Show that no root of the polynomial $x^5 + 21x^4 - 14x^3 + 28x^2 - 7x + 42$ is constructible.

Question

Show that no root of the polynomial $$x^5 + 21x^4 - 14x^3 + 28x^2 - 7x + 42$$ is constructible.

Is it enough to say the degree is 5 is not a power of 2?

Answer 1

I will prove the following assertion. If $f(x)\in\mathbb{Q}[x]$ is an irreducible polynomial of degree $n$ such that $n$ has an odd prime factor, then the roots of $f(x)$ are not constructible.

The proof is done by proving the contrapositive statement. We just have to show that the minimal polynomial of any constructible number $\alpha$ has degree $2^k$ for some integer $k\ge 0$. However, this is simply because there exist field extensions $$\Bbb{Q}=F_0<F_1<F_2<\ldots < F_k=\Bbb{Q}(\alpha)$$ such that $[F_i:F_{i-1}]=2$ for $i=1,2,\ldots,k$. Therefore, $[F_k:\Bbb Q]=2^k$, proving the claim.

The converse doesn't hold. The roots of the irreducible polynomial $x^4-x-1$ are not constructible. This is because the Galois group of $x^4-x-1$ is the symmetric group $S_4$, which is not a $2$-group. (See example 3.2 of this article.)