TITLE: Dominant finite morphism and finite algebraic extension
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: 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 $\mathbb{Z}$.