Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / B / 7

In the typical case, the line through $P,Q$, or their tangent when they coincide, is $v=mu+c$. Its intersections with $D$ have first coordinates given byTheir product is $c^2$. The third intersection is $-(P+Q)$, which has the same first coordinate as $P+Q$. ConsequentlyPassing to the quotient by rational squares gives $q(P)q(Q)q(P+Q)=1$. Every square class is its own inverse, soThe excluded cases have their specified special values; the source asks only for this chord-and-tangent argument away from them.

For an affine point of $C$ with $x\ne 0$, the first coordinate of its image is $(y/x{)}^2$. Thus its image has trivial square class whenever that coordinate is nonzero. The identity and $T=(0,0)$ map to the identity. If a point maps to $T_1=(0,0)$, it has $y=0$ and $x^2+ax+b=0$, forcing $b_1=a^2-4b$ to be a rational square; in that case $q(T_1)=1$ too. Hence $\varphi (C(\mathbb{Q}))\subseteq \ker q$.

Conversely, let $(u,v)\in D(\mathbb{Q})$ have $q(u,v)=1$ and $u\ne 0$. Write $u=t^2$ with $t\in {\mathbb{Q}}^{\ast }$. The equation of $D$ givesSet $x=(u-a+v/t)/2$ and $y=tx$. Then $x$ solves $x^2+(a-u)x+b=0$ and is nonzero because $b\ne 0$. Hence $y^2=u{x}^2=x(x^2+ax+b)$, so $(x,y)\in C(\mathbb{Q})$. Also $x-b/x=v/t$, giving $\varphi (x,y)=(u,v)$.

The identity is an image of the identity. Finally, $T_1\in \ker q$ exactly when $b_1$ is a rational square; then either nonzero root $x=(-a\pm \sqrt{b_1})/2$ gives $(x,0)\in C(\mathbb{Q})$ mapping to $T_1$. This covers all exceptional points and proves

Here $a=1$, $b=-2$ and the two-isogenous curve is $D:v^2=u(u^2-2u+9)$. Let ${\delta }_C$ and ${\delta }_D$ be the first-coordinate square-class maps, using the coefficient $b$ or $b_1$ at the point $(0,0)$. The same chord argument shows that both maps are homomorphisms.

For an equation $y^2=x(x^2+ax+b)$ with integer coefficients, its first-coordinate square class has a square-free representative dividing $b$, including a possible sign. Indeed, at a prime not dividing $b$, a negative valuation of $x$ must be even because $x^3$ dominates; a positive valuation must be even because $x^2+ax+b$ is a unit. Thus only primes dividing $b$ can occur to odd order. For $C$, the possibilities are $1,-1,2,-2$. The classes $1,-2$ occur at the identity and $(0,0)$. If class two occurred, write $x=2M^2/e^2$ with coprime integers $M,e$. Substituting in the equation would give a rational square, necessarily an integer square,Modulo three, the right side is two whenever $M,e$ are not both divisible by three: this holds whether just one or both are units. That is impossible. Since the image is a subgroup containing $-2$, exclusion of two also excludes minus one. Therefore $|{\delta }_C(C(\mathbb{Q}))|=2$.

For $D$, the quadratic $u^2-2u+9=(u-1{)}^2+8$ is positive, so every nonzero first coordinate is positive. Its square class can only be one or three. Both occur, with three supplied by $(3,6)$, so $|{\delta }_D(D(\mathbb{Q}))|=2$.

Let $\psi :D\to C$ be the dual two-isogeny, with $\psi \varphi =[2]$. The kernel identities of part (ii), applied to both curves and the usual coordinate scaling, give $[C:\psi D]=2$ and $[D:\varphi C]=2$. Moreover $\ker \psi =\{O,T_1\}$ lies in $\varphi C$, since $b_1=9$ is a square. Thus $[\psi D:2C]=[D:\varphi C]=2$, andThis is a rank-zero two-isogeny descent for roots zero one and minus two. The rational two-torsion has four points, since the cubic factors as $x(x-1)(x+2)$. By the Mordell-Weil theorem, the last index is $2^r|C(\mathbb{Q})[2]|=2^{r+2}$. Therefore