# Kernel of an étale self-map of an elliptic curve and the Galois group.

This question originates from Hartshorne, Chapter IV, Problem 4.8. Fix some algebraically closed field $k$ of characteristic zero, and let $E \to \text{Spec }k$ be an elliptic curve. I want to compute the étale fundamental group $$\pi_1^{\text{ét}}(E) = \lim_{\leftarrow} \text{Gal}(K'/K),$$ where $K$ is the function field of $E$ and the limit runs over all Galois extensions of $K$ such that the corresponding curve $E'$ is étale over $E$. Hartshorne gives a long hint for this, and my question essentially comes from trying to justify everything in it.

I can prove that any such étale map $f:E' \to E$ is dominated by the degree morphism $n_E:E\to E,$ where $n = \text{deg } f$. Further the degree morphism is also étale as it is unramified (the kernel has the same cardinality as its degree since we are in characteristic zero) and it is flat (its a finite surjective map of nonsingular varieties). So now it suffices to only consider the system given by the $n_E$, for each $n$. What I do not understand is how I am supposed to identify $\text{Gal}(K'/K)$ with $\text{Ker }n_E$. I believe this is true by analogy with covering spaces, but I do not see it explicitly in this context.

Is there an easy (i.e. no étale-sites and such) explanation for this?

For each $p\in\mathrm{Ker}\,n_E$, let $T_p:E\to E$ be translation by $p$ (that is, $T_p(x)=x+p$). Then $n_E\circ T_p(x)=n_E(x+p)=n_E(x)+n_E(p)=n_E(x)$, so $T_p$ gives an automorphism of the covering $n_E:E\to E$. If $K'/K$ is the extension of function fields corresponding to this covering, then $T_p$ induces an automorphism of $K'/K$, taking a rational function $f$ to $f\circ T_p$.
The map $p\mapsto T_p$ is a group homomorphism $\mathrm{Ker}\, n_E\to\mathrm{Gal}(K'/K)$. This homomorphism is injective because $p\neq 0$ implies $T_p\neq \mathrm{Id}$, and it follows that the homomorphism is surjective because $[K':K]=\#\mathrm{Ker}\, n_E=n^2$.