$f: X \to Y$ is finite between irreducible projectives. How to see the degree of a rational function in $K(X)$ over $K(Y)$? Suppose that $f : X \to Y$ is a finite unramified map between irreducible projective varities. Let $a \in K[U]$, for $U = f^{-1}(V)$ the preimage of some affine neighborhood in $Y$. For $y \in Y$, some comment in Shaferevich suggests that it is possible to see the degree of $a$ in $K(X)$ over $K(Y)$ by looking at the values that it takes on the points $f^{-1}(y)$. 
EDIT: I have fixed the proof below and posted it as a (partial) answer. I would be very interested in other ways to think about this. 
Here is my thought process so far:
Suppose that the regular function $a$ takes on all distinct values at the $f^{-1}(y)$. Let $r = \deg f$.


*

*Consider the fiber of $X$ above $y$. The coordinate ring of this is $k[U] \otimes_{k[V]} k[V] / m_y = k[U] / (k[U] m_y) \cong k^r$, where r is the degree of the map $f$.

*Using finiteness of the map, assume that $a$ satisfies some monic degree $n$ polynomial $f$ in $k[V][T]$. Then $\bar{a} \in k^r$ satisfies a degree $n$ polynomial $\bar{f}$ in $k[T]$, obtained by substituting the coefficients in $k[Y]$ of $f$ with their evaluations at the point $y$. 

*But each coordinate of $a$ is therefore a root of $\bar{f}$, a nonzero degree $n$ polynomial because $f$ was monic, which means $a$ can have at most $n$ roots. Since there are $r$ distinct values of $a$ among $f^{-1}(y)$ then this implies that $n \geq r$.

*Now we have the following commutative algebra situation: An integral extension of domains $A/B$, and an element $a \in A$, with the property that its minimal monic polynomial with coefficients in $B$ has degree $\geq r$. I would like to argue that the minimal polynomial for $a$ in $Frac(B)[T]$ also has degree $\geq r$, which will imply that $Frac(A) = Frac(B)(a)$. Here I am stuck... I imagine that there is some convenient way to move between these two minimal polynomials, but I am not seeing it. (The integrality theory I vaguely remember really only works when we are studying extensions of a PID. Oh - in the one dimensional case I would be tempted to use Noether normalization and recast both $A$ and $B$ as extensions of $k[t]$, but I don't think that will work in general.)
 A: I think I got it: you build the field-theoretic minimal polynomial of $a$ by looking at the product of $(T - \sigma a)$, where $\sigma a$ runs over its Galois orbits (a finite set). Since Galois conjugates of something integral remain integral, the coefficients of this minimal polynomial are all integral. So if we assume that $Y$ is normal, then we can conclude that the minimal polynomial of $a$ is amenable to the technique in steps 1-3, which implies that $a$ generates a field extension of $deg \geq r$, hence $K(Y)(a) = K(X)$.
Small comment on what I was writing before: I don't think it makes sense to think about the minimal polynomial in $B[T]$, since that ring is probably not a PID.
Here is the full argument:
Suppose that the regular function $a$ takes on all distinct values at the $f^{-1}(y)$. Let $r = \deg f$. Assume further that $Y$ is normal.


*

*Consider the fiber of $X$ above $y$. The coordinate ring of this is $k[U] \otimes_{k[V]} k[V] / m_y = k[U] / (k[U] m_y) \cong k^r$, where r is the degree of the map $f$.

*Using finiteness of the map, assume that $a$ satisfies some monic degree $n$ polynomial $f$ in $k[V][T]$. Then $\bar{a} \in k^r$ satisfies a degree $n$ polynomial $\bar{f}$ in $k[T]$, obtained by substituting the coefficients in $k[Y]$ of $f$ with their evaluations at the point $y$. 

*But each coordinate of $a$ is therefore a root of $\bar{f}$, a nonzero degree $n$ polynomial because $f$ was monic, which means $a$ can have at most $n$ roots. Since there are $r$ distinct values of $a$ among $f^{-1}(y)$ then this implies that $n \geq r$.

*Now we have the following commutative algebra situation: An integral extension of domains $A/B$, and an element $a \in A$, with the property that any polynomial with coefficients in $B$ that has a root $a$ has degree $\geq r$. I would like to argue that the minimal polynomial for $a$ in $Frac(B)[T]$ also has degree $\geq r$, which will imply that $Frac(A) = Frac(B)(a)$.

*Since $Y$ is normal, so is $V$. Hence in step 4 we can also assume that $B$ is an integrally closed domain. Let $g = \Pi (T - \sigma a)$, where $\sigma a$ runs over a choice of representatives for the Galois orbits of $a$. Then $g$ is the minimal polynomial for $a$, and moreover all of its coefficients are integral over $B$ and in $Frac(B)$ (since it is Galois invariant). By $B$ integrally closed, all of its coefficients are in $B$. Hence $g \in B[T]$, and is monic. By the steps 1 through 3, it follows that $\deg g \geq r$. Hence $a$ generates the field extension.
