---
title: Galois Groups of Polynomials
module: Galois Theory
moduleNumber: 11
lessonNumber: 4
order: 1104
summary: >
  Ordering the roots of a separable polynomial embeds its Galois group in the
  symmetric group $S_n$, and the group is transitive exactly when the polynomial
  is irreducible. The discriminant decides membership in $A_n$; for cubics and
  quartics the resolvent cubic pins the group down; and reduction modulo a prime
  produces elements of prescribed cycle type, the standard tool for computing
  Galois groups over $\mathbb{Q}$.
topics: [Galois Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 14 — Galois Theory; §14.6 Galois Groups of Polynomials"
  - book: Dummit & Foote
    ref: "§14.8 Computation of Galois Groups over Q"
draft: false
---

Galois's original object was the group of permutations of the roots of a polynomial. The
[Galois correspondence](/abstract-algebra/galois-theory/the-galois-correspondence)
recovers that viewpoint concretely: label the $n$ roots of a separable polynomial, and
each automorphism of the splitting field permutes them, embedding the Galois group as a
subgroup of the [symmetric group](/abstract-algebra/groups-and-symmetry/dihedral-and-symmetric-groups)
$S_n$. Which subgroup it is can be read from two computable invariants — the discriminant
and, for quartics, a resolvent cubic — and settled over $\mathbb{Q}$ by reducing
modulo primes.

## The Galois group as a permutation group

Let $f(x) \in F[x]$ be separable of degree $n$ with distinct roots $\alpha_1, \dots,
\alpha_n$ in its splitting field $K$. An automorphism $\sigma \in \operatorname{Gal}(K/F)$
sends each root to another root, and since $K = F(\alpha_1, \dots, \alpha_n)$, the whole
of $\sigma$ is determined by the permutation it induces. Fixing an ordering of the roots
gives an injective homomorphism $\operatorname{Gal}(K/F) \hookrightarrow S_n$.

> **Definition (Galois group of a polynomial).** The **Galois group of $f$** is the
> Galois group of its splitting field, viewed via a labeling of the roots as a subgroup of
> $S_n$. A different labeling replaces the subgroup by a conjugate.

Irreducibility corresponds to a group-theoretic property of the action.

> **Proposition (Transitivity $\Leftrightarrow$ irreducibility).** The Galois group of $f$
> acts transitively on the roots of $f$ if and only if $f$ is irreducible over $F$.

If $f$ is irreducible, any two roots $\alpha_i, \alpha_j$ have the same minimal polynomial
$f$, so an isomorphism $F(\alpha_i) \cong F(\alpha_j)$ fixing $F$ extends to an
automorphism of $K$ carrying $\alpha_i$ to $\alpha_j$. Conversely a factorization of $f$
splits the roots into blocks no automorphism can cross.[^df-transitive]

$$
% caption: An automorphism of the splitting field permutes the roots; for an irreducible
% cubic the group acts transitively, and a generator can cycle the three roots
% $a_1 \to a_2 \to a_3 \to a_1$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, circle, inner sep=1.5pt] (a1) at (90:1.7)  {$a_1$};
\node[draw, circle, inner sep=1.5pt] (a2) at (210:1.7) {$a_2$};
\node[draw, circle, inner sep=1.5pt] (a3) at (330:1.7) {$a_3$};
\draw[->, acc, thick] (a1) to[bend left=22] (a2);
\draw[->, acc, thick] (a2) to[bend left=22] (a3);
\draw[->, acc, thick] (a3) to[bend left=22] (a1);
\node[font=\footnotesize, acc] at (0,0) {$3$-cycle};
\end{tikzpicture}
$$

Because the Galois group embeds in $S_n$, its order divides $n!$, and transitivity forces
$n$ to divide the order — a first constraint on which groups can occur. The extreme case
is when nothing constrains the roots at all: over the field of rational functions in the
coefficients, the group is the full symmetric group.

> **Theorem (Generic Galois group).** The general polynomial $x^n - s_1 x^{n-1} + \cdots +
> (-1)^n s_n$ over $F(s_1, \dots, s_n)$, with the $s_i$ the elementary symmetric functions
> of indeterminate roots, is separable with Galois group $S_n$.

Over $\mathbb{Q}$ this is the typical case: most polynomials of degree $n$ have Galois
group $S_n$.[^df-generic]

## The discriminant and the alternating group

The subgroup $A_n$ of even permutations is detected by a single element of the splitting
field.

> **Definition (Discriminant).** For roots $\alpha_1, \dots, \alpha_n$, the
> **discriminant** is
> $$
> D = \prod_{i < j} (\alpha_i - \alpha_j)^2, \qquad \sqrt{D} = \prod_{i < j} (\alpha_i - \alpha_j).
> $$
> $D$ is symmetric in the roots, hence lies in $F$; $\sqrt{D}$ changes sign under odd
> permutations and is fixed under even ones.

The number $D$ vanishes exactly when two roots coincide, so $D \neq 0$ is separability.
The quantity $\sqrt{D}$ is the same object used to define the
[sign homomorphism](/abstract-algebra/subgroups-and-quotients/composition-series-and-the-alternating-group):
a permutation fixes it precisely when it is even. This gives the discriminant test.

> **Proposition (Discriminant test).** In characteristic not $2$, the Galois group of a
> separable $f \in F[x]$ is contained in $A_n$ if and only if $D$ is a square in $F$.
> Equivalently, $\sqrt{D} \in F$.

Since $\sqrt{D}$ always lies in the splitting field, $F(\sqrt{D})$ is the fixed field of
$\operatorname{Gal}(f) \cap A_n$ — the unique quadratic subextension detecting parity.

$$
% caption: For an irreducible polynomial the discriminant splits the possibilities: a
% square discriminant forces the group into $A_n$, a non-square keeps a transposition in
% it. For an irreducible cubic this is the whole classification.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=30mm, minimum height=10mm, align=center] (root) at (0,2.4) {$D$ a square in $F$?};
\node[draw, minimum width=24mm, minimum height=10mm, align=center, draw=acc] (yes) at (-2.6,0) {yes:\\$G = A_3$};
\node[draw, minimum width=24mm, minimum height=10mm, align=center] (no) at (2.6,0) {no:\\$G = S_3$};
\draw[->, acc, thick] (root) -- node[left, font=\footnotesize, black]{even} (yes);
\draw[->, black, thick] (root) -- node[right, font=\footnotesize]{has a swap} (no);
\end{tikzpicture}
$$

## Cubics

A cubic $f(x) = x^3 + ax^2 + bx + c$ becomes the depressed form $y^3 + py + q$ after $x =
y - a/3$, with the same splitting field and discriminant. Its discriminant is
$$
D = -4p^3 - 27q^2.
$$
The classification is complete once irreducibility and the square-class of $D$ are known.

- **Reducible.** A linear-times-quadratic factorization gives group $1$ or $\mathbb{Z}/2$;
  three linear factors give the trivial group.
- **Irreducible, $D$ a square.** The splitting field has degree $3$; the group is $A_3
  \cong \mathbb{Z}/3$, and adjoining a single root already splits $f$.
- **Irreducible, $D$ not a square.** The splitting field has degree $6$; the group is
  $S_3$, and the splitting field is $F(\theta, \sqrt{D})$ for any root $\theta$.

> **Worked example.** For $x^3 - 2$ over $\mathbb{Q}$, already depressed with $p = 0$,
> $q = -2$, the discriminant is
> $$
> D = -4p^3 - 27q^2 = -27 \cdot 4 = -108,
> $$
> not a rational square, and $x^3 - 2$ has no rational root, so it is irreducible. The
> group is therefore $S_3$, and the splitting field $\mathbb{Q}(\sqrt[3]{2}, \sqrt{-108}) =
> \mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$ has degree $6$ — the correspondence worked out in
> [the Galois correspondence](/abstract-algebra/galois-theory/the-galois-correspondence).

Cardano's explicit root formulas are the computational face of this solvability
— see [solvability by radicals](/abstract-algebra/galois-theory/solvability-by-radicals-and-the-quintic).

## Quartics and the resolvent cubic

A quartic $f(x) = x^4 + ax^3 + bx^2 + cx + d$ depresses to $y^4 + py^2 + qy + r$. Its
Galois group is one of the five **transitive subgroups** of $S_4$, since transitivity is
forced by irreducibility.

$$
% caption: The five transitive subgroups of $S_4$ that can occur as the Galois group of an
% irreducible quartic, ordered by containment: $S_4$ and $A_4$ at the top, the Sylow
% $2$-subgroup $D_8$, the cyclic $C_4$, and the normal Klein four-group $V$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node (S4) at (0,3.2) {$S_4$};
\node (A4) at (-1.8,2) {$A_4$};
\node (D8) at (1.8,2) {$D_8$};
\node (V)  at (0,0.7) {$V$};
\node (C4) at (3,0.7) {$C_4$};
\draw[black] (A4) -- (S4);
\draw[black] (D8) -- (S4);
\draw[black] (V) -- (A4);
\draw[black] (V) -- (D8);
\draw[black] (C4) -- (D8);
\end{tikzpicture}
$$

To distinguish them, form the three elements $\theta_1 = (\alpha_1 + \alpha_2)(\alpha_3 +
\alpha_4)$, $\theta_2 = (\alpha_1 + \alpha_3)(\alpha_2 + \alpha_4)$, $\theta_3 = (\alpha_1 +
\alpha_4)(\alpha_2 + \alpha_3)$. The symmetric group only permutes these three among
themselves, so their elementary symmetric functions lie in $F$: they are the roots of the
**resolvent cubic**.

> **Definition (Resolvent cubic).** For the depressed quartic $y^4 + py^2 + qy + r$, the
> resolvent cubic is
> $$
> h(x) = x^3 - 2p\,x^2 + (p^2 - 4r)\,x + q^2,
> $$
> whose roots are $\theta_1, \theta_2, \theta_3$. Its discriminant equals that of the
> quartic, and its splitting field is a subfield of the quartic's, so $\operatorname{Gal}(h)$
> is a quotient of $\operatorname{Gal}(f)$.

Reading the group off $h$ combines its factorization with the square-class of $D$.

$$
% caption: The resolvent cubic decides the quartic Galois group: its factorization type
% plus whether the discriminant is a square selects among $S_4$, $A_4$, $V$, $D_8$, $C_4$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, minimum width=34mm, minimum height=9mm, align=center] (h) at (0,4.2) {resolvent cubic $h$};
\node[draw, minimum width=26mm, minimum height=8mm, align=center] (irr) at (-3.2,2.6) {$h$ irreducible};
\node[draw, minimum width=26mm, minimum height=8mm, align=center] (red) at (3.2,2.6) {$h$ reducible};
\draw[->, black] (h) -- (irr);
\draw[->, black] (h) -- (red);
% irreducible branch
\node[draw, draw=acc, minimum width=17mm, align=center] (s4) at (-4.6,1) {$D$ non-sq.\\$S_4$};
\node[draw, draw=acc, minimum width=17mm, align=center] (a4) at (-2.2,1) {$D$ square\\$A_4$};
\draw[->, black] (irr) -- (s4);
\draw[->, black] (irr) -- (a4);
% reducible branch
\node[draw, draw=acc, minimum width=15mm, align=center] (v) at (1.4,1) {splits\\$V$};
\node[draw, draw=acc, minimum width=28mm, align=center] (dc) at (4.6,1) {one root in $F$\\$D_8$ or $C_4$};
\draw[->, black] (red) -- (v);
\draw[->, black] (red) -- (dc);
\end{tikzpicture}
$$

The last case is resolved by one more factorization: when exactly one $\theta_i$ lies in
$F$, the group is $D_8$ or $C_4$, and $F(\sqrt{D})$ is the fixed field of
$\operatorname{Gal}(f) \cap A_4$. Since $D_8 \cap A_4 = V$ is transitive on the roots while
$C_4 \cap A_4$ is not, the group is $D_8$ exactly when the quartic stays irreducible over
$F(\sqrt{D})$.[^df-quartic]

| Resolvent cubic $h$ | Discriminant $D$ | Galois group $G$ |
| --- | --- | --- |
| irreducible | not a square | $S_4$ |
| irreducible | a square | $A_4$ |
| splits completely | a square | $V$ |
| one root in $F$ | not a square | $D_8$ or $C_4$ |

### Worked example: $x^4 - 2$ over $\mathbb{Q}$

> **Worked example.** The polynomial $f(x) = x^4 - 2$ is Eisenstein at $2$, hence
> irreducible, and is already depressed with $p = 0$, $q = 0$, $r = -2$. The resolvent
> cubic is
> $$
> h(x) = x^3 - 2p\,x^2 + (p^2 - 4r)\,x + q^2 = x^3 + 8x = x(x^2 + 8),
> $$
> with a single rational root $\theta_1 = 0$ (the roots $\pm\sqrt{-8}$ are imaginary), so
> the group is $D_8$ or $C_4$. The discriminant is
> $$
> D = 16p^4 r - 4p^3 q^2 - 128 p^2 r^2 + 144\,pq^2 r - 27 q^4 + 256 r^3 = 256\,(-2)^3 = -2048,
> $$
> not a rational square, and $\mathbb{Q}(\sqrt{D}) = \mathbb{Q}(\sqrt{-2})$. Over
> $\mathbb{Q}(\sqrt{-2})$ the quartic stays irreducible: the real field
> $\mathbb{Q}(\sqrt[4]{2})$ does not contain $\sqrt{-2}$, so each root keeps degree $4$
> over $\mathbb{Q}(\sqrt{-2})$. By the transitivity criterion the group is $D_8$.
>
> The splitting field is $\mathbb{Q}(\sqrt[4]{2}, i)$ of degree $8$, and the Galois group
> is generated by $\sigma : \sqrt[4]{2} \mapsto i\sqrt[4]{2}$ (a $4$-cycle of the roots,
> fixing $i$) and complex conjugation $\tau$ (a transposition of the two imaginary roots),
> with $\sigma^4 = \tau^2 = 1$ and $\tau\sigma\tau = \sigma^{-1}$ — the dihedral relations.
> The four roots sit at the vertices of a square in $\mathbb{C}$: $\sigma$ rotates it a
> quarter turn, $\tau$ reflects it across the real axis.

## Computing Galois groups by reduction modulo $p$

For an integer polynomial, reducing modulo a prime turns a Galois-group computation into
polynomial factorization over a
[finite field](/abstract-algebra/galois-theory/finite-fields). Take $f \in \mathbb{Z}[x]$
separable, with nonzero discriminant $D$. For any prime $p \nmid D$, the reduction
$\bar{f} \in \mathbb{F}_p[x]$ is separable, and its Galois group over $\mathbb{F}_p$ is
cyclic, generated by the Frobenius. The cycle type of that generator is dictated by the
factorization of $\bar{f}$.

> **Theorem (Cycle types from reduction).** For a prime $p$ not dividing the discriminant
> of $f \in \mathbb{Z}[x]$, the Galois group of $f$ over $\mathbb{Q}$ contains a
> permutation whose cycle type is $(n_1, n_2, \dots, n_k)$, where the $n_i$ are the degrees
> of the irreducible factors of $\bar{f} = f \bmod p$.

The Frobenius acts transitively on the roots of each irreducible factor $\bar{f}_i$, so it
contributes a cycle of length $\deg \bar{f}_i$; assembling the factors gives the full cycle
type. Different primes populate the Galois group with different cycle types, and once
enough cycle types are collected, the group is often forced.[^df-cycletype]

For example, $f(x) = x^5 - x - 1$ has discriminant $2869 = 19 \cdot 151$, so any prime
other than $19, 151$ is usable.

- **Mod $2$:** $x^5 - x - 1 \equiv (x^2 + x + 1)(x^3 + x + 1)$, giving a $(2,3)$ cycle.
  Its cube is a transposition.
- **Mod $3$:** $x^5 - x - 1$ is irreducible, giving a $5$-cycle. This also proves $f$
  irreducible over $\mathbb{Z}$.

Since a transitive subgroup of $S_5$ containing a transposition and a $5$-cycle is all of
$S_5$, the Galois group of $x^5 - x - 1$ over $\mathbb{Q}$ is $S_5$. The same method builds
polynomials with $S_n$ as Galois group for every $n$, by prescribing factorization types
modulo $2$, $3$, $5$ that force a transposition and an $n$-cycle.

## The fundamental theorem of algebra, group-theoretically

The permutation viewpoint yields a proof that $\mathbb{C}$ is algebraically closed from two
analytic facts: every odd-degree real polynomial has a real root (intermediate value
theorem), and every complex number has a complex square root. Given $f \in \mathbb{R}[x]$,
let $K$ be the splitting field of $f$ over $\mathbb{R}$ adjoined with $i$, a Galois
extension with group $G$. A Sylow $2$-subgroup $P$ has odd index, so its fixed field has
odd degree over $\mathbb{R}$; the odd-degree fact forces that fixed field to be
$\mathbb{R}$, so $G = P$ is a $2$-group. Then $\operatorname{Gal}(K/\mathbb{C})$ is also a
$2$-group, and a $2$-group with a nontrivial part would have a subgroup of index $2$,
producing a quadratic extension of $\mathbb{C}$ — impossible. Hence $K = \mathbb{C}$.[^df-fta]
This argument of Artin reduces a theorem of analysis to the
[structure of $p$-groups](/abstract-algebra/products-and-group-structure/nilpotent-and-solvable-groups),
using analysis only for the two starting facts.

[^df-transitive]: **Dummit & Foote**, _Abstract Algebra_, §14.6 — the Galois group of a separable polynomial embeds in $S_n$ by its action on the roots, and acts transitively iff the polynomial is irreducible.
[^df-generic]: **Dummit & Foote**, _Abstract Algebra_, §14.6, Theorem 32 — the general polynomial over $F(s_1, \dots, s_n)$ has Galois group $S_n$; over $\mathbb{Q}$ the generic polynomial of degree $n$ realizes $S_n$.
[^df-quartic]: **Dummit & Foote**, _Abstract Algebra_, §14.6, Propositions 33–34 and the quartic classification — the discriminant test for $A_n$, the resolvent cubic $h(x) = x^3 - 2px^2 + (p^2 - 4r)x + q^2$, and the case analysis distinguishing $S_4, A_4, V, D_8, C_4$.
[^df-cycletype]: **Dummit & Foote**, _Abstract Algebra_, §14.8, Corollary 41 — for $p$ not dividing the discriminant, the Galois group over $\mathbb{Q}$ contains an element with cycle type equal to the degrees of the irreducible factors of $f \bmod p$; the worked example $x^5 - x - 1$ has group $S_5$.
[^df-fta]: **Dummit & Foote**, _Abstract Algebra_, §14.6, Theorem 35 (Fundamental Theorem of Algebra) — Artin's proof via a Sylow $2$-subgroup of the Galois group, using only the intermediate value theorem and complex square roots.