## 15.107 Miscellaneous on branches

Some results related to branches of local rings as defined in Section 15.106.

Lemma 15.107.1. Let $A$ and $B$ be domains and let $A \to B$ be a ring map. Assume $A \to B$ has additionally at least one of the following properties

1. it is the localization of an étale ring map,

2. it is flat and the localization of an unramified ring map,

3. it is flat and the localization of a quasi-finite ring map,

4. it is flat and the localization of an integral ring map,

5. it is flat and there are no nontrivial specializations between points of fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$,

6. $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ maps the generic point to the generic point and there are no nontrivial specializations between points of fibres, or

7. exactly one point of $\mathop{\mathrm{Spec}}(B)$ is mapped to the generic point of $\mathop{\mathrm{Spec}}(A)$.

Then $A \cap J$ is nonzero for every nonzero ideal $J$ of $B$.

Proof. Proof in case (7). Let $K$, resp. $L$ be the fraction field of $A$, resp. $B$. By Algebra, Lemma 10.30.7 we see that the unique point of $\mathop{\mathrm{Spec}}(B)$ which maps to the generic point $(0) \in \mathop{\mathrm{Spec}}(A)$ is $(0) \in \mathop{\mathrm{Spec}}(B)$. We conclude that $B \otimes _ A K$ is a ring with a unique prime ideal whose residue field is $L$ (in fact it is equal to $L$ but we do not need this). Choose $b \in J$ nonzero. Then $b$ maps to a unit of $L$. Hence $b$ maps to a unit of $B \otimes _ A K$ (Algebra, Lemma 10.19.2). Since $B \otimes _ A K = \mathop{\mathrm{colim}}\nolimits _{f \in A \setminus \{ 0\} } B_ f$ we see that $b$ maps to a unit of $B_ f$ for some $f \in A$ nonzero. This means that $b b' = f^ n$ for some $b' \in B$ and $n \geq 1$. Thus $f^ n \in A \cap J$ as desired.

In the rest of the proof, we show that each of the other assumptions imply (7). Under assumptions (1) – (5), the ring map $A \to B$ is flat and hence $A \to B$ is injective (since flat local homomorphisms are faithfully flat by Algebra, Lemma 10.39.17). Hence the generic point of $\mathop{\mathrm{Spec}}(B)$ maps to the generic point of $\mathop{\mathrm{Spec}}(A)$. Now, if there are no nontrivial specializations between points of fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$, then of course this generic point of $\mathop{\mathrm{Spec}}(B)$ has to be the unique point mapping to the generic point of $\mathop{\mathrm{Spec}}(A)$. So (6) implies (7). Finally, to finish we show that in cases (1) – (5) there are no nontrivial specializations between the points of fibres of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. Namely, see Algebra, Lemma 10.36.20 for the integral case, Algebra, Definition 10.122.3 for the quasi-finite case, and use that unramified and étale ring maps are quasi-finite (Algebra, Lemmas 10.151.6 and 10.143.6). $\square$

Lemma 15.107.2. Let $A \to B$ be a ring map. Let $\mathfrak q \subset B$ be a prime ideal lying over the prime $\mathfrak p \subset A$. Assume

1. $A$ is a domain,

2. $A_\mathfrak p$ is geometrically unibranch,

3. $A \to B$ is unramified at $\mathfrak q$, and

4. $A_\mathfrak p \to B_\mathfrak q$ is injective.

Then there exists a $g \in B$, $g \not\in \mathfrak q$ such that $B_ g$ is étale over $A$.

Proof. By Algebra, Proposition 10.152.1 after replacing $B$ by a principal localization, we can find a standard étale ring map $A \to B'$ and a surjection $B' \to B$. Denote $\mathfrak q' \subset B'$ the inverse image of $\mathfrak q$. We will show that $B' \to B$ is injective after possibly replacing $B'$ by a principal localization.

In this paragraph we reduce to the case that $B'$ is a domain. Since $A$ is a domain, the ring $B'$ is reduced, see Algebra, Lemma 15.42.1. Let $K$ be the fraction field of $A$. Then $B' \otimes _ A K$ is étale over a field, hence is a finite product of fields, see Algebra, Lemma 10.143.4. Since $A \to B'$ is étale (hence flat) the minimal primes of $B'$ are lie over $(0) \subset A$ (by going down for flat ring maps). We conclude that $B'$ has finitely many minimal primes, say $\mathfrak r_1, \ldots , \mathfrak r_ r \subset B'$. Since $A_\mathfrak p$ is geometrically unibranch and $A \to B'$ étale, the ring $B'_{\mathfrak q'}$ is a domain, see Lemmas 15.106.8 and 15.106.7. Hence $\mathfrak q' \supset \mathfrak r_ i$ for exactly one $i = i_0$. Choose $g' \in B'$, $g' \not\in \mathfrak r_{i_0}$ but $g' \in \mathfrak r_ i$ for $i \not= i_0$, see Algebra, Lemma 10.15.2. After replacing $B'$ and $B$ by $B'_{g'}$ and $B_{g'}$ we obtain that $B'$ is a domain.

Assume $B'$ is a domain, in particular $B' \subset B'_{\mathfrak q'}$. If $B' \to B$ is not injective, then $J = \mathop{\mathrm{Ker}}(B'_{\mathfrak q'} \to B_\mathfrak q)$ is nonzero. By Lemma 15.107.1 applied to $A_\mathfrak p \to B'_{\mathfrak q'}$ we find a nonzero element $a \in A_\mathfrak p$ mapping to zero in $B_\mathfrak q$ contradicting assumption (4). This finishes the proof. $\square$

Lemma 15.107.3. Let $(A, \mathfrak m)$ be a geometrically unibranch local domain. Let $A \to B$ be an injective local homomorphism of local rings, which is essentially of finite type. If $\mathfrak m B$ is the maximal ideal of $B$ and the induced extension of residue fields is separable, then $A \to B$ is the localization of an étale ring map.

Proof. We may write $B = C_\mathfrak q$ where $A \to C$ is a finite type ring map and $\mathfrak q \subset C$ is a prime ideal lying over $\mathfrak m$. By Algebra, Lemma 10.151.7 the ring map $A \to C$ is unramified at $\mathfrak q$. By Algebra, Proposition 10.152.1 after replacing $C$ by a principal localization, we can find a standard étale ring map $A \to C'$ and a surjection $C' \to C$. Denote $\mathfrak q' \subset C'$ the inverse image of $\mathfrak q$ and set $B' = C'_{\mathfrak q'}$. Then $B' \to B$ is surjective. It suffices to show that $B' \to B$ is also injective.

Since $A$ is a domain, the rings $C'$ and $B'$ are reduced, see Algebra, Lemma 15.42.1. Since $A$ is geometrically unibranch, the ring $B'$ is a domain, see by Lemmas 15.106.8 and 15.106.7. If $B' \to B$ is not injective, then $A \cap \mathop{\mathrm{Ker}}(B' \to B)$ is nonzero by Lemma 15.107.1 which contradicts the assumption that $A \to B$ is injective. $\square$

Lemma 15.107.4. 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 15.106.5 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

$R = (A' \otimes _ k B')_{ \mathfrak m_{A'} \otimes _ k B' + A' \otimes _ k \mathfrak m_{B'}}$

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

$R = (A' \otimes _ k B')_{ \mathfrak m_ A \otimes _ k B + A \otimes _ k \mathfrak m_ B}$

by Algebra, Lemma 10.41.11. It follows that

$(A \otimes _ k B)_{ \mathfrak m_ A \otimes _ k B + A \otimes _ k \mathfrak m_ B} \longrightarrow R$

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 15.106.5 once again we conclude that $C$ has a unique prime ideal. $\square$

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