Galois theory essentials

In one sentence

For a normal and separable extension the subgroups of the Galois group correspond exactly to the intermediate fields, inclusion-reversing — and the whole subject is the consequences of that dictionary.

Why the exam asks it

Because it turns questions about infinitely many intermediate fields into questions about finitely many subgroups, and because "Galois" has two hypotheses that options drop. $\mathbb{Q}(\sqrt[3]2)$ is not Galois, and the group of order $3$ people expect does not exist.

The idea

$L/K$ finite is Galois if it is normal and separable. Then $\operatorname{Gal}(L/K)$, the automorphisms of $L$ fixing $K$ pointwise, has order exactly $[L:K]$.

In characteristic $0$ and over finite fields every extension is separable, so in this syllabus Galois effectively means normal, that is a splitting field.

Fundamental theorem. For $L/K$ Galois with $G = \operatorname{Gal}(L/K)$, there is an inclusion-reversing bijection

$$\{\text{subgroups } H \le G\} \longleftrightarrow \{\text{fields } K \subseteq F \subseteq L\},$$

given by $H \mapsto L^H$ and $F \mapsto \operatorname{Gal}(L/F)$, with $[L:F] = |H|$ and $[F:K] = [G:H]$. Moreover $F/K$ is normal $\iff$ $H \trianglelefteq G$, and then $\operatorname{Gal}(F/K)\cong G/H$.

The smallest example showing everything is $\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}$, degree $4$, with group the Klein four-group. Its three subgroups of order $2$ match the three intermediate fields $\mathbb{Q}(\sqrt2)$, $\mathbb{Q}(\sqrt3)$, $\mathbb{Q}(\sqrt6)$. The third one is the detail worth noticing: $\sqrt6 = \sqrt2\sqrt3$ is fixed by the automorphism negating both square roots, which is why it is there and why the count comes out at three rather than two.

Since $V_4$ is abelian, every subgroup is normal, so every intermediate field is normal over $\mathbb{Q}$.

The non-normal example

$x^3-2$ has splitting field $\mathbb{Q}(\sqrt[3]2,\omega)$ of degree $6$, with $\operatorname{Gal} \cong S_3$ — the group permutes the three roots and every permutation occurs.

$\mathbb{Q}(\sqrt[3]2)$ corresponds to a subgroup of order $2$, of index $3$. That subgroup is not normal in $S_3$, which is precisely why $\mathbb{Q}(\sqrt[3]2)/\mathbb{Q}$ is not a normal extension. The dictionary translates one fact into the other exactly.

Note also $\operatorname{Gal}(\mathbb{Q}(\sqrt[3]2)/\mathbb{Q})$ is trivial, not of order $3$: any automorphism must send $\sqrt[3]2$ to a root of $x^3-2$ inside the field, and only one is there. So $|\operatorname{Gal}| = 1 \ne 3 = [L:K]$, and the failure of that equality is the failure of being Galois.

Standard Galois groups

• Cyclotomic: $\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n)^\times$, of order $\varphi(n)$. For $n = 5$ this is cyclic of order $4$; for $n = 8$ it is the Klein group, so it is not always cyclic.
• Finite fields: $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is cyclic of order $n$, generated by Frobenius. Its subgroup lattice is the divisor lattice of $n$, which is the subfield picture from the previous chapter seen through the dictionary.
• Degree 2: always $C_2$.
• Splitting field of an irreducible cubic: $S_3$, or $C_3$ when the discriminant is a square.

Solvability by radicals

A polynomial is solvable by radicals $\iff$ its Galois group is a solvable group. $S_n$ is solvable for $n \le 4$ and not for $n \ge 5$, because $A_5$ is simple and non-abelian. Hence the general quintic has no formula in radicals, and specific quintics with group $S_5$ — such as $x^5-6x+3$ — have no radical solution.

This is where the group theory chapters pay off: simplicity of $A_5$ was proved by counting conjugacy classes, and it is the reason for this theorem.

Where intuition breaks

"Every finite extension is Galois." Needs normal and separable. $\mathbb{Q}(\sqrt[3]2)/\mathbb{Q}$ is neither Galois nor normal.

"$|\operatorname{Gal}(L/K)| = [L:K]$." Only when Galois. For $\mathbb{Q}(\sqrt[3]2)$ the degree is $3$ and the group is trivial.

"The correspondence preserves inclusions." It reverses them. Bigger field, smaller group; the whole field corresponds to the identity.

"Every intermediate field is normal over $K$." Only those matching normal subgroups. $\mathbb{Q}(\sqrt[3]2)$ inside the degree-$6$ splitting field is the counterexample, and it is the same fact as its subgroup not being normal.

"$\operatorname{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is cyclic." It is $(\mathbb{Z}/n)^\times$, which is cyclic exactly when $n$ is $1,2,4,p^k$ or $2p^k$. For $n=8$ it is $V_4$.

"$\mathbb{Q}(\sqrt2,\sqrt3)$ has two intermediate fields." Three — $\sqrt6$ is easy to miss, and the group tells you to look for it.

"Separability is a real constraint here." In characteristic $0$ and over finite fields, no. It bites only in characteristic $p$ with infinite base fields, such as $\mathbb{F}_p(t^{1/p})/\mathbb{F}_p(t)$.

"Solvable group means solvable equation of any degree." Yes — and the point is that $S_5$ is not solvable, not that high degree is inherently unsolvable. $x^5-2$ has a solvable group and is solvable by radicals.

The exam's angle

1. Compute the degree first; the group order equals it if Galois. A mismatch tells you the extension is not.
2. Identify the group by its action on roots. For a splitting field of a degree-$n$ irreducible, the group embeds in $S_n$, and the degree pins which subgroup.
3. Count intermediate fields by counting subgroups. That is the whole point of the dictionary, and it catches $\mathbb{Q}(\sqrt6)$.
4. Normal subgroup $\iff$ normal intermediate extension. Questions about normality of a subfield are group questions.
5. For solvability, name the group and ask whether it is solvable — $S_n$ for $n\le4$ yes, $n \ge 5$ no.

The night before

• Galois $=$ normal $+$ separable. In characteristic $0$, Galois $=$ splitting field.
• $|\operatorname{Gal}(L/K)| = [L:K]$ iff Galois. $\mathbb{Q}(\sqrt[3]2)$: degree $3$, group trivial.
• Fundamental theorem: inclusion-reversing bijection; $[L:L^H] = |H|$; $F/K$ normal $\iff$ $H \trianglelefteq G$, with $\operatorname{Gal}(F/K)\cong G/H$.
• $\mathbb{Q}(\sqrt2,\sqrt3)$: degree $4$, group $V_4$, three intermediate fields including $\mathbb{Q}(\sqrt6)$.
• $x^3-2$: splitting field degree $6$, group $S_3$; $\mathbb{Q}(\sqrt[3]2)$ matches a non-normal subgroup of order $2$.
• Cyclotomic: $(\mathbb{Z}/n)^\times$, order $\varphi(n)$; cyclic only for $n = 1,2,4,p^k,2p^k$.
• Finite fields: $\operatorname{Gal}$ cyclic of order $n$, generated by Frobenius; subgroups match divisors.
• Solvable by radicals $\iff$ solvable Galois group. $S_n$ solvable for $n\le4$ only, because $A_5$ is simple.