Finite map between affine varieties implies finitely generated. In the Shaferevich's Basic Algebraic Geometry, he defined a map $f:X \to Y$ between affine varieties is finite if $k[X]$ is integral over $k[Y]$. Then, in the theorem 1.13 of his book (in page 62), he try to show that finiteness is a local property as below. 

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

To prove this, he assume a finite cover of $Y$ consisting of principal open set $D(g_{\alpha})$. Let $V_{\alpha} = f^{-1}(D(g_{\alpha}))$ Then, he says that the assumption implies $k[V_{\alpha}] = k[X][1/g_{\alpha}]$ has a finite ``basis" over $k[D(g_{\alpha})]=k[Y][1/g_{\alpha}].$
However, I don't understand why there is such a finite ``basis". (First of all, I think generators are more applicable in this sense since we don't know whether $k[X][1/g_{\alpha}]$ is a free-module over $k[Y][1/g_{\alpha}]$ or not. But this is minor point so skip it.) The major reason I doubt the statement is that there exists an integral extension which is infinitely generated. (For example, think about $k[\{\sqrt[p]{2}\}_{p\text{ is prime}}]$) How can he assume that $k[X][1/g_{\alpha}]$ is finitely generated module over $k[Y][1/g_{\alpha}]$ based on the assumption that $f|_{V}$ is finite map?
For your information, in page 60, he define the finite map $f:X \to Y$ be a map such that $k[X]$ is integral over $k[X]$. 
Edit: Last time I wrote down the last sentence as
``the finite map $f:X \to Y$ be a map such that $f^{\ast}:k[Y] \to k[X]$ is finite." It's wrong; in page 60, he wrote down $f$ is finite if $k[Y] \to k[X]$ induces an integral extension. Sorry for the confusion.
 A: It is enough to prove that $k[Y]_{g_\alpha}\to k[X]_{g_\alpha}$ is of finite type. We'll show first that $k[Y]\to k[X]$ is of finite type and then that the property of being finite type is preserved under localization.
First, we note that every affine variety over a field $k$ is of finite type over that field: the closed immersion $X\to \Bbb A^n_k$ gives a surjection $k[x_1,\cdots,x_n]\to k[X]$, which demonstrates that $k[X]$ is of finite type over $k$. We may then observe that $k[Y]\to k[X]$ is of finite type by the general fact that if we have a composition of ring maps $A\to B\to C$ with $A\to C$ finite type, $B\to C$ must also be finite type. (Alternatively, adjoin all the variables to $k[Y]$ that you need to surject on to $k[X]$, and then quotient by the necessary relations and quotient by all the non-constant functions on $Y$.)
Finally, if a ring map $A\to B$ is of finite type, then the map $A[1/f]\to B[1/f]$ is again of finite type: we localize the surjection $A[x_1,\cdots,x_n]\to B$ at $f$, and as localization is exact, we see that it preserves the surjection.
Alternatively, we may also note that $V_\alpha$ and $D(g_\alpha)$ are affine varieties in their own right: if we have a closed immersion $\phi:Y\to \Bbb A^n$, then we get a closed immersion $(\phi,1/g_\alpha): D(g_\alpha) \to \Bbb A^{n+1}$ where the equations cutting out the image are all the equations of $Y$ plus the equation $x_{n+1}g_\alpha=1$. So we could skip the last part of the proof involving localizations.
