Lemma 51.11.2. Let $A \to B$ be a finite homomorphism of Noetherian rings. Let $I \subset A$ be an ideal and set $J = IB$. Let $M$ be a finite $B$-module. If $A$ is universally catenary, then $s_{B, J}(M) = s_{A, I}(M)$.
Proof. Let $\mathfrak p \subset \mathfrak q \subset A$ be primes with $I \subset \mathfrak q$ and $I \not\subset \mathfrak p$. Since $A \to B$ is finite there are finitely many primes $\mathfrak p_ i$ lying over $\mathfrak p$. By Algebra, Lemma 10.72.11 we have
\[ \text{depth}(M_\mathfrak p) = \min \text{depth}(M_{\mathfrak p_ i}) \]
Let $\mathfrak p_ i \subset \mathfrak q_{ij}$ be primes lying over $\mathfrak q$. By going up for $A \to B$ (Algebra, Lemma 10.36.22) there is at least one $\mathfrak q_{ij}$ for each $i$. Then we see that
\[ \dim ((B/\mathfrak p_ i)_{\mathfrak q_{ij}}) = \dim ((A/\mathfrak p)_\mathfrak q) \]
by the dimension formula, see Algebra, Lemma 10.113.1. This implies that the minimum of the quantities used to define $s_{B, J}(M)$ for the pairs $(\mathfrak p_ i, \mathfrak q_{ij})$ is equal to the quantity for the pair $(\mathfrak p, \mathfrak q)$. This proves the lemma. $\square$
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