The Stacks project

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). $\square$

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. $\square$

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. $\square$

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

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