Some question about map of varieties X,Y are varieties, and $A(X),A(Y)$ are coordinate rimg ,respectively. If$f:X\rightarrow Y$ is a finite surjective map, that is $f^{-1}(p)$ is finite for all $p\in Y$, then $A(X)$ is finitely generated $A(Y)$- module??
 A: No.
Take for $X$ the hyperbola $xy=1$ in $ \mathbb  A_k^2$ ($k$ an algebraically closed field), for $Y$ the affine line $ \mathbb  A_k^1$ and for $f:X \to Y$ the projection $(x,y)\mapsto x$.  The fibers are finite: they consist of exactly one point, except for the origin whose fiber is empty.
However the  corresponding ring morphism  $f^\ast :\mathcal O(Y)\to \mathcal O(Y)$ is the inclusion 
$f^\ast :k[X]\to k[X,Y]/(XY-1)=k[X,X^{-1}]$, and $k[X,X^{-1}]$ is not a finitely generated module over the ring $k[X]$.
This is why algebraic geometers do not call morphisms with finite fibers "finite", but call them "quasi-finite" (up to some technical nitpicking). A finite morphism $X\to Y$ is defined (in the case of affine varieties) by the  stronger property of module-finiteness of $\mathcal O(X)$ over $\mathcal O(Y)$.
Edit: An answer to Sang's actual question! I had overlooked that Sang wanted to know about surjective morphisms with finite fibers. The answer is still "no".
I am going to exhibit a surjective morphism of varieties $f:X \to Y$ with finite fibers whose corresponding ring morphism  $f^\ast :\mathcal O(Y)\to \mathcal O(Y)$ is not module-finite.        
The variety $Y$ is still the affine line $ \mathbb  A_k^1$ with coordinate $x$. In $ \mathbb  A_k^2$ with coordinates $\xi,\eta$, consider the  equation $\xi=\eta^2$; it
defines a parabola $X'$ whose ring of functions is $k[\xi, \eta]/(\xi-\eta^2)=k[x,y]=k[y]$  (since $x=y^2$).
Now take for $X$ the punctured parabola $X=X'\setminus \{P\}$ where $P$ is for example the point $(1,1)$. This is still an affine variety, whose ring of functions is the localized ring $k[y,\frac {1}{y-1}]$
We again take for $f$ the projection $f:X\to Y:(x,y) \mapsto x$ : it is surjective with finite fibers, as required by Sang.
However the corresponding ring map $f^\ast: \mathcal O(Y)\to \mathcal O(X)$ is 
$k[x]\to k[y,\frac {1}{y-1}]: x\mapsto y^2$
and it is not module-finite.
