Rank of elliptic curve over quadratic extension $K=\Bbb{Q}(\sqrt{D})$ is calculated by a formula $rank(E/K)=rank(E/\Bbb{Q})+rank(E_D/\Bbb{Q})$(this is easy to prove).
Thus rank over quadratic field can be calculated by the information of the rank over $\Bbb{Q}$.
But what is known about the rank over $\Bbb{Q}(2^{1/3})$ ?
For example, can we calculate a rank of $E:y^2=x^3+x$ over
$\Bbb{Q}(2^{1/3})$ ?
Best Answer
I'll address the question of generalising the formula $\mathrm{rank}(E/K) = \mathrm{rank}(E/\mathbb Q) + \mathrm{rank}(E_D/\mathbb Q)$.
In general, if $E/K$ is an elliptic curve, then there is an abelian variety $$A := \mathrm{Res}_{K/\mathbb Q}(E)$$ called the restriction of scalars of $E$ with the property that $$A(\mathbb Q) \cong E(K).$$
If $K/\mathbb Q$ is quadratic, then there is an isogeny $A\simeq E \times E_D$ and, since the rank is an isogeny invariant, you recover your formula.
If $K = \mathbb Q(2^{1/ 3})$, then $A$ is isogenous to $E\times B_2$, where $B_2$ is an abelian surface over $\mathbb Q$ that can be described as follows. Let $B = \mathbb Z[\zeta_3]\otimes_{\mathbb Z} E$, an abelian surface over $\mathbb Q$ with $\mathrm{Aut}_{\overline{\mathbb Q}}(B) \supset \langle \zeta_3\rangle$. In particular, $B$ admits cubic twists, and $B_2$ is the cubic twist corresponding to the extension $\mathbb Q(2^{1/ 3})$.
A lot of this and more can be deduced from this paper of Mazur–Rubin–Silverberg.