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Let $R,\mathfrak{m}$ be a Cohen Macaulay local ring and $M$ be an $R$ module such that $\mathfrak{m}\in Ass(M)$. i.e., the maximal ideal $\mathfrak{m}$ is an associated prime of $M$.

Now suppose $Hom(R/\mathfrak{m},M)=0$ then does it necessarily imply that $M=0$?

dongrugose
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1 Answers1

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Yes. By assumption you find some $m \in M$ with $\operatorname{Ann}(m) = \mathfrak m$. In particular $m \neq 0 $ and $\mathfrak mm=0$, i.e. $m$ gives rise to a non-zero map $R/\mathfrak m \to M$.

Actually $\mathfrak m \in \operatorname{Ass} M$ and $\operatorname{Hom}(R/\mathfrak m,M)=0$ imply everything since they are contradictory.

MooS
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