# Dominant finite morphism and finite algebraic extension

I don't know how to prove the following proposition.

If two varieties $X$ and $Y$ are irreducible, a morphism $\phi: X \rightarrow Y$ is dominant and finite, then $K(X)$ is a finite algebraic extension of $\phi^*K(Y)$.

Here, $\phi^*: K[Y] \rightarrow K[X]$ sends a function $f \in K[Y]$ to $f \circ \phi \in K[X]$. A morphism $\phi$ of irreducible varieties is called dominant if $\phi(X)$ is dense in $Y$. It is called finite if $K[X]$ is an integral over $\phi^*K[Y]$.

Is the following more general statement true?

$R_1$ is a domain, which is integral over its subdomain $R_2$. $F_1$ and $F_2$ are respective fields of fractions. Then $F_1/F_2$ is a finite algebraic extension.

This extension must be algebraic. But I think the finiteness is not so obvious.

-
Note that $R_2 \otimes_{R_1} F_1$ is a field (it's integral over the field $F_1$ because localization preserves integrality). This is thus $F_2$. Since this is finite over $F_1$ as base-change preserves finiteness, QED. –  Akhil Mathew Jul 10 '11 at 14:15
(As Qiaochu observes, the more general statement is false: the above comment proves it only when $R_2/R_1$ is finite. Otherwise, you could take for instance $\mathbb{Z}$ and its integral closure in $\overline{\mathbb{Q}}$.) –  Akhil Mathew Jul 10 '11 at 14:22
@Akhil Mathew: Thank you very much! –  ShinyaSakai Jul 12 '11 at 14:49
That isn't the correct definition of finite; you need that $K[X]$ is a finitely-generated module over $K[Y]$ (which implies, but is strictly stronger than, integral), and that gives you the correct general statement. The general statement as you've written it is false; just take $R_1$ to be the algebraic integers and $R_2$ to be $\mathbb{Z}$.
Thank you very much! Now I know what's wrong with the too general statement. The definiton for finiteness I have written is applicable for affine varieties, I think. But I still don't understand why "$K[X]$ is a finitely-generated module over $K[Y]$" is stronger than "$K[X]$ is integral over $K[Y]$"... –  ShinyaSakai Jul 12 '11 at 14:54
@ShinyaSakai: For example, the integral closure of $k[y]$ in its fraction field is integral over $k[y]$ but not a finitely-generated module. –  Qiaochu Yuan Jul 12 '11 at 14:58
Thank you very much. I think I misunderstood your meaning. However, in page 31 of James Humphreys' Algebraic Linear Algebra (GTM21), I see "Let $\phi: X \rightarrow Y$ be a morphism of affine varieties. If $K[Y]$ is integral over the subring $\phi^*K[Y]$, we call $\phi$ finite. " I don't know if these two definitions coincide under the condition of affine varieties. –  ShinyaSakai Jul 13 '11 at 15:58