# Finding an elliptic curve with Frobenius trace zero

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.

• Please update your question and define $N(t)$... Thanks! Maybe also where (which text) your theorem 4.2 appears? Dec 20, 2019 at 4:23
• The screenshot is from 'Nonsingular Plane Cubic Curves over Finite Fields' by Schoof, but the proof is on 'Abelian varieties over finite fields' by Waterhouse.
– José
Dec 20, 2019 at 12:33

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.

• Thank you. So really there isn't a finite field el. cur. that isn't a twist?
– José
Dec 20, 2019 at 19:36
• No idea of what you mean, clarify Dec 20, 2019 at 20:33
• I misinterpreted what you said. My doubt is if it is there is one such curve on the form $y^2 = x^3 + ax^2 + bx +c$ such that all the constants are in $\mathbb{F}_p$? At least one of them must be in $\mathbb{F}_q \setminus \mathbb{F}_p$.
– José
Dec 22, 2019 at 19:31
• I added a paragraph about why it can't Dec 22, 2019 at 21:53
• Why does $\varphi^2-t\varphi+p=0$ imply $(\varphi^{2n})^2 - t_{2n} \varphi^{2n}+p^{2n} = 0$? I'm assuming $t_{2n}$ is the Frobenius trace of $\varphi^{2n}$. What am i missing here?
– José
Dec 23, 2019 at 1:15