Lemma 66.14.2. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.

If $|X|$ is a singleton then $X$ is a scheme.

If $|X|$ is a singleton and $X$ is reduced, then $X \cong \mathop{\mathrm{Spec}}(k)$ for some field $k$.

Lemma 66.14.2. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$.

If $|X|$ is a singleton then $X$ is a scheme.

If $|X|$ is a singleton and $X$ is reduced, then $X \cong \mathop{\mathrm{Spec}}(k)$ for some field $k$.

**Proof.**
Assume $|X|$ is a singleton. It follows immediately from Theorem 66.10.2 that $X$ is a scheme, but we can also argue directly as follows. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$. Set $R = U \times _ X U$. Then $U$ and $R$ have finitely many points by Lemma 66.4.5 (and the definition of a decent space). All of these points are closed in $U$ and $R$ by Lemma 66.12.1. It follows that $U$ and $R$ are affine schemes. We may shrink $U$ to a singleton space. Then $U$ is the spectrum of a henselian local ring, see Algebra, Lemma 10.152.10. The projections $R \to U$ are étale, hence finite étale because $U$ is the spectrum of a $0$-dimensional henselian local ring, see Algebra, Lemma 10.152.3. It follows that $X$ is a scheme by Groupoids, Proposition 39.23.9.

Part (2) follows from (1) and the fact that a reduced singleton scheme is the spectrum of a field. $\square$

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