normal domain implies s two

Given: The objects and hypotheses in the statement. The Axiom of Choice is assumed; 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.

A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are: Assume the Axiom of Choice. Let $A$ be a domain. Then the following are equivalent: 1. $A$ is integrally closed. 2. For every prime ideal $\mathfrak{p}$ of $A$, the localisation $A_{\mathfrak{p}}$ is integrally closed. 3. For every maximal ideal $\mathfrak{m}$ of $A$, the localisation $A_{\mathfrak{m}}$ is integrally closed.

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 $\mathrm{depth}(M)=0⟺\mathfrak{m}\in {\mathrm{Ass}}_R(M)\mathrm{.}$

Depth drops by one after quotienting by a regular element: Let $R$ be Noetherian, let $M$ be finite, let $I$ lie in the Jacobson radical, and let $x\in I$ be $M$-regular. Then ${\mathrm{depth}}_I(M\mathrm{/}xM)={\mathrm{depth}}_I(M)-1.$

Krull's height theorem: Let $R$ be a Noetherian commutative ring, let $I=(x_1,\dots ,x_n)$ be an ideal generated by $n\ge 1$ elements, and let $\mathfrak{p}$ be a prime ideal minimal over $I$. Then $\mathrm{ht}(\mathfrak{p})\le n$.