
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 $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ Y \to g_*\mathcal{O}_ V)$ and $Z'$ is cut out by $\mathcal{I}' = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to (V \times _ Y X \to X)_*\mathcal{O}_{V \times _ Y X})$, 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. $\square$

Comment #3833 by slogan_bot on

Suggested slogan: "Schme theoretic image of a quasi-compact morphism commutes with flat base change"

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