Processing math: 100%

The Stacks project

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


Comments (0)


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.