Converse of the theorem "If a real number $c$ is constructible, then $c$ is algebraic of degree a power of 2 over the field $\mathbb{Q}$ of rationals"

Question

In Thomas W. Hungerford's ALGEBRA book Chapter V, Proposition 1.16 (page # 239) states that

If a real number $c$ is constructible, then $c$ is algebraic of degree a power of 2 over the field $\mathbb{Q}$ of rationals.

Our Galois theory lecturer said that the converse of this theorem is not true. That is there is at least one algebraic number (say $c_{0}$) which is algebraic of degree a power of 2 over the field $\mathbb{Q}$ of rationals but it is not constructible. So I tried to find such an algebraic number. But I'm having trouble finding a counterexample. Can any one please give me a hint or an idea ?

Any hints/ideas are much appreciated. Thanks in advance for any replies.

Answer 1

You need to find a minimal polynomial of degree a power of 2, but with splitting field degree having other prime factors.

Here is an example (from here): $x^4 + x + 1$

On MO there is another example.

Answer 2

Note that the following theorem holds: Let $c\in\mathbb R$ be an algebraic number over $\mathbb Q$, and let $K$ be the splitting field of the minimal polynomial of $c$ over $\mathbb Q$. The following are equivalent:

1. $c$ is constructible;

2. There is a chain of fields $\mathbb Q= F_0\leq F_1\leq\ldots\leq F_k=\mathbb Q(c)$ such that $|F_i:F_{i-1}|=2$, for all $1\leq i\leq k$;

3. $|K:\mathbb Q|= 2^n$, for some $n$.

The proof of this theorem uses some Galois theory.

So, for a counterexample find a polynomial of degree $2^k$ whose splitting field is not of degree $2^n$ over $\mathbb Q$.