# Module M/IM of finite length $\implies$ Ring A/I of finite length

This question is due to a proof in an algebra book (on the topic of dimension theory) which I don't fully understand (specifically, the proof of Thm 6.9b) in Kommutative Algebra by Ischebeck). It may have a very simple answer which I currently don't see.

Let A be a semilocal, noetherian ring, $M$ a finite A-module with $\mathrm{Ann}_A(M)=0$ and $a_1,\ldots,a_s\in \mathrm{Jac}(A)$ with $l_A(M/(a_1,\ldots,a_s)M)\lt\infty.$ Why is $l_A(A/(a_1,\ldots,a_s))\lt\infty$ ?

(where $\mathrm{Ann}_A(M)=\{x\in A\mid xM=0\}$, $\mathrm{Jac}(A)=\bigcap\limits_{\text{m is maximal ideal in } A} m$, $l_X(Y)=$ length of $Y$ as an $X$-module)

My ideas: I know that from $l_A(M/(a_1,\ldots,a_s)M)\lt \infty$ it follows that the ring $A/\mathrm{Ann}_A(M/(a_1,\ldots,a_s)M)$ is of finite length, because it can be embedded (as an $A$-module) in $M/(a_1,\ldots,a_s)M.$ But I don't know if/why $\mathrm{Ann}_A(M/(a_1,\ldots,a_s)M)=(a_1,\ldots,a_s)$.

Also I don't know if $A$ being semilocal or $a_1,\ldots,a_s\in \mathrm{Jac}(A)$ instead of $a_1,\ldots,a_s\in A$ is of any relevance to the question.

As further information, the final goal is to show that the degree of the polynomial p(n), which for large n gives the length $l_{A/Jac(A)}(M/Jac(A)^nM)$, is lesser or equal the minimum number s for which $l(M/(a_1,..,a_s)M)<∞$ as above.

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It is always useful to mention the title and author of a book one talks about—saying «an algebra book» is too vague! – Mariano Suárez-Alvarez♦ Apr 22 '12 at 23:08
The proof is the one of Thm 6.9b) in Kommutative Algebra by Ischebeck, a German textbook which probably noone here knows. :) The goal is to show that the degree of the polynomial p(n), which for large n gives the length $l_{A/Jac(A)}(M/Jac(A)^nM)$, is lesser or equal the minimum number s for which $l(M/(a_1,..,a_s)M)<\infty$ as above – juffo Apr 23 '12 at 1:09

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It is best if you edit the body of the question and add all that information there. – Mariano Suárez-Alvarez♦ Apr 23 '12 at 1:15

## 3 Answers

Suppose that $A$ is a noetherian commutative ring acting faithfully on a module $M$, and let $I$ be an ideal of $A$. It is not true in general that $A/I$ acts faithfully on $M/IM$, but one can show (e.g. using the Artin--Rees lemma) that the map $A/I \to End(M/IM)$ has nilpotent kernel.

In particular, if $M/IM$ has finite length, then $End(M/IM)$ also has finite length, and hence $A/I$ modulo a nilpotent ideal has finite length. This last statement in turn suffices to imply that $A/I$ itself is of finite length. (I didn't think about whether there is a more direct proof in your particular situation.)

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(I fixed a typo) – Mariano Suárez-Alvarez♦ Apr 23 '12 at 2:27
Thank you for your answer. Could you elaborate a bit on how to show that the kernel is nilpotent, i.e. on which ideal and submodule to apply Artin-Rees? – juffo Apr 23 '12 at 19:31

$\mathrm{Supp}(M/IM)=V(I+\mathrm{Ann}M)=V(I)=\mathrm{Supp}(R/I)$. Since $M/IM$ is of finite length, $\mathrm{Supp}(M/IM)$ consists of maximal ideals, so the same is true for $\mathrm{Supp}(R/I)$. Thus $R/I$ is of finite length. The second equality above follows because $\mathrm{Ann}M=0$.

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This looks like a perfect answer ! – QiL'8 Jan 3 at 11:37

The answer Matt E gave is absolutely right.

I gave the proof that the kernel is nilpotent in other way.

Since $M$ is finite over $A$, say $M$ is generated by $m_1,\ldots,m_n$, then we have an injective map

$$A\to \bigoplus_{i=1}^nM,a\mapsto (am_1,\ldots,am_n)$$

So we have a map $A/I\to \bigoplus_{i=1}^nM/IM$. Let the image of $a$ in $A/I$ be in the kernel, i.e., $aM\subset IM$, so by Nakayama's lemma, there is an $i\in I$ such that $(a^n+i)M=0$, because $M$ is a faithful $A$-module, we obtain that $a^n+i=0$, hence $a^n\in I$. Because $A/I$ is Noetherian, the kernel (denoted by $K$) is finitely generated, thus $K$ is nilpotent, i.e., there exists an $l$ such that $K^l=0$.

Okay, replacing $A/I$ by $A$, we have the conditions: $A$ is a Noetherian semilocal ring, $A/K$ is of finite length and $K$ is nilpotent. We want to show $A$ itself is of finite length.

The condition $A/K$ is of finite length implies that $A/K$ is an Artinian ring. Since $K$ is nilpotent, this implies that $A$ itself is an Artinian ring (dim=0,and Noetherian)! Hence $A$ is of finite length.

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