Show $[\mathbb{Q}(\cos\frac{\theta}{3}, \cos\theta): \mathbb{Q}(\cos\theta)]=2$ implies that $\theta$ is trisectable using straightedge and compass.

Question

Show $[\mathbb{Q}(\cos\frac{\theta}{3}, \cos\theta): \mathbb{Q}(\cos\theta)]=2$ implies that $\theta$ is trisectable using straightedge and compass.

Proof: $[\mathbb{Q}(\cos\frac{\theta}{3}, \cos\theta): \mathbb{Q}(\cos\theta)]=2$ implies that $\mathbb{Q}(\cos\frac{\theta}{3}, \cos\theta)=\mathbb{Q}(\sqrt{\alpha})$ for some $\alpha \in \mathbb{Q}(\cos\theta)$. Let $P=\{0, 1, \cos\theta\}$, then clearly $\mathbb{Q}(P)=\mathbb{Q}(\cos\theta)$, hence $[\mathbb{Q}(\sqrt{\alpha}): \mathbb{Q}(P)]=[\mathbb{Q}(\cos\frac{\theta}{3}, \cos\theta): \mathbb{Q}(\cos\theta)]=2$, thus $\sqrt{\alpha}\in \mathbb{Q}(P)^{py}$, where $\mathbb{Q}(P)^{py}$ is the Pythagorean closure. Therefore $\mathbb{Q}(\sqrt{\alpha}) \subseteq \mathbb{Q}(P)^{py}$.

Now, $\cos\frac{\theta}{3}\in \mathbb{Q}(\cos\frac{\theta}{3}, \cos\theta)=\mathbb{Q}(\sqrt{\alpha})$, hence $\cos\frac{\theta}{3} \in \mathbb{Q}(P)^{py}$, meaning $\cos\frac{\theta}{3}$ is constructible from $P$, and it follows that $\theta$ can be trisected. QED.

I am a bit skeptical about my application of the theorem involving the Pythagorean closure $\mathbb{Q}(P)^{py}$. Please verify the proof if you can-many thanks in advance!

Answer 1

This is just a special case of: If $a_1,\ldots,a_n$ are given (or constructible) and $[F:\Bbb Q(a_1,\ldots, a_n)]=2$, then every $a\in F$ is constructible from the givens. One can use Pythagorean closure as you did, but might as well use directly that quadratic equations (with constructible coefficients) can be solved with straight-edge and compasses.