Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 21 / 1

The Mordell-Weil theorem gives $E(\mathbb{Q})\cong E(\mathbb{Q}{)}_{\mathrm{tors}}\oplus {\mathbb{Z}}^r$. Move the rational point of order two to $(0,0)$ and clear denominators to obtain an integral Weierstrass equation of an elliptic curveThe two-isogeny formula gives $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$ and dual isogenies $\varphi :E\to E^{\prime }$, $\hat{\varphi }:E^{\prime }\to E$ with $\hat{\varphi }\varphi =[2]$. Compute the torsion subgroup using reduction of torsion points on an elliptic curve at several good primes, followed by division polynomials of an elliptic curve to identify the possible points.

For the free part, use two-isogeny descent. The two-torsion square-class homomorphism isin ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, with kernel $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$; define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly. The prime-support bound in two-isogeny descent restricts the first image to signed square-free divisors of $b$ and the second to those of $a^2-4b$. For each candidate $d$, solve its two-isogeny descent quarticTest local solubility at the real place and the relevant p-adic fields, discard impossible classes, and search the surviving coverings for rational points. A point with $UV\ne 0$ gives $x=d(U/V{)}^2$, $y=dUN/V^3$. Include the identity and rational 2-torsion cases separately. Once both rational images are determined, the two-isogeny rank formula givesThe point searches give explicit candidates for a basis. Their canonical height pairing can certify independence; Mordell-Weil saturation and effective height bounds then certify that they generate the full free part. Rank alone does not certify generators of the whole Mordell-Weil group. If the rational images are not fully determined, the procedure instead supplies lower and upper bounds on the rank of an elliptic curve.

The obstruction is the local-to-global gap in isogeny descent. A two-isogeny descent quartic can have points over $\mathbb{R}$ and every p-adic field without a rational point. Local tests compute an isogeny Selmer group, which can strictly contain the rational-point quotient. For an isogeny of elliptic curves $\varphi :E\to E^{\prime }$, the comparison isThe last group is the kernel of $\varphi :\mathrm{Sha}(E/\mathbb{Q})\to \mathrm{Sha}(E^{\prime }/\mathbb{Q})$, so it records elements of the Tate–Shafarevich group invisible to all local tests. Searching a locally soluble covering which has no rational points never succeeds. Thus a Selmer group of an elliptic curve gives an upper bound and found independent points give a lower bound, and these need not coincide. Higher descent can improve the bounds, but this particular procedure has no unconditional guarantee of closing every gap.

For $E:y^2=x^3-17x$, the two-isogeny formula gives $E^{\prime }:Y^2=X^3+68X$. By the prime-support bound in two-isogeny descent, the image of the two-torsion square-class homomorphism on $E$ lies in $\{1,-1,17,-17\}$. The identity gives $1$, $T=(0,0)$ gives $-17$, and $P=(-1,4)$ gives $-1$, since $(-1{)}^3-17(-1)=16$. The image is a subgroup of the square-class group, so their product $17$ occurs as well; explicitly $P+T=(17,68)$. HenceOn $E^{\prime }$, a nonzero affine point has $X>0$, because $Y^2=X(X^2+68)$. The only candidate square classes are therefore $1,2,17,34$. The identity gives $1$, $T^{\prime }=(0,0)$ gives $[68]=17$, and $(2,12)$ gives $2$, since $2^3+68\cdot 2=144$. Their product $34$ occurs; adding $T^{\prime }$ gives $(34,-204)$. ThusBoth images have order four. The two-isogeny rank formula provesThis is the rank-two elliptic curve with cubic x cubed minus seventeen x; all candidate classes are represented, so no unresolved local-to-global gap in isogeny descent remains in this computation.