# The Stacks Project

## Tag 0356

Lemma 25.12.6. 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 $I \subset \mathcal{O}_X$ be the ideal sheaf of $Z$. For any affine opens $U \subset X$, $\mathop{\rm Spec}(B) = V \subset Y$ 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 25.12.2, and Algebra, Lemma 10.16.2). Hence $f$ factors through $Z$ by Lemma 25.4.6. $\square$

The code snippet corresponding to this tag is a part of the file schemes.tex and is located in lines 2112–2120 (see updates for more information).

\begin{lemma}
\label{lemma-map-into-reduction}
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 $Y \to X_{red} \to X$.
\end{lemma}

\begin{proof}
Assume $f(Y) \subset Z$ (set theoretically).
Let $I \subset \mathcal{O}_X$ be the ideal sheaf of $Z$.
For any affine opens $U \subset X$, $\Spec(B) = V \subset Y$
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 \ref{lemma-reduced}, and
Algebra, Lemma \ref{algebra-lemma-Zariski-topology}).
Hence $f$ factors through
$Z$ by Lemma \ref{lemma-characterize-closed-subspace}.
\end{proof}

## Comments (2)

Comment #651 by Anfang on June 3, 2014 a 10:35 am UTC

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 (site) on June 4, 2014 a 8:50 pm UTC

Thanks! This is fixed here.

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