Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 32 (Elliptic Curves)

For the Weierstrass equation of an elliptic curve in this part, $a_1=a_2=a_6=0$, $a_3=1$, $a_4=-1$. HenceThe elliptic-curve discriminant is thereforeEvery prime other than $37$ has unit discriminant, so this integral equation gives good reduction. At $37$ the valuation is one. An Admissible change of Weierstrass coordinates changes discriminant valuations by a multiple of twelve; consequently no integral model can lower this value to zero. Thus the equation is minimal there and the reduction is genuinely bad, rather than an artifact of a nonminimal model. Explicitly $(5,18)$ is singular on the reduced affine cubic: both partial derivatives vanish there. The set of bad reduction primes is

Over ${\mathbb{F}}_2$, both $x=0,1$ give $x^3-x=0$, and both $y=0,1$ give $y^2+y=0$. There are four affine rational points, together with the point $O$ at infinity. Over ${\mathbb{F}}_3$, each $x=0,1,2$ gives $x^3-x=0$, while precisely $y=0,2$ solve $y^2+y=0$. ThusBoth reductions are nonsingular by the discriminant calculation, so these are the point counts of the reduced elliptic curves, with their group laws.

Use reduction of torsion points on an elliptic curve, which is injective on torsion of order prime to the residue characteristic. Any two-primary torsion point of an elliptic curve must inject into the group of order seven at three, so is zero. Any three-primary torsion point of an elliptic curve injects into the group of order five at two, so is zero. For a prime $q\ne 2,3$, its primary torsion injects into both reduced groups and would have order dividing both five and seven, so is also zero. Decomposing the cyclic group generated by any torsion point of an elliptic curve into its primary parts shows that $E(\mathbb{Q})$ has no nonidentity torsion at all. This is the two-prime bound on rational elliptic torsion, with the residue-characteristic exceptions handled explicitly. Since $P=(0,0)$ is not $O$,

The inverse on this Weierstrass equation of an elliptic curve is $(x,y)\mapsto (x,-y-1)$. A line $y=\lambda x+\nu$ through the points being added has third intersection $R$ with $x(R)={\lambda }^2-x_1-x_2$, and the sum is $-R$.

At $P=(0,0)$, implicit differentiation gives $\lambda =(3x^2-1)/(2y+1)=-1$. The tangent is $y=-x$. Substitution into the equation gives $x^2=x^3$, so its third intersection is $(1,-1)$, whose inverse is $(1,0)$. Next the line through $(0,0)$ and $(1,0)$ is $y=0$; its remaining intersection is $(-1,0)$, whose inverse is $(-1,-1)$. Consequently

The good-reduction map respects elliptic addition. Since the finite groups at two and three have orders five and seven, respectively, $5P$ reduces to $O$ at two and $7P$ reduces to $O$ at three. These multiples are not $O$, since $P$ has infinite order. The formal kernel of a minimal Weierstrass equation gives, for every nonidentity point in the reduction kernel at $p$,Thus both coordinates of $5P$ have negative two-adic valuation, and both coordinates of $7P$ have negative three-adic valuation. In particularThere is also an elementary way to see that both coordinates must fail integrality here. An integral affine pair cannot reduce to the point at infinity. If one coordinate had negative valuation and the other nonnegative valuation, the equation would give different valuations on its two sides. When both are negative the lowest-valuation terms give $2v_p(y)=3v_p(x)$, so both are indeed negative.

For a direct numerical confirmation, continuing the chord-and-tangent group law calculation givesThe original printed part repeats its request about $7P$ at the end; the two distinct requested multiples are $5P$ and $7P$.

An isogeny of elliptic curves over $k$ is a nonconstant $k$-morphism $\phi :E_1\to E_2$ which sends $O_1$ to $O_2$. A nonconstant morphism of smooth projective curves is finite and surjective. Its compatibility with the elliptic curve group law from Riemann-Roch can be seen without assuming it in the definition.

Identify a point $P$ with the degree-zero divisor class $[P-O_1]$. A finite morphism pushes forward divisors, and pushes a principal divisor to the principal divisor of the norm of its rational function. It therefore induces a group homomorphism on degree-zero Picard groups of a curve. On geometric points, the pushforward of $[P-O_1]$ is $[\phi (P)-O_2]$. Since point addition is addition of these divisor classes, this provesThe equality is defined over $k$, so in particular it holds on the groups of $k$-rational points and after every field extension. The argument also covers inseparable isogenies of elliptic curves; separability is not part of the definition.

The squares in ${\mathbb{F}}_5$ are $0,1,4$. The following table counts affine solutions at each $x$, with one point at infinity to be added:Hence $\#E_1({\mathbb{F}}_5)=\#E_2({\mathbb{F}}_5)=8$. In characteristic five, a finite point has order two precisely when $y=0$. The first cubic has three roots $0,1,4$, giving four elements killed by two, including $O$. The second cubic has only the root $3$, giving two elements killed by two. An isomorphism of groups would preserve these counts, so the groups are not isomorphic.

More precisely $(2,1)\in E_1$ doubles to $(0,0)$ and has order four; the independent two-torsion point $(1,0)$ gives $C_4\times C_2$. On $E_2$, the point $(0,1)$ doubles to $(4,1)$, which doubles to $(3,0)$; it therefore has order eight. Thus the eight-point isogenous elliptic curves over F5 have point groups

Put $u=x_1-1$, $v=y_1$. Then $v^2=u(u^2+3u+2)$. After using the curve equation, the displayed rational functions becomeSet $X_0=X+2=u+3+2/u$. Direct substitution into $v^2=u(u^2+3u+2)$ givesover ${\mathbb{F}}_5$. Replacing $X_0$ by $X+2$ makes this $Y^2=X^3-X+1$. This verifies the target equation explicitly, and identifies the functions with the two-isogeny formula after translating the source's two-torsion point to zero and then translating the target coordinate.

We must also check the omitted affine points. At $T=(1,0)$, $v$ is a local parameter, and $u=v^2/(x_1(x_1+1))$ has order two, since $x_1(x_1+1)=2$ at $T$. Thus $X$ has a pole of order two and $Y$ one of order three. At $O_1$, the usual pole orders on a smooth cubic in Weierstrass form give poles two and three as well: the correction terms have higher order and do not cancel their leading terms. There are no other poles. A rational map from a smooth curve into a projective curve extends across these points; the indicated pole orders show that both $T$ and $O_1$ map to $O_2$. Hence this is a nonconstant identity-preserving morphism, and part (i) makes it an isogeny of elliptic curves.

The pullback of the pole divisor $2O_2$ of $X$ is $2O_1+2T$. Degrees of divisors multiply by the morphism degree, so $2\mathrm{deg}\phi =4$. We obtainThe degree is prime to five, so the elliptic isogeny is separable. Equivalently $dX/Y=d{x}_1/y_1$ is a nonzero pullback of an invariant differential.

For a Short Weierstrass form $y^2=x^3+Ax+B$, use $c_4=-48A$ and $\Delta =-16(4A^3+27B^2)$. Both curves have $A=-1$, so $c_4=48=3$ in ${\mathbb{F}}_5$. For $E_1$, $B=0$ gives ${\Delta }_1=64=4$; for $E_2$, $B=1$ gives ${\Delta }_2=-368=2$. ThusThe j-invariant of an elliptic curve is preserved by every isomorphism over the algebraic closure. Since these values differ, the two curves are not geometrically isomorphic, even though the preceding part gives a degree-two elliptic isogeny.

Choose an integral Minimal Weierstrass equation whose discriminant is a $p$-adic unit. Given $P=[X:Y:Z]\in E({\mathbb{Q}}_p)$, scale the coordinates so that $X,Y,Z\in {\mathbb{Z}}_p$ and at least one is a unit. DefineAnother primitive representative differs by a unit scalar, so the projective reduction is well defined. The reduced cubic is smooth by good reduction.

To prove the group homomorphism property, view this smooth projective cubic with its section $O$ as the integral elliptic model $\mathcal{E}/{\mathbb{Z}}_p$. Its addition law is defined over ${\mathbb{Z}}_p$, by adding degree-zero divisor classes, and thus is a regular morphism $\mathcal{E}\times \mathcal{E}\to \mathcal{E}$. Properness identifies $E({\mathbb{Q}}_p)$ with the integral sections just described. Reducing the addition morphism is the group operation of the smooth special fiber. ConsequentlyThis is the good-reduction map respects elliptic addition argument; it covers tangent, vertical and infinity cases where reducing a single affine slope formula would involve division by a nonunit. Smoothness also makes reduction surjective: at each reduced point a partial derivative is a unit, and the Hensel lemma lifts that point in a local affine chart.

Near $O$, put $t=-x/y$ and $w=-1/y$, or projectively $t=-X/Y$, $w=-Z/Y$. The integral Weierstrass equation of an elliptic curve becomesThe linear coefficient of $w$ at the origin is a unit, so successive coefficient comparison gives a unique series $w(t)\in {\mathbb{Z}}_p[[t]]$ beginning $t^3$. Expressing the addition morphism in the local parameters of its two inputs givesIts identity, inverse, commutativity and associativity identities follow from the corresponding elliptic group identities. This one-dimensional commutative formal group law is the formal group of an elliptic curve $\hat{E}$ attached to the model.

A point reducing to $O$ lies in the chart with $Y$ a unit and has $t,w\in p{\mathbb{Z}}_p$. The implicit equation determines $w$ uniquely from $t$. Conversely every $t\in p{\mathbb{Z}}_p$ gives the convergent solution $w(t)$ and the point $[-t:1:-w(t)]$; $t=0$ gives $O$. Hence the parameter is a bijection from the reduction kernel to $p{\mathbb{Z}}_p$. The series $F$ converges there and describes its addition. ThereforeFor nonzero $t$ with $v_p(t)=r>0$, $w(t)=t^3$ times a unit, so $v_p(x)=-2r$ and $v_p(y)=-3r$. These are the formal kernel of a minimal Weierstrass equation valuations used earlier.

Write $E[q^{\infty }]$ for the points killed by some power of $q$, rather than any notion of all points which are $q$-divisible. On the reduction kernel, formal multiplication isSince $q\ne p$, its linear coefficient is a unit. Recursive inversion gives an integral inverse series, convergent on $p{\mathbb{Z}}_p$; equivalently the Hensel lemma gives a unique solution to $[q^m](t)=s$ for every $s\in p{\mathbb{Z}}_p$. Thus multiplication by $q^m$ is bijective on the kernel, proving that its $q$-primary torsion is zero. The reduction group homomorphism is therefore injective on $E({\mathbb{Q}}_p)[q^{\infty }]$.

For surjectivity on this torsion, take a reduced point $\tilde{P}$ killed by $q^m$. Lift it to $P\in E({\mathbb{Q}}_p)$ using smooth reduction. Then $q^{m}P$ belongs to the kernel, so choose its unique kernel division point $R$ with $q^{m}R=q^{m}P$. The point $P-R$ is killed by $q^m$ and reduces to $\tilde{P}$. This proves the prime-to-p torsion lifts uniquely at good reduction statement:The right side is a subgroup of a finite group. Thus the left side is finite and has exactly the requested order. No claim that the whole formal kernel is torsion-free at $p=2$ is needed.

Translate the rational point of order two to $(0,0)$ and complete the square. After clearing denominators by admissible scaling, writeThe companion curve is $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$. The two-isogeny formula gives maps $\varphi :E\to E^{\prime }$ and $\psi :E^{\prime }\to E$ satisfying $\psi \varphi =[2]$.

For the Weak Mordell-Weil theorem in this setting, use the two-torsion square-class homomorphismand its counterpart ${\alpha }^{\prime }$ on $E^{\prime }$. These are homomorphisms: a nonvertical line intersecting the cubic in three points has the product of their $x$-coordinates a square, by comparison of the constant term of the intersection cubic. For a line through $(0,0)$, the product of the other two coordinates is $b$, explaining the special value there. The kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$, respectively, as follows by solving the elliptic isogeny preimage quadratic and testing its discriminant for a square.

The prime-support bound in two-isogeny descent makes both images finite. Indeed at a prime not dividing $b$, a negative valuation of $x$ makes the leading cubic term dominate, giving $2v(y)=3v(x)$; a positive valuation makes $x^2+ax+b$ a unit, giving $2v(y)=v(x)$. In either case $v(x)$ is even. Thus the square class of $x$ is supported only on primes dividing $b$, together with a possible sign. The same argument uses $b^{\prime }=a^2-4b$ on $E^{\prime }$. Since $2E=\psi \varphi E$, the two finite elliptic isogeny quotients imply that $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite.

Finite generation still needs a descent on height. Let $h_x$ be the logarithmic height of the $x$-coordinate. The duplication formula has a degree-four rational $x$-coordinate map with coprime numerator and denominator. Polynomial height estimates, with the resultant supplying the lower bound, give the height growth under a morphism of the projective line relation $h_x(2P)=4h_x(P)+O(1)$. Thus the convergent telescoping limitdefines the canonical height of an elliptic curve, without assuming finite generation. It is nonnegative, satisfies $\hat{h}(2P)=4\hat{h}(P)$ and the height parallelogram identity, and differs from $h_x/2$ by a bounded amount. Consequently its bounded subsets of rational points are finite by the Northcott theorem.

Choose the finite list $R_i$ of representatives modulo $2E(\mathbb{Q})$ and let $H={\max }_i\hat{h}(R_i)$. For $P=2Q+R_i$, the canonical height pairing inequality givesRepeated halving brings the height below $H+1$. There are only finitely many such terminal points, and recovering $P$ expresses it as an integer combination of those points and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation, so

Retain $E:y^2=x(x^2+ax+b)$ and put $b^{\prime }=a^2-4b$. The explicit two-isogeny formula iswith the maps extended at the kernel points. Direct substitution verifies the equations and $\psi \varphi =[2]$.

Here is a concrete kernel test for the descent homomorphisms. A preimage of $(X,Y)\in E^{\prime }$ has $x$ satisfying $x^2+(a-X)x+b=0$, whose discriminant is $(X-a{)}^2-4b$. The target equation says $Y^2=X((X-a{)}^2-4b)$, so for nonexceptional points this quadratic has a rational solution precisely when $X$ is a square. The resulting $y$ follows from the displayed elliptic isogeny formula. For a nonzero target two-torsion point, the preimage quadratic has a double root $x=(X-a)/2$, and the source equation gives $y^2=x^{2}X$, so the same square condition still applies. At $(0,0)\in E^{\prime }$, a rational preimage exists precisely when $b^{\prime }$ is a square, matching the special square-class value ${\alpha }^{\prime }((0,0))=[b^{\prime }]$. The dual argument gives the other kernel. ThusLet $\delta =[\ker \psi :\ker \psi \cap \varphi E(\mathbb{Q})]$. The quotient map induced by $\psi$ has this kernel, givingIf $b^{\prime }$ is a rational square, $E$ has four rational two-torsion points and $\delta =1$; otherwise it has two and $\delta =2$. In either case $\delta \#E(\mathbb{Q})[2]=4$. Combining with finite generation therefore proves the two-isogeny rank formulaThis derivation checks the factor four, including curves with full rational two-torsion.

To compute these images, enumerate the signed square-free divisors $d$ of $b$ and $b^{\prime }$. Substituting $x=d(U/V{)}^2$ yields the quartic covering in a two-isogeny descentThe class $d$ occurs precisely when this covering has a rational point, with the identity and two-torsion handled by the boundary cases. Test at the real place and at primes dividing $2b{b}^{\prime }$ to rule out candidates, and search for rational points to prove that surviving classes occur. The locally soluble classes form a finite elliptic isogeny Selmer group of an elliptic curve, giving an upper rank bound. rational points and independent height pairings give a lower bound. When the bounds agree the rank is proved. Local solubility alone does not prove a rational point exists: if a gap remains, a higher descent or further information about the covering obstructions is needed. This is why the procedure usually computes the rank but is not an unjustified universal termination claim.

For a rank-zero example, take $E_0:y^2=x^3+x$, with rational two-torsion $(0,0)$. Its two-isogeny formula companion is $E_0^{\prime }:y^2=x^3-4x$. The prime-support bound in two-isogeny descent allows only $\pm 1$ on $E_0$; all nonzero affine $x$ are positive because $x(x^2+1)$ is a square. Hence $A_0=\{1\}$. For $E_0^{\prime }$ the possible classes are $\{1,-1,2,-2\}$, all realized: $O$ gives $1$, $(0,0)$ gives $[-4]=-1$, and $(2,0),(-2,0)$ give $2,-2$. ThereforeThis is the rank-zero elliptic curve with cubic x cubed plus x certificate; no unresolved locally soluble covering is being counted as soluble.

For a positive-rank example, take $E_1:y^2=x^3-2x$. Its companion is $E_1^{\prime }:y^2=x^3+8x$. The first image is contained in $\{1,-1,2,-2\}$. The identity gives $1$, $(0,0)$ gives $-2$, and $(2,2)$ gives $2$; multiplicativity of the square-class group homomorphism also gives $-1$. Thus $A_1$ has all four classes. On $E_1^{\prime }$, the same prime-support bound allows $\pm 1,\pm 2$, but $x(x^2+8)$ is nonnegative only for $x\ge 0$, eliminating the negative classes. The identity gives $1$ and $(0,0)$ gives $[8]=2$, so $B_1=\{1,2\}$. ConsequentlyThe rank-one elliptic curve with cubic x cubed minus twice x therefore supplies the required positive-rank example. The exhibited point $(2,2)$ is itself nontorsion: doubling gives $(9/4,-21/8)$, whose nonintegral coordinates contradict the Nagell–Lutz theorem if it were torsion. No assertion that this point generates the full free part is needed for the rank calculation.