The Stacks project

Lemma 67.13.10. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$ be a point. Assume

  1. $|f|(|Y|)$ is contained in $\{ x\} \subset |X|$,

  2. $Y$ is reduced, and

  3. $x$ can be represented by a quasi-compact monomorphism $x : \mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field (for example if $X$ is decent).

Then $f$ factors through the residual space $Z_ x = \mathop{\mathrm{Spec}}(k)$ of $X$ at $x$.

Proof. By Lemma 67.13.7 we have $Z_ x = \mathop{\mathrm{Spec}}(k)$.

Preliminary remark: since $\mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism, it suffices to find a surjective ├ętale morphism $Y' \to Y$ such that $Y' \to X$ factors through $Z_ x$. A remark here is that $Y'$ is reduced as well.

After replacing $X$ by a quasi-compact open neighbourhood of $x$, we may assume $X$ quasi-compact. By Lemma 67.8.3, $x$ is a point of $T \subset U \subset X$ where $T \to U$ (resp. $U \to X$) is a closed (resp. open) immersion, and $T$ is a scheme. By Properties of Spaces, Lemma 65.4.9, $f$ factors through $U$, so we may assume $U = X$. Then $f$ factors through $T$ because $Y$ is reduced, see Properties of Spaces, Lemma 65.12.4. So we may assume that $X = T$ is a scheme. By our preliminary remark we may assume $Y$ is a scheme too. This reduces us to Morphisms, Lemma 29.57.1. $\square$


Comments (3)

Comment #7760 by Laurent Moret-Bailly on

Apologies: when I wrote my previous comment, I had missed this section. The example in my comment is of course a "counterexample" to this lemma in the non-decent case.

Comment #7761 by Laurent Moret-Bailly on

Here is a different proof where we only need to be a "decent point" (rather than being decent): We may assume quasi-compact. By 67.8.3, is a point of where (resp. ) is a closed (resp. open) immersion, and is a scheme. By 65.4.9, factors through , so we may assume . Then factors through because is reduced (consider the ideal sheaf of ). So we may assume that is a scheme.


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