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. $\square$

Comment #651 by Anfang on

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 #4900 by Lukas Sauer on

Typo. For any affine open $U \sub X, instead of For any affine opens. Comment #4903 by Lukas Sauer on Ah. "opens" refers to both $U$ and $V$. Excuse me. Comment #4904 by Lukas Sauer on 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. ## Post a comment Your email address will not be published.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).

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