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

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.