The Stacks project

Proof. Note that $R_{\mathfrak p}^ h \otimes _ R S$ is quasi-finite over $R_{\mathfrak p}^ h$ at the prime ideal corresponding to $\mathfrak q$, see Lemma 10.122.6. Hence the localization $S'$ of $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ is henselian and finite over $R_{\mathfrak p}^ h$, see Lemma 10.153.4. As a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.155.8 we see that $S_\mathfrak q^ h$ is the henselization of $R_{\mathfrak p}^ h \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$. Thus $S' = S_\mathfrak q^ h$ by the uniqueness result of Lemma 10.154.7. $\square$

Proof. This is a special case of Lemma 10.156.1. $\square$

Proof. Since $R \to S$ is quasi-finite at $\mathfrak q$ we see that the extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is finite, see Definition 10.122.3 and Lemma 10.122.2. Hence $\kappa _1^{sep}$ is a separable algebraic closure of $\kappa (\mathfrak p)$ (small detail omitted). In particular Lemma 10.155.10 does really apply. Next, the compositum of $\kappa (\mathfrak p)$ and $\kappa _1^{sep}$ in $\kappa _2^{sep}$ is separably algebraically closed and hence equal to $\kappa _2^{sep}$. We conclude that $\kappa _2^{sep}/\kappa _1^{sep}$ is finite. By construction the extension $\kappa _2^{sep}/\kappa _1^{sep}$ is purely inseparable. The ring map $R_{\mathfrak p}^{sh} \to S_{\mathfrak q}^{sh}$ is indeed local and induces the residue field extension $\kappa _2^{sep}/\kappa _1^{sep}$ which is indeed finite purely inseparable.

Note that $R_{\mathfrak p}^{sh} \otimes _ R S$ is quasi-finite over $R_{\mathfrak p}^{sh}$ at the prime ideal $\mathfrak q'$ given in the statement of the lemma, see Lemma 10.122.6. Hence the localization $S'$ of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ at $\mathfrak q'$ is henselian and finite over $R_{\mathfrak p}^{sh}$, see Lemma 10.153.4. Note that the residue field of $S'$ is $\kappa _2^{sep}$ as the map $\kappa _1^{sep} \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak q) \to \kappa _2^{sep}$ is surjective by the discussion in the previous paragraph. Furthermore, as a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.155.12 we see that $S_{\mathfrak q}^{sh}$ is the strict henselization of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ at $\mathfrak q'$. Thus $S' = S_\mathfrak q^{sh}$ by the uniqueness result of Lemma 10.154.7. $\square$

Proof. This is a special case of Lemma 10.156.3. $\square$

Proof. Any maximal ideal of $D$ lies over the maximal ideal of $B$ by going up for the integral ring map $B \to D$ (Lemma 10.36.22). Now $D/\mathfrak m_ B D = \kappa (\mathfrak m_ B) \otimes _ A C = \kappa (\mathfrak m_ B) \otimes _{\kappa (\mathfrak m_ A)} C/\mathfrak m_ A C$. The spectrum of $C/\mathfrak m_ A C$ consists of a single point, namely $\mathfrak m_ C$. Thus the spectrum of $D/\mathfrak m_ B D$ is the same as the spectrum of $\kappa (\mathfrak m_ B) \otimes _{\kappa (\mathfrak m_ A)} \kappa (\mathfrak m_ C)$ which is a single point by our assumption that either $\kappa (\mathfrak m_ C)/\kappa (\mathfrak m_ A)$ or $\kappa (\mathfrak m_ B)/\kappa (\mathfrak m_ A)$ is purely inseparable. This proves that $D$ is local and that the ring maps $B \to D$ and $C \to D$ are local. $\square$

Proof. Write $D = B \otimes _ A C$. Denote $\mathfrak p_ A = \mathop{\mathrm{Ker}}(A \to \kappa )$ and similarly for $\mathfrak p_ B$, $\mathfrak p_ C$, and $\mathfrak p_ D$. Denote $\kappa _ A \subset \kappa $ the separable algebraic closure of $\kappa (\mathfrak p_ A)$ in $\kappa $ and similarly for $\kappa _ B$, $\kappa _ C$, and $\kappa _ D$. Denote $A^{sh}$ the strict henselization of $A_{\mathfrak p_ A}$ constructed using the separable algebraic closure $\kappa _ A/\kappa (\mathfrak p_ A)$. Similarly for $B^{sh}$, $C^{sh}$, and $D^{sh}$. We obtain the commutative diagram of the lemma from the functoriality of Lemma 10.155.10.

Consider the map

we obtain from the commutative diagram. If $A \to B$ is quasi-finite at $\mathfrak p_ B = \mathop{\mathrm{Ker}}(B \to \kappa )$, then the ring map $C \to D$ is quasi-finite at $\mathfrak p_ D$ by Lemma 10.122.6. Hence by Lemma 10.156.3 (and Lemma 10.36.13) the ring map $c$ is a homomorphism of finite $C^{sh}$-algebras and

for some primes $\mathfrak q$ and $\mathfrak r$. Since

we conclude that source and target of $c$ are both localizations of $B \otimes _ A C^{sh}$ (compatibly with the map). Hence it suffices to show that $B^{sh} \otimes _{A^{sh}} C^{sh}$ is local (small detail omitted). This follows from Lemma 10.156.5 and the fact that $A^{sh} \to B^{sh}$ is finite with purely inseparable residue field extension by the already used Lemma 10.156.3. This proves case (1) of the lemma.

In case (2) write $B = \mathop{\mathrm{colim}}\nolimits B_ i$ as a filtered colimit of quasi-finite $A$-algebras. We correspondingly get $D = \mathop{\mathrm{colim}}\nolimits D_ i$ with $D_ i = B_ i \otimes _ A C$. Observe that $B^{sh} = \mathop{\mathrm{colim}}\nolimits B_ i^{sh}$. Namely, the ring $\mathop{\mathrm{colim}}\nolimits B_ i^{sh}$ is a strictly henselian local ring by Lemma 10.154.8. Also $\mathop{\mathrm{colim}}\nolimits B_ i^{sh}$ is a filtered colimit of étale $B$-algebras by Lemma 10.154.4. Finally, the residue field of $\mathop{\mathrm{colim}}\nolimits B_ i^{sh}$ is a separable algebraic closure of $\kappa (\mathfrak p_ B)$ (details omitted). Hence we conclude that $B^{sh} = \mathop{\mathrm{colim}}\nolimits B_ i^{sh}$, see discussion following Definition 10.155.3. Similarly, we have $D^{sh} = \mathop{\mathrm{colim}}\nolimits D_ i^{sh}$. Then we conclude by case (1) because

since filtered colimit commute with tensor products.

Case (3). We may replace $A$, $B$, $C$ by their localizations at $\mathfrak p_ A$, $\mathfrak p_ B$, and $\mathfrak p_ C$. Thus (3) follows from (2).

Since an integral ring map is a filtered colimit of finite ring maps, we see that (4) follows from (2) as well. $\square$