Let $E/\mathbb{Q}$ be an elliptic curve with good reduction at two distinct primes $p, \ell$. Suppose the mod $\ell$ Galois representation associated to $E$ is surjective. Let $K=\mathbb{Q}(E[\ell])$ be the field cut out by the representation. Then $G=\mathrm{Gal}(K/\mathbb{Q}) \simeq \mathrm{GL}_2(\mathbb{F}_\ell)$.
In a 2002 paper by Duke and Tóth (MR1969646), a recipe is given for constructing a matrix $\sigma_p$ representing the conjugacy class of Frobenius at $p$ inside $G$. Since $p$ is unramified in $K/\mathbb{Q}$, we should be able to determine the decomposition behavior of $p$ in $K$ using this matrix in two ways. In particular, let $g$ denote the number of distinct primes above $p$ in $K$.
- Let $d$ be the multiplicative order of $\sigma_p$. Then $d$ corresponds to the inertial degree of every prime above $p$, so $g=\frac{|G|}{d}$.
- Let $\mathcal{C} \subset G$ denote the conjugacy class of $\sigma_p$ in $G$. Since the primes above $p$ are all conjugate, we have $g=|\mathcal{C}$|.
However, in practice one can write down the matrix $\sigma_p$ and directly compute (e.g. using Sage) all of the quantities above, and methods (1) and (2) often yield different values. Here is a concrete example:
Let $E$ be the elliptic curve with Cremona label 43a1. Then $E$ has surjective mod 5 Galois representation and good (supersingular) reduction at $p=2$ (with $a_p(E)=-2)$. The recipe of Duke-Toth yields the matrix $$\left(\begin{matrix}1 & 1 \\ -1 & 1\end{matrix}\right).$$ Considered as an element of $\mathrm{GL}_2(\mathbb{F}_5)$, this matrix has multiplicative order $d=4$ and lives in a conjugacy class of cardinality $|\mathcal{C}|=30$. The cardinality of $\mathrm{GL}_2(\mathbb{F}_5)$ is $480$. But clearly 480/4=120, not 30.
At first I thought something might be going wrong since $p=2$, $E$ is supersingular at $2$, and $a_p(E) \neq 0$. But it's easy enough to find plenty of other examples where $p, \ell$ are both larger, and where $E$ has good ordinary reduction.
What explains this discrepancy?
