Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 28 / 1 / i

The Hasse theorem for elliptic curves statesWe prove it by the positive degree of an isogeny. Let $\pi$ be the $q$-power Frobenius isogeny, of degree $q$. Its fixed points are precisely $E({\mathbb{F}}_q)$, so they form the kernel of $1-\pi$. Since the differential of $\pi$ is zero, the differential of $1-\pi$ is the identity; thus $1-\pi$ is a separable isogeny. Its degree is the number of its geometric kernel points. ThereforeWe use the standard degree parallelogram law, together with $\mathrm{deg}[m]=m^2$: degree, extended by zero at the zero endomorphism, is a positive quadratic form on the endomorphism ring of an elliptic curve. Put $t=1+q-N$, the Trace of Frobenius. Polarizing the degree form gives, for every pair of integers $m,n$,Indeed, the values on $1$ and $\pi$ are $1$ and $q$, and the cross coefficient is fixed by $\mathrm{deg}(1-\pi )=1+q-t$. These are standard degree facts about isogenies of elliptic curves, which suffice without an assumption about characteristic or ordinariness.

For $n\ne 0$, division by $n^2$ shows $x^2-tx+q\ge 0$ at every rational $x=m/n$. Density of the rational numbers and continuity of this polynomial make it nonnegative on the real line. Evaluating at its minimum $x=t/2$ gives $q-t^2/4\ge 0$, whence $|t|\le 2\sqrt{q}$. Since $N=q+1-t$, this proves the estimate. This is the degree-form proof of the Hasse bound; it proves both sides simultaneously by positivity of one degree form.