Depth at a prime is bounded by local support dimension

Put $A=R_{\mathfrak{p}}$. Let $N$ be a nonzero finite $A$-module and let $x$ lie in the maximal ideal and act injectively on $N$. By [L3], $N\mathrm{/}xN\mathrm{\ne }0$. Localization and [L3] give ${\mathrm{Supp}}_A(N\mathrm{/}xN)={\mathrm{Supp}}_A(N)\cap V_A(x)$: outside $V(x)$ the quotient is zero, while at a support prime containing $x$ its localized nonzero finite module cannot equal its $x$-multiple.

Let $\mathfrak{q}$ be minimal in ${\mathrm{Supp}}_A(N)$. By [L1], it is minimal over ${\mathrm{Ann}}_A(N)$. In $A_{\mathfrak{q}}$ the only support prime of $N_{\mathfrak{q}}$ is $\mathfrak{q}{A}_{\mathfrak{q}}$; [L1] therefore makes $\sqrt{\mathrm{Ann}(N_{\mathfrak{q}})}=\mathfrak{q}{A}_{\mathfrak{q}}$. This maximal ideal is finitely generated, so some power annihilates $N_{\mathfrak{q}}$. If $x\in \mathfrak{q}$, a power of multiplication by $x$ would be zero on the nonzero module $N_{\mathfrak{q}}$, contradicting injectivity after localization. Hence no minimal support prime contains $x$.

Take any strict chain ${\mathfrak{r}}_0⊊\cdots ⊊{\mathfrak{r}}_s$ in ${\mathrm{Supp}}_A(N\mathrm{/}xN)$. Since ${\mathfrak{r}}_0$ contains $\mathrm{Ann}N$, the finite minimal-prime decomposition [L2] gives a minimal support prime $\mathfrak{q}\subseteq {\mathfrak{r}}_0$. Because $x\in {\mathfrak{r}}_0$ but $x\mathrm{\notin }\mathfrak{q}$ by step 2.1, this inclusion is strict. Prepending $\mathfrak{q}$ gives a chain of length $s+1$ in ${\mathrm{Supp}}_A(N)$. Thus $\dim {\mathrm{Supp}}_A(N\mathrm{/}xN)\le \dim {\mathrm{Supp}}_A(N)-1$.

Conversely, by [L4] choose a parameter tuple of length $e=\dim {\mathrm{Supp}}_A(N\mathrm{/}xN)$ for $N\mathrm{/}xN$. Prepend $x$. The resulting quotient of $N$ is the same finite-length module, so [L4] gives $\dim {\mathrm{Supp}}_A(N)\le e+1$. Together with step 3.1, this shows that quotienting by an $N$-regular element in the maximal ideal lowers support dimension by exactly one.

Starting with $M_{\mathfrak{p}}$, take any regular sequence of length $r$ in the maximal ideal. Its successive quotients are nonzero by [L5], so step 4.1 applies $r$ times and shows $r\le \dim {\mathrm{Supp}}_A(M_{\mathfrak{p}})$. Taking the supremum over all regular sequences, exactly as depth is defined in [L5], proves the inequality. If the dimension is zero, only the empty regular sequence can occur, and both sides are zero.