Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 26 / 2

Bad reduction of an elliptic curve must be tested on a Minimal Weierstrass equation, not inferred merely from primes dividing the displayed elliptic-curve discriminant. For an odd prime $p$, write $v_p(D)=2k+ϵ$ with $ϵ\in \{0,1\}$. The coordinate change $x=p^{2k}X$, $y=p^{3k}Y$ gives the same form with parameter $D/p^{2k}$. If $ϵ=0$, that parameter is a unit at $p$ and its discriminant is a unit, so reduction is good.

If $ϵ=1$, the new discriminant valuation is six. Discriminants of isomorphic Weierstrass models differ by a twelfth power under a coordinate change, so their valuations differ by a multiple of twelve. An integral model has nonnegative discriminant valuation; consequently a model of valuation six cannot be replaced by one of valuation zero. Reduction is bad. At $p=2$, oddness of $D$ makes the original discriminant valuation six, and the same argument gives bad reduction of an elliptic curve there as well. ThusThese are the bad primes of a congruent number curve with odd parameter. If $D$ is squarefree this simplifies to the primes dividing $2D$. Squarefreeness is not stated in the PDF: for example, $D=9$ gives a curve isomorphic to $E_1$ by $x=9X,y=27Y$, so it has good reduction at three although the displayed discriminant is divisible by three.

At every good prime we have $p\ne 2$. After removing any even p-adic valuation in the parameter as in part (a), use a smooth reduction $y^2=x^3-d^{2}x$ with $d\in {\mathbb{F}}_p^{\times }$. Let $\chi$ be the Legendre symbol, extended by $\chi (0)=0$. Counting the two, one or zero y-values above each x givesIf $p\equiv 3(\mathrm{mod}4)$, then $\chi (-1)=-1$. The terms for $x$ and $-x$ cancel, including zeros, because the cubic is odd. Hence $a_p=0$.

For the converse, the four distinct points $O,(0,0),(d,0),(-d,0)$ form the full rational 2-torsion subgroup of the reduced curve. Its group order is therefore divisible by four. If $p\equiv 1(\mathrm{mod}4)$, zero trace would give group order $p+1\equiv 2(\mathrm{mod}4)$, a contradiction. ConsequentlyThis is the vanishing trace criterion for a congruent number curve. The Hasse theorem for elliptic curves statesFor good primes that divide a nonsquarefree displayed parameter, the tilde denotes the smooth reduction of a minimal model, not the singular special fibre of that nonminimal equation.

We first determine the torsion. The rational 2-torsion points are $O,(0,0),(5,0),(-5,0)$. At the good primes three and seven, part (b) gives group orders four and eight. Prime-to-residue-characteristic torsion injects under reduction: an element in the kernel has a formal parameter $t$ of positive valuation, and multiplication by an integer prime to the residue characteristic has series $mt+O(t^2)$ with unit leading coefficient, so cannot kill a nonzero such parameter. At seven this excludes rational 3-primary torsion; at three it excludes 7-primary torsion and every odd-primary component other than the already excluded 3-primary component. Thus all torsion is 2-primary and injects at three, bounding its order by four. Consequently

For the rank, use two-isogeny descent withThe two-isogeny formula givesThey extend at the exceptional points, have kernels $\{O,(0,0)\}$ on their respective curves, and satisfy $\psi \varphi =[2]$. These assertions can be checked by substitution; for example the composed x-coordinate is $(x^2+25{)}^2/[4x(x^2-25)]$, the duplication x-coordinate. The maps respect addition because they are morphisms fixing $O$: pushforward of divisor classes is additive, takes principal divisors to principal divisors by the norm of a rational function, and under the identification in part 1(a) sends $[(P)-(O)]$ to $[(\varphi P)-(O)]$, and likewise for $\psi$.

Define the two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$ by $\alpha (x,y)=[x]$, $\alpha (O)=1$, and $\alpha ((0,0))=[-25]=[-1]$. Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[100]=1$. To see the homomorphism property, a nonvertical line $y=mx+c$ cuts either curve in three points whose x-coordinates have product $c^2$. The chord-and-tangent law therefore makes their square classes multiply to one. When the line passes through $(0,0)$, the two remaining x-coordinates have product equal to the coefficient of $x$, exactly explaining the assigned value there. Vertical lines give the inverse-point relation. The same argument includes repeated tangent intersections.

The kernels are $\ker \alpha =\psi E^{\prime }(\mathbb{Q})$ and $\ker {\alpha }^{\prime }=\varphi E(\mathbb{Q})$. Here is an explicit check, not just an invocation of the descent correspondence. The x-coordinate of every nonexceptional $\psi$-image is a square. Conversely, if $(x,y)\in E$ has $x=s^2\ne 0$, put $X=2x+2y/s$ and $Y=2sX$. Then $X^2-4xX+100=0$, so $(X,Y)\in E^{\prime }$ and the displayed $\psi$ maps it to $(x,y)$. For $E^{\prime }$, every nonexceptional $\varphi$-image has $X=(y/x{)}^2$; if $X=s^2\ne 0$, put $x=(X+Y/s)/2$, $y=sx$. Substitution gives a point of $E$ mapping to $(X,Y)$. The point $(0,0)\in E^{\prime }$ has square class one and equals $\varphi (5,0)$; the point $(0,0)\in E$ has nonsquare class $-1$. Together with the origins these checks handle every exception.

The prime-support bound in two-isogeny descent can be read directly from $y^2=x(x^2+B)$. At a prime not dividing the integral coefficient $B$, positive valuation of $x$ gives $2v(y)=v(x)$, while negative valuation gives $2v(y)=3v(x)$; either way the valuation of $x$ is even. Thus the first image has only the candidate classes $1,-1,5,-5$. All four occur, already at the four torsion points. Hence $\#\alpha (E)=4$.

On $E^{\prime }$, nonzero real x-coordinates are positive since $X(X^2+100)$ has the sign of $X$. Prime support leaves only $1,2,5,10$. The point $(5,25)$ supplies class five. Class two would give the quartic covering in a two-isogeny descentIndeed write $X=2(U/V{)}^2$ and clear the square factor in the curve equation. If $5\nmid U$, the right side is two modulo five, impossible for a square. If $5\mid U$, then $5\nmid V$ and $5\mid W$; division by twenty-five gives $(W/5{)}^2\equiv 2V^4\equiv 2(\mathrm{mod}5)$, again impossible. This is the five-adic obstruction to square class two on the curve with coefficient one hundred. Because the image is a subgroup and contains five, class ten is excluded as well. Therefore ${\alpha }^{\prime }(E^{\prime })=\{1,5\}$ has size two.

Now $[E:\psi E^{\prime }]=4$. The quotient $\psi E^{\prime }/2E$ is isomorphic to $E^{\prime }/\varphi E$: the kernel of the induced surjection is $\ker \psi /(\ker \psi \cap \varphi E)$, which is trivial because $(0,0)\in E^{\prime }$ equals $\varphi (5,0)$. Thus $[E:2E]=4\cdot 2=8$. Finite generation follows from the Mordell-Weil theorem, whose height-descent proof is supplied in the essay below. A finitely generated group with the torsion already found has $[E:2E]=2^{r+2}$, so $r=1$. HenceThe point $(-4,6)$ has infinite order, since it is not one of the torsion points. No assertion that this particular point generates the free factor is needed.

Let $P_0=(-4,6)$, the infinite-order point established in part (c). Every positive multiple $n{P}_0$ is rational and is neither $O$ nor 2-torsion: either possibility would force some nonzero multiple of $P_0$ to vanish. Write its coordinates as $(x,y)$, so $y\ne 0$ and $x\notin \{0,5,-5\}$.

The stated algebraic identity gives positive rational sidesThey satisfy $a^2+b^2=c^2$, and their area isby the curve equation. This is the infinitely many rational right triangles from a nontorsion point construction for the congruent number elliptic curve.

The infinitely many points cannot produce only finitely many triangles. For a fixed ratio $r=c/b$, the positive number $X=|x|$ satisfies $X^2-10rX+25=0$, so there are at most two choices for $|x|$, then at most two signs of $x$ and two signs of $y$. Thus the map to ordered triangles is finite-to-one. Permuting the two legs introduces only another finite ambiguity. Consequently there are infinitely many distinct rational right triangles of area five. For $P_0$ itself the sides are $(3/2,20/3,41/6)$.