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I have the following proplem: Let $k$ be an algebraically closed field and let $X,Y$ be schemes of finte type over $k$.

Now let $f:X\to Y$ be a morphism of schemes that is not surjective.

Question: Is there a closed point in $Y\setminus f(X)$?

What I tried so far: I assumed $X,Y$ are affine and tried to play a little bit on the ring level using that the coordinate rings are jacobian rings. But I haven't got very far yet. Any help would be appreciated.

Maik Pickl
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1 Answers1

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By Chevalley's theorem, $f$ is surjective if and only if $f$ is surjective on closed points (see there).

Assume all closed points of $Y$ are in $f(X)$. Let $y \in Y$ be a closed point. The scheme-theoretic fiber $X_y $ of $y$ is nonempty and is a closed subscheme of $X$. Thus $X_y$ contains a closed point $x$ and by definition we have $f(x)=y$. But $x$ is also a closed point of $X$. Consequently, $f$ is surjective on closed points.

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BrL
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  • Thanks for the link! I'm not really sure what you are proving, though. You prove something like "If all the closed point are in the image, then $f$ is surjective on closed points". But this is rather trivial. What I would need is "If $f$ is surjective on closed points, then it is surjective." – Maik Pickl Jun 07 '16 at 15:32
  • You are right, I was not precise enough. I wrote that if every closed point is in the image then every closed point is the image of a closed point (which is stronger). This is what I called "$f$ is surjective on closed points". Then the link precicely says that surjectivity and surjectivity on closed points are equivalent for morphisms between schemes of finite type over a field. – BrL Jun 07 '16 at 15:37