If $M$ is a finitely generated $A$-module, where $A$ is noetherian, and if $I$ is an ideal of $A$ such that $Supp_A(M) \subseteq \mathcal{V}(I)$, I want to show that $I^nM = 0$ for some $n$. An obvious approach is to use the fact that $A$ is noetherian to show that there is an $n$ such that $I^n = I^{n+k}$ for all $k$, since $I \subseteq I^2 \subseteq I^3 \subseteq \ldots$ is an ascending chain of ideals. Also, since $M$ is finitely generated, we have $Supp_A(M) = \mathcal{V}(Ann_A(M))$, we have the fact that every prime ideal which contains the annihilator of $M$ also contains $I$. I'm not quite sure how to use all this information to show that $I^n$ annihilates $M$. Now that I type it, it doesn't seem true, since $I^n$ would have to be contained in $Ann_A(M)$. Any hints are appreciated.
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It's actually $...\subset I^3\subset I^2\subset I$ – Maxime Ramzi Oct 01 '18 at 05:21
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You know that every prime that contains the annihilator contains $I$, therefore the intersection of primes that contain $I$ ($=\sqrt{I}$) is contained in $\sqrt{Ann_A(M)}$. Now use that $I$ is finitely generated ($A$ is noetherian) – Maxime Ramzi Oct 01 '18 at 05:24
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@Max Oh, I think I see. So since each generator $a_i$ of I is in $\sqrt{Ann_A(M)}$, I can say that $a_i^{n_i}$ annihilates $M$, so if I let $n$ be the product of the $n_i$, $I^n$ will annihilate $M$? – Wyatt Gregory Oct 01 '18 at 07:04
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I think the sum of the $n_i$ suffices but yes that's the idea – Maxime Ramzi Oct 01 '18 at 08:02