How does $\text{Gal}(L/K)$ act on the automorphism group of an elliptic curve?

Question

Let $L/K$ be a finite Galois extension of number fields; I'm interested mainly in the case $K = \Bbb{Q}$ and $L= \Bbb{Q}(\sqrt{d})$. Let $X$ be an elliptic curve over $K$ and $\text{Aut}(X_L)$ the group of $L$ automorphisms of $X_{L}$. Now I have read many times that we can consider the Galois cohomology group $H^1(\text{Gal}(L/K),\text{Aut}(X_L))$, but however:

Doesn't this require the Galois group to act on $\text{Aut}(X_L)$? I can't seem to see what the action is. I looked in Silverman and Tate's Arithmetic of Elliptic curves, but it I can't find it stated anywhere for what the action is.

Answer 1

An element of $\operatorname{Gal}(L/K)$ induces an automorphism of $X_L$ through its action on the plane over $L$. I imagine the intended action of a Galois element is via conjugation by the automorphism it induces.

Answer 2

The curve $E$ is defined over $K$, so $E_L := \text{Spec } L \times_{\text{Spec } K} E.$ Thus there is an action of Gal$(L/K)$ on $E_L$, via its action on the first factor. (This is the scheme-theoretic interpretation of the action on coefficients discussed by Hurkyl). This not an action on $E_L$ as an $L$-scheme, since Gal$(L/K)$ acts non-trivially on $L$.

But now, if $\sigma$ some automorphism of $E_L$ in the category of $L$-schemes, then the composite $g \sigma g^{-1}$ will be another automorphism of $E_L$ in the category of $L$-schemes (i.e. it lies over the identity map on Spec $L$).

This is the definition of the Gal$(L/K)$-action on Aut$(E_L)$.