serre r zero s one characterises reducedness

Given: The objects and hypotheses in the statement. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j\ge 0$, condition $(R_j)$ means that $R_{\mathfrak{p}}$ is regular whenever $\mathrm{ht}\mathfrak{p}\le j$. Condition $(S_j)$ means that $\mathrm{depth}{R}_{\mathfrak{p}}\ge \min \{j,\dim R_{\mathfrak{p}}\}$ for every prime $\mathfrak{p}$. A finite module $M$ satisfies $(S_j)$ if ${\mathrm{depth}}_{R_{\mathfrak{p}}}{M}_{\mathfrak{p}}\ge \min \{j,\dim {\mathrm{Supp}}_{R_{\mathfrak{p}}}{M}_{\mathfrak{p}}\}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+\mathrm{\infty }$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_j)$ conditions, and the zero ring satisfies both families vacuously.

The local depth-zero associated-prime criterion: Let $(R,\mathfrak{m})$ be a Noetherian local ring and let $M\mathrm{\ne }0$ be a finite $R$-module. Then $\mathrm{depth}(M)=0⟺\mathfrak{m}\in {\mathrm{Ass}}_R(M)\mathrm{.}$

Zero divisors on a module over a Noetherian ring are the union of its associated primes: Let $R$ be a Noetherian commutative ring and let $M$ be a left $R$-module. Then the set of zero divisors on $M$ is ${\bigcup }_{\mathfrak{p}\in {\mathrm{Ass}}_R(M)}\mathfrak{p}\mathrm{.}$ If $M$ is finitely generated, this is a finite union.

A nonzero module over a Noetherian ring has an associated prime: Let $R$ be a Noetherian commutative ring and let $M$ be a nonzero left $R$-module. Then ${\mathrm{Ass}}_R(M)$ is nonempty.

A Noetherian ring has finitely many minimal prime ideals: Let $R$ be a Noetherian commutative ring. Then $R$ has only finitely many minimal prime ideals. This theorem inherits only the dependent-choice cost already recorded in the cited Noetherian-induction corollary.

A radical ideal in a Noetherian ring is the intersection of its minimal primes: Assume Dependent Choice. Let $R$ be a Noetherian commutative ring and let $I⊴R$ be a radical ideal. Then there exist finitely many prime ideals ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_m$ minimal over $I$ such that $I={\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_m\mathrm{.}$ When $I=R$, this is the empty intersection.

Associated primes commute with localization for finite modules: Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$-module, and let $S\subseteq R$ be multiplicative. Then ${\mathrm{Ass}}_{S^{-1}R}(S^{-1}M)=\{S^{-1}\mathfrak{p}:\mathfrak{p}\in {\mathrm{Ass}}_R(M),\mathfrak{p}\cap S=\mathrm{\varnothing }\}\mathrm{.}$

An ideal contained in a finite union of prime ideals lies in one of them: Let $R$ be a commutative ring, let $I⊴R$ be an ideal, and let ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_n$ be prime ideals with $n\ge 1$. If $I\subseteq {\mathfrak{p}}_1\cup \cdots \cup {\mathfrak{p}}_n,$ then $I\subseteq {\mathfrak{p}}_i$ for some $i$.

Minimal support primes of a finite module are associated: Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$-module. If $\mathfrak{p}$ is minimal in ${\mathrm{Supp}}_R(M)$, then $\mathfrak{p}\in {\mathrm{Ass}}_R(M)\mathrm{.}$