Lemma 15.105.8. Let $A \to B$ be a local homomorphism of local rings which is the localization of a smooth ring map.

1. The number of geometric branches of $A$ is equal to the number of geometric branches of $B$.

2. If $A \to B$ induces a purely inseparable extension of residue fields, then the number of branches of $A$ is the number of branches of $B$.

Proof. We will use that smooth ring maps are flat (Algebra, Lemma 10.137.10), that localizations are flat (Algebra, Lemma 10.39.18), that compositions of flat ring maps are flat (Algebra, Lemma 10.39.4), that base change of a flat ring map is flat (Algebra, Lemma 10.39.7), that flat local homomorphisms are faithfully flat (Algebra, Lemma 10.39.17), that (strict) henselization is flat (Lemma 15.45.1), and Going down for flat ring maps (Algebra, Lemma 10.39.19).

Proof of (2). Let $A^ h$, $B^ h$ be the henselizations of $A$, $B$. Then $B^ h$ is the henselization of $A^ h \otimes _ A B$ at the unique maximal ideal lying over $\mathfrak m_ B$, see Algebra, Lemma 10.155.8. Thus we may and do assume $A$ is henselian. Since $A \to B \to B^ h$ is flat, every minimal prime of $B^ h$ lies over a minimal prime of $A$ and since $A \to B^ h$ is faithfully flat, every minimal prime of $A$ does lie under a minimal prime of $B^ h$; in both cases use going down for flat ring maps. Therefore it suffices to show that given a minimal prime $\mathfrak p \subset A$, there is at most one minimal prime of $B^ h$ lying over $\mathfrak p$. After replacing $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ we may assume that $A$ is a domain; the $A$ is still henselian by Algebra, Lemma 10.156.2. By Lemma 15.105.3 we see that the integral closure $A'$ of $A$ in its field of fractions is a local domain. Of course $A'$ is a normal domain. By Algebra, Lemma 10.163.9 we see that $A' \otimes _ A B^ h$ is a normal ring (the lemma just gives it for $A' \otimes _ A B$, to go up to $A' \otimes _ A B^ h$ use that $B^ h$ is a colimit of étale $B$-algebras and use Algebra, Lemma 10.37.17). By Algebra, Lemma 10.156.5 we see that $A' \otimes _ A B^ h$ is local (this is where we use the assumption on the residue fields of $A$ and $B$). Hence $A' \otimes _ A B^ h$ is a local normal ring, hence a local domain. Since $B^ h \subset A' \otimes _ A B^ h$ by flatness of $A \to B^ h$ we conclude that $B^ h$ is a domain as desired.

Proof of (1). Let $A^{sh}$, $B^{sh}$ be strict henselizations of $A$, $B$. Then $B^{sh}$ is a strict henselization of $A^ h \otimes _ A B$ at a maximal ideal lying over $\mathfrak m_ B$ and $\mathfrak m_{A^ h}$, see Algebra, Lemma 10.155.12. Thus we may and do assume $A$ is strictly henselian. Since $A \to B \to B^{sh}$ is flat, every minimal prime of $B^{sh}$ lies over a minimal prime of $A$ and since $A \to B^{sh}$ is faithfully flat, every minimal prime of $A$ does lie under a minimal prime of $B^{sh}$; in both cases use going down for flat ring maps. Therefore it suffices to show that given a minimal prime $\mathfrak p \subset A$, there is at most one minimal prime of $B^{sh}$ lying over $\mathfrak p$. After replacing $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ we may assume that $A$ is a domain; then $A$ is still strictly henselian by Algebra, Lemma 10.156.4. By Lemma 15.105.5 we see that the integral closure $A'$ of $A$ in its field of fractions is a local domain whose residue field is a purely inseparable extension of the residue field of $A$. Of course $A'$ is a normal domain. By Algebra, Lemma 10.163.9 we see that $A' \otimes _ A B^{sh}$ is a normal ring (the lemma just gives it for $A' \otimes _ A B$, to go up to $A' \otimes _ A B^{sh}$ use that $B^{sh}$ is a colimit of étale $B$-algebras and use Algebra, Lemma 10.37.17). By Algebra, Lemma 10.156.5 we see that $A' \otimes _ A B^{sh}$ is local (since $A \subset A'$ induces a purely inseparable residue field extension). Hence $A' \otimes _ A B^{sh}$ is a local normal ring, hence a local domain. Since $B^{sh} \subset A' \otimes _ A B^{sh}$ by flatness of $A \to B^{sh}$ we conclude that $B^{sh}$ is a domain as desired. $\square$

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