Lemma 68.13.7. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The residual space $Z_ x$ of $X$ at $x$ is isomorphic to the spectrum of a field if and only if $x$ can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field. If $X$ is decent, this holds for all $x \in |X|$.

**Proof.**
Since $Z_ x \to X$ is a monomorphism, if $Z_ x = \mathop{\mathrm{Spec}}(k)$ for some field $k$, then $x$ is represented by the monomorphism $\mathop{\mathrm{Spec}}(k) = Z_ x \to X$. Conversely, if $\mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism which represents $x$, then $Z_ x \times _ X \mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k)$ is a monomorphism whose source is nonempty by Properties of Spaces, Lemma 66.4.3. Hence $Z_ x \times _ X \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(k)$ by Morphisms of Spaces, Lemma 67.10.8. Hence we get a monomorphism $\mathop{\mathrm{Spec}}(k) \to Z_ x$. This is an isomorphism by Lemma 68.13.4. The final statement follows from Lemma 68.11.1.
$\square$

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