Lemma 10.156.6. Let $A \to B$ and $A \to C$ be ring maps. Let $\kappa $ be a separably algebraically closed field and let $B \otimes _ A C \to \kappa $ be a ring homomorphism. Denote

\[ \xymatrix{ B^{sh} \ar[r] & (B \otimes _ A C)^{sh} \\ A^{sh} \ar[u] \ar[r] & C^{sh} \ar[u] } \]

the corresponding maps of strict henselizations (see proof). If

$A \to B$ is quasi-finite at the prime $\mathfrak p_ B = \mathop{\mathrm{Ker}}(B \to \kappa )$, or

$B$ is a filtered colimit of quasi-finite $A$-algebras, or

$B_{\mathfrak p_ B}$ is a filtered colimit of quasi-finite algebras over $A_{\mathfrak p_ A}$, or

$B$ is integral over $A$,

then $B^{sh} \otimes _{A^{sh}} C^{sh} \to (B \otimes _ A C)^{sh}$ is an isomorphism.

**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

\[ c : B^{sh} \otimes _{A^{sh}} C^{sh} \to D^{sh} = (B \otimes _ A C)^{sh} \]

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

\[ B^{sh} = (B \otimes _ A A^{sh})_{\mathfrak q} \quad \text{and}\quad D^{sh} = (D \otimes _ C C^{sh})_{\mathfrak r} = (B \otimes _ A C^{sh})_{\mathfrak r} \]

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

\[ B^{sh} \otimes _{A^{sh}} C^{sh} = (B \otimes _ A A^{sh})_{\mathfrak q} \otimes _{A^{sh}} C^{sh} = \text{a localization of } B \otimes _ A C^{sh} \]

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

\[ D^{sh} = \mathop{\mathrm{colim}}\nolimits D_ i^{sh} = \mathop{\mathrm{colim}}\nolimits B_ i^{sh} \otimes _{A^{sh}} C^{sh} = B^{sh} \otimes _{A^{sh}} C^{sh} \]

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.
$\square$

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