From Silverman's AEC page 323 (Chapter 10, section 3 on Homogeneous Spaces), the proof of the following fact (Proposition 3.2, part (a)):
Let $E/K$ be an elliptic curve and let $C/K$ be a homogeneous space for it. Fix a point $p_0 \in C$, define a map $$ \theta: E \rightarrow C, \theta(P) = p_0+P. $$ for all $P \in E$.
The proof says that if the point $p_0$ is fixed by the Galois group, $G_{\bar{K}/K}$ then $\theta$ is defined over $K$. Hence $\theta$ is defined over $K(p_0)$. Further, it says that
Since the action of $E$ on $C$ is simple transitive, degree of $\theta$ is one hence it's an isomorphism.
Could someone help elaborate on the statement in bold, which is the only part of the proof I don't understand. As far as I insert, showing the degree of the map $\theta$ means that $[\bar{K}(E) : \phi^*(\bar{K}(C))] =1$, but I don't know how to relate it with the action of $E$ on $C$.
Any help towards the clarification will be appreciated.
Thank you.
A simply transitive group action of a group $G$ on a set $X$ is one so that for any $x,y\in X$, there is exactly one $g\in G$ so that $gx=y$. Thus the fiber of $\theta$ over any point is a single point, and so the map must be of separable degree one. This is enough to prove the claim in the case when the underlying field has characteristic zero, since the only nontrivial inseparable extensions are in characteristic $p$. This may suffice for some readers, since this is the setting in which Silverman uses these results later on in the book (Mordell-Weil is about abelian varieties over number fields, aka finite extensions of $\Bbb Q$).
In characteristic $p$, there is more to say. It may be that a map of curves with separable degree one is not an isomorphism: the $k$-linear Frobenius is an example. On the other hand, over a perfect field, this is the only thing that can go wrong. By tag 0CCZ, if $f:X\to Y$ is a purely inseparable map of proper regular curves over $k$ with $X$ smooth over $k$, then $f:X\to Y$ is uniquely isomorphic to $F_{X/k}^n:X\to X^{(p^n)}$, the $n$-fold relative Frobenius over $k$. If $k$ is perfect, then $X\cong X^{(p)}$ since the absolute Frobenius on $k$ is an isomorphism and $X^{(p)}\to k$ is defined to be the base change of $X\to k$ along the absolute Frobenius on $k$. This implies that given a map $E\to C$ as in the problem, we may factor this map as $E\to E^{(p^n)}\to C$ by factoring the field extension $K(C)\subset K(E)$ in to separable and inseparable parts followed by identifying $E^{(p^n)}$ with $E$ to find a separable map $E\to C$ with all the same properties. (Note that the base field was assumed perfect in the introduction to the section, so we can use this argument.)
This doesn't completely make the issue go away, though. For instance, on any elliptic curve $X$ over a perfect field with Hasse invariant 0 the map $X\times X\to X$ given by $(x_1,x_2)\mapsto x_1+[p]x_2$ fits the definition of a homogeneous space but not the conclusions of the theorem as written. My guess is that this was an oversight, but it's not mentioned in the errata. If anyone feels inclined to email Professor Silverman and obtain an explanation, I would welcome hearing about it in the comments.