Lemma 15.44.2. 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 10.112.1. 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 10.143.6) 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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