Associated-prime localizations detect elements and have depth zero

Let $a\in R$ be nonzero. The cyclic module $Ra$ is nonzero and finite, so [F1] supplies a nonzero $b=ra\in Ra$ with prime annihilator $\mathfrak{q}={\mathrm{Ann}}_R(b)$. Because $b$ is also an element of $R$, this $\mathfrak{q}$ belongs to $\mathrm{Ass}(R)$. If $a\mathrm{/}1=0$ in $R_{\mathfrak{q}}$, some $s\mathrm{\notin }\mathfrak{q}$ would satisfy $sa=0$ and hence $sb=sra=0$, contrary to ${\mathrm{Ann}}_R(b)=\mathfrak{q}$. Thus every nonzero $a$ survives at one associated-prime localization, proving injectivity.

Conversely fix any $\mathfrak{q}\in \mathrm{Ass}(R)$ and choose $0\mathrm{\ne }b\in R$ with ${\mathrm{Ann}}_R(b)=\mathfrak{q}$. The element $b\mathrm{/}1$ is nonzero in $R_{\mathfrak{q}}$: otherwise a denominator outside $\mathfrak{q}$ would annihilate $b$. Its annihilator after localization is $\mathfrak{q}{R}_{\mathfrak{q}}$, the maximal ideal. Thus every member of that maximal ideal is a zerodivisor on the nonzero element $b\mathrm{/}1$, so no one-term regular sequence exists there and depth is zero. AC is inherited through [F1]; the localization argument itself makes only one witness selection at a time.