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 $X$ 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.

## Comments (1)

Comment #8585 by Dan B on

The phrase "acting on X by translations" should say "acting on Y by translations".

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