Can finite modules have non-finite rank, and vice versa?

Stream: maths

[Eric Wieser]

docs#finite_dimensional.rank_lt_aleph_0 tells us that finite and free modules have finite rank.

[Eric Wieser]

Are there obvious counterexamples (finite modules with infinite rank) that show that the module.free assumption is needed?

[Jireh Loreaux]

This isn't something I know much about, but I suspect there is a counterexample for rings not satisfying the strong rank condition. If the ring satisfies that then the result holds without the module.free assumption using docs#linear_independent_le_span.

[Johan Commelin]

Here's an example that addresses the "vice versa" in the title of the thread:
Consider $ℚ$ as a $ℤ$-module. This is not free, and not finite (i.e., not finitely generated as module). But if you have $x, y ∈ ℚ$, then there exist $a, b ∈ ℤ$ such that $ax + by = 0$. Indeed, if $x = p/q$ and $y = r/s$, then take $a = rq$ and $b = -ps$. Then we get $rq*(p/q) - ps*(r/s) = rp - pr = 0$. Hence $ℚ$ has rank 1.

[Reid Barton]

So docs#linear_independent_le_span implies docs#finite_dimensional.rank_lt_aleph_0 with the module.free assumption from the latter dropped, right? Because both of them assume the strong rank condition.

[Reid Barton]

For (noncommutative) rings in general (without the strong rank condition) it can fail. Even the ring $R$ itself can contain an infinite $R$-linearly independent subset.

[Reid Barton]

The easiest example is the free noncommutative $k$-algebra on countably many generators.

[Junyan Xu]

I think the ring of column finite matrices here gives another example: apparently it's isomorphic to the direct product of countably infinitely many copies of itself (as module over itself), which contains the countably infinite direct sum.

However, it's not possible for a module over an arbitrary nontrivial ring to possess simultaneously a finite basis and an infinite basis, by the argument here, but note it's slightly incomplete; we also need that when |B| is finite the union is also finite.

[Riccardo Brasca]

I've asked this on mathoverflow months ago to address a comment in mathlib (maybe the same you are thinking about).

[Junyan Xu]

The answer also follows from this which I just came across. We may assume the finitely many generators are linearly independent, because we may replace the original module by a finite free module that surjects onto it, and any lifts of the independent elements will stay independent.
(Edit: this is basically what the answer in Riccardo's link says.)

[Eric Wieser]

Riccardo Brasca said:

I've asked this on mathoverflow months ago to address a comment in mathlib (maybe the same you are thinking about).

I didn't see the comment, but I'm pretty certain the context is identical!

[Eric Wieser]

Reid Barton said:

So docs#linear_independent_le_span implies docs#finite_dimensional.rank_lt_aleph_0 with the module.free assumption from the latter dropped, right? Because both of them assume the strong rank condition.

#18792

[Eric Wieser]

Johan Commelin said:

Hence $ℚ$ has rank 1.

Does module.rank ℤ ℚ = 1 belong in mathlib somewhere? Is there some is_fraction_ring generalization?

[Kevin Buzzard]

docs#module.rank

[Kevin Buzzard]

I guess if R is commutative then the rank of any localisation of R is 1

[Reid Barton]

At most one :)

[Kevin Buzzard]

Oh yes, if the localisation kills a component then the rank is 0? For example if we localise $R:=\Z\times\Z$ at $\{(1,0)\}$ then no subset of positive size is $R$-independent because every element is killed by $(0,1)$.

[Johan Commelin]

Or you just invert 0. That'll give you something of rank 0 :smiley:

[Kevin Buzzard]

Yeah, that's killing all the components :-) (I chose the more complex example to show that it wasn't just inverting zero which caused problems)

[Johan Commelin]

In other words: rank is really a more local notion

[Kevin Buzzard]

Yeah, in some sense the definition we have is "wrong" if Spec(R) has more than one irreducible component. If R=Z^2 again and M and N are the localisations of R at the two nontrivial idempotents then it seems to me that M and N both have rank 0 but their direct sum has rank 1.

[Kevin Buzzard]

But really their ranks should be something like (0,1) and (1,0)