Let $E / \mathbb{Q}$ be the elliptic curve given by $y^{2} = x^{3} - x$. I would like to know explicitly what the field of all $2$-power torsion looks like, as well as the image in $\mathrm{GL}(T_{2}(E))$ of the $2$-adic Galois representation (and I would be interested in analogous descriptions for $\ell$-adic Galois representations with $\ell \neq 2$ as well). I'm sure this could be computed using Sage or Magma, but I thought I'd ask to see if it's already written down somewhere.

Here's what I do know:

1) Since $E$ has complex multiplication with $\mathrm{End}_{\mathbb{Q}(i)}(E) \cong \mathbb{Z}[i]$, and $j(E) \in \mathbb{Q}$, the dyadic torsion field must be an abelian extension of $\mathbb{Q}(i)$ ramified only above the prime $(1 + i)$. In fact, it must be the maximal such abelian extension.

2) The dyadic torsion field must contain all $2$-power roots of unity.

3) The division field $\mathbb{Q}(E[8])$ is $\mathbb{Q}(\zeta_{16}, 2^{1/4})$, and the image of $\mathrm{Gal}(\bar{\mathbb{Q}} / \mathbb{Q}(\zeta_{8}))$ modulo $8$ is generated by the matrices $$\left[ {\begin{array}{cc} 1 & 4 \\ 4 & 1 \end{array} } \right], \left[ {\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array} } \right] \in \mathrm{SL}_{2}(\mathbb{Z} / 8\mathbb{Z})$$ with respect to a suitable basis.

[EDIT: I corrected a mistake I noticed in my claim (3) above.]


1 Answer 1


I don't know that this is written down anywhere, but it's possible. It is known in general that $GL(T_{\ell}(E))$ is contained in the normalizer of $R_{\ell}^{\times}$, where $R_{\ell} = \mathbb{Z}[i] \otimes \mathbb{Z}_{\ell}$. (This follows for example from Corollary 2 on page 502 of Serre and Tate's ''Good reduction of abelian varieties'' Annals paper.) Moreover, if $\ell$ is sufficiently large, we have that the $\ell$-adic image equals the normalizer (in $GL_{2}(\mathbb{Z}_{\ell})$) of $R_{\ell}^{\times}$.

For the case of the particular curve at hand, we have that the normalizer of $R_{2}^{\times}$ is $$ G = \left\{ \begin{bmatrix} a & b \\ \mp b & \pm a \end{bmatrix} \in GL_{2}(\mathbb{Z}_{2}) : a, b \in \mathbb{Z}_{2} \right\}. $$ This is not the $2$-adic image, however, because the $2$-adic representation has index $4$ in $G$.

To see the index is at least $4$, observe that $E : y^{2} = x^{3} - x$ has full rational $2$-torsion, and also that $-I$ is not in the $2$-adic image (since the $4$-torsion field is generated by the $x$-coordinates of $4$-torsion points).

To see the index is at most $4$, I cheat a little bit. I had already computed the $2$-adic image for $E' : y^{2} = x^{3} + 3x$ using the usual Frattini subgroup argument, this shows that the $2$-adic image for this curve is all of $G$. However, these two curves are isomorphic over $\mathbb{Q}[\sqrt[4]{3}]$, and so the $2$-adic image for $E$ has index at most $4$. More precisely, the image is $$ \left\{ g \in G : g \bmod 4 \in \left\langle \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}, \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix} \right\rangle \right\}. $$

  • $\begingroup$ Thanks! I'm still not sure exactly what the field extension would look like, but maybe some class field theory computations would help with that. $\endgroup$ Dec 10, 2015 at 10:03

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