Essentially an angle trisector would allow you to construct a root of $4x^3 - 3x - a$ for any constructible $a$ with $|a| \le 1$. By scaling and translation, it would seem this means that using a compass, straightedge, and angle trisector, one can construct any cubic field extension with positive discriminant.

It isn't at all obvious to me, though, whether or not there might be some trick which would allow constructing roots of cubic equations with negative discriminant - and that's exactly the case which $\sqrt[3]{2}$ falls into. For example, might it be possible to construct an extension of $\mathbb{Q}(\sqrt[3]{2})$ using quadratic and positive-discriminant cubic extensions, then refactor the extension of $\mathbb{Q}$ so that all of the cubic extensions have positive discriminant?

So, to recap the title question: is it known whether it is possible to construct $\sqrt[3]{2}$ using compass, straightedge, and angle trisector? And more generally, are all cubic field extensions constructible with these tools?

  • $\begingroup$ A supposed 'proof' can be found here, but because $2^{\frac{1}{3}}$ is not a Euclidean number, I'd be wary when reading this. $\endgroup$ – user472341 Sep 16 '17 at 0:16
  • $\begingroup$ Both doubling a cube and trisecting an angle can be done with compass and a marked straight edge, so can also be done with origami $\endgroup$ – Henry Sep 16 '17 at 0:27

No. (Or, to be cautious, I think not.)

If you can trisect an angle you can construct a $7$- and a $13$-gon, but can't duplicate the cube.

See A. M. Gleason, Angle trisection, the heptagon, and the triskaidecagon, Amer Math. Monthly 95 (1988), 185–194. Addendum, p. 911. MR 90c:51023, discussed at http://www.ams.org/notices/200910/rtx091001236p.pdf

(But I don't think there's a proof in that article.)

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