Proposition 10

Let $A$ be a Noetherian ring, $E$ a finitely generated $A$-module and $F$ an $A$-module. Then

By hypothesis, $E$ is isomorphic to an $A$-module of the form $A^n/R$, hence $\operatorname{Hom}_A(E, F)$ is isomorphic to a submodule of $\operatorname{Hom}_A(A^n, F)$ and the latter is isomorphic to $F^n$; now, $\operatorname{Ass}(F^n) = \operatorname{Ass}(F)$ (no. 1, Corollary 1 to Proposition 3); thus $\operatorname{Ass}(\operatorname{Hom}_A(E, F)) \subset \operatorname{Ass}(F)$. On the other hand,

for every prime ideal $p$ of $A$, $\operatorname{Hom}_{A_p}(E_p, F_p)$ is isomorphic to $(\operatorname{Hom}_A(E, F))_p$ (Chapter 11, § 2, no. 7, Proposition 19), whence our assertion immediately; then we conclude from Theorem 2 that

Conversely, let $p$ be a prime ideal of $A$ belonging to $\operatorname{Ass}(F) \cap \operatorname{Supp}(E)$. By definition, $F$ contains a submodule isomorphic to $A/p$. On the other hand, since $E$ is finitely generated and $E, \neq 0$, we know that there exists a homomorphism $w \neq 0$ from $E$ to $A/p$ (Chapter II, § 4, no. 4, Proposition 20). As there exists an injective homomorphism $j$ from $A/p$ to $F$, $j \circ w \in \operatorname{Hom}(E, F)$ and $j \circ w \neq 0$. On the other hand, the relation $aw = 0$ for some $a \in A$ is equivalent to $a \in p$, the annihilator of every element $\neq 0$ of $A/p$ being $p$; then certainly $p \in \operatorname{Ass}(\operatorname{Hom}_A(E, F))$.