The lattice of submodules

Stream: maths

[Antoine Chambert-Loir]

Let $R$ be a ring and $M$ be an $R$-module. The lattice of submodules of $M$ satisfies docs#Submodule.mem_sup:
For submodules $p,q$ of $M$ and $x\in M$, one has $x\in p\sqcup q$ iff there exists $y\in p$ and $z\in q$ such that $x=y+z$.

An analogous property holds in certain lattices (in particular distributive lattices, see docs#DistribLattice): if $L$ is such a lattice, the conditions $x\leq p \sqcup q$ is equivalent to the existence of $y\leq p$ and $z\leq q$ such that $x=y\sqcup z$. (In a distributive lattice, one can set $y=x\sqcap p$ and $z=x\sqcap q$.)

But the lattice of submodules is not distributive in not general (take three lines in a plane), hence I wonder whether anybody is aware of some property that would encompass both cases.

[Andrew Yang]

I'm no expert in lattice theory but sounds like docs#IsModularLattice.

[Antoine Chambert-Loir]

Thanks for the link. Indeed, we have docs#IsModularLattice.inf_sup_inf_assoc, but that rewrites $x \sqcap p \sqcup x \sqcap q$ to the unsymmetric $x\sqcap ((p \sqcap x) \sqcup q)$ which equals $x$ iff $x \leq (p \sqcap x)\sqcup q$…

[Junyan Xu]

Subgroups of a nonabelian group still form a modular lattice, but doesn't satisfy the analogue of mem_sup ...

[Antoine Chambert-Loir]

Good point!

[Yaël Dillies]

I'm very certain people, eg Dilworth, have already sorted this out 100 years ago. I'm not sure what condition exactly it is, though

[Yaël Dillies]

Antoine Chambert-Loir said:

the conditions $x≤ p ⊔ q$ is equivalent to the existence of $y≤ p$ and $z≤ q$ such that $x=y⊔ z$.

It really looks like docs#SupIrred and docs#SupPrime, but it's not quite it

[Kevin Buzzard]

I am confused by this question. In the theory of submodules it's not true that X <= Y + Z implies that X can be written Y' + Z' with Y' <= Y etc. However all its elements can be. But this is a different claim because elements are not the same as sets so the lattice framework doesn't apply. So what exactly is this question asking?

[Yaël Dillies]

Kevin Buzzard said:

elements are not the same as sets so the lattice framework doesn't apply

Actually, I don't think that's true. Elements of rings become atoms of their lattice of submodules, and that's a lattice theoretic concept.

[Yaël Dillies]

You can rephrase Antoine's statement as "If $x$ is an atom and $x ≤ p ⊔ q$, then there exist atoms $y ≤ p$ and $z ≤ q$ such that $x ≤ y ⊔ z$"

[Antoine Chambert-Loir]

That's why I wrote “analogous”. (Using it, I could prove both cases along the same lines.)

[Antoine Chambert-Loir]

Moreover, elements correspond to monogeneous submodules and are not exactly atoms.

[Yaël Dillies]

Sorry, I guess they are only atoms when the scalars are a field

[Kevin Buzzard]

OK well if there is a way of interpreting the above in terms of lattices, is there still a mathematics question left in this thread and, if so, what is it?

[Yaël Dillies]

The question is "What is the mathematical name (if any) of the lattices that are such that $x\leq p \sqcup q$ is equivalent to the existence of $y\leq p$ and $z\leq q$ such that $x=y\sqcup z$?"

[Andrew Yang]

I don't think this holds for the lattice of submodules?

[Yaël Dillies]

I agree. Counterexample: The submodules of $ℝ²$ generated by $(1, 0), (0, 1), (1, 1)$

[Mitchell Lee]

Yaël Dillies said:

The question is "What is the mathematical name (if any) of the lattices that are such that $x\leq p \sqcup q$ is equivalent to the existence of $y\leq p$ and $z\leq q$ such that $x=y\sqcup z$?"

The condition holds for a lattice $L$ if and only if $L$ is distributive.

Proof

The backward direction is easy. For the forward direction, let $a, b, c \in L$. To prove that $L$ is distributive, it suffices to show that

$$a \sqcap (b \sqcup c) \leq (a \sqcap b) \sqcup (a \sqcap c).$$

We have $a \sqcap (b \sqcup c) \leq b \sqcup c$, so by the condition, there exist $y \leq b$ and $z \leq c$ such that $a \sqcap (b \sqcup c) = y \sqcup z$. This yields $y \leq y \sqcup z = a \sqcap (b \sqcup c) \leq a$ and similarly $z \leq y \sqcup z = a \sqcap (b \sqcup c) \leq a$. Hence $y \leq a \sqcap b$ and $z \leq a \sqcap c$, so

$$a \sqcap (b \sqcup c) = y \sqcup z \leq (a \sqcap b) \sqcup (a \sqcap c)$$

as desired.

[Dean Young]

The lattice of submodules is a (modular) quantale action. In the attachment the condition of being locally principal is related to the condition that the first isomorphism theorem holds given modularity, which is true for the lattice arising from a ring or ring action (action under tensor, i.e. module).

IdealLatticesandLocallyPrincipalIdeals

Professor Derksen was very generous that summer - he let me explore whatever I wanted to. I decided it was interesting to see what conditions could be expressed in ideal theoretic terms, taking inspiration from 1st and 2nd semester commutative algebra courses (Atiyah & MacDonald, Altman & Kleiman, and Mel Hochster's notes).

In sum: it was Dilworth who found that, while the condition of being a principal ideal was lost in the associated quantale structure, the condition of being locally principal was not. In the paper, the condition on a quantale which generalizes locally principal is called Dilworth principal, and it is shown to be equivalent to two other conditions given modularity.

[Dean Young]

Yaël Dillies said:

The question is "What is the mathematical name (if any) of the lattices that are such that $x\leq p \sqcup q$ is equivalent to the existence of $y\leq p$ and $z\leq q$ such that $x=y\sqcup z$?"

The main insight here is about how the condition of being locally principal can be expressed in terms of the lattice structure because of an equivalence with the projection formula given modularity. This insight allows one to proceed with ideal theory by replacing principal ideals with locally principal ideals, which are related to "saturated maps" mentioned in the paper in how the first isomorphism theorem holds. In general (without modularity), some of the three conditions are stronger than others (strictly), which is why only one of them is called Dilworth principal.

After this theory is developed you can impose that each submodule is the join of locally principal submodules, which is more natural because of how the submodule lattice (which is a quantale action as defined in the attachment) is not distributive in general. Instead, there is a functor from quantales to distributive lattices which roughly consists of taking Hom with the Sierpinski lattice (two element lattice) [[-,𝕊],𝕊] twice in the category of join lattices (any frame is a quantale under the meet operation, including the two-element one). This gives a functor into distributive lattices, in which the mentioned condition holds, but it also suggests what is better about a certain modified question, in which elements (and principal ideals) are replaced with locally principal ideals:

Theorem: Let $[M]$ be a quantale action arising from a Noetherian module $M$ over a Noetherian commutative ring (lattice of submodules). For submodules $p$ and $q$ of $M$ i.e. $p, q : [M]$, and a locally principal submodule $x : [M]$, $x \leq p + q$ if and only if one can Zariski locally find $y_D$ and $z_D$ such that $y_D \leq p_D$, $z_D \leq q_D$ and $x_D=y_D + z_D$ in the quantale action arising from $M_D$.

The answer is not so different from the one @Kevin Buzzard suggests, since locally one can do the same thing. But the exciting thing is that the result can be expressed in terms of quantales and their actions.

[Dean Young]

(deleted)

[Dean Young]

Screenshot-2024-06-18-at-10.37.50PM.png
Screenshot-2024-06-18-at-10.37.57PM.png
Screenshot-2024-06-18-at-10.38.20PM.png

[Dean Young]

Yaël Dillies said:

I'm very certain people, eg Dilworth, have already sorted this out 100 years ago. I'm not sure what condition exactly it is, though

So the condition is "distributive" if you want to answer the question, and "locally" principal if you want to modify the question to be true for any quantale-action arising from a module over a commutative ring (however you may need the Noetherian condition, at least at first).

[Dean Young]

Here is the main lemma in the attachment:

Screenshot-2024-06-18-at-11.12.39PM.png

Screenshot-2024-06-18-at-11.13.00PM.png

[Dean Young]

Yaël Dillies said:

Kevin Buzzard said:

elements are not the same as sets so the lattice framework doesn't apply

Actually, I don't think that's true. Elements of rings become atoms of their lattice of submodules, and that's a lattice theoretic concept.

So, from the perspective of the above, "atom" is replaced with "atom locally", which can still be expressed in the quantale/quantale-action language.

[Dean Young]

One more comment is about the compactness condition and the "locally adic topology" on a commutative unitial ring $A$: the Noetherian condition is somehow related to the condition that a certain topology on $A$ -- one in which the preimage of an open ideal is closed under the homomorphism $A \rightarrow A_{\mathfrak{p}}$ and $A \rightarrow A_{D}$ -- it is related to the compactness condition on that topology. See Mel Hochster's thesis, I think it's the "patch topology" and that the adic topology forms a spectral space?

[Dean Young]

Kevin Buzzard said:

I am confused by this question. In the theory of submodules it's not true that X <= Y + Z implies that X can be written Y' + Z' with Y' <= Y etc. However all its elements can be. But this is a different claim because elements are not the same as sets so the lattice framework doesn't apply. So what exactly is this question asking?

So in sum, the condition of being principal has this problem, but locally principal is equivalent to one that can be expressed in pure ideal theoretic terms.

[Dean Young]

https://github.com/leanprover-community/mathlib4/blob/8666bd82efec40b8b3a5abca02dc9b24bbdf2652/Mathlib/Order/Category/Frm.lean#L31-L32

https://github.com/leanprover-community/mathlib4/blob/8666bd82efec40b8b3a5abca02dc9b24bbdf2652/Mathlib/Topology/Category/Locale.lean#L24-L25

[Dean Young]

Idea bin

[Junyan Xu]

Junyan Xu said:

Subgroups of a nonabelian group still form a modular lattice, but doesn't satisfy the analogue of mem_sup ...

This is actually wrong as pointed out by Wikipedia. The failure of modularity is definitely connected to the non-existence of $y\in p$ and $z\in q$ such that $x=y+z$ in the original post, so maybe there's indeed a closer connection.

Speaking of modular lattices, maybe you'd be interested in reviewing #17457 ...