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Tag 081I

Chapter 28: Morphisms of Schemes > Section 28.24: Flat morphisms

Lemma 28.24.15. Let $f : X \to Y$ be a flat morphism of schemes. Let $g : V \to Y$ be a quasi-compact morphism of schemes. Let $Z \subset Y$ be the scheme theoretic image of $g$ and let $Z' \subset X$ be the scheme theoretic image of the base change $V \times_Y X \to X$. Then $Z' = f^{-1}Z$.

Proof. Recall that $Z$ is cut out by $\mathcal{I} = \mathop{\rm Ker}(\mathcal{O}_Y \to g_*\mathcal{O}_V)$ and $Z'$ is cut out by $\mathcal{I}' = \mathop{\rm Ker}(\mathcal{O}_X \to (V \times_Y X \to X)_*\mathcal{O}_{V \times_Y X})$, see Lemma 28.6.3. Hence the question is local on $X$ and $Y$ and we may assume $X$ and $Y$ affine. Note that we may replace $V$ by $\coprod V_i$ where $V = V_1 \cup \ldots \cup V_n$ is a finite affine open covering. Hence we may assume $g$ is affine. In this case $(V \times_Y X \to X)_*\mathcal{O}_{V \times_Y X}$ is the pullback of $g_*\mathcal{O}_V$ by $f$. Since $f$ is flat we conclude that $f^*\mathcal{I} = \mathcal{I}'$ and the lemma holds. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4497–4503 (see updates for more information).

    \begin{lemma}
    \label{lemma-flat-base-change-scheme-theoretic-image}
    Let $f : X \to Y$ be a flat morphism of schemes. Let $g : V \to Y$ be a
    quasi-compact morphism of schemes. Let $Z \subset Y$ be the scheme theoretic
    image of $g$ and let $Z' \subset X$ be the scheme theoretic image of the
    base change $V \times_Y X \to X$. Then $Z' = f^{-1}Z$.
    \end{lemma}
    
    \begin{proof}
    Recall that $Z$ is cut out by
    $\mathcal{I} = \Ker(\mathcal{O}_Y \to g_*\mathcal{O}_V)$
    and $Z'$ is cut out by
    $\mathcal{I}' = \Ker(\mathcal{O}_X \to
    (V \times_Y X \to X)_*\mathcal{O}_{V \times_Y X})$, see
    Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}.
    Hence the question is local on $X$ and $Y$ and we may assume $X$ and $Y$
    affine. Note that we may replace $V$ by $\coprod V_i$ where
    $V = V_1 \cup \ldots \cup V_n$ is a finite affine open covering.
    Hence we may assume $g$ is affine. In this case
    $(V \times_Y X \to X)_*\mathcal{O}_{V \times_Y X}$ is the pullback
    of $g_*\mathcal{O}_V$ by $f$. Since $f$ is flat we conclude that
    $f^*\mathcal{I} = \mathcal{I}'$ and the lemma holds.
    \end{proof}

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