The Stacks project

Lemma 85.23.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume

  1. $f$ is representable by algebraic spaces,

  2. $f$ is a monomorphism,

  3. the inclusion $Y_{red} \to Y$ factors through $f$, and

  4. $f$ is locally of finite type or $Y$ is locally Noetherian.

Then $f$ is a closed immersion.

Proof. Assumptions (2) and (3) imply that $X_{red} = X \times _ Y Y_{red} = Y_{red}$. We will use this without further mention.

If $Y' \to Y$ is an étale morphism of formal algebraic spaces over $S$, then the base change $f' : X \times _ Y Y' \to Y'$ satisfies conditions (1) – (4). Hence by Lemma 85.23.4 we may assume $Y$ is an affine formal algebraic space.

Say $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda $ as in Definition 85.5.1. Then $X_\lambda = X \times _ Y Y_\lambda $ is an algebraic space endowed with a monomorphism $f_\lambda : X_\lambda \to Y_\lambda $ which induces an isomorphism $X_{\lambda , red} \to Y_{\lambda , red}$. Thus $X_\lambda $ is an affine scheme by Limits of Spaces, Proposition 68.15.2 (as $X_{\lambda , red} \to X_\lambda $ is surjective and integral). To finish the proof it suffices to show that $X_\lambda \to Y_\lambda $ is a closed immersion which we will do in the next paragraph.

Let $X \to Y$ be a monomorphism of affine schemes such that $X_{red} = X \times _ Y Y_{red} = Y_{red}$. In general, this does not imply that $X \to Y$ is a closed immersion, see Examples, Section 108.34. However, under our assumption (4) we know that in the previous parapgrah either $X_\lambda \to Y_\lambda $ is of finite type or $Y_\lambda $ is Noetherian. This means that $X \to Y$ corresponds to a ring map $R \to A$ such that $R/I \to A/IA$ is an isomorphism where $I \subset R$ is the nil radical (ie., the maximal locally nilpotent ideal of $R$) and either $R \to A$ is of finite type or $R$ is Noetherian. In the first case $R \to A$ is surjective by Algebra, Lemma 10.126.9 and in the second case $I$ is finitely generated, hence nilpotent, hence $R \to A$ is surjective by Nakayama's lemma, see Algebra, Lemma 10.20.1 part (11). $\square$

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Your email address will not be published.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