Consider a number field $K$ and a finite Galois field extension $L/K$. Let $E$ be an elliptic curve over $K$ and consider the abelian group


$$E(L)\otimes L^{\times}.$$


Every element $g$ in $\text{Gal}(L/K)$ acts on it by $g(P\otimes \alpha) = g(P)\otimes g(\alpha)$ for $P\in E(L)$ and $\alpha\in L^{\times}$.



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> **Problem 1.1** Do we have $(E(L)\otimes L^{\times})^{\text{Gal}(L/K)} = E(K)\otimes K^{\times}$?
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(This question came up when others and I were trying our hand at the proof of the weak Mordell-Weil theorem. It has nothing to do with it, it just came up).


Also, what about $H^1$? That is



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> **Problem 1.2** Is there an analog of Hilbert's Theorem 90 so that $H^1(\text{Gal}(L/K), E(L)\otimes L^{\times})=0$?
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Now write $E(L)\_{tf}$ for the maximal torsion-free quotient of $E(L)$.



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> **Problem 2.1** Do we have $(E(L)\_{tf}\otimes L^{\times})^{\text{Gal}(L/K)} = E(K)\_{tf}\otimes K^{\times}$?
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> **Problem 2.2** Do we have $H^1(\text{Gal}(L/K),E(L)\_{tf}\otimes L^{\times})=0$?
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Galois invariants and tensor products
Consider $E:y^2=x^3-3$ over $\mathbb Q$. Then $E(\mathbb Q)$ is trivial, hence so is $E(\mathbb Q)\otimes\mathbb Q^\times$. On the other hand, over $L=\mathbb Q(\sqrt{-3})$ we have a point $P=(0,\sqrt{-3})\in E(L)$ which is mapped to its opposite by the complex conjugation. The same is true of $\sqrt{-3}\in L$, therefore $P\otimes\sqrt{-3}\in(E(L)\otimes L^{\times})^{\text{Gal}(L/K)}$, and we can check this element is nontrivial.


An analogous example for torsion-free parts is given by the point $P=(0,\sqrt{-2})\in E(L)$ for $E:y^2=x^3+x-2$ and $L=\mathbb Q(\sqrt{-2})$. $E(\mathbb Q)\cong\mathbb Z/2$ and $P$ is a point of infinite order, so $P\otimes\sqrt{-2}$ will be nontrivial in $E(L)\_{tf}\otimes L^\times$.


I don't know any result for $H^1$ but I doubt there is anything as simple as Hilbert's 90. As a Galois representation, $E(L)$ can have quite nontrivial structure, even after tensoring with $\mathbb Q$, but I don't have an explicit counterexample at hand. (edit: actually see Chris Wuthrich's comment below though)


