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

Let $A$ be a commutative ring. I have a short question about the small result (Proposition 2.5 of Lang's book on Algebra, pg. 418) that if $M$ is an $A$ module, and $a\in A$, then $a_M$ defined by $x\mapsto ax$ for $x\in M$, (i.e. the left-multiplication map) is locally nilpotent if and only if $a$ lies in every prime ideal $p$ such that $M_p\neq 0$. Here $M_p$ is the localization of $M$ by $A\setminus p$.

In the converse statement, if $a_M$ is not locally nilpotent, then there is $x\in M$ such that $a^nx\neq 0$ for all $n\geq 0$. Let $S=\{1,a,a^2,\dots\}$ be a multiplicative set, so I know there exists a prime $p$ maximal among ideals not intersecting $S$. Why does it follow that is $(Ax)_p\neq 0$, so that $M_p\neq 0$? Lang states this in one line, so I feel it must be obvious.

share|cite|improve this question
Sorry I don't understand something, what is $M_p$? – user38268 Feb 19 '12 at 5:50
@BenjaminLim You can find the definition of $M_p$ in the relevant Wikipedia article or in chapter $3$ of Atiyah and Macdonald's textbook on commutative algebra. In short, $M_p$ is known as the "localization of $M$ at the prime ideal $p$" and is defined as the set of equivalence classes of ordered pairs $(x,s)$, $x\in M$, $s\in A - p$, where the equivalence relation is given by $(x,s)\equiv (y,t)$ if and only if there exists $r\not\in p$ such that $r(tx-sy)=0$. The set $M_p$ is an abelian group and and an $A_p$-module in natural ways. – Amitesh Datta Feb 19 '12 at 6:26
up vote 1 down vote accepted

For any $A$-module $M$ let $S(M)$ be its support and $\mathfrak a(M)$ its annihilator. Let $LN$ be the set of those $a$ in $A$ such that $a_M$ is locally nilpotent. We want to prove $$ LN=\bigcap_{\mathfrak p\in S(M)}\ \mathfrak p.\qquad\qquad(1) $$ The main lemma will be:

$(2)$ If $M$ is finitely generated, then $S(M)=V(\mathfrak a)$, with $\mathfrak a:=\mathfrak a(M)$.

Recall: $V(\mathfrak a):=\{\mathfrak p\in\text{Spec}(A)\ |\ \mathfrak p\supset\mathfrak a\}$.

Proof of $(2)$. Assume $\mathfrak p\supset\mathfrak a$. We must show $M_{\mathfrak p}\neq0$. Suppose by contradiction $M_{\mathfrak p}=0$. Let $x_1,\dots,x_n$ be generators of $M$. For each $i=1,\dots,n$ there is an $s_i$ which is $A$ but not in $\mathfrak p$ such that $s_ix_i=0$. Then $s_1\cdots s_n$ is in $\mathfrak a$ but not in $\mathfrak p$, contradiction.

Conversely, assume $M_{\mathfrak p}\neq0$. Then there is an $x$ in $M$ such that $sx\neq0$ for all those $s$ in $A$ which are not in $\mathfrak p$. Let $\alpha\in\mathfrak a$. As $\alpha x=0$, the element $\alpha$ is in $\mathfrak p$.

Proof of $(1)$. Let $(M_i)$ be a family of finitely generated submodules of $M$ whose sum is $M$. Put $\mathfrak a_i:=\mathfrak a(M_i)$.

By definition we have $$ LN=\bigcap_i\ r(\mathfrak a_i). $$ Recall the equality $$ r(\mathfrak a_i)=\bigcap_{\mathfrak p\in V(\mathfrak a_i)}\ \mathfrak p. $$ where $r(\mathfrak a_i)$ is the radical of $\mathfrak a_i$. By $(2)$, we have $$ \bigcap_{\mathfrak p\in V(\mathfrak a_i)}\ \mathfrak p=\bigcap_{\mathfrak p\in S(M_i)}\ \mathfrak p. $$ For formal reasons we have $$ \bigcap_i\ \bigcap_{\mathfrak p\in S(M_i)}\ \mathfrak p=\bigcap_{\mathfrak p\in\bigcup S(M_i)}\ \mathfrak p. $$ Now $(1)$ follows from the fact that $S(M)$ is the union of the $S(M_i)$.

share|cite|improve this answer
Thanks Pierre-Yves. I hadn't been on in a while, and did not realize this was posted until now. :) – Vika Apr 13 '12 at 8:03

An useful observation, which may help you conclude.

If the map $\alpha:m\in M\mapsto am\in $ is locally nilpotent, then the map $1-\alpha:M\to M$ is invertible: you can use the geometric series to define its inverse.

share|cite|improve this answer
So the inverse of $1-\alpha$ is $1+\alpha+\alpha^2+\cdots$. How does this show $(Ax)_p\neq 0$? – Vika Feb 19 '12 at 5:34

I think I understand this now. It follows from the definitions that if $x$ is an element of $M$ with $\frak{a}$ its annihilator, and $p$ a prime ideal of $A$, then $(Ax)_p\neq 0$ iff $p\supset\frak{a}$.

Now $S$ is disjoint from the annihilator of $x$, so the annihilator of $x$ is contained in some ideal maximal among those not intersecting $S$, which happens to be prime. Take $p$ to be this prime ideal, and so $(Ax)_p\neq 0$.

share|cite|improve this answer
If this is wrong, please correct me of course. – Vika Feb 19 '12 at 6:43

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.