The Stacks project

Proof. We will use the criterion of Lemma 15.50.7. Let $\mathfrak q \subset R^ h$ be a prime and set $\mathfrak p = R \cap \mathfrak q$. Set $\mathfrak q_1 = \mathfrak q$ and let $\mathfrak q_2, \ldots , \mathfrak q_ t$ be the other primes of $R^ h$ lying over $\mathfrak p$, so that $R^ h \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i)$, see Lemma 15.45.13. Using that $(R^ h)^\wedge = R^\wedge $ (Lemma 15.45.3) we see

Hence $(R^ h)^\wedge \otimes _{R^ h} \kappa (\mathfrak q_ i)$ is geometrically regular over $\kappa (\mathfrak p)$ by assumption. Since $\kappa (\mathfrak q_ i)$ is separable algebraic over $\kappa (\mathfrak p)$ it follows from Algebra, Lemma 10.166.6 that $(R^ h)^\wedge \otimes _{R^ h} \kappa (\mathfrak q_ i)$ is geometrically regular over $\kappa (\mathfrak q_ i)$.

Let $\mathfrak r \subset R^{sh}$ be a prime and set $\mathfrak p = R \cap \mathfrak r$. Set $\mathfrak r_1 = \mathfrak r$ and let $\mathfrak r_2, \ldots , \mathfrak r_ s$ be the other primes of $R^{sh}$ lying over $\mathfrak p$, so that $R^{sh} \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , s} \kappa (\mathfrak r_ i)$, see Lemma 15.45.13. Then we see that

Note that $R^\wedge \to (R^{sh})^\wedge $ is formally smooth in the $\mathfrak m_{(R^{sh})^\wedge }$-adic topology, see Lemma 15.45.3. Hence $R^\wedge \to (R^{sh})^\wedge $ is regular by Proposition 15.49.2. We conclude that $(R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r_ i)$ is regular over $\kappa (\mathfrak p)$ by Lemma 15.41.4 as $R^\wedge \otimes _ R \kappa (\mathfrak p)$ is regular over $\kappa (\mathfrak p)$ by assumption. Since $\kappa (\mathfrak r_ i)$ is separable algebraic over $\kappa (\mathfrak p)$ it follows from Algebra, Lemma 10.166.6 that $(R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r_ i)$ is geometrically regular over $\kappa (\mathfrak r_ i)$. $\square$