---
title: Solvability by Radicals and the Quintic
module: Galois Theory
moduleNumber: 11
lessonNumber: 5
order: 1105
summary: >
  A polynomial is solvable by radicals exactly when its Galois group is solvable.
  Cyclic extensions are radical extensions once roots of unity are present, which
  turns a radical tower into a solvable subnormal series. Since $S_n$ is solvable
  only for $n \le 4$, the general quintic has no radical formula, and an explicit
  quintic with Galois group $S_5$ has roots provably not expressible in radicals.
topics: [Galois Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 14 — Galois Theory; §14.7 Solvable and Radical Extensions: Insolvability of the Quintic"
  - book: Judson
    ref: "Ch. 23 — Galois Theory; §23.3 Applications"
draft: false
---

The quadratic formula, and its harder cousins for the cubic and quartic, express roots
using field operations and root extractions. Galois's theorem explains exactly when such a
formula exists: the roots of $f$ are expressible in radicals if and only if the
[Galois group](/abstract-algebra/galois-theory/galois-groups-of-polynomials) of $f$ is a
[solvable group](/abstract-algebra/products-and-group-structure/nilpotent-and-solvable-groups).
The word "solvable" for groups is not a coincidence — it was named for this. Because $S_5$
is not solvable, the general quintic has no radical formula, and a specific quintic over
$\mathbb{Q}$ with group $S_5$ exhibits a polynomial whose roots provably cannot be written
in radicals. We work over a field of characteristic $0$ throughout.

## Radical extensions

Solving by radicals means climbing a tower where each step adjoins a root of an element
already built.

> **Definition (Radical extension).** An element $\alpha$ is **expressible by radicals**
> over $F$ if it lies in a field $K$ reached by a chain of simple radical extensions
> $$
> F = K_0 \subseteq K_1 \subseteq \cdots \subseteq K_s = K, \qquad K_{i+1} = K_i(\sqrt[n_i]{a_i}),
> $$
> where $a_i \in K_i$ and $\sqrt[n_i]{a_i}$ is a root of $x^{n_i} - a_i$. A polynomial is
> **solvable by radicals** if all its roots are.

The chain formalizes "a succession of field operations and root extractions." For example,
the constructible number for the regular $17$-gon lies in a tower of four square-root
extensions, which is why the $17$-gon is constructible by straightedge and compass — only
square roots appear.

$$
% caption: A radical tower: each step adjoins an $n_i$th root of an element of the field
% below it, so every element of the top field is built from the base by successive root
% extractions.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (K0) at (0,0)   {$F = K_0$};
\node (K1) at (0,1.3) {$K_1$};
\node (K2) at (0,2.6) {$K_2$};
\node (Kd) at (0,3.9) {$\vdots$};
\node (Ks) at (0,5.2) {$K_s = K$};
\draw[->, acc, thick] (K0) -- node[right, font=\footnotesize, black]{adjoin an $n_0$th root} (K1);
\draw[->, acc, thick] (K1) -- node[right, font=\footnotesize, black]{adjoin an $n_1$th root} (K2);
\draw[->, acc, thick] (K2) -- (Kd);
\draw[->, acc, thick] (Kd) -- (Ks);
\end{tikzpicture}
$$

## Cyclic extensions are radical extensions

The link between radicals and groups runs through cyclic extensions, once the base field
has enough
[roots of unity](/abstract-algebra/galois-theory/cyclotomic-and-abelian-extensions). Roots
of unity are themselves radicals ($\zeta_n = \sqrt[n]{1}$), so they can always be adjoined
at the start of any radical tower.

> **Proposition (Radical $\Rightarrow$ cyclic).** If $F$ contains the $n$th roots of unity
> (and $\operatorname{char} F \nmid n$), then $F(\sqrt[n]{a})/F$ is cyclic of degree
> dividing $n$.

An automorphism $\sigma$ sends $\sqrt[n]{a}$ to $\zeta_\sigma \sqrt[n]{a}$ for some root of
unity $\zeta_\sigma$, and $\sigma \mapsto \zeta_\sigma$ is an injective homomorphism into
the cyclic group $\mu_n$, so the Galois group is cyclic.[^df-radcyc] The converse recovers
a radical from a cyclic extension using a **Lagrange resolvent**.

> **Definition (Lagrange resolvent).** For a generator $\sigma$ of a cyclic
> $\operatorname{Gal}(K/F)$ of order $n$, an $n$th root of unity $\zeta$, and $\beta \in
> K$, the resolvent is
> $$
> (\beta, \zeta) = \beta + \zeta\,\sigma(\beta) + \zeta^2 \sigma^2(\beta) + \cdots + \zeta^{n-1}\sigma^{n-1}(\beta).
> $$

Applying $\sigma$ multiplies the resolvent by $\zeta^{-1}$, so $\sigma\bigl((\beta,
\zeta)^n\bigr) = (\beta, \zeta)^n$, which places $(\beta, \zeta)^n = a$ in $F$. By
[independence of characters](/abstract-algebra/galois-theory/the-galois-correspondence)
some $\beta$ gives a nonzero resolvent, and it lies in no proper subfield, so
$K = F\bigl((\beta, \zeta)\bigr) = F(\sqrt[n]{a})$.

> **Proposition (Cyclic $\Rightarrow$ radical).** Any cyclic extension of degree $n$ over a
> field containing the $n$th roots of unity (with $\operatorname{char} F \nmid n$) is
> $F(\sqrt[n]{a})$ for some $a \in F$.

These two propositions are the basis of Kummer theory: with roots of unity present, cyclic
and simple-radical extensions coincide.[^df-kummer]

## Galois's theorem

A [solvable group](/abstract-algebra/products-and-group-structure/nilpotent-and-solvable-groups)
is one with a subnormal series
$$
1 = G_s \trianglelefteq G_{s-1} \trianglelefteq \cdots \trianglelefteq G_0 = G, \qquad G_i / G_{i+1} \text{ cyclic}.
$$
This is the shape of the Galois-group side of a radical tower.

> **Theorem (Galois).** A polynomial $f$ over a field of characteristic $0$ is solvable by
> radicals if and only if its Galois group is a solvable group.

The two directions each translate one tower into the other.

- **Solvable by radicals $\Rightarrow$ solvable group.** A radical tower can be enlarged to
  a Galois radical tower whose successive extensions are cyclic (adjoin roots of unity,
  take the Galois closure). The corresponding subgroups $G_i = \operatorname{Gal}(L/K_i)$
  form a chain with cyclic quotients $G_i/G_{i+1} \cong \operatorname{Gal}(K_{i+1}/K_i)$, so
  $G$ is solvable. The Galois group of $f$ is a quotient of $G$, hence solvable.
- **Solvable group $\Rightarrow$ solvable by radicals.** Take a subnormal series with cyclic
  quotients for $G = \operatorname{Gal}(K/F)$ and its fixed fields $F = K_0 \subseteq \cdots
  \subseteq K_s = K$. After adjoining the needed roots of unity, each $K_{i+1}/K_i$ is
  cyclic, hence a simple radical extension. So every root of $f$ lies in a radical
  tower.[^df-galois]

$$
% caption: Galois's theorem matches a radical tower of fields (left) with a subnormal
% series of the Galois group (right): each cyclic quotient $G_i / G_{i+1}$ corresponds to a
% single radical step, so the tower exists exactly when the series does.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% fields
\node (K0) at (0,0)   {$F = K_0$};
\node (K1) at (0,1.4) {$K_1$};
\node (K2) at (0,2.8) {$K_2$};
\node (K3) at (0,4.2) {$K_s = K$};
\draw[black] (K0) -- (K1) -- (K2);
\draw[black, dashed] (K2) -- (K3);
% groups
\node (G0) at (5,0)   {$G = G_0$};
\node (G1) at (5,1.4) {$G_1$};
\node (G2) at (5,2.8) {$G_2$};
\node (G3) at (5,4.2) {$1$};
\draw[black] (G0) -- node[right, font=\footnotesize]{cyclic} (G1);
\draw[black] (G1) -- node[right, font=\footnotesize]{cyclic} (G2);
\draw[black, dashed] (G2) -- (G3);
% correspondence
\draw[<->, acc, dashed] (K0) -- (G0);
\draw[<->, acc, dashed] (K1) -- (G1);
\draw[<->, acc, dashed] (K2) -- (G2);
\draw[<->, acc, dashed] (K3) -- (G3);
\end{tikzpicture}
$$

## Degrees two through four are solvable

For low degree the symmetric group is solvable, so a radical formula must exist. The
composition series exhibit the cyclic quotients that Galois's theorem converts into
radicals.

| Degree | Group | Subnormal series | Formula |
| --- | --- | --- | --- |
| $2$ | $S_2$ | $1 \trianglelefteq S_2$ | quadratic formula |
| $3$ | $S_3$ | $1 \trianglelefteq A_3 \trianglelefteq S_3$ | Cardano |
| $4$ | $S_4$ | $1 \trianglelefteq V \trianglelefteq A_4 \trianglelefteq S_4$ | Ferrari |

> **Worked example.** Read the solvability of $S_4$ off its composition factors. The
> series $1 \trianglelefteq V \trianglelefteq A_4 \trianglelefteq S_4$ has quotients
> $$
> S_4/A_4 \cong \mathbb{Z}/2, \qquad A_4/V \cong \mathbb{Z}/3, \qquad V \cong \mathbb{Z}/2 \times \mathbb{Z}/2,
> $$
> of orders $2, 3, 4$. Refining the last step through an order-$2$ subgroup of $V$ splits
> it into two $\mathbb{Z}/2$ quotients, so every composition factor is cyclic of prime
> order. Hence $S_4$ is solvable, and Galois's theorem turns each factor into one radical
> step — Ferrari's formula.

For the cubic $y^3 + py + q$, the series $1 \trianglelefteq A_3 \trianglelefteq S_3$
predicts adjoining a cube root of unity and forming Lagrange resolvents. With $\rho$ a
primitive cube root of unity and roots $\alpha, \beta, \gamma$, the resolvents $\theta_1 =
\alpha + \rho\beta + \rho^2\gamma$ and $\theta_2 = \alpha + \rho^2\beta + \rho\gamma$
satisfy $\theta_1^3, \theta_2^3 \in \mathbb{Q}(\sqrt{D}, \rho)$, yielding Cardano's
formula: with $A = \sqrt[3]{-\tfrac{27}{2}q + \tfrac{3}{2}\sqrt{-3D}}$ and $B$ its partner
chosen so $AB = -3p$, the roots are $\tfrac{1}{3}(A + B)$ and its $\rho$-twists. The
quartic reduces to its
[resolvent cubic](/abstract-algebra/galois-theory/galois-groups-of-polynomials): the
quotient $S_4/V \cong S_3$ is the cubic's group, and the final $V$ step contributes the
two square roots that solve for the four roots. This is Ferrari's method.

## The quintic is not solvable

At degree $5$ the pattern stops, because the alternating group becomes simple.

> **Theorem (Insolvability of $A_n$ and $S_n$).** For $n \geq 5$, the group $A_n$ is
> [simple](/abstract-algebra/group-actions-and-sylow/automorphisms-and-simple-groups) and
> non-abelian, so it has no subnormal series with abelian quotients. Hence $A_n$ and $S_n$
> are not solvable.

Simplicity blocks the descent: the only normal subgroups of $A_5$ are $1$ and $A_5$
itself, and $A_5$ is non-abelian, so any attempt at a subnormal series with cyclic
quotients stalls at $A_5$ and never reaches $1$. The derived subgroup $[A_5, A_5] = A_5$,
so the derived series is constant. Contrast $S_4$, whose derived series $S_4 \supseteq A_4
\supseteq V \supseteq 1$ terminates.

$$
% caption: $S_4$ is solvable because its derived series descends to $1$; $S_5$ is not,
% because $A_5$ is a simple non-abelian group equal to its own commutator subgroup, so the
% descent stalls there.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{B23A48}
% solvable S4
\node (s4) at (0,3) {$S_4$};
\node (a4) at (0,2) {$A_4$};
\node (v) at (0,1) {$V$};
\node (one) at (0,0) {$1$};
\draw[->, acc, thick] (s4) -- (a4);
\draw[->, acc, thick] (a4) -- (v);
\draw[->, acc, thick] (v) -- (one);
\node[font=\footnotesize, acc, anchor=west] at (0.4,1.5) {reaches $1$};
% non-solvable S5
\begin{scope}[xshift=5cm]
\node (s5) at (0,3) {$S_5$};
\node (a5) at (0,1.7) {$A_5$};
\draw[->, red, thick] (s5) -- (a5);
\draw[->, red, thick] (a5) to[out=-40,in=-110,looseness=6] (a5);
\node[font=\footnotesize, red, anchor=west] at (0.6,1.0) {commutator is itself};
\node[font=\footnotesize, red, anchor=west] at (0.4,2.4) {stalls};
\end{scope}
\end{tikzpicture}
$$

Combining with the fact that the generic degree-$n$ polynomial has group $S_n$ gives the
classical impossibility result.

> **Corollary (Abel–Ruffini).** The general polynomial of degree $n \geq 5$ cannot be
> solved by radicals. There is no formula in the coefficients, using field operations and
> root extractions, for the roots of the general quintic.

## An explicit unsolvable quintic

The general result concerns the generic polynomial; a specific rational quintic with a
non-solvable Galois group provides an explicit example.

> **Claim.** The polynomial $f(x) = x^5 - 6x + 3 \in \mathbb{Q}[x]$ has Galois group $S_5$
> and is therefore not solvable by radicals.

The argument combines irreducibility, a cycle count, and a real-root count.

- **A $5$-cycle.** $f$ is
  [Eisenstein](/abstract-algebra/factorization-and-polynomials/irreducibility-criteria-and-groebner)
  at $3$, hence irreducible, so the splitting field has degree divisible by $5$ and the
  Galois group $G \leq S_5$ has order divisible by $5$. An element of order $5$ in $S_5$ is
  a $5$-cycle, so $G$ contains one.
- **A transposition.** Evaluating gives $f(-2) = -17$, $f(0) = 3$, $f(1) = -2$, $f(2) =
  23$, so $f$ has real roots in $(-2, 0)$, $(0, 1)$, $(1, 2)$. The derivative $f'(x) = 5x^4 -
  6$ has only two real zeros, so $f$ has exactly three real roots and one conjugate pair
  of complex roots. Complex conjugation restricts to an automorphism that fixes the three
  real roots and swaps the two complex ones — a transposition in $G$.

$$
% caption: The graph of $f(x) = x^5 - 6x + 3$ crosses the axis three times, giving three
% real roots and one conjugate pair of complex roots; complex conjugation swaps the pair
% and acts as a transposition on the roots.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{B23A48}
% axes
\draw[black, ->] (-2.4,0) -- (2.4,0) node[right, font=\footnotesize]{$x$};
\draw[black, ->] (0,-1.6) -- (0,2.0) node[above, font=\footnotesize]{$y$};
% quintic-like curve crossing axis 3 times (schematic)
\draw[acc, very thick] plot[smooth, tension=0.8] coordinates
  {(-2.0,-1.4) (-1.5,0.55) (-1.0,1.1) (-0.3,0.75) (0.5,0.02) (1.0,-0.5) (1.4,0.05) (1.9,1.6)};
% real roots
\fill[red] (-1.68,0) circle (2pt);
\fill[red] (0.51,0) circle (2pt);
\fill[red] (1.30,0) circle (2pt);
\node[red, font=\footnotesize, anchor=north] at (-1.68,-0.1) {real};
\node[red, font=\footnotesize, anchor=north west] at (1.30,-0.15) {real};
\node[black, font=\footnotesize, anchor=west] at (0.5,1.55) {$2$ complex roots};
\end{tikzpicture}
$$

A transitive subgroup of $S_5$ that contains a transposition and a $5$-cycle is all of
$S_5$: the $5$-cycle and one transposition generate $S_5$. Since $S_5$ is not solvable, the
roots of $x^5 - 6x + 3$ cannot be written in radicals.[^df-quintic] The same construction
prescribing factorization types modulo small primes builds quintics with group $S_5$ at
will, and the reduction-mod-$p$ criterion — cycle types from the factorization of $f
\bmod p$ — identifies them.

## Three classical impossibilities

Galois theory converts three classical impossibilities into statements about groups. The
[non-constructibility](/abstract-algebra/field-theory/straightedge-and-compass-constructions)
of doubling the cube and trisecting the angle is a degree obstruction; the constructible
polygons are governed by
[cyclotomic Galois groups](/abstract-algebra/galois-theory/cyclotomic-and-abelian-extensions);
and the quintic's lack of a formula is the non-solvability of $S_5$. In each case a
long-standing question about numbers is settled by translating it into the finite group
attached to the extension.

[^df-radcyc]: **Dummit & Foote**, _Abstract Algebra_, §14.7, Proposition 36 — with the $n$th roots of unity in $F$, $F(\sqrt[n]{a})/F$ is cyclic of degree dividing $n$, via $\sigma \mapsto \zeta_\sigma$ into $\mu_n$.
[^df-kummer]: **Dummit & Foote**, _Abstract Algebra_, §14.7, Proposition 37 and the Lagrange resolvent construction — every cyclic extension of degree $n$ over a field with the $n$th roots of unity is a simple radical extension; the two propositions form the base of Kummer theory.
[^df-galois]: **Dummit & Foote**, _Abstract Algebra_, §14.7, Lemma 38 and Theorem 39 — a radical tower can be refined to a Galois tower with cyclic steps, giving the equivalence of solvability by radicals with solvability of the Galois group.
[^df-quintic]: **Dummit & Foote**, _Abstract Algebra_, §14.7, Corollary 40 and the worked example $x^5 - 6x + 3$ — $S_n$ is not solvable for $n \geq 5$, and the explicit quintic has Galois group $S_5$ from a $5$-cycle and a transposition, hence is not solvable by radicals.