Lemma 67.14.5. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Consider a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(k) \ar[rr] \ar[rd] & & X \ar[ld] \\ & S }$

Assume that the image point $s \in S$ of $\mathop{\mathrm{Spec}}(k) \to S$ is a closed point and that the field extension $k/\kappa (s)$ is finite. Then $\mathop{\mathrm{Spec}}(k) \to X$ is finite morphism. If $\kappa (s) = k$ then $\mathop{\mathrm{Spec}}(k) \to X$ is closed immersion.

Proof. By Lemma 67.14.4 the image point $x \in |X|$ is closed. Let $Z \subset X$ be the reduced closed subspace with $|Z| = \{ x\}$ (Properties of Spaces, Lemma 65.12.3). Note that $Z$ is a decent algebraic space by Lemma 67.6.5. By Lemma 67.14.2 we see that $Z = \mathop{\mathrm{Spec}}(k')$ for some field $k'$. Of course $k \supset k' \supset \kappa (s)$. Then $\mathop{\mathrm{Spec}}(k) \to Z$ is a finite morphism of schemes and $Z \to X$ is a finite morphism as it is a closed immersion. Hence $\mathop{\mathrm{Spec}}(k) \to X$ is finite (Morphisms of Spaces, Lemma 66.45.4). If $k = \kappa (s)$, then $\mathop{\mathrm{Spec}}(k) = Z$ and $\mathop{\mathrm{Spec}}(k) \to X$ is a closed immersion. $\square$

Comment #7752 by Laurent Moret-Bailly on

Typos in statement: "is finite morphism", "is closed immersion".

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