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Let $E$ be an elliptic curve defined over $\mathbb{F}_q$. For a prime $\ell \neq q$, we have that the $\ell$-torsion subgroup $E[\ell] \cong (\mathbb{Z}/\ell \mathbb{Z})^2$. As can be easily seen, there exist pairs of curves over $\mathbb{F}_q$ with the same trace of Frobenius, but whose Frobenius actions on $E[\ell]$ are different, more or less because the $\ell$-torsion lies in different extensions of $\mathbb{F}_q$.

However, what if we know the $j$-invariant of $E$ in addition to the trace of Frobenius $a_q(E) \in \mathbb{Z}$? Can we then draw conclusions about the eigenvalues $\lambda_1, \lambda_2 \in \overline{\mathbb{F}_{\ell}}$ of $F$? This paper relates these traces to values of hypergeometric functions over finite fields, in some cases. Does anyone know of any other results? In particular I am very interested in a simple method for deciding whether $F$ acts by scalar multiplication on $E[\ell]$.

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As can be easily seen, there exist pairs of curves over $\mathbb{F}_q$ with the same trace of Frobenius, but whose Frobenius actions on $E[\ell]$ are different, more or less because the $\ell$-torsion lies in different extensions of $\mathbb{F}_q$.

No, this can not happen. If two curves have the same trace of Frobenius, then, by Tate's isogeny theorem, their Tate modules are isomorphic as Galois representations. In particular, the Frobenius actions on the $E[\ell]$ must be the same. If you are only interested in the eigenvalues, a much simpler argument shows that the characteristic polynomial of Frobenius acting on $E$ is $T^2 - aT + q$, hence the eigenvalues are determined by the trace of Frobenius.

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  • $\begingroup$ Consider the curves $E$ and $E'$ defined over $\mathbb{F}_7$ by the equations $y^2 = x^3 + 2$ and $y^2 = x^3 + 3x + 2$, respectively. Then $E(\mathbb{F}_7) \cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$ while $E'(\mathbb{F}_7) \cong \mathbb{Z}/9\mathbb{Z}$. In both cases the trace of Frobenius is $-1$, so (taking $\ell=3$) your claim does not appear to hold water... $\endgroup$ – mlbaker Jun 24 '15 at 22:23
  • $\begingroup$ To be clear, I'm viewing the action of the Frobenius on $E[\ell]$ as a conjugacy class of $\mathrm{GL}(2,\mathbb{F}_\ell)$. In this sense the two examples above really are "different". $\endgroup$ – mlbaker Jun 24 '15 at 22:40

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