The Galois Group of an Extension

The Galois group is the organizing structure behind the modern proof that the general quintic has no solution by radicals: solvability by radicals corresponds precisely to the Galois group of the splitting field being a solvable group, a criterion due to Galois himself. In number theory, the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ is a central object of study, acting on Galois representations that encode deep arithmetic information used in the proof of Fermat's Last Theorem. In algebraic geometry and topology, Galois groups of function field extensions classify branched covering spaces, linking algebra to covering-space theory. In coding theory and cryptography, Galois groups of finite fields (generated by the Frobenius automorphism) underlie the structure of finite field arithmetic used in Reed–Solomon codes and elliptic-curve cryptosystems. In computer algebra systems, algorithms that compute Galois groups of specific polynomials (via resolvent polynomials and permutation group recognition) are used to decide solvability and to simplify radical expressions symbolically.

:::quiz[For $K = \mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$ over $F=\mathbb{Q}$, what is $|\mathrm{Aut}(K/F)|$?]

• 3, matching $[K:F]$
• 1, since $K$ contains only one of the three roots of $x^3-2$ ✓
• 6, since $x^3-2$ has three roots
• Infinite

:::quiz[What is the key finiteness hypothesis in the theorem $|\mathrm{Aut}(K/F)| \le [K:F]$?]

• $F$ must be algebraically closed
• $K$ must be uncountable
• $[K:F]$ must be finite ✓
• $K$ must equal $F$

:::quiz[For the splitting field $K=\mathbb{Q}(\sqrt[3]2,\omega)$ of $x^3-2$ over $\mathbb{Q}$, the Galois group is isomorphic to]

• The trivial group
• $\mathbb{Z}/6\mathbb{Z}$
• $S_3$, of order 6 ✓
• $S_5$

:::faq[Can $\mathrm{Aut}(K/F)$ be trivial even when $[K:F]>1$?] Yes, and this is common rather than exceptional. Whenever $K$ does not contain all the roots of the minimal polynomials of its generators over $F$, many candidate images for a generator are simply unavailable inside $K$, shrinking the group. The clean example is $K=\mathbb{Q}(\sqrt[3]2)$ over $\mathbb{Q}$: the only cube root of 2 lying in this real field is $\sqrt[3]2$ itself, forcing every automorphism to fix it, hence to be the identity. This is exactly the phenomenon that motivates requiring $K$ to be a splitting field (normality) when one wants the automorphism group to be as large as possible.

:::faq[Does the bound $|\mathrm{Aut}(K/F)|\le [K:F]$ hold in every characteristic?] Yes, the inequality itself holds in every characteristic, including positive characteristic, as long as $[K:F]$ is finite; the proof given uses only general field homomorphism properties and the tower law, neither of which depends on characteristic. What can fail in positive characteristic is the equality case: even for a splitting field of an irreducible polynomial, if the polynomial is inseparable (has repeated roots in its splitting field, which can only happen in characteristic $p$), the number of distinct roots is strictly less than the degree, and the equality $|\mathrm{Aut}(K/F)|=[K:F]$ can genuinely fail. This is why the definition of a Galois extension requires separability in addition to normality.

Ask the tutor: Why does the Galois group of the splitting field of an irreducible polynomial always act transitively on its roots?

Think of $F$ as a fixed floor and $K$ as a room built on top of it by adjoining new elements — typically roots of polynomials. An element $\sigma \in \mathrm{Aut}(K/F)$ is a rigid motion of the room that leaves the floor perfectly still (every point of $F$ stays put) but is allowed to shuffle the newly adjoined furniture, as long as it respects all the algebraic joints (addition, multiplication) holding the room together. The key structural fact is that this shuffling is never arbitrary. If $\alpha \in K$ is a root of a polynomial $f(x) \in F[x]$, applying $\sigma$ to the equation $f(\alpha) = 0$ and using that $\sigma$ fixes the coefficients of $f$ shows $f(\sigma(\alpha)) = 0$ too. So $\sigma$ can only move $\alpha$ to another root of the same polynomial. This is why the Galois group naturally acts on finite sets of roots, turning an infinite-looking automorphism group into something computable: a subgroup of a symmetric group on finitely many roots. Consider $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ over $F = \mathbb{Q}$, the splitting field of $x^3 - 2$, where $\omega = e^{2\pi i/3}$. The three roots $\sqrt[3]{2}, \omega\sqrt[3]{2}, \omega^2\sqrt[3]{2}$ sit at the vertices of an equilateral triangle in $\mathbb{C}$ (once we embed $K$ there), and automorphisms fixing $\mathbb{Q}$ permute these three roots while respecting the field structure — visibly the symmetries of the triangle.

Example 1: $K = \mathbb{Q}(\sqrt[3]{2})$ over $F = \mathbb{Q}$ — a genuinely asymmetric extension. Here $x^3 - 2$ is irreducible over $\mathbb{Q}$ (Eisenstein at $p=2$), so $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$, and $\mathbb{Q}(\sqrt[3]2) \subset \mathbb{R}$. By the theorem, any $\sigma \in \mathrm{Aut}(K/\mathbb{Q})$ sends $\alpha := \sqrt[3]2$ to a root in $K$ of the minimal polynomial $x^3 - 2$. The three complex roots of $x^3-2$ are $\alpha, \omega\alpha, \omega^2\alpha$ where $\omega = e^{2\pi i/3}$ is a primitive cube root of unity. But $\omega\alpha$ and $\omega^2\alpha$ are non-real (since $\omega \notin \mathbb{R}$ and $\alpha \ne 0$), while $K = \mathbb{Q}(\sqrt[3]2) \subset \mathbb{R}$. So the only root of $x^3-2$ actually lying in $K$ is $\alpha$ itself. Hence any $\sigma \in \mathrm{Aut}(K/\mathbb{Q})$ must send $\alpha \mapsto \alpha$. Since $K = \mathbb{Q}(\alpha)$ and $\sigma$ fixes $\mathbb{Q}$ and fixes $\alpha$, and every element of $K$ is a $\mathbb{Q}$-polynomial expression in $\alpha$ (namely $K = \{a+b\alpha+c\alpha^2 : a,b,c \in \mathbb{Q}\}$), $\sigma$ fixes every element of $K$. So $\sigma = \mathrm{id}_K$. Therefore

This is a striking illustration that the inequality in the theorem is often strict: $K/\mathbb{Q}$ has a nontrivial degree but a trivial automorphism group, precisely because $K$ does not contain enough roots of the defining polynomial (it is not a splitting field, only contains one of three roots). This phenomenon — the group being too small to "see" the full symmetry of the polynomial — is exactly what the notion of a Galois Extension is built to rule out.

Example 2: $K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ over $F = \mathbb{Q}$ — the full splitting field. Now let $K$ be the splitting field of $x^3-2$ over $\mathbb{Q}$, i.e. $K = \mathbb{Q}(\alpha, \omega)$ with $\alpha = \sqrt[3]2$, $\omega = e^{2\pi i/3}$, so $\omega^2+\omega+1=0$. First compute $[K:\mathbb{Q}]$. We have $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$ as above. The minimal polynomial of $\omega$ over $\mathbb{Q}(\alpha) \subset \mathbb{R}$ is $x^2+x+1$ (it is irreducible over $\mathbb{Q}(\alpha)$ since its roots $\omega,\omega^2$ are non-real, hence not in $\mathbb{Q}(\alpha)$, hence it has no linear factor over $\mathbb{Q}(\alpha)$, hence — being degree 2 — it is irreducible there). So $[\mathbb{Q}(\alpha,\omega):\mathbb{Q}(\alpha)] = 2$, and by the tower law

By the theorem, $|\mathrm{Aut}(K/\mathbb{Q})| \le 6$. We now exhibit 6 distinct automorphisms, which forces equality. Any $\sigma \in \mathrm{Aut}(K/\mathbb{Q})$ must send $\alpha$ to a root of $x^3-2$ lying in $K$ — and now all three roots $\alpha, \omega\alpha, \omega^2\alpha$ lie in $K$, since $K$ contains both $\alpha$ and $\omega$. Likewise $\sigma$ must send $\omega$ to a root of $x^2+x+1$ lying in $K$, namely $\omega$ or $\omega^2$ (both already in $K$). Since $K = \mathbb{Q}(\alpha,\omega)$, an $F$-automorphism of $K$ is completely determined once we specify $\sigma(\alpha)$ and $\sigma(\omega)$, because every element of $K$ is a $\mathbb{Q}$-polynomial combination of $\alpha$ and $\omega$. There are at most $3 \times 2 = 6$ candidate pairs $(\sigma(\alpha), \sigma(\omega))$. We verify each is realized by an actual automorphism, by explicitly constructing the six maps as compositions of two basic automorphisms:

• $\tau$: the field automorphism with $\tau(\alpha) = \omega\alpha$, $\tau(\omega) = \omega$ (a "rotation" of the roots of $x^3-2$).
• $\rho$: the field automorphism with $\rho(\alpha) = \alpha$, $\rho(\omega) = \omega^2$ (complex conjugation restricted to $K$, since $K\subset \mathbb{C}$ and conjugation fixes $\alpha\in\mathbb R$ and swaps $\omega\leftrightarrow\bar\omega=\omega^2$).

That $\tau,\rho$ exist and are well-defined field automorphisms of $K$ follows from the isomorphism-extension theorem underlying Uniqueness of Splitting Fields: an assignment of each root of $x^3-2$ (resp. $x^2+x+1$) to a root of the same polynomial extends to an automorphism of the splitting field. One computes $\tau^3 = \mathrm{id}$, $\rho^2 = \mathrm{id}$, and the relation $\rho\tau\rho^{-1} = \tau^{-1}$ (check on generators: $\rho\tau\rho^{-1}(\alpha) = \rho\tau(\alpha) = \rho(\omega\alpha) = \omega^2\alpha = \tau^{-1}(\alpha)$, and both sides fix $\omega$ appropriately). The six elements

are pairwise distinct (they act differently on the pair $(\alpha,\omega)$: e.g. $\tau^k$ sends $\alpha \mapsto \omega^k\alpha$ for $k=0,1,2$, distinguishing the first three, and multiplying by $\rho$ flips $\omega \mapsto \omega^2$, distinguishing the last three from the first three). Hence $|\mathrm{Aut}(K/\mathbb{Q})| \ge 6$, and combined with the theorem's bound $|\mathrm{Aut}(K/\mathbb{Q})| \le 6$, we get equality:

The isomorphism with $S_3$ is visible from the action on the three roots $\{\alpha,\omega\alpha,\omega^2\alpha\}$: $\tau$ acts as a 3-cycle, $\rho$ acts as a transposition (fixing $\alpha$, swapping $\omega\alpha \leftrightarrow \omega^2\alpha$), and $\langle \tau,\rho\rangle$ with these relations is exactly the presentation of $S_3$ (equivalently the dihedral group of order 6).

:::counterexample[Why finiteness of $[K:F]$ is needed] The theorem requires $[K:F] < \infty$. Drop this hypothesis and the bound $|\mathrm{Aut}(K/F)|\le[K:F]$ becomes meaningless or false in spirit, because it is possible for $|\mathrm{Aut}(K/F)|$ to be infinite while comparing cardinals loses all content, but more importantly the proof method itself breaks down: the induction on $[K:F]$ requires a well-founded finite descent, which is unavailable for infinite degree. Concretely, take $F = \mathbb{Q}$ and $K = \mathbb{Q}(x)$, wait — we need a field extension, so instead take $K = \overline{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$, with $F=\mathbb{Q}$. Here $[K:F]=\infty$. Yet $\mathrm{Aut}(K/F)$ is uncountably infinite (it contains, for instance, an automorphism swapping any two conjugate roots of any irreducible polynomial while fixing a chosen set of others, and by a Zorn's-lemma argument these can be combined in uncountably many consistent ways). There is no finite bound to state, and the inductive proof — which crucially used the strict decrease $[K:F(\alpha)] < [K:F]$ to invoke the inductive hypothesis on a smaller finite number — has no base case to anchor to and no well-ordering to descend along. Even more simply: take $F$ any field and $K = F(t)$ a purely transcendental extension by an indeterminate $t$. Then $[K:F] = \infty$, and $\mathrm{Aut}(K/F)$ contains the map $t \mapsto t+1$, whose powers $t \mapsto t+n$ for every integer $n$ give infinitely many distinct elements — again illustrating that without finiteness of $[K:F]$, there is no meaningful finite bound, and the entire counting argument (which partitioned $\mathrm{Aut}(K/F)$ into finitely many finite pieces $S_i$) fails to terminate.