Number of maximal submodules of a finitely generated module

Question

Let $R$ be a commutative Noetherian local ring and $M$ be a finitely generated $R$-module. Is the number of maximal submodules of $M$ finite?

Answer 1

You can view $V=M/mM$ as a finite dimension space over $k=R/m$ (the residual field).

The set of $R$-submodules of $M$ containing $mM$ is by Nakayama in bijection with the set of $k$-subspaces of $V$.

I am assuming $V$ has dimension $\geq 2$. ( if the dimension is $1$, $M=Ra$ for some $a \in M$ has exactly one maximal subspace, $ma$).

So if $V$ has infinitely many maximal strict subspaces (ie $k$ is infinite), $M$ has infinitely many maximal submodules.

To prove the converse, we need to show that maximal submodules of $M$ always contain $mM$. Now, if $N$ is a maximal submodule of $M$ not containing $mM$, $M=N+mM$ and by Nakayama we get a contradiction again.

As a conclusion: $M$ has finitely many maximal submodules iff the residual field is finite.