Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 125 / 2 / i / Solution

For an elliptic curve over ${\mathbb{F}}_q$, where $q$ is a prime power, the Hasse theorem for elliptic curves isWe prove it from the degree form on elliptic-curve endomorphisms, including the required degree facts.

A nonzero homomorphism of elliptic curves is a finite surjective morphism; its degree is the degree of the induced extension of function fields. Set the degree of the zero homomorphism to $0$. The degree of an isogeny is also the degree of the pullback of any point divisor: locally, a finite map of smooth curves gives a module of rank equal to the function-field degree, and its fibre length is the sum of the local multiplicities. For a composition, the tower law for function fields givesIf the isogeny of elliptic curves is separable, its pullback of a nonzero invariant differential on an elliptic curve is nonzero. Translation invariance implies that its differential is nonzero at every point, so each fibre point has multiplicity $1$. Every fibre is a translate of the kernel. ThereforeFor an inseparable isogeny, multiplicities must be retained; counting geometric kernel points alone would be incorrect.

Here is a divisor proof of the degree parallelogram law. On $E\times E$, let $\Delta$ be the diagonal and ${\Delta }^{-}$ the locus $P=-Q$. A Weierstrass $x$-function has a double pole at $O$, and the equation $x(P)=x(Q)$ describes $P=Q$ or $P=-Q$. ConsequentlyThis remains valid in characteristic $2$ using a general Weierstrass equation of an elliptic curve: the $x$-map still has degree $2$ and its two points are exchanged by negation. The divisor identity gives an identity of line bundles. Pull it back by $P\mapsto (\alpha (P),\beta (P))$ and take degrees. The diagonal is the inverse image of $O$ under subtraction and the antidiagonal under addition, soUsing line bundles makes this pullback argument valid even when $\alpha =\beta$, $\alpha =-\beta$ or one map is zero, when directly substituting into the rational function would fail.

Let $d_n=\mathrm{deg}[n]$. Applying the degree parallelogram law to $[n]$ and $[1]$ gives $d_{n+1}+d_{n-1}=2d_n+2$, with $d_0=0,d_1=1$. Induction givesMore generally, the same recurrence and polarization show that the degree restricted to the subgroup generated by two endomorphisms is a quadratic form: its mixed coefficient is determined by their sum or difference. This can be checked directly by the second-difference recurrence in each integer variable and the parallelogram identity in the two diagonal directions.

Let $\pi$ be the $q$-power Frobenius isogeny of an elliptic curve. It has degree $q$: it is purely inseparable, has one geometric point in each fibre, and sends a local parameter at the rational point $O$ to its $q$th power, giving fibre multiplicity $q$. Its pullback of any differential is zero. On the other hand, the standard addition rule for an invariant differential on an elliptic curve givesHence $1-\pi$ is separable, and its kernel consists exactly of the points fixed by Frobenius, namely $E({\mathbb{F}}_q)$. Put $N=\#E({\mathbb{F}}_q)$ and $t=q+1-N$. Thenand the quadratic-form calculation gives, for all $m,n\in \mathbb{Z}$,If $t^2>4q$, the real polynomial $X^2-tX+q$ is negative on an open interval. That interval contains a rational $n/m$ with $m\ne 0$, contradicting the displayed nonnegativity after multiplication by $m^2$. Thus $t^2\le 4q$, which is precisely the Hasse theorem for elliptic curves.