Solvability by Radicals

Galois theory provides the definitive answer to the ancient problem of solving polynomial equations by radicals: a polynomial is solvable by radicals if and only if its Galois group is a solvable group.

Radical Extensions

Definition8.3Radical extension

A field extension $K/F$ is a radical extension if there exists a tower $F = F_0 \subset F_1 \subset \cdots \subset F_m = K$ where each $F_{i+1} = F_i(\alpha_i)$ with $\alpha_i^{n_i} \in F_i$ for some $n_i \geq 1$ . Each step adjoins an $n_i$ -th root of an element of the previous field.

A polynomial $f \in F[x]$ is solvable by radicals if its splitting field is contained in some radical extension of $F$ .

The Main Theorem

Theorem8.4Galois's criterion for solvability

Let $F$ be a field of characteristic 0 and $f \in F[x]$ a polynomial with splitting field $K$ and Galois group $G = \mathrm{Gal}(K/F)$ . Then $f$ is solvable by radicals if and only if $G$ is a solvable group (has a composition series with abelian factors).

ExampleSolvable and unsolvable polynomials

Quadratic $ax^2 + bx + c$ : $G \leq S_2$ , always solvable. Solution: $x = (-b \pm \sqrt{b^2-4ac})/2a$ .
Cubic: $G \leq S_3$ , always solvable ( $S_3$ is solvable). Cardano's formula.
Quartic: $G \leq S_4$ , always solvable ( $S_4$ is solvable). Ferrari's formula.
Generic quintic: $G = S_5$ , which is not solvable ( $A_5$ is simple and non-abelian). No formula exists.
$x^5 - 4x + 2$ over $\mathbb{Q}$ : $G = S_5$ (3 real roots, 2 complex; $G$ acts transitively and contains a transposition and a 5-cycle). Unsolvable by radicals.

Why $S_5$ Is Not Solvable

RemarkThe proof of unsolvability

$S_n$ is not solvable for $n \geq 5$ because:

$A_n$ is simple for $n \geq 5$ (no normal subgroups except $\{e\}$ and $A_n$ ).
The only normal subgroup of $S_n$ containing $A_n$ is $S_n$ itself.
Therefore the composition series of $S_n$ is $\{e\} \triangleleft A_n \triangleleft S_n$ , with factors $A_n$ and $\mathbb{Z}/2\mathbb{Z}$ .
Since $A_n$ is simple and non-abelian (for $n \geq 5$ ), this series has a non-abelian factor, so $S_n$ is not solvable.

This algebraic fact, combined with the existence of polynomials of degree $\geq 5$ with Galois group $S_n$ , proves the Abel-Ruffini theorem.

RemarkSpecific solvable quintics

Not all quintics are unsolvable. $x^5 - 2$ has Galois group $\mathbb{Z}/5\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$ (the Frobenius group of order 20), which is solvable. Its roots are $\sqrt[5]{2}\zeta_5^k$ , expressible using radicals.

Solvability by Radicals

Radical Extensions

The Main Theorem

Why S5S_5S5​ Is Not Solvable

Why $S_5$ Is Not Solvable