Tate–Shafarevich group and $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \operatorname{Sha}(E/L)$

Question

$\DeclareMathOperator\Sha{Sha}\DeclareMathOperator\Gal{Gal}$Let $L/K$ be a quadratic extension of number field $K$. Let $\sigma$ be a generator of $\Gal(L/K)$.

Let $E/K$ be an elliptic curve defined over $K$ and $\Sha(E/K)$ be its Tate–Shafarevich group. $\Gal(L/K)$ acts on $\Sha(E/L)$ naturally (cf. How Galois group acts on Tate-Shafarevich group?).

There is a canonical isomorphism $\tau: E(L)\cong E_D(L), (x,y)\mapsto (x,y/\sqrt{D})$ .

My goal is to prove $\tau$ induces $(1-\sigma)\Sha(E/L)\cong 　\operatorname{trace}\Sha(E_D/L)$. This isomorphism appears in p219 of Yu - On Tate-Shafarevich groups over galois extensions.

To prove this isomorphism, it is enough to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$.

There exists an isomorphism $\phi: \Sha(E/L)\cong \Sha(E_D/L)$ induced by $\tau$ though I cannot write down the map between them.

We can check $\sigma \tau=-\tau \sigma$ because we can calculate its coordinates explicitly, but I cannot calculate both $\sigma \phi$ and $\phi \sigma$, so I'm having difficulty to prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$. How can I overcome this trouble and prove $\sigma \phi(C)=-\phi \sigma(C)$ for all $C \in \Sha(E/L)$ ?

Another approach of proving $(1-\sigma)\Sha(E/L)\cong \operatorname{trace}\Sha(E_D/L)$ is also appreciated.

Answer 1

Let $\sigma$ be the non-trivial element of the Galois group of the quadratic extension $L/K$. Let $\phi \colon E \to E_D$ be the isomorphism defined over $L$.

First, if $P \in E(\bar L)$ then $\sigma\bigl( \phi(P)\bigr) = - \phi\bigl( \sigma(P)\bigr)$. This is easy to check on the coordinates of $P=(x,y)$ by $$\sigma\bigl(\phi(x,y)\bigr) = \sigma\bigl( x,y/\sqrt{D}\bigr) =\bigl( \sigma(x), \sigma(y)/\sigma(\sqrt{D})\bigr) = \bigl(\sigma(x), -\sigma(y)/\sqrt{D}\bigr) = - \bigl( \sigma(x), \sigma(y)/\sqrt{D}\bigr) = - \phi\bigl(\sigma(x),\sigma(y)\bigr)=-\phi\bigl(\sigma(x,y)\bigr).$$

Now $\phi$ induces a map $\phi_*\colon H^1(L,E)\to H^1(L,E_D)$. Let $\xi\in H^1(L,E)$ and let $g\in G_L$. Then write $*$ for the action of $\operatorname{Gal}(L/K)$ on these cohomology groups. We obtain $$\begin{align*}\bigl(\sigma * \phi_*(\xi )\bigr)(g) &= \sigma\Bigl( \phi_*(\xi)(\sigma^{-1} g \sigma)\Bigr) &&\text{by def of $*$}\\ &=\sigma\Bigl( \phi\bigl(\xi(\sigma^{-1} g \sigma)\bigr)\Bigr) &&\text{by def of $\phi_*$}\\ &=-\phi\Bigl(\sigma\bigl(\xi(\sigma^{-1} g\sigma)\bigr)\Bigr) &&\text{by the above}\\ &=-\phi \bigl( (\sigma * \xi)(g)\bigr) &&\text{by def of $*$} \\ &=\bigl(-\phi_*(\sigma * \xi)\bigr) (g) &&\text{by def of $\phi_*$.} \end{align*} $$ As $\sigma*\phi_*(\xi)=-\phi_*(\sigma* \xi)$ holds for all $\xi\in H^1(L,E)$ it also holds for its subgroup $Ш(E/L)$. Now $(1+\sigma)*\bigl(\phi_*(\xi)\bigr) = \phi_*\bigl((1-\sigma)*\xi\bigr)$ shows that $\phi$ induces an isomorphism between $N\bigl( Ш(E_D/L)\bigr)$ and $(1-\sigma)\,Ш(E/L)$.