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If $f:X\to Y$ is a dominant(i.e.$f(X)$ is dense in $Y$) regular map of affine varieties, then $f$ is called a finite map if $k[X]$ is integral over $f^*k[Y]$.

My question is: if $Z\subset X$ is a closed subset of $X$, then how to show the restiction $f|_Z: Z\to \overline{f(Z)}$ is still a finite map?

(Note: This result is supposed to prove the fact that "a finite map takes closed sets to closed sets"(cf.Page60, Basic Algebraic Geometry 1, by Shafarevich), so please do not use the latter fact in the answer.)

Hang
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    The answer is yes. Also, if you haven't realized, finite maps are closed, so you do not have to take closures. – Mohan Apr 17 '16 at 13:27

2 Answers2

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Regular map $f:X\to Y$ of affine varieties correspond bijectively to morphisms of $k$-algebras $\phi: k[Y]\to k[X]$.
The map $f$ is dominant iff $\phi$ is injective, and is (by definition) finite iff $\phi$ makes $k[X]$ a finite $k[Y]$- module i.e. if $k[X]$ is a module of finite type over $\phi(k[Y])$ .
[The definition in Shafarevich is equivalent but confuses the issue with irrelevant integrality conditions.]
For any subset $Z\subset X$ the induced morphism $\bar \phi:k[Y]/\phi ^{-1}(I(Z))\to k[X]/I(Z) $ is also injective and module-finite so that the corresponding geometric restriction map map $\operatorname {res}(f): Z\to \overline {f(Z)}$ is dominant and finite, just as $f$ was.

NB
Notice that, in conformity with your request, I have never mentioned closedness of any map.

Georges Elencwajg
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  • Thank you. May I ask just one more question? Why does $\phi^{-1}(I(Z))$ corresponds exactly to $\overline{f(Z)}$? I will appreciate you if you can help me grasp it. – Hang Apr 20 '16 at 12:23
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    The map $\phi$ is just composition with $f$ so that a regular function $u:Y\to k$ is zero on $f(Z)$ iff $u\circ f=\phi(u)$ is zero on $Z$. Hence $I(f(Z))=\phi^{-1}(I(Z))$.Finally $\overline {f(Z)}=V(I(f(Z)))=V(\phi^{-1}(I(Z)))$ , which means exactly that $\overline {f(Z)}$ corresponds to $\phi^{-1}(I(Z))$. And this has nothing to do with the Nullstellensatz! – Georges Elencwajg Apr 20 '16 at 13:43
  • This formula you are asking about doesn't seem to be mentioned in Shafarevich, nor in other elementary books like Fulton, Hassett, Hulek,...This is very regrettable. I advise you to try to understand it and most importantly learn it by heart [ even if I'm the only teacher remaining in the world recommending that procedure in 2016 :-)] – Georges Elencwajg Apr 20 '16 at 14:06
  • Thank you very much! Your comment helps me understand better. And, I will definitely learn it by heart. :) – Hang Apr 21 '16 at 14:38
  • I don't get why the textbook doesn't state this as a proposition and prove it. Instead in another proof, it states "This is clearly a finite map". Not clear to me at all, which is why I looked for a forum like this one. – hlcrypto123 Oct 23 '20 at 00:30
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A dominant map of affine varieties corresponds to an inclusion of rings $R \subseteq S$, where $S$ is integral over $R$. You can identify $X, Y$ with the maximal spectra of $S, R$, and $f: X \rightarrow Y$ is the contraction map $\mathfrak n \mapsto \mathfrak n \cap R$.

A closed subset $Z$ of $X$ corresponds to a radical ideal $J$ of $S$. Specifically, $J$ is the intersection of all the maximal ideals in $Z$. Finite morphisms of varieties are closed, so $f(Z)$ corresponds to a radical ideal $I$ of $R$. One can identify $Z$ with the maximal spectrum of $S/J$, and $f(Z)$ with the maximal spectrum of $R/I$.

Concretely, $f(Z)$ is the set of all $\mathfrak n \cap R : \mathfrak n \in Z$, and so $I$ is equal to $$\bigcap\limits_{\mathfrak m \in f(Z)} \mathfrak m = \bigcap\limits_{\mathfrak n \in Z} (\mathfrak n \cap R) = R \cap \bigcap\limits_{\mathfrak n \in Z} \mathfrak n = R \cap J$$

The composition $Z \rightarrow X \rightarrow f(Z)$ then corresponds to the ring homomorphism $\pi: R/I \rightarrow S/J$. This is well defined and injective. Since $S$ is integral over $R$, $S/J$ is integral over the image of $R/I$, hence $Z \rightarrow f(Z)$ is a finite, dominant morphism of varieties.

D_S
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  • Thanks for your answer. But I was wondering if you can prove it without the fact you quoted : "Finite morphisms of varieties are closed". Because the question I ask is aimed to show this fact. – Hang Apr 18 '16 at 06:38
  • I'll think about it. But, it's probably easier and less technical if you just show directly that a finite morphism of varieties is closed. It's equivalent to a standard argument from commutative algebra. I can show you how to do that if you want. – D_S Apr 18 '16 at 14:33