$f\in K[X]\subseteq \mathbb{C}[X]$ with degree $p\geq 5$ irreducible is solvable if and only if $N=K[\alpha_i,\alpha_j]$ for every $i,j$

Question

Let $K\subseteq \mathbb{C}$ be a field. Let $f\in K[X]$ irreducible of prime degree $p\geq 5$. Let $\alpha_1,\cdots ,\alpha_p$ be its roots, and let $N=K[\alpha_1,\cdots ,\alpha_p]$ be its splitting field. Prove that $f$ is solvable if and only if $N=K[\alpha_i,\alpha_j]$ for every $1\leq i<j\leq p$.

Answer 1

Let me do one direction, and leave you a hint for the other direction:

Denote the Galois group of $f$ by $G$, then since $f$ is irreducible, $$ 5\mid |G| $$ Since $G<S_5$, it follows that $G$ must be one of the following:

1. $C_5 \cong \langle (12345)\rangle$
2. $D_5 \cong \langle (12345), (13)(45)\rangle$
3. $F_{20} \cong \langle (12345),(2354)\rangle$
4. $A_5$
5. $S_5$

The first three of these are solvable, and so $f$ is solvable iff $|G| \leq 20$. Now, $$ [K(\alpha_1):K] = \deg(f) = 5 $$ and since $\alpha_2$ satisfies the polynomial $f(x) \in K(\alpha_1)[x]$, $$ [K(\alpha_1,\alpha_2):K] \leq 5\times 4 = 20 $$ Hence, if $N = K(\alpha_1,\alpha_2)$, then $|G| = [N:K] \leq 20$, and is hence solvable.

For the other direction, you will need to carefully examine the index $[G:Gal_F(N)]$ where $F = K(\alpha_1,\alpha_2)$ and prove in each case that $K\subset F$ is a normal extension.