The Stacks project

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

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 ├ętale 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 ├ętale 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GS6. Beware of the difference between the letter 'O' and the digit '0'.