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. $\square$

There are also:

• 2 comment(s) on Section 64.12: Reduced spaces

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).