Lemma 64.12.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. There exists a unique closed subspace $Z \subset X$ with the following properties: (a) we have $|Z| = T$, and (b) $Z$ is reduced.

**Proof.**
Let $U \to X$ be a surjective étale morphism, where $U$ is a scheme. Set $R = U \times _ X U$, so that $X = U/R$, see Spaces, Lemma 63.9.1. As usual we denote $s, t : R \to U$ the two projection morphisms. By Lemma 64.4.5 we see that $T$ corresponds to a closed subset $T' \subset |U|$ such that $s^{-1}(T') = t^{-1}(T')$. Let $Z' \subset U$ be the reduced induced scheme structure on $T'$. In this case the fibre products $Z' \times _{U, t} R$ and $Z' \times _{U, s} R$ are closed subschemes of $R$ (Schemes, Lemma 26.18.2) which are étale over $Z'$ (Morphisms, Lemma 29.35.4), and hence reduced (because being reduced is local in the étale topology, see Remark 64.7.3). Since they have the same underlying topological space (see above) we conclude that $Z' \times _{U, t} R = Z' \times _{U, s} R$. Thus we can apply Lemma 64.12.2 to obtain a closed subspace $Z \subset X$ whose pullback to $U$ is $Z'$. By construction $|Z| = T$ and $Z$ is reduced. This proves existence. We omit the proof of uniqueness.
$\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: