Finding an elliptic curve with Frobenius trace zero

Question

The following theorem by Waterhouse lists all values of the Frobenius trace such that there is a corresponding elliptic curve over $\mathbb{F}_q$, $q = p^n$, $p$ prime.

The thing is, i couldn't find a single curve with $n$ even, $t = 0$ and $p \ne 1 $mod $4$ at the same time. I iterated over thousands of curves with low $q$ checking for those conditions and not one of them appeared, while i could find examples of every other condition this way. Condition (iii3) must be exceedingly rare.

Can you give me an example of such curve?

EDIT: N(t) is the number of curves with Frobenius trace $t$. $N(t) \ne 0 \Leftrightarrow$ there is at least one curve.

Answer 1

Using that $Tr(\phi_q)=0\implies \Bbb{Z}[\phi_q] \cong \Bbb{Z}[i p^{n/2}] \subset \Bbb{Z}[i]\implies $ the curve is probably a reduction of $y^2=x^3+x$, also the dual endomorphism of $\phi_q$ is $-\phi_q$ thus the curve is supersingular, and this

I obtained the Magma code

        K<a>:= GF(7^2);   E:=EllipticCurve([K|1, 0]);  // y^2=x^3+x
        T :=Twists(E); // the twists of E, ie. the curves isomorphic over F_{q^r} but not over F_q
        P<t> := PolynomialRing(K);
        C := T[2];

        C;
        "trace of the Frobenius ";
        #K +1 - #C;
        "minimal polynomial of a";
        (t-a)*(t-a^7);

Result

        Elliptic Curve defined by y^2 = x^3 + a*x over GF(7^2)
        trace of the Frobenius
        0
        minimal polynomial of a
        t^2 + 6*t + 3

The obtained curve is not defined over $\Bbb{F}_p$ and is not a quadratic twist of $y^2=x^3+x$ ie. not isomorphic to $dy^2=x^3+x$ that's why it was hard to find by hand.

Any $E$ satisfying your requirement can't be defined over $\Bbb{F}_p$ : if it is there is the Frobenius $\varphi(x,y)=(x^p,y^p)\in End(E)$, the main theorem is that there is a dual endomorphism such that $\varphi^*\varphi = p\in End(E)$ and $t=\varphi+\varphi^* \in \Bbb{Z}\subset End(E)$, the minimal polynomial of $\varphi$ is $(X-\varphi)(X-\varphi^*)=X^2-t X+p$ (it means that $\varphi^2-t\varphi+p=0\in End(E)$), thus the minimal polynomial of $\varphi^{n}$ is a quadratic polynomial too, $X^2-t_{n} X+p^{n}\in \Bbb{Z}[x]$. Your assumption is that for some $n$, $t_{2n}=0$, it means $\varphi^{2n}$ is a root of $X^2+p^{2n}$ thus we can identity it with $\pm ip^n$. There is no quadratic integer whose square is $\pm i p^n$, contradicting that $\varphi^n$ is the root of a quadratic polynomial $\in \Bbb{Z}[x]$. Thus $E$ isn't defined over $\Bbb{F}_{p^n}$, contradiction.