Lemma 26.12.7 (0356)—The Stacks project

Lemma 26.12.7. Let $X$ be a scheme. Let $Z \subset X$ be a closed subscheme. Let $Y$ be a reduced scheme. A morphism $f : Y \to X$ factors through $Z$ if and only if $f(Y) \subset Z$ (set theoretically). In particular, any morphism $Y \to X$ factors as $Y \to X_{red} \to X$.

Proof. Assume $f(Y) \subset Z$ (set theoretically). Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal sheaf of $Z$. For any affine opens $U \subset X$, $\mathop{\mathrm{Spec}}(B) = V \subset Y$ with $f(V) \subset U$ and any $g \in \mathcal{I}(U)$ the pullback $b = f^\sharp (g) \in \Gamma (V, \mathcal{O}_ Y) = B$ maps to zero in the residue field of any $y \in V$. In other words $b \in \bigcap _{\mathfrak p \subset B} \mathfrak p$. This implies $b = 0$ as $B$ is reduced (Lemma 26.12.2, and Algebra, Lemma 10.17.2). Hence $f$ factors through $Z$ by Lemma 26.4.6. $\square$

Comments (8)

Comment #651 by Anfang on June 03, 2014 at 10:35

Typo. In the proof, it should be that $b \in \bigcap_{\mathfrak p \subset B} \mathfrak p$ , not $g \in \bigcap_{\mathfrak p \subset B} \mathfrak p$ .

Comment #662 by Johan on June 04, 2014 at 20:50

Thanks! This is fixed here.

Comment #4900 by Lukas Sauer on January 28, 2020 at 08:56

Typo. For any affine open $U \sub X, instead of For any affine opens.

Comment #4903 by Lukas Sauer on January 29, 2020 at 04:55

Ah. "opens" refers to both $U$ and $V$ . Excuse me.

Comment #4904 by Lukas Sauer on January 29, 2020 at 05:09

However, it should be $\mathcal I \subset \mathcal O_X$ instead of $I \subset \mathcal O_X$ for the sake of consistent notation within the proof.

Comment #5177 by Johan on June 09, 2020 at 13:38

Thanks and fixed here.

Comment #8526 by Elías Guisado on June 26, 2023 at 11:35

In the statement, I think we should say "let $Z\subset X$ be a reduced closed subscheme" or something like that.

Comment #8527 by Johan on June 26, 2023 at 13:27

No, it is fine as is.

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2 comment(s) on Section 26.12: Reduced schemes

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