Lemma 67.14.6. Let $S$ be a scheme. Suppose $X$ is a decent algebraic space over $S$. Let $x \in |X|$ be a closed point. Then $x$ can be represented by a closed immersion $i : \mathop{\mathrm{Spec}}(k) \to X$ from the spectrum of a field.

Proof. We know that $x$ can be represented by a quasi-compact monomorphism $i : \mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field (Definition 67.6.1). Let $U \to X$ be an étale morphism where $U$ is an affine scheme. As $x$ is closed and $X$ decent, the fibre $F$ of $|U| \to |X|$ over $x$ consists of closed points (Lemma 67.12.1). As $i$ is a monomorphism, so is $U_ k = U \times _ X \mathop{\mathrm{Spec}}(k) \to U$. In particular, the map $|U_ k| \to F$ is injective. Since $U_ k$ is quasi-compact and étale over a field, we see that $U_ k$ is a finite disjoint union of spectra of fields (Remark 67.4.1). Say $U_ k = \mathop{\mathrm{Spec}}(k_1) \amalg \ldots \amalg \mathop{\mathrm{Spec}}(k_ r)$. Since $\mathop{\mathrm{Spec}}(k_ i) \to U$ is a monomorphism, we see that its image $u_ i$ has residue field $\kappa (u_ i) = k_ i$. Since $u_ i \in F$ is a closed point we conclude the morphism $\mathop{\mathrm{Spec}}(k_ i) \to U$ is a closed immersion. As the $u_ i$ are pairwise distinct, $U_ k \to U$ is a closed immersion. Hence $i$ is a closed immersion (Morphisms of Spaces, Lemma 66.12.1). This finishes the proof. $\square$

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