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If $R$ is Noetherian and $M$ and $N$ are finitely generated $R$-modules, show that $$\operatorname{Ass}\operatorname{Hom}_R(M,N)=\operatorname{Supp}M\cap \operatorname{Ass}N$$ where $\operatorname{Supp}M$ is the set of all primes containing the annihilator of $M$.

Taking $M=R/I$, and setting $(0:_NI)=\{n\in N\mid In=0\}$, show that $\operatorname{Hom}_R(M,N)=(0:_NI)$, and thus $$\operatorname{Ass}(0:_NI)=\operatorname{Ass}N\cap\{P\subset R\mid P\text{ is a prime ideal and }I\subset P\}.$$

This is Exercise 3.3 in GTM150 (page 109) and I have no idea about how to solve the problem and make no progress.

user26857
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Let ${\frak p} \in {\rm Ass} \, {\rm Hom}_R(M,N)$. Then ${\frak p} R_{\frak p} \in {\rm Ass}_{R_{\frak p}} {\rm Hom}_{R_{\frak p}}(M_{\frak p},N_{\frak p}) $. Setting $\dfrac{R_{\frak p}}{{\frak p} R_{\frak p}}:=\kappa(p)$ one has $$0 \neq{\rm Hom_{R_{\frak p}}(\kappa({\frak p}), \, }{\rm Hom}_{R_{\frak p}}(M_{\frak p},N_{\frak p}) \cong {\rm Hom}_{R_{\frak p}} (\kappa({\frak p}) \otimes_{R_{\frak p}} M_{\frak p} , N_{\frak p} ).$$ Thus $ \kappa({\frak p}) \otimes_{R_{\frak p}} M_{\frak p} $ is a nonzero vector space over $\kappa({\frak p})$. Therefore $M_{\frak p} \neq 0$ and so that ${\frak p }\in \rm Supp \, M$ and ${\rm Hom}_{R_{\frak p}} (\kappa({\frak p}) , N_{\frak p} ) \neq 0$ which shows that ${\frak p} R_{\frak p} \in {\rm Ass}_{R_{\frak p}} N_{\frak p}$ and therefore ${\frak p} \in {\rm Ass}_{R} N $. The reverse inclusion holds similarly.

Note: We didn't use the finiteness of $N$, so it can be omitted.

Mohammad Bagheri
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