Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 125 / 5

Start with an elliptic curve having a rational point $T$ of order $2$. Move that point to $(0,0)$ and take an integral modelPut $b^{\prime }=a^2-4b$ andThe two-isogeny formula and its dual areThey extend to the projective curves, with kernels $\{O,T\}$ and $\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)\in E^{\prime }$. Substitution verifies their target equations, and the elliptic-curve addition formula verifies $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$. The division by $4,8$ in the dual identifies the twice-transformed curve, with coefficients $4a,16b$, with $E$.

Define the two-torsion square-class homomorphismsQuestion 1(ii) proves these are homomorphisms, since $f^{\prime }(0)=b$. Their kernels areFor completeness, the first coordinate of the dual is $Y^2/(4X^2)$, so a nonexceptional dual image has square $x$-coordinate. Conversely, if $x=u^2\ne 0$, the preimage equation iswhose discriminant is $16(x^2+ax+b)=16(y/u{)}^2$. The roots $X=2x+a\pm 2y/u$ are rational and nonzero. Choosing $Y=\pm 2uX$ gives a point on $E^{\prime }$, and a choice of sign makes its dual image exactly $(x,y)$. The point $T$ has a rational dual preimage precisely when $b$ is a square, as seen from the roots $a\pm 2\sqrt{b}$ of the nonzero two-torsion polynomial on $E^{\prime }$. The identity is already a dual image. Applying this argument to the transformed curve proves the second kernel assertion as well, because scaling an $x$-coordinate by $4$ does not change its square class.

The two-isogeny descent is finite because an image square class has a signed square-free integer representative dividing $b$. Indeed, if $\ell \nmid b$ and $v_{\ell }(x)>0$, the other factor $x^2+ax+b$ is a unit, so $v_{\ell }(y^2)=v_{\ell }(x)$ forces an even valuation. If $v_{\ell }(x)<0$, the term $x^2$ dominates that factor and $v_{\ell }(y^2)=3v_{\ell }(x)$, again forcing even valuation. Thus odd valuations can occur only at primes dividing $b$. The same argument applies to $b^{\prime }$ on $E^{\prime }$.

For each candidate signed squarefree divisor $d\mid b$, put $x=d(U/V{)}^2$, with coprime integers $U,V$. The curve equation becomes the two-isogeny descent quarticA solution with $UV\ne 0$ yields $x=d{U}^2/V^2$ and $y=dUN/V^3$. Conversely, every point in that class gives such a primitive integer solution, since a rational square root of an integer is integral. The boundary solutions $U=0$ and $V=0$ account respectively for the classes of $T$ and $O$. Rational solutions prove that a class occurs; a real or congruence obstruction excludes it. Merely finding local solutions everywhere does not automatically prove a rational solution.

To extract the rank, write $A=\alpha (E(\mathbb{Q}))$ and $A^{\prime }={\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))$. The Mordell-Weil theorem givesThe isogeny factorization gives the index $|A|$ for the first quotient by $\hat{\varphi }{E}^{\prime }$. For the remaining index, apply $\hat{\varphi }$ to $E^{\prime }/\varphi E$. Its kernel has size $2/e$, whereThe equality follows because $\hat{\varphi }(\varphi P)=2P$. Consequentlyand cancellation yields the two-isogeny rank formulaThis formula is valid whether there is just one rational nonzero two-torsion point or all three.

For the first curve, $a=-1,b=1$, and the isogenous curve isThe only candidate square classes on $E$ are $1,-1$. Since $x^2-x+1>0$ for every real $x$, a real affine point has $x\ge 0$; the exceptional torsion class is $[b]=1$. ThusOn $E^{\prime }$, the candidates are $1,-1,3,-3$. They all occur: the identity gives $1$, $T^{\prime }$ gives $-3$, $(3,6)$ gives $3$, and $(-1,2)$ gives $-1$. Hence $|A^{\prime }|=4$, andThe product in the rank formula is $1\cdot 4=4$, not $1\cdot 2$; all four classes on the companion curve are essential.

For the second curve, $a=5,b=-6$, andThe candidate classes on $E$ are $\{\pm 1,\pm 2,\pm 3,\pm 6\}$. The identity, $T$, $(2,4)$ and $(-3,6)$ show thatThis is a subgroup of order $4$. Its other coset is $\{3,6,-1,-2\}$, so it suffices to exclude the representative $3$.

The modulo-eight obstruction to a two-isogeny descent class uses the corresponding two-isogeny descent quartic,If both $U,V$ are odd, its right side is $6$ modulo $8$. If $U$ is odd and $V$ even, it is $3$ or $7$ modulo $8$. If $U$ is even and $V$ odd, it is $6$ or $2$ modulo $8$. These are all primitive parity cases, and none is a square modulo $8$. Hence the class $3$ is impossible. Since $A$ is a subgroup, every class in its coset is impossible, and thereforeOn the companion curve, the only candidates are $\pm 1,\pm 7$. Its quadratic factor is $(X-5{)}^2+24>0$, so every nonzero real affine $X$ is positive. The exceptional value $[49]$ is $1$, and the point $(7,14)$ supplies the class $7$. ThusThe rank formula gives $2^r=4\cdot 2/4=2$, soEvery included class has an explicit rational representative and every excluded coset has a proved real or congruence obstruction, so these are exact ranks rather than bounds obtained from a point search.