 Solution

Write $T=(\theta ,0)$, so $f(\theta )=0$, $K=\mathbb{Q}(\theta )$, and $f^{\prime }(\theta )\ne 0$. Let $[a]$ denote the class of $a\in K^{\ast }$ modulo squares. The required two-torsion square-class homomorphism isThe 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,If none of these points is $T$, evaluating at $\theta$ givesand 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 $x_1,x_2$ satisfyEvaluating at $\theta$ now givesThus $\alpha (T)\alpha (P_1)\alpha (P_2)=[f^{\prime }(\theta ){]}^2=1$, 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.