The Stacks project

Lemma 86.18.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of locally Noetherian formal algebraic spaces which is representable by algebraic spaces, étale, and surjective. Then $f$ is rig-surjective.

Proof. Let $p : \text{Spf}(R) \to Y$ be an adic morphism where $R$ is a complete discrete valuation ring. Let $Z = \text{Spf}(R) \times _ Y X$. Then $Z \to \text{Spf}(R)$ is representable by algebraic spaces, étale, and surjective. Hence $Z$ is nonempty. Pick a nonempty affine formal algebraic space $V$ and an étale morphism $V \to Z$ (possible by our definitions). Then $V \to \text{Spf}(R)$ corresponds to $R \to A^\wedge $ where $R \to A$ is an étale ring map, see Formal Spaces, Lemma 85.15.13. Since $A^\wedge \not= 0$ (as $V \not= \emptyset $) we can find a maximal ideal $\mathfrak m$ of $A$ lying over $\mathfrak m_ R$. Then $A_\mathfrak m$ is a discrete valuation ring (More on Algebra, Lemma 15.44.4). Then $R' = A_\mathfrak m^\wedge $ is a complete discrete valuation ring (More on Algebra, Lemma 15.43.5). Applying Formal Spaces, Lemma 85.5.10. we find the desired morphism $\text{Spf}(R') \to V \to Z \to X$. $\square$


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Your email address will not be published.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