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. $X$ is decent.

Then $f$ factors through the residual space $Z_ x$ of $X$ at $x$. In other words, if $x : \mathop{\mathrm{Spec}}(k) \to X$ is a monomorphism from the spectrum of a field representing $x$, then $f$ factors through $x$.

Proof. By Lemma 67.13.7 we can represent $x$ by a monomorphism $x : \mathop{\mathrm{Spec}}(k) \to X$ as in the last sentence and then $Z_ x = \mathop{\mathrm{Spec}}(k)$.

Preliminary remark: since $Z_ x \to X$ is a monomorphism, it suffices to find a surjective étale morphism $Y' \to Y$ such that $Y' \to X$ factors through $Z_ x$. A remark here is that $Y'$ is reduced as well.

Let $Z \subset X$ be the reduced closed subspace with $|Z| = \overline{\{ x\} }$ (Properties of Spaces, Lemma 65.12.3). Note that $Z$ is a decent algebraic space by Lemma 67.6.5. By Properties of Spaces, Lemma 65.12.4 we see that $f$ factors through $Z$. Thus we may replace $X$ by $Z$ and assume $x$ is a generic point of $|X|$. Then $x$ is in the schematic locus of $X$ by Theorem 67.10.2 and hence we reduce to the case where $X$ is a scheme. By our preliminary remark we may assume $Y$ is a scheme too. This reduces us to Morphisms, Lemma 29.57.1. $\square$

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 $x$ to be a "decent point" (rather than $X$ being decent): We may assume $X$ quasi-compact. By 67.8.3, $x$ is a point of $T\subset U\subset X$ where $T\subset U$ (resp. $U\subset X$) is a closed (resp. open) immersion, and $T$ is a scheme. By 65.4.9, $f$ factors through $U$, so we may assume $U=X$. Then $f$ factors through $T$ because $Y$ is reduced (consider the ideal sheaf of $T$). So we may assume that $X=T$ is a scheme.

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