ag.algebraic geometry - Generic fiber of morphism between non-singular curves

Here is a complete proof: as remarked in the answer by Norondion, we can reduce to
the case when $C_1 rightarrow C_2$ is generically separable, i.e. $k(C_1)$ is separable
over $k(C_2)$. Let $A subset k(C_1)$ be a finite type $k$-algebra consisting of the regular
functions on some non-empty affine open subset $U$ of $C_2$ (it doesn't matter which one
you choose), so that $k(C_2)$ is the fraction field of $A$.

By the primitive element theorem, we may write $k(C_1) = k(C_2)[alpha]$, where
$alpha$ satisfies some polynomial $f(alpha) = alpha^n + a_{n-1}alpha^{n-1} + cdots
+ a_1 alpha + a_0 = 0,$ for some $a_i$ in $K(C_2)$.

Now the $a_i$ can be written as fractions involving elements of $A$, i.e. each
$a_i = b_i/c_i$ for some $b_i,c_i in A$ (with $c_i$ non-zero). We may replace
$A$ by $A[c_0^{-1},ldots,c_{n-1}^{-1}]$ (this corresponds to puncturing $U$ at
the zeroes of the $c_i$), and thus assume that in fact the $a_i$ lie in $A$.

The ring $A[alpha]$ is now integral over $A$, and of course has fraction field equal
to $k(C_2)[alpha] = k(C_1)$. It need not be that $A[alpha]$ is integrally closed,
though. We are going to shrink $U$ further so we can be sure of this.

By separability of $k(C_1)$ over $k(C_2)$, we know that the discriminant $Delta$
of $f$ is non-zero, and so replacing $A$ by $A[Delta^{-1}]$ (i.e. shrinking $U$
some more) we may assume that $Delta$ is invertible in $A$ as well.

It's now not hard to prove that $A[alpha]$ is integrally closed over $A$. Thus
$text{Spec }A[alpha]$ is the preimage of $U$ in $C_1$ (in a map of smooth curves,
taking preimages of an affine open precisely corresponds to taking the integral closure
of the corresponding ring).

In other words, restricted to $U subset C_2$, the map has the form
$text{Spec }A[alpha] rightarrow text{Spec }A,$ or, what is the same,
$text{Spec }A[x]/(f(x)) rightarrow text{Spec A}$.

Now if you fix a closed point $mathfrak m in text{Spec }A,$ the fibre over this point
is equal to $text{Spec }(A/mathfrak m)[x]/(overline{f}(x)) = k[x]/(overline{f}(x)),$
where here $overline{f}$ denotes the reduction of $f$ mod $mathfrak m$.
(Here is where we use that $k$ is algebraically closed, to deduce that $A/mathfrak m = k,$
and not some finite extension of $k$.)

Now we arranged for $Delta$ to be in $A^{times}$, and so $bar{Delta}$ (the reduction
of $Delta$ mod $mathfrak m$, or equivalently, the discriminant of $bar{f}$)
is non-zero, and so $k[x]/(bar{f}(x))$ is just a product of copies of $k$,
as many as equal to the degree of $f$, which equals the degree of $k(C_1)$ over
$k(C_2)$. Thus $text{Spec }k[x]/(bar{f}(x))$ is a union of that many points,
which is what we wanted to show.