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

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.

Let $\pi ,{\pi }^{\prime }$ be the $p$-power Frobenius isogenies of $E,E^{\prime }$. Since $\psi$ is defined over ${\mathbb{F}}_p$, it commutes with Frobenius:Taking degrees, using multiplicativity and cancelling the nonzero degree of an isogeny $\mathrm{deg}\psi$, givesBoth differences are separable, so the preceding proof identifies their degrees with rational point counts. Therefore

Isogenous elliptic curves can have different rational point groups. Here is an explicit example. Over ${\mathbb{F}}_7$, takeThe mapextends across $O$ and $(0,0)$ to a degree-$2$ isogeny of elliptic curves with those two points as its kernel. Substitution verifies the target equation, or this follows from the two-isogeny formula with $a=0,b=-1$. Both curves are smooth modulo $7$.

For $x=0,1,\dots ,6$, the numbers of affine points with that abscissa on $E$ are $(1,1,0,0,2,2,1)$, and on $E^{\prime }$ they are $(1,0,2,2,0,0,2)$. Adding the point at infinity gives $8$ on each. The first curve has four rational points of order dividing $2$, from $O$ and the three roots $0,1,-1$. The second has only two: its quadratic factor $X^2+4$ has no root modulo $7$, since $3$ is not a square. By the classification of finite abelian groups,They are thus isogenous with equal orders and nonisomorphic groups.