2
$\begingroup$

I've asked several times about properties of the lattice of submodules/ideals of modules/rings with specific properties.

This time I wonder about what interesting properties can one see in the lattice of submodules of a noetherian module and a finitely generated module.

For instance I read Lasker-Noether theorem which states (in the module formulation):

Every submodule of a finitely generated module over a Noetherian ring is a finite intersection of primary submodules.

In general, I don't have the intuition to foresee what will be the properties of these lattices. Let me give a preliminary list in any case to try to be more concrete.

Properties of lattices (wikipedia's classification of lattices)

ok: Complete, Modular,Algebraic lattice,Arguesian lattice

no: Distributive, Complemented

Metric lattice ??

Projective lattice ??

...

So do the lattices of submodules of noetherian/finitely generated modules verify any of the above properties? Is there something else interesting that can be said about then?

Any references are welcome.

Another way of looking at the problem:

Given the lattice of submodules $L(M)$ of a module $M$, how to know if it is noetherian/finitely generated?

M is Noetherian $\iff$ every ascending chain of submodules stabilizes.

$M$ finitely generated $\iff$ every ascending chain of submodules with union M stabilizes.

$\endgroup$
  • 1
    $\begingroup$ It's always modular, but in general not distributive. That's true for every ring, by the way. $\endgroup$ – Arnaud D. Apr 4 '18 at 19:56
  • 1
    $\begingroup$ And, to be sure, the lattice of submodules of any module is modular (finite generation doesn't matter.) $\endgroup$ – rschwieb Apr 4 '18 at 20:17
2
$\begingroup$

I can strengthen both of rschwieb's properties:

The lattice of submodules is always Arguesian and algebraic.

This is a strengthening since Arguesian lattices are modular and algebraic lattices are complete.

An Arguesian lattice is a lattice satisfying a complicated identity that comes from projective geometry. The lattice of subspaces of a projective space is Arguesian iff the space satisfies Desargue's law. Since the lattice of subspaces is always modular, there exist modular non-Arguesian lattices.

An algebraic lattice is a complete lattice in which every element is the join of compact elements. Here is an example of a complete lattice that is not algebraic.

$\endgroup$
  • $\begingroup$ it seems that you exhaust the branch (on the wikipedia link i posted) of complete lattices with this answer. Then i wonder if arguesian lattices have anything to do with projective ones... $\endgroup$ – Javier Apr 4 '18 at 22:06
  • $\begingroup$ +1 Interesting. I did not know enough to say about algebraic lattices, and I'd never heard of an Arguesian lattice. $\endgroup$ – rschwieb Apr 4 '18 at 23:54
2
$\begingroup$

The lattice of submodules of any module over any ring is modular and complete.

The lattice of a finitely generated/Noetherian module needs not be distributive. Example: Let $F$ be a field (finite, if you like) and $M=F\times F$ be a vector space of dimension $2$. The lattice of submodules has the diamond lattice, which is not distributive.

The lattice need not be complemented. Example: let $F$ be a field (finite, if you like), and $R=F[x]/(x^2)=M$. There is precisely one nontrivial submodule and it is not complemented.

$\endgroup$
  • 1
    $\begingroup$ I scanned through this chart at wikipedia, but I could not find anything further to say beyond what I said. $\endgroup$ – rschwieb Apr 4 '18 at 20:28
  • $\begingroup$ thanks for your answer. on the other hand, is there any property that looking at the lattice of submodules tells me that there is an underlying noetherian module? or at least a property that is shared by some modules (including noetherian) and not the rest? $\endgroup$ – Javier Apr 4 '18 at 20:51
  • 1
    $\begingroup$ If every member of the lattice of submodules of $M$ is finitely generated, then $M$ is Noetherian. If every increasing chain of members the lattice stabilies, $M$ is Noetherian. But you knew these already, right? $\endgroup$ – rschwieb Apr 4 '18 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.