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. 



Trace of Frobenius over $F_q$
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 $\pm 1$.


