Lemma 2

Suppose that $A$ is semi-local; let $q$ be an ideal of definition of $A$, and let $\rho : A \to B$ be a ring homomorphism making $B$ into a finitely generated $A$-module. Suppose that, for every prime ideal (necessarily minimal) $p$ of $A$ such that $\dim(A/p) = \dim(A)$, there exists a unique prime ideal $r$ of $B$ lying over $p$ and that the canonical homomorphism $\rho_p : A_p \to B_r$ is bijective. Then one has $\dim_A(B) = \dim(A)$ and $e_q^A(B) = e_q^A(A)$.

Let $\mathfrak{S}_A$ (resp. $\mathfrak{S}_B$) be the set of prime ideals $p$ of $A$ such that

one has $\mathfrak{S}_A \neq \varnothing$. Let $p \in \mathfrak{S}_A$; by assumption there exists a prime ideal of $B$ lying over $p$. Then one has $\rho^{-1}(pB) = p$ (II, § 2, no. 5, cor. 3 to prop. 11), and

by th. 1, b) and c) of § 2, no. 3. Consequently, one has

This implies $\mathfrak{S}_A \subset \mathfrak{S}_B$ and $\dim(A) = \dim_A(B)$. Conversely, if $p \in \mathfrak{S}_B$, one has the inequalities

whence $p \in \mathfrak{S}_A$ and $\mathfrak{S}_B = \mathfrak{S}_A$. By prop. 3 of no. 1 and its corollary, one has

by Lemma 1, one has $\operatorname{long}_{A_p}(A_p \otimes_A B) = 1$ for every $p \in \mathfrak{S}_A$, whence $e_q^A(A) = e_q^A(B)$.