite field to be prime since it works just as well over any other finite field

[Violeta Hernández]

Oh I just said that so I could use that $q$ is prime

[Aaron Liu]

uhh

[Violeta Hernández]

I'm not even sure if I do

[Violeta Hernández]

I might rewrite that if I don't

[Aaron Liu]

if it's algebraic over a finite field then it's also algebraic over the prime field

[Violeta Hernández]

yeah

[Aaron Liu]

just by transitivity

[Violeta Hernández]

I mean, $q$ is the characteristic of the field

[Aaron Liu]

yeah

[Violeta Hernández]

I wrote it like that so I wouldn't have to explicitly mention "... and let $q$ be the characteristic of the field"

[Aaron Liu]

sure I guess

[Violeta Hernández]

again I'll probably rewrite this anyways

[Violeta Hernández]

hmm, is this part true if you don't have $q$ prime?
image.png

[Aaron Liu]

no

[Violeta Hernández]

well there you go

[Violeta Hernández]

that's why I wrote it like that

[Violeta Hernández]

I guess I can make it clearer by saying "Without loss of generality, we make $q$ prime"

[Violeta Hernández]

Violeta Hernández said:

hmm, is this part true if you don't have $q$ prime?
image.png

Actually, are you sure that this isn't true in general? Like, if you take the direct limit of $\mathbb Z/2^n\mathbb Z$ and you skip $n=1$, you still get the same group. Is it not the same here?

[Aaron Liu]

oh right of course

[Aaron Liu]

oh yeah it's true

[Aaron Liu]

why did I think it's not

[Kevin Buzzard]

Yeah you always get ZHat for any finite field (although I bet we don't have this in mathlib because we don't have ZHat)

[Violeta Hernández]

That's tragic, this is the result that everything else depends on :sob:

[Aaron Liu]

a subgroup of $\widehat{\mathbb{Z}}$ with finite index must be $\widehat{\mathbb{Z}}$

[Aaron Liu]

Violeta Hernández said:

That's tragic, this is the result that everything else depends on :sob:

we just need abelian

[Aaron Liu]

right

[Violeta Hernández]

Yes, but is there any easy way to know that this galois group is abelian without knowing that it's the profinite integers?

[Aaron Liu]

I don't think anything specifically ended up depending on having $\widehat{\mathbb{Z}}$ specifically

[Aaron Liu]

Violeta Hernández said:

Yes, but is there any easy way to know that this galois group is abelian without knowing that it's the profinite integers?

well all the finite extensions are abelian

[Aaron Liu]

and it's a limit

[Kevin Buzzard]

Hmm, an open subgroup is iso to ZHat.

[Kevin Buzzard]

Violeta Hernández said:

Yes, but is there any easy way to know that this galois group is abelian without knowing that it's the profinite integers?

Yes, just show all finite extensions are abelian

[Aaron Liu]

Kevin Buzzard said:

Hmm, an open subgroup is iso to ZHat.

add some topological magic

[Kevin Buzzard]

There are non-open finite index subgroups of an infinite product of $\Z_{37}$ s but I don't know immediately if the same is true for ZHat

[Violeta Hernández]

Kevin Buzzard said:

Yes, just show all finite extensions are abelian

What's the exact result here? If every finite extension is abelian, then every algebraic extension is abelian?

[Aaron Liu]

are there any open subgroups

[Aaron Liu]

Violeta Hernández said:

Kevin Buzzard said:

Yes, just show all finite extensions are abelian

What's the exact result here? If every finite extension is abelian, then every algebraic extension is abelian?

I think it's "if every finite subextension is abelian then it's abelian"

[Aaron Liu]

and maybe add some conditions

[Aaron Liu]

ok how about this

[Aaron Liu]

let $K/\mathbb{F}_p$ be an algebraic extension and let $\sigma, \tau \in \text{Gal}(K/\mathbb{F}_p)$ such that $\sigma \circ \tau \ne \tau \circ \sigma$

[Aaron Liu]

then there exists $x \in K$ such that $\sigma(\tau(x)) \ne \tau(\sigma(x))$

[Aaron Liu]

now restrict to the field generated by $x$ which is a finite field

[Aaron Liu]

I guess this also works for $K/F$ where $F$ is algebraic over a finite field

[Aaron Liu]

you can see how this argument might generalize

[Aaron Liu]

or I guess maybe you can't I don't know what you're thinking

[Violeta Hernández]

Hmm, sorry to change the topic, but I'm trying to figure out how the $\alpha_p$ existence argument can be generalized

[Aaron Liu]

what's the argument again

[Violeta Hernández]

I think it was basically this: if $\mathcal F$ has a finite subfield whose multiplicative order is a multiple of $p$, then it can't have all $p$-th roots, and neither can any extension of it

[Aaron Liu]

no

[Aaron Liu]

the algebraic closure is an extension that does have all $p$-th roots

[Violeta Hernández]

neither can any extension whose degree isn't divisible by $p$

[Violeta Hernández]

Was it that, then?

[Aaron Liu]

a finite extensions then?

[Violeta Hernández]

perhaps

[Aaron Liu]

or

[Aaron Liu]

idk

[Aaron Liu]

maybe it's "an extension whose galois group has no subgroup of order divisible by $p$ (or maybe it's of index divisible by $p$?)"

[Aaron Liu]

some of these groups are pretty big though

[Aaron Liu]

like no finite subgroups and no finite index subgroups

[Aaron Liu]

so maybe not that

[Violeta Hernández]

What even is the theorem statement we want to prove? I mean, what we want to show is that $\mathcal P_{\omega^{\omega^k}}$ does not have all $p_{k+1}$-th roots, but surely there's something more general to be said?

[Aaron Liu]

we just need to show it isn't $p$-th degree closed

[Aaron Liu]

then the root thing follows

[Violeta Hernández]

in the sense of, all $p$-th degree polynomials have a root?

[Aaron Liu]

yeah we want to show that that's not the case

[Aaron Liu]

and that would suffice for our particular use case

[Violeta Hernández]

That seems reasonable

[Aaron Liu]

well I guess the thing is that finite subfields of $P_{\omega^{\omega^k}}$ have order not divisible by any prime at least $p_k$ (or maybe it's $p_{k+1}$)

[Violeta Hernández]

their orders are a power of 2

[Aaron Liu]

the log2

[Aaron Liu]

take the log2

[Violeta Hernández]

yeah that

[Violeta Hernández]

Actually, why was that the case?

[Aaron Liu]

because the degrees

[Violeta Hernández]

Elaborate?

[Aaron Liu]

uhm

[Violeta Hernández]

No wait I think I remember

[Aaron Liu]

no irreducible degree $p$ polynomial has a root (because we didn't adjoin any yet)

[Violeta Hernández]

Wait, really?

[Violeta Hernández]

is it that easy?

[Aaron Liu]

that's my intuition

[Violeta Hernández]

because we do know that there are irreducible polynomials of every degree

[Aaron Liu]

are they degree $p$ or are they every degree

[Aaron Liu]

yeah

[Aaron Liu]

so we just need to show that adjoining the root of another polynomial doesn't "accidentally" solve any irreducible polynomials of degree $p$

[Aaron Liu]

probably by restricting to the finite field generated by all the relevant numbers

[Violeta Hernández]

Violeta Hernández said:

because we do know that there are irreducible polynomials of every degree

Wait, isn't this actually a consequence of that same theorem from earlier "every finite separable extension is simple"

[Aaron Liu]

what the reasoning here

[Violeta Hernández]

Irreducible polynomials over a finite field, I should specify

[Violeta Hernández]

the reason being, the extension $\mathbb F_{q^n}/\mathbb F_q$ is finite and separable, so it has a primitive element $\alpha$, whose minimal polynomial is irreducible over $\mathbb F_q$ and has degree $n$

[Aaron Liu]

or you can just count them with necklace polynomials

[Violeta Hernández]

necklace polynomials?

[Violeta Hernández]

can you buy those on Amazon

[Aaron Liu]

you can probably get them for free in the OEIS

[Violeta Hernández]

Aaron Liu said:

so we just need to show that adjoining the root of another polynomial doesn't "accidentally" solve any irreducible polynomials of degree $p$

I think the question to ask is. Say $p$ is irreducible over $\mathcal F$, and $\mathcal G$ is an extension. What conditions on $\mathcal G$ guarantee that $p$ is still irreducible over $\mathcal G$?

[Aaron Liu]

hmmmmmmmm

[Aaron Liu]

is this even true

[Violeta Hernández]

Not in general, obviously

[Violeta Hernández]

$\mathcal G$ could be the splitting field of $p$

[Aaron Liu]

well no because all the algebraic extensions are galois

[Aaron Liu]

yeah

[Violeta Hernández]

But if the extension is of a lower degree than $p$ then that should work I think?

[Aaron Liu]

so if it has one root it has all the roots

[Violeta Hernández]

Actually, if the degree of the extension is not divisible by the degree of $p$, that should also work

[Violeta Hernández]

assuming also that the extension is abelian

[Aaron Liu]

but I guess it could still factor without giving a root

[Violeta Hernández]

Aaron Liu said:

but I guess it could still factor without giving a root

Is that possible?

[Violeta Hernández]

I had not considered that

[Aaron Liu]

I feel like if it factors then it has to do so in some symmetric way

[Aaron Liu]

and the degree is prime

[Aaron Liu]

but why must it be symmetrically

[Aaron Liu]

maybe I can use the galois group

[Aaron Liu]

do we know the galois group is cyclic or just abelian

[Violeta Hernández]

I don't know

[Violeta Hernández]

Can you give me an example of an irreducible polynomial over a field that factors but does not split over some extension?

[Aaron Liu]

$X^3-2$ over $\mathbb{Q}$ factors but does not split over $\mathbb{Q}(\sqrt[3]{2})$

[Violeta Hernández]

hey this seems relevant

[Violeta Hernández]

https://math.stackexchange.com/questions/86778/why-does-an-irreducible-polynomial-split-into-irreducible-factors-of-equal-degre

[Aaron Liu]

it does seem relevant

[Violeta Hernández]

Aaron Liu said:

$X^3-2$ over $\mathbb{Q}$ factors but does not split over $\mathbb{Q}(\sqrt[3]{2})$

that has no roots over $\mathbb Q$ so it can't factor

[Aaron Liu]

yes it's irreducible over $\mathbb{Q}$

[Aaron Liu]

but over $\mathbb{Q}(\sqrt[3]{2})$ it's $(X-\sqrt[3]{2})\left(X^2+\sqrt[3]{2}X+(\sqrt[3]{2})^2\right)$

[Aaron Liu]

factors but does not split

[Violeta Hernández]

ah

[Violeta Hernández]

non galois extensions smh

[Aaron Liu]

ok fine

[Violeta Hernández]

I'll be back in like 30 mins

[Aaron Liu]

do you want me to figure something out in the meantime

[Violeta Hernández]

sure

[Aaron Liu]

what do I figure out

[Violeta Hernández]

nvm I'm really tired

[Violeta Hernández]

going to sleep

[Violeta Hernández]

will think about this tomorrow

[Aaron Liu]

gn

[Aaron Liu]

ok I think I got it

[Aaron Liu]

let $L/K$ be a galois extension and let $p \in K[X]$

[Aaron Liu]

assume $p$ is irreducible over $K$

[Aaron Liu]

let $G=\text{Gal}(L/K)$ be the galois group

[Aaron Liu]

then $G$ acts on $L$ by the action so then $G$ acts on $L[X]$ since taking the polynomial ring is functorial

[Aaron Liu]

let $p_1, p_2, \dots, p_n$ be the irreducible factors of $p$ over $L[X]$ (counting multiplicities) so that $p_1p_2 \dots p_n=p$

[Aaron Liu]

then $p_1, p_2, \dots, p_n$ are all distinct and form an orbit under the action by $G$

[Aaron Liu]

first that they're an orbit

[Aaron Liu]

since if they aren't an orbit then they are multiple orbits and you can partition them based on their orbits

[Aaron Liu]

so let $p_1,\dots,p_k$ be one of these orbits, chosen arbitrarily

[Aaron Liu]

then the product $p_1 \dots p_k$ is invariant under the galois action and therefore lies in $K[X]$

[Aaron Liu]

but then it's a nontrivial factor of $p$, which was supposed to be irreducible

[Aaron Liu]

and then we know that all the $p_n$ are distinct

[Aaron Liu]

since given any $\sigma \in G$ applying the action $\sigma(p) = p$ so $\sigma(p_1)\sigma(p_2)\dots\sigma(p_n)=p_1p_2 \dots p_n$

[Aaron Liu]

so $\sigma$ defines a permutation of the $p_i$

[Aaron Liu]

so all the $p_i$ in the same orbit have the same multiplicity

[Aaron Liu]

and we showed that there's only one orbit so they all have the same multiplicity

[Aaron Liu]

and if they all have multiplicity at least two then $p$ is not squarefree so not irreducible

[Aaron Liu]

since all the $p_i$ are an orbit under the action by galois group in particular they all have the same degree

[Aaron Liu]

so applying this to the case where $p$ has prime degree

[Aaron Liu]

the irreducible factors $p_i$ are all of the same degree $d$ and there's $n$ many of them

[Aaron Liu]

so the prime degree of $p$ is equal to $dn$

[Aaron Liu]

so $d=1$ and it splits or $n=1$ and it's irreducible

[Violeta Hernández]

I think this is correct?

[Violeta Hernández]

So the claim is, if $L/K$ is galois, and $p:K[X]$ is irreducible of prime degree, then $p$ is either irreducible or splits in $L$?

[Aaron Liu]

yes that's correct

[Aaron Liu]

if $p$ is irreducible then its irreducible factors over the extension are an orbit under the galois group

[Violeta Hernández]

the Galois group of the splitting field is transitive on the roots of $p$

[Aaron Liu]

I think that's the same thing

[Aaron Liu]

remind me what transitive is again

[Aaron Liu]

wait which splitting field are we looking at here

[Violeta Hernández]

the one of $p$

[Violeta Hernández]

Aaron Liu said:

remind me what transitive is again

single orbit

[Aaron Liu]

then it's not what I proved

[Aaron Liu]

not exactly I don't think

[Aaron Liu]

wait which galois group are we looking at

[Violeta Hernández]

the one of the splitting field I think?

[Aaron Liu]

ok

[Aaron Liu]

what I have said is:

[Aaron Liu]

for an arbitrary galois extension $L/K$ and an arbitrary monic polynomial $p \in K[X]$ irreducible in $K[X]$

[Aaron Liu]

the polynomial $p$ has an image in $L[X]$ which may factor

[Aaron Liu]

then the monic irreducible factors of $p$ in $L[X]$ have no duplicates

[Aaron Liu]

and they form an orbit for the action of $\text{Gal}(L/K)$ on $L[X]$

[Aaron Liu]

this specializes to the case where $p \in K[X]$ has prime degree in which case over $L[X]$ it must either remain irreducible or split into linear factors

[Violeta Hernández]

oh wait yeah I see what you proved

[Aaron Liu]

that's great

[Violeta Hernández]

Aaron Liu said:

and if they all have multiplicity at least two then $p$ is not squarefree so not irreducible

Why can't $p$ has a square root in $L$ but not in $K$?

[Aaron Liu]

you can factor out one copy of each

[Aaron Liu]

then the result is invariant under galois

[Aaron Liu]

ok I think this proof is actually in three steps

[Aaron Liu]

1. prove that everything in the same orbit has the same multiplicity
2. prove that there is only one orbit
3. prove that everything has multiplicity one

[Violeta Hernández]

I feel like the argument on Math SE was simpler

[Aaron Liu]

sure

[Aaron Liu]

you can use that too

[Violeta Hernández]

though I do like the extended results you proved

[Violeta Hernández]

actually wait

[Violeta Hernández]

Math SE used the hypothesis that $L$ was a subfield of the splitting field of $p$

[Aaron Liu]

ok

[Violeta Hernández]

but that's not needed, right?

[Aaron Liu]

I don't think so

[Aaron Liu]

unless you have a problem with my proof

[Violeta Hernández]

your proof seems fine

[Violeta Hernández]

I think I need to review it a bit more carefully

[Violeta Hernández]

hi! sorry i was out of commission for a while due to a headache but i'm back

[Violeta Hernández]

Isn't it possible to more easily show that the multiplicity of the irreducible factors must be one?

[Violeta Hernández]

it feels like if it weren't, it'd violate separability

[Violeta Hernández]

Actually here's a question. An algebraic extension $L/K$ is separable when the minimal polynomial of every element in $L$ is separable. Does that also mean that every irreducible polynomial in $K[X]$ is separable?

[Violeta Hernández]

nvm i'm dumb

[Violeta Hernández]

Ok so your result shows that $p$ must factorize as the product of distinct polynomials over $L$, even though the polynomial might not even be separable?

[Violeta Hernández]

That's kind of insane

[Violeta Hernández]

I think we can cheapen out and just prove the part that all the factors are of the same degree

[Violeta Hernández]

That should be enough for what we're doing

[Violeta Hernández]

In fact, $L/K$ doesn't even need to be Galois

[Violeta Hernández]

Only normal

[Violeta Hernández]

Not sure if you need Galois for the part about the factors having multiplicity one

[Violeta Hernández]

(deleted)

[Violeta Hernández]

Let $L/K$ be a normal extension, let $p:K[X]$ be irreducible over $K$. Suppose $p$ factorizes as $\prod_i q_i$ over $L$. Let $\overline K\supseteq L$ be an algebraic closure of $K$.

Let $\alpha\in\overline K$ be a root of $f_i$, let $\beta\in\overline K$ be a root of $f_j$. By transitivity, there exists $\sigma\in\textrm{Gal}(\overline K/K)$ such that $\sigma(\alpha)=\beta$. By normality, $\sigma|_L$ is an automorphism in $L$. Thus, $\sigma(f_i)$ is a polynomial in $L$ with $\beta$ as a root, meaning $f_j|\sigma(f_i)$, and so $\deg(f_j)\le\deg(f_i)$. Since this is true for any $i, j$, we conclude all factors have the same degree.

[Violeta Hernández]

I believe this same argument shows that $\textrm{Gal}(L/K)$ is transitive on the $q_i$, which can be then used as in your argument to show all factors have multiplicity one

[Violeta Hernández]

Though I think the last step of showing the multiplicity 1 does need a Galois extension

[Violeta Hernández]

Since you use the fact that the fixed field of $\textrm{Gal}(L/K)$ is $K$ which is probably not true in general?

[Violeta Hernández]

I've gotten horribly sidetracked, we were talking about nimbers, right?

[Violeta Hernández]

Ok so, let $f : K[X]$ be irreducible over $K$, with prime degree $p$. Let $L/K$ be a finite normal extension, whose degree is not divisible by $p$. If $f$ were to split over $L$, we could take a root $\alpha$, so the intermediate field $K(\alpha)$ would have degree $p$, which is impossible. Since $f$ must be the product of irreducible factors of the same degree over $L$, we conclude that it must be irreducible over $L$.

[Violeta Hernández]

So back to nimbers. We can prove that every finite subfield of $\mathcal P_{\kappa_{p}}$ has size $2^n$ for $n$ a $(p-1)$-smooth number. (How again?) If $f:\mathbb F_2[X]$ is an irreducible polynomial of degree $p$ (which we know exists), then $f$ must be irreducible over any finite subfield of $\mathcal P_{\kappa_{p}}$, hence over $\mathcal P_{\kappa_{p}}$. And this implies that $\mathcal P_{\kappa_p}$ did not have all $p$-th roots, how again?

[Violeta Hernández]

This all feels so overcomplicated

[Violeta Hernández]

It should not be this hard to prove that "not every single polynomial of the form $x^p-a$ has a root"

[Violeta Hernández]

In fact, it feels like you'd need a specific kind of field for this property to even be false

[Violeta Hernández]

ok actually

[Violeta Hernández]

I have a way simpler proof for the claim we actually want

[Violeta Hernández]

and of course it uses primitive roots once again

[Aaron Liu]

Hi I woke up

[Violeta Hernández]

Claim: let $p$ be a prime. Let $\mathcal F$ be a finite field such that $p\mid |\mathcal F|-1$, and let $\mathcal G$ be a finite extension whose degree does not divide $p$. Then:

a) there exists an element in $\mathcal F$ without a $p$-th root
b) that element does not have a $p$-th root in $\mathcal G$ either.

[Aaron Liu]

it doesn't have a root in $\mathcal{F}$ twice? surely one of those was supposed to be $\mathcal{G}$?

[Violeta Hernández]

For a, you can just take any primitive root $\xi$. The $p$-th powers of units are all in the subgroup of $\mathcal F^*$ with index $p$, which does not contain $\xi$.

[Aaron Liu]

yes

[Violeta Hernández]

In fact, the elements in $\mathcal F$ without $p$-th roots are exactly of the form $\xi^k$ for $p\nmid k$

[Aaron Liu]

correct

[Violeta Hernández]

For b, write $n=\frac{|\mathcal G|-1}{|\mathcal F|-1}$. Let $\xi$ be a primitive root of $\mathcal G$. Then $\xi^n$ is a primitive root of $\mathcal F$.

[Aaron Liu]

that sounds not wrong

[Violeta Hernández]

By LTE, $p\nmid n$, meaning that $\xi^{nk}$ is a $p$-th root in $\mathcal G$ exactly when $p\mid k$, which is when it would be a $p$-th root in $\mathcal F$.

[Aaron Liu]

I'll trust you on that

[Violeta Hernández]

Now, there's two claims I'm still not sure about

[Violeta Hernández]

Considering all the finite subfields of $\mathcal P_{\kappa_p}$, why are the log2 of their orders all $(p-1)$-smooth? And why can we make them divisible by an arbitrary $(p-1)$-smooth integer?

[Violeta Hernández]

I get the intuition of "we can just keep taking extensions of lower order" but how do we actually formalize this?

[Aaron Liu]

well

[Aaron Liu]

hmmmm

[Aaron Liu]

ok which one do you want first

[Violeta Hernández]

The first one?

[Aaron Liu]

the log2 are all smooth

[Violeta Hernández]

yeah

[Aaron Liu]

ok pick a finite subfield

[Aaron Liu]

and look at its intersection with $P_\omega$

[Violeta Hernández]

the log2 of the order is 2-smooth, sure

[Aaron Liu]

then look at its intersection with $P_{\omega^\omega}$

[Aaron Liu]

this is some extension of the previous field

[Violeta Hernández]

hmm actually check this out

[Violeta Hernández]

any finite subfield is a subfield of the subfield generated by $\{2, 4, \ldots, \omega, \omega^3, \ldots, \omega^{\omega}, \omega^{\omega^2}, \ldots\}$ stopping at some appropriate point at every ellipsis

[Aaron Liu]

sounds about right

[Violeta Hernández]

and this subfield can be split up into extensions that are all of degree less than $p$

[Aaron Liu]

yes

[Violeta Hernández]

...this feels a bit crude, admittedly?

[Aaron Liu]

it works

[Violeta Hernández]

like, how do we know what the appropriate point to stop at each ellipsis is?

[Violeta Hernández]

I feel like there might be an easier inductive argument

[Violeta Hernández]

both to explain and to formalize

[Aaron Liu]

just stop once you've covered the finite subfield

[Aaron Liu]

it's finite so you have to cover it eventually

[Violeta Hernández]

I think that might not be enough

[Violeta Hernández]

Since you also want $\alpha_{p_k}$ to be added before you add $\omega^{\omega^k}$, just to make sure that the extension is no bigger than it needs to be

[Aaron Liu]

since it's finite there's some maximum $\omega^{\omega^k}$ it's larger than

[Aaron Liu]

so just cover the field + all the alphas up to that point

[Aaron Liu]

still finite

[Violeta Hernández]

Still, I'm interested in how an inductive argument would look

[Violeta Hernández]

Say that a field $\mathcal F$ of characteristic $q\ne 0$ is $k$-smooth when every finite subfield has order $q^r$ for $r$ a $k$-smooth number. Let $\mathcal G$ be a finite extension whose degree is also $k$-smooth. Does this imply that the field $\mathcal G$ is $k$-smooth?

[Aaron Liu]

probably?

[Violeta Hernández]

If the extension is separable, the answer is yes

[Violeta Hernández]

By the primitive element theorem, we can write $\mathcal G=\mathcal F(\alpha)$. Let $\mathcal G_0\subseteq\mathcal G$ be finite. We can write $\mathcal G_0\subseteq\mathcal F_0(\alpha)$ for some finite subfield $\mathcal F_0\subseteq \mathcal F$. Define $\mathcal F_1$ as the subfield of $\mathcal F$ generated by $\mathcal F_0$ and the coefficients of the minimal polynomial of $\alpha$. Then $\mathcal G_0\subseteq F_1(\alpha)$. Since $[\mathcal F_1(\alpha):\mathcal F_1] = [\mathcal G:\mathcal F]$ is $k$-smooth, as is $\mathcal F_1$, we're done.

[Aaron Liu]

Violeta Hernández said:

We can write $\mathcal G_0\subseteq\mathcal F_0(\alpha)$ for some finite subfield $\mathcal F_0\subseteq \mathcal F$.

why

[Violeta Hernández]

The inverse limit of $k$-smooth fields is $k$-smooth, which gives us our induction

[Violeta Hernández]

Aaron Liu said:

Violeta Hernández said:

We can write $\mathcal G_0\subseteq\mathcal F_0(\alpha)$ for some finite subfield $\mathcal F_0\subseteq \mathcal F$.

why

Every element in $\mathcal G_0$ can be written as an $\mathcal F$-linear combination of powers of $\alpha$, so just take those coefficients

[Aaron Liu]

so just because you have finitely many elements to generate the field doesn't mean that the resulting field is finite

[Violeta Hernández]

You are so right

[Violeta Hernández]

I think you once again need that "algebraic over its prime field" hypothesis

[Aaron Liu]

ok we can add that

[Violeta Hernández]

I think another relevant result is this

[Violeta Hernández]

Let $\mathcal G/\mathcal F$ be an algebraic extension. Then $[\mathcal G(\alpha):\mathcal G]$ divides $[\mathcal F(\alpha):\mathcal F]$.

[Violeta Hernández]

...I think? Might be missing hypotheses.

[Violeta Hernández]

An algebraic normal extension, sorry

[Violeta Hernández]

Because, if $m_{\mathcal F}$ and $m_{\mathcal G}$ are the minimal polynomials of $\alpha$ over the respective fields, then $m_{\mathcal G}\mid m_{\mathcal F}$, and $m_{\mathcal G}$ is irreducible

[Violeta Hernández]

So by a result we proved earlier, $\deg(m_{\mathcal G})\mid\deg(m_{\mathcal F})$

[Aaron Liu]

oh that's yeah makes sense

[Violeta Hernández]

This is relevant since it shows that e.g. if $\mathcal F\subseteq\mathcal P_{\omega}$, then $[\mathcal F(\omega):\mathcal F]$ is a multiple of $3$

[Aaron Liu]

Violeta Hernández said:

So by a result we proved earlier, $\deg(m_{\mathcal G})\mid\deg(m_{\mathcal F})$

does this work with just a normal extension that's not necessarily separable

[Violeta Hernández]

that result didn't require separability

[Aaron Liu]

Violeta Hernández said:

Let $L/K$ be a normal extension, let $p:K[X]$ be irreducible over $K$. Suppose $p$ factorizes as $\prod_i q_i$ over $L$. Let $\overline K\supseteq L$ be an algebraic closure of $K$.

Let $\alpha\in\overline K$ be a root of $f_i$, let $\beta\in\overline K$ be a root of $f_j$. By transitivity, there exists $\sigma\in\textrm{Gal}(\overline K/K)$ such that $\sigma(\alpha)=\beta$. By normality, $\sigma|_L$ is an automorphism in $L$. Thus, $\sigma(f_i)$ is a polynomial in $L$ with $\beta$ as a root, meaning $f_j|\sigma(f_i)$, and so $\deg(f_j)\le\deg(f_i)$. Since this is true for any $i, j$, we conclude all factors have the same degree.

explain the transitivity?

[Violeta Hernández]

Violeta Hernández said:

This is relevant since it shows that e.g. if $\mathcal F\subseteq\mathcal P_{\omega}$, then $[\mathcal F(\omega):\mathcal F]$ is a multiple of $3$

In other words, for any $(p-1)$-smooth $k$, we can create a subfield $\mathcal F$ such that $k\mid\log_2|\mathcal F|$ by just tacking enough of these quadratic, cubic, etc. extensions (they might have a higher degree than that, but that doesn't matter)

[Aaron Liu]

well yeah

[Aaron Liu]

just take the prescribed number of quadratic, cubic, quintic etc. extensions in that order

[Violeta Hernández]

Aaron Liu said:

explain the transitivity?

If $p:K[X]$ is irreducible but splits over $L$, then $\textrm{Gal}(L/K)$ acts transitively on the roots of $p$

[Aaron Liu]

since you're closed under small degree polynomials it should work

[Aaron Liu]

and you never run out of irreducible polynomials

[Violeta Hernández]

Funny how the existence of $\alpha_p$ seems to be the most involved part of this entire argument

[Violeta Hernández]

Aaron Liu said:

and you never run out of irreducible polynomials

Ohh you're right actually

[Violeta Hernández]

silly me

[Aaron Liu]

what happened

[Aaron Liu]

did you make a silly

[Violeta Hernández]

No I just realized that your argument is simpler than what I was doing

[Aaron Liu]

what were you doing

[Violeta Hernández]

My idea to create a subfield of order divisible by $\prod p_i^{n_i}$ was to first adjoin $2,4,\ldots, 2^{n_1}$, then $\omega,\omega^3,\ldots,\omega^{3^{n_2}}$, and so on

[Aaron Liu]

oh

[Violeta Hernández]

And I guess that works, but your argument feels more elegant

[Violeta Hernández]

just adjoin roots of irreducible polynomials, since we know they exist in the subfield

[Violeta Hernández]

No need to explicitly specify which

[Violeta Hernández]

And in fact your argument shows that there are subfields of any $(p-1)$-smooth order

[Violeta Hernández]

which is nice

[Aaron Liu]

Violeta Hernández said:

And in fact your argument shows that there are subfields of any $(p-1)$-smooth order

yes that's what I claimed above

[Violeta Hernández]

yeah well I agree with you now haha

[Violeta Hernández]

just took me a bit

[Violeta Hernández]

Ok so, the claim is that $\mathcal P_{\kappa_p}$ does not contain all $p$-th roots. For this, we first take a subfield of order $2^{2^{p-1}}$. Any primitive root of it can not have a $p$-th root in a finite subextension, thus in any subextension, thus in $\mathcal P_{\kappa_p}$.

[Violeta Hernández]

Nice!

[Violeta Hernández]

Does this seem reasonable?
image.png

[Aaron Liu]

seems reasonable

[Aaron Liu]

but why does $[\mathcal{F}_1(\alpha):\mathcal{F}_1]=[\mathcal{G}:\mathcal{F}]$ again

[Violeta Hernández]

because $\mathcal F_1$ contains the minimal polynomial of $\alpha$ over $\mathcal F$

[Aaron Liu]

ok

[Violeta Hernández]

made a typo at the end I meant $\mathcal G_0\subseteq\mathcal F_1(\alpha)$

[Aaron Liu]

yeah I noticed

[Aaron Liu]

Aaron Liu said:

ok

No actually I don't get it anymore

[Violeta Hernández]

• $[\mathcal G:\mathcal F]=[\mathcal F(\alpha):\mathcal F]$ is the degree of the minimal polynomial of $\alpha$ over $\mathcal F$.
• $[\mathcal F_1(\alpha):\mathcal F_1]$ is the degree of the minimal polynomial of $\alpha$ over $\mathcal F_1$.

But these are the same, because the minimal polynomial over $\mathcal F$ is explicitly contained in $\mathcal F_1$.

[Aaron Liu]

I think it makes sense now

[Violeta Hernández]

What about this?
image.png

[Violeta Hernández]

I think this should be our last "field theory prerequisite" theorem

[Violeta Hernández]

Actually

[Violeta Hernández]

Is this somehow useful in finding an "explicit" description of $\alpha_p$?

[Violeta Hernández]

Can we describe it as the smallest primitive root in some specific subfield?

[Violeta Hernández]

I guess the problem is that it's easy to find elements without p-th roots but how do we know we have the smallest one?

[Violeta Hernández]

I guess the problem is that it's easy to find elements without p-th roots but how do we know we have the smallest one?

[Violeta Hernández]

Actually, here's something cool

[Violeta Hernández]

$\alpha_q$ must be between $2^{\omega^k}$ and $2^{\omega^{k+1}}$, where $p_k$ is the largest prime divisor of $q-1$

[Violeta Hernández]

oh my god I'm actually correct

[Violeta Hernández]

the table of values is in fact consistent with this

[Violeta Hernández]

anyways, the reason is simple

[Violeta Hernández]

any finite subfield of $2^{\omega^k}$ is $(p_k-1)$-smooth and thus contains all $p_k$-th roots

[Violeta Hernández]

on the other hand, we can use the trick from before to find some nimber below $2^{\omega^{k+1}}$ without a $p$-th root

[Violeta Hernández]

I wonder how much of Lenstra's algorithm we can figure out on our own

[Violeta Hernández]

We can actually find a better bound

[Violeta Hernández]

$\alpha_q<2^{\omega^k p_k^{d}}$ where $p_k^d$ is the largest power of the largest prime dividing $q-1$

[Violeta Hernández]

for exactly the same reasons

[Violeta Hernández]

the case $q=101$ seems to be the first where the leading term doesn't just use $d=1$

[Violeta Hernández]

Ok I can't help but think that this whole argument about smooth groups could be done much easier

[Violeta Hernández]

so, the Galois groups of the fields we're constructing always look like $\prod_i \mathbb Z_{p_i^{n_i}}$, right? Where we interpret $\mathbb Z_{p^\infty}$ as the $p$-adics

[Violeta Hernández]

presumably, we can show that the only possible finite indices of subgroups are $\prod_i{p_i}^{m_i}$ for $m_i\le n_i$

[Violeta Hernández]

This would give us an even more general result, characterizing the sizes of the finite subfields of any field $\mathcal P_{x}$ below $\omega^{\omega^\omega}$

[Violeta Hernández]

Oh yeah

[Violeta Hernández]

Lenstra states, without proof really, that for $q=p^n$, $\mathcal P_{\kappa_q}$ is the set of nimbers whose minimal polynomial over $\mathbb F_2$ has a degree not divisible by $q$, all of whose prime factors are less or equal than $p$

[Violeta Hernández]

because of the primitive element theorem, this claim is equivalent to the one we were making about subfields

[Violeta Hernández]

Though actually I think what I was doing with smooth fields can be straightforwardly generalized to prove this other claim

[Violeta Hernández]

Violeta Hernández said:

so, the Galois groups of the fields we're constructing always look like $\prod_i \mathbb Z_{p_i^{n_i}}$, right? Where we interpret $\mathbb Z_{p^\infty}$ as the $p$-adics

How can we show this?

[Violeta Hernández]

e.g. suppose we take a cubic extension of $\mathcal P_{\omega}$, which has Galois group $\mathbb Z_{2^\infty}$. Why is the Galois group of the extension $\mathbb Z_{2^\infty}\times\mathbb Z_3$?

[Aaron Liu]

It's the limit of all the galois groups for all the finite subextensions

[Kevin Buzzard]

There is only one cubic extension by Galois theory and that's the Galois group by Galois theory. Galois theory translates questions about fields into questions about groups. By $\Z_p$ I mean the $p$-adics and by changing notation you're making things more confusing. You are asking "suppose I have a closed subgroup $Q$ of $\prod_p\Z_p$ which is contained in $X:=\prod_{p>2}\Z_p$ and such that $X/Q$ has size 3. Why is $\prod_p\Z_p/Q$ equal to $\Z_2\times\Z/3\Z$?" and the answer is "because the only possibility for $Q$ is $3\Z_3\times\prod_{p>3}\Z_p$"

[Aaron Liu]

Why do we need to show the absolute galois group of a finite field is $\widehat{\mathbb{Z}}$ what do we need it for

[Violeta Hernández]

The notation change was inspired by me coming across this Wiki article: https://en.wikipedia.org/wiki/Supernatural_number

[Violeta Hernández]

It feels like the linked paper might be potentially relevant? Apparently it talks about algebraic extensions of finite fields which is exactly what we're dealing with

[Django Peeters]

Some authors use $\mathbb{Z}_p$ for $\mathbb{Z}/p\mathbb{Z}$. The CoFSG uses it too.

[Violeta Hernández]

It's very frustrating that I want to use both $\mathbb Z_p$ in the finite field sense and in the $p$-adic sense

[Kevin Buzzard]

Django Peeters said:

Some authors use $\mathbb{Z}_p$ for $\mathbb{Z}/p\mathbb{Z}$. The CoFSG uses it too.

I know, but here we are dealing with p-adic integers as well as Z/p and there's no "other" notation for p-adic integers that people who use Z_p for Z/p have, so we're forced into the notation I use here.

[Aaron Liu]

Why do we need p-adics

[Kevin Buzzard]

Violeta Hernández said:

It's very frustrating that I want to use both $\mathbb Z_p$ in the finite field sense and in the $p$-adic sense

Just use $\mathbb{F}_p$, it's standard for everyone other than for the finite group theory people

[Aaron Liu]

Kevin Buzzard said:

Violeta Hernández said:

It's very frustrating that I want to use both $\mathbb Z_p$ in the finite field sense and in the $p$-adic sense

Just use $\mathbb{F}_p$, it's standard for everyone other than for the finite group theory people

What do the finite group theory people do

[Kevin Buzzard]

Aaron Liu said:

Why do we need p-adics

Just a convenient way of talking about extensions. You don't need p-adics, you can work with the systems without ever taking the limits, this is what everyone did until 1930, if you want to work with one hand tied behind your back then go ahead, indeed the proof of the fundamental theorem of infinite Galois theory is "prove it for finite extension and then take a limit". The disadvantage is that now you're carrying around the system instead.

[Violeta Hernández]

So $\mathbb F_p$ is the finite field and $\mathbb Z_p$ is the $p$-adics? That's only marginally less confusing

[Kevin Buzzard]

Aaron Liu said:

Kevin Buzzard said:

Violeta Hernández said:

It's very frustrating that I want to use both $\mathbb Z_p$ in the finite field sense and in the $p$-adic sense

Just use $\mathbb{F}_p$, it's standard for everyone other than for the finite group theory people

What do the finite group theory people do

They use $\mathbb{Z}_p$ for $\mathbb{F}_p$ (the finite group theorists in my department do it too, I think they all do it, it confuses the undergraduates).

[Aaron Liu]

no can't we just work with the galois group without ever proving that it's the p-adics specifically?

[Kevin Buzzard]

Violeta Hernández said:

So $\mathbb F_p$ is the finite field and $\mathbb Z_p$ is the $p$-adics? That's only marginally less confusing

Well that's what Wiles uses in his FLT paper so I'll take it on authority that that's a sensible choice of notation.

[Kevin Buzzard]

Aaron Liu said:

no can't we just work with the galois group without ever proving that it's the p-adics specifically?

Sure, you can just carry around the fact that for all finite fields there's a unique extension of any positive integer degree -- that's exactly carrying around the system instead of working with the limit.

[Aaron Liu]

I thought the galois group was the limit

[Aaron Liu]

It's the limit of all the finite subgroups

[Kevin Buzzard]

And now you have to find a language for working with this field which you seem to be interested in, which in my language is "the extension corresponding to $\prod_{p>2}\Z_p$" and in the system language is "the union of $\mathbb{F}_{2^{2^n}}$ and now chase around lots of diagrams to make sure that this makes sense because in this language it is not obvious"

[Aaron Liu]

It's actually "the subfield $P_\omega$ of the nimbers"

[Aaron Liu]

I don't think we need diagrams for that

[Aaron Liu]

everything is a subfield of the big field

[Kevin Buzzard]

Well above there are all manner of questions about controlling maps from $\mathbb{F}_{2^{2^a}}$ to $\mathbb{F}_{2^{2^b}}$ and how many they are and are they compatible etc, which was what prompted me to try and explain the limit language instead of the system language.

[Aaron Liu]

I think we figured that out?

[Aaron Liu]

It just seems like a lot of work to me that you have to define the profinite integers and then prove it's the absolute galois group of the finite fields

[Aaron Liu]

but maybe it's actually not that much work

[Kevin Buzzard]

It might depend on how many decades of experience you have thinking about it all in terms of limits...

[Aaron Liu]

I don't know why that would be

[Kevin Buzzard]

I didn't have to "define the profinite integers", I have been comfortable with them since before you were born.

[Aaron Liu]

but if we have to get it into Lean then we need to tell Lean what the profinite integers are

[Kevin Buzzard]

Aah getting into Lean is a different question. The vast majority of the discussion in this thread is not about getting it into Lean, it's about understanding the mathematics. To understand the mathematics the limit picture is by far the best idea because the limit is doing a lot of the bookkeeping for you, it's turning it into topology. When formalizing it's a completely reasonable idea to try and understand the ideas mathematically in the limit case first and then laboriously translate them back down into the system language, but of course what you should do for the good of mathlib is to define ZHat and work with that anyway.

[Aaron Liu]

ahhh I thought most of this was going into Lean

[Kevin Buzzard]

my impression of the thread above (which I'm only 1/2-paying attention to) is that you are still struggling with the mathematics to a large extent.

[Aaron Liu]

if it's about understanding the mathematics then yeah I think the limit picture is good

[Aaron Liu]

Aaron Liu said:

It's the limit of all the finite subgroups

wait ZHat doesn't have any finite subgroups

maybe it's the finite index subgroups?

[Aaron Liu]

or the finite quotients?

[Aaron Liu]

it's the finite somethings

[Kevin Buzzard]

Yeah, limits map down to the things they're a limit of, so it's finite quotients in group theory

[Django Peeters]

Kevin Buzzard said:

I didn't have to "define the profinite integers", I have been comfortable with them since before you were born.

How did you become so comfortable with profinite groups? Is it similar to going from the rationals to the reals before uni?

[Kevin Buzzard]

I'm a 57 year old researcher in representations of infinite Galois groups so I just have decades of practice, is probably the most reasonable answer.

[Kevin Buzzard]

I think that doing a lot of exercises about p-adic integers when I was a Masters student helped a great deal.

[Aaron Liu]

Kevin Buzzard said:

my impression of the thread above (which I'm only 1/2-paying attention to) is that you are still struggling with the mathematics to a large extent.

Here's something I'm still not sure if it was proved correctly

We have $\mathcal{F}$ a field algebraic over $\mathbb{F}_2$ and $p$ an odd prime. All polynomials of degree $<p$ with coefficients in $\mathcal{F}$ split into linear factors over $\mathcal{F}$. We have $\mathcal{F}'/\mathcal{F}$ a degree $p$ extension. Then $\mathcal{F}'$ is generated as a field extension over $\mathcal{F}$ by a single element $\alpha \in \mathcal{F}'$ for which $\alpha^p \in \mathcal{F}$ (I'm pretty sure the proof of this part is correct). Need to show that $\alpha$ does not have a $p$-th root in $\mathcal{F}'$. I think the divisibility arguments above are correct or close to correct but maybe you have a better way (that hopefully doesn't involve LTE which I'm still not sure I understand)

[Aaron Liu]

in case that's not enough I think we also have: every polynomial with coefficients in $\mathbb{F}_2$ irreducible over $\mathbb{F}_2$ with prime degree $>p$ is also irreducible over $\mathcal{F}$

[Kevin Buzzard]

Why do you need this?

[Kevin Buzzard]

The hypothesis applied to (X^p-1)/(X-1) implies that F has p distinct p'th roots of unity and so the existence of alpha follows immediately from Kummer theory

[Kevin Buzzard]

You want to rule out $\beta\in F'$ with $\beta^p\in F'$ but not $F$ and $\beta^{p^2}\in F$. Say this happened. Then if $L$ is the finite subfield of $$F'$ generated by $\beta$ and the $p$'th roots of 1, and if $K=L\cap F$ (which also contains the $p$'th roots of 1) then you have $\beta\in L$ and $\beta^p\in L$ but not $K$ and $\beta^{p^2}\in K$, so $\beta$ is an element of order $p^2$ in the finite cyclic group $L^\times/K^\times$, meaning that the $p$-adic valuation of $|L^\times|$ is at least two more than the $p$-adic valuation of $|K^\times|$ so by some messy divisibility calculation we must have $p^2|[L:K]$. On the other hand $K\subseteq K(\beta^p)\subseteq K(\beta)$ and each of these subextensions has degree at most $p$ so the extension is exactly $p^2$. We now have some commutative square of fields $K\subset L\subseteq F'$ and $K\subseteq F\subseteq F'$ with $F'=LF$ and $K=L\cap F$ and $[L:K]=p^2$ but $[F':F]=p$ and this contradicts Galois theory because $F'/K$ is Galois with procyclic Galois group so you can't have a subgroup of this group of index $p^2$ and a subgroup of order $p$ which generate. This proof is pretty horrible though. Why do you need this?

[Aaron Liu]

Kevin Buzzard said:

Why do you need this?

Having calculated $\alpha$ the next extension is to take a $p$-th root of $\alpha$, but not if it already has a $p$-th root so we need to show it doesn't

[Kevin Buzzard]

I don't like my proof because it relies at the end of the day on this numerical calculation involving sizes of finite fields. I am using that if $q>1$ is a power of 2 which is 1 mod $p$ then I can read off the $p$-adic valuation of $q^n-1$ from the $p$-adic valuation of $n$ and the $p$-adic valuation of $q-1$ (you just add them). This is slightly fiddly and is the key input into the construction of the $p$-adic logarithm a.k.a. the proof that $\Z_p^\times$ is procyclic for $p>2$ (or put in a less pretentious way, that $(\Z/p^n\Z)^\times$ is cyclic for all $n>1$ when $p>2$, given that it is cyclic for $n=1$). I would like to see an argument which avoids descending to the finite field case and relying on this crutch, using instead that yuo know the absolute Galois group of $F$ is procyclic. But I couldn't immediately find one.

[Aaron Liu]

Kevin Buzzard said:

I am using that if $q>1$ is a power of 2 which is 1 mod $p$ then I can read off the $p$-adic valuation of $q^n-1$ from the $p$-adic valuation of $n$ and the $p$-adic valuation of $q-1$ (you just add them). This is slightly fiddly

I think that's LTE which I was trying to avoid since I'm not used to it

[Aaron Liu]

it's in mathlib already so maybe I just got to get used to it

[Kevin Buzzard]

I don't know what you mean by LTE. The proof is some weird induction on $n$ depending on whether it's a multiple of $p$ or not (I guess it's induction on $v(n)$, with $v$ the $p$-adic valuation). You use the binomial expansion to show that if $p>2$ and $v(a)=n>0$ then $v((1+a)^p-1)$ is $n+1$, and that $v((1+a)^m-1)$ is $n$ if $m$ is prime to $p$.

[Aaron Liu]

https://en.wikipedia.org/wiki/Lifting-the-exponent_lemma

[Kevin Buzzard]

You have to be careful at 2 because $v_2(1+2)=1$ but $v_2((1+2)^2-1)$ is 3, which is what stops (Z/8Z)^* from being cyclic. Yeah, I would rather avoid using that lemma because it forces you to descend to the finite case and then go up again, which seems a bit ridiculous.

[Kevin Buzzard]

Aaron Liu said:

https://en.wikipedia.org/wiki/Lifting-the-exponent_lemma

"Its main use is in [mathematical olympiads]" -- funny, I don't remember seeing the p-adic logarithm in mathematical olympiads.

[Kevin Buzzard]

I was completely confused by your references to "LTE" which I'd never heard before; I had assumed you were talking sarcastically about the liquid tensor experiment, which is what LTE usually refers to on this forum.

[Kevin Buzzard]

This claim on Wikipedia that it was first published in 2006 is complete nonsense.

[Aaron Liu]

Kevin Buzzard said:

I was completely confused by your references to "LTE" which I'd never heard before; I had assumed you were talking sarcastically about the liquid tensor experiment, which is what LTE usually refers to on this forum.

Blame @Violeta Hernández she used it first

Violeta Hernández said:

If $n$ is prime and $n \mid |F|-1$ then $n \mid k$ by LTE

[Violeta Hernández]

I learned about LTE on math olympiads, and I presume whoever wrote that article was writing it from that perspective

[Joseph Myers]

The basic idea of lifting the exponent dates back at least as far as Gauss (Disquisitiones Arithmeticae, art. 86-87), even though someone removed the reference to Gauss from the Wikipedia article.

[Django Peeters]

Django Peeters said:

Do you know how we could get this on the OEIS?

Done!

[Violeta Hernández]

What's the link?

[Django Peeters]

You can search "Lenstra excess" on the OEIS, it leads to this.

[Violeta Hernández]

(Hi everyone! Sorry for the inactivity. I've had a bunch of schoolwork, might only be free again until December)