Lemma 15.106.4. Let $k$ be an algebraically closed field. Let $A$, $B$ be strictly henselian local $k$-algebras with residue field equal to $k$. Let $C$ be the strict henselization of $A \otimes _ k B$ at the maximal ideal $\mathfrak m_ A \otimes _ k B + A \otimes _ k \mathfrak m_ B$. Then the minimal primes of $C$ correspond $1$-to-$1$ to pairs of minimal primes of $A$ and $B$.

**Proof.**
First note that a minimal prime $\mathfrak r$ of $C$ maps to a minimal prime $\mathfrak p$ in $A$ and to a minimal prime $\mathfrak q$ of $B$ because the ring maps $A \to C$ and $B \to C$ are flat (by going down for flat ring map Algebra, Lemma 10.39.19). Hence it suffices to show that the strict henselization of $(A/\mathfrak p \otimes _ k B/\mathfrak q)_{ \mathfrak m_ A \otimes _ k B + A \otimes _ k \mathfrak m_ B}$ has a unique minimal prime ideal. By Algebra, Lemma 10.156.4 the rings $A/\mathfrak p$, $B/\mathfrak q$ are strictly henselian. Hence we may assume that $A$ and $B$ are strictly henselian local domains and our goal is to show that $C$ has a unique minimal prime. By Lemma 15.105.5 the integral closure $A'$ of $A$ in its fraction field is a normal local domain with residue field $k$. Similarly for the integral closure $B'$ of $B$ into its fraction field. By Algebra, Lemma 10.165.5 we see that $A' \otimes _ k B'$ is a normal ring. Hence its localization

is a normal local domain. Note that $A \otimes _ k B \to A' \otimes _ k B'$ is integral (hence gong up holds – Algebra, Lemma 10.36.22) and that $\mathfrak m_{A'} \otimes _ k B' + A' \otimes _ k \mathfrak m_{B'}$ is the unique maximal ideal of $A' \otimes _ k B'$ lying over $\mathfrak m_ A \otimes _ k B + A \otimes _ k \mathfrak m_ B$. Hence we see that

by Algebra, Lemma 10.41.11. It follows that

is integral. We conclude that $R$ is the integral closure of $(A \otimes _ k B)_{ \mathfrak m_ A \otimes _ k B + A \otimes _ k \mathfrak m_ B}$ in its fraction field, and by Lemma 15.105.5 once again we conclude that $C$ has a unique prime ideal. $\square$

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