The Stacks project

Lemma 58.10.3. Let $X$ be a scheme. Let $U \subset X$ be a dense open. Assume

  1. the underlying topological space of $X$ is Noetherian, and

  2. for every $x \in X \setminus U$ the punctured spectrum of the strict henselization of $\mathcal{O}_{X, x}$ is connected.

Then $\textit{FÉt}_ X \to \textit{Fét}_ U$ is fully faithful.

Proof. Let $Y_1, Y_2$ be finite étale over $X$ and let $\varphi : (Y_1)_ U \to (Y_2)_ U$ be a morphism over $U$. We have to show that $\varphi $ lifts uniquely to a morphism $Y_1 \to Y_2$ over $X$. Uniqueness follows from Lemma 58.10.1.

Let $x \in X \setminus U$ be a generic point of an irreducible component of $X \setminus U$. Set $V = U \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By our choice of $x$ this is the punctured spectrum of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$. By Lemma 58.10.2 we can extend the morphism $\varphi _ V : (Y_1)_ V \to (Y_2)_ V$ uniquely to a morphism $(Y_1)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})} \to (Y_2)_{\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})}$. By Limits, Lemma 32.20.3 we find an open $U \subset U'$ containing $x$ and an extension $\varphi ' : (Y_1)_{U'} \to (Y_2)_{U'}$ of $\varphi $. Since the underlying topological space of $X$ is Noetherian this finishes the proof by Noetherian induction on the complement of the open over which $\varphi $ is defined. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 58.10: Local connectedness

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Your email address will not be published.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 0BQF. Beware of the difference between the letter 'O' and the digit '0'.