I am currently reading Basic Algebraic Geometry by Shafarevich. I have two questions about finite maps on affine and quasi-projective varieites.

For a finite map $f:X\rightarrow Y$ between affine varieties $X$ and $Y$, we say that $f$ is finite if $f(X)$ is dense in $Y$ and $k[X]$ is integral over $k[Y]$ (where $k[Y]$ has been identified with a subring of $k[X]$).

Question 1

The first question is about the proof of Theorem 1.13. The statement of the theorem is as follows.

If $f:X\rightarrow Y$ is a regular map of affine varieties, and every $x\in Y$ has an affine neighbourhood $U\ni x$ such that $V = f^{-1}(U)$ is affine and $f:V\rightarrow U$ is finite, then $f$ itself is finite.

The proof itself is not so hard, but I find myself unable to understand the first step, which is to assume that each neighbourhood $U$ in the statement of the theorem above is a principal open set.

As far as I understand it, the neighbourhood $U$ of $x$ may be decomposed into a finite union of principal open sets $U_1,\ldots, U_r$ and for each $U_i$, the set $V_i := f^{-1}(U_i)$ is also a principal open set in $X$. However, I fail to understand why $f$ must be finite when restricted to any of these sets $V_i$.

Question 2

The definition of finite maps on quasi-projective varieties follows the "local definition" above. More precisely,

A map $f: X \rightarrow Y$ of quasi-projective varieties is finite if any point $y\in Y$ has an affine neighbourhood $U$ such that $V = f^{-1}(U)$ is affine and $f:V \rightarrow U$ is a finite map between affine varieties.

This definition makes sense given the theorem above, but it got me wondering regarding the following fairly easy fact about finite maps between affine varieties.

Fact: If $X,Y$ are affine varieties and $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ are finite maps, then so is their composition $g\circ f$.

It's not clear to me if the above holds for finite maps between quasiprojective varieties as defined above. The "obvious" proof again runs into a similar obstacle where one needs to show that a restriction of a finite map to some open affine subset is still finite.

Possibly I am missing some simple clean fact about finite maps. If so, I would like to know it.


For question (1), what you are trying to show is that if you have a finite ring map $A \to B$ and some element $f \in A$ then $A_f \to B_f$ is finite (meaning invert the multiplicative subset of powers of $f$ in both rings). That is, we have some $A$-module surjection $A^n \twoheadrightarrow B$. Since localization is exact, and it commutes with direct sums, we get a surjection $A_f^n \twoheadrightarrow B_f$, as desired.

In general you're coming upon the idea of properties of morphisms that are affine-local on the target, where their truth can be verified either by checking every affine open on the target, or by checking it on a cover of the target by affine opens. This idea is discussed very cleanly in Vakil's Rising Sea notes section 5.3.

  • $\begingroup$ Thanks Adam. I looked at Vakil's notes and it seems like reading Chapter 5 requires understanding quite a bit of material related to Schemes etc. that I don't know about. Do you know of a reference that explains this in the setting of affine or quasiprojective varieties? Thanks again! $\endgroup$ – stupidqnasker Dec 20 '20 at 8:16
  • 1
    $\begingroup$ I don't personally know of any, though I admit I'm not that well read (in particular I've never read Shafarevich). In my opinion if you're already worried about whether or not a morphism is affine and working with other subtle properties of morphisms you're pretty close to scheme theory anyways, and depending on your situation it might be worthwhile to take the plunge, though admittedly it takes a lot of work to get moving there. $\endgroup$ – Adam K Dec 20 '20 at 23:31
  • $\begingroup$ In this particular case, if you can make sense of what the affine communication lemma in Vakil 5.3 is saying, you can try and prove it in language that you're comfortable with. It really is just about refining covers and using basic properties of what a basis for a topology gets you $\endgroup$ – Adam K Dec 20 '20 at 23:47
  • $\begingroup$ I'm accepting this answer as it answered the first part of my question. Thanks Adam. $\endgroup$ – stupidqnasker Dec 26 '20 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.