2 added 832 characters in body

I'm not sure what to say except that

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 yoursdivisible 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 square root $\pm$ switches $\lambda_{+}$ and $\lambda_{-}$ while preserving the value of $a$, and the out square root 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 $\pm 1$.

1

I'm not sure what to say except that I just did the computation and got Mazur's result, not yours. 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 square root switches $\lambda_{+}$ and $\lambda_{-}$ while preserving the value of $a$, and the out square root 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 $\pm 1$.