Contents / Abstract Algebra / Field Extensions and Galois Theory

Chapter 8

Field Extensions and Galois Theory

Field extensions and the tower law proved via the product basis; algebraic elements, minimal polynomials and splitting fields; the Galois group with |Gal| = [K:F] exactly for normal separable extensions; the fundamental theorem as an inclusion-reversing correspondence, worked completely for Q(sqrt2,sqrt3) and for the splitting field of x^3-2; finite and cyclotomic extensions; ruler-and-compass impossibility by degree; and solvability by radicals with Abel-Ruffini.

Introduction

Every earlier chapter of this subject built one algebraic object at a time: a group in Groups and their structure, a ring in Rings, a quotient in Ideals and quotient rings, a field in Integral domains and fields. This chapter is where those objects start talking to each other. Galois theory is a dictionary. On one side of it sit the fields sandwiched between a base field $F$ and a larger field $K$ — a collection that is hard to survey directly, because a field is an infinite object and there is no obvious way to list all the subfields of one. On the other side sit the subgroups of a single finite group. The dictionary says these two collections are the same collection, read in opposite directions, and it converts a question nobody can answer by inspection into a question a first course in group theory can answer completely.

The payoff is not abstract. Three questions that resisted mathematics for two thousand years fall out of it. Can a cube of volume $2$ be constructed with ruler and compass? Can an arbitrary angle be trisected? Is there a formula, built from the coefficients using only arithmetic and $n$ th roots, that solves every quintic? The answers are no, no and no, and in each case the proof is the same two-step move: attach a number — a degree, or a group — to the construction being demanded, show that the number must have a certain shape if the construction exists, and then compute that the number does not have that shape. Nothing about the impossibility is visible in the geometry or in the polynomial. It is visible in the arithmetic of degrees and in the subgroup structure of $S_{5}$ .

The chapter runs in five movements. The first is linear algebra in disguise: a field containing $F$ is a vector space over $F$ , its dimension is called the degree, and the tower law says degrees multiply up a chain of extensions. That one multiplicative identity is the computational engine of everything that follows, and it is what kills the cube and the trisection before we have said the word "group". The second movement identifies which elements can be reached at finite cost — the algebraic ones — and shows that adjoining one of them is nothing more exotic than forming a quotient ring $F [x] / (m)$ by an irreducible polynomial, exactly as in Ideals and quotient rings. The third builds the symmetry group. The fourth is the Fundamental Theorem, which makes the dictionary precise and, crucially, inclusion-reversing: bigger subgroup, smaller fixed field. The fifth cashes everything in for Abel–Ruffini.

A word about rigour and its limits. Four results in this chapter — the primitive element theorem, Artin's lemma on linear independence of characters, the Fundamental Theorem itself, and Galois's solvability criterion — have proofs that are genuinely longer than a chapter of this size can carry honestly. Where that happens the argument is given as a proof sketch that states at its head exactly which input is being assumed and proves everything downstream of it. What is never done here is to assert a result and move on as if it were obvious. A reader should always be able to see the join.

Throughout, all fields have characteristic $0$ unless a statement says otherwise; this makes every algebraic extension separable and removes a hypothesis that would otherwise clutter every theorem. The one place characteristic $p$ appears — finite fields — is flagged when it arrives.

8.1Field extensions and the tower law

The definition is almost a non-definition: an extension is a field sitting inside a bigger field. What makes the notion useful is the observation, one line long, that the bigger field is a vector space over the smaller one, so the entire apparatus of Linear independence and bases applies to it.

Definition 8.1 (Field extension, degree, finite extension). A field extension $K / F$ is a field $K$ together with a subfield $F \subseteq K$ . We call $F$ the base field and $K$ the top field.

Since $K$ is closed under addition and under multiplication by elements of $F$ , it is a vector space over $F$ . Its dimension is the degree of the extension, written

[K : F] = dim_{F} K .

The extension is finite if $[K : F] < \infty$ , and infinite otherwise.

The vector space structure is worth pausing on, because it is doing more work than it looks. The scalars are the elements of $F$ ; the vectors are the elements of $K$ ; and the vector addition and the scalar multiplication are just the field addition and field multiplication of $K$ , restricted. The axioms of a vector space are then a strict subset of the field axioms of $K$ , so there is nothing to check. What we have gained is a number, $[K : F]$ , attached to a pair of fields, and numbers can be compared, multiplied and divided in ways that fields cannot.

Note that $[K : F] = 1$ exactly when $K = F$ : a one-dimensional $F$ -space containing $1$ is spanned by $1$ , hence equals $F$ .

Example 8.2 (Three degrees computed straight from the definition). Find $[C : R]$ , $[Q (i) : Q]$ and $[R : Q]$ .

Solution.

Every complex number is $a + bi$ with $a, b \in R$ uniquely, so ${1, i}$ spans $C$ over $R$ and is independent (if $a + bi = 0$ with $a, b$ real then $a = b = 0$ ). Hence $[C : R] = 2$ .
The same basis works over $Q$ : $Q (i) = {a + bi : a, b \in Q}$ , which is closed under multiplication because $i^{2} = - 1$ folds back into $Q$ , and closed under inversion because $(a + bi)^{- 1} = (a - bi) / (a^{2} + b^{2})$ . So $[Q (i) : Q] = 2$ .
$[R : Q]$ is infinite. A finite-dimensional $Q$ -space is countable, being the set of finite rational combinations of finitely many vectors, and $R$ is uncountable. So no finite spanning set exists.

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The counting argument in part 3 is worth keeping: cardinality alone rules out finiteness, without mentioning a single transcendental number. It also shows that "infinite degree" is the generic situation and finiteness is the special, useful case.

Now the central computational fact. Given a chain $F \subseteq E \subseteq K$ , there are three degrees in sight, and they are not independent.

Theorem 8.3 (Tower law). Let $F \subseteq E \subseteq K$ be fields. Then

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

in the sense that $K / F$ is finite if and only if both $K / E$ and $E / F$ are finite, and in that case the degrees satisfy the displayed identity. Explicitly, if ${α_{1}, \dots, α_{m}}$ is a basis of $E$ over $F$ and ${β_{1}, \dots, β_{n}}$ is a basis of $K$ over $E$ , then the $mn$ products

{α_{i} β_{j} : 1 \leq i \leq m, 1 \leq j \leq n}

form a basis of $K$ over $F$ .

Proof. Assume first that $m = [E : F]$ and $n = [K : E]$ are finite, with bases as stated. We prove the products span and are independent; the count $mn$ then gives the identity.

Spanning. Let $γ \in K$ . Since the $β_{j}$ span $K$ over $E$ , there are $c_{j} \in E$ with

γ = j = 1 \sum n c_{j} β_{j} .

Each coefficient $c_{j}$ lies in $E$ , and the $α_{i}$ span $E$ over $F$ , so there are $a_{ij} \in F$ with $c_{j} = \sum_{i = 1}^{m} a_{ij} α_{i}$ . Substituting,

γ = j = 1 \sum n (i = 1 \sum m a_{ij} α_{i}) β_{j} = i, j \sum a_{ij} α_{i} β_{j},

an $F$ -linear combination of the products. So they span $K$ over $F$ .

Independence. Suppose $\sum_{i, j} a_{ij} α_{i} β_{j} = 0$ with all $a_{ij} \in F$ . Group the sum by $j$ :

j = 1 \sum n (= c_{j} \in E \sum_{i = 1}^{m} a_{ij} α_{i}) β_{j} = 0.

Each bracket $c_{j}$ lies in $E$ , because $a_{ij} \in F \subseteq E$ and $α_{i} \in E$ . This is an $E$ -linear relation among the $β_{j}$ , which are independent over $E$ , so every $c_{j} = 0$ . But $c_{j} = \sum_{i} a_{ij} α_{i} = 0$ is an $F$ -linear relation among the $α_{i}$ , which are independent over $F$ , so every $a_{ij} = 0$ . Hence the $mn$ products are independent over $F$ .

So the products are a basis and $[K : F] = mn$ , finite.

The converse. Suppose $[K : F] = N < \infty$ . Then $E$ is an $F$ -subspace of $K$ , so $[E : F] \leq N$ is finite. And any set of elements of $K$ independent over $E$ is independent over $F$ — an $F$ -relation is in particular an $E$ -relation, since $F \subseteq E$ — so $[K : E] \leq N$ is finite too. Now the first part applies. Finally, if either $K / E$ or $E / F$ is infinite then $K / F$ cannot be finite, by the same two observations run backwards.∎

The proof is short, but the statement it proves is the one to internalise: a basis of the top over the middle, times a basis of the middle over the bottom, is a basis of the top over the bottom. Concretely, ${1, 2}$ over $Q$ times ${1, 3}$ gives ${1, 2, 3, 6}$ , which is exactly the basis one would write down by hand for $Q (2, 3)$ .

Corollary 8.4 (Intermediate degrees divide). If $F \subseteq E \subseteq K$ with $[K : F]$ finite, then both $[E : F]$ and $[K : E]$ divide $[K : F]$ . In particular, if $[K : F]$ is prime then there is no field strictly between $F$ and $K$ .

Proof. By Theorem Tower law, $[K : F] = [K : E] [E : F]$ with both factors positive integers, so each divides the product. If $[K : F] = p$ is prime, then $[E : F] \in {1, p}$ ; the first case gives $E = F$ and the second forces $[K : E] = 1$ , i.e. $E = K$ . So no strictly intermediate field exists.∎

This is the first genuine obstruction in the chapter and the template for all the later ones. It does not construct anything; it rules things out by arithmetic. A field of degree $5$ over $Q$ simply has no subfield of degree $2$ , no matter how the field was built, because $2 ∤ 5$ .

Definition 8.5 (Adjoining elements; simple extensions). Let $K / F$ be an extension and $α_{1}, \dots, α_{r} \in K$ . Then $F (α_{1}, \dots, α_{r})$ denotes the smallest subfield of $K$ containing $F$ and all the $α_{i}$ — equivalently, the intersection of all such subfields, which is again a field. An extension of the form $F (α)$ , generated by a single element, is called simple, and $α$ is a primitive element for it.

Example 8.6 (A tower computed twice, two different ways). Compute $[Q (2, 3) : Q]$ , and exhibit an explicit $Q$ -basis.

Solution.

Build the tower $Q \subseteq Q (2) \subseteq Q (2, 3)$ .
Bottom step. $x^{2} - 2$ is irreducible over $Q$ (a rational root would have to be $\pm 1$ or $\pm 2$ , and none works), so $[Q (2) : Q] = 2$ with basis ${1, 2}$ .
Top step. We must decide whether $x^{2} - 3$ stays irreducible over $Q (2)$ , i.e. whether $3 \in Q (2)$ . Suppose $3 = a + b 2$ with $a, b \in Q$ . Squaring, $3 = a^{2} + 2 b^{2} + 2 ab 2$ . Since $2 \in / Q$ , the irrational part must vanish: $2 ab = 0$ . If $b = 0$ then $3 = a$ is rational, false. If $a = 0$ then $3 = 2 b^{2}$ , so $b^{2} = 3/2$ ; writing $b = p / q$ in lowest terms gives $2 p^{2} = 3 q^{2}$ , and comparing the exponent of $3$ on each side (odd on the right, even on the left) is a contradiction. So $3 \in / Q (2)$ and the top step has degree $2$ .
Tower law. $[Q (2, 3) : Q] = 2 \cdot 2 = 4$ , and the product basis promised by Theorem Tower law is ${1, 2} \cdot {1, 3} = {1, 2, 3, 6}$ .
Check by the other route. Build instead through $Q (3)$ : by the symmetric argument each step again has degree $2$ , and the product is again $4$ . The tower law guarantees the two routes agree, which is a useful consistency check whenever a degree is computed by a choice of path.

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Intuition. Think of $[K : F]$ as how many independent dials you need to specify an element of $K$ using only numbers from $F$ . A complex number needs two real dials, so $[C : R] = 2$ . An element of $Q (2, 3)$ needs four rational dials — the coefficients on $1, 2, 3, 6$ — so the degree is $4$ .

The tower law is then the statement that dials compose the way coordinates do. If each floor of a building has $m$ rooms and each room has $n$ shelves, there are $mn$ shelves: to name a shelf you name a room and then a shelf within it. The proof is literally that bookkeeping, with "room" replaced by "basis vector of $E$ over $F$ " and "shelf" by "basis vector of $K$ over $E$ ".

Exercises for §8.1 →

8.2Algebraic elements, minimal polynomials and splitting fields

Degree is a crude measure until we can compute it, and we can compute it because of one theorem: adjoining a root of an irreducible polynomial of degree $n$ multiplies the degree by exactly $n$ . That theorem turns every degree computation into a question about irreducibility of polynomials, which the techniques of Polynomial rings — Eisenstein, reduction mod $p$ , the rational root test — already answer.

Definition 8.7 (Algebraic and transcendental elements). Let $K / F$ be an extension and $α \in K$ . Then $α$ is algebraic over $F$ if $f (α) = 0$ for some nonzero $f \in F [x]$ , and transcendental over $F$ otherwise. The extension $K / F$ is algebraic if every element of $K$ is algebraic over $F$ .

The whole notion is relative. The number $π$ is transcendental over $Q$ — a hard theorem of Lindemann, which we shall quote once and never prove — but it is algebraic over $Q (π)$ , being a root of $x - π$ . "Algebraic" always means algebraic over something.

Definition 8.8 (Minimal polynomial). Let $α$ be algebraic over $F$ . The minimal polynomial of $α$ over $F$ , written $m_{α, F} (x)$ or just $m (x)$ , is the monic polynomial in $F [x]$ of least degree having $α$ as a root. Its degree is called the degree of $α$ over $F$ .

Theorem 8.9 (The minimal polynomial generates the vanishing ideal). Let $α \in K$ be algebraic over $F$ and let

I = {f \in F [x] : f (α) = 0} .

Then $I$ is a nonzero ideal of $F [x]$ , and $I = (m)$ where $m = m_{α, F}$ . Consequently:

$m$ is unique;
$m$ is irreducible over $F$ ;
$f (α) = 0$ for $f \in F [x]$ if and only if $m ∣ f$ ;
if $g \in F [x]$ is monic, irreducible and $g (α) = 0$ , then $g = m$ .

Proof. Write $ev_{α} : F [x] \to K$ for the evaluation map $f \mapsto f (α)$ . It is a ring homomorphism, and $I = ker (ev_{α})$ , so $I$ is an ideal; it is nonzero because $α$ is algebraic. By the theorem Polynomial rings over a field are principal ideal domains, $I = (g)$ for some nonzero $g$ , which we may take monic after scaling by a unit of $F$ .

Every nonzero element of $I$ is a multiple of $g$ and so has degree at least $de g g$ ; and $g$ itself lies in $I$ . So $g$ is a monic element of $I$ of least degree, which is the definition of $m$ . Uniqueness follows: two monic generators of the same principal ideal in a domain differ by a unit, and two monic polynomials differing by a constant are equal. This proves (1) and, with $I = (m)$ , also (3).

Irreducibility. Suppose $m = g h$ with $de g g, de g h < de g m$ . Then $0 = m (α) = g (α) h (α)$ in the field $K$ , which has no zero divisors, so $g (α) = 0$ or $h (α) = 0$ . Either way $I$ contains a nonzero polynomial of degree strictly less than $de g m$ , contradicting minimality. This is (2).

Claim (4). If $g$ is monic irreducible with $g (α) = 0$ then $m ∣ g$ by (3); since $g$ is irreducible its only monic divisors are $1$ and $g$ , and $m \neq = 1$ because $m (α) = 0$ while $1 \neq = 0$ . So $g = m$ .∎

Point (4) is the one used constantly in practice: to find a minimal polynomial you do not have to search for the smallest degree. You produce any monic irreducible polynomial killing $α$ , by whatever trick, and it is automatically the minimal one. Eisenstein's criterion is usually the trick.

Theorem 8.10 (Structure of a simple algebraic extension). Let $α \in K$ be algebraic over $F$ with minimal polynomial $m$ of degree $n$ . Then

F (α) = F [α] ≅ F [x] / (m),

the isomorphism being the one induced by $x \mapsto α$ and fixing $F$ . Moreover $[F (α) : F] = n$ , and the power basis

{1, α, α^{2}, \dots, α^{n - 1}}

is an $F$ -basis of $F (α)$ .

Proof. The evaluation map $ev_{α} : F [x] \to K$ has image $F [α]$ , the ring of polynomial expressions in $α$ , and kernel $(m)$ by Theorem The minimal polynomial generates the vanishing ideal. By the First isomorphism theorem for rings,

F [x] / (m) ≅ F [α] .

Since $m$ is irreducible and $F [x]$ is a principal ideal domain, $(m)$ is a maximal ideal, so by the theorem Maximal ideals and quotient fields the quotient $F [x] / (m)$ is a field. Hence $F [α]$ is a field containing $F$ and $α$ . Any field containing $F$ and $α$ contains every polynomial expression in $α$ , hence contains $F [α]$ ; so $F [α]$ is the smallest such field, i.e. $F [α] = F (α)$ .

For the basis, work in $F [x] / (m)$ . Given $f \in F [x]$ , divide with remainder: $f = q m + r$ with $r = 0$ or $de g r < n$ . Then $f \equiv r$ , so the classes of $1, x, \dots, x^{n - 1}$ span the quotient. They are independent: a nontrivial $F$ -relation $\sum_{k < n} c_{k} x^{k} \equiv 0$ means $m$ divides a nonzero polynomial of degree less than $n$ , impossible. Transporting along the isomorphism, ${1, α, \dots, α^{n - 1}}$ is a basis of $F (α)$ , and $[F (α) : F] = n = de g m$ .∎

This theorem is the reason the subject is computable. It says that the exotic-looking field $Q (32)$ is nothing but $Q [x] / (x^{3} - 2)$ — a quotient ring built by the machinery of Ideals and quotient rings — and that arithmetic in it is polynomial arithmetic reduced mod $x^{3} - 2$ . It also says that the extension depends only on the minimal polynomial, not on which root was chosen: $Q (32)$ and $Q (ω 32)$ are abstractly isomorphic, both being $Q [x] / (x^{3} - 2)$ , even though they are different subfields of $C$ . That single observation is the seed of the entire Galois group: an automorphism is exactly a choice of which root to send $α$ to.

Example 8.11 (Degree of $32$ , and arithmetic in $Q (32)$ ). Compute $[Q (32) : Q]$ and express $(1 + 32)^{- 1}$ in the power basis.

Solution.

$x^{3} - 2$ satisfies Eisenstein's criterion at $p = 2$ : the leading coefficient $1$ is not divisible by $2$ , the others ( $0, 0, - 2$ ) are, and $4 ∤ - 2$ . So it is irreducible over $Q$ , and being monic with $32$ as a root it is the minimal polynomial by Theorem The minimal polynomial generates the vanishing ideal(4).
By Theorem Structure of a simple algebraic extension, $[Q (32) : Q] = 3$ with basis ${1, 32, 34}$ .
For the inverse, run the Euclidean algorithm on $x^{3} - 2$ and $x + 1$ . Division gives $x^{3} - 2 = (x + 1) (x^{2} - x + 1) - 3$ , so $3 = (x + 1) (x^{2} - x + 1) - (x^{3} - 2)$ .
Reduce modulo $x^{3} - 2$ and substitute $x = 32$ : $3 = (1 + 32) (34 - 32 + 1)$ , hence

(1 + 32)^{- 1} = \frac{1}{3} (34 - 32 + 1) .

5. Check: this is a genuine

Q

-combination of the power basis, as Theorem Structure of a simple algebraic extension guarantees it must be. The Euclidean algorithm is the general inversion routine here, and it works precisely because

g cd (m, f) = 1

whenever

f (α) \neq = 0

.□

Proposition 8.12 (Finite implies algebraic, but not conversely). Every finite extension is algebraic. The converse fails: there exist algebraic extensions of infinite degree.

Proof. Finite $\Rightarrow$ algebraic. Let $[K : F] = n$ and $α \in K$ . The $n + 1$ elements $1, α, α^{2}, \dots, α^{n}$ lie in an $F$ -space of dimension $n$ , so by the theorem More vectors than the dimension are dependent they satisfy a nontrivial $F$ -linear relation $\sum_{k = 0}^{n} c_{k} α^{k} = 0$ . The polynomial $\sum c_{k} x^{k}$ is nonzero and kills $α$ , so $α$ is algebraic over $F$ of degree at most $n$ .

The converse fails. Let $A \subseteq C$ be the set of all numbers algebraic over $Q$ ; it is a field by Proposition The algebraic elements form a field, and every element of it is algebraic over $Q$ by construction, so $A / Q$ is an algebraic extension. But it is not finite: for each $n$ , $x^{n} - 2$ is irreducible over $Q$ by Eisenstein at $2$ , so $n 2 \in A$ has degree exactly $n$ over $Q$ , and $[A : Q] \geq [Q (n 2) : Q] = n$ for every $n$ by Corollary Intermediate degrees divide. A degree exceeding every integer is infinite.∎

Proposition 8.13 (The algebraic elements form a field). Let $K / F$ be an extension and let $A = {α \in K : α is algebraic over F}$ . Then $A$ is a subfield of $K$ containing $F$ . In particular, sums, products, differences and quotients of algebraic numbers are algebraic.

Proof. Every $a \in F$ is a root of $x - a$ , so $F \subseteq A$ . Let $α, β \in A$ , of degrees $m$ and $n$ over $F$ , with $β \neq = 0$ where needed.

Consider the tower $F \subseteq F (α) \subseteq F (α, β)$ . The bottom step has degree $m$ by Theorem Structure of a simple algebraic extension. For the top step, $β$ is algebraic over $F$ , hence over the larger field $F (α)$ — the same polynomial works — and its minimal polynomial over $F (α)$ divides its minimal polynomial over $F$ by Theorem The minimal polynomial generates the vanishing ideal(3), so $[F (α, β) : F (α)] \leq n$ . By Theorem Tower law,

[F (α, β) : F] \leq mn < \infty.

So $F (α, β)$ is a finite extension of $F$ , and by Proposition Finite implies algebraic, but not conversely every one of its elements is algebraic over $F$ . Since $α + β$ , $α - β$ , $α β$ and $α / β$ all lie in $F (α, β)$ , they all lie in $A$ . Hence $A$ is a subfield.∎

This proof deserves a second look, because it is a model of what the degree function buys. Proving directly that $2 + 35$ is algebraic means producing a polynomial that kills it — possible, but a computation. The argument above never produces one. It traps the number inside a field of finite dimension and lets Proposition Finite implies algebraic, but not conversely do the rest. Existence without construction, bought with a dimension count.

Definition 8.14 (Splitting field). Let $f \in F [x]$ be nonconstant. An extension $K / F$ is a splitting field of $f$ over $F$ if

$f$ factors into linear factors in $K [x]$ , say $f = c (x - α_{1}) \dots (x - α_{n})$ , and
$K = F (α_{1}, \dots, α_{n})$ — no smaller subfield of $K$ containing $F$ does the job.

Theorem 8.15 (Existence of splitting fields). Every nonconstant $f \in F [x]$ has a splitting field $K$ over $F$ , and $[K : F] \leq (de g f)!$ .

Proof. Induct on $n = de g f$ . If $n = 1$ then $f$ is already linear and $K = F$ works, with $[K : F] = 1 = 1!$ .

Let $n > 1$ and let $p$ be an irreducible factor of $f$ in $F [x]$ , of degree $d \leq n$ . Kronecker's construction: set $F_{1} = F [x] / (p)$ . As in the proof of Theorem Structure of a simple algebraic extension, $(p)$ is maximal, so $F_{1}$ is a field; the composite $F ↪ F [x] \to F_{1}$ is a ring map from a field, hence injective, so we may regard $F \subseteq F_{1}$ . Writing $α$ for the class of $x$ , we get $p (α) = 0$ in $F_{1}$ and $[F_{1} : F] = d \leq n$ .

In $F_{1} [x]$ we may now divide: $f = (x - α) g$ with $de g g = n - 1$ . By the inductive hypothesis applied to $g$ over $F_{1}$ , there is a splitting field $K$ of $g$ over $F_{1}$ with $[K : F_{1}] \leq (n - 1)!$ . Then $f$ splits completely in $K$ , and $K$ is generated over $F$ by $α$ together with the roots of $g$ — that is, by all the roots of $f$ — so $K$ is a splitting field of $f$ over $F$ . By Theorem Tower law,

[K : F] = [K : F_{1}] [F_{1} : F] \leq (n - 1)! \cdot n = n! .

∎

Remark. Splitting fields are also unique up to an isomorphism fixing $F$ . The proof is the same induction run with an isomorphism carried along: an isomorphism $F \to F^{'}$ carrying $f$ to $f^{'}$ extends to the simple extensions by sending a root of an irreducible factor to a root of its image, because both simple extensions are quotients $F [x] / (p)$ and $F^{'} [x] / (p^{'})$ of corresponding rings. We use uniqueness below — it is what lets us say "the splitting field" — but do not repeat the induction.

The bound $n!$ is attained, and understanding when it is not attained is most of the content of a Galois group computation. For $x^{2} - 2$ the bound gives $2$ and the truth is $2$ . For $(x^{2} - 2) (x^{2} - 3)$ , of degree $4$ , the bound gives $24$ and the truth is $4$ : the roots are heavily constrained by each other. For $x^{3} - 2$ the bound gives $6$ and the truth is $6$ , as the next example shows.

Example 8.16 (The splitting field of $x^{3} - 2$ over $Q$ ). Identify the splitting field $K$ of $x^{3} - 2$ over $Q$ and compute $[K : Q]$ .

Solution.

Over $C$ the roots are $32$ , $ω 32$ and $ω^{2} 32$ , where $ω = e^{2 πi /3}$ is a primitive cube root of unity. So $K = Q (32, ω 32, ω^{2} 32)$ , which equals $Q (32, ω)$ , since $ω = (ω 32) / 32$ is recoverable from two of the roots.
Bottom step. $[Q (32) : Q] = 3$ , by Eisenstein at $2$ as in the earlier example.
Top step. The minimal polynomial of $ω$ over $Q$ is $x^{2} + x + 1$ (it is the quotient $(x^{3} - 1) / (x - 1)$ , is monic, has $ω$ as a root, and is irreducible since its roots are not real). Over $Q (32)$ it either stays irreducible or splits. But $Q (32) \subseteq R$ and $ω \in / R$ , so $ω \in / Q (32)$ and the polynomial stays irreducible: $[K : Q (32)] = 2$ .
Tower law. $[K : Q] = 3 \cdot 2 = 6$ .
Cross-check with the other route. Going up through $Q (ω)$ instead: $[Q (ω) : Q] = 2$ , and $x^{3} - 2$ remains irreducible over $Q (ω)$ because a root would generate a subfield of degree $3$ inside a field of degree $2$ , impossible by Corollary Intermediate degrees divide. So the degree is $2 \cdot 3 = 6$ again. Note that $6 = 3!$ , the maximum allowed by Theorem Existence of splitting fields.

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Figure 8.17. The tower of subfields inside the splitting field of

x^{3} - 2

. Degrees are marked on the edges and multiply along any upward path:

3 \times 2 = 2 \times 3 = 6

Example 8.18 (The minimal polynomial of $2 + 3$ , and a primitive element). Find the minimal polynomial of $α = 2 + 3$ over $Q$ , and deduce that $Q (α) = Q (2, 3)$ .

Solution.

Square: $α^{2} = 5 + 26$ , so $α^{2} - 5 = 26$ . Square again: $α^{4} - 10 α^{2} + 25 = 24$ , i.e. $α^{4} - 10 α^{2} + 1 = 0$ .
So $α$ is a root of $f (x) = x^{4} - 10 x^{2} + 1$ , and $de g m_{α, Q} \leq 4$ .
On the other hand $α \in Q (2, 3)$ , so $Q (α)$ is an intermediate field of an extension of degree $4$ , and $[Q (α) : Q] \in {1, 2, 4}$ by Corollary Intermediate degrees divide.
Rule out the small cases. $α \in / Q$ since $α^{2} = 5 + 26$ is irrational. If $[Q (α) : Q] = 2$ then $α^{2} \in Q (α)$ would force $6 \in Q (α)$ and $Q (α) = Q (6)$ ; but then $α = 2 + 3 \in Q (6) \subseteq R$ , and squaring $α - 2 = 3$ inside $Q (6)$ would put $2$ in $Q (6)$ , whence $3 \in Q (6)$ too and $Q (6)$ would have degree $4$ — false.
So $[Q (α) : Q] = 4$ , forcing $m_{α, Q} = f$ (a monic degree- $4$ polynomial killing $α$ must be the minimal one once we know the degree is $4$ ), and $f$ is therefore irreducible over $Q$ .
Finally $Q (α) \subseteq Q (2, 3)$ and both have degree $4$ over $Q$ , so they are equal: $α$ is a primitive element for $Q (2, 3)$ .

□

Remark (The primitive element theorem, assumed). Step 6 is an instance of a general fact: every finite extension in characteristic $0$ is simple. The proof shows that among the finitely many candidates $α + c β$ with $c \in F$ , all but finitely many generate; since $Q$ is infinite, one exists. We assume this theorem where convenient below — it is what lets us always picture a Galois extension as $F (α)$ with $G$ permuting the conjugates of $α$ — but the results of this chapter can all be proved without it, at the cost of longer arguments.

Exercises for §8.2 →

8.3Ruler-and-compass constructions

Before any group theory, the tower law alone settles two of the three classical Greek problems. The strategy is to translate "constructible with ruler and compass" into "reachable by a chain of degree- $2$ steps", and then to notice that such a chain forces the degree to be a power of $2$ .

Definition 8.19 (Constructible number). Start with the points $(0, 0)$ and $(1, 0)$ in the plane. A point is constructible if it can be obtained from these by finitely many steps, each of which draws a line through two already-constructed points, or a circle centred at a constructed point through another constructed point, and then takes an intersection point of two such drawn objects. A real number $α$ is constructible if $(α, 0)$ is a constructible point.

Theorem 8.20 (Constructible numbers have degree a power of two). If $α \in R$ is constructible, then $α$ is algebraic over $Q$ and

[Q (α) : Q] = 2^{k} for some integer k \geq 0.

Proof. Let $P_{0} = {(0, 0), (1, 0)}$ and let $P_{0}, P_{1}, \dots, P_{N}$ record the points after each construction step, with $α$ appearing as a coordinate at stage $N$ . Let $F_{j} \subseteq R$ be the field generated over $Q$ by all coordinates of points constructed by stage $j$ , so $F_{0} = Q$ .

Each step raises the degree by at most a factor of $2$ . A line through two points with coordinates in $F_{j}$ has an equation $a x + b y = c$ with $a, b, c \in F_{j}$ . A circle centred at a point of $F_{j}$ through a point of $F_{j}$ has an equation $x^{2} + y^{2} + d x + ey + f = 0$ with $d, e, f \in F_{j}$ . Now examine the three kinds of intersection:

Line with line. Solving two linear equations over $F_{j}$ by elimination gives a solution with coordinates in $F_{j}$ : degree $1$ .
Line with circle. Solve the line for one variable and substitute, obtaining a quadratic over $F_{j}$ in the other. Its roots lie in $F_{j}$ or in a quadratic extension $F_{j} (D)$ , $D \in F_{j}$ the discriminant.
Circle with circle. Subtracting the two circle equations cancels $x^{2} + y^{2}$ and leaves a linear equation over $F_{j}$ — the radical axis. So this case reduces to the previous one.

In every case, $F_{j + 1} = F_{j}$ or $F_{j + 1} = F_{j} (D)$ with $[F_{j + 1} : F_{j}] = 2$ .

Assemble the tower. We obtain a chain $Q = F_{0} \subseteq F_{1} \subseteq \dots \subseteq F_{N}$ with each step of degree $1$ or $2$ , and $α \in F_{N}$ . By Theorem Tower law applied repeatedly,

[F_{N} : Q] = j = 0 \prod N - 1 [F_{j + 1} : F_{j}] = 2^{k}

for some $k \geq 0$ . In particular $F_{N} / Q$ is finite, so $α$ is algebraic by Proposition Finite implies algebraic, but not conversely, and $Q (α)$ is an intermediate field of $F_{N} / Q$ . By Corollary Intermediate degrees divide, $[Q (α) : Q]$ divides $2^{k}$ and is therefore itself a power of $2$ .∎

Corollary 8.21 (The cube cannot be doubled and the angle cannot be trisected).

There is no ruler-and-compass construction producing a cube of twice the volume of a given cube; equivalently, $32$ is not constructible.
There is no ruler-and-compass construction trisecting an arbitrary given angle; in particular the angle $6 0^{\circ}$ cannot be trisected.

Proof. (1) Doubling a unit cube means constructing a side $α$ with $α^{3} = 2$ , i.e. $α = 32$ . Its minimal polynomial over $Q$ is $x^{3} - 2$ (Eisenstein at $2$ ), so $[Q (32) : Q] = 3$ by Theorem Structure of a simple algebraic extension. But $3$ is not a power of $2$ , so Theorem Constructible numbers have degree a power of two forbids it.

(2) An angle $θ$ is constructible precisely when $cos θ$ is, since $(cos θ, sin θ)$ and the foot of its perpendicular are mutually constructible. The angle $6 0^{\circ}$ is constructible ( $cos 6 0^{\circ} = 1/2$ ). Suppose $2 0^{\circ}$ were constructible and put $β = cos 2 0^{\circ}$ . The triple-angle identity $cos 3 θ = 4 cos^{3} θ - 3 cos θ$ with $θ = 2 0^{\circ}$ gives

\frac{1}{2} = 4 β^{3} - 3 β, i.e. 8 β^{3} - 6 β - 1 = 0.

Set $γ = 2 β$ ; then $γ^{3} - 3 γ - 1 = 0$ . A rational root of $x^{3} - 3 x - 1$ would be $\pm 1$ by the rational root test, and neither works ( $1 - 3 - 1 = - 3$ , $- 1 + 3 - 1 = 1$ ), so this cubic has no rational root; a cubic with no root in $Q$ is irreducible over $Q$ , because any factorisation of a cubic has a linear factor. Hence $[Q (γ) : Q] = 3$ , and $Q (β) = Q (γ)$ since $γ = 2 β$ . So $cos 2 0^{\circ}$ has degree $3$ , not a power of $2$ , and is not constructible by Theorem Constructible numbers have degree a power of two.∎

Remark (Squaring the circle, and the converse). The third classical problem, squaring the circle, needs $π$ and hence $π$ to be constructible. Theorem Constructible numbers have degree a power of two would then make $π$ algebraic, contradicting Lindemann's theorem that $π$ is transcendental — a genuinely analytic result which we quote and do not prove.

Note also that Theorem Constructible numbers have degree a power of two is only a necessary condition. Degree $2^{k}$ does not imply constructible: one needs the whole Galois closure to have $2$ -power degree. A root of an irreducible quartic with Galois group $S_{4}$ has degree $4$ and is not constructible, because $∣ S_{4} ∣ = 24$ is not a power of $2$ .

Example 8.22 (Is a regular $9$ -gon constructible?). Decide whether the regular nonagon can be constructed with ruler and compass.

Solution.

A regular $9$ -gon is constructible exactly when the angle $36 0^{\circ} /9 = 4 0^{\circ}$ is, equivalently when $2 0^{\circ}$ is (bisection is constructible, and doubling an angle is constructible).
By Corollary The cube cannot be doubled and the angle cannot be trisected(2), $cos 2 0^{\circ}$ has degree $3$ over $Q$ .
Since $3$ is not a power of $2$ , Theorem Constructible numbers have degree a power of two rules it out. The regular $9$ -gon is not constructible.
Contrast with the regular $17$ -gon, which is constructible: $[Q (ζ_{17}) : Q] = φ (17) = 16 = 2^{4}$ , and the extension is abelian, which as we shall see in Theorem The cyclotomic Galois group supplies the required chain of quadratic steps.

□

Exercises for §8.3 →

8.4Automorphisms and the Galois group

We now build the symmetry group. The idea is already visible in Theorem Structure of a simple algebraic extension: the field $F (α)$ knows only the relation $m (α) = 0$ , so any other root of $m$ would serve equally well as a generator. Relabelling the roots is a symmetry, and the collection of all such relabellings is a group.

Definition 8.23 ( $F$ -automorphism and the automorphism group). Let $K / F$ be an extension. An $F$ -automorphism of $K$ is a field isomorphism $σ : K \to K$ which fixes $F$ pointwise, i.e. $σ (a) = a$ for every $a \in F$ . Under composition these form a group,

Aut (K / F) = {σ : K \to K field isomorphism, σ ∣_{F} = id_{F}} .

Two consequences of "fixes $F$ pointwise" are used so often they should be automatic. First, $σ$ is $F$ -linear: $σ (a x + b y) = σ (a) σ (x) + σ (b) σ (y) = aσ (x) + bσ (y)$ for $a, b \in F$ . So $Aut (K / F)$ sits inside the group of invertible $F$ -linear maps of the vector space $K$ . Second, $σ$ commutes with every polynomial with coefficients in $F$ .

Proposition 8.24 (Automorphisms permute roots, and there are at most $[K : F]$ of them). Let $K / F$ be a finite extension and $σ \in Aut (K / F)$ .

If $f \in F [x]$ and $α \in K$ satisfies $f (α) = 0$ , then $f (σ α) = 0$ . Hence $σ$ permutes the roots in $K$ of every polynomial over $F$ , and in particular maps each $α$ to a root of its own minimal polynomial $m_{α, F}$ .
$σ$ is determined by its values on any generating set: if $K = F (α_{1}, \dots, α_{r})$ and $σ, τ$ agree on each $α_{i}$ , then $σ = τ$ .
$∣ Aut (K / F) ∣ \leq [K : F]$ .

Proof. (1) Write $f = \sum_{k} c_{k} x^{k}$ with $c_{k} \in F$ . Since $σ$ is a ring homomorphism fixing each $c_{k}$ ,

f (σ α) = k \sum c_{k} (σ α)^{k} = k \sum σ (c_{k}) σ (α^{k}) = σ (k \sum c_{k} α^{k}) = σ (f (α)) = σ (0) = 0.

The set of roots of $f$ in $K$ is finite, and $σ$ is injective, so $σ$ restricts to a permutation of it.

(2) The set ${x \in K : σ (x) = τ (x)}$ is a subfield of $K$ (it is closed under the field operations because $σ, τ$ are homomorphisms). It contains $F$ and every $α_{i}$ , hence contains $F (α_{1}, \dots, α_{r}) = K$ .

(3) Write $K = F (α_{1}, \dots, α_{r})$ — possible since $K / F$ is finite, taking a basis as generators — and build the tower $F = K_{0} \subseteq K_{1} \subseteq \dots \subseteq K_{r} = K$ with $K_{i} = K_{i - 1} (α_{i})$ . Let $d_{i} = [K_{i} : K_{i - 1}]$ , so $\prod d_{i} = [K : F]$ by Theorem Tower law. An automorphism $σ$ is determined by the tuple $(σ α_{1}, \dots, σ α_{r})$ by (2). Given the choices already made on $K_{i - 1}$ , part (1) applied over $K_{i - 1}$ forces $σ α_{i}$ to be a root of the minimal polynomial of $α_{i}$ over $K_{i - 1}$ , a polynomial of degree $d_{i}$ , so there are at most $d_{i}$ choices at stage $i$ . Multiplying the counts, $∣ Aut (K / F) ∣ \leq \prod_{i} d_{i} = [K : F]$ .∎

The bound in part (3) is often strict, and the two ways it can fail are exactly the two hypotheses in the next definition. It can fail because the minimal polynomial has roots outside $K$ , so the candidate targets are not available — this is the failure of normality. Or it can fail because the minimal polynomial has repeated roots, so there are fewer distinct targets than the degree — this is the failure of separability, and it cannot happen in characteristic $0$ .

Definition 8.25 (Normal, separable, Galois). Let $K / F$ be a finite extension.

$K / F$ is normal if every irreducible polynomial in $F [x]$ that has one root in $K$ splits completely in $K$ .
$K / F$ is separable if the minimal polynomial over $F$ of every element of $K$ has no repeated roots in a splitting field. (Automatic in characteristic $0$ , and for finite fields.)
$K / F$ is a Galois extension if it is finite, normal and separable. In that case we write $Gal (K / F) = Aut (K / F)$ and call it the Galois group.

Example 8.26 (An extension that is not normal). Show that $Q (32) / Q$ is not normal, and compute $Aut (Q (32) / Q)$ .

Solution.

The polynomial $x^{3} - 2$ is irreducible over $Q$ and has the root $32 \in Q (32)$ . Its other two roots $ω 32, ω^{2} 32$ are non-real, while $Q (32) \subseteq R$ . So it does not split: the extension is not normal.
By Proposition Automorphisms permute roots, and there are at most $[K : F]$ of them(1), any $σ \in Aut$ must send $32$ to a root of $x^{3} - 2$ lying inside the field. The only such root is $32$ itself.
By part (2) of the same proposition, $σ$ is then the identity. So $Aut (Q (32) / Q)$ is trivial, of order $1$ .
Compare with the degree: $1 < 3 = [Q (32) : Q]$ . The inequality of Proposition Automorphisms permute roots, and there are at most $[K : F]$ of them(3) is strict, and normality is precisely what failed. A degree- $3$ extension with only one symmetry is "rigid": the field cannot tell its generator apart from anything, because there is nothing to swap it with.

□

Theorem 8.27 (Characterisations of a Galois extension). Let $K / F$ be a finite extension with $G = Aut (K / F)$ . The following are equivalent.

$K / F$ is Galois (normal and separable).
$∣ G ∣ = [K : F]$ .
$K^{G} := {x \in K : σ (x) = x for all σ \in G}$ equals $F$ .
$K$ is the splitting field over $F$ of some separable polynomial in $F [x]$ .

In particular, for a Galois extension $∣ Gal (K / F) ∣ = [K : F]$ , and for any finite extension $∣ Aut (K / F) ∣$ divides $[K : F]$ , being the index-free order of a group acting with $K^{G}$ as base field.

Proof. Proof sketch. The one input assumed is Artin's lemma: if $G$ is a finite group of automorphisms of a field $K$ and $E = K^{G}$ , then $[K : E] = ∣ G ∣$ . (Its proof uses Dedekind's linear independence of characters to bound $[K : E] \leq ∣ G ∣$ , together with a direct construction for the reverse inequality; it is a page of linear algebra over $K$ and is the one genuine omission here.) Everything else we prove.

$(2) \Rightarrow (3)$ . Let $E = K^{G}$ . Then $F \subseteq E$ , since every $σ \in G$ fixes $F$ . Artin's lemma gives $[K : E] = ∣ G ∣ = [K : F]$ by (2). But $F \subseteq E \subseteq K$ , so $[K : F] = [K : E] [E : F]$ by Theorem Tower law, forcing $[E : F] = 1$ , i.e. $E = F$ .

$(3) \Rightarrow (2)$ . Artin's lemma applied to $G$ gives $[K : K^{G}] = ∣ G ∣$ , and $K^{G} = F$ by (3), so $∣ G ∣ = [K : F]$ .

$(3) \Rightarrow (1)$ . Let $α \in K$ and let ${α_{1}, \dots, α_{r}}$ be the (distinct) elements of the orbit $G α$ . Form

g (x) = i = 1 \prod r (x - α_{i}) .

Each $σ \in G$ permutes the orbit, hence permutes the factors, hence fixes every coefficient of $g$ ; so the coefficients lie in $K^{G} = F$ and $g \in F [x]$ . Now $g (α) = 0$ , so $m_{α, F} ∣ g$ . Conversely each $α_{i} = σ α$ is a root of $m_{α, F}$ by Proposition Automorphisms permute roots, and there are at most $[K : F]$ of them(1), so $g ∣ m_{α, F}$ ; both being monic, $m_{α, F} = g$ . But $g$ splits in $K$ by construction and has distinct roots by construction. Since $α$ was arbitrary, $K / F$ is normal and separable.

$(1) \Rightarrow (4)$ . Let $α_{1}, \dots, α_{s}$ generate $K$ over $F$ and let $f$ be the product of the distinct minimal polynomials $m_{α_{i}, F}$ . Each factor splits in $K$ by normality and is separable by separability; distinct irreducible monic polynomials share no root, so $f$ is separable. And $K$ is generated by the roots of $f$ because it is already generated by the $α_{i}$ . So $K$ is the splitting field of the separable $f$ .

$(4) \Rightarrow (2)$ . This is the counting step, proved by induction on $[K : F]$ exactly as in Proposition Automorphisms permute roots, and there are at most $[K : F]$ of them(3), but with the inequality now an equality at each stage: when $K$ is a splitting field of a separable $f$ , an embedding of $K_{i - 1}$ fixing $F$ extends to $K_{i} = K_{i - 1} (α_{i})$ in exactly $d_{i} = [K_{i} : K_{i - 1}]$ ways, because the minimal polynomial of $α_{i}$ over $K_{i - 1}$ divides $f$ , hence splits in $K$ (normality) with $d_{i}$ distinct roots (separability), each giving a distinct extension. Multiplying gives $∣ G ∣ = \prod d_{i} = [K : F]$ .

Finally, for an arbitrary finite $K / F$ with $E = K^{G}$ , Artin's lemma gives $∣ G ∣ = [K : E]$ , which divides $[K : F]$ by Corollary Intermediate degrees divide.∎

The equivalence $(2) \Leftrightarrow (3)$ is worth restating in words, because it is the sentence the whole chapter turns on. An extension is Galois exactly when the only elements fixed by all the symmetries are the ones we started with. If some extra element is fixed by everything, the group is too small to see all of $K$ , and the dictionary of the next section would have gaps.

Example 8.28 (The Klein four group: $Gal (Q (2, 3) / Q)$ ). Compute the Galois group of $K = Q (2, 3)$ over $Q$ explicitly, as a group of maps.

Solution.

$K$ is the splitting field of $(x^{2} - 2) (x^{2} - 3)$ , which is separable (four distinct roots, characteristic $0$ ), so $K / Q$ is Galois by Theorem Characterisations of a Galois extension(4), and $∣ G ∣ = [K : Q] = 4$ .
By Proposition Automorphisms permute roots, and there are at most $[K : F]$ of them, any $σ \in G$ satisfies $σ (2) = \pm 2$ and $σ (3) = \pm 3$ , and is determined by that pair of signs. There are $4$ sign patterns and $∣ G ∣ = 4$ , so all four occur:

id, σ : {2 \mapsto - 2 3 \mapsto 3 τ : {2 \mapsto 2 3 \mapsto - 3 σ τ : {2 \mapsto - 2 3 \mapsto - 3 .

3. Each of

σ, τ, σ τ

squares to the identity (flipping a sign twice restores it), and they commute. A group of order

4

with every non-identity element of order

2

is the Klein four group:

G ≅ V_{4} ≅ Z /2 \times Z /2

, not

Z /4

. 4. Note the action on

6 = 23

σ (6) = - 6

τ (6) = - 6

, and

σ τ (6) = (- 2) (- 3) = 6

. So

6

is fixed exactly by

{id, σ τ}

— a fact we shall need in the next section. 5. Check (3) of Theorem Characterisations of a Galois extension: an element

a + b 2 + c 3 + d 6

fixed by

σ

has

b = d = 0

, and if also fixed by

τ

has

c = 0

. So

K^{G} = Q

, as required.□

Example 8.29 ( $S_{3}$ : the Galois group of the splitting field of $x^{3} - 2$ ). Compute $Gal (K / Q)$ for $K = Q (32, ω)$ .

Solution.

$K$ is the splitting field of the separable polynomial $x^{3} - 2$ , so it is Galois and $∣ G ∣ = [K : Q] = 6$ by Theorem Characterisations of a Galois extension.
Every $σ \in G$ permutes the three roots $α_{1} = 32$ , $α_{2} = ω 32$ , $α_{3} = ω^{2} 32$ , and is determined by that permutation, since the roots generate $K$ . This gives an injective homomorphism $G ↪ S_{3}$ .
$∣ G ∣ = 6 = ∣ S_{3} ∣$ , so the injection is an isomorphism: $G ≅ S_{3}$ .
Concretely, $σ : 32 \mapsto ω 32, ω \mapsto ω$ has order $3$ (it cycles the roots), and complex conjugation $τ$ fixes $32$ and sends $ω \mapsto ω^{2}$ , so it is a transposition of order $2$ . Then $G = ⟨ σ, τ ⟩$ with $τ σ τ^{- 1} = σ^{- 1}$ , the standard presentation of $S_{3}$ .
This is the smallest example of a non-abelian Galois group, and the source of every interesting phenomenon in the next section: $S_{3}$ has non-normal subgroups, and they will correspond to non-normal intermediate fields.

□

Figure 8.30. The subgroup lattice of

S_{3} ≅ Gal (Q (32, ω) / Q)

: three conjugate subgroups of order

2

generated by transpositions, one normal subgroup

A_{3}

of order

3

, and the two trivial ends.

Exercises for §8.4 →

8.5The fundamental theorem

Everything so far has been preparation. We now state and use the dictionary.

Definition 8.31 (Fixed field). Let $K$ be a field and $H$ a group of automorphisms of $K$ . The fixed field of $H$ is

K^{H} = {x \in K : σ (x) = x for all σ \in H} .

Lemma 8.32 (Fixed sets are fields, and the two maps are mutually decreasing). Let $K / F$ be Galois with $G = Gal (K / F)$ .

For any subgroup $H \leq G$ , the set $K^{H}$ is a field with $F \subseteq K^{H} \subseteq K$ .
For any intermediate field $E$ with $F \subseteq E \subseteq K$ , the set $Gal (K / E)$ is a subgroup of $G$ .
Both assignments reverse inclusion: $H_{1} \leq H_{2} \Rightarrow K^{H_{2}} \subseteq K^{H_{1}}$ , and $E_{1} \subseteq E_{2} \Rightarrow Gal (K / E_{2}) \leq Gal (K / E_{1})$ .
$E \subseteq K^{Gal (K / E)}$ and $H \leq Gal (K / K^{H})$ always.

Proof. (1) If $x, y \in K^{H}$ and $σ \in H$ , then $σ (x \pm y) = σ x \pm σ y = x \pm y$ and $σ (x y) = σ (x) σ (y) = x y$ , and for $x \neq = 0$ , $σ (x^{- 1}) = σ (x)^{- 1} = x^{- 1}$ . Also $σ (0) = 0$ , $σ (1) = 1$ . So $K^{H}$ is a subfield. Every $σ \in H \leq G$ fixes $F$ pointwise, so $F \subseteq K^{H}$ .

(2) $K / E$ is again finite; an $E$ -automorphism of $K$ fixes $E \supseteq F$ pointwise, hence is an $F$ -automorphism, so $Gal (K / E) \subseteq G$ as a set. It is closed under composition and inverses (both of which still fix $E$ ), and contains the identity, so it is a subgroup.

(3) If $H_{1} \leq H_{2}$ then a point fixed by everything in $H_{2}$ is in particular fixed by everything in $H_{1}$ : $K^{H_{2}} \subseteq K^{H_{1}}$ . If $E_{1} \subseteq E_{2}$ then an automorphism fixing $E_{2}$ pointwise fixes $E_{1}$ pointwise: $Gal (K / E_{2}) \leq Gal (K / E_{1})$ .

(4) Every element of $E$ is fixed by every automorphism that fixes $E$ pointwise — that is the definition — so $E \subseteq K^{Gal (K / E)}$ . Symmetrically every $σ \in H$ fixes $K^{H}$ pointwise by definition of $K^{H}$ , so $H \leq Gal (K / K^{H})$ .∎

Part (3) is the inclusion-reversal, and it is the single most misremembered fact in the subject, so it is worth saying why it must go that way. Asking more symmetries to fix a point is a stronger demand, so it leaves fewer points standing. The biggest group $G$ fixes only $F$ ; the smallest group ${1}$ fixes all of $K$ . Any picture in which the two lattices run parallel is wrong.

Part (4) gives two inclusions for free. The content of the Fundamental Theorem is that, for a Galois extension, both are equalities.

Theorem 8.33 (Fundamental theorem of Galois theory). Let $K / F$ be a Galois extension with group $G = Gal (K / F)$ , $∣ G ∣ = [K : F] = n$ . Then the maps

E ⟼ Gal (K / E), H ⟼ K^{H}

are mutually inverse, inclusion-reversing bijections between the set of intermediate fields $F \subseteq E \subseteq K$ and the set of subgroups $H \leq G$ . Moreover, for corresponding $E$ and $H = Gal (K / E)$ :

$K / E$ is Galois, with $[K : E] = ∣ H ∣$ ;
$[E : F] = [G : H]$ , the index of $H$ in $G$ ;
$σ H σ^{- 1}$ corresponds to $σ (E)$ , for each $σ \in G$ .

Proof. Proof sketch. Assumed: Artin's lemma, exactly as in Theorem Characterisations of a Galois extension ( $[K : K^{H}] = ∣ H ∣$ for a finite group $H$ of automorphisms of $K$ ). Given it, the rest is a short argument, which we give in full.

$K / E$ is Galois, and $K^{Gal (K / E)} = E$ . By Theorem Characterisations of a Galois extension(4), $K$ is the splitting field over $F$ of a separable $f \in F [x]$ . Then $K$ is also the splitting field of the same $f$ regarded in $E [x]$ , and $f$ is still separable, so $K / E$ is Galois by the same criterion. Applying part (3) of that theorem to $K / E$ gives $K^{Gal (K / E)} = E$ . So $H \mapsto K^{H}$ is a left inverse of $E \mapsto Gal (K / E)$ ; in particular $E \mapsto Gal (K / E)$ is injective. Claim (1), $[K : E] = ∣ Gal (K / E) ∣$ , is then Theorem Characterisations of a Galois extension(2) applied to $K / E$ .

$Gal (K / K^{H}) = H$ . Let $H \leq G$ and put $E = K^{H}$ . By Lemma Fixed sets are fields, and the two maps are mutually decreasing(4), $H \leq Gal (K / E)$ . Compare orders: Artin's lemma gives $[K : E] = ∣ H ∣$ , while the previous paragraph gives $[K : E] = ∣ Gal (K / E) ∣$ . So $∣ H ∣ = ∣ Gal (K / E) ∣$ , and a subgroup of the same finite order as the group containing it is that group: $H = Gal (K / E)$ . Hence the two maps are mutually inverse bijections. Inclusion-reversal is Lemma Fixed sets are fields, and the two maps are mutually decreasing(3).

Claim (2). By Theorem Tower law and (1),

[E : F] = \frac{[ K : F ]}{[ K : E ]} = \frac{∣ G ∣}{∣ H ∣} = [G : H],

using Lagrange's theorem from Subgroups and quotient groups for the last equality.

Claim (3). Let $σ \in G$ and $x \in E$ . For $τ \in H$ we have $σ τ σ^{- 1} (σ x) = σ τ (x) = σ (x)$ , so $σ H σ^{- 1}$ fixes $σ (E)$ pointwise; thus $σ H σ^{- 1} \leq Gal (K / σ E)$ . Conjugation is a bijection of $G$ , so the two groups have equal order $∣ H ∣ = [K : E] = [K : σ E]$ — the last equality because $σ$ restricts to an $F$ -isomorphism $E \to σ E$ , which preserves degree. Equal finite orders and one contained in the other give equality.∎

Proposition 8.34 (Normal subgroups correspond to normal subextensions). With notation as in Theorem Fundamental theorem of Galois theory, let $E \leftrightarrow H$ correspond. Then

E / F is a normal (hence Galois) extension ⟺ H ⊴ G,

and in that case restriction $σ \mapsto σ ∣_{E}$ induces an isomorphism

Gal (E / F) ≅ G / H .

Proof. ( $\Leftarrow$ ) Suppose $H ⊴ G$ . For every $σ \in G$ , claim (3) of Theorem Fundamental theorem of Galois theory says $σ (E)$ corresponds to $σ H σ^{- 1} = H$ , and the correspondence is a bijection, so $σ (E) = E$ . Now let $p \in F [x]$ be irreducible with a root $α \in E$ . Since $K / F$ is normal, $p$ splits in $K$ ; let $β \in K$ be any root. By Theorem Characterisations of a Galois extension applied to $K / F$ , the orbit argument there shows $m_{α, F}$ has as its roots exactly the orbit $G α$ , so $β = σ α$ for some $σ \in G$ ; hence $β \in σ (E) = E$ . So $p$ splits in $E$ and $E / F$ is normal. Separability is inherited from $K / F$ , so $E / F$ is Galois.

( $\Rightarrow$ ) Suppose $E / F$ is normal. For $σ \in G$ and $α \in E$ , $σ α$ is a root of $m_{α, F}$ , which splits in $E$ by normality, so $σ α \in E$ : that is, $σ (E) \subseteq E$ , and by degree equality $σ (E) = E$ . Then claim (3) gives $σ H σ^{- 1} = Gal (K / σ E) = Gal (K / E) = H$ for every $σ$ , i.e. $H ⊴ G$ .

The quotient. Assume $H ⊴ G$ . By the previous paragraph every $σ \in G$ maps $E$ onto $E$ , so restriction gives a map

r : G \to Gal (E / F), r (σ) = σ ∣_{E},

which is a group homomorphism because restricting a composite is the composite of the restrictions. Its kernel is ${σ : σ ∣_{E} = id_{E}} = Gal (K / E) = H$ . It is surjective: given $ρ \in Gal (E / F)$ , the extension theorem used in Theorem Characterisations of a Galois extension $(4) \Rightarrow (2)$ — an isomorphism of base fields extends to splitting fields — extends $ρ$ to an automorphism of $K$ , which lies in $G$ and restricts to $ρ$ . By the First isomorphism theorem from Group homomorphisms,

G / H = G / ker r ≅ im r = Gal (E / F) .

∎

Example 8.35 (The full correspondence for $Q (2, 3)$ ). Write down every intermediate field of $K = Q (2, 3)$ over $Q$ , matched to its subgroup.

Solution.

From the example The Klein four group, $G = {id, σ, τ, σ τ} ≅ V_{4}$ . Its subgroups are: ${id}$ , the three cyclic groups $⟨ σ ⟩, ⟨ τ ⟩, ⟨ σ τ ⟩$ of order $2$ , and $G$ . That is five subgroups, so by Theorem Fundamental theorem of Galois theory there are exactly five intermediate fields, three of them strictly between $Q$ and $K$ .
The two ends are forced: $K^{{id}} = K$ and $K^{G} = Q$ .
$K^{⟨ σ ⟩}$ : $σ$ fixes $3$ and negates $2$ (and hence $6$ ), so on $a + b 2 + c 3 + d 6$ it forces $b = d = 0$ . The fixed field is $Q (3)$ .
$K^{⟨ τ ⟩} = Q (2)$ by the mirror computation, and $K^{⟨ σ τ ⟩} = Q (6)$ , using step 4 of the example The Klein four group: $σ τ$ fixes $6$ and negates both $2$ and $3$ .
Check the degrees against claim (2): each order- $2$ subgroup has index $2$ , and indeed each of $Q (2), Q (3), Q (6)$ has degree $2$ over $Q$ .
Check the reversal: ${id} \leq ⟨ σ ⟩ \leq G$ on the group side corresponds to $K \supseteq Q (3) \supseteq Q$ on the field side — the chain runs downwards as the groups run upwards.
Every subgroup of the abelian group $V_{4}$ is normal, so by Proposition Normal subgroups correspond to normal subextensions every intermediate field is Galois over $Q$ — as it must be, each being a splitting field of $x^{2} - d$ .

□

Figure 8.36. The Galois correspondence for

Q (2, 3) / Q

. On the left the lattice of intermediate fields; on the right the lattice of subgroups of

V_{4}

, drawn with the whole group at the bottom so that corresponding nodes sit at the same height. Here

s

negates

2

and

t

negates

3

. Going up on the left is going down on the right.

Example 8.37 (Intermediate fields of the splitting field of $x^{3} - 2$ ). How many fields lie strictly between $Q$ and $K = Q (32, ω)$ , and which of them are normal over $Q$ ?

Solution.

$G ≅ S_{3}$ by the example $S_{3}$ : the Galois group of the splitting field of $x^{3} - 2$ . The subgroups of $S_{3}$ are ${e}$ , three subgroups of order $2$ generated by the transpositions, one subgroup $A_{3}$ of order $3$ , and $S_{3}$ : six in all.
By Theorem Fundamental theorem of Galois theory there are six intermediate fields, so four lie strictly between.
$A_{3}$ has index $2$ , so its fixed field has degree $2$ : it is $Q (ω)$ , the elements fixed by the $3$ -cycles.
Each order- $2$ subgroup has index $3$ , giving a cubic field. The subgroup generated by complex conjugation fixes $32$ , so its fixed field is $Q (32)$ ; the two conjugate subgroups give $Q (ω 32)$ and $Q (ω^{2} 32)$ .
Normality: $A_{3} ⊴ S_{3}$ (index $2$ ), so $Q (ω) / Q$ is Galois — correct, it is the splitting field of $x^{2} + x + 1$ . The three order- $2$ subgroups are not normal (they are conjugate to each other, and a normal subgroup is its own only conjugate), so by Proposition Normal subgroups correspond to normal subextensions the three cubic fields are not normal over $Q$ — correct, as the example An extension that is not normal already showed for $Q (32)$ .
The quotient check: $Gal (Q (ω) / Q) ≅ S_{3} / A_{3} ≅ Z /2$ , of order $2$ , matching $[Q (ω) : Q] = 2$ .

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Pitfall. The correspondence counts intermediate fields only for Galois extensions. The extension $Q (32) / Q$ has $∣ Aut ∣ = 1$ , whose subgroup lattice has one element, yet the extension has two intermediate fields ( $Q$ and itself). There is no contradiction and no bijection: the hypothesis fails. Always verify "normal and separable" — in practice, "is it a splitting field?" — before counting subgroups.

Example 8.38 (Using the correspondence in reverse). An extension $K / Q$ is Galois with $Gal (K / Q) ≅ Z /12$ . How many intermediate fields are there, and how many are of degree $3$ over $Q$ ?

Solution.

A cyclic group of order $n$ has exactly one subgroup for each divisor of $n$ , so $Z /12$ has subgroups of orders $1, 2, 3, 4, 6, 12$ : six subgroups. By Theorem Fundamental theorem of Galois theory there are exactly six intermediate fields.
Degree $3$ over $Q$ means index $3$ by claim (2), i.e. a subgroup of order $12/3 = 4$ . A cyclic group has exactly one subgroup of order $4$ , so there is exactly one such field.
Sanity check on the reversal: the subgroup of order $4$ is small-ish, so its fixed field is largish — degree $12/4 = 3$ ... and indeed the smallest subgroup ${e}$ has the largest fixed field $K$ , of degree $12$ .
Every subgroup of an abelian group is normal, so all six intermediate fields are Galois over $Q$ .

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Exercises for §8.5 →

8.6Finite fields and cyclotomic extensions

Two families of Galois extensions can be described completely, and between them they cover most explicit computations one ever does. In both the Galois group turns out to be abelian, which is why both families are "easy".

Theorem 8.39 (The Galois group of a finite field is cyclic, generated by Frobenius). Let $p$ be prime and $q = p^{n}$ . The extension $F_{q} / F_{p}$ is Galois of degree $n$ , and

Gal (F_{p^{n}} / F_{p}) = ⟨ φ ⟩ ≅ Z / n Z, φ (x) = x^{p},

where $φ$ is the Frobenius automorphism.

Proof. $F_{q}$ is Galois over $F_{p}$ . The field $F_{q}$ has $q$ elements; its multiplicative group has order $q - 1$ , so $x^{q - 1} = 1$ for every $x \neq = 0$ by Lagrange, and hence $x^{q} = x$ for every $x \in F_{q}$ including $0$ . So all $q$ elements are roots of $x^{q} - x$ , a polynomial of degree $q$ , which therefore splits in $F_{q}$ with $q$ distinct roots and no smaller field contains them all. Thus $F_{q}$ is the splitting field of the separable polynomial $x^{q} - x$ over $F_{p}$ , hence Galois by Theorem Characterisations of a Galois extension(4), and $[F_{q} : F_{p}] = n$ since $∣ F_{q} ∣ = p^{n} = p^{[F_{q} : F_{p}]}$ by counting coordinates in a basis.

$φ$ is an automorphism fixing $F_{p}$ . Multiplicativity is clear. Additivity is the "freshman's dream": in characteristic $p$ ,

(x + y)^{p} = k = 0 \sum p (k p) x^{k} y^{p - k} = x^{p} + y^{p},

because for $0 < k < p$ the binomial coefficient $(k p) = \frac{p !}{k ! ( p - k )!}$ has a factor $p$ in the numerator that is cancelled by nothing in the denominator, so it vanishes mod $p$ . Thus $φ$ is a ring homomorphism; it is injective since a field homomorphism is, and injective on a finite set means bijective. By Fermat's little theorem $a^{p} = a$ for $a \in F_{p}$ , so $φ$ fixes $F_{p}$ pointwise.

$φ$ has order exactly $n$ . We have $φ^{k} (x) = x^{p^{k}}$ . Now $φ^{n} (x) = x^{p^{n}} = x^{q} = x$ for all $x$ , so $φ^{n} = id$ and the order divides $n$ . If $φ^{k} = id$ with $0 < k < n$ , then every one of the $q = p^{n}$ elements of $F_{q}$ is a root of $x^{p^{k}} - x$ , a nonzero polynomial with at most $p^{k} < p^{n}$ roots — a contradiction. So $φ$ has order $n$ .

Conclusion. $⟨ φ ⟩$ is a cyclic subgroup of order $n$ inside $Gal (F_{q} / F_{p})$ , which has order $[F_{q} : F_{p}] = n$ . Equal orders force equality.∎

Corollary 8.40 (Subfields of a finite field). The subfields of $F_{p^{n}}$ containing $F_{p}$ are in bijection with the positive divisors $d ∣ n$ , one subfield $F_{p^{d}}$ for each. In particular $F_{p^{n}}$ has exactly $τ (n)$ such subfields, where $τ$ counts divisors.

Proof. By Theorem The Galois group of a finite field is cyclic, generated by Frobenius the group is cyclic of order $n$ , and a cyclic group of order $n$ has exactly one subgroup of order $m$ for each $m ∣ n$ and no others. By Theorem Fundamental theorem of Galois theory the intermediate fields are in bijection with these subgroups. The subgroup $⟨ φ^{d} ⟩$ has order $n / d$ , so by claim (2) its fixed field has degree $[G : ⟨ φ^{d} ⟩] = d$ over $F_{p}$ , i.e. has $p^{d}$ elements. As $d$ runs over the divisors of $n$ , so does $n / d$ , giving exactly $τ (n)$ subfields.∎

Example 8.41 (Subfields of $F_{2^{12}}$ ). List the subfields of $F_{4096}$ and give the order of the Galois group of $F_{4096}$ over $F_{16}$ .

Solution.

$4096 = 2^{12}$ , so $n = 12$ and the divisors are $1, 2, 3, 4, 6, 12$ — six of them.
By Corollary Subfields of a finite field the subfields are $F_{2}, F_{4}, F_{8}, F_{16}, F_{64}, F_{4096}$ .
$F_{16} = F_{2^{4}}$ , and $[F_{2^{12}} : F_{2^{4}}] = 12/4 = 3$ by Theorem Tower law. The extension is Galois (it is again a splitting field of $x^{q} - x$ ), so its group has order $3$ and is cyclic, generated by $φ^{4}$ .
Note $F_{32}$ is not a subfield, since $5 ∤ 12$ — the divisor condition is the whole content.

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Theorem 8.42 (The cyclotomic Galois group). Let $ζ_{n} = e^{2 πi / n}$ be a primitive $n$ th root of unity. Then $Q (ζ_{n}) / Q$ is Galois, and

Gal (Q (ζ_{n}) / Q) ≅ (Z / n Z)^{\times}, σ_{k} (ζ_{n}) = ζ_{n}^{k},

an abelian group of order $φ (n)$ . Consequently $[Q (ζ_{n}) : Q] = φ (n)$ , and the minimal polynomial of $ζ_{n}$ is the $n$ th cyclotomic polynomial $Φ_{n}$ , of degree $φ (n)$ .

Proof. Proof sketch. Assumed: $Φ_{n}$ is irreducible over $Q$ . (The standard proof shows that if $ζ$ is a root of an irreducible factor $f$ and $p ∤ n$ is prime, then $ζ^{p}$ is also a root of $f$ , by reducing mod $p$ and using that $x^{n} - 1$ is separable mod $p$ ; iterating over the prime factors of any $k$ coprime to $n$ shows $f$ has all $φ (n)$ primitive roots as roots, so $f = Φ_{n}$ . It is a page, and we take it.) Everything else is proved here.

Galois. $Q (ζ_{n})$ contains $ζ_{n}^{k}$ for every $k$ , hence all $n$ roots of $x^{n} - 1$ , and is generated by them; $x^{n} - 1$ has no repeated roots in characteristic $0$ since it is coprime to its derivative $n x^{n - 1}$ . So $Q (ζ_{n})$ is the splitting field of a separable polynomial: Galois, by Theorem Characterisations of a Galois extension(4).

The map. By the assumed irreducibility, $[Q (ζ_{n}) : Q] = de g Φ_{n} = φ (n)$ using Theorem Structure of a simple algebraic extension. Any $σ \in G$ sends $ζ_{n}$ to a root of $Φ_{n}$ , i.e. to some $ζ_{n}^{k}$ with $g cd (k, n) = 1$ , and is determined by that $k$ since $ζ_{n}$ generates. Define $Ψ (σ) = k mod n$ . This is a homomorphism: if $σ (ζ) = ζ^{k}$ and $ρ (ζ) = ζ^{l}$ , then

(σ ρ) (ζ) = σ (ζ^{l}) = (σ ζ)^{l} = ζ^{k l},

so $Ψ (σ ρ) = k l = Ψ (σ) Ψ (ρ)$ in $(Z / n)^{\times}$ . It is injective, because $k \equiv 1$ means $σ$ fixes the generator, hence is the identity. Finally both groups have order $φ (n)$ — the left by the degree count, the right by definition of Euler's totient — so $Ψ$ is an isomorphism. Abelian-ness is inherited from $(Z / n)^{\times}$ .∎

Example 8.43 (Two cyclotomic degrees and a Galois group). Compute $[Q (ζ_{12}) : Q]$ and $[Q (ζ_{15}) : Q]$ , and identify $Gal (Q (ζ_{5}) / Q)$ up to isomorphism.

Solution.

$φ (12) = φ (4) φ (3) = 2 \cdot 2 = 4$ , so $[Q (ζ_{12}) : Q] = 4$ by Theorem The cyclotomic Galois group. (Indeed $Φ_{12} (x) = x^{4} - x^{2} + 1$ .)
$φ (15) = φ (3) φ (5) = 2 \cdot 4 = 8$ , so $[Q (ζ_{15}) : Q] = 8$ .
For $n = 5$ : $Gal ≅ (Z /5)^{\times}$ , of order $φ (5) = 4$ . The element $2$ has powers $2, 4, 3, 1$ mod $5$ , so it generates: the group is cyclic, $≅ Z /4 Z$ , not $V_{4}$ .
Consequence via Theorem Fundamental theorem of Galois theory: $Z /4$ has exactly three subgroups, so $Q (ζ_{5})$ has exactly three intermediate fields, the middle one being the unique quadratic subfield $Q (5)$ .

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Example 8.44 (Why the regular $17$ -gon is constructible). Explain, using the cyclotomic Galois group, why $ζ_{17}$ is constructible.

Solution.

$[Q (ζ_{17}) : Q] = φ (17) = 16 = 2^{4}$ , and $G = Gal ≅ (Z /17)^{\times}$ is cyclic of order $16$ .
A cyclic group of order $16$ has a chain of subgroups ${e} = H_{4} < H_{3} < H_{2} < H_{1} < H_{0} = G$ with $∣ H_{i} ∣ = 2^{4 - i}$ , each of index $2$ in the next.
By Theorem Fundamental theorem of Galois theory this gives a chain of fields $Q = E_{0} \subset E_{1} \subset E_{2} \subset E_{3} \subset E_{4} = Q (ζ_{17})$ with $[E_{i + 1} : E_{i}] = 2$ at every step.
Each degree- $2$ step is obtained by adjoining a square root (complete the square on the quadratic minimal polynomial), and square roots are constructible with ruler and compass. So $ζ_{17}$ — hence the regular $17$ -gon — is constructible, which is Gauss's theorem of 1796.
The contrast with the $9$ -gon in the example Is a regular $9$ -gon constructible? is exactly the contrast between $φ (17) = 16$ and $φ (9) = 6$ : a power of $2$ against a non-power of $2$ .

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Exercises for §8.6 →

8.7Solvability by radicals and Abel–Ruffini

The last movement converts "is there a formula?" into "is the group solvable?". A formula by radicals is a finite recipe of arithmetic and root extractions, and each root extraction, we shall see, contributes exactly one abelian layer to the group. Galois's theorem says that is the whole story.

Definition 8.45 (Radical extension and solvability by radicals). An extension $K / F$ is radical if there is a tower

F = F_{0} \subseteq F_{1} \subseteq \dots \subseteq F_{m} = K, F_{i + 1} = F_{i} (α_{i}) with α_{i}^{n_{i}} \in F_{i}

for some positive integers $n_{i}$ — each step adjoins an $n_{i}$ th root of something already present. A polynomial $f \in F [x]$ is solvable by radicals if its splitting field is contained in some radical extension of $F$ .

Definition 8.46 (Solvable group). A finite group $G$ is solvable if there is a chain

{1} = G_{0} ⊴ G_{1} ⊴ \dots ⊴ G_{m} = G

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

Proposition 8.47 (Solvability passes to subgroups and quotients). Let $G$ be a solvable finite group. Then every subgroup of $G$ is solvable, and every quotient $G / N$ by a normal subgroup is solvable. Conversely, if $N ⊴ G$ with $N$ and $G / N$ both solvable, then $G$ is solvable.

Proof. Subgroups. Let $H \leq G$ and let ${1} = G_{0} ⊴ \dots ⊴ G_{m} = G$ be a solvable chain. Put $H_{i} = H \cap G_{i}$ . Then $H_{i} ⊴ H_{i + 1}$ : for $h \in H_{i + 1}$ , conjugation by $h$ preserves $H$ and preserves $G_{i}$ (since $h \in G_{i + 1}$ and $G_{i} ⊴ G_{i + 1}$ ), so it preserves the intersection. The inclusion $H_{i + 1} ↪ G_{i + 1}$ sends $H_{i}$ into $G_{i}$ , hence induces an injective homomorphism

H_{i + 1} / H_{i} ↪ G_{i + 1} / G_{i},

injective because $h \in H_{i + 1}$ mapping into $G_{i}$ lies in $H \cap G_{i} = H_{i}$ . A subgroup of an abelian group is abelian, so each $H_{i + 1} / H_{i}$ is abelian and $H$ is solvable.

Quotients. Let $π : G \to G / N$ be the projection and set $\overset{ˉ}{G}_{i} = π (G_{i})$ . Images of normal subgroups under a surjection onto the relevant piece are normal, so $\overset{ˉ}{G}_{i} ⊴ \overset{ˉ}{G}_{i + 1}$ , and $π$ induces a surjection $G_{i + 1} / G_{i} ↠ \overset{ˉ}{G}_{i + 1} / \overset{ˉ}{G}_{i}$ . A quotient of an abelian group is abelian, so the chain ${1} = \overset{ˉ}{G}_{0} ⊴ \dots ⊴ \overset{ˉ}{G}_{m} = G / N$ witnesses solvability.

Converse. Splice a solvable chain for $N$ with the preimage under $π$ of a solvable chain for $G / N$ . The preimages are normal in one another with the same quotients, by the Correspondence theorem for groups, and the bottom of the second chain is $N$ , the top of the first.∎

Theorem 8.48 (Galois's solvability criterion). Let $F$ have characteristic $0$ and let $f \in F [x]$ be nonconstant with splitting field $K$ . Then

f is solvable by radicals ⟺ Gal (K / F) is a solvable group .

Proof. Proof sketch. Assumed: the Fundamental Theorem (Theorem Fundamental theorem of Galois theory) and the following two facts about cyclic extensions, both standard and both proved with the assumed input of Hilbert's Theorem 90 or a Lagrange-resolvent computation: (a) if $F$ contains a primitive $n$ th root of unity and $α^{n} \in F$ , then $F (α) / F$ is Galois with cyclic group embedding into $Z / n$ ; (b) conversely a cyclic Galois extension of degree $n$ , over a base containing the $n$ th roots of unity, is $F (α)$ with $α^{n} \in F$ . Given these, the argument is as follows.

( $\Rightarrow$ ) Suppose $K \subseteq L$ with $L / F$ radical, built by adjoining $α_{i}$ with $α_{i}^{n_{i}} \in F_{i}$ . Enlarging $L$ if necessary, adjoin a primitive $N$ th root of unity $ζ$ for $N = \prod n_{i}$ first; the step $F \subseteq F (ζ)$ is Galois with abelian group by Theorem The cyclotomic Galois group, so it contributes one abelian layer. Each remaining step is now cyclic by (a). We may also enlarge $L$ to be normal over $F$ (take the splitting field of the product of the minimal polynomials of the $α_{i}$ ; the enlargement is still radical because the conjugates of a radical tower form radical towers). Applying Theorem Fundamental theorem of Galois theory to the tower $F \subseteq F (ζ) \subseteq \dots \subseteq L$ turns it into a chain of subgroups of $Gal (L / F)$ , each normal in the next with abelian quotient by Proposition Normal subgroups correspond to normal subextensions. Hence $Gal (L / F)$ is solvable. Finally $Gal (K / F)$ is a quotient of $Gal (L / F)$ — again by Proposition Normal subgroups correspond to normal subextensions, $K / F$ being normal — and quotients of solvable groups are solvable by Proposition Solvability passes to subgroups and quotients.

( $\Leftarrow$ ) Suppose $G = Gal (K / F)$ is solvable, with chain ${1} = G_{0} ⊴ \dots ⊴ G_{m} = G$ . Refine it so every quotient is cyclic of prime order — possible because a finite abelian group has such a chain. Adjoin a primitive $N$ th root of unity with $N = ∣ G ∣$ to get $F^{'} = F (ζ)$ and $K^{'} = K (ζ)$ ; then $Gal (K^{'} / F^{'})$ embeds in $G$ , hence is solvable by Proposition Solvability passes to subgroups and quotients. Translate its cyclic chain through Theorem Fundamental theorem of Galois theory into a tower of fields with each step cyclic of prime degree, and apply (b) to write each step as the adjunction of a single radical. The result exhibits $K^{'} \supseteq K$ inside a radical extension of $F$ .∎

Lemma 8.49 ( $S_{n}$ is not solvable for $n \geq 5$ ). For $n \geq 5$ , the alternating group $A_{n}$ is simple and non-abelian, and consequently $S_{n}$ is not solvable.

Proof. Proof sketch. Assumed: $A_{5}$ is simple. (The usual proof lists the conjugacy classes of $A_{5}$ — sizes $1, 15, 20, 12, 12$ — and observes that no union containing $1$ and summing to a divisor of $60$ other than $1$ or $60$ exists; simplicity of $A_{n}$ for $n > 5$ then follows by an induction on $n$ using that a nontrivial normal subgroup must contain a $3$ -cycle.) Granting simplicity, here is the deduction.

Suppose $S_{n}$ were solvable for some $n \geq 5$ . By Proposition Solvability passes to subgroups and quotients, the subgroup $A_{n} \leq S_{n}$ would be solvable, so there would be a chain ${1} = G_{0} ⊴ \dots ⊴ G_{m} = A_{n}$ with abelian quotients and $G_{m - 1} \neq = A_{n}$ (taking the chain without repetitions, with $m \geq 1$ ). Then $G_{m - 1}$ is a proper normal subgroup of $A_{n}$ , so by simplicity $G_{m - 1} = {1}$ , and the chain reads ${1} ⊴ A_{n}$ with quotient $A_{n} / {1} ≅ A_{n}$ abelian.

But $A_{n}$ is not abelian for $n \geq 5$ : take $σ = (123)$ and $τ = (345)$ , both even. Then $σ τ$ sends $1 \mapsto 2$ , while $τ σ$ sends $1 \mapsto 2$ as well — so compare on $3$ : $σ τ (3) = σ (4) = 4$ whereas $τ σ (3) = τ (1) = 1$ . Hence $σ τ \neq = τ σ$ . Contradiction, so $S_{n}$ is not solvable.∎

Corollary 8.50 (Abel–Ruffini: the general quintic has no radical formula). There is no formula, built from the coefficients using field operations and $n$ th roots, that solves every quintic equation over $Q$ . Concretely, the polynomial

f (x) = x^{5} - 6 x + 3 \in Q [x]

is irreducible with Galois group $S_{5}$ , and is therefore not solvable by radicals.

Proof. We verify the three claims about $f$ and then conclude.

Irreducible. Eisenstein at $p = 3$ : the non-leading coefficients $0, 0, 0, - 6, 3$ are all divisible by $3$ , the leading coefficient $1$ is not, and $9 ∤ 3$ . So $f$ is irreducible over $Q$ .

Exactly three real roots. $f^{'} (x) = 5 x^{4} - 6$ has the two real zeros $x = \pm (6/5)^{1/4} \approx \pm 1.047$ , so $f$ has exactly two turning points and at most three real roots. It has at least three by the Intermediate Value Theorem, since

f (- 2) = - 17 < 0, f (0) = 3 > 0, f (1) = - 2 < 0, f (2) = 23 > 0,

giving sign changes on $(- 2, 0)$ , $(0, 1)$ and $(1, 2)$ . So there are exactly three real roots and hence one conjugate pair of non-real roots.

The group is $S_{5}$ . Let $G = Gal (K / Q) \leq S_{5}$ , acting on the five roots. Since $f$ is irreducible, $G$ is transitive on the roots, so $5 ∣ ∣ G ∣$ by the orbit–stabiliser theorem from Groups and their structure; by Cauchy's theorem $G$ contains an element of order $5$ , which in $S_{5}$ must be a $5$ -cycle. Complex conjugation restricts to an element of $G$ which fixes the three real roots and swaps the two non-real ones: a transposition. Now a $5$ -cycle and any transposition generate $S_{5}$ — relabel so the transposition is $(12)$ and a power of the $5$ -cycle is $(12345)$ ; conjugating $(12)$ repeatedly by it yields $(23), (34), (45)$ , and adjacent transpositions generate $S_{5}$ . Hence $G = S_{5}$ .

Conclusion. $S_{5}$ is not solvable by Lemma $S_{n}$ is not solvable for $n \geq 5$ , so by Theorem Galois's solvability criterion $f$ is not solvable by radicals. A general formula in the coefficients would in particular solve $f$ by radicals, so no such formula exists.∎

Figure 8.51. The quintic

x^{5} - 6 x + 3

crosses the axis three times, so it has three real roots and one conjugate pair of complex roots. That pair is what supplies a transposition in the Galois group, which together with a

5

-cycle forces the group to be all of

S_{5}

Example 8.52 (A quintic that *is* solvable by radicals). Show that $x^{5} - 2$ is solvable by radicals, and compute the order of its Galois group over $Q$ .

Solution.

The roots are $ζ_{5}^{k} 52$ for $k = 0, \dots, 4$ , so the splitting field is $K = Q (52, ζ_{5})$ , which is manifestly contained in a radical extension: adjoin $ζ_{5}$ (a root of $x^{5} - 1$ , and $1 \in Q$ ), then $52$ . So $x^{5} - 2$ is solvable by radicals by Definition Radical extension and solvability by radicals.
Degree: $[Q (ζ_{5}) : Q] = φ (5) = 4$ and $[Q (52) : Q] = 5$ (Eisenstein at $2$ ). Both $4$ and $5$ divide $[K : Q]$ by Corollary Intermediate degrees divide, so $20 ∣ [K : Q]$ ; and $[K : Q] \leq 4 \cdot 5 = 20$ by Theorem Tower law. Hence $[K : Q] = 20$ and $∣ Gal (K / Q) ∣ = 20$ .
Consistency with Theorem Galois's solvability criterion: a group of order $20 = 2^{2} \cdot 5$ is solvable (indeed it has a normal Sylow $5$ -subgroup with abelian quotient of order $4$ ), matching step 1.
Moral: "quintic" does not mean "unsolvable". The obstruction is the group, not the degree, and only the general quintic — the one whose group is the full $S_{5}$ — is beyond radicals.

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Example 8.53 (Reading solvability off a degree- $6$ example). The splitting field of $x^{3} - 2$ over $Q$ has group $S_{3}$ . Is $x^{3} - 2$ solvable by radicals, and does that agree with Cardano?

Solution.

$S_{3}$ has the chain ${e} ⊴ A_{3} ⊴ S_{3}$ . The quotients are $A_{3} ≅ Z /3$ and $S_{3} / A_{3} ≅ Z /2$ , both abelian, so $S_{3}$ is solvable by Definition Solvable group.
By Theorem Galois's solvability criterion, $x^{3} - 2$ is solvable by radicals.
Explicitly it is: the roots are $32, ω 32, ω^{2} 32$ , visibly radical expressions, and the general cubic is handled by Cardano's formula — which is exactly the two-layer structure ( $Z /2$ then $Z /3$ ) made explicit as "take a square root, then a cube root".
The same argument works for every $n \leq 4$ : $S_{2}, S_{3}, S_{4}$ are all solvable ( $S_{4}$ via ${e} ⊴ V_{4} ⊴ A_{4} ⊴ S_{4}$ , with quotients $V_{4}, Z /3, Z /2$ ), which is why formulas exist up to the quartic and stop there.

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Intuition. Picture solving by radicals as opening a combination lock whose dial you may turn only in "abelian" moves. Adjoining an $n$ th root is one such move: once the $n$ th roots of unity are around, the $n$ possible values of $n a$ differ by a root of unity, so the symmetries of that single step form a cyclic group, the most abelian thing there is.

A radical formula is therefore a finite stack of abelian moves, and the group it can reach is a group built out of abelian layers — a solvable group. The group $S_{5}$ contains $A_{5}$ , which is simple and non-abelian: a solid block with no seams to peel. No stack of abelian layers can be that block, so no formula can reach the roots. The impossibility is not about the difficulty of the algebra; it is about a structural feature of one particular group of order $60$ .

Exercises for §8.7 →

Common Mistakes

Running the correspondence the wrong way. Bigger subgroup means smaller fixed field. The whole group $G$ fixes only the base $F$ ; the trivial group fixes all of $K$ . If your answer has $G \leftrightarrow K$ , you have reversed Lemma Fixed sets are fields, and the two maps are mutually decreasing.
Counting subgroups of a non-Galois extension. The bijection of Theorem Fundamental theorem of Galois theory needs normal and separable. $Q (32) / Q$ has a trivial automorphism group and two intermediate fields; no contradiction, just a failed hypothesis.
Confusing "algebraic" with "finite". Every finite extension is algebraic, but $\overline{Q} / Q$ is algebraic of infinite degree, by Proposition Finite implies algebraic, but not conversely.
Using $de g f$ instead of $[K : F]$ for $∣ Gal ∣$ . The order of the Galois group is the degree of the splitting field, not the degree of the polynomial. For $x^{3} - 2$ the polynomial has degree $3$ but the group has order $6$ .
Assuming $Q (α, β)$ has degree $de g α \cdot de g β$ . That product is only an upper bound (Proposition The algebraic elements form a field). It is achieved when the degrees are coprime or the extensions are "independent", and fails when they overlap: $[Q (2, 8) : Q] = 2$ , not $4$ .
Reading Theorem Constructible numbers have degree a power of two as an equivalence. Degree $2^{k}$ is necessary, not sufficient; the Galois closure must also have $2$ -power degree.
Declaring all quintics unsolvable. Only those with non-solvable Galois group. $x^{5} - 2$ has a group of order $20$ and is perfectly solvable by radicals; $x^{5} - 6 x + 3$ has group $S_{5}$ and is not.
Forgetting separability in characteristic $p$ . Everything here assumed characteristic $0$ (or finite fields). In characteristic $p$ an extension such as $F_{p} (t^{1/ p}) / F_{p} (t)$ has degree $p$ and only the identity automorphism, and none of the counting works.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter