Here's the big idea. A field extension is a hard thing to see: infinitely many elements, and subfields hiding all over the place. But its symmetries (the automorphisms that fix the base field) form a finite group, and for the nice extensions, that group knows everything. Every intermediate field corresponds to a subgroup, degrees become indices, and "Galois over the base" becomes "normal subgroup." Field questions turn into group questions, and group questions you already know how to answer.
Let $K/F$ be a field extension. $\operatorname{Aut}(K)$ is the group of all field automorphisms of $K$ (ring isomorphisms $K \to K$), under composition. We care about the ones that leave $F$ alone.
$\operatorname{Aut}(K/F)$, the automorphism group of the extension $K/F$, is the subgroup of $\operatorname{Aut}(K)$ consisting of automorphisms that fix $F$ pointwise: $\sigma(a) = a$ for all $a \in F$.
Same word as on Automorphisms for groups: a structure-preserving bijection from something to itself. Here it preserves $+$ and $\times$.
Two facts make these groups computable. An automorphism is pinned down by what it does to generators, and it can only send a root to another root of the same polynomial.
If $K = F(\alpha_1, \dots, \alpha_m)$, then any $\sigma \in \operatorname{Aut}(K/F)$ is uniquely determined by its values on the generators $\sigma(\alpha_1), \dots, \sigma(\alpha_m)$. Furthermore, if $\alpha \in K$ is algebraic over $F$ with minimal polynomial $f \in F[x]$, then $f(\sigma(\alpha)) = 0$ for all $\sigma \in \operatorname{Aut}(K/F)$.
Proof. If $\sigma, \tau \in \operatorname{Aut}(K/F)$ agree on every $\alpha_i$, the set $\{x \in K : \sigma(x) = \tau(x)\}$ is a subfield of $K$ (closed under $+, -, \times, \div$ because $\sigma, \tau$ are homomorphisms) containing $F$ and every $\alpha_i$, so it is all of $K$. For the second part, write $f = \sum a_i x^i$ with $a_i \in F$. Then $f(\sigma(\alpha)) = \sum a_i \sigma(\alpha)^i = \sum \sigma(a_i)\sigma(\alpha^i) = \sigma(f(\alpha)) = \sigma(0) = 0$. $\blacksquare$
Find $\operatorname{Aut}(\mathbb{C}/\mathbb{R})$ and $\operatorname{Aut}(\mathbb{Q}(\sqrt[3]2)/\mathbb{Q})$.
$\mathbb{C} = \mathbb{R}(i)$, and $\sigma(i)$ must be a root of $x^2 + 1$: $\pm i$. Both work (the identity and complex conjugation), so $\operatorname{Aut}(\mathbb{C}/\mathbb{R}) = \{\operatorname{id}, z \mapsto \bar z\} \cong C_2$.
$\sigma(\sqrt[3]2)$ must be a root of $x^3 - 2$ that lies in $\mathbb{Q}(\sqrt[3]2) \subseteq \mathbb{R}$. The only real root is $\sqrt[3]2$ itself. So $\sigma = \operatorname{id}$, and the group is trivial, even though the degree is $3$.
So an automorphism group can be smaller than the degree. It can never be bigger.
If $K/F$ is a finite separable extension, then $|\operatorname{Aut}(K/F)| \le [K:F]$.
(For a simple extension $K = F(\alpha)$ of degree $n$, this is just the theorem above: $\sigma$ is determined by $\sigma(\alpha)$, which is one of the at most $n$ roots of $f_\alpha$ in $K$. The general case reduces to this one.)
Count some automorphisms yourself. Choose where the generators go; the widget follows every root to where it lands, as a permutation.
The extensions where the bound is reached are the ones with "enough symmetry."
A finite extension $K/F$ is called Galois if $|\operatorname{Aut}(K/F)| = [K:F]$.
If $K/F$ is Galois, $\operatorname{Aut}(K/F)$ is denoted $\operatorname{Gal}(K/F)$ and called the Galois group of the extension.
So $\mathbb{C}/\mathbb{R}$ is Galois ($2 = 2$) and $\mathbb{Q}(\sqrt[3]2)/\mathbb{Q}$ isn't ($1 < 3$). The difference: $\mathbb{C}$ contains both roots of $x^2 + 1$, while $\mathbb{Q}(\sqrt[3]2)$ is missing two of the three roots of $x^3 - 2$, so automorphisms have nowhere to send $\sqrt[3]2$. That's the whole story:
A finite separable extension $K/F$ is Galois if and only if $K$ is the splitting field of some polynomial $f \in F[x]$.
The proof of uniqueness of splitting fields on Splitting Fields & Algebraic Closures built isomorphisms one root at a time, with a free choice of target root at each step. When the target field is $K$ itself, those choices are the automorphisms, and there are exactly $[K:F]$ of them.
"Galois" is a property of the extension, not of the top field. $\mathbb{Q}(\sqrt[3]2, \zeta_3)$ is Galois over $\mathbb{Q}$ (it's a splitting field), and it's Galois over $\mathbb{Q}(\sqrt[3]2)$ too (it splits $x^2 + \sqrt[3]2 x + \sqrt[3]4$ there). But $\mathbb{Q}(\sqrt[3]2)$ is not Galois over $\mathbb{Q}$. A tower of Galois extensions need not be Galois.
From subgroups back to fields: every group of automorphisms fixes something.
For any subgroup $H \le \operatorname{Aut}(K)$, the fixed field of $H$ is
$$K_H = \{ \alpha \in K : \sigma(\alpha) = \alpha \text{ for all } \sigma \in H \}.$$
($K_H$ is a field because the elements where $\sigma$ agrees with the identity form a subfield, and intersecting over $\sigma \in H$ keeps it one.) Bigger groups fix less, so $H_1 \le H_2$ gives $K_{H_1} \supseteq K_{H_2}$. That order-reversal is the heart of the theorem.
Suppose $K/F$ is a Galois extension with Galois group $G = \operatorname{Gal}(K/F)$. Then:
Part 3 is why normal subgroups and quotient groups exist, historically: Galois invented the idea of a normal subgroup to describe exactly this.
Click any subgroup on the left (drawn upside down, with $\{1\}$ on top) or any field on the right. Its partner lights up, along with the degrees and whether the bottom piece is Galois.
Use the Fundamental Theorem to find every intermediate field of $\mathbb{Q}(\sqrt2, \sqrt3)/\mathbb{Q}$.
$K = \mathbb{Q}(\sqrt2, \sqrt3)$ is the splitting field of $(x^2 - 2)(x^2 - 3)$, so it's Galois of degree $4$. An automorphism sends $\sqrt2 \mapsto \pm\sqrt2$ and $\sqrt3 \mapsto \pm\sqrt3$; that's $4$ choices, so all of them occur: $G = \{1, \sigma, \tau, \sigma\tau\}$ with $\sigma$ flipping $\sqrt2$ and $\tau$ flipping $\sqrt3$. Every element has order $\le 2$, so $G \cong V_4$.
$V_4$ has exactly three subgroups of order $2$, so $K$ has exactly three intermediate fields of degree $2$: $\langle\sigma\rangle$ fixes $\sqrt3$, $\langle\tau\rangle$ fixes $\sqrt2$, and $\langle\sigma\tau\rangle$ fixes $\sqrt2\sqrt3 = \sqrt6$. So the intermediate fields are $\mathbb{Q}$, $\mathbb{Q}(\sqrt2)$, $\mathbb{Q}(\sqrt3)$, $\mathbb{Q}(\sqrt6)$, $K$, and there are no others. (Try proving "no others" without Galois theory. It's not fun.)
The correspondence flips sizes. A small subgroup has a big fixed field: $|H| = [K:K_H]$ measures how far $K_H$ is from the top, not from the bottom. If you match "order 2 subgroup" with "degree 2 field over $F$," you've matched upside down; an order-$2$ subgroup of a group of order $8$ fixes a field of degree $4$ over $F$.
Évariste Galois worked this out as a teenager and wrote the last of it down the night before a duel, at twenty. He lost the duel. The math is fine.
Five classic computations, each using the same moves: find the degree, list where generators can go, recognize the group.
Find $\operatorname{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q})$.
The minimal polynomial is $\Phi_5(x) = x^4 + x^3 + x^2 + x + 1$, so the degree is $4$. Its roots are $\zeta_5, \zeta_5^2, \zeta_5^3, \zeta_5^4$, all in $\mathbb{Q}(\zeta_5)$, so $\mathbb{Q}(\zeta_5)$ is the splitting field of $\Phi_5$: Galois. The Galois group has order $4$, and automorphisms are determined by $\sigma_a(\zeta_5) = \zeta_5^a$ for $a \in \{1, 2, 3, 4\}$.
$\sigma_2$ has order $4$: $\sigma_2^k(\zeta_5) = \zeta_5^{2^k}$, and $2^k \bmod 5$ runs $2, 4, 3, 1$. So this group is cyclic, $\operatorname{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q}) \cong C_4$, with one intermediate field (fixed by $\langle \sigma_4 \rangle$, i.e. by complex conjugation): $\mathbb{Q}(\zeta_5 + \zeta_5^{-1}) = \mathbb{Q}(\sqrt5)$.
Cyclotomic fields give Abelian Galois groups. Splitting fields of $x^n - 2$ don't:
Find the Galois group of the splitting field of $x^3 - 2$ over $\mathbb{Q}$.
The roots are $\sqrt[3]{2}, \sqrt[3]{2}\zeta_3, \sqrt[3]{2}\zeta_3^2$. The splitting field is $K = \mathbb{Q}(\sqrt[3]{2}, \zeta_3)$, and $[K:\mathbb{Q}] = 6$ (Algebraic Extensions). So $\operatorname{Gal}(K/\mathbb{Q})$ has order $6$.
Every automorphism permutes the $3$ roots, and is determined by what it does to them (they generate $K$), so $G \hookrightarrow S_3$. Since $|G| = 6 = |S_3|$, we have $\operatorname{Gal}(K/\mathbb{Q}) \cong S_3$.
One degree higher, the group becomes the symmetry group of a square.
Find the Galois group of the splitting field of $x^4 - 2$ over $\mathbb{Q}$.
The roots are $\pm \sqrt[4]{2}, \pm i\sqrt[4]{2}$. The splitting field is $K = \mathbb{Q}(\sqrt[4]{2}, i)$ with $[K:\mathbb{Q}] = 8$ (an exercise on Splitting Fields). Any $\sigma \in \operatorname{Gal}(K/\mathbb{Q})$ is determined by $\sigma(\sqrt[4]{2}) \in \{\pm\sqrt[4]{2}, \pm i\sqrt[4]{2}\}$ and $\sigma(i) \in \{\pm i\}$. That's at most $4 \cdot 2 = 8$ choices, and $|G| = 8$, so every choice is an automorphism.
Let $r$ send $\sqrt[4]2 \mapsto i\sqrt[4]2$ and fix $i$, and let $s$ be complex conjugation. On the roots, listed in order around the circle $\sqrt[4]2, i\sqrt[4]2, -\sqrt[4]2, -i\sqrt[4]2$, $r$ is a $4$-cycle (rotation) and $s$ swaps the second and fourth (a reflection). They generate the $8$ automorphisms, with $r^4 = s^2 = 1$ and $srs = r^{-1}$: $\operatorname{Gal}(K/\mathbb{Q}) \cong D_4$, the symmetries of a square whose corners are the four roots.
Examples 2 and 3 found their groups by counting choices for generators. For a quintic that's hopeless, so instead we trap the group between two facts: one from the degree, one from the real roots.
Find the Galois group $G$ of the splitting field $K$ of $f(x) = x^5 - 4x + 2$ over $\mathbb{Q}$.
By Eisenstein's criterion ($p = 2$), $f$ is irreducible. So a root $\alpha$ has $[\mathbb{Q}(\alpha):\mathbb{Q}] = 5$, and the Tower Law says $5$ divides $[K:\mathbb{Q}] = |G|$. By Cauchy's theorem, $G$ contains an element of order $5$; acting on the $5$ roots, it's a $5$-cycle.
Calculus shows $f$ has exactly $3$ real roots (graph below: $f' = 5x^4 - 4$ has only two real zeros, so $f$ turns around only twice, and the signs $f(-2) < 0 < f(-1)$, $f(1) < 0 < f(2)$, $f(0) > 0$ give three crossings), so the other $2$ are non-real complex conjugates. Complex conjugation restricted to $K$ fixes the $3$ real roots and swaps the $2$ non-real ones: a transposition.
Any subgroup of $S_5$ containing a $5$-cycle and a transposition must be $S_5$ (true for $S_p$, $p$ prime). Hence $G \cong S_5$.
Finally, finite fields, where the Galois group is as simple as it could possibly be.
Find $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$.
Let $K = \mathbb{F}_{p^n}$ and $F = \mathbb{F}_p$. $K/F$ is the splitting field of $x^{p^n} - x$ (Finite Fields), so it is Galois with $[K:F] = n$. The Frobenius automorphism $\sigma(\alpha) = \alpha^p$ fixes $F$ (Fermat: $a^p = a$ in $\mathbb{F}_p$).
If $\sigma^k = \operatorname{id}$, then every $\alpha \in K$ is a root of $x^{p^k} - x$, which has at most $p^k$ roots, so $p^n \le p^k$, i.e. $k \ge n$. So $\sigma$ has order $n = |G|$, and the Galois group is cyclic of order $n$, generated by Frobenius: $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong C_n$.
Example 4 is the famous one. $S_5$ is not a solvable group, and Solvability by Radicals turns that into: the roots of $x^5 - 4x + 2$ cannot be written with $+, -, \times, \div$, and $n$-th roots.
Read the subgroup lattice, read off the fields.
Let $K$ be the splitting field of $x^3 - 2$ over $\mathbb{Q}$, with $G = \operatorname{Gal}(K/\mathbb{Q}) \cong S_3$. Which intermediate fields are Galois over $\mathbb{Q}$?
The subgroups of $S_3$ are $\{1\}$, three subgroups of order $2$, $A_3$, and $S_3$. The normal ones are $\{1\}$, $A_3$, $S_3$, so the Galois intermediate fields are $K$, $K_{A_3}$, and $\mathbb{Q}$.
$[K_{A_3}:\mathbb{Q}] = [S_3:A_3] = 2$, and $\zeta_3$ is fixed by the $3$-cycles (they send $\sqrt[3]2 \mapsto \zeta_3\sqrt[3]2$ while fixing $\zeta_3$), so $K_{A_3} = \mathbb{Q}(\zeta_3)$: the splitting field of $x^2 + x + 1$. The three fields of degree $3$, $\mathbb{Q}(\sqrt[3]2)$, $\mathbb{Q}(\zeta_3\sqrt[3]2)$, $\mathbb{Q}(\zeta_3^2\sqrt[3]2)$, correspond to the non-normal order-$2$ subgroups, and none is Galois.
Show that $\operatorname{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q})$ is cyclic of order $6$, and find the degrees of its intermediate fields. How many are there?
$\mathbb{Q}(\zeta_7)$ is the splitting field of $\Phi_7$, of degree $\varphi(7) = 6$. The automorphisms are $\sigma_a\colon \zeta_7 \mapsto \zeta_7^a$, $a = 1, \dots, 6$. Powers of $3$ mod $7$ are $3, 2, 6, 4, 5, 1$, so $\sigma_3$ has order $6$ and the group is $C_6$.
$C_6$ has one subgroup of each order $1, 2, 3, 6$, so there are exactly $4$ intermediate fields, of degrees $6, 3, 2, 1$ over $\mathbb{Q}$. (The cubic one is $\mathbb{Q}(\zeta_7 + \zeta_7^{-1}) = \mathbb{Q}(\cos\frac{2\pi}{7})$, fixed by conjugation $\sigma_6$; the quadratic one is $\mathbb{Q}(\sqrt{-7})$.) All are Galois over $\mathbb{Q}$, since every subgroup of an Abelian group is normal.
Let $K = \mathbb{Q}(\sqrt[4]2, i)$ with $G \cong D_4$ as in Example 3. Find the fixed field of $\langle r \rangle$ and of $\langle s \rangle$. Which is Galois over $\mathbb{Q}$?
$r$ fixes $i$, and $|\langle r\rangle| = 4$, so its fixed field has degree $[G:\langle r\rangle] = 2$ over $\mathbb{Q}$ and contains $\mathbb{Q}(i)$, which already has degree $2$: $K_{\langle r\rangle} = \mathbb{Q}(i)$. Since $\langle r \rangle$ has index $2$, it's normal, and $\mathbb{Q}(i)/\mathbb{Q}$ is Galois (it splits $x^2 + 1$).
$s$ (complex conjugation) fixes the real number $\sqrt[4]2$; $|\langle s\rangle| = 2$, so the fixed field has degree $4$ over $\mathbb{Q}$, and it contains $\mathbb{Q}(\sqrt[4]2)$ of degree $4$: $K_{\langle s\rangle} = \mathbb{Q}(\sqrt[4]2)$. $\langle s \rangle$ is not normal ($rsr^{-1} = r^2s \neq s$), and indeed $\mathbb{Q}(\sqrt[4]2)$ contains only two of the four roots of $x^4 - 2$, so it isn't Galois.
Example 1 was no accident. Every cyclotomic field has the same kind of Galois group.
Prove that $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$ for every $n \ge 1$. In particular it is Abelian.
Proof. Let $\zeta = \zeta_n$ and $K = \mathbb{Q}(\zeta)$. Every $n$-th root of unity is a power of $\zeta$, so $K$ is the splitting field of $x^n - 1$: Galois, with $|G| = [K:\mathbb{Q}] = \varphi(n)$ (Cyclotomic Polynomials).
Each $\sigma \in G$ sends $\zeta$ to a root of its minimal polynomial $\Phi_n$, i.e. $\sigma(\zeta) = \zeta^{a}$ for a unique $a \in (\mathbb{Z}/n\mathbb{Z})^\times$. Define $\theta\colon G \to (\mathbb{Z}/n\mathbb{Z})^\times$, $\sigma \mapsto a$.
So $\theta$ is an isomorphism. $\blacksquare$