Let $E/\mathbb{Q}$ be an elliptic curve defined by $E: y^2 = x^3 + b$ (where $b \in \mathbb{Q}$).

Let $E_D$ be an elliptic curve defined by $E_D: y^2 = x^3 + bD^2$.

$E$ and $E_D$ are isomorphic over $\mathbb{Q}(D^{1/3})$.

Is there a known formula for $\text{rank}(E/\mathbb{Q}(D^{1/3}))$ in terms of $\text{rank}(E/\mathbb{Q})$ and $\text{rank}(E_D/\mathbb{Q})$?

cf. For a quadratic twist, the well-known formula $\text{rank}(E_D/\mathbb{Q}(\sqrt{D})) = \text{rank}(E/\mathbb{Q}) + \text{rank}(E_D/\mathbb{Q})$ is proved by decomposing $E(\mathbb{Q})$ into plus and minus parts by the Galois action of $\text{Gal}(\mathbb{Q}(\sqrt{D})/\mathbb{Q})$, or by calculating the kernel and cokernel of the trace map $E(\mathbb{Q}(\sqrt{D})) \to E(\mathbb{Q})$.

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    $\begingroup$ $\def\rank{\operatorname{rank}}\def\QQ{\mathbb{Q}}$Seems more likely that there's a formula for $\rank E(\QQ(\zeta,D^{1/3}))$ in terms of $\rank E(\QQ(\zeta))$ and $\rank E_D(\QQ(\zeta))$, where $\zeta$ is a primitive cube root of 1. But without $\zeta$ in your ground field, the Galois action won't decompose. $\endgroup$ Dec 5, 2023 at 11:56

1 Answer 1


There is a formula but it involves both cubic twists. Let $E: y^2 = x^3+B$ be an elliptic curve over $\mathbb{Q}$ with $j=0$ as the one in the question. Let $D$ be a cubefree integer. Set $E_1: y^2=x^3+D^2\,B$ and $E_2:y^2=x^3+D^4\,B$. Both $E_1$ and $E_2$ become isomorphic to $E$ over $L=\mathbb{Q}\bigl(\sqrt[3]{D}\bigr)$. I believe the formula is

$\DeclareMathOperator{\rk}{rk}$ $\rk E(L) = \rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q}).$

This can easily checked to be the case for the given curve and some small $D$.

Let $K=L(\mu_3)$ be the Galois closure of $L$ and let $F=\mathbb{Q}(\mu_3)$. Denote by $G$ the cyclic Galois group of $K/F$ and write $\chi$ and $\bar \chi$ for the two non-trivial characters of $G$.

First the motivation: The $L$-function associated to the $G_F$-representation $T_p E\otimes \mathbb{C}[G]$ is the $L$-function of $E/K$. Let $\psi$ be the Grössencharacter associated to $E$, then this $L$-function splits into six $L$-functions, namely those for $\psi$, $\bar\psi$, $\psi\chi$, $\overline{\psi\chi}$, $\psi\bar{\chi}$, and $\bar{\psi}\chi$. The first two give the $L$-function of $E/F$, the middle two the $L$-function of $E_1/F$ and the last two the $L$-function of $E_2/F$. The Birch and Swinnerton-Dyer conjecture now implies that $\rk E(K) = \rk E(F)+\rk E_1(F)+\rk E_2(F)$.

We can prove this directly by looking at the $G$-action on $E(K)$. Write $g$ for the element of $G$ that sends $\alpha\in L$ with $\alpha^3 =D$ to $\zeta\cdot \alpha$ where $\zeta^3=1$. Let $[\zeta]\in \operatorname{End}(E)$ be the element of order $3$ given by $[\zeta](x,y) = (\zeta \cdot x, y)$. Define $E(K)_i = \bigl\{ P \in E(K)\, :\, g(P) = [\zeta]^i(P)\bigr\}$. Then $E(K)_0 = E(F)$ and the map $E_1(F) \to E(K)$ sending $(x,y)$ to $(x/\alpha^2, y/\alpha^3)$ has image equal to $E(K)_1$. Similar $(x,y)\mapsto (x/\alpha^4,y/\alpha^6)$ brings $E_2(F)$ to $E(K)_2$.

The kernel of the map from $E(F)\oplus E_1(F)\oplus E_2(F)$ to $E(K)$ is contained in the finite subgroup $E(F)[\zeta-1]$. For any $P\in E(K)$ we have $3P = (1+g+g^2)(P) + (1+[\zeta]g+[\zeta]^2g^2)(P) + (1+[\zeta]^2g+[\zeta]^4g^2)(P)$ which belongs to the sum $E(F)+E(K)_1+E(K)_2$. Thherefore the cokernel is also finite. Hence $\rk E(K) = \rk E(F)+\rk E_1(F)+\rk E_2(F)$.

Since $E$ has complex multiplication by an order in $F$, the rank of $E(F)$ is twice the rank of $E(\mathbb{Q})$; and similarly for $E_1$ and $E_2$. The representation $E(K)\otimes\mathbb{C}$ of the Galois group (equal to $S_3$) of $K/\mathbb{Q}$ splits as $\mathbb{1}^a + \epsilon^a+\rho^b$ where $\epsilon$ is the irreducible non-trivial $1$-dimensional and $\rho$ is the irreducible $2$-dimensional representation of $S_3$. Therefore $\rk E(L) = a+b = \tfrac{1}{2} (2a+2b) = \tfrac{1}{2} \rk E(K) = \tfrac{1}{2}\cdot \bigl( \rk E(F) + \rk E_1(F) + \rk E_2(F) \bigr) =\rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q})$.

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    $\begingroup$ Note that $T_pE \otimes \chi$ is not $T_p E_1$ and so there is no reason to expect that $E_1$ and $E_2$ have the same rank, and they indeed do not have that as seen in examples. $\endgroup$ Dec 5, 2023 at 13:58
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    $\begingroup$ As a further challenge, I suggest making it $B^2$ in $E$ so that the torsion subgroup is $Z/3Z$ and then play with $D$ to find a cubic field $L$ over which $E(L)$ would have as high rank as possible. From the top of my head, rank $9$ should be possible in reasonable time (say, $9=6+2+1$). I would use this paper to construct $E$: arxiv.org/pdf/1604.02693.pdf. $\endgroup$ Jan 20 at 1:16
  • $\begingroup$ Grossly miscalculated in my original estimate. According to Z3 records, we can produce $E(L)$ with rank at least $21=11+10+0$. $\endgroup$ Jan 20 at 11:16

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