Lemma 68.14.4. 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 algebraic. Then the image $x$ of $\mathop{\mathrm{Spec}}(k) \to X$ is a closed point of $|X|$.

**Proof.**
Suppose that $x \leadsto x'$ for some $x' \in |X|$. Choose an étale morphism $U \to X$ where $U$ is a scheme and a point $u' \in U'$ mapping to $x'$. Choose a specialization $u \leadsto u'$ in $U$ with $u$ mapping to $x$ in $X$, see Lemma 68.12.2. Then $u$ is the image of a point $w$ of the scheme $W = \mathop{\mathrm{Spec}}(k) \times _ X U$. Since the projection $W \to \mathop{\mathrm{Spec}}(k)$ is étale we see that $\kappa (w) \supset k$ is finite. Hence $\kappa (w) \supset \kappa (s)$ is algebraic. Hence $\kappa (u) \supset \kappa (s)$ is algebraic. Thus $u$ is a closed point of $U$ by Morphisms, Lemma 29.20.2. Thus $u = u'$, whence $x = x'$.
$\square$

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