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Let $E$ be an elliptic curve defined over $\mathbf{Q}$, and let $K$ be a Galois number field. The Galois group $G=\mathrm{Gal}(K/\mathbf{Q})$ acts on the Mordell-Weil group $E(K)$ and thus on the finite-dimensional complex vector space $E(K) \otimes \mathbf{C}$, giving rise to an Artin representation of $G$.

Does every irreducible Artin representation of $G$ appear in the Mordell-Weil group $E(K) \otimes \mathbf{C}$ for some elliptic curve $E$ over $\mathbf{Q}$?

For example, the answer is yes for the trivial representation, since there are elliptic curves $E$ whose Mordell-Weil group $E(\mathbf{Q})$ is infinite. If $K$ is a quadratic field and $\chi$ is the corresponding quadratic character, then again the answer is yes, by considering the quadratic twist of an elliptic curve $E/\mathbf{Q}$ of rank $\geq 1$.

Related to this question is the following result: for every elliptic curve $E/\mathbf{Q}$, the vector space $E(\overline{\mathbf{Q}}) \otimes \mathbf{C}$ is infinite dimensional. In other words the rank of $E(K)$ gets arbitrarily large if we let $K$ vary over all number fields. (Marc Hindry once gave me an argument for proving this using height functions.) But I don't even know whether, for a fixed $K$, there always exists an elliptic curve $E/\mathbf{Q}$ such that the rank of $E(K)$ is greater than the rank of $E(\mathbf{Q})$. This would be a first step towards answering the above question.

The same questions make sense for elliptic curves over arbitrary global fields.

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    $\begingroup$ If the degree of $K$ is small, like $<8$, you should be able to pick a point $P\in\mathbb{P}^2(K)$ with well-chosen coordinates and then find a cubic $E$ that passes through $P$ (and its conjugates). That should answer it for a few representation, but won't get much further. $\endgroup$
    – Chris Wuthrich
    Commented Sep 7, 2017 at 13:34
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    $\begingroup$ If $K = \mathbb{Q}(\sqrt{d})$, then ${\rm rank} E(K) = {\rm rank} E(\mathbb{Q}) + {\rm rank} E_{d}(\mathbb{Q})$, where $E_{d}$ is the quadratic twist of $E$ by $d$. Hence, you can find an $E$ so that ${\rm rank} E(K) > {\rm rank} E(\mathbb{Q})$ just by taking any elliptic curve $F/\mathbb{Q}$ with positive rank and setting $E = F_{d}$. $\endgroup$
    – Jeremy Rouse
    Commented Sep 7, 2017 at 16:15
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    $\begingroup$ As Hindry said, one can use a height argument to get $\text{rank}\,E(\overline{\mathbb Q})=\infty$. But it's easier to just find points in lots of disjoint quadratic fields. Not that this answers your question. $\endgroup$
    – Joe Silverman
    Commented Jul 17, 2021 at 10:43

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