As the ring $A/p$ is an integral domain, the annihilator of an element $\neq 0$ of $A/p$ is $p$.

Let $a = \operatorname{Ann}(x)$ ($x \in M, x \neq 0$) be such a maximal element; it is sufficient to show that $a$ is prime. As $x \neq 0$, $a \neq A$. Let $b, c$ be elements of $A$ such that $bc \in a$ and $c \notin a$. Then $cx \neq 0$, $b \in \operatorname{Ann}(cx)$ and $a \subset \operatorname{Ann}(cx)$. As $a$ is maximal, $\operatorname{Ann}(cx) = a$, whence $b \in a$, so that $a$ is prime.

If $M = \{0\}$, clearly $\operatorname{Ass}(M)$ is empty (without any hypothesis on $A$). If $M \neq \{0\}$, the set of ideals of the form $\operatorname{Ann}(x)$, where $x \in M$ and $x \neq 0$, is non-empty and consists of ideals $\# A$; as $A$ is Noetherian, this set has a maximal element; then it suffices to apply Proposition 2.

If $a$ belongs to a prime ideal $p \in \operatorname{Ass}(M)$, then $p = \operatorname{Ann}(x)$, where $x \in M, x \neq 0$; whence $ax = 0$ and the homothety with ratio $a$ is not injective. Conversely, if $ax = 0$ for some $x \in M$ such that $x \# 0$, then $Ax \# \{0\}$, whence $\operatorname{Ass}(x) \neq \varnothing$ (Corollary 1). Let $p \in \operatorname{Ass}(Ax)$; then obviously $p \in \operatorname{Ass}(M)$ and $p = \operatorname{Ann}(bx)$, where $b \in A$; whence $a \in p$, since $abx = 0$.

It may be reduced to the case where $I$ is finite by means of (1), then to the case where $\operatorname{Card}(I) = 2$ by induction on $\operatorname{Card}(I)$. Then let $I = \{i, j\}$, where $i \neq j$; as $M/M_i$ is isomorphic to $M_j$, $\operatorname{Ass}(M) \subset \operatorname{Ass}(M_i) \cup \operatorname{Ass}(M_j)$ (Proposition 3); moreover, $\operatorname{Ass}(M_i)$ and $\operatorname{Ass}(M_j)$ are contained in $\operatorname{Ass}(M)$ (Proposition 3), whence the result.

Let $\mathscr{E}$ be the set of submodules $P$ of $M$ such that $\operatorname{Ass}(P) \subset \operatorname{Ass}(M) - \Phi$. Formula (1) shows that the set $\mathscr{E}$, ordered by inclusion, is inductive; moreover, $\{0\} \in \mathscr{E}$ and hence $\mathscr{E} \neq \varnothing$. Let $N$ be a maximal element of $\mathscr{E}$. Then $\operatorname{Ass}(N) \subset \operatorname{Ass}(M) - \Phi$. We shall see that $\operatorname{Ass}(M/N) \subset \Phi$, which, by Proposition 3, will complete the proof. Let $p \in \operatorname{Ass}(M/N)$; then $M/N$ contains a submodule $F/N$ isomorphic to $A/p$. By Propositions 1 and 3, $\operatorname{Ass}(F) \subset \operatorname{Ass}(N) \cup \{p\}$. Since $N$ is maximal in $\mathscr{E}$, $F \notin \mathscr{E}$ and hence $p \in \Phi$.

Recall (Chapter II, § 2, no. 5, Proposition 11) that the mapping $p \mapsto S^{-1}p$ is a bijection of $\Phi$ onto the set of prime ideals of $S^{-1}A$. If $p \in \operatorname{Ass}_A(M) \cap \Phi$, $p$ is the annihilator of a monogenous submodule $N$ of $M$; then $S^{-1}p$ is the annihilator of the monogenous submodule $S^{-1}N$ of $S^{-1}M$ (Chapter II, § 2, no. 4, formula (9)) and hence $S^{-1}\mathfrak{p} \in \mathrm{Ass}_{S^{-1}A}(S^{-1}M)$. Conversely, suppose that $\mathfrak{p} \in \Phi$ is finitely generated and such that $S^{-1}\mathfrak{p}$ is associated with $S^{-1}M$; then there exists $x \in M$ and $t \in S$ such that $S^{-1}\mathfrak{p}$ is the annihilator of $x/t$. Let $(a_i)_{1 \leq i \leq n}$ be a system of generators of $\mathfrak{p}$; then $(a_i/1)(x/t) = 0$ and hence there exists $s_i \in S$ such that $s_i a_i x = 0 \ (1 \leq i \leq n)$. Let us write $s = s_1 s_2 \ldots s_n$; for all $a \in \mathfrak{p}$, $s a x = 0$, whence $p \subset \mathrm{Ann}(s x)$; on the other hand, if $b \in A$ satisfies $b s x = 0$, then $b/1 \in S^{-1}\mathfrak{p}$ by definition, whence $b \in \mathfrak{p}$. Then $\mathfrak{p} = \mathrm{Ann}(s x)$ and $\mathfrak{p} \in \mathrm{Ass}_A(M)$.

If A is not Noetherian, the mapping $p \mapsto S^{-1}p$ of $\mathrm{Ass}_A(M) \cap \Phi$ to $\mathrm{Ass}_{S^{-1}A}(S^{-1}M)$ is not necessarily surjective (Exercise 1).

By Proposition 4 of no. 1, there exists a submodule $N'$ of M which satisfies the relations $\mathrm{Ass}(N') = \mathrm{Ass}(M) - Y$ and $\mathrm{Ass}(M/N') = \Psi$. We need to prove $N = N'$. Consider the commutative diagram

where $p, u, v$ are the canonical homomorphisms. We shall show that $S^{-1}p$ and $v$ are injective, which will prove that $u$ and $p$ have the same kernel and hence $N = N'$.

As $\mathrm{Ass}(N') \cap \Psi = \varnothing$, every element of $\mathrm{Ass}(N')$ meets S. Then $\mathrm{Ass}_{S^{-1}A}(S^{-1}N') = \varnothing$ (Corollary to Proposition 5), whence $S^{-1}N' = \{0\}$ (no. 1, Corollary 1 to Proposition 2), which proves that $S^{-1}p$ is injective (Chapter II, §2, no. 4, Theorem 1). On the other hand, if $x$ belongs to the kernel K of $v$, then $\mathrm{Ann}(x) \cap S \neq \varnothing$ (Chapter II, §2, no. 2, Proposition 4); hence $\mathrm{Ass}(K) = \varnothing$ since $\mathrm{Ass}(K) \subset \mathrm{Ass}(M/N') = \Psi$; we deduce that $K = \{0\}$ (no. 1, Corollary 1 to Proposition 2) and $v$ is injective.

(i) Every prime ideal $\mathfrak{p}$ of A containing an element $\in \operatorname{Ass}(M)$ belongs to $\operatorname{Supp}(M)$.

(ii) Conversely, if A is Noetherian, every ideal $\mathfrak{p} \in \operatorname{Supp}(M)$ contains an element $\in \operatorname{Ass}(M)$.

If $\mathfrak{p}$ contains an element $q$ of $\operatorname{Ass}(M)$, then $q \cap (A - \mathfrak{p}) = \varnothing$ and hence, if we write $S = A - \mathfrak{p}$, $S^{-1}\mathfrak{p}$ is a prime ideal associated with $S^{-1}M = M$, (no. 2, Proposition 5) and a fortiori $M, \# 0$, hence $\mathfrak{p} \in \operatorname{Supp}(M)$. Conversely, if A is Noetherian, so is A, (Chapter II, § 2, no. 4, Corollary 2 to Proposition 10). If $M_{\mathfrak{p}} \neq 0$, then $\operatorname{Ass}_{A_{\mathfrak{p}}}(M_{\mathfrak{p}}) \neq 0$ (no. 1, Corollary 1 to Proposition 2) and hence there exists $q \in \operatorname{Ass}_A(M)$ such that $q \cap (A - \mathfrak{p}) = \varnothing$ (no. 2, Corollary to Proposition 5).

We know that the nilradical of A is the intersection of the minimal elements of $\operatorname{Spec}(A) = \operatorname{Supp}(A)$ (Chapter II, § 2, no. 6, Proposition 13).

Let $\mathscr{G}$ be the set of submodules of M which have a composition series with the property of the statement. As $\mathscr{G}$ is non-empty (for $\{0\}$ belongs to $\mathscr{G}$) and M is Noetherian, $\mathscr{G}$ has a maximal element N. If $M \neq N$, then $M/N \neq 0$ and hence $\operatorname{Ass}(M/N) \neq \varnothing$ (no. 1, Corollary 1 to Proposition 2); $M/N$ therefore contains a submodule $N'/N$ isomorphic to an A-module of the form $A/\mathfrak{p}$, where $\mathfrak{p}$ is prime; then by definition $N' \in \mathscr{G}$, which contradicts the maximal character of N. Then necessarily $N = M$.

(4)

the minimal elements of these three sets are the same and coincide with the minimal elements of the set of prime ideals containing Ann(M).

The inclusion Ass(M) $\subset \{p_0, \ldots, p_{n-1}\}$ follows immediately from Propositions 1 and 3 of no. 1. For $0 \leq i \leq n - 1$,

(Chapter II, § 4, no. 4, Example), whence $p_i \in \operatorname{Supp}(M_i) \subset \operatorname{Supp}(M)$ (Chapter II, § 4, no. 4, Proposition 16), which shows the inclusion

Corollary 1 to Proposition 7 of no. 3 shows that Ass(M) and Supp(M) have the same minimal elements and (4) shows that these are just the minimal elements of $\{p_0, \ldots, p_{n-1}\}$. The last assertion then follows from Chapter II, § 4, no. 4, Proposition 17.

Under the conditions of Theorem 2, the set $\{p_0, \ldots, p_{n-1}\}$ is not necessarily determined uniquely by M; in particular it may be distinct from Ass(M) (Exercise 6).

(a) there exists an element x $\# O$ of M such that $ax = 0$. (b) for all $a \in a$, there exists an element x $\neq O$ of M such that $ax = 0$; (c) there exists $p \in \operatorname{Ass}(M)$ such that $a \subset p$.

Clearly (a) implies (b). By virtue of no. 1, Corollary 2 to Proposition 2, condition (b) means that the ideal a is contained in the union of the prime ideals associated with M and hence in one of them since Ass(M) is finite (Chapter 11, § 1, no. 1, Proposition 2); thus (b) implies (c). Finally, if there exists $p \in \operatorname{Ass}(M)$ such that $a \subset p$, p is the annihilator of an element $x \neq 0$ of M (no. 1, Definition 1) and $ax = 0$; thus (c) implies (a).

This intersection is also that of the minimal elements of Supp(M) (no. 3, Corollary 1 to Proposition 7) and to say that a is contained in this intersection is equivalent to saying that $V(a) \supset \operatorname{Supp}(M)$ in the notation of Chapter II, § 4; the conclusion then follows from Chapter 11, § 4, no. 4, Corollary 2 to Proposition 17.

The condition of the statement is equivalent to saying that for all $x \in M$ there exists $n(x) > 0$ such that $(Aa)^{n(x)}(Ax) = 0$; by Proposition 9 this means also that $a$ belongs to all the prime ideals associated with the submodule $Ax$ of $M$; the corollary then follows from the fact that $\operatorname{Ass}(M)$ is the union of the $\operatorname{Ass}(Ax)$ where $x$ runs through $M$ (no. 1, formula (1)).

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 $p \in \operatorname{Ass}(\operatorname{Hom}_A(E, F))$.