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
Comments (0)