Lemma 10.63.19. Let $R$ be a ring. Let $M$ be an $R$-module. If $R$ is Noetherian the map

is injective.

Lemma 10.63.19. Let $R$ be a ring. Let $M$ be an $R$-module. If $R$ is Noetherian the map

\[ M \longrightarrow \prod \nolimits _{\mathfrak p \in \text{Ass}(M)} M_{\mathfrak p} \]

is injective.

**Proof.**
Let $x \in M$ be an element of the kernel of the map. Then if $\mathfrak p$ is an associated prime of $Rx \subset M$ we see on the one hand that $\mathfrak p \in \text{Ass}(M)$ (Lemma 10.63.3) and on the other hand that $(Rx)_{\mathfrak p} \subset M_{\mathfrak p}$ is not zero. This contradiction shows that $\text{Ass}(Rx) = \emptyset $. Hence $Rx = 0$ by Lemma 10.63.7.
$\square$

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