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Let $K$ be a quadratic number field. Let $\sigma$ be a generator of Galois group of $K/\Bbb{Q}$. Let $E$ be an elliptic curve defined over $\mathbb{Q}$.

Let $\mathrm{Sha}(E/K)$ denote the Tate-Shafarevich group of $E/K$. Can we explicitly write down the norm map of the Tate-Shafarevich group $\mathrm{Sha}(E/K) \to \mathrm{Sha}(E/\mathbb{Q})$?

(Reference: What's the Hilbert class field of an elliptic curve?)

At first, I thought that the map $[C] \mapsto [C] + [C]^{\sigma}$ would provide the desired map. However, it does not commute with the Galois action, meaning that $([C] + [C]^{\sigma} )^{\sigma} \neq [C]^{\sigma} + [C]$ in general, thus this is not well-defined.

Thank you for your help.

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    $\begingroup$ Why should you ever have a $\neq$ in you statement? -- The norm on Sha is induced by the corestriction on the Weil-Châtelet group $H^1(K,E) \to H^1(\mathbb{Q}, E)$. I am certain that it is equal to your sum of torsors. Though I doubt you want to calculate a sum of torsors instead of working with cohomology if you want to be "explicit". $\endgroup$
    – Chris Wuthrich
    Commented Jun 11, 2023 at 15:28
  • $\begingroup$ I thought contradiction occurs as described below if we admit trace map in my question. Let $E/\Bbb{Q}$ be an elliptic curve such that $Sha(E/\Bbb{Q})=Sha(E/\Bbb{Q})[2]$. Let fix an arbitrary quadratic number field $K$. If trace map in my question exists, $ker(tr\mid{Sha(E/K)[2]})=Sha(E/\Bbb{Q})[2]$ where $tr \mid{Sha(E/K)[2]}$ is a restriction map of trace to ${Sha(E/K)[2]}$. This is because $a+a^{\sigma}=0$ is equivalent to $a^{\sigma}=a$ in $2-$part. This implies $Sha(E/K)[2]$ is bounded, but it is known that $Sha(E/K)[2]$ can be arbitrary large under quadratic extension. $\endgroup$
    – Duality
    Commented Jun 11, 2023 at 17:54
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    $\begingroup$ In general, let $E/k$ be an elliptic curve and $K/k$ a Galois extension with group $G$. Then it is not true that the kernel of corestriction in Sha for $E/K$ is isomorphic to the Sha for $E/k$. It is also not true that the $G$-fixed part of Sha for $E/K$ is Sha for $E/k$; though the latter is true on the $p$-primary parts for $p$ not dividing the group order of $G$. This becomes clear if you'd use classical results from group cohomology like the inflation-restriction five term sequence. $\endgroup$
    – Chris Wuthrich
    Commented Jun 11, 2023 at 18:40
  • $\begingroup$ Thank you very much. Oh, $\ker(\mathrm{tr})$ is not as simple as I initially thought. Are there any known results for controlling $\ker(\mathrm{tr})$? For example, $\ker(\mathrm{tr})$ is embedded into a Galois cohomology group. For the restriction map $\mathrm{Sha}(E/\mathbb{Q})\to \mathrm{Sha}(E/K)$, the corresponding inflation-restriction cohomology sequence is $H^1(\mathrm{Gal}(K/\mathbb{Q}),E(K))\to H^1(G_{\mathbb{Q}},E)\to H^1(G_K,E)$. We can embed $\ker(\mathrm{res})$ into the finite Galois cohomology $H^1(\mathrm{Gal}(K/\mathbb{Q}),E(K))$. $\endgroup$
    – Duality
    Commented Jun 12, 2023 at 4:55
  • $\begingroup$ However, when it comes to $\mathrm{tr}:\mathrm{Sha}(E/K)\to \mathrm{Sha}(E/\mathbb{Q})$, I do not have an appropriate cohomological sequence that controls $\ker(\mathrm{tr})$. $\endgroup$
    – Duality
    Commented Jun 12, 2023 at 4:55

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