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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.

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Sorry I don't understand something, what is $M_p$? –  fpqc 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
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3 Answers

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 $S(M_i)$.

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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
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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.

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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
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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 $(Ax)_p\neq 0$.

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