Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

EDIT:

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 $J^{k}M=0$?

share|improve this question
add comment

1 Answer 1

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.)

share|improve this answer
    
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
add comment

Your Answer

 
discard

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.