# Trace of Frobenius over $F_q$

Let $q_0$ be a prime and $q$ = $q_0^n$. Let $a(F_q/F_{q_0})$ denote any integer which is trace of Frobenius over the field $F_q$ for some elliptic curve which can be defined over $F_{q_0}$.

It is stated in Mazur's paper (Rational isogenies of Primes degree, Inventiones mathematicae, 1978) that $a(F_3^{12}/F_3)$ = 658, -1358, +1458.

I can get only $\pm$ 1458 or $\pm$ 729 or 0 if i use the following thm.

(It might be a simple answer. But, I don't see that how 658, - 1358 occurs).

Theorem: Let $q_0$ be a prime and $q$ = $q_0^n$. Then there exists an elliptic curve $E$ dened over $F_q$ such that the trace of the Frobenius equals to $\beta$ ($\beta$ $\leq$ $\lfloor 2\sqrt{q} \rfloor$) if and only if one of the following cases occur:

(i) $q_0$ does not divide $\beta$,

(ii). $q$ is a square (i.e. $n$ is even) and

$\beta$ = $\pm 2 \sqrt{q}$ or

$\beta$ = $\pm \sqrt{q}$ ($q_0$ $\not\equiv$ 1 (mod 3)) or

$\beta$= 0 and ($q_0$ $\not\equiv$ 1 (mod 4)),

(iii) $q$ is not a square (i.e. $n$ is odd) and

$\beta$ = 0 or $\beta$ = $\pm$ $q_0^{n+1/2}$ and $q_0$ = 2,3.

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## 1 Answer

I just did the computation directly and got Mazur's result (details below). I am not sure how you are trying to use the result you cite. Your result describes which numbers occur as traces of Frobenius on elliptic curves defined over $\mathbb{F}_{3^{12}}$, but doesn't single out which of those curves will be obtained from curves definable over $\mathbb{F}_3$. So the numbers $1458$, $658$ and $-1358$ do indeed match the criteria of your theorem. The first one is $2 \times \sqrt{3^{12}}$, case 2, and the other two are not divisible by $3$, case $1$. Of course there are tons of other numbers which match the criteria of your theorem; they will be traces of Frobenius on elliptic curves defined over $\mathbb{F}_{3^{12}}$ that don't descend to $\mathbb{F}_3$.

The way I will do the computation is to first list all the curves definable over $\mathbb{F}_3$, and then work out the trace of $3^{12}$-th power Frobenius.

Let $E$ be an elliptic curve defined over $\mathbb{F}_3$. Let $a$ be the trace of $3$rd power Frobenius. By the result you cite, $a$ is one of $\{ \pm 3, \pm 2, \pm 1, 0 \}$. As yet, I haven't used the extension field $\mathbb{F}_{3^{12}}$ anywhere.

Denote the eigenvalues of $3$rd power Frobenius by $\lambda_{\pm}$. They are the roots of $$x^2 - ax + 3 =0.$$

Using the values of $a$ above, I get that they are $$\pm \frac{-3 \pm \sqrt{-3}}{2},\ \pm (-1 \pm \sqrt{-2}),\ \pm \frac{-1 \pm \sqrt{-11}}{2},\ \pm \sqrt{-3}.$$ In each case, the inner $\pm$ switches $\lambda_{+}$ and $\lambda_{-}$ while preserving the value of $a$, and the outer $\pm$ negates $a$.

If $3$rd power Frobenius has eigenvalues $\lambda_{+}$ and $\lambda_{-}$ then $3^{12}$ Frobenius has eigenvalues $\lambda_+^{12}$ and $\lambda_-^{12}$, and hence trace $\lambda_{+}^{12} + \lambda_{-}^{12}$.

Applying this to each of the terms above I get $$\begin{matrix} 729+729 &= 1458 & (329 + 460 \sqrt{-2}) + (329 - 460 \sqrt{-2}) &= 658 \\ (-679+80 \sqrt{-11}) + (-679 - 80 \sqrt{-1}) &= -1358 & 729+729 &= 1458 \end{matrix}.$$

In summary, the term $1458$ arises from the elliptic curves where $3$rd power Frobenius has trace $\pm 3$ or $0$, the term $658$ arises from the elliptic curves where $3$rd power Frobenius has trace $\pm 2$ and the term $-1358$ arises from the elliptic curves where $3$rd power Frobenius has trace $\pm 1$.

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Dear David, I need Mazur's result, not 729 or 0. Thanks for your immediate reply. –  Srilakshmi Mar 18 '12 at 15:14
I think I now understand where you are confused. See if the first paragraph above clears it up. –  David Speyer Mar 18 '12 at 15:18
Yes. Your edited answer clears up my confusion. Thanks. –  Srilakshmi Mar 18 '12 at 15:29