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Let $L/K$ be a quadratic number field extension. Let $\operatorname{Sha}(E/L)$ be Tate-Shafarevich group of elliptic curve $E/L$. How $\sigma \in \operatorname{Gal}(L/K)$ acts on $\operatorname{Sha}(E/L)$?

I think the action is given by $[C/L]^{\sigma}=[C^{\sigma}]$ for $[C]\in \operatorname{Sha}(E/L)$, where $C^{\sigma} $ denotes a curve whose coefficients are transported by $\sigma$.

Once I could prove $[C/L]^{\sigma}\in \operatorname{Sha}(E/L)$, this is clearly a group action easily.

To prove $[C/L]^{\sigma}\in \operatorname{Sha}(E/L)$, 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 $K$.

$f: C \cong C^{\sigma}$ over algebraic closure, thus 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 $\sigma \in \operatorname{Gal}(L/K)$ on $\operatorname{Sha}(E/L)$ correct ? If so, why the last composite defined over $K$ ?

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    $\begingroup$ This action should only exist if $E$ itself is defined over $K$. In that case, one can simply apply $\sigma$ to the coefficients of polynomial equations defining the action of $E$ on $C$. $\endgroup$
    – Will Sawin
    Commented Jun 7, 2023 at 16:34
  • $\begingroup$ Thank you very much, if $φ: E×C\to C, (P,p)\to φ(P,p)$, you explains, $φ^{\sigma}: E×C^{\sigma}\to C^{\sigma}, (P,p) \to φ^{\sigma}(P,p)$ gives the action of $E$ to $C^{\sigma}$. Indeed, this is defined over $K$ because it is Galois invariant under the action of $Gal(L/K)$ ! Could you please tell me why the $φ(P,p)$ is polynomial ? $\endgroup$
    – Duality
    Commented Jun 7, 2023 at 17:42

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