depth two excludes finite punctured extension

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

total ring of fractions: For a nonzero commutative ring $R$, let $S$ be the set of its nonzerodivisors, meaning elements whose multiplication maps on $R$ are injective. Its total ring of fractions is $Q(R)=S^{-1}R$. The set $S$ is multiplicative since composites of injective multiplication maps are injective. The natural map $R\to Q(R)$ is injective: $a\mathrm{/}1=0$ implies $sa=0$ for some $s\in S$, hence $a=0$. Set $Q(0)=0$. For a domain this recovers the fraction field; for a ring with zero divisors it need not be a field.

The three Depth Lemma inequalities: Let $(R,\mathfrak{m})$ be a Noetherian local ring and $0\to A\to B\to C\to 0$ a short exact sequence of finite $R$-modules. With $a={\mathrm{depth}}_R(A)$, $b={\mathrm{depth}}_R(B)$, and $c={\mathrm{depth}}_R(C)$, $b\ge \min \{a,c\},a\ge \min \{b,c+1\},c\ge \min \{a-1,b\}\mathrm{.}$ The last inequality is vacuous when $a=0$.

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

For a finite module, support is the set of primes containing the annihilator: If $M$ is a finitely generated left $R$-module, then ${\mathrm{Supp}}_R(M)=\{\mathfrak{p}:{\mathrm{Ann}}_R(M)\subseteq \mathfrak{p}\}\mathrm{.}$

Assuming the Axiom of Choice, Nakayama's lemma: Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I⊴R$ satisfy $I\subseteq J(R)$, and let $M$ be a finitely generated left $R$-module. If $IM=M$, then $M=0$.