= Solution
Translate the <rational point> of order two to $(0,0)$ and complete the square. After clearing denominators by admissible scaling, write
$$
E:y^2=x(x^2+ax+b),\qquad a,b\in\mathbb Z,\qquad b(a^2-4b)\ne0.
$$
The companion curve is $E':Y^2=X(X^2-2aX+a^2-4b)$. The <two-isogeny formula> gives maps $\phi:E\to E'$ and $\psi:E'\to E$ satisfying $\psi\phi=[2]$.
For the <Weak Mordell-Weil theorem> in this setting, use the <two-torsion square-class homomorphism>
$$
\alpha(O)=1,\quad\alpha((0,0))=[b],\quad\alpha((x,y))=[x]\in\mathbb Q^*/\mathbb Q^{*2},
$$
and its counterpart $\alpha'$ on $E'$. These are homomorphisms: a nonvertical line intersecting the cubic in three points has the product of their $x$-coordinates a square, by comparison of the constant term of the intersection cubic. For a line through $(0,0)$, the product of the other two coordinates is $b$, explaining the special value there. The kernels are $\psi E'(\mathbb Q)$ and $\phi E(\mathbb Q)$, respectively, as follows by solving the <elliptic isogeny> preimage quadratic and testing its discriminant for a square.
The <prime-support bound in two-isogeny descent> makes both images finite. Indeed at a <prime> not dividing $b$, a negative <valuation> of $x$ makes the leading cubic term dominate, giving $2v(y)=3v(x)$; a positive <valuation> makes $x^2+ax+b$ a unit, giving $2v(y)=v(x)$. In either case $v(x)$ is even. Thus the <square class> of $x$ is supported only on primes dividing $b$, together with a possible sign. The same argument uses $b'=a^2-4b$ on $E'$. Since $2E=\psi\phi E$, the two finite <elliptic isogeny> quotients imply that $E(\mathbb Q)/2E(\mathbb Q)$ is finite.
Finite generation still needs a descent on height. Let $h_x$ be the <logarithmic height> of the $x$-coordinate. The duplication formula has a degree-four rational $x$-coordinate map with coprime numerator and denominator. Polynomial height estimates, with the resultant supplying the lower bound, give the <height growth under a morphism of the projective line> relation $h_x(2P)=4h_x(P)+O(1)$. Thus the convergent telescoping limit
$$
\widehat h(P)=\tfrac12\lim_{n\to\infty}4^{-n}h_x(2^nP)
$$
defines the <canonical height of an elliptic curve>, without assuming finite generation. It is nonnegative, satisfies $\widehat h(2P)=4\widehat h(P)$ and the <height parallelogram identity>, and differs from $h_x/2$ by a bounded amount. Consequently its bounded subsets of <rational points> are finite by the <Northcott theorem>.
Choose the finite list $R_i$ of representatives modulo $2E(\mathbb Q)$ and let $H=\max_i\widehat h(R_i)$. For $P=2Q+R_i$, the <canonical height pairing> inequality gives
$$
\widehat h(Q)=\tfrac14\widehat h(P-R_i)\le\tfrac12\bigl(\widehat h(P)+H\bigr).
$$
Repeated halving brings the height below $H+1$. There are only finitely many such terminal points, and recovering $P$ expresses it as an integer combination of those points and the $R_i$. This proves the <canonical-height proof of Mordell-Weil finite generation>, so
$$
\boxed{E(\mathbb Q)\cong\mathbb Z^{g_E}\oplus T,\qquad T\text{ finite}.}
$$
Back to article page
