Finiteness of associated primes

Remarks

The proof builds a chain of submodules $0 = M_0 \subsetneq M_1 \subsetneq \cdots \subsetneq M_n = M$ with $M_i / M_{i-1} \cong R/P_i$ for prime ideals $P_i$. Each step picks an associated prime of $M/M_{i-1}$, which exists when this module is nonzero, and the chain stops because $M$ is a Noetherian module. Then $\operatorname{Ass}_R(M) \subseteq \set{P_1, \dots, P_n}$, because associated primes of an extension lie among those of the two pieces. For example, $\operatorname{Ass}_\Z(\Z/n\Z)$ is the finite set of ideals $(p)$ for the primes $p$ dividing $n \ge 2$. Finite generation is needed: the $\Z$-module $\bigoplus_p \Z/p\Z$, over all primes $p$, has every $(p)$ as an associated prime.

Going deeper

For a student

Associated primes of a nonzero module also exist, so $\operatorname{Ass}_R(M)$ is finite and nonempty when $M \ne 0$. With the result that zero divisors are the union of associated primes, this lets one use prime avoidance to find non-zero-divisors outside finitely many primes. This is a key step in building regular sequences and in the theory of depth.