Galois Theory Fundamentals

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 $\mathbb{Z}/2\mathbb{Z}$.

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:

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

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

2. $[E : K] = |\mathrm{Gal}(E/K)|$ and $[K : F] = [\mathrm{Gal}(E/F) : \mathrm{Gal}(E/K)]$.

3. $K/F$ is Galois if and only if $\mathrm{Gal}(E/K) \trianglelefteq \mathrm{Gal}(E/F)$In which case $\mathrm{Gal}(K/F) \cong \mathrm{Gal}(E/F) / \mathrm{Gal}(E/K)$.

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:

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

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 $S_5$. $\blacksquare$

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 $\mathrm{Gal}(x^3 - 3x + 1) = A_3 \cong \mathbb{Z}/3\mathbb{Z}$.

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

• 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 $[E:\mathbb{Q}]=6$).