The ancient problem of solving polynomial equations by radicals — expressing roots using only the four arithmetic operations and the extraction of $n$-th roots — runs into a wall at degree five. The quadratic formula has been known since antiquity. The cubic and quartic formulas, discovered in sixteenth-century Italy, required enormous ingenuity. But every attempt to find a similar formula for the general quintic failed. By the early nineteenth century, Abel had proved that no such formula exists, and Galois had uncovered the reason: whether a polynomial can be solved by radicals depends on a certain group of symmetries attached to it. That group is the Galois group.

The insight driving Galois's theory is that the roots of a polynomial are not isolated numbers; they are related by algebraic dependencies, and any algebraic relation that holds among the roots must be respected by any automorphism that fixes the ground field. The Galois group captures exactly these symmetries. It is a finite group that encodes the arithmetic of the splitting field, and the interplay between its subgroups and the intermediate fields of the extension is one of the deepest correspondences in all of algebra.

[example: Roots of Unity] Before introducing the formalism, consider the polynomial $f(x) = x^4 - 1$ over $\mathbb{Q}$. Its roots are $1, -1, i, -i$. The splitting field is $\mathbb{Q}(i)$, since $i$ generates all four roots: $1 = i^4$, $-1 = i^2$, $-i = i^3$. Any automorphism $\sigma$ of $\mathbb{Q}(i)$ fixing $\mathbb{Q}$ must send $i$ to another root of $x^2 + 1$, which means either $\sigma(i) = i$ or $\sigma(i) = -i$. There are exactly two such automorphisms: the identity and complex conjugation. The Galois group $\operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q})$ is therefore a group of order two, isomorphic to $\mathbb{Z}/2\mathbb{Z}$. Notice that this group also permutes the four roots of $x^4 - 1$: the identity fixes all four, while complex conjugation swaps $i \leftrightarrow -i$ and fixes $1$ and $-1$. Any algebraic relation among $1, -1, i, -i$ with rational coefficients — such as $i \cdot (-i) = 1$ — is preserved by both automorphisms. [/example]

## Definition

The key objects in Galois theory are field extensions, and the Galois group is defined for a specific class of them. To understand why the definition takes its particular form, we need to understand what can go wrong with arbitrary extensions.

Consider the extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$. An automorphism fixing $\mathbb{Q}$ must send $\sqrt[3]{2}$ to another cube root of $2$. The real cube roots of $x^3 - 2$ are just $\sqrt[3]{2}$ itself, so no nontrivial automorphism of $\mathbb{Q}(\sqrt[3]{2})$ fixes $\mathbb{Q}$. The group of such automorphisms has order one, far too small to contain the information one would want. The problem is that $\mathbb{Q}(\sqrt[3]{2})$ does not contain the other two cube roots of $2$ — namely $\omega\sqrt[3]{2}$ and $\omega^2\sqrt[3]{2}$, where $\omega = e^{2\pi i/3}$ — so there are no automorphisms to permute them. The correct setting requires a splitting field, and the extension must also be separable. When both conditions are met, the extension is called Galois.

[definition: Galois Extension] Let $k$ be a field and $K/k$ a finite field extension. The extension $K/k$ is **Galois** if it is both normal and separable: - $K/k$ is **normal** if $K$ is the splitting field of some polynomial $f \in k[x]$. - $K/k$ is **separable** if every element $\alpha \in K$ has a separable minimal polynomial over $k$, meaning $\operatorname{min}(\alpha, k)$ has no repeated roots in $\overline{k}$. [/definition]

Normality has an equivalent reformulation: $K/k$ is normal if and only if every irreducible polynomial in $k[x]$ that has at least one root in $K$ splits completely in $K[x]$. This reformulation makes clear that normality is a global condition — it is not enough for $K$ to contain one root of an irreducible polynomial; it must contain all of them.

Separability is automatic over fields of characteristic zero (such as $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$) and over finite fields, so in practice most extensions encountered in a first course in Galois theory are automatically separable. Normality is the structural requirement: it ensures that the splitting field contains all the conjugates of every element.

With the notion of a Galois extension in place, the Galois group is the automorphism group of the extension.

[definition: Galois Group] Let $K/k$ be a Galois extension. The **Galois group** of $K/k$, denoted $\operatorname{Gal}(K/k)$, is the group of all field automorphisms of $K$ that fix $k$ pointwise: \begin{align*} \operatorname{Gal}(K/k) := \{ \sigma \in \operatorname{Aut}(K) : \sigma(\alpha) = \alpha \text{ for all } \alpha \in k \}. \end{align*} The group operation is composition of automorphisms. [/definition]

[remark: Fixing the base field] The condition $\sigma(\alpha) = \alpha$ for all $\alpha \in k$ means that $\sigma$ is a $k$-algebra automorphism of $K$: it is a ring automorphism of $K$ that is $k$-linear. Since every ring automorphism of a field is injective and $K$ is finite-dimensional over $k$, this forces $\sigma$ to be bijective as well. So elements of $\operatorname{Gal}(K/k)$ are automatically $k$-linear bijections $K \to K$ that respect the field operations. [/remark]

The most important structural fact about the Galois group is its size.

[quotetheorem:3325]

This theorem is the foundation of everything that follows. It says that the Galois group is exactly as large as the degree of the extension — large enough to "see" every dimension of $K$ as a $k$-vector space. The proof proceeds via the primitive element theorem (which guarantees $K = k(\alpha)$ for some $\alpha$, since the extension is separable and finite) and the observation that automorphisms are determined by where they send $\alpha$, which must go to a root of $\operatorname{min}(\alpha, k)$, a separable polynomial whose degree equals $[K:k]$.

## The Fundamental Theorem of Galois Theory

The real power of the Galois group emerges from a perfect correspondence between its subgroups and the intermediate fields of the extension. This correspondence is not merely a coincidence of cardinality; it is a deep structural duality that reverses inclusion.

To state the correspondence, we need to define the fixed field of a subgroup.

The Galois correspondence works by reversing the direction of inclusion: large subgroups of $G$ fix fewer elements and so correspond to small intermediate fields, while small subgroups fix more and correspond to large intermediate fields. To make this precise, we ask: given a subgroup $H \le \operatorname{Gal}(K/k)$, which elements of $K$ survive every symmetry in $H$? These are precisely the elements that $H$ cannot distinguish from their neighbors in $K$ — the elements $\alpha$ such that every $\sigma \in H$ sends $\alpha$ back to itself. Collecting them gives the fixed field, the largest intermediate field on which $H$ acts as the identity.

[definition: Fixed Field] Let $K/k$ be a Galois extension and $H \le \operatorname{Gal}(K/k)$ a subgroup. The **fixed field** of $H$ is \begin{align*} K^H := \{ \alpha \in K : \sigma(\alpha) = \alpha \text{ for all } \sigma \in H \}. \end{align*} [/definition]