Galois Theory

Study Sheet

Galois Theory

Where the two halves meet: symmetry groups of fields, the fundamental theorem, and the unsolvable quintic

Automorphisms of Fields

Concept
Field automorphisms fixing a base field

An automorphism of a field EE is an isomorphism σ:E→E\sigma:E\to E. For an extension E/FE/F, an FF-automorphism also fixes every element of FF. If α∈E\alpha\in E is a root of f∈F[x]f\in F[x], then σ(α)\sigma(\alpha) is again a root of ff: FF-automorphisms PERMUTE the roots of polynomials over FF.

Applying σ\sigma to f(α)=0f(\alpha)=0 gives f(σ(α))=0f(\sigma(\alpha))=0, because σ\sigma respects sums and products and leaves the coefficients (in FF) alone. So an FF-automorphism can only move a root to another root of the same polynomial. If EE is generated by roots, an automorphism is completely determined by where it sends them, which turns field questions into questions about permutations — groups.

Example
The automorphisms of Q(2)\mathbb Q(\sqrt2)

A Q\mathbb Q-automorphism must send 2\sqrt2 to a root of x2−2x^2-2: either 2\sqrt2 or −2-\sqrt2. So there are exactly two: the identity and σ(a+b2)=a−b2\sigma(a+b\sqrt2)=a-b\sqrt2. For example σ(3+52)=3−52\sigma(3+5\sqrt2)=3-5\sqrt2, and σ\sigma respects products: σ((1+2)2)=σ(3+22)=3−22=(1−2)2\sigma\big((1+\sqrt2)^2\big)=\sigma(3+2\sqrt2)=3-2\sqrt2=(1-\sqrt2)^2.

Example
Only one automorphism of Q(23)\mathbb Q(\sqrt[3]2)

A Q\mathbb Q-automorphism of Q(23)\mathbb Q(\sqrt[3]2) sends 23\sqrt[3]2 to a root of x3−2x^3-2 lying in Q(23)\mathbb Q(\sqrt[3]2). But that field is real, and x3−2x^3-2 has only one real root. So the only automorphism is the identity — even though the degree is 33.

Side note
Where it goes wrong

An automorphism cannot send a root anywhere it likes: it must go to a root of the SAME minimal polynomial. In Q(2,3)\mathbb Q(\sqrt2,\sqrt3) no automorphism sends 2\sqrt2 to 3\sqrt3, since 3\sqrt3 is not a root of x2−2x^2-2.

Try it
Try it: complex conjugation

Complex conjugation z↦zˉz\mapsto\bar z is an R\mathbb R-automorphism of C\mathbb C. Which roots of x2+1x^2+1 does it swap, and what is (2+3i)(1−i)‾\overline{(2+3i)(1-i)}?

Show the answer

It swaps ii and −i-i. Since (2+3i)(1−i)=5+i(2+3i)(1-i)=5+i, its conjugate is 5−i\mathbf{5-i}, which equals (2−3i)(1+i)(2-3i)(1+i) — conjugation respects products.

The Galois Group

Concept
The Galois group of an extension

The FF-automorphisms of EE form a group Gal⁡(E/F)\operatorname{Gal}(E/F) under composition. When EE is the splitting field of a polynomial with distinct roots (a Galois extension), ∣Gal⁡(E/F)∣=[E:F]|\operatorname{Gal}(E/F)|=[E:F]. The Galois group of a polynomial is the Galois group of its splitting field; it acts faithfully on the roots, so it is a subgroup of SnS_n.

Here the two halves of the course meet. A field extension now has a group attached to it, the group of its symmetries, and the size of that group equals the degree of the extension. Computing a Galois group means deciding which permutations of the roots are realized by automorphisms.

Example
Gal⁡(Q(2,3)/Q)≅Z2×Z2\operatorname{Gal}(\mathbb Q(\sqrt2,\sqrt3)/\mathbb Q)\cong\mathbb Z_2\times\mathbb Z_2

An automorphism sends 2↦±2\sqrt2\mapsto\pm\sqrt2 and 3↦±3\sqrt3\mapsto\pm\sqrt3 independently, giving 44 automorphisms — matching [Q(2,3):Q]=4[\mathbb Q(\sqrt2,\sqrt3):\mathbb Q]=4. Each non-identity one has order 22, so the group is Z2×Z2\mathbb Z_2\times\mathbb Z_2, not Z4\mathbb Z_4.

Example
The Galois group of x3−2x^3-2 is S3S_3

The splitting field has degree 66, so the Galois group has order 66 and sits inside S3S_3, which also has order 66. So it is all of S3S_3: every permutation of the three roots 23,23ω,23ω2\sqrt[3]2,\sqrt[3]2\omega,\sqrt[3]2\omega^2 comes from an automorphism. It is non-abelian.

Example
Finite fields: the Frobenius map

Gal⁡(Fpn/Fp)\operatorname{Gal}(\mathbb F_{p^n}/\mathbb F_p) is cyclic of order nn, generated by the Frobenius automorphism a↦apa\mapsto a^p (it respects sums by the freshman's dream). For F8/F2\mathbb F_8/\mathbb F_2 the group is Z3\mathbb Z_3: a↦a2↦a4↦a8=aa\mapsto a^2\mapsto a^4\mapsto a^8=a.

Side note
Where it goes wrong

∣Gal⁡(E/F)∣=[E:F]|\operatorname{Gal}(E/F)|=[E:F] only for Galois extensions. Q(23)\mathbb Q(\sqrt[3]2) has degree 33 but only 11 automorphism, because it is not the splitting field of x3−2x^3-2.

Try it
Try it: a cyclotomic Galois group

What is the order of the Galois group of Φ5=x4+x3+x2+x+1\Phi_5=x^4+x^3+x^2+x+1 over Q\mathbb Q?

Show the answer

Its splitting field is Q(ζ5)\mathbb Q(\zeta_5), of degree φ(5)=4\varphi(5)=4, so the group has order 4\mathbf4 (it is cyclic, ≅U(5)≅Z4\cong U(5)\cong\mathbb Z_4).

The Fundamental Theorem of Galois Theory

Concept
Subgroups correspond to intermediate fields

For a finite Galois extension E/FE/F with group GG, the maps H↦{x∈E:σ(x)=x ∀σ∈H}H\mapsto\{x\in E:\sigma(x)=x\ \forall\sigma\in H\} (the fixed field) and K↦Gal⁡(E/K)K\mapsto\operatorname{Gal}(E/K) are inverse bijections between subgroups H≤GH\le G and intermediate fields F⊆K⊆EF\subseteq K\subseteq E. They reverse inclusion, and [E:K]=∣H∣[E:K]=|H|, [K:F]=[G:H][K:F]=[G:H]. Moreover K/FK/F is Galois exactly when HH is normal in GG.

This is the theorem the whole course has been building towards. Questions about fields between FF and EE — how many there are, their degrees, which are themselves splitting fields — become questions about the subgroups of one finite group, which the first half taught you to answer. Bigger subgroups fix smaller fields.

Example
The intermediate fields of Q(2,3)\mathbb Q(\sqrt2,\sqrt3)

G=Z2×Z2G=\mathbb Z_2\times\mathbb Z_2 has 55 subgroups: the trivial one, three of order 22, and GG. So there are 55 intermediate fields: Q(2,3)\mathbb Q(\sqrt2,\sqrt3), the three quadratic fields Q(2)\mathbb Q(\sqrt2), Q(3)\mathbb Q(\sqrt3), Q(6)\mathbb Q(\sqrt6), and Q\mathbb Q.

Example
The intermediate fields for x3−2x^3-2

S3S_3 has 66 subgroups, so the splitting field has 66 intermediate fields: Q\mathbb Q, EE, the quadratic field Q(ω)\mathbb Q(\omega) (fixed by A3A_3, normal — and indeed Q(ω)/Q\mathbb Q(\omega)/\mathbb Q is Galois), and three cubic fields Q(23)\mathbb Q(\sqrt[3]2), Q(23ω)\mathbb Q(\sqrt[3]2\omega), Q(23ω2)\mathbb Q(\sqrt[3]2\omega^2) fixed by the three subgroups of order 22, which are not normal.

Side note
Where it goes wrong

The correspondence reverses inclusion: the WHOLE group corresponds to the SMALLEST field FF, and the trivial subgroup to the largest field EE. Mixing this up gives degrees upside down; use [E:K]=∣H∣[E:K]=|H| to keep track.

Try it
Try it: count intermediate fields

The Galois group of Q(ζ7)/Q\mathbb Q(\zeta_7)/\mathbb Q is cyclic of order 66. How many intermediate fields are there, counting Q\mathbb Q and Q(ζ7)\mathbb Q(\zeta_7)?

Show the answer

A cyclic group of order 66 has one subgroup for each divisor of 66: 1,2,3,61,2,3,6 — so 4\mathbf4 intermediate fields.

Solvability by Radicals

Concept
Galois's criterion

A polynomial over Q\mathbb Q can be solved by radicals (its roots expressed using +,−,×,÷+,-,\times,\div and nn-th roots) exactly when its Galois group is a solvable group. S2,S3,S4S_2,S_3,S_4 are solvable, which is why there are formulas for degrees 2,3,42,3,4; S5S_5 is not, because A5A_5 is simple.

Adjoining an nn-th root produces an extension whose Galois group is abelian (given the roots of unity), so a radical formula builds the splitting field through a chain of abelian steps — and the Galois group must break down into abelian pieces, which is what "solvable" means. An irreducible quintic whose Galois group is S5S_5 therefore has no formula in radicals. The Sylow and simple-group results of the first half are exactly what proves S5S_5 is not solvable.

Example
A quintic with no radical formula

f=x5−6x+3f=x^5-6x+3 is irreducible (Eisenstein, p=3p=3) and has exactly 33 real roots, so complex conjugation swaps the other two: the Galois group contains a transposition. It also contains a 55-cycle (since 55 divides the degree of the splitting field). A transposition and a 55-cycle generate S5S_5, so the Galois group is S5S_5, of order 120120, and the roots cannot be written with radicals.

Example
A solvable quintic

x5−2x^5-2 is solved by radicals: its roots are 25 ζ5k\sqrt[5]2\,\zeta_5^k. Its Galois group has order 2020 and is solvable, consistent with the criterion.

Side note
Where it goes wrong

The theorem does not say quintics are never solvable — x5−2x^5-2 and x5−1x^5-1 are. It says there is no single formula for ALL quintics, and that particular ones, like x5−6x+3x^5-6x+3, cannot be solved by radicals at all.

Try it
Try it: why degree 44 is fine

Why does every polynomial of degree at most 44 have a solution in radicals?

Show the answer

Its Galois group is a subgroup of S4S_4 (or smaller), and S4S_4 is solvable (S4⊃A4⊃V4⊃{e}S_4\supset A_4\supset V_4\supset\{e\} with abelian quotients); subgroups of solvable groups are solvable. So the criterion is met.

Why It Works

If f=∑akxkf=\sum a_kx^k with ak∈Fa_k\in F and f(α)=0f(\alpha)=0, then 0=σ(0)=σ(∑akαk)=∑σ(ak)σ(α)k=∑akσ(α)k=f(σ(α))0=\sigma(0)=\sigma\big(\sum a_k\alpha^k\big)=\sum\sigma(a_k)\sigma(\alpha)^k=\sum a_k\sigma(\alpha)^k=f(\sigma(\alpha)), using that σ\sigma fixes each aka_k.

If E=F(α)E=F(\alpha) and mm is the minimal polynomial of α\alpha, an FF-automorphism is determined by σ(α)\sigma(\alpha), which must be a root of mm in EE. There are at most deg⁡m=[E:F]\deg m=[E:F] such roots, with equality when mm splits into distinct linear factors in EE.

If H≤GH\le G has fixed field KK, then E/KE/K is Galois with group HH, so [E:K]=∣H∣[E:K]=|H|. The tower law gives [K:F]=[E:F][E:K]=∣G∣∣H∣=[G:H][K:F]=\frac{[E:F]}{[E:K]}=\frac{|G|}{|H|}=[G:H] — Lagrange's theorem and the tower law are two views of the same count.