Lemma 15.43.2. If $A \to B$ is an étale ring map and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p \subset A$, then $\dim (A_{\mathfrak p}) = \dim (B_{\mathfrak q})$.

**Proof.**
Namely, because $A_{\mathfrak p} \to B_{\mathfrak q}$ is flat we have going down, and hence the inequality $\dim (A_{\mathfrak p}) \leq \dim (B_{\mathfrak q})$, see Algebra, Lemma 10.111.1. On the other hand, suppose that $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ n$ is a chain of primes in $B_{\mathfrak q}$. Then the corresponding sequence of primes $\mathfrak p_0 \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ n$ (with $\mathfrak p_ i = \mathfrak q_ i \cap A_{\mathfrak p}$) is chain also (i.e., no equalities in the sequence) as an étale ring map is quasi-finite (see Algebra, Lemma 10.142.6) and a quasi-finite ring map induces a map of spectra with discrete fibres (by definition). This means that $\dim (A_{\mathfrak p}) \geq \dim (B_{\mathfrak q})$ as desired.
$\square$

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