Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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

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.)

share|cite|improve this answer
(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 $\mathrm{Ann}M=0$.

share|cite|improve this answer
This looks like a perfect answer ! – user18119 Jan 3 '13 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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