Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 22 / 2

Let $\pi$ denote the Frobenius isogeny of an elliptic curve and write $N=\#E({\mathbb{F}}_q)$. Its degree of an isogeny is $q$. The isogeny of elliptic curves $1-\pi$ has differential equal to the identity because $d\pi =0$, so it is a separable isogeny. Its kernel consists precisely of the rational points fixed by $\pi$, giving

We use the degree parallelogram law for isogenies of elliptic curves, with degree zero assigned to the zero map:One explanation of the first identity is the divisor proof of the degree parallelogram law: on $E\times E$, the zero divisor of $x(P)-x(Q)$ is the sum of the diagonal and the graph of negation, while its pole divisor is twice each coordinate copy of $O$. Pulling the associated line bundle identity back by $(f,g)$ and taking degrees gives the identity, including exceptional cases by the line-bundle formulation. This works for a general Weierstrass equation of an elliptic curve, including characteristic two; $x$ is the quotient coordinate for negation. Polarization therefore makes degree a quadratic form on the endomorphism ring of an elliptic curve.

Set $a=q+1-N$, the Trace of Frobenius. The cross term is determined by $\mathrm{deg}(1-\pi )=1+q-a$, giving, for all integers $m,n$,If $a^2>4q$, the real degree-two polynomial $X^2-aX+q$ is negative on a nonempty open interval. That interval contains a rational $m/n$, contradicting the displayed nonnegativity after multiplication by $n^2$. Thus $a^2\le 4q$. Hasse's bounds areThe argument proves the Hasse theorem for elliptic curves without assuming its bound in advance.

The squares in the finite field ${\mathbb{F}}_7$ are $0,1,2,4$. For $x=0,1,\dots ,6$, the values of $x^3-x+1$ and the numbers of possible $y$ are respectivelyAdding $O$ yields $\#E({\mathbb{F}}_7)=12$. The Trace of Frobenius is $a=8-12=-4$. The trace of the square of an elliptic-curve endomorphism is $a^2-2q=16-14=2$, so the elliptic-curve point count over a finite field givesFor example, this trace identity follows from ${\pi }^2-[a]\pi +[q]=0$ and $\mathrm{tr}[q]=2q$.

One suitable second elliptic curve isHere $4\cdot 3^3+27\cdot 6^2\equiv 2(\mathrm{mod}7)$, so its elliptic-curve discriminant is nonzero. Its complete list of rational points is $O,(3,0),(6,3),(6,4)$, obtained by checking the seven $x$-values. For $P=(6,3)$ the tangent slope is $1$ in ${\mathbb{F}}_7$, and the elliptic-curve addition formula gives $2P=(3,0)$. Consequently $P$ has order four and $E^{\prime }({\mathbb{F}}_7)\cong \mathbb{Z}/4\mathbb{Z}$.