Zero divisors are the union of associated primes

Remarks

If $P = \operatorname{Ann}(m)$ is associated, then each $r \in P$ kills $m \ne 0$, so $r$ is a zero divisor. Conversely, if $rm = 0$ with $m \ne 0$, the set of annihilators of nonzero elements that contain $\operatorname{Ann}(m)$ has a maximal member, since $R$ is Noetherian. A maximal annihilator $\operatorname{Ann}(m')$ is prime: if $ab \in \operatorname{Ann}(m')$ and $b m' \ne 0$, then $\operatorname{Ann}(b m') \supseteq \operatorname{Ann}(m') + (a)$, so $a \in \operatorname{Ann}(m')$. For example, an integer $r$ is a zero divisor on $\Z/12\Z$ if and only if $\gcd(r, 12) > 1$, that is, $r \in (2) \cup (3)$. Taking $M = R$, the set of zero divisors of $R$, together with $0$, is a finite union of prime ideals.

Going deeper

For a student

Combine it with finiteness of associated primes and prime avoidance. This shows: if an ideal $I$ consists of zero divisors on a finitely generated module $M$, then $I$ kills a nonzero element of $M$. This is the basic step in the theory of regular sequences and depth. It is also used to show that $R$ has a non-zero-divisor in an ideal not contained in any associated prime of $R$.