I would like to go back to a quite delicate question of commutative algebra, that of associated prime ideals of modules. In most textbooks (Bourbaki, Matsumura…), this concept is considered for modules over a noetherian ring, while it is also necessary to consider it in a greater generality for some applications in algebraic geometry. For my book, (Mostly) commutative algebra (Springer Nature, 2021), I preferred to introduce the general concept (§6.5), because I observed that the initial proofs are in fact easier. In yesterday's class (Cohomology of coherent sheaves, 2nd year of Master course at Université Paris Cité), some remarks of a student, Elias Caeiro, helped me simplify two steps of the treatment I proposed in my book.
Definition. — Let $A$ be a ring and let $M$ be an $A$-module. Say that a prime ideal $P$ of $A$ is associated to $M$ if there exists an element $m\in M$ such that $P$ is minimal among all prime ideals containing $\ann_A(m)$.
We write $\Ass_A(M)$ (sometimes spelt out as “assassin”) for the set of all associated prime ideals of $M$.
(Here, $\ann_A(m)$ is the annihilator of $m$, the ideal of all $a\in A$ such that $am=0$.)
There is a geometric way to intepret this definition: it means that in the spectrum $\Spec(A)$, the irreducible closed set $V(P)$ (of which $P$ is the generic point) is an irreducible component of $V(\ann_A(m))$. Thanks to this remark, associated prime ideals are compatible with localisation: \[ \Ass_{S^{-1}A}(S^{-1}A) = \Ass_A(M) \cap \Spec(S^{-1}A), \] where $\Spec(S^{-1}A)$ is identified as the subset of $\Spec(A)$ consisting of prime ideals $P$ which are disjoint from $S$. In particular, $P$ is associated to $M$ if and only if the maximal ideal $PA_P$ of the local ring $A_P$ is associated to the module $M_P$.
Here is what the associated prime ideals mean, from the point view of module theory.
Proposition. — Let $a\in A$.
a) The multiplication by $a$ is injective in $M$ if and only if $a$ does not belong to any associated prime ideal of $M$.
b) The localized module $M_a$ is zero if and only if $a$ belongs to all associated prime ideals of $M$.
c) In particular, $M=0$ if and only if $\Ass_A(M)=\emptyset$.
Proof. — a) If $a$ belongs to the associated prime ideal $P$, then $a$ belongs to the associated prime ideal $PA_P$ of $M_P$, which means that there exists $m\in M$ such that $PA_P$ is the only prime ideal containing $\ann_{A_P}(m)$. Consequently, $a$ is nilpotent modulo $\ann_{A_P}(m)$ and there exists $n\geq 0$ and $b\in A\setminus P$ such that $a^nb\in\ann_A(m)$. Take a minimal such $n$. Since $b\notin P$, one has $n\geq 1$; then $a^{n-1}b m\neq0$, while $a\cdot a^{n-1}bm=0$ and the homothety $(a)_M$ is not injective. Conversely, if $(a)_M$ is not injective, take $m\neq0$ in $M$ such that $am=0$; the annihilator $\ann_A(m)$ is not equal to $A$, hence $V(\ann_A(m))\neq \emptyset$; take an irreducible component of this closed subset — equivalently a minimal prime ideal $P$ among those containing $\ann_A(m)$; one has $a\in \ann_A(m)$, hence $a\in P$.
b) follows from c), with $a=1$.
c) The module $M$ is zero if and only if the multiplication by $0$ is injective on $M$. By a), this is equivalent to the fact that $\Ass_A(M)$ is empty.
Corollary. — A prime ideal $P$ is in the support of $M$ if and only if it contains some associated prime ideal. The prime ideal $P$ belongs to the support of $M$ if and only if $M_P\neq0$, if and only if $\Ass_{A_P}(M_P)$ is not empty, if and only if there exists an associated prime ideal of $M$ which belongs to $\Spec(A_P)$, that is, is contained in $P$.
For noetherian rings, one has the following characterization of associated prime ideals, which is usually taken at their definition.
Theorem. — Let $A$ be a noetherian ring and $M$ be an $A$-module. A prime ideal $P$ of $A$ is associated to $M$ if and only if there exists $m\in M$ such that $P=\ann_A(m)$.
If $P=\ann_A(m)$, then $P$ is associated to $M$. Conversely, let $m\in M$ and let $P$ be a minimal prime ideal of $A$ among those containing $\ann_A(m)$. We first assume that $A$ is local with maximal ideal $P$; then $P$ is the only prime ideal of $A$ that contains $\ann_A(m)$, which implies that any element of $P$ is nilpotent modulo $\ann_A(m)$. Since $P$ is finitely generated (because $A$ is noetherian), there exists an integer $n$ such that $P^n\subseteq \ann_A(m)$. Take a minimal such $n$. Since $\ann_A(m)\subseteq P$, one has $n\geq 1$; then $P^{n-1}\not\subseteq\ann_A(m)$ so that there exists $b\in P^{n-1}$ such that $bm\neq0$. Then $ab\in P^n$ for every $a\in P$, so that $P\subseteq \ann_A(bm)$, and $\ann_A(bm)\subseteq P$ because $bm\neq0$. Consequently, $P=\ann_A(bm)$. In the general case, we use the case of a local ring to obtain $m\in M$ such that $\ann_{A_P}(m/1)=PA_P$. Consequently, $\ann_A(m)\subseteq P$, and for every $a\in P$, there exists $b\notin P$ such that $abm=0$. Using that $P$ is finitely generated, one finds $b\notin P$ such that $abm=0$ for every $a\in P$; then $\ann_A(bm)=P$, as was to be shown.
From that point on, both presentations converge. One deduces from the preceding theorem that if $A$ is noetherian and $M$ is finitely generated, there exists a composition series $0=M_0\subseteq M_1 \subseteq \dots \subseteq M_n=M$, with successive quotients $M_k/M_{k-1}$ of the form $A/P_k$, for some prime ideals $P_k$ of $A$, and then $\Ass_A(M)$ is contained in $\{P_1,\dots,P_n\}$, in view of the following lemma. In particular, $\Ass_A(M)$ is finite.
Lemma. — Let $M$ be an $A$-module and let $N$ be a submodule of $M$; then $ \Ass_A(N)\subseteq \Ass_A(M)\subseteq \Ass_A(N)\cup \Ass_A(M/N)$.
The first inclusion $\Ass_A(N)\subseteq \Ass_A(M)$ follows from the definition. Let us prove the second one. Let $P\in\Ass_A(M)$ and let $m\in M$ be such that $P$ is a minimal prime ideal of $A$ among those containing $\ann_A(m)$. Let $m'$ be the image of $M$ in $M/N$. If $P$ contains $\ann_A(m')$, then $P$ is also minimal among such prime ideals, hence $P\in\Ass_A(M/N)$. Otherwise, there exists $b\in \ann_A(m')$ such that $b\notin P$. Let us prove that $P$ is minimal among the prime ideals containing $\ann_A(bm)$. First of all, let $a\in\ann_A(bm)$; then $abm=0$, hence $ab\in P$, hence $a\in P$ since $b\notin P$. Since $\ann_A(m)\subseteq\ann_A(bm)$, it also follows that $P$ is minimal among the prime ideals containing $\ann_A(bm)$. Since $b\in\ann_A(m')$, one has $bm'=0$, hence $bm\in N$ and $P\in\Ass_A(N)$.
