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Given a quasi-finite (the each fiber is a finite set) morphism between two affine varieties (in the sense of the zero set of polynomials): $\phi:X\to Y$.

What can we say about the induced ring homomorphism $\phi^*:A(Y)\to A(X)$ as well as the relation between $A(Y),A(X)$? More precisely, I know if $\phi$ is finite (quasi-finite+proper) iff $\phi^*: A(Y)\to A(X)$ is finite, can we say something like this when $\phi$ is quasi-finite?

Moreover, if $\phi:X\to Y$ is a morphism between two affine varieties with the same dimension, is $\phi$ quasi-finite? Or what additional condition do we need to add so that $\phi$ is quasi-finite?

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The following modified form of Zariski's Main Theorem can help:

EGA IV3, Theorem 8.12.6 (pg 45 in link): Suppose $Y$ is quasicompact and quasiseparated. If $f:X\to Y$ is quasi-finite, separated, and of finite presentation, then $f$ factors as $X\hookrightarrow X'\to Y$ where the first morphism is an open immersion and the second is finite.

As all affine schemes are qcqs and all affine maps are separated, we may apply the theorem assuming the ring map is of finite presentation (which will be the case if both $X,Y$ are closed subvarieties of some finite-dimensional affine space as it seems that you assume). Here, as finite morphisms are affine, $X'$ is again affine and thus we may observe that $A(Y)\to A(X)$ is the composition of a finite ring map with a localization.

As to your final question about morphisms between varieties of the same dimension, your hypotheses are rather thin at the moment: you certainly need that the image of your subvariety can't be contained in any lower-dimensional subvariety, for instance. This doesn't completely do it, though - consider $\Bbb A^2\to \Bbb A^2$ by $(x,y)\mapsto (x,xy)$, which is even dominant but has an infinite fiber over the origin. You may find a reasonable list of conditions at Stacks, but it might not be everything you hoped for.

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  • $\begingroup$ But I am asking for the affine varieties in the sense of the zero set of polynomials, which are not schemes. I wonder if the theorem you mentioned is built on my setting? $\endgroup$ – 6666 Nov 25 '19 at 13:35
  • $\begingroup$ There's a fully faithful functor taking zero sets of polynomials to schemes, which means the stated result holds in your setting as well. The functor is particularly easy to describe in your case, even: $V(f_1,\cdots,f_m)\subset \Bbb A^n_k$ gets sent to $\operatorname{Spec} k[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$. (Or, $Y$ gets sent to $\operatorname{Spec} A(Y)$.) $\endgroup$ – KReiser Nov 25 '19 at 18:31

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