Lemma 15.45.7. Given any local ring R we have \dim (R) = \dim (R^ h) = \dim (R^{sh}).
Proof. Since R \to R^{sh} is faithfully flat (Lemma 15.45.1) we see that \dim (R^{sh}) \geq \dim (R) by going down, see Algebra, Lemma 10.112.1. For the converse, we write R^{sh} = \mathop{\mathrm{colim}}\nolimits R_ i as a directed colimit of local rings R_ i each of which is a localization of an étale R-algebra. Now if \mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ n is a chain of prime ideals in R^{sh}, then for some sufficiently large i the sequence
R_ i \cap \mathfrak q_0 \subset R_ i \cap \mathfrak q_1 \subset \ldots \subset R_ i \cap \mathfrak q_ n
is a chain of primes in R_ i. Thus we see that \dim (R^{sh}) \leq \sup _ i \dim (R_ i). But by the result of Lemma 15.44.2 we have \dim (R_ i) = \dim (R) for each i and we win. \square
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