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This question came from a proof in Algebraic Geometry by Hartshorne (Chapt3, Corollary 9.6)

To be precise, Let $f:X \to Y$ be a flat morphism of schemes of finite type over a field $k$. Then is it true that the image of closed point of $X$ is also a closed point of $Y$?

Of course, one can restrict to affine schemes, say $X=\rm{Spec}A, Y=\rm{Spec}B$, and $\phi :B \to A$ is flat. Is there any lying over (or going up) property of flat map as it is in the case of integral extension? (If it has such property, one can prove the claim without difficulty).

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  • $\begingroup$ If $f$ is faithfully flat then it is even surjective on closed points, but I don't know about $f$ flat... $\endgroup$ – Zhen Lin Mar 5 '12 at 7:51
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This is just a consequence of Nullstellensatz (the flatness is useless). If $x\in X$ is a closed point, then $k(x)$ is a finite extension of $k$ and so is $k(f(x))$ (being a subextension of $k(x)$). This is enough to show that $f(x)$ is a closed point.

If $X, Y$ are not finite type over a field, this is false even if $f$ is finite type (and faithfully flat if you like): consider a DVR $R$ with uniformizing element $\pi$, and let $f : X=\mathbb A^1_R \to Y=\mathrm{Spec}(R)$ be the canonical morphism. It is finite type and faithfully flat. The polynomial $1-\pi T\in R[T]$ defines a closed point of $X$ whose image by $f$ is the generic point of $Y$.

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    $\begingroup$ Great! Thank you so much, I cannot appreciate more! $\endgroup$ – Li Zhan Mar 5 '12 at 19:23

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