Why does a mapping of curves have degree equal to size of kernel?

Question

I have seen in Silverman that the degree of a morphism of curves $f: C \rightarrow C'$ is defined as $\deg(f) := [K(C) : f^{*}K(C')]$. But in some notes I read recently from a colleague, the degree of such a map is $\deg(f) = |\text{ker}(f)|$. This made me think of other occurences where I've seen something similar to "$\deg(f) = |f^{-1}(y)|$ for almost all $y \in C'$".

Can someone please explain the last two concepts - how they follow from the Silverman definition?

Edit: I meant to say that $C, C'$ are abelian varieties. So take them to be elliptic curves I guess.

Answer 1

Notice that 2) is proper to elliptic curve because usually there is no group structure on a curve. In fact projective algebraic groups are torus (this is a standard argument).

To see 2) $\Leftrightarrow$ 3) : Let $f : E \to E'$ be an algebraic morphism of groups between elliptic curves. Then, any fiber is a translate of $\ker(f)$ by definition, in particular its cardinality coincide with the definition 3) (and in this case, the cardinality of the fibers is constant).

Equivalence between 1) and 3) is standard and probably in any book of algebraic geometry (e.g Hartshorne or Shafarevich).

Remark : For the definition used in 3), the "almost" is necessary, because in general we need to count the fiber with multiplicities, e.g $f : \mathbb P^1 \to \mathbb P^1, z \mapsto z^2$ has degree $2$ even if $f^{-1}(0) = \{0\}$.

Answer 2

They are not always the same. In general your map $\phi: E_1 \rightarrow E_2$ can be inseparable (meaning that the extension $\phi^{*}K(E_2) \hookrightarrow K(E_1)$ is not separable). In that case we still have (see Silverman) $$\#\phi^{-1}(y) = \deg_s(\phi)$$ for almost all $y \in E_2$, where $\deg_s(\phi)$ is the separable degree of $\phi$.