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. $\square$

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).