## 70.16 Spaces smooth over fields

This section is the analogue of Varieties, Section 33.25.

Lemma 70.16.1. Let $k$ be a field. Let $X$ be an algebraic space smooth over $k$. Then $X$ is a regular algebraic space.

Proof. Choose a scheme $U$ and a surjective étale morphism $U \to X$. The morphism $U \to \mathop{\mathrm{Spec}}(k)$ is smooth as a composition of an étale (hence smooth) morphism and a smooth morphism (see Morphisms of Spaces, Lemmas 65.39.6 and 65.37.2). Hence $U$ is regular by Varieties, Lemma 33.25.3. By Properties of Spaces, Definition 64.7.2 this means that $X$ is regular. $\square$

Lemma 70.16.2. Let $k$ be a field. Let $X$ be an algebraic space smooth over $\mathop{\mathrm{Spec}}(k)$. The set of $x \in |X|$ which are image of morphisms $\mathop{\mathrm{Spec}}(k') \to X$ with $k' \supset k$ finite separable is dense in $|X|$.

Proof. Choose a scheme $U$ and a surjective étale morphism $U \to X$. The morphism $U \to \mathop{\mathrm{Spec}}(k)$ is smooth as a composition of an étale (hence smooth) morphism and a smooth morphism (see Morphisms of Spaces, Lemmas 65.39.6 and 65.37.2). Hence we can apply Varieties, Lemma 33.25.6 to see that the closed points of $U$ whose residue fields are finite separable over $k$ are dense. This implies the lemma by our definition of the topology on $|X|$. $\square$

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).