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Let $A \not= \{0 \}$ be a Noetherian commutative ring and let $M$ be an $A$-module. Prove that the canonical homomorphism $$M \to \bigoplus_{P \in \text{Ass}(M)} M_p$$ is injective.

My question is, how does this homomorphism look like? I can't figure it out even though it says canonical. My first thought is to prove that the kernel is trivial, is this a good approach?

user26857
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  • don't you need $\prod$, instead of $\bigoplus$? – Krish Dec 30 '14 at 17:26
  • @Krish, it actually says $\bigoplus$ in the question i got. –  Dec 30 '14 at 17:29
  • I'm not sure about $\bigoplus$. I couldn't find any canonical map in this case. on the other hand $\prod$ is more natural. may be it was a typo. – Krish Dec 30 '14 at 18:44
  • @Krish I will look this up and leave a comment later. –  Dec 30 '14 at 18:49
  • @Krish There was another assumption which was not in the question first, namely that $M$ is finitely generated, then your comments makes sense :). May you answer my comment below please? –  Dec 30 '14 at 21:14
  • if the module $M$ is finitely generated, then $\text{Ass}R(M)$ is a finite set. so there is no difference between $\bigoplus$ and $\prod$ and you are done. if $M$ is not finitely generated then also the map $M\to\prod\limits{{\mathfrak p}\in\text{Ass}R(M)}M{\mathfrak p}$ is injective. see the answer by Hanno below. – Krish Dec 31 '14 at 04:20
  • All we need to solve this question can be found here. – user26857 Jan 11 '15 at 23:27

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If $M$ is finitely generated, then $\text{Ass}_R(M)$ is finite, hence $\bigoplus\limits_{{\mathfrak p}\in\text{Ass}_R(M)} M_{\mathfrak p}\cong\prod\limits_{{\mathfrak p}\in\text{Ass}_R(M)}M_{\mathfrak p}$ and the desired map is the product of the localization maps $M\to M_{\mathfrak p}$. If $M$ is not finitely generated and therefore $\text{Ass}_R(M)$ not necessarily finite, this only gives a morphism $$(\ddagger)\quad M\to\prod\limits_{{\mathfrak p}\in\text{Ass}_R(M)}M_{\mathfrak p}.$$ In any case, $(\ddagger)$ is injective: Check first that the kernel of $M\to M_{\mathfrak p}$ consists of those $m\in M$ for which $\text{Ann}_R(m)\not\subset{\mathfrak p}$. Hence, to prove injectivity you only have to find, for any $0\neq m\in M$, an associated prime ${\mathfrak p}\in\text{Ass}_R(M)$ such that $\text{Ann}_R(m)\subset{\mathfrak p}$, and such exists because any ideal which is maximal among the annihilators of non-zero elements of $M$ is associated.

Hanno
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  • Hmpf, sorry, my fault! Luckily the rest stays the same :) Thanks! – Hanno Dec 30 '14 at 17:07
  • @Hanno My question was how the morphism looks like? What is it given by? –  Dec 30 '14 at 18:06
  • @Sodan: for each prime ideal $\mathfrak{p}$ in Ass $M,$ consider the canonical map $ \phi_{\mathfrak{p}} :M \rightarrow M_{\mathfrak{p}}.$ then take the map $M\to\prod\limits_{{\mathfrak p}\in\text{Ass}R(M)}M{\mathfrak p}$ defined by $m \mapsto \prod\limits_{{\mathfrak p}\in\text{Ass}R(M)} \phi{\mathfrak{p}}(m).$ – Krish Dec 30 '14 at 18:14
  • @Krish Why do you have a product sign? Why isnt it a direct sum sign? What am I missing. –  Dec 30 '14 at 18:37
  • Detail for the last part: if $\text{Ann}(x)$ is maximal and $ab\in\text{Ann}(x)$ then $abx=0$. If $bx=0$ then $b\in\text{Ann}(x)$. if $bx\neq0$ then $a\in\text{Ann}(bx)$. But $\text{Ann}(bx)\supseteq\text{Ann}(x)$ which is maximal so that $\text{Ann}(bx)=\text{Ann}(x)$. Hence $a\in\text{Ann}(x)$. – Gabriel Soranzo Apr 05 '21 at 17:56