[Expose VIII, Corollaire 4.3, SGA1] and [IV, Corollaire 2.3.12, EGA]

Lemma 29.25.12. Let $f : X \to Y$ be a quasi-compact, surjective, flat morphism. A subset $T \subset Y$ is open (resp. closed) if and only $f^{-1}(T)$ is open (resp. closed). In other words, $f$ is a submersive morphism.

Proof. The question is local on $Y$, hence we may assume that $Y$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ as a finite union of affine opens. Then $f' : X' = X_1 \amalg \ldots \amalg X_ n \to Y$ is a surjective flat morphism of affine schemes. Note that for $T \subset Y$ we have $(f')^{-1}(T) = f^{-1}(T) \cap X_1 \amalg \ldots \amalg f^{-1}(T) \cap X_ n$. Hence, $f^{-1}(T)$ is open if and only if $(f')^{-1}(T)$ is open. Thus we may assume both $X$ and $Y$ are affine.

Let $f : \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be a surjective morphism of affine schemes corresponding to a flat ring map $A \to B$. Suppose that $f^{-1}(T)$ is closed, say $f^{-1}(T) = V(J)$ for $J \subset B$ an ideal. Then $T = f(f^{-1}(T)) = f(V(J))$ is the image of $\mathop{\mathrm{Spec}}(B/J) \to \mathop{\mathrm{Spec}}(A)$ (here we use that $f$ is surjective). On the other hand, generalizations lift along $f$ (Lemma 29.25.9). Hence by Topology, Lemma 5.19.6 we see that $Y \setminus T = f(X \setminus f^{-1}(T))$ is stable under generalization. Hence $T$ is stable under specialization (Topology, Lemma 5.19.2). Thus $T$ is closed by Algebra, Lemma 10.41.5. $\square$

## Comments (4)

Comment #2873 by Ko Aoki on

Typo in the proof: The orientation of the morphism between $\mathop{\rm Spec} A$ and $\mathop{\rm Spec} B$ is reversed halfway.

Comment #3546 by on

Some references: [SGA1, Exposé VIII, Corollaire 4.3] and [EGAIV$_2$, Corollaire 2.3.12].

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• 4 comment(s) on Section 29.25: Flat morphisms

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