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Let $R$ be a local ring with maximal ideal $\mathfrak{m}$ and let $M$ be an $R$-module. For a prime ideal $\mathfrak{p}\in R$, $M_p$ shall denote the localization of $M$ at $p$.

I've read the following and I can't see why it's true:

If $M_p=0$ for any prime ideal in $R$ that is distinct from $\mathfrak{m}$, then $M$ is a module of finite length.

I just know that if $M_m=0$ as well, that $M=0$.

Bernard
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2 Answers2

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Hint: Show that for any $m\in M$, $\text{Ann}_R(m)$ is not contained in any non-maximal prime - consider what it means for $m$ to map to $0$ in $M_{\mathfrak p}$ for that. Deduce that there exists $l\gg 0$ (depending on $m$) such that ${\mathfrak m}^l m = \{0\}$ (note that $Rm\subset M$ is a module over $R/\text{Ann}_R(m)$, which is Artinian local by the first step). Finally, using finite generation of $M$ conclude that there's some $l\gg 0$ such that ${\mathfrak m}^l M = 0$, and build a finite filtration from that.

Hanno
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  • If $m$ is mapped to 0 in $M_\mathfrak{p}$, than there exists a $s\in R\setminus \mathfrak{p}$ with $ms=0$ and if $Ann_R(m)\subseteq \mathfrak{p}$ for a non-maximal prime, than $m$ would not be send to $0$ in $M_\mathfrak{p}$ so $M_\mathfrak{p}\not=0$. A contradiction. I can't see the conclusion so far....maybe I am missing a known result –  Nov 20 '17 at 09:32
  • @Jules I added some details - let me know if it helps. – Hanno Nov 21 '17 at 19:24
  • First of all, thank you for your patience. I still don't know about the existence of the powers $l$. But the rest is clear: If $\mathfrak{m}^l m={0}$ for some $l>>0$ for any $m\in M$, then it is clearly also true for the generators of $M$ and so there is some $l>>0$ with $\mathfrak{m}^lM=0$. And this gives us a finite filtration: $0=\mathfrak{m}^l M\subseteq \mathfrak{m}^{l-1} M\subseteq .... \subseteq M$, so it lacks to show that this gives us a composition series. But this is done in another post. See: –  Nov 22 '17 at 10:59
  • See: https://math.stackexchange.com/questions/570883/a-noetherian-module-annihilated-by-a-power-of-maximal-ideal-must-has-finite-leng –  Nov 22 '17 at 11:04
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    @Jules The ring $R/\text{Ann}_R(m)$ is local artinian, and hence the maximal ideal is nilpotent (in general, the nilpotents are the intersection of all prime ideals, but there's only the maximal ideal here). Does that help? – Hanno Nov 22 '17 at 22:42
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Since everything is noetherian, finite length is the same as artinian. You can also do the following: $\{\mathfrak m\}=\operatorname {Supp} M=V(\operatorname {Ann} M)$, hence the formal Nullstellensatz yields $\mathfrak m =\sqrt{\operatorname {Ann} M} $. Hence $R/\operatorname {Ann}M $ is artinian and thus $M$ is also artinian, since it is finitely generated over an artinian ring.

Note that the converse is also true. If a finitely generated module over a noetherian local ring is artinian, then it is supported on the maximal ideal. Indeed, say $M$ is generated by $n$ elements, then these give rise to an injection $R/\operatorname{Ann} M \hookrightarrow \bigoplus_{i=1}^n M$. Thus $R/\operatorname{Ann} M$ is artinian, hence zero-dimensional. Hence the only prime ideal of $R/\operatorname{Ann} M$ is $\mathfrak m$, i.e. $\operatorname {Supp} M=V(\operatorname {Ann} M)=\{\mathfrak m\}$

MooS
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