Doubling the cube, field extensions and minimal polynoms I know the proof of the "Doubling the cube problem". What is used there is the fact that if a number $a$ is constructible then $[\mathbb{Q}(a):\mathbb{Q}]$ is a power of $2$.
I found in a German textbook the remark that the inversion is not correct:
If $[\mathbb{Q}(a):\mathbb{Q}]$ is a power of $2$, then $a$ is not necessarily constructible.
Do you know an example for an $a$ where $[\mathbb{Q}(a):\mathbb{Q}]$ is a power of $2$ and which is not constructible – or a textbook with an example?
 A: Cox's Galois Theory, Example 10.1.13 (p.263) gives the example $f(x)=x^4-4x^2+x+1$. It is irreducible, so if $f(a)=0$ then $[\mathbb{Q}(a):\mathbb{Q}]=4$. However, the splitting field of $f$ over $\mathbb{Q}$ has degree 24.
Cox also shows that $a$ is constructible if and only if the splitting field of $f$ over $\mathbb{Q}$ has degree a power of 2, where $f$ is the minimal polynomial of $a$. (Theorem 10.1.12, p.262). Therefore the above $a$ is not constructible.
A: The set of constructible numbers is the smallest extension of $\Bbb Q$ where each positive number has a square root. (That's essentially what straightedge and compass constructions are able to do: field operations and square roots.) Every constructible number can be described by using rational numbers, field operations and square roots.
I can't construct a concrete example of the following off the top of my head. But it is relatively well-known that there are irreducible degree 8 polynomials that aren't solvable through radicals. A root of such a polynomial won't be expressible with only the rational numbers, field operations and square roots. Thus it won't be constructible. Yet the field extension you get from adjoining such a root to $\Bbb Q$ has degree $8$.
A: The formulas solving the quartic involve taking cube roots in general. Some auxiliary cubic equation is needed in the resolution of a quartic in general.
