For me, an affine variety is an irreducible closed subset of some $A_k^n$. A quasi-affine variety is a non-empty open subset of an affine variety. A projective variety is an irreducible closed subset of some $P_K^n$. A quasi-projective variety is a non-empty open subset of a projective variety. A variety is everyone of the above: an affine variety, or a quasi-affine variety, a projective variety, or a quasi projective variety.

If $X\subseteq A_K^n$ is an affine variety, then the quotient $A(X)=K[x_1,\dots,x_n]/I(X)$ is the affine coordinate ring of $X$, where $I(X)$ is the ideal of the polynomials that vanish on $X$.

If $X$ and $Y$ are affine varieties, $f\colon X\to Y$ is said a finite morphism if $f$ is dominant (i.e. $f(X)$ is dense in $Y$) and $A(X)$ is integral over its subring $A(Y)$.

If $X$ and $Y$ are varieties, then $f\colon X\to Y$ is said a finite morpshism if exists a finite open cover $(U\alpha)_\alpha$ of $Y$, i.e. $Y=\bigcup_{\alpha} U_\alpha$, where $\forall \alpha$ $U_\alpha$ is affine, i.e. $U_\alpha$ is isomorphic to an affine variety, and also $f^{-1}(U_\alpha)$ is affine, and $f\colon f^{-1}(U_\alpha)\to U_\alpha$ is a finite morpshism in the sense of affine varieties (of the previous definition).

For completeness: if $X$ and $Y$ are varieties, then a map $f\colon X\to Y$ is said a morphism if $f$ is continuous and for every open subset $U$ of $Y$ and for every map $g\colon U\to K$ regular in $U$, then $g\circ f: f^{-1}(U)\to K$ is regular in $f^{-1}(U)$. ($K$ is an algebraically closed field).

How can i show that the composition of a finite morphism between varieties is a finite morphism?

I'am able to show this if $f\colon X\to Y$ and $g\colon Y \to Z$ are morphism of affine varieties, since i use the fact that if $A\leq B\leq C$ are commutative rings whit (the same) unit, then if $C$ is integral over $B$ and $B$ is integral over $A$, then $C$ is integral over $A$.

But i don't see how to show this in the general case with the second definition.

EDIT 21/02/18 I will ad some results about finite morpshisms of affine variety that may help:

1) A finite morphism of affine varieties is surjective

2) A finite morphism of affine varieties is a closed map

3) A finite morphism of affine varieties has all fiber finite

4) The restriction of a finite morpshism to a subvariety of the domain is also a finite morphism onto the image.


1 Answer 1


I'm not sure I understand the difficulty. If $Z=\bigcup_{\alpha\in I}U_\alpha$ is a finite affine open cover, then $Y=\bigcup_{\alpha\in I}g^{-1}(U_\alpha)$ is also a finite affine open cover and $$g:g^{-1}(U_\alpha)\to U_\alpha\;\mbox{ and } \;f:f^{-1}(g^{-1}(U_{\alpha}))\to g^{-1}(U_\alpha)$$ are finite morphisms of affine varieties for all $\alpha\in I$. You say you've shown that the composition of finite morphisms of affine varieties is finite, so $$g\circ f:f^{-1}g^{-1}(U_\alpha)\to U_\alpha$$ is finite for all $\alpha\in I$. Hence $g\circ f$ is a finite morphism.

Am I missing something?

  • $\begingroup$ Why $f^{-1}(g^{-1}(U_\alpha))$ is affine? I know that exists $Z=\bigcup_{\alpha\in I}U_\alpha$ finite affine open cover such that $Y=\bigcup_{\alpha\in I}g^{-1}(U_\alpha)$ is also finite affine oper cover and $g:g−1(Uα)→Uα$ is a finite morpshism of affine varieties. Then i know that exists $Y=\bigcup_{\beta\in J}V_\beta$ finite affine oper cover such that $X=\bigcup_{\beta\in J}f^{-1}(V_\beta)$ is also finite affine oper cover and $f:f^{-1}(V_\beta)\to V_\beta$ is a finite morphism of affine varieties. But now i don't know how to go on. $\endgroup$
    – Minato
    Feb 17, 2018 at 8:52
  • $\begingroup$ Your post does not indicate that the definition holds for a particular open cover. In fact, you completely failed to include a quantified of any kind in your definition. I interpreted the definition be that the condition was true for all open covers, which is what I would typically expect (and yields the result you want tho prove, making it a good definition). $\endgroup$
    – David Hill
    Feb 17, 2018 at 21:28
  • $\begingroup$ I suggest you review the definition and update your post to indicate whether there exists an open cover such that the conditions hold, or the conditions hold for all open covers. $\endgroup$
    – David Hill
    Feb 17, 2018 at 21:30
  • $\begingroup$ 'quantifier' . . . stupid auto-correct. $\endgroup$
    – David Hill
    Feb 18, 2018 at 2:07
  • $\begingroup$ You are right, i wasn't enough accurate. Sorry. $\endgroup$
    – Minato
    Feb 18, 2018 at 9:13

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