\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

The Stacks project

Lemma 60.14.5. 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 $\kappa (s) \subset k$ 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 60.14.4 the image point $x \in |X|$ is closed. Let $Z \subset X$ be the reduced closed subspace with $|Z| = \{ x\} $ (Properties of Spaces, Lemma 58.12.4). Note that $Z$ is a decent algebraic space by Lemma 60.6.5. By Lemma 60.14.2 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 59.45.4). If $k = \kappa (s)$, then $\mathop{\mathrm{Spec}}(k) = Z$ and $\mathop{\mathrm{Spec}}(k) \to X$ is a closed immersion. $\square$


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