Lemma 67.13.7. 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 65.4.3. Hence $Z_ x \times _ X \mathop{\mathrm{Spec}}(k) = \mathop{\mathrm{Spec}}(k)$ by Morphisms of Spaces, Lemma 66.10.8. Hence we get a monomorphism $\mathop{\mathrm{Spec}}(k) \to Z_ x$. This is an isomorphism by Lemma 67.13.4. The final statement follows from Lemma 67.11.1.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)