Example 68.13.11. Here is a counter example to Lemmas 68.13.9 and 68.13.10 in case $X$ is neither locally Noetherian nor decent. Let $k$ be a field. Let $G$ be an infinite profinite group. Let $Y$ be $G$ viewed as a zero-dimensional affine $k$-group scheme, i.e., $Y = \mathop{\mathrm{Spec}}(\text{locally constant maps } G \to k)$. Let $\Gamma $ be $G$ viewed as a discrete $k$-group scheme, acting on $Y$ by translations. Put $X = Y/\Gamma $. This is a one-point algebraic space, with projection $q : Y \to X$. Let $e \in G$ be the origin (any element would do), and view it as a $k$-point of $Y$. We get a $k$-point $x : \mathop{\mathrm{Spec}}(k) \to X$ which is a monomorphism since it is a section of $X \to \mathop{\mathrm{Spec}}(k)$. We claim that (although $Y$ is affine and reduced and $|X| = \{ x\} $), the morphism $q$ does not factor through any morphism $\mathop{\mathrm{Spec}}(K) \to X$, where $K$ is a field. Otherwise it would factor through $x$ by Properties of Spaces, Lemma 66.4.11. Now the pullback of $q$ by $x$ is $\Gamma \to \mathop{\mathrm{Spec}}(k)$, with the projection $\Gamma \to Y$ being the orbit map $g \mapsto g \cdot e$. The latter has no section, whence the claim.
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8585 by Dan B on
Comment #9161 by Stacks project on