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. $\square$

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