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Tag 0356

Chapter 25: Schemes > Section 25.12: Reduced schemes

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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