Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 (Elliptic Curves)

The Hasse theorem for elliptic curves states that every elliptic curve $E/{\mathbb{F}}_q$ satisfiesLet $\pi$ be the Frobenius isogeny and put $a=\mathrm{tr}(\pi )$. Its fixed points are exactly $E({\mathbb{F}}_q)$, so separability of $1-\pi$ givesThe degree of an isogeny is a nonnegative quadratic form on the endomorphism ring, and for all integers $m,n$,If $a^2>4q$, this quadratic polynomial has two real roots and takes a negative value at some rational $m/n$ between them, hence after clearing denominators at some integer pair $(m,n)$. Therefore $a^2\le 4q$, and substituting $a=q+1-\#E({\mathbb{F}}_q)$ proves the bound. This is the degree-form proof of the Hasse bound.

For $x=0,1,2$, the value of $x^3-x-1$ is the nonsquare $2$ in ${\mathbb{F}}_3$. Thus the point at infinity is the only rational point andThe eigenvalues of the Frobenius isogeny are the rootsof $T^2-3T+3$. The elliptic-curve point count over a finite field givesThe last term vanishes exactly when $r\pi /6\equiv \pi /2(\mathrm{mod}\pi )$. Hence

If $p\mid D$, square-freeness gives the cuspidal reduction $y^2=x^3$. Its nonsingular group is isomorphic to the additive group ${\mathbb{F}}_p$, so it is cyclic of order $p$.

Suppose $p\nmid D$. The reduction is an elliptic curve. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is a quadratic nonresidue. Pairing $x$ with $-x$ in the quadratic-character sum and usingshows that the two contributions cancel. Therefore $\#\tilde{E}({\mathbb{F}}_p)=p+1$. The three roots $0,D,-D$ are distinct, so the full 2-torsion is rational. A cyclic group has at most two elements killed by $2$, hence $\tilde{E}({\mathbb{F}}_p)$ is noncyclic. Thus

A one-dimensional commutative formal group law over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfyingAn isomorphism $u:F\to G$ is a series $u(T)\in TR[[T]]$ with a compositional inverse and

The multiplication series is defined recursively by $[0{]}_F=0$, $[1{]}_F=T$ and $[n+1{]}_F=F([n{]}_F,T)$, with the formal inverse handling negative $n$. Its linear term isBy the invertible morphism criterion for formal group laws, it is an isomorphism exactly when its linear coefficient $n$ is a unit of $R$. Indeed, when $n\in R^{\times }$, recursive coefficient comparison constructs a unique compositional inverse; applying the morphism identity for $[n{]}_F$ shows that the inverse also respects $F$. Conversely, an invertible series must have a unit linear coefficient. Therefore

A positive integer $D$ is a congruent number when it is the area of a right triangle with positive rational side lengths. For a point $P=(x,y)$ with $y\ne 0$ onthe formulasgive $a^2+b^2=c^2$ and $ab/2=D$, after changing signs if necessary. Conversely, a rational right triangle of area $D$ givesso these constructions are inverse up to the usual sign choices.

It remains to distinguish torsion. If an odd prime $\ell$ divided the order of a rational torsion point, choose by the Dirichlet theorem on primes in arithmetic progressions a good prime $p\equiv 3(\mathrm{mod}4)$ for which $p\not\equiv -1(\mathrm{mod}\ell )$. Part (a) gives $\#E_D({\mathbb{F}}_p)=p+1$, while part (b), applied to the formal group of an elliptic curve, makes reduction injective on $\ell$-power torsion because $\ell$ is a unit in ${\mathbb{Z}}_p$. This is impossible. Similarly, a good prime $p\equiv 3(\mathrm{mod}8)$ shows that the rational $2$-primary torsion has order at most four. Sincealready form the full rational 2-torsion,The triangle construction uses exactly the points with $y\ne 0$, which are therefore nontorsion. By the Mordell-Weil theorem, such a point exists exactly when the free part has positive rank. Hence

On $y^2=x^3+5x^2+4x$, the slope through $P_1=(-1,0)$ and $P_2=(-2,2)$ is $-2$. The elliptic-curve addition formula givesFor doubling $P_2$, the tangent slope isand hence

The elliptic-curve discriminant is supported at $2$ and $3$, so $5$ and $7$ are primes of good reduction. Direct point counts giveFor example, summing $1+{\chi }_p(x(x+1)(x+4))$ over $x\in {\mathbb{F}}_p$ and adding the point at infinity gives these values.

The reduction of torsion points on an elliptic curve at the two primes shows that $|E(\mathbb{Q}{)}_{\mathrm{tors}}|$ divides eight. Part (a) shows that $P_2$ has order four, and $P_1$ is an independent point of order two because it does not lie in $⟨P_2⟩$. They already generate eight points, so

Use the two-descent on an elliptic curve map associated with the three rational roots $0,-1,-4$,with the standard limiting values at the 2-torsion. Only the square classes of $-1$, $2$ and $3$ can occur, because all other numerator and denominator valuations in the three factors $x,x+1,x+4$ are even. Checking solubility over $\mathbb{R}$, ${\mathbb{Q}}_2$ and ${\mathbb{Q}}_3$ leaves exactlyThese four classes are represented respectively by $O$, $P_1$, $P_2$ and $P_1+P_2$. Thus $|E(\mathbb{Q})/2E(\mathbb{Q})|=4$. Since part (b) givesthe quotient by doubling has order $2^{r+2}$. Therefore

Let the common difference of $r^2,s^2,1,t^2$ be $d$. ThenFor $x=-2s^2$ and $y=2rst$,so $(-2s^2,2rst)\in E(\mathbb{Q})$.

Part (c) says every rational point is one of the eight torsion points from part (b). Among their $x$-coordinates, the only negative value of the form $-2s^2$ with $s\ne 0$ is $-2$, arising from $P_2$ or $-P_2$. Thus $s^2=1$, and the four-term arithmetic progression has common difference zero. ConsequentlyClearing denominators gives Euler's corresponding result for integer squares.

Kummer theory begins with the exact sequenceGalois cohomology and Hilbert theorem 90 identifyso cyclic extensions of exponent dividing $n$ are described by adjoining $n$th roots when $K$ contains ${\mu }_n$.

For an elliptic curve $E/K$, multiplication by $n$ gives the Kummer exact sequence of an elliptic curveIts connecting homomorphism is the injective Kummer map of an elliptic curvewhere $nQ=P$. Passing to the finite division field of an elliptic curve $K(E[n])$ makes $E[n]$ constant. Rational functions whose divisors are $n(T)-n(O)$ then express the classes through finitely many elements of $K(E[n]{)}^{\times }/K(E[n]{)}^{\times n}$.

Let $S$ contain the primes above $n$, the primes of bad reduction of an elliptic curve and the finitely many primes introduced by these functions. The local theory of good reduction shows that every Kummer class coming from $E(K)$ is unramified outside $S$, so it lies in an S-unramified power class group. Such a group is finite: valuations outside $S$ vanish modulo $n$, the ideal class group is finite, and the Dirichlet unit theorem makes the group of $S$-units modulo $n$th powers finite. The kernel created by passing to $K(E[n])$ is finite by finite-group Galois cohomology. HenceThis is the Kummer-theoretic proof of the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the Mordell-Weil theorem: $E(K)$ is a finitely generated abelian group.

For $P=[x_0:\cdots :x_N]\in {\mathbb{P}}^N(\mathbb{Q})$, choose coprime integer coordinates and define the projective heightWrite the morphism $F:{\mathbb{P}}^1\to {\mathbb{P}}^1$ as $F=[F_0:F_1]$, where $F_0,F_1\in \mathbb{Z}[X,Y]$ are homogeneous of degree $d$ with no common zero. Bounding each polynomial by the sum of the absolute values of its coefficients gives

Because $F_0$ and $F_1$ have no common projective zero, the Projective Nullstellensatz gives an integer $m\ge d$ and homogeneous polynomials $A_{ij}$ such that suitable nonzero integer multiples of $X^m$ and $Y^m$ lie in the ideal $(F_0,F_1)$. Evaluating at primitive coordinates and using the same coefficient bound givesafter absorbing the bounded common divisor of $F_0(x,y)$ and $F_1(x,y)$ into $C$. ThereforeThis is height growth under a morphism of the projective line.

Let $x:E\to {\mathbb{P}}^1$ be the $x$-coordinate map and define the naive logarithmic height byThe duplication formula induces a degree-four morphism on the $x$-line, so part (a) givesuniformly in $P$. The telescoping sequence $4^{-n}h(2^{n}P)$ is therefore Cauchy. Define the canonical height of an elliptic curve bySumming the geometric error series proves that $|h(P)-\hat{h}(P)|$ is bounded. If another function has this bounded-difference property and scales by four under doubling, evaluating the bounded difference at $2^{n}P$ and dividing by $4^n$ proves uniqueness.

The addition law and part (a) give the approximate parallelogram identityReplace $P,Q$ by $2^{n}P,2^{n}Q$, divide by $4^n$, and let $n\to \infty$. The error disappears, leaving the exact parallelogram lawTogether with $\hat{h}(2P)=4\hat{h}(P)$, induction on $|m|$ yieldsfor every integer $m$.

Let $T$ have order $m>0$. Property (ii) gives $\hat{h}(T)=0$, sinceIt also givesThereforefor every rational torsion point of an elliptic curve. The construction in part (i), quadraticity in part (ii), and this identity establish existence; the bounded-difference and doubling argument in part (i) establishes uniqueness.

Since $\hat{h}-h$ is bounded, a bound on $\hat{h}(P)$ bounds the projective height of $x(P)$. The Northcott theorem gives only finitely many possible rational $x$-coordinates, and each has at most two points above it. Hence

By the Mordell-Weil theorem, $E(\mathbb{Q})\cong {\mathbb{Z}}^r\oplus E(\mathbb{Q}{)}_{\mathrm{tors}}$. The canonical height is a positive-definite quadratic form on the lattice ${\mathbb{Z}}^r$, so canonical-height lattice-point growth givesIf $r=0$, this count is bounded; if $r=1$, it is $O(\sqrt{B})$. Consequently