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Suppose $R$ is a commutative local ring and $M$ is a finitely generated module. Prove that $M$ is artinian and noetherian if and only if $\mathfrak{m}^nM=0$ for some $n$.

For the forward direction (which should be done), note that $M$ has finite length as it is artinian and noetherian. If no power of $\mathfrak{m}$ kills $M$, then I obtain a descending chain of submodules of $M$ of the form $\mathfrak{m}^iM$ - which must stabilize, say at $n$. Then $\mathfrak{m}^{n+1}M=\mathfrak{m}^nM$ but this is $\mathfrak{m}(\mathfrak{m}^n M)=\mathfrak{m}^nM$ and $\mathfrak{m}^nM$ is finitely generated - being the submodule of the noetherian module $M$. Then by Nakayama, we have $\mathfrak{m}^nM=0$, which I believe works.

However, the the reverse direction remains unclear to me. I feel as though the reverse direction should follow simply using something like $R/\mathfrak{m}$ is a field and $M/\mathfrak{m}M$ is a vector spaces over this field. I know that $M$ is finitely generated and hence we get a finite amount of generators for $M/\mathfrak{m}M$ but I am unsure of where to go from there or how $\mathfrak{m}^nM=0$ comes in. But I feel I am missing something obvious. How do I proceed from here?

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Your proof that “artinian implies $\mathfrak{m}^nM=0$ for some $n$” is good.

For the converse direction, we can use induction. The case $n=0$ is obvious. Suppose we know the statement for all modules $N$ having $\mathfrak{m}^{n-1}N=0$ and suppose $\mathfrak{m}^{n}M=0$. Then we can consider $N=\mathfrak{m}M$, which is artinian and noetherian by induction hypothesis. Moreover $M/N$ is a finitely generated vector space over $R/\mathfrak{m}$. You can surely finish up.

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  • $\begingroup$ $M/N$ is a finitely generated vector space over $R/\mathfrak{m}$. Then it is finite dimensional so it is artinian and noetherian. The exact sequence $0 \to N \to M \to M/N \to 0$ tells us that since $N,M/N$ are artinian and noetherian, then so too is $M$. Correct? $\endgroup$ – NPH Dec 23 '16 at 2:41
  • $\begingroup$ @NPH Yes, sure! Together with the fact that subspaces and submodules in $M/N$ are the same. $\endgroup$ – egreg Dec 23 '16 at 9:01

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