 Solution

Retain $E:y^2=x(x^2+ax+b)$ and put $b^{\prime }=a^2-4b$. The explicit two-isogeny formula iswith the maps extended at the kernel points. Direct substitution verifies the equations and $\psi \varphi =[2]$.

Here is a concrete kernel test for the descent homomorphisms. A preimage of $(X,Y)\in E^{\prime }$ has $x$ satisfying $x^2+(a-X)x+b=0$, whose discriminant is $(X-a{)}^2-4b$. The target equation says $Y^2=X((X-a{)}^2-4b)$, so for nonexceptional points this quadratic has a rational solution precisely when $X$ is a square. The resulting $y$ follows from the displayed elliptic isogeny formula. For a nonzero target two-torsion point, the preimage quadratic has a double root $x=(X-a)/2$, and the source equation gives $y^2=x^{2}X$, so the same square condition still applies. At $(0,0)\in E^{\prime }$, a rational preimage exists precisely when $b^{\prime }$ is a square, matching the special square-class value ${\alpha }^{\prime }((0,0))=[b^{\prime }]$. The dual argument gives the other kernel. ThusLet $\delta =[\ker \psi :\ker \psi \cap \varphi E(\mathbb{Q})]$. The quotient map induced by $\psi$ has this kernel, givingIf $b^{\prime }$ is a rational square, $E$ has four rational two-torsion points and $\delta =1$; otherwise it has two and $\delta =2$. In either case $\delta \#E(\mathbb{Q})[2]=4$. Combining with finite generation therefore proves the two-isogeny rank formulaThis derivation checks the factor four, including curves with full rational two-torsion.

To compute these images, enumerate the signed square-free divisors $d$ of $b$ and $b^{\prime }$. Substituting $x=d(U/V{)}^2$ yields the quartic covering in a two-isogeny descentThe class $d$ occurs precisely when this covering has a rational point, with the identity and two-torsion handled by the boundary cases. Test at the real place and at primes dividing $2b{b}^{\prime }$ to rule out candidates, and search for rational points to prove that surviving classes occur. The locally soluble classes form a finite elliptic isogeny Selmer group of an elliptic curve, giving an upper rank bound. rational points and independent height pairings give a lower bound. When the bounds agree the rank is proved. Local solubility alone does not prove a rational point exists: if a gap remains, a higher descent or further information about the covering obstructions is needed. This is why the procedure usually computes the rank but is not an unjustified universal termination claim.