The Stacks project

Proof. We have seen this for the henselization in Lemma 15.51.7. To prove it for the strict henselization, it suffices to show that the formal fibres of $R^{sh}$ have $P$, see Lemma 15.51.4. 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 property $P$ holds for $\kappa (\mathfrak p) \to (R^{sh})^\wedge \otimes _ R \kappa (\mathfrak p)$ by (C) and our assumption on $R$. Using property (B), using the decomposition above, and looking at local rings we conclude that property $P$ holds for $\kappa (\mathfrak p) \to (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r)$. Since $\kappa (\mathfrak r)/\kappa (\mathfrak p)$ is separable algebraic, it follows from (E) that $P$ holds for $\kappa (\mathfrak r) \to (R^{sh})^\wedge \otimes _{R^{sh}} \kappa (\mathfrak r)$. $\square$