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Let $R$ be a Noetherian ring , $M$ a finitely generated $R$-module and let $J$ be an ideal such that $Supp(M) \subset V(J)$ where $V(J) = \{P \in Spec(R) : P \supseteq J\}$. How to show there exists some $k>0$ such that $J^{k}M=0$?

I know that when $M$ is finitely generated we have $Supp(M)=V(ann(M))$ but I still don't see it. Can you please help?


So far I've figured it out that $J \subseteq \sqrt{Ann(M)}$. Now since $R$ is Noetherian then $J$ is finitely generated, say by $b_{1},b_{2},...,b_{t}$. For each $b_{i} \in \langle b_{1},..,b_{t} \rangle$ we have $b_{i}^{n_{i}} M =0$ for some natural $n_{i}$. Can we simply take then $k=n_{1} + n_{2}+...+n_{t}$ then by the binomial theorem $J^{k}M=0$?

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up vote 3 down vote accepted

Some hints:

  1. You want to show that $J^k M = 0$. Rewrite this as a relationship between $J$ and $ann(M)$.

  2. From what you've written, you see that $V(ann(M)) \subset V(J)$. What relationship does this imply about the ideals $ann(M)$ and $J$?

  3. Compare 1. and 2.

(Incidentally, for step 3. to succeed, it may be that you need to assume $R$ Noetherian, or at least that $J$ is finitely generated.)

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thanks, can you please have a look at what I wrote? – user6495 May 25 '11 at 1:27
@user6495: That looks good. A slightly cleaner way to phrase it is to say that because $J \subset \sqrt{Ann(M)}$ and $J$ is f.g., we have $J^k \subset Ann(M)$ for some large enough $k$ (by the same binomial theorem argument), and hence $J^k M = 0$. (But of course this is just a reformulation of what you argued.) Regards, – Matt E May 25 '11 at 1:33
thanks a lot! – user6495 May 25 '11 at 1:35

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