Abstract Algebra — 05 · Field Extensions and Galois Theory

Post 04 ended by building new fields: $\mathbb{C} = \mathbb{R}[x]/(x^2+1)$ and $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2+x+1)$. This post studies such constructions systematically. A bigger field $E$ containing $F$ is a vector space over $F$, and its dimension, the degree, turns out to be a very strong invariant: a single divisibility argument proves that no ruler-and-compass construction doubles a cube or trisects a 60° angle. Galois theory then attaches to a polynomial a finite group, the permutations of its roots that respect all algebraic relations, and the Fundamental Theorem matches the subfields of the splitting field with the subgroups of that group. Everything from posts 01–04 comes together here: groups, normal subgroups and quotients, group actions, Sylow-type counting, irreducible polynomials and quotient rings. The chapter ends with the reason there is no quintic formula.

Series map: 00 Integers, Relations & Modular Arithmetic → 01 Groups & Subgroups → 02 Cosets, Lagrange & Quotient Groups → 03 Homomorphisms, Group Actions & Sylow → 04 Rings, Ideals & Fields → 05 Field Extensions & Galois Theory (this post).

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Learning Outcome

• Compute the degree of a field extension, and prove and use the tower law.
• Find minimal polynomials, and prove that $F(\alpha) \cong F[x]/(m_\alpha)$ with degree $\deg m_\alpha$.
• Show that $\mathbb{Q}(\sqrt2, \sqrt3)$ has degree 4 and is generated by $\sqrt2 + \sqrt3$.
• Prove that doubling the cube and trisecting a 60° angle are impossible with ruler and compass.
• Build splitting fields, and prove that there is one finite field of each prime-power order, up to isomorphism.
• Define separable, normal and Galois extensions and the Galois group.
• State the Fundamental Theorem of Galois Theory and draw the subfield/subgroup lattices for $\mathbb{Q}(\sqrt2, \sqrt3)$ and for the splitting field of $x^3 - 2$.
• Define solvable groups and outline why the roots of $x^5 - 4x + 2$ cannot be written with radicals.

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Chapter 1 — Field extensions and degree

Terminology

• Extension — a pair of fields $F \subseteq E$, written $E/F$ ("$E$ over $F$"). $F$ is the base field.
• Degree — $[E : F] = \dim_F E$, the dimension of $E$ as a vector space over $F$. The extension is finite if the degree is finite.
• Generated extension — for $\alpha \in E$, $F(\alpha)$ is the smallest subfield of $E$ containing $F$ and $\alpha$. Likewise $F(\alpha_1, \dots, \alpha_n)$. An extension $F(\alpha)/F$ is simple and $\alpha$ is a primitive element.
• Intermediate field — a field $K$ with $F \subseteq K \subseteq E$.

Examples: $[\mathbb{C} : \mathbb{R}] = 2$ with basis $1, i$. $[\mathbb{Q}(\sqrt2) : \mathbb{Q}] = 2$ with basis $1, \sqrt2$, since $\mathbb{Q}(\sqrt2) = \{a + b\sqrt2 : a, b \in \mathbb{Q}\}$ and $\sqrt2 \notin \mathbb{Q}$. $[\mathbb{R} : \mathbb{Q}]$ is infinite (a countable field cannot span the uncountable $\mathbb{R}$ with finitely many vectors). A field with $q$ elements that contains $\mathbb{F}_p$ has $q = p^n$ where $n = [E : \mathbb{F}_p]$, because an $n$-dimensional vector space over $\mathbb{F}_p$ has $p^n$ elements.

The tower law

Theorem (tower law). If $F \subseteq K \subseteq E$ are fields, then

$$[E : F] = [E : K]\,[K : F],$$

in the sense that the left side is finite exactly when both factors are, and then equality holds.

Proof (finite case). Let $\alpha_1, \dots, \alpha_m$ be a basis of $K$ over $F$ and $\beta_1, \dots, \beta_n$ a basis of $E$ over $K$. The claim is that the $mn$ products $\alpha_i\beta_j$ form a basis of $E$ over $F$.

• Spanning. Any $e \in E$ is $\sum_j k_j \beta_j$ with $k_j \in K$, and each $k_j = \sum_i f_{ij}\alpha_i$ with $f_{ij} \in F$. So $e = \sum_{i,j} f_{ij}\,\alpha_i\beta_j$.
• Independence. If $\sum_{i,j} f_{ij}\alpha_i\beta_j = 0$, regroup as $\sum_j \big(\sum_i f_{ij}\alpha_i\big)\beta_j = 0$. The inner sums are in $K$ and the $\beta_j$ are independent over $K$, so each inner sum is $0$. The $\alpha_i$ are independent over $F$, so every $f_{ij} = 0$. ∎

Corollaries. (1) If $K$ is intermediate in a finite extension $E/F$, then $[K : F]$ divides $[E : F]$. (2) If $[E : F]$ is prime, there is no intermediate field other than $F$ and $E$. (3) If $\alpha \in E$, then $[F(\alpha) : F]$ divides $[E : F]$.

Trick: most degree computations are "bound above by a tower, bound below by divisibility". If $[K : F] = 2$ and $[L : F] = 3$ are both inside $E$, then $6 \mid [E : F]$.

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Chapter 2 — Algebraic elements and minimal polynomials

Terminology

• Algebraic — $\alpha \in E$ is algebraic over $F$ if $f(\alpha) = 0$ for some nonzero $f \in F[x]$. Otherwise $\alpha$ is transcendental over $F$.
• Algebraic extension — every element of $E$ is algebraic over $F$.
• Minimal polynomial — the monic polynomial $m_\alpha \in F[x]$ of least degree with $m_\alpha(\alpha) = 0$.

$\sqrt2$, $i$, $\sqrt[3]{2}$ and $e^{2\pi i/5}$ are algebraic over $\mathbb{Q}$. Hermite proved in 1873 that $e$ is transcendental and Lindemann in 1882 that $\pi$ is transcendental; both proofs are analytic and are not needed below.

The minimal polynomial

Theorem. Let $\alpha$ be algebraic over $F$. Then:

1. $m_\alpha$ exists, is unique, and is irreducible over $F$;
2. for $f \in F[x]$: $f(\alpha) = 0$ if and only if $m_\alpha \mid f$;
3. $F(\alpha) \cong F[x]/(m_\alpha)$, so $[F(\alpha) : F] = \deg m_\alpha$, with basis $1, \alpha, \dots, \alpha^{n-1}$ where $n = \deg m_\alpha$.

Proof. Evaluation $\mathrm{ev}_\alpha : F[x] \to E$, $f \mapsto f(\alpha)$, is a ring homomorphism. Its kernel $I$ is nonzero because $\alpha$ is algebraic, and $F[x]$ is a PID (post 04), so $I = (m)$ for a unique monic $m$; this $m$ has least degree among nonzero elements of $I$, so $m = m_\alpha$, and statement 2 is "$f \in I \iff m_\alpha \mid f$". The image of $\mathrm{ev}_\alpha$ is a subring of the field $E$, hence a domain, so by the First Isomorphism Theorem $F[x]/(m_\alpha)$ is a domain: $(m_\alpha)$ is prime, so $m_\alpha$ is irreducible. Then $(m_\alpha)$ is maximal (post 04, Chapter 6), so the image $F[\alpha]$ is a field containing $F$ and $\alpha$; it is therefore $F(\alpha)$. The basis claim is the division algorithm: every class has a unique representative of degree $< n$. ∎

So to find $[F(\alpha) : F]$: find a polynomial with root $\alpha$ and prove it irreducible with the tests of post 04.

Common pitfall: the minimal polynomial depends on the base field. $m_{\sqrt2}$ over $\mathbb{Q}$ is $x^2 - 2$; over $\mathbb{R}$ it is $x - \sqrt2$.

Arithmetic in $\mathbb{Q}(\sqrt[3]{2})$. Let $a = \sqrt[3]{2}$, with $m_a = x^3 - 2$ (Eisenstein at 2), so $[\mathbb{Q}(a) : \mathbb{Q}] = 3$ with basis $1, a, a^2$. To invert $1 + a$, use $(1 + a)(1 - a + a^2) = 1 + a^3 = 3$:

$$\frac{1}{1 + \sqrt[3]{2}} = \frac{1 - \sqrt[3]{2} + \sqrt[3]{4}}{3}.$$

Finite implies algebraic. If $[E : F] = n$ and $\alpha \in E$, the $n + 1$ elements $1, \alpha, \dots, \alpha^n$ are linearly dependent over $F$, and a dependence relation is a nonzero polynomial with root $\alpha$. Consequently, if $\alpha$ and $\beta$ are algebraic then $F(\alpha, \beta)$ is finite (tower law), so $\alpha \pm \beta$, $\alpha\beta$ and $\alpha/\beta$ are algebraic too: the algebraic numbers form a field.

Worked example: Q(√2, √3)

Degree. $[\mathbb{Q}(\sqrt2) : \mathbb{Q}] = 2$. Next, $\sqrt3 \notin \mathbb{Q}(\sqrt2)$: if $(a + b\sqrt2)^2 = 3$ with $a, b \in \mathbb{Q}$, then $a^2 + 2b^2 = 3$ and $2ab = 0$. $b = 0$ gives $a^2 = 3$ and $a = 0$ gives $b^2 = 3/2$, neither possible in $\mathbb{Q}$. So $x^2 - 3$ stays irreducible over $\mathbb{Q}(\sqrt2)$, and by the tower law

$$[\mathbb{Q}(\sqrt2, \sqrt3) : \mathbb{Q}] = [\mathbb{Q}(\sqrt2, \sqrt3) : \mathbb{Q}(\sqrt2)]\,[\mathbb{Q}(\sqrt2) : \mathbb{Q}] = 2 \cdot 2 = 4,$$

with basis $1, \sqrt2, \sqrt3, \sqrt6$ (products of the two bases, as in the proof of the tower law).

A primitive element. Let $\theta = \sqrt2 + \sqrt3$. Then $\theta^2 = 5 + 2\sqrt6$, so $(\theta^2 - 5)^2 = 24$, that is

$$\theta^4 - 10\theta^2 + 1 = 0.$$

Also $\theta^3 = 11\sqrt2 + 9\sqrt3$, which gives $\sqrt2 = (\theta^3 - 9\theta)/2$ and $\sqrt3 = (11\theta - \theta^3)/2$. So $\mathbb{Q}(\theta)$ contains $\sqrt2$ and $\sqrt3$, hence $\mathbb{Q}(\theta) = \mathbb{Q}(\sqrt2, \sqrt3)$ has degree 4. A degree-4 monic polynomial with root $\theta$ must then be the minimal polynomial: $m_\theta = x^4 - 10x^2 + 1$, irreducible over $\mathbb{Q}$ (its four roots are $\pm\sqrt2 \pm \sqrt3$).

This is an instance of the primitive element theorem: every finite separable extension (Chapter 5; in particular every finite extension of $\mathbb{Q}$) is simple.

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Chapter 3 — Ruler and compass

Start with the points $0$ and $1$ in the plane $\mathbb{C}$. A ruler draws the line through two constructed points; a compass draws a circle with a constructed centre through a constructed point. New points are intersections of lines and circles. A real number is constructible if it is a coordinate of a constructible point.

Key observation. If all points so far have coordinates in a field $K$, then:

• the intersection of two lines solves a linear system, so it has coordinates in $K$;
• an intersection involving a circle solves a quadratic equation over $K$, so it has coordinates in $K$ or in some $K(\sqrt d)$ with $d \in K$, an extension of degree 2.

Theorem. If $\alpha$ is constructible, then $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ is a power of 2.

Proof. By the observation, $\alpha$ lies in the top field of a tower $\mathbb{Q} = K_0 \subset K_1 \subset \cdots \subset K_r$ with each $[K_{i+1} : K_i] = 2$. By the tower law $[K_r : \mathbb{Q}] = 2^r$, and $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ divides it. ∎

Doubling the cube. A cube of volume 2 needs side $\sqrt[3]{2}$, whose minimal polynomial $x^3 - 2$ has degree 3. Not a power of 2, so impossible.

Trisecting 60°. A 60° angle is constructible (equilateral triangle). Trisecting it means constructing $c = \cos 20^\circ$. From $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ with $\theta = 20^\circ$ and $\cos 60^\circ = \tfrac12$:

$$4c^3 - 3c = \tfrac12 \quad\Longleftrightarrow\quad 8c^3 - 6c - 1 = 0.$$

By the rational root test, a rational root would be one of $\pm1, \pm\tfrac12, \pm\tfrac14, \pm\tfrac18$; substituting gives $1, -3, -3, 1, -\tfrac{19}{8}, \tfrac38, -\tfrac{111}{64}, -\tfrac{17}{64}$, never $0$. A cubic without rational roots is irreducible, so $[\mathbb{Q}(c) : \mathbb{Q}] = 3$ and the trisection is impossible. (Some angles, such as 90°, can be trisected; the theorem is that no method works for every angle.)

Squaring the circle. A square of area $\pi$ needs side $\sqrt\pi$. If $\sqrt\pi$ were constructible it would be algebraic, so $\pi$ would be algebraic, contradicting Lindemann.

Watch out: "degree a power of 2" is necessary, not sufficient. The exact criterion is that the splitting field of the minimal polynomial has degree a power of 2. The Gauss–Wantzel theorem answers the polygon question: a regular $n$-gon is constructible if and only if $n$ is a power of 2 times a product of distinct Fermat primes $3, 5, 17, 257, 65537$.

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Chapter 4 — Splitting fields and finite fields

Splitting fields

A splitting field of a nonconstant $f \in F[x]$ is an extension $E/F$ in which $f$ factors into linear factors, $f = c(x - \alpha_1)\cdots(x - \alpha_n)$, and which is generated by the roots: $E = F(\alpha_1, \dots, \alpha_n)$.

Theorem. Every nonconstant $f \in F[x]$ has a splitting field $E$, with $[E : F] \le n!$ where $n = \deg f$. Any two splitting fields of $f$ over $F$ are isomorphic by an isomorphism that fixes $F$.

(Existence: take an irreducible factor $p$ of $f$; in $F[x]/(p)$ it has a root (post 04), so $f$ gains a linear factor over an extension of degree $\le n$; repeat on the quotient, whose degree is $n - 1$. Uniqueness is proved by the same induction, extending an isomorphism one root at a time.)

Examples:

• $x^2 - 2$ over $\mathbb{Q}$: splitting field $\mathbb{Q}(\sqrt2)$, degree 2.
• $x^4 - 10x^2 + 1$: roots $\pm\sqrt2 \pm \sqrt3$, splitting field $\mathbb{Q}(\sqrt2, \sqrt3)$, degree 4.
• $x^3 - 2$: roots $\alpha$, $\alpha\omega$, $\alpha\omega^2$ with $\alpha = \sqrt[3]{2}$ and $\omega = e^{2\pi i/3}$, a root of $x^2 + x + 1$. The splitting field is $E = \mathbb{Q}(\alpha, \omega)$ (it contains $\omega = \alpha\omega/\alpha$). $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 3$, and $\omega \notin \mathbb{Q}(\alpha) \subset \mathbb{R}$, so $[E : \mathbb{Q}(\alpha)] = 2$. Hence $[E : \mathbb{Q}] = 6 = 3!$, the maximum.

Finite fields

Let $E$ be a finite field. Its characteristic is a prime $p$ (post 04), so it contains $\mathbb{F}_p$ and has $p^n$ elements, $n = [E : \mathbb{F}_p]$.

Lemma (freshman's dream). In a commutative ring of characteristic $p$, $(a + b)^p = a^p + b^p$.

Proof. Expand with the binomial theorem. For $0 < k < p$, $\binom{p}{k} = \frac{p!}{k!(p-k)!}$ is divisible by $p$: $p$ divides the numerator but not the denominator. So all middle terms vanish. ∎

Hence the Frobenius map $\mathrm{Fr}(a) = a^p$ is a ring homomorphism. On a field it is injective (post 04: homomorphisms out of fields are injective), so on a finite field it is an automorphism, and it fixes exactly the elements of $\mathbb{F}_p$ (the roots of $x^p - x$; there are at most $p$ of them, and all of $\mathbb{F}_p$ qualifies by Fermat's little theorem from post 00).

Theorem (existence and uniqueness of finite fields). For every prime $p$ and $n \ge 1$ there is a field with $q = p^n$ elements, and any two such fields are isomorphic. It is written $\mathbb{F}_q$.

Proof. Existence. Let $E$ be a splitting field of $g = x^q - x$ over $\mathbb{F}_p$, and $S$ the set of roots of $g$ in $E$. The derivative $g' = qx^{q-1} - 1 = -1$ has no common root with $g$, so the roots are distinct and $|S| = q$. $S$ is a subfield: if $a^q = a$ and $b^q = b$, then $(a + b)^q = a^q + b^q = a + b$ (apply the lemma $n$ times), $(ab)^q = ab$, and $(a^{-1})^q = a^{-1}$. A subfield containing all roots is the whole splitting field, so $E = S$ has $q$ elements.

Uniqueness. In any field $K$ with $q$ elements, the multiplicative group has order $q - 1$, so $a^{q-1} = 1$ for $a \ne 0$ by Lagrange, and $a^q = a$ for all $a$. So $K$ consists of the $q$ roots of $x^q - x$, and is a splitting field of $x^q - x$ over $\mathbb{F}_p$. Splitting fields are unique up to isomorphism. ∎

Two further facts, stated without proof: the multiplicative group $\mathbb{F}_q^\times$ is cyclic (post 04's $\mathbb{F}_8$ table showed a generator), and $\mathbb{F}_{p^d}$ sits inside $\mathbb{F}_{p^n}$ exactly when $d \mid n$.

Worked example. Over $\mathbb{F}_2$:

$$\begin{aligned} x^4 - x &= x\,(x + 1)\,(x^2 + x + 1), \\ x^8 - x &= x\,(x + 1)\,(x^3 + x + 1)\,(x^3 + x^2 + 1). \end{aligned}$$

The first line lists the minimal polynomials of the 4 elements of $\mathbb{F}_4$; the second those of the 8 elements of $\mathbb{F}_8$. The two cubics are the two irreducible cubics over $\mathbb{F}_2$, each with three roots in $\mathbb{F}_8$. No quadratic appears in the second line, matching $2 \nmid 3$: $\mathbb{F}_4 \not\subseteq \mathbb{F}_8$.

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Chapter 5 — Galois extensions and the Galois group

Terminology

• $F$-automorphism of $E$ — a field automorphism $\sigma : E \to E$ with $\sigma(a) = a$ for every $a \in F$. They form a group $\operatorname{Aut}(E/F)$.
• Fixed field — for a subgroup $H \le \operatorname{Aut}(E)$, $E^H = \{ e \in E : \sigma(e) = e \text{ for all } \sigma \in H \}$.
• Separable polynomial — one with no repeated root in its splitting field. An algebraic extension is separable if every minimal polynomial is separable.
• Normal extension — an algebraic $E/F$ such that every irreducible polynomial in $F[x]$ with one root in $E$ splits completely in $E$.
• Galois extension — a finite extension that is normal and separable. Its Galois group is $\operatorname{Gal}(E/F) = \operatorname{Aut}(E/F)$.
• Galois group of a polynomial — $\operatorname{Gal}(E/F)$ for $E$ a splitting field of the (separable) polynomial.

Separability is automatic in the cases you meet most. Over a field of characteristic 0, and over any finite field, every irreducible polynomial is separable. (An irreducible $f$ has a repeated root only if $\gcd(f, f') \ne 1$, which forces $f' = 0$; in characteristic 0 that is impossible for nonconstant $f$.) The standard inseparable example is $x^p - t$ over the field $\mathbb{F}_p(t)$ of rational functions: it is irreducible (Eisenstein at the prime $t$), but over its splitting field it equals $(x - t^{1/p})^p$.

Automorphisms permute roots

Lemma. If $f \in F[x]$, $\alpha \in E$ is a root of $f$, and $\sigma \in \operatorname{Aut}(E/F)$, then $\sigma(\alpha)$ is a root of $f$.

Proof. If $f = \sum c_k x^k$ with $c_k \in F$, then $f(\sigma(\alpha)) = \sum c_k \sigma(\alpha)^k = \sigma\big(\sum c_k\alpha^k\big) = \sigma(0) = 0$, since $\sigma$ fixes each $c_k$. ∎

So if $E = F(\alpha_1, \dots, \alpha_n)$ is a splitting field of $f$, an $F$-automorphism is determined by how it permutes $\alpha_1, \dots, \alpha_n$, and $\operatorname{Aut}(E/F)$ is (isomorphic to) a subgroup of $S_n$, an action on the roots in the sense of post 03.

Theorem (characterisations of Galois extensions). For a finite extension $E/F$ the following are equivalent:

1. $E/F$ is Galois (normal and separable);
2. $E$ is the splitting field over $F$ of a separable polynomial;
3. $\lvert\operatorname{Aut}(E/F)\rvert = [E : F]$;
4. the fixed field of $\operatorname{Aut}(E/F)$ is exactly $F$.

In general $\lvert\operatorname{Aut}(E/F)\rvert \le [E : F]$ and equality is what "Galois" means in practice.

Examples.

• $\operatorname{Gal}(\mathbb{C}/\mathbb{R}) = \{\mathrm{id}, \text{complex conjugation}\} \cong \mathbb{Z}/2\mathbb{Z}$.
• $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not Galois. An automorphism must send $\sqrt[3]{2}$ to a root of $x^3 - 2$ inside $\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{R}$, and the only real root is $\sqrt[3]{2}$ itself. So $\operatorname{Aut}$ is trivial, of order $1 < 3$. It fails normality: $x^3 - 2$ has one root in the field but does not split there.
• $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)$ is cyclic of order $n$, generated by Frobenius. ($\mathrm{Fr}^k$ fixes exactly the roots of $x^{p^k} - x$, so $\mathrm{Fr}$ has order exactly $n$; and the group has order $[\mathbb{F}_{p^n} : \mathbb{F}_p] = n$ because $\mathbb{F}_{p^n}$ is the splitting field of the separable $x^{p^n} - x$.)

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Chapter 6 — The Fundamental Theorem of Galois Theory

Theorem (Fundamental Theorem of Galois Theory). Let $E/F$ be a finite Galois extension with group $G = \operatorname{Gal}(E/F)$. Then the maps

$$H \longmapsto E^H \qquad\text{and}\qquad K \longmapsto \operatorname{Gal}(E/K)$$

are mutually inverse bijections between subgroups $H \le G$ and intermediate fields $F \subseteq K \subseteq E$, and:

1. they reverse inclusion: $H_1 \le H_2$ if and only if $E^{H_1} \supseteq E^{H_2}$;
2. degrees match indices: $[E : E^H] = |H|$ and $[E^H : F] = [G : H]$;
3. $K/F$ is Galois if and only if $\operatorname{Gal}(E/K)$ is a normal subgroup of $G$, and then $\operatorname{Gal}(K/F) \cong G / \operatorname{Gal}(E/K)$.

So the lattice of intermediate fields is the lattice of subgroups turned upside down: the big field $E$ matches the trivial group, the base $F$ matches all of $G$. Part 3 is where normal subgroups and quotient groups (post 02) acquire their original meaning.

Worked example 1: Q(√2, √3)

$E = \mathbb{Q}(\sqrt2, \sqrt3)$ is the splitting field of the separable $(x^2 - 2)(x^2 - 3)$, so it is Galois of degree 4 and $G$ has 4 elements. Each automorphism sends $\sqrt2 \mapsto \pm\sqrt2$ and $\sqrt3 \mapsto \pm\sqrt3$, and all four sign choices occur. Let

$$\sigma : \sqrt2 \mapsto -\sqrt2,\ \sqrt3 \mapsto \sqrt3, \qquad \tau : \sqrt2 \mapsto \sqrt2,\ \sqrt3 \mapsto -\sqrt3.$$

Each has order 2 and they commute, so $G = \{\mathrm{id}, \sigma, \tau, \sigma\tau\} \cong \mathbb{Z}/2 \times \mathbb{Z}/2$. The subgroups and their fixed fields:

Subgroup	Order	Fixed field	Degree over $\mathbb{Q}$
$\{\mathrm{id}\}$	1	$\mathbb{Q}(\sqrt2, \sqrt3)$	4
$\langle\sigma\rangle$	2	$\mathbb{Q}(\sqrt3)$	2
$\langle\tau\rangle$	2	$\mathbb{Q}(\sqrt2)$	2
$\langle\sigma\tau\rangle$	2	$\mathbb{Q}(\sqrt6)$	2
$G$	4	$\mathbb{Q}$	1

($\sigma\tau(\sqrt6) = (-\sqrt2)(-\sqrt3) = \sqrt6$.) By the theorem these are all the intermediate fields: exactly three quadratic subfields, which is not obvious without Galois theory. $G$ is abelian, so every subgroup is normal and every intermediate field is Galois over $\mathbb{Q}$.

Worked example 2: the splitting field of x³ − 2

$E = \mathbb{Q}(\alpha, \omega)$ with $\alpha = \sqrt[3]{2}$, of degree 6 (Chapter 4). It is Galois, so $|G| = 6$, and $G$ acts on the three roots $\alpha, \alpha\omega, \alpha\omega^2$, so $G$ is a subgroup of $S_3$ of order 6: $G \cong S_3$. Concretely an automorphism is fixed by $\alpha \mapsto \alpha\omega^a$ ($a \in \{0, 1, 2\}$) and $\omega \mapsto \omega$ or $\omega^2$. Let

$$\sigma : \alpha \mapsto \alpha\omega,\ \omega \mapsto \omega \quad(\text{order } 3), \qquad \tau : \alpha \mapsto \alpha,\ \omega \mapsto \omega^2 \quad(\text{complex conjugation, order } 2).$$

Composing right to left (apply $\tau$ first in $\sigma\tau$): $\sigma\tau$ sends $\alpha \mapsto \alpha\omega$ and $\omega \mapsto \omega^2$, so $\sigma\tau(\alpha\omega^2) = \alpha\omega \cdot \omega^4 = \alpha\omega^2$. Similarly $\sigma^2\tau$ fixes $\alpha\omega$. The six subgroups of $S_3$ and their fixed fields:

Subgroup	Order	Normal?	Fixed field	Degree over $\mathbb{Q}$
$\{\mathrm{id}\}$	1	yes	$\mathbb{Q}(\alpha, \omega)$	6
$\langle\tau\rangle$	2	no	$\mathbb{Q}(\alpha)$	3
$\langle\sigma\tau\rangle$	2	no	$\mathbb{Q}(\alpha\omega^2)$	3
$\langle\sigma^2\tau\rangle$	2	no	$\mathbb{Q}(\alpha\omega)$	3
$\langle\sigma\rangle = A_3$	3	yes	$\mathbb{Q}(\omega)$	2
$S_3$	6	yes	$\mathbb{Q}$	1

Each order-2 subgroup fixes one real or complex cube root, so its fixed field contains that root; the degree count $[G : H] = 3 = [\mathbb{Q}(\alpha\omega^k) : \mathbb{Q}]$ makes it equal to $\mathbb{Q}(\alpha\omega^k)$. $\sigma$ fixes $\omega$, and $[G : \langle\sigma\rangle] = 2 = [\mathbb{Q}(\omega) : \mathbb{Q}]$.

Part 3 of the theorem, checked on this example: $A_3 = \langle\sigma\rangle$ is normal, and $\mathbb{Q}(\omega)$ is Galois over $\mathbb{Q}$ (it is the splitting field of $x^2 + x + 1$), with $\operatorname{Gal}(\mathbb{Q}(\omega)/\mathbb{Q}) \cong S_3/A_3 \cong \mathbb{Z}/2$. The three subgroups of order 2 are not normal (they are conjugate, like the three Sylow 2-subgroups of $S_3$ in post 03), and correspondingly $\mathbb{Q}(\sqrt[3]{2})$ and its two conjugate fields are not Galois over $\mathbb{Q}$.

Common pitfall: the correspondence reverses order. A bigger subgroup fixes more conditions and so has a smaller fixed field. Draw the two lattices with $E$ at the top and $\{\mathrm{id}\}$ at the top.

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Chapter 7 — Solvable groups and the quintic

Solvable groups

A group $G$ is solvable if there is a chain of subgroups

$$\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_k = G$$

with each $G_i$ normal in $G_{i+1}$ and each quotient $G_{i+1}/G_i$ abelian.

• Abelian groups are solvable ($k = 1$).
• $S_3$: $\{e\} \trianglelefteq A_3 \trianglelefteq S_3$ with quotients $\mathbb{Z}/3$ and $\mathbb{Z}/2$.
• $S_4$: $\{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$ with quotients $\mathbb{Z}/2 \times \mathbb{Z}/2$, $\mathbb{Z}/3$, $\mathbb{Z}/2$. So every subgroup of $S_4$ is solvable too (subgroups and quotients of solvable groups are solvable).
• $S_5$ is not solvable. Its only normal subgroups are $\{e\}$, $A_5$, $S_5$, and $A_5$ is simple and non-abelian (post 02). A chain would have to pass through $A_5$, and the step below $A_5$ would need a normal subgroup of $A_5$ with abelian quotient; the only candidates are $A_5$ itself (quotient trivial, no progress) and $\{e\}$ (quotient $A_5$, not abelian). Equivalently, the commutator subgroup of $A_5$ is $A_5$, so the derived series $S_5 \supset A_5 \supset A_5 \supset \cdots$ never reaches $\{e\}$.

Radicals

A polynomial $f \in \mathbb{Q}[x]$ is solvable by radicals if its roots can be written starting from rational numbers using $+, -, \times, \div$ and $k$-th roots. The quadratic formula shows this for degree 2; Cardano's and Ferrari's formulas do it for degrees 3 and 4.

Theorem (Galois, characteristic 0). $f$ is solvable by radicals if and only if its Galois group is solvable.

(Idea of "only if": adjoining a $k$-th root to a field that already contains the $k$-th roots of unity gives a Galois extension with cyclic group; adjoining roots of unity gives an abelian group. A tower of such steps gives, through part 3 of the Fundamental Theorem, a chain of normal subgroups with abelian quotients.)

Since $S_4$ is solvable and the Galois group of a degree-$n$ polynomial embeds in $S_n$, every polynomial of degree $\le 4$ is solvable by radicals. The Abel–Ruffini theorem says the general quintic is not: the generic polynomial $x^5 + a_4x^4 + \cdots + a_0$ with independent indeterminate coefficients has Galois group $S_5$ over $\mathbb{Q}(a_0, \dots, a_4)$, which is not solvable, so there is no formula in radicals valid for all quintics. Galois theory gives more: specific quintics with rational coefficients have unsolvable roots.

A concrete unsolvable quintic

Lemma. Let $p$ be prime and $f \in \mathbb{Q}[x]$ irreducible of degree $p$ with exactly two non-real roots. Then $\operatorname{Gal}(f) \cong S_p$.

Proof. View $G = \operatorname{Gal}(f)$ as a subgroup of $S_p$ acting on the $p$ roots. For a root $\alpha$, $[\mathbb{Q}(\alpha) : \mathbb{Q}] = p$ divides $[E : \mathbb{Q}] = |G|$, so by Cauchy's theorem (post 03) $G$ has an element of order $p$; in $S_p$ the elements of order $p$ are the $p$-cycles. Complex conjugation restricts to an element of $G$ that swaps the two non-real roots and fixes the $p - 2$ real ones: a transposition. For prime $p$, a $p$-cycle and a transposition generate all of $S_p$. ∎

Theorem. $f(x) = x^5 - 4x + 2$ is not solvable by radicals over $\mathbb{Q}$.

Proof. Irreducible: Eisenstein at $p = 2$ ($2 \mid -4$, $2 \mid 2$, $4 \nmid 2$, leading coefficient 1).

Exactly three real roots: $f'(x) = 5x^4 - 4$ vanishes only at $x = \pm c$ with $c = (4/5)^{1/4} \approx 0.9457$. So $f$ increases, then decreases on $[-c, c]$, then increases. The local maximum $f(-c) \approx 5.03 > 0$ and the local minimum $f(c) \approx -1.03 < 0$, while $f \to -\infty$ as $x \to -\infty$ and $f \to +\infty$ as $x \to +\infty$. So $f$ crosses zero once on each of the three monotone pieces: three real roots, approximately $-1.5185$, $0.5085$ and $1.2436$, and the other two roots are a complex-conjugate pair.

By the lemma with $p = 5$, $\operatorname{Gal}(f) \cong S_5$, which is not solvable. By Galois's theorem, $f$ is not solvable by radicals. ∎

Watch out: "not solvable by radicals" does not mean the roots are hard to compute. Numerically they are easy (the values above). It means there is no finite expression in $+, -, \times, \div, \sqrt[k]{\ }$ for them.

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Practice Exercises

1. Find $[\mathbb{Q}(\sqrt2, \sqrt[3]{2}) : \mathbb{Q}]$.

Solution

$\mathbb{Q}(\sqrt2)$ has degree 2 and $\mathbb{Q}(\sqrt[3]{2})$ has degree 3, both intermediate, so 2 and 3 divide the degree, and so does 6. On the other hand $[\mathbb{Q}(\sqrt2, \sqrt[3]{2}) : \mathbb{Q}(\sqrt[3]{2})] \le 2$ (root of $x^2 - 2$), so the degree is at most $2 \cdot 3 = 6$. Hence it is exactly 6.

2. Find the minimal polynomial of $\sqrt2 + i$ over $\mathbb{Q}$.

Solution

Let $\theta = \sqrt2 + i$. Then $\theta^2 = 1 + 2\sqrt2\,i$, so $(\theta^2 - 1)^2 = -8$, giving $\theta^4 - 2\theta^2 + 9 = 0$. As in the $\sqrt2 + \sqrt3$ example, $\mathbb{Q}(\theta) = \mathbb{Q}(\sqrt2, i)$, which has degree 4 ($i \notin \mathbb{Q}(\sqrt2) \subset \mathbb{R}$). So $m_\theta = x^4 - 2x^2 + 9$.

3. Show that the regular 7-gon is not constructible, using that $2\cos(2\pi/7)$ is a root of $x^3 + x^2 - 2x - 1$.

Solution

Rational roots could only be $\pm 1$, which give $-1$ and $1$, so the cubic has no rational root and is irreducible. Then $[\mathbb{Q}(2\cos(2\pi/7)) : \mathbb{Q}] = 3$, not a power of 2, so $\cos(2\pi/7)$ is not constructible, and neither is the 7-gon. (Consistent with Gauss–Wantzel: 7 is prime but not a Fermat prime.)

4. List all subfields of $\mathbb{F}_{64}$ and draw their inclusions. What is $\operatorname{Gal}(\mathbb{F}_{64}/\mathbb{F}_2)$?

Solution

$64 = 2^6$ and the divisors of 6 are 1, 2, 3, 6, so the subfields are $\mathbb{F}_2, \mathbb{F}_4, \mathbb{F}_8, \mathbb{F}_{64}$. Inclusions: $\mathbb{F}_2 \subset \mathbb{F}_4 \subset \mathbb{F}_{64}$ and $\mathbb{F}_2 \subset \mathbb{F}_8 \subset \mathbb{F}_{64}$, while $\mathbb{F}_4 \not\subseteq \mathbb{F}_8$. The Galois group is cyclic of order 6, generated by Frobenius; its subgroups (orders 1, 2, 3, 6) match the subfields $\mathbb{F}_{64}, \mathbb{F}_8, \mathbb{F}_4, \mathbb{F}_2$ in that order.

5. In $\mathbb{Q}(\sqrt2, \sqrt3)$, which automorphism fixes $\sqrt6$, and what is the fixed field of $\{\mathrm{id}, \sigma\}$? Why does the Fundamental Theorem guarantee that $\mathbb{Q}(\sqrt2 + \sqrt3)$ is the whole field without computing $\theta^3$?

Solution

$\sigma\tau$ fixes $\sqrt6$ (both signs flip). $\sigma$ fixes $\sqrt3$, so $E^{\langle\sigma\rangle} \supseteq \mathbb{Q}(\sqrt3)$, and both have degree 2, so they are equal. For $\theta = \sqrt2 + \sqrt3$: the non-identity automorphisms send $\theta$ to $-\sqrt2 + \sqrt3$, $\sqrt2 - \sqrt3$, $-\sqrt2 - \sqrt3$, all different from $\theta$. So only the trivial subgroup fixes $\theta$, and $\mathbb{Q}(\theta)$ corresponds to $\{\mathrm{id}\}$, i.e. it is all of $E$.

6. Explain why every polynomial of degree at most 4 over $\mathbb{Q}$ is solvable by radicals, but this argument says nothing about degree 5.

Solution

The Galois group of a degree-$n$ polynomial is a subgroup of $S_n$. For $n \le 4$, $S_n$ is solvable (chains above), and subgroups of solvable groups are solvable, so Galois's theorem applies. For $n = 5$ the group may be a solvable subgroup of $S_5$ (for example $x^5 - 2$ is solvable by radicals, since $\sqrt[5]{2}$ is already a radical), or all of $S_5$, as for $x^5 - 4x + 2$. The degree alone does not decide.

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Cheat-sheet

Concept	Key fact
Degree	$[E : F] = \dim_F E$
Tower law	$[E : F] = [E : K][K : F]$
Minimal polynomial	monic, irreducible, divides every $f$ with $f(\alpha) = 0$
Simple extension	$F(\alpha) \cong F[x]/(m_\alpha)$, degree $\deg m_\alpha$
$\mathbb{Q}(\sqrt2, \sqrt3)$	degree 4, primitive element $\sqrt2 + \sqrt3$, min poly $x^4 - 10x^2 + 1$
Constructible	degree over $\mathbb{Q}$ is a power of 2 (necessary)
Impossible constructions	$\sqrt[3]{2}$ (degree 3), $\cos 20^\circ$ (root of $8x^3 - 6x - 1$), $\sqrt\pi$
Splitting field	exists, unique up to isomorphism, degree at most $n!$
Finite fields	one field of each order $p^n$: the roots of $x^{p^n} - x$
Frobenius	$a \mapsto a^p$; generates $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p) \cong \mathbb{Z}/n$
Galois extension	normal and separable; $\lvert\operatorname{Aut}(E/F)\rvert = [E : F]$
Fundamental Theorem	subgroups ↔ intermediate fields, order-reversing; normal ↔ Galois
$x^3 - 2$	splitting field degree 6, group $S_3$, subfield $\mathbb{Q}(\omega)$ ↔ $A_3$
Solvable group	chain of normal steps with abelian quotients; $S_4$ yes, $S_5$ no
Galois's criterion	solvable by radicals ⇔ solvable Galois group (char 0)
Unsolvable quintic	$x^5 - 4x + 2$: Eisenstein at 2, three real roots, group $S_5$

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What's next

This post completes the core course: from the integers and modular arithmetic (post 00), through groups, actions and Sylow theory (posts 01–03), rings and fields (post 04), to Galois theory. The table below lists the planned extensions. Each builds directly on the material here, so the series can grow without renumbering the existing posts.

Planned future extensions

Post	Topic	What it covers
06	Modules and linear algebra over rings	Modules, free modules, the structure theorem for finitely generated modules over a PID; the finite abelian groups of post 03 and the rational and Jordan canonical forms as two special cases
07	Representation theory of finite groups	Group representations, Maschke's theorem, Schur's lemma, characters and orthogonality, character tables of $S_3$, $S_4$ and $A_5$
08	Algebraic number theory	Rings of integers, Dedekind domains, how unique factorisation of ideals repairs $\mathbb{Z}[\sqrt{-5}]$, the class group, cyclotomic fields
09	Commutative algebra and a first look at algebraic geometry	Noetherian rings, the Hilbert basis theorem, localisation, the Nullstellensatz, varieties and their coordinate rings
10	Algebra in cryptography	Finite-field arithmetic in practice (AES uses $\mathbb{F}_{2^8}$), Diffie–Hellman and discrete logarithms, elliptic-curve groups, lattices and lattice-based cryptography

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References

• D. S. Dummit and R. M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004 — Chapters 13 (field extensions, constructions, splitting fields, finite fields) and 14 (Galois theory, finite fields, Galois groups of polynomials, solvability by radicals).
• M. Artin, Algebra, 2nd edition, Pearson, 2011 — Chapters 15 (fields, ruler and compass) and 16 (Galois theory, quintic equations).
• I. Stewart, Galois Theory, 4th edition, CRC Press, 2015 — a readable, historically informed development, including Abel–Ruffini.
• J. S. Milne, Fields and Galois Theory, course notes (free).
• E. Artin, Galois Theory, Notre Dame Mathematical Lectures 2, 1942 (Dover reprint) — the classic linear-algebra approach to the Fundamental Theorem.
• MIT OpenCourseWare, 18.702 Algebra II (Spring 2011) — lecture notes on fields and Galois theory.
• K. Conrad, Expository papers ("blurbs") — short notes on Galois theory, including Galois groups of cubics and quartics.

Note: this post is in progress. Planned additions: a full proof of the Fundamental Theorem (Artin's lemma on fixed fields), cyclotomic extensions and the constructible 17-gon, the discriminant as a test for $\operatorname{Gal} \subseteq A_n$, and a worked Cardano solution of a cubic read through its Galois group.

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