Galois Theory
Galois Theory
Galois theory — perhaps the pinnacle of undergraduate algebra — studies field extensions through their symmetry groups. The Galois group $\text{Gal}(K/F)$ consists of all field automorphisms of $K$ fixing $F$ pointwise. The Galois correspondence bijectively pairs subgroups of $\text{Gal}(K/F)$ with intermediate fields. This correspondence resolves ancient questions: which polynomial equations are solvable by radicals, and which geometric constructions are impossible with compass and straightedge.
Galois Group & Galois Correspondence
For a Galois extension $K/F$ (normal and separable), $|\text{Gal}(K/F)|=[K:F]$. The Galois correspondence: (i) $H\leq\text{Gal}(K/F)\longleftrightarrow E=K^H=\{x\in K:\sigma(x)=x\,\forall\sigma\in H\}$ (fixed field). (ii) $[K:E]=|H|$ and $[E:F]=[G:H]$. (iii) $H\trianglelefteq G\iff E/F$ is Galois. Consequence: if $f$ has $n$ distinct roots in $K$, then $\text{Gal}(f/F)\leq S_n$ (permutes the roots).
Solvability by Radicals
$f\in F[x]$ is solvable by radicals iff its Galois group $G=\text{Gal}(f/F)$ is a solvable group (has a subnormal series $G=G_0\supset G_1\supset\cdots\supset G_k=\{e\}$ with $G_i/G_{i+1}$ abelian). $S_n$ is solvable for $n\leq 4$ (explaining the quadratic/cubic/quartic formulas) but $A_5\leq S_5$ is simple (non-trivial non-abelian simple group), so $S_5$ is not solvable — the general quintic has no radical formula (Abel-Ruffini theorem, 1824).
Example 1
Compute $\text{Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$.
Solution: $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$, $[K:\mathbb{Q}]=4$. Automorphisms must permute the roots of $x^2-2$ and $x^2-3$. Four automorphisms: $e$: identity; $\sigma$: $\sqrt{2}\mapsto-\sqrt{2},\sqrt{3}\mapsto\sqrt{3}$; $\tau$: $\sqrt{2}\mapsto\sqrt{2},\sqrt{3}\mapsto-\sqrt{3}$; $\sigma\tau$: $\sqrt{2}\mapsto-\sqrt{2},\sqrt{3}\mapsto-\sqrt{3}$. Each has order 2: $G\cong\mathbb{Z}_2\times\mathbb{Z}_2$ (Klein four-group). Subfields correspond to subgroups: $K^{\langle\sigma\rangle}=\mathbb{Q}(\sqrt{3})$, $K^{\langle\tau\rangle}=\mathbb{Q}(\sqrt{2})$, $K^{\langle\sigma\tau\rangle}=\mathbb{Q}(\sqrt{6})$.
Example 2
Show the regular 17-gon is constructible (Gauss).
Solution: Gauss proved (1796): a regular $n$-gon is constructible iff $n=2^k p_1\cdots p_m$ where $p_i$ are distinct Fermat primes. $17=F_2=2^{2^2}+1$ is Fermat prime. The splitting field of $x^{17}-1$ over $\mathbb{Q}$ has Galois group $\mathbb{Z}_{16}$ (cyclic of order $\phi(17)=16$); since $16=2^4$ is a power of 2, the tower of subfields corresponds to a tower of degree-2 extensions — each step constructible by quadratic formula. Hence $\cos(2\pi/17)$ is constructible.
Practice
- Compute $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ for $\zeta_n=e^{2\pi i/n}$ and show it is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$.
- Prove the Galois group of a separable polynomial permutes its roots faithfully.
- Show the splitting field of $x^3-2$ over $\mathbb{Q}$ has Galois group $S_3$ and find all six intermediate fields.
- Verify Abel-Ruffini for $f(x)=x^5-10x+5$ by computing its Galois group is $S_5$.
Show Answer Key
1. Each $\sigma\in\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is determined by $\sigma(\zeta_n)=\zeta_n^a$ with $\gcd(a,n)=1$. The map $\sigma\mapsto a\bmod n$ is an isomorphism $\text{Gal}\cong(\mathbb{Z}/n\mathbb{Z})^*$.
2. The Galois group permutes roots of the minimal polynomial of each element. If $\sigma(\alpha)=\alpha$ for all roots $\alpha$, then $\sigma=\text{id}$ (since the splitting field is generated by these roots). So the action on roots is faithful.
3. Splitting field of $x^3-2$ is $\mathbb{Q}(\sqrt[3]{2},\omega)$ where $\omega=e^{2\pi i/3}$. Degree $=[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$, so $|\text{Gal}|=6$, hence $\text{Gal}\cong S_3$. The six intermediate fields: $\mathbb{Q}(\sqrt[3]{2})$, $\mathbb{Q}(\omega\sqrt[3]{2})$, $\mathbb{Q}(\omega^2\sqrt[3]{2})$ (degree 3, fixed by reflections), $\mathbb{Q}(\omega)$ (degree 2, fixed by $A_3$), plus $\mathbb{Q}$ and the full field.
4. $f(x)=x^5-10x+5$ is irreducible by Eisenstein ($p=5$). $f$ has exactly 3 real roots and 2 complex conjugate roots. A degree-5 irreducible polynomial with exactly 2 complex roots has Galois group $S_5$ (it contains a 5-cycle from irreducibility and a transposition from complex conjugation, which generate $S_5$). Since $S_5$ is not solvable, $f$ is not solvable by radicals.