Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / B / 6

We use two-isogeny descent, stating the arithmetic results involved. Fordefine the two-torsion square-class homomorphism ${\alpha }_E:E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$ by ${\alpha }_E(o)=1$, ${\alpha }_E((0,0))=b$, and ${\alpha }_E((x,y))=x$ otherwise. Its kernel is the image of the dual two-isogeny formula; the analogous map ${\alpha }_{E^{\prime }}$ has kernel the image of the forward two-isogeny formula. The two-isogeny rank formula isThe images have signed squarefree representatives supported on primes dividing $b$ and $a^2-4b$, respectively. This prime-support bound in two-isogeny descent follows by valuation comparison: if a prime does not divide $b$, the valuations of $x$ at that prime must be even, whether positive or negative. A candidate squarefree class $d$ is represented precisely when its quartic covering in a two-isogeny descenthas the corresponding rational solution, apart from the separately included identity and two-torsion cases; clearing denominators permits coprime integers $U,V$.

Here $a=5,b=-5$, soOn $E$ the possible classes are $1,-1,5,-5$. They all occur: the identity gives one, $T=(0,0)$ gives $-5$, and $(-1,3)$ gives $-1$. Since the image is a group, their product also gives five. Thus its image has order four.

On $E^{\prime }$, the quadratic factor $(X-5{)}^2+20$ is strictly positive for real $X$. Every nonzero rational $X$-coordinate is therefore positive. The prime-support bound in two-isogeny descent restricts its image to $1,3,5,15$. The identity gives one and $(0,0)$ gives $45$, that is class five.

Class three is impossible. Its quartic covering in a two-isogeny descent would requireModulo three this is $N^2\equiv 2U^2{V}^2$, so three divides $N$ and at least one of $U,V$. They cannot both be divisible. If $3\mid U$ and $3\nmid V$, the right side has valuation exactly one, from $15V^4$. If $3\mid V$ and $3\nmid U$, it again has valuation exactly one, from $3U^4$. Either case contradicts its being a square. Class fifteen is then impossible too, since multiplying it by the known image class five would give class three.

Hence the second image is exactly $\{1,5\}$, of order two. The rank formula givesThis proves the rank-one elliptic curve with coefficients five and minus five calculation, including the local obstruction that removes the spurious extra descent classes.

For $E:y^2=x^3+px$ with $p>0$, the factor $x^2+p$ is strictly positive on the real line. Thus every nonzero rational $x$-coordinate is positive. In the two-isogeny descent map, prime support restricts its square class to $1$ or $p$. Both are realized by the identity and $(0,0)$, although only the upper bound is needed:The two-isogenous curve is $E^{\prime }:Y^2=X^3-4pX$. Its two-torsion square-class homomorphism has signed squarefree representatives supported on the primes dividing $4p$. For odd $p$ these are amongHence the second image has at most eight elements. At $p=2$ the list reduces to $1,-1,2,-2$, so the bound by eight remains valid. The two-isogeny rank formula therefore givesIn fact the same simple count yields $r\le 1$ when $p=2$. The positivity restriction on the first descent image is essential for the stated bound of two for all primes.

Use the two-isogeny formula formula with $a=\alpha$ and $b=-\alpha$. Its companion isBoth curves are nonsingular because $\alpha \ne 0,-4$. With $b^{\prime }=\alpha (\alpha +4)$, chooseEach map extends across its exceptional affine points, and $\hat{\varphi }\varphi =[2]$. On $D$, the first coordinate of the dual can also be written $v^2/(4u^2)$.

Let $P=(1,1)$ and solve $\hat{\varphi }(u,v)=P$. The first coordinate gives $v^2=4u^2$. Since $u\ne 0$, the equation of $D$ then givesThe second coordinate selects the signs. The two rational preimages are exactlySubstituting into the dual formula checks that both map to $(1,1)$. Their coordinates are nonexceptional under the assumptions, and the degree-two map has no further preimages.

The two-torsion square-class homomorphism ${\alpha }_D$ has kernel $\varphi (C(\mathbb{Q}))$, and on these two points it is their nonzero $u$-coordinate modulo squares. Thus $Q_1$ is in the image of $\varphi$ exactly when $\alpha$ is a rational square, and $Q_2$ is in that image exactly when $\alpha +4$ is a rational square.

If $P=2R$, then $\varphi (R)$ is one of the two preimages, because $\hat{\varphi }\varphi (R)=2R=P$. Conversely, if $Q_j=\varphi (R)$, applying the dual shows $P=2R$. ThereforeFinally, if an integer equals $(a/b{)}^2$ in lowest terms, then $b^2\mid a^2$ forces $b=1$. Thus rational squareness of these integer parameters is exactly integer squareness, andThis is the two-isogeny halving criterion for the alpha family.