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Set $k$ be an algebraically closed field. Let $X$ and $Y$ be two $k$-varieties.(one can assume they are projective) A question I have met many times is that: if $f$ is a morphism from $X$ to $Y$, then can we know $\operatorname{dim} f(X)\leq \operatorname{dim} X$?

I know the conclusion is right if $f$ is open. But most of the times we don't have this property.

If we know $X, Y$ are projective, then $f$ is proper, hence closed, but I can't see the dimension of $f(X)$.

Could you give some helpful properties?(references are also welcome) Thanks!

Richard
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  • If you take $Y$ to be the scheme theoretic image of $f$, then $f$ is dominant. The question about dimension can be handled locally, so you may as well consider a finite-type extension of $k$-algebras. Here, you might apply Atiyah-Macdonald Ch. 5 Ex. 20 (which is based on Noether Normalization). – yearning4pi Dec 18 '21 at 03:40
  • https://math.stackexchange.com/questions/2890249/surjective-closed-morphism-of-schemes-induces-inequality-of-dimensions – Rishi Sonthalia Aug 02 '22 at 20:50
  • https://math.stackexchange.com/questions/95670/surjective-morphism-of-affine-varieties-and-dimension – Rishi Sonthalia Aug 02 '22 at 20:50

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