Galois Theory Fundamentals | Mathematics

13.1 Automorphisms and the Galois Group

Let $E/F$ be a field extension. An $F$ -automorphism of $E$ is an automorphism $\sigma : E \to E$ That fixes $F$ pointwise (i.e., $\sigma(c) = c$ for all $c \in F$ ).

The set of all $F$ -automorphisms of $E$ forms a group under composition, called the Galois group Of $E/F$ Denoted $\mathrm{Gal}(E/F)$ .

Example. $\mathrm{Gal}(\mathbb{C}/\mathbb{R}) = \{id, \sigma\}$ where $\sigma(a + bi) = a - bi$ . This is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ .

13.2 The Fundamental Theorem of Galois Theory

A finite extension $E/F$ is Galois if $|\mathrm{Gal}(E/F)| = [E : F]$ Or equivalently, If $E$ is the splitting field of a separable polynomial over $F$ .

Theorem 13.1 (Fundamental Theorem of Galois Theory). Let $E/F$ be a Galois extension. Then:

There is an inclusion-reversing bijection between intermediate fields $F \subseteq K \subseteq E$ and subgroups $H \subseteq \mathrm{Gal}(E/F)$ Given by:

$K \mapsto \mathrm{Gal}(E/K)$ .
$H \mapsto E^H = \{x \in E : \sigma(x) = x\ \mathrm{for\ all\ }\sigma \in H\}$ .

$[E : K] = |\mathrm{Gal}(E/K)|$ and $[K : F] = [\mathrm{Gal}(E/F) : \mathrm{Gal}(E/K)]$ .
$K/F$ is Galois if and only if $\mathrm{Gal}(E/K) \trianglelefteq \mathrm{Gal}(E/F)$ In which case $\mathrm{Gal}(K/F) \cong \mathrm{Gal}(E/F) / \mathrm{Gal}(E/K)$ .

13.3 Worked Example

Problem. Find the Galois group of $x^3 - 2$ over $\mathbb{Q}$ .

Solution. The roots of $x^3 - 2$ are $\sqrt[3]{2}$ , $\omega\sqrt[3]{2}$ , $\omega^2\sqrt[3]{2}$ Where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity. The splitting field is $E = \mathbb{Q}(\sqrt[3]{2}, \omega)$ . We have $[E : \mathbb{Q}] = [E : \mathbb{Q}(\sqrt[3]{2})] \cdot [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 2 \cdot 3 = 6$ .

The Galois group $\mathrm{Gal}(E/\mathbb{Q})$ acts as permutations of the three roots, so $\mathrm{Gal}(E/\mathbb{Q}) \cong S_3$ .

The subgroup lattice of $S_3$ corresponds to the lattice of intermediate fields:

$\{e\} \leftrightarrow E$
$A_3 = \langle (1\ 2\ 3) \rangle \leftrightarrow \mathbb{Q}(\omega)$
$\langle (1\ 2) \rangle \leftrightarrow \mathbb{Q}(\omega^2 \sqrt[3]{2})$
$\langle (1\ 3) \rangle \leftrightarrow \mathbb{Q}(\omega \sqrt[3]{2})$
$\langle (2\ 3) \rangle \leftrightarrow \mathbb{Q}(\sqrt[3]{2})$
$S_3 \leftrightarrow \mathbb{Q}$ $\blacksquare$

13.4 Solvability by Radicals

Definition. A polynomial $f \in F[x]$ is solvable by radicals if its roots can be expressed Using field operations and radicals (nth roots).

Theorem 13.2. A polynomial $f \in \mathbb{Q}[x]$ is solvable by radicals if and only if its Galois Group is a solvable group.

Corollary 13.3 (Abel-Ruffini Theorem). The general polynomial of degree 5 is not solvable by Radicals.

Proof. The symmetric group $S_5$ is not solvable (its only normal series is $S_5 \triangleright A_5 \triangleright \{e\}$ And $A_5 / \{e\} \cong A_5$ is non-abelian). The Galois group of $x^5 - x - 1$ (and many other quintics) Over $\mathbb{Q}$ is $S_5$ . $\blacksquare$

13.5 The Discriminant and Galois Groups

The discriminant of $f(x) = (x - \alpha_1)\cdots(x - \alpha_n)$ is

$\Delta = \prod_{i \lt j} (\alpha_i - \alpha_j)^2$

The discriminant is a symmetric function of the roots, so $\Delta \in \mathbb{Q}$ when $f \in \mathbb{Q}[x]$ .

Proposition 13.4. Let $G = \mathrm{Gal}(f) \leq S_n$ . Then $G \leq A_n$ (i.e., $G$ is contained in the Alternating group) if and only if $\Delta$ is a perfect square in the base field.

Proof. The Galois group acts on $\delta = \prod_{i \lt j}(\alpha_i - \alpha_j)$ by permutation. For any $\sigma \in G$ , $\sigma(\delta) = \mathrm{sgn}(\sigma) \cdot \delta$ . If $\sigma \in A_n$ $\sigma(\delta) = \delta$ ; if $\sigma \notin A_n$ , $\sigma(\delta) = -\delta$ .

If $G \leq A_n$ Then $\delta$ is fixed by all of $G$ So $\delta \in F$ Hence $\Delta = \delta^2$ is a square. Conversely, if $\Delta$ is a square in $F$ Then $\delta \in F$ (or $-\delta \in F$ ), so $\delta$ is fixed By $G$ Meaning every element of $G$ acts as an even permutation. $\blacksquare$

Example. The discriminant of $x^3 - 3x + 1$ is $\Delta = 81 = 9^2$ A perfect square. Therefore $\mathrm{Gal}(x^3 - 3x + 1) \leq A_3 \cong \mathbb{Z}/3\mathbb{Z}$ . Since the polynomial is irreducible, The Galois group is transitive, so $\mathrm{Gal}(x^3 - 3x + 1) = A_3 \cong \mathbb{Z}/3\mathbb{Z}$ .

13.6 Worked Example: Galois Group of a Quartic

Problem. Determine the Galois group of $f(x) = x^4 - 2$ over $\mathbb{Q}$ .

Solution

Solution. The roots are $\alpha_1 = \sqrt[4]{2}$ , $\alpha_2 = i\sqrt[4]{2}$ , $\alpha_3 = -\sqrt[4]{2}$ $\alpha_4 = -i\sqrt[4]{2}$ . The splitting field is $E = \mathbb{Q}(\sqrt[4]{2}, i)$ .

$[\mathbb{Q}(\sqrt[4]{2}) : \mathbb{Q}] = 4$ (since $x^4 - 2$ is irreducible by Eisenstein). $i \notin \mathbb{Q}(\sqrt[4]{2}) \subset \mathbb{R}$ So $[E : \mathbb{Q}(\sqrt[4]{2})] = 2$ . Thus $[E : \mathbb{Q}] = 8$ .

The Galois group has order $8$ . It is generated by: $\sigma: \sqrt[4]{2} \mapsto i\sqrt[4]{2},\ i \mapsto i$ (order $4$ ) $\tau: \sqrt[4]{2} \mapsto \sqrt[4]{2},\ i \mapsto -i$ (order $2$ )

We check: $\tau\sigma\tau^{-1}(\sqrt[4]{2}) = \tau(i\sqrt[4]{2}) = -i\sqrt[4]{2} = \sigma^{-1}(\sqrt[4]{2})$ . So $\tau\sigma\tau^{-1} = \sigma^{-1}$ The defining relation of $D_4$ .

Therefore $\mathrm{Gal}(E/\mathbb{Q}) \cong D_4$ (dihedral group of order $8$ ). $\blacksquare$

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Cross-References

Additional Results: Extends the group theory toolkit with Cauchy’s theorem and the structure theorem for abelian groups, which underpin the classification of Galois groups.

Intuition

Galois theory reveals the deep connection between field extensions and group theory. The fundamental theorem establishes a correspondence between intermediate fields of a field extension and subgroups of its Galois group, with inclusion-reversing properties. This transforms questions about the solvability of polynomial equations into questions about the structure of groups. A polynomial is solvable by radicals precisely when its Galois group is a solvable group, meaning it has a chain of normal subgroups with abelian quotients. Since the symmetric group on five or more elements is not solvable, the general quintic equation cannot be solved by radicals, answering a question that had remained open for centuries.

Common Pitfalls

Splitting field degree vs.\ polynomial degree. The degree $[E:\mathbb{Q}]$ of a splitting field is not always equal to the degree of the polynomial; it equals the order of the Galois group, which can be larger (e.g.\ $x^3-2$ has degree 3 but $[E:\mathbb{Q}]=6$ ).

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