= Solution
Write $T=(\theta,0)$, so $f(\theta)=0$, $K=\mathbb Q(\theta)$, and $f'(\theta)\ne0$. Let $[a]$ denote the class of $a\in K^*$ modulo squares. The required <two-torsion square-class homomorphism> is
$$
\boxed{\alpha(O)=1,\qquad\alpha(P)=[x(P)-\theta]\ (P\ne O,T),\qquad\alpha(T)=[f'(\theta)].}
$$
The last value is needed in the domain only if $T\in E(\mathbb Q)$. Every displayed nonexceptional value is nonzero, because an affine point with $x=\theta$ necessarily has $y=0$ and equals $T$.
Consider a nonvertical line $y=mx+c$ whose three intersection points have $x$-coordinates $x_1,x_2,x_3$, counted with multiplicity. Since $f$ is monic,
$$
f(x)-(mx+c)^2=\prod_{i=1}^3(x-x_i).
$$
If none of these points is $T$, evaluating at $\theta$ gives
$$
\prod_{i=1}^3(x_i-\theta)=(m\theta+c)^2,
$$
and therefore the product of their $\alpha$-values is $1$ in the <square-class group>. Tangencies are included by repeated factors. Since negation preserves the $x$-coordinate, this identity says $\alpha(P+Q)=\alpha(P)\alpha(Q)$.
If the line passes through $T$, put $f(x)=(x-\theta)g(x)$ and write the line as $y=m(x-\theta)$. The other two intersection abscissae $x_1,x_2$ satisfy
$$
g(x)-m^2(x-\theta)=(x-x_1)(x-x_2).
$$
Evaluating at $\theta$ now gives
$$
(x_1-\theta)(x_2-\theta)=g(\theta)=f'(\theta).
$$
Thus $\alpha(T)\alpha(P_1)\alpha(P_2)=[f'(\theta)]^2=1$, as required. This covers a tangent at a different point whose third intersection is $T$ as well.
The tangent at $T$ is vertical. For a vertical chord or tangent the two affine intersections are $P,-P$, so their product of <square classes> is $\alpha(P)^2=1$; in particular $\alpha(T)^2=1$ agrees with $2T=O$. Finally, adding $O$ changes neither side. These cases exhaust the <chord-and-tangent group law>, proving that the displayed map is a <group homomorphism>.
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