Lemma 51.11.3. Let $A$ be a Noetherian ring which has a dualizing complex. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $A', M'$ be the $I$-adic completions of $A, M$. Let $\mathfrak p' \subset \mathfrak q'$ be prime ideals of $A'$ with $\mathfrak q' \in V(IA')$ lying over $\mathfrak p \subset \mathfrak q$ in $A$. Then

$\text{depth}_{A_{\mathfrak p'}}(M'_{\mathfrak p'}) \geq \text{depth}_{A_\mathfrak p}(M_\mathfrak p)$

and

$\text{depth}_{A_{\mathfrak p'}}(M'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) = \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q)$

Proof. We have

$\text{depth}(M'_{\mathfrak p'}) = \text{depth}(M_\mathfrak p) + \text{depth}(A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'}) \geq \text{depth}(M_\mathfrak p)$

by flatness of $A \to A'$, see Algebra, Lemma 10.163.1. Since the fibres of $A \to A'$ are Cohen-Macaulay (Dualizing Complexes, Lemma 47.23.2 and More on Algebra, Section 15.51) we see that $\text{depth}(A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'}) = \dim (A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'})$. Thus we obtain

\begin{align*} \text{depth}(M'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) & = \text{depth}(M_\mathfrak p) + \dim (A'_{\mathfrak p'}/\mathfrak p A'_{\mathfrak p'}) + \dim ((A'/\mathfrak p')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \dim ((A'/\mathfrak p A')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \end{align*}

Second equality because $A'$ is catenary and third equality by More on Algebra, Lemma 15.43.1 as $(A/\mathfrak p)_\mathfrak q$ and $(A'/\mathfrak p A')_{\mathfrak q'}$ have the same $I$-adic completions. $\square$

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