Solvability by Radicals

You learned the quadratic formula as a child: x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Its whole magic is that it builds the roots out of the coefficients using nothing but addition, subtraction, multiplication, division, and a square root. The Renaissance found the same kind of formula for cubics (Cardano) and quartics (Ferrari) — messier, nested radicals, but still a formula. For centuries everyone assumed a degree-5 version was just around the corner. It is not, and the reason is one of the most beautiful theorems in all of mathematics.

We say a polynomial is solvable by radicals when its roots can be written as an expression in the coefficients using +,\, -,\, \times,\, \div and n-th roots \sqrt[n]{\ }. The question "does a radical formula exist?" sounds like it lives in the world of algebra-with-symbols. Galois's staggering insight was to translate it, without loss, into a question about a finite group:

A polynomial f over a field of characteristic 0 is solvable by radicals if and only if its Galois group \mathrm{Gal}(f) is a solvable group.
A group is solvable when it has a chain of subgroups G = G_0 \trianglerighteq G_1 \trianglerighteq \dots \trianglerighteq G_m = \{e\} with each quotient G_i / G_{i+1} abelian.

Two completely different meanings of the word "solvable" — a formula in radicals exists and a group breaks into abelian pieces — turn out to be the same condition. That coincidence of names is not an accident; it is Galois's theorem, and this page is about the bridge between the two shores.

A radical tower: building the roots one root at a time

What does "expressible in radicals" mean, precisely, in the language of fields? It means you can reach a field containing all the roots by a tower of extensions, where each step just adjoins one radical. Start with the base field F (holding the coefficients) and climb:

F = F_0 \subseteq F_1 \subseteq F_2 \subseteq \dots \subseteq F_m, \qquad F_{i+1} = F_i\!\left(\sqrt[n_i]{a_i}\right),\ \ a_i \in F_i.

Each new floor F_{i+1} is obtained from the one below by throwing in a single n_i-th root of something already present. A tower like this is called a radical tower, and "solvable by radicals" means exactly: all the roots of f live somewhere in the top field F_m. Writing a root in radicals — \sqrt{2 + \sqrt[3]{5}}, say — is nothing but a recipe for climbing such a tower.

Here is Galois's translation, step by step. Adjoining a single n-th root (once the base contains the right roots of unity) is the "simplest possible" extension — its own little Galois group is cyclic, hence abelian. Stacking radical steps therefore builds the Galois group out of abelian layers, glued top to bottom. By the Galois correspondence, that tower of fields mirrors a chain of subgroups with abelian quotients — which is precisely the definition of a solvable group. The mirror below shows the two chains side by side.

Why quadratics, cubics and quartics all have formulas

The general degree-n polynomial (coefficients treated as independent unknowns) has Galois group the full symmetric group S_n — every permutation of the roots is allowed, because generic roots satisfy no special relations. So Galois's criterion turns "is there a formula in degree n?" into "is S_n a solvable group?" And for the small cases it is:

S_2 has order 2; it is already abelian, so the chain S_2 \trianglerighteq \{e\} does the job. Quadratics are solvable — the schoolroom formula.
S_3 (order 6) has the chain S_3 \trianglerighteq A_3 \trianglerighteq \{e\} with quotients \mathbb{Z}_2 and \mathbb{Z}_3, both abelian. Cubics are solvable — Cardano's formula.
S_4 (order 24) has the beautiful chain S_4 \;\trianglerighteq\; A_4 \;\trianglerighteq\; V_4 \;\trianglerighteq\; \{e\}, with quotients S_4/A_4 \cong \mathbb{Z}_2, A_4/V_4 \cong \mathbb{Z}_3, and V_4/\{e\} \cong V_4 — every one abelian. Quartics are solvable — Ferrari's formula.

That last chain leans on the little accident that A_4 contains the normal Klein four-group V_4 — a coincidence special to degree 4. It is the last time the luck holds. The pattern that has carried us from 2 to 4 is about to break, and it breaks at exactly 5: S_5 is not a solvable group, because sitting inside it is A_5, a non-abelian simple group with no further normal subgroups to continue the chain. That collapse is the whole story of the next page.

This is the misconception that trips up almost everyone the first time. "Solvable by radicals" is a statement about the existence of a formula — a way to write every root in terms of the coefficients using arithmetic and n-th roots. It says nothing about whether roots exist. Every non-constant polynomial has all its roots in the complex numbers — that is guaranteed no matter what the Galois group is. And you can always find those roots numerically, to as many decimal places as you like. What can fail is only the existence of a closed-form radical expression.

A second trap: solvability is a property of the specific polynomial's Galois group, not of the degree by itself. There are perfectly ordinary quintics that are solvable by radicals — x^5 - 2 has roots \sqrt[5]{2}\,\zeta^k (a radical times a root of unity), and its Galois group is solvable. Insolvability is about the general quintic, or particular ones whose group is S_5 or A_5. Degree alone is not destiny; the group is.

A technical fussiness worth knowing. The clean statement "adjoining an n-th root gives a cyclic (abelian) Galois group" needs the base field to already contain the n-th roots of unity \zeta_n. Without them, F(\sqrt[n]{a}) may not even be a normal extension. The standard fix is to adjoin the roots of unity first: the extension F(\zeta_n)/F is itself abelian (its Galois group embeds in the units (\mathbb{Z}/n)^\times), so it is a harmless first floor of the tower that doesn't spoil solvability. With the roots of unity in hand, each subsequent radical step is genuinely cyclic, and Galois's dictionary reads perfectly. In characteristic 0 this costs nothing and is always available.