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Let $A$ be a Noetherian ring and $M$ an $A$-module. I want to show that $M=0$ if $M_P = 0$ for each $P \in \text{Ass}(M)$.

Here is my attempt at a solution:

Assume for a contradiction that $M \not= 0$. Then there exist nonzero elements $x \in M$, so the set of annihilators $$\left\{ \text{Ann}(x) \mid 0 \not= x \in M \right\} $$ is nonempty. Because $A$ is Noetherian, this set of ideals has a maximal element; and it is known that a maximal element of such a set of annihilators is contained in $\text{Ass}(M)$. Therefore, there exists an $x$ such that $P = \text{Ann}(x) \in \text{Ass}(M)$.

Now consider the element $x/1 \in M_P$. By assumption we must have $x/1 = 0$, and by definition this means that there exists a $y \in A\setminus P$ such that $yx = 0$. But then $y \in \text{Ann}(x) = P$. So $y$ is both in $P$ and not in $P$, a contradiction. This proves that $M = 0$.

Is this proof correct, or have I done something wrong?

user26857
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Andrea
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    What you have shown is that if the module is not zero, then there is an associated prime $\mathfrak p$ such that $M_\mathfrak p$ is not zero. That's right. – Pedro Jan 09 '15 at 02:41
  • Also, you may use the following http://math.stackexchange.com/q/1085752/84157. – Youngsu Jan 09 '15 at 04:13
  • This question and the linked one follow easily from http://math.stackexchange.com/questions/1071542/if-p-in-operatornamesuppm-prove-that-p-contains-a-prime-ideal-q-with – user26857 Jan 09 '15 at 06:03

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