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I am reading Proposition 8, page 9, of the book LOCAL ALGEBRA by Serre.

Consider that $A$ is a commutative ring with identity and noetherian and $M$ a finitely generated $A$-module.

For each $x \in {A}$ we define the endomorphism $x_M: M \longrightarrow {M}$ by $x_M(m) = xm$. The following are equivalent:

i) $x_M$ is nilpotent

ii) $x \in \displaystyle \bigcap_{P \in {\mathrm{Ass}(M)}}P$

Proof:

$i) \Rightarrow {ii)}$ By contradiction.

Suppose $x \not \in {P}$ for some $P \in {Ass(M)}$.

Now since $P \in {Ass(M)}$ (since $P \in {Ass(M)}$ then there is a $m\in {M}$ such that $P = Ann(m)$, this means that $xm \neq {0}$) then there is a submodule $N$ of $M$ which is isomorphic to $A / P$. Consider the constraint $x_M |_N: N \rightarrow {M}$ of $x_M$ which is also nilpotent and we know that $m\in {N}$ ( it is known that $N = <m>$), then we have that there is a positive integer $r$such that $(xm) ^ r = 0$ then $x ^ r \in {Ann (m^r)}$ . I can't get to the contradiction, I'm waiting for a suggestion.

Thank you

user26857
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Tomais
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1 Answers1

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It's essentially the same proof as the fact that the nilpotents of a ring $R$ are precisely the elements contained in every (minimal) prime of $R$.

An associated prime of $M$ is of the form $P = \operatorname{Ann}(m)$ for some $m \in M$. If multplication by $x \in R$ defines a nilpotent endomorphism $x_M: M \rightarrow M$, then $x^n$ acts as $0$ on $M$ for some $n$. So $x^n m = 0m = 0$. Since $x^n \in \operatorname{Ann}(m)$, and $\operatorname{Ann}(m)$ is prime by assumption, also $x \in \operatorname{Ann}(m)$.

Badam Baplan
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