Let $L/K$ be a quadratic number field extension. Let $Sha(E/L)$$\operatorname{Sha}(E/L)$ be Tate-Shafarevich group of elliptic curve $E/L$. How $\sigma \in Gal(L/K)$$\sigma \in \operatorname{Gal}(L/K)$ acts on $Sha(E/L)$$\operatorname{Sha}(E/L)$?
I think the action is given by $[C]^{\sigma}=[C^{\sigma}]$$[C/L]^{\sigma}=[C^{\sigma}]$ for $[C]\in Sha(E/L)$$[C]\in \operatorname{Sha}(E/L)$, where $C^{\sigma} $ denotes a curve whose coefficients are transported by $\sigma$.
Once I could prove $[C]^{\sigma}\in Sha(E/L)$$[C/L]^{\sigma}\in \operatorname{Sha}(E/L)$, this is clearly a group action easily. To
To prove $[C/L]^{\sigma}\in Sha(E/L)$$[C/L]^{\sigma}\in \operatorname{Sha}(E/L)$, what we should do is to prove $C/L$ is $E/L$-torsor,in other words, there is a simply transitive algebraic group action $E/L×C^{\sigma} \to C^{\sigma}$ defined over $K$.
Since $f: C \cong C^{\sigma}$over over algebraic closure, wethus we can define a map $E/L×C^{\sigma} \to C^{\sigma}$ by composition of $id×f^{-1}$ and $φ:E/L×C \to C$ (this is simply transitive and defined over $K$) and $f$.
This composition is indeed transitive, but I cannot prove this is defined over $K$.
Is my definition of action of $\sigma \in Gal(L/K)$on$\sigma \in \operatorname{Gal}(L/K)$ on $Sha(E/L)$$\operatorname{Sha}(E/L)$ correct ? If so, why the last composite defined over $K$ ?
