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.

Generalization of [Expose I, Theorem 9.5 part (ii), SGA1]

**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. $\square$

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