# Cubic twist of elliptic curves and its rank

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 $$E(\mathbb{Q}(\sqrt{D})) \to E(\mathbb{Q})$$.

• $\def\rank{\operatorname{rank}}\def\QQ{\mathbb{Q}}$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. Dec 5, 2023 at 11:56

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 $$\rk E(L) = a+b = \tfrac{1}{2} (2a+2b) = \tfrac{1}{2} \rk E(K) = \tfrac{1}{2}\cdot \bigl( \rk E(F) + \rk E_1(F) + \rk E_2(F) \bigr) =\rk E(\mathbb{Q}) + \rk E_1(\mathbb{Q}) + \rk E_2(\mathbb{Q})$$.

• 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. Dec 5, 2023 at 13:58
• 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. Jan 20 at 1:16
• Grossly miscalculated in my original estimate. According to Z3 records, we can produce $E(L)$ with rank at least $21=11+10+0$. Jan 20 at 11:16