Map of smooth curves and its separability degree I'm interested in a proof of the following fact from Silverman: Arithmetic of Elliptic Curves:
Let $\Phi: C_1 \rightarrow C_2$ be a nonconstant map of smooth curves. Then for all but finitely many  $Q \in C_2$:
$ \# \Phi^{-1}(Q) = deg_s (\Phi)$,
where $ deg_s (\Phi)$ is the separability degree of $\Phi$. 
In particular, I'd prefer to see a proof not using the notion of schemes as a) I've not worked with schemes yet, b) I need this statement for a presentation where most the audience isn't familiary with the notion of schemes.
Also I'd love to see a geometric interpretation of the statement (why only for nearly all points).
Thanks a lot in advance!
 A: You can't really avoid the commutative algebra. You can avoid talking about schemes, but you're not getting out of the commutative algebra.
This follows from two things: first, from the famous $n = \sum ef$ formula: if $B/A$ is a finite separable extension of Dedekind domains, of degree $n$, and $\mathfrak p$ is a prime of $A$ which decomposes in $B$ as
$$\mathfrak p B = \mathfrak q_1^{e_1} \dots  \mathfrak q_r^{e_r},$$
then $n = \sum_{i=0}^r e_if_i$, where $f_i = [B/\mathfrak q_i : A/\mathfrak p]$ is the degree of the residue class field extension. Second, it follows from the fact that finitely many primes of $A$ ramify in $B$.
Now suppose that $C_1$ and $C_2$ are smooth affine curves over an algebraically closed field $k$, with coordinate rings $B$ and $A$ respectively. The rings $B$ and $A$ are Dedekind domains and they are $k$-algebras. Their nonzero primes correspond to points on the respective curves. Since $k$ is algebraically closed the residue class field extensions $B/\mathfrak q$ of $A/\mathfrak p$ is trivial (both identify with $k$). It follows that whenever $\mathfrak p \subseteq A$ is not ramified, we have $r=n$.
For a morphism of projective curves, we just use an open affine cover to reduce it to the affine case.
Remark that this can fail badly for non-smooth curves. For instance, if $C$ is the nodal cubic $y^2 = x^3-x^2$, the rational parametrization $y=t(t^2-1),x = t^2-1$ is a morphism $\mathbf A^1 \to C$ which has degree $2$, but almost all fibres have a single point in them.
