Course 311: Hilary Term 2006 Part IV: Introduction to Galois Theory

D. R. Wilkins

Copyright \begin{matrix}c \\ ◯ \end{matrix} David R. Wilkins 1997–2006

Contents

4 Introduction to Galois Theory 2 4.1 Polynomial Rings . . . . . . . . . . . . . . . . . . . . . . . . . 2 4.2 Gauss’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 5 4.3 Eisenstein’s Irreducibility Criterion . . . . . . . . . . . . . . . 6 4.4 Field Extensions and the Tower Law . . . . . . . . . . . . . . 6 4.5 Algebraic Field Extensions . . . . . . . . . . . . . . . . . . . . 8 4.6 Algebraically Closed Fields . . . . . . . . . . . . . . . . . . . . 11 4.7 Ruler and Compass Constructions . . . . . . . . . . . . . . . . 11 4.8 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . 16 4.9 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . 19 4.10 Separability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 4.11 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 4.12 The Primitive Element Theorem . . . . . . . . . . . . . . . . . 25 4.13 The Galois Group of a Field Extension . . . . . . . . . . . . . 25 4.14 The Galois correspondence . . . . . . . . . . . . . . . . . . . . 28 4.15 Quadratic Polynomials . . . . . . . . . . . . . . . . . . . . . . 30 4.16 Cubic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . 30 4.17 Quartic Polynomials . . . . . . . . . . . . . . . . . . . . . . . 32 4.18 The Galois group of the polynomial x^{4} − 2 . . . . . . . . . . . 34 4.19 The Galois group of a polynomial . . . . . . . . . . . . . . . . 35 4.20 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4.21 Solvable polynomials and their Galois groups . . . . . . . . . . 39 4.22 A quintic polynomial that is not solvable by radicals . . . . . 43

4 Introduction to Galois Theory

4.1 Polynomial Rings

Let R be a unital commutative ring. The set of all polynomials

c_{0} + c_{1}x + c_{2}x^{2} + ⋯ + c_{n}x^{n}

in an indeterminate x with coefficients c_{0}, …, c_{n} in the ring R themselves constitute a ring, which we shall denote by R[x]. If the coefficient c_{n} of highest power of x is non-zero then the polynomial is said to be of degree n, and the coefficient c_{n} is referred to as the leading coefficient of the polynomial. The polynomial is said to be monic if the leading coefficient c_{n} is equal to the multiplicative identity element 1 of the ring R.

Two polynomials with coefficients in the ring R are equal if and only if they are of the same degree and corresponding coefficients are equal. Polynomials may be added, subtracted and multiplied in the usual fashion.

We now consider various properties of polynomials whose coefficients belong to a field K (such as the field of rational numbers, real numbers or complex numbers).

Lemma 4.1 Let K be a field, and let f ∈ K[x] be a non-zero polynomial with coefficients in K. Then, given any polynomial h ∈ K[x], there exist unique polynomials q and r in K[x] such that h = fq + r and either r = 0 or else deg r < deg f.

Proof If deg h < deg f then we may take q = 0 and r = h. In general we prove the existence of q and r by induction on the degree deg h of h. Thus suppose that deg h ≥ deg f and that any polynomial of degree less than deg h can be expressed in the required form. Now there is some element c of K for which the polynomials h(x) and cf(x) have the same leading coefficient. Let h_{1}(x) = h(x) − cx^{m}f(x), where m = deg h − deg f. Then either h_{1} = 0 or deg h_{1} < deg h. The inductive hypothesis then ensures the existence of polynomials q_{1} and r such that h_{1} = fq_{1} + r and either r = 0 or else deg r < deg f. But then h = fq + r, where q(x) = cx^{m} + q_{1}(x). We now verify the uniqueness of q and r. Suppose that fq + r = f\overline{q} + \overline{r}, where \overline{q}, \overline{r} ∈ K[x] and either \overline{r} = 0 or deg \overline{r} < deg f. Then (q − \overline{q})f = r − \overline{r}. But deg((q − \overline{q})f) ≥ deg f whenever q ≠ \overline{q}, and deg(r − \overline{r}) < deg f whenever r ≠ \overline{r}. Therefore the equality (q − \overline{q})f = r − \overline{r} cannot hold unless q = \overline{q} and r = \overline{r}. This proves the uniqueness of q and r.

Any polynomial f with coefficients in a field K generates an ideal (f) of the polynomial ring K[x] consisting of all polynomials in K[x] that are divisible by f.

Lemma 4.2 Let K be a field, and let I be an ideal of the polynomial ring K[x]. Then there exists f ∈ K[x] such that I = (f), where (f) denotes the ideal of K[x] generated by f.

Proof If I = {0} then we can take f = 0. Otherwise choose f ∈ I such that f ≠ 0 and the degree of f does not exceed the degree of any non-zero polynomial in I. Then, for each h ∈ I, there exist polynomials q and r in K[x] such that h = fq + r and either r = 0 or else deg r < deg f. (Lemma 4.1). But r ∈ I, since r = h − fq and h and f both belong to I. The choice of f then ensures that r = 0 and h = qf. Thus I = (f).

Definition Polynomials f_{1}, f_{2}, …, f_{k} with coefficients in some field K. are said to be coprime if there is no non-constant polynomial that divides all of them.

Theorem 4.3 Let f_{1}, f_{2}, …, f_{k} be coprime polynomials with coefficients in some field K. Then there exist polynomials g_{1}, g_{2}, …, g_{k} with coefficients in K such that

f_{1}(x)g_{1}(x) + f_{2}(x)g_{2}(x) + ⋯ + f_{k}(x)g_{k}(x) = 1.

Proof Let I be the ideal in K[x] generated by f_{1}, f_{2}, …, f_{k}. It follows from Lemma 4.2 that the ideal I is generated by some polynomial d. Then d divides all of f_{1}, f_{2}, …, f_{k} and is therefore a constant polynomial, since these polynomials are coprime. It follows that I = K[x]. But the ideal I of K[x] generated by f_{1}, f_{2}, …, f_{k} coincides with the subset of K[x] consisting of all polynomials that may be represented as finite sums of the form

f_{1}(x)g_{1}(x) + f_{2}(x)g_{2}(x) + ⋯ + f_{k}(x)g_{k}(x)

for some polynomials g_{1}, g_{2}, …, g_{k}. It follows that the constant polynomial with value 1 may be expressed as a sum of this form, as required.

Definition A non-constant polynomial f with coefficients in a field K is said to be irreducible over K if it is not divisible by any non-constant polynomial of lower degree with coefficients in K.

Any polynomial with coefficients in a field K may be factored as a product of irreducible polynomials. This is easily proved by induction on the degree of the polynomial, for if a non-constant polynomial is not itself irreducible, then it can be factored as a product of polynomials of lower degree.

Lemma 4.4 Let K be a field. Then the ring K[x] of polynomials with coefficients in K contains infinitely many irreducible polynomials.

Proof Let f_{1}, f_{2}, …, f_{k} ∈ K[x] be irreducible polynomials, and let

g = f_{1}f_{2} ⋯ f_{k} + 1.

Then g is not divisible by f_{1}, f_{2}, …, f_{k}, and therefore no irreducible factor of g is divisible by any of f_{1}, f_{2}, …, f_{k}. It follows that K[x] must contain irreducible polynomials distinct from f_{1}, f_{2}, …, f_{k}. Thus the number of irreducible polynomials in K[x] cannot be finite.

The proof of Lemma 4.4 is a direct analogue of Euclid’s proof of the existence of infinitely many prime numbers.

Proposition 4.5 Let f, g and h be polynomials with coefficients in some field K. Suppose that f is irreducible over K and that f divides the product gh. Then either f divides g or else f divides h.

Proof Suppose that f does not divide g. We must show that f divides h. Now the only polynomials that divide f are constant polynomials and multiples of f. No multiple of f divides g. Therefore the only polynomials that divide both f and g are constant polynomials. Thus f and g are coprime. It follows from Proposition 4.3 that there exist polynomials u and v with coefficients in K such that 1 = ug + vf. Then h = ugh + vfh. But f divides ugh + vfh, since f divides gh. It follows that f divides h, as required.

Proposition 4.6 Let K be a field, and let (f) be the ideal of K[x] generated by an irreducible polynomial f with coefficients in K. Then K[x]/(f) is a field.

Proof Let I = (f). Then the quotient ring K[x]/I is commutative and has a multiplicative identity element I +1. Let g ∈ K[x]. Suppose that I +g ≠ I. Now the only factors of f are constant polynomials and constant multiples of f, since f is irreducible. But no constant multiple of f can divide g, since g ∉ I. It follows that the only common factors of f and g are constant polynomials. Thus f and g are coprime. It follows from Proposition 4.3 that there exist polynomials h, k ∈ K[x] such that fh + gk = 1. But then (I +k)(I +g) = I +1 in K[x]/I, since fh ∈ I. Thus I +k is the multiplicative inverse of I + g in K[x]/I. We deduce that every non-zero element of K[x]/I is invertible, and thus K[x]/I is a field, as required.

4.2 Gauss’s Lemma

We shall show that a polynomial with integer coefficients is irreducible over ℚ if and only if it cannot be expressed as a product of polynomials of lower degree with integer coefficients.

Definition A polynomial with integer coefficients is said to be primitive if there is no prime number that divides all the coefficients of the polynomial

Lemma 4.7 (Gauss’s Lemma) Let g and h be polynomials with integer coefficients. If g and h are both primitive then so is gh.

Proof Let g(x) = b_{0} + b_{1}x + b_{2}x^{2} + ⋯ + b_{r}x^{r} and h(x) = c_{0} + c_{1}x + c_{2}x^{2} + ⋯ + c_{s}x^{s}, and let g(x)h(x) = a_{0} + a_{1}x + a_{2}x^{2} + ⋯ + a_{r+s}x^{r+s}. Let p be a prime number. Then the polynomials g and h must both have at least one coefficient that is not divisible by p. Let j and k be the smallest values of i for which p does not divide b_{i} and c_{i} respectively. Then a_{j+k} −b_{j}c_{k} is divisible

by p, since a_{j+k} − b_{j}c_{k} = ∑_{i=0}^{j−1} b_{i}c_{j+k−i} + ∑_{i=0}^{k−1} b_{j+k−i}c_{i}, where p divides b_{i} for all

i < j and p divides c_{i} for all i < k. But p does not divide b_{j}c_{k} since p does not divide either b_{j} or c_{k}. Therefore p does not divide the coefficient a_{j+k} of gh. This shows that the polynomial gh is primitive, as required.

Proposition 4.8 A polynomial with integer coefficients is irreducible over the field ℚ of rational numbers if and only if it cannot be factored as a product of polynomials of lower degree with integer coefficients.

Proof Let f be a polynomial with integer coefficients. If f is irreducible over ℚ then f clearly cannot be factored as a product of polynomials of lower degree with integer coefficients. Conversely suppose that f cannot be factored in this way. Let f(x) = g(x)h(x), where g and h are polynomials with rational coefficients. Then there exist positive integers r and s such that the polynomials rg(x) and sh(x) have integer coefficients. Let the positive integers u and v be the highest common factors of the coefficients of the polynomials rg(x) and sh(x) respectively. Then rg(x) = ug_{∗}(x) and sh(x) = vh_{∗}(x), where g_{∗} and h_{∗} are primitive polynomials with integer coefficients. Then (rs)f(x) = (uv)g_{∗}(x)h_{∗}(x). We now show that f(x) = mg_{∗}(x)h_{∗}(x) for some integer m. Let l be the smallest divisor of rs such that lf(x) = mg_{∗}(x)h_{∗}(x) for some integer m. We show that l = 1. Suppose that it were the case that l > 1. Then there would exist a prime factor p of l. Now p could not divide m, since otherwise (l/p)f(x) = (m/p)g_{∗}(x)h_{∗}(x), which contradicts the definition of l. Theorefore p would have to divide each coefficient of g_{∗}(x)h_{∗}(x), which is impossible, since it follows from Gauss’s Lemma (Lemma 4.7) that the product g_{∗}h_{∗} of the primitive polynomials g_{∗} and h_{∗} is itself a primitive polynomial. Therefore l = 1 and f(x) = mg_{∗}(x)h_{∗}(x). Now f does not factor as a product of polynomials of lower degree with integer coefficients. Therefore either deg f = deg g_{∗} = deg g, or else deg f = deg h_{∗} = deg h, Thus f is irreducible over ℚ, as required.

4.3 Eisenstein’s Irreducibility Criterion

Proposition 4.9 (Eisenstein’s Irreducibility Criterion) Let

f(x) = a_{0} + a_{1}x + a_{2}x^{2} + ⋯ + a_{n}x^{n}

be a polynomial of degree n with integer coefficients, and let p be a prime number. Suppose that

• p does not divide a_{n},

• p divides a_{0}, a_{1}, …, a_{n−1},

• p^{2} does not divide a_{0}.

Then the polynomial f is irreducible over the field ℚ of rational numbers.

Proof Suppose that f(x) = g(x)h(x), where g and h are polynomials with integer coefficients. Let g(x) = b_{0} + b_{1}x + b_{2}x^{2} + ⋯ + b_{r}x^{r} and h(x) = c_{0} +c_{1}x+c_{2}x^{2} +⋯+c_{s}x^{s}. Then a_{0} = b_{0}c_{0}. Now a_{0} is divisible by p but is not divisible by p^{2}. Therefore exactly one of the coefficients b_{0} and c_{0} is divisible by p. Suppose that p divides b_{0} but does not divide c_{0}. Now p does not divide all the coefficients of g(x), since it does not divide all the coefficients of f(x). Let j be the smallest value of i for which p does not divide b_{i}. Then p divides

a_{j} − b_{j}c_{0}, since a_{j} − b_{j}c_{0} = ∑_{i=0}^{j−1} b_{i}c_{j−i} and b_{i} is divisible by p when i < j. But

b_{j}c_{0} is not divisible by p, since p is prime and neither b_{j} nor c_{0} is divisible by p. Therefore a_{j} is not divisible by p, and hence j = n and deg g ≥ n = deg f. Thus deg g = deg f and deg h = 0. Thus the polynomial f does not factor as a product of polynomials of lower degree with integer coefficients, and therefore f is irreducible over ℚ (Proposition 4.8).

4.4 Field Extensions and the Tower Law

Let K be a field. An extension L: K of K is an embedding of K in some larger field L.

Definition Let L: K and M: K be field extensions. A K-homomorphism θ: L → M is a homomorphism of fields which satisfies θ(a) = a for all a ∈ K. A K-monomorphism is an injective K-homomorphism. A K-isomorphism is a bijective K-homomorphism. A K-automorphism of L is a K-isomorphism mapping L onto itself.

Two extensions L_{1}: K and L_{2}: K of a field K are said to be K-isomorphic (or isomorphic) if there exists a K-isomorphism φ: L_{1} → L_{2} between L_{1} and L_{2}.

If L: K is a field extension then we can regard L as a vector space over the field K. If L is a finite-dimensional vector space over K then we say that the extension L: K is finite. The degree [L: K] of a finite field extension L: K is defined to be the dimension of L considered as a vector space over K.

Proposition 4.10 (The Tower Law) Let M: L and L: K be field extensions. Then the extension M: K is finite if and only if M: L and L: K are both finite, in which case [M: K] = [M: L][L: K].

Proof Suppose that M: K is a finite field extension. Then L, regarded as a vector space over K, is a subspace of the finite-dimensional vector space M, and therefore L is itself a finite-dimensional vector space over K. Thus L: K is finite. Also there exists a finite subset of M which spans M as a vector space over K, since M: K is finite, and this finite subset must also span M over L, and thus M: L must be finite.

Conversely suppose that M: L and L: K are both finite extensions. Let x_{1}, x_{2}, …, x_{m} be a basis for L, considered as a vector space over the field K, and let y_{1}, y_{2}, …, y_{n} be a basis for M, considered as a vector space over the field L. Note that m = [L: K] and n = [M: L]. We claim that the set of all products x_{i}y_{j} with i = 1, 2, …, m and j = 1, 2, …, n is a basis for M, considered as a vector space over K.

First we show that the elements x_{i}y_{j} are linearly independent over K.

Suppose that ∑_{i=1}^{m} ∑_{j=1}^{n} λ_{ij}x_{i}y_{j} = 0, where λ_{ij} ∈ K for all i and j. Then

∑_{i=1}^{m} λ_{ij}x_{i} ∈ L for all j, and y_{1}, y_{2}, …, y_{n} are linearly independent over L,

and therefore ∑_{i=1}^{m} λ_{ij}x_{i} = 0 for j = 1, 2, …, n. But x_{1}, x_{2}, …, x_{m} are linearly

independent over K. It follows that λ_{ij} = 0 for all i and j. This shows that the elements x_{i}y_{j} are linearly independent over K.

Now y_{1}, y_{2}, …, y_{n} span M as a vector space over L, and therefore any

element z of M can be written in the form z = ∑_{j=1}^{n} μ_{j}y_{j}, where μ_{j} ∈ L for all j. But each μ_{j} can be written in the form μ_{j} = ∑_{i=1}^{m} λ_{ij}x_{i}, where λ_{ij} ∈ K

for all i and j. But then z = ∑_{i=1}^{m} ∑_{j=1}^{n} λ_{ij}x_{i}y_{j}. This shows that the products

x_{i}y_{j} span M as a vector space over K, and thus

{x_{i}y_{j} : 1 ≤ i ≤ m and 1 ≤ j ≤ n}

is a basis of M, considered as a vector space over K. We conclude that the extension M: K is finite, and

[M: K] = mn = [M: L][L: K],

as required.

Let L: K be a field extension. If A is any subset of L, then the set K ∪ A generates a subfield K(A) of L which is the intersection of all subfields of L that contain K ∪ A. (Note that any intersection of subfields of L is itself a subfield of L.) We say that K(A) is the field obtained from K by adjoining the set A.

We denote K({α_{1}, α_{2}, …, α_{k}}) by K(α_{1}, α_{2}, …, α_{k}) for any finite subset {α_{1}, α_{2}, …, α_{k}} of L. In particular K(α) denotes the field obtained by adjoining some element α of L to K. A field extension L: K is said to be simple if there exists some element α of L such that L = K(α).

4.5 Algebraic Field Extensions

Definition Let L: K be a field extension, and let α be an element of L. If there exists some non-zero polynomial f ∈ K[x] with coefficients in K such that f(α) = 0, then α is said to be algebraic over K; otherwise α is said to be transcendental over K. A field extension L: K is said to be algebraic if every element of L is algebraic over K.

Lemma 4.11 A finite field extension is algebraic.

Proof Let L: K be a finite field extension, and let n = [L: K]. Let α ∈ L. Then either the elements 1, α, α^{2}, …, α^{n} are not all distinct, or else these elements are linearly dependent over the field K (since a linearly independent subset of L can have at most n elements.) Therefore there exist c_{0}, c_{1}, c_{2}, …, c_{n} ∈ K, not all zero, such that

c_{0} + c_{1}α + c_{2}α^{2} + ⋯ + c_{n}α^{n} = 0.

Thus α is algebraic over K. This shows that the field extension L: K is algebraic, as required.

Definition A polynomial f with coefficients in some field or unital ring is said to be monic if its leading coefficient (i.e., the coefficient of the highest power of x occurring in f(x) with a non-zero coefficient) is equal to 1.

Lemma 4.12 Let K be a field and let α be an element of some extension field L of K. Suppose that α is algebraic over K. Then there exists a unique irreducible monic polynomial m ∈ K[x], with coefficients in K, characterized by the following property: f ∈ K[x] satisfies f(α) = 0 if and only if m divides f in K[x].

Proof Let I = {f ∈ K[x] : f(α) = 0}. Then I is a non-zero ideal of K[x]. Now there exists some polynomial m with coefficients in K which generates the ideal I (Lemma 4.2). Moreover, by dividing m by its leading coefficient, if necessary, we can ensure that m is a monic polynomial. Then f ∈ K[x] satisfies f(α) = 0 if and only if m divides f.

Suppose that m = gh where g, h ∈ K[x]. Then 0 = m(α) = g(α)h(α). But then either g(α) = 0, in which case m divides g, or else h(α) = 0, in which case m divides h. The polynomial m is thus irreducible over K.

The polynomial m is uniquely determined since if some monic polynomial \overline{m} also satisfies the required conditions then m and \overline{m} divide one another and therefore \overline{m} = m.

Definition Let K be a field and let L be an extension field of K. Let α be an element of L that is algebraic over K. The minimum polynomial m of α over K is the unique irreducible monic polynomial m ∈ K[x] with coefficients in K characterized by the following property: f ∈ K[x] satisfies f(α) = 0 if and only if m divides f in K[x].

Note that if f ∈ K[x] is an irreducible monic polynomial, and if α is a root of f in some extension field L of K, then f is the minimum polynomial of α over K.

Theorem 4.13 A simple field extension K(α): K is finite if and only if α is algebraic over K, in which case [K(α): K] is the degree of the minimum polynomial of α over K.

Proof Suppose that the field extension K(α): K is finite. It then follows from Lemma 4.11 that α is algebraic over K.

Conversely suppose that α is algebraic over K. Let R = {f(α) : f ∈ K[x]}. Now f(α) = 0 if and only if the minimum polynomial m of α over K divides f. It follows that f(α) = 0 if and only if f ∈ (m), where (m) is the ideal of K[x] generated by m. The ring homomorphism from K[x] to R that sends f ∈ K[x] to f(α) therefore induces an isomorphism between the quotient ring K[x]/(m) and the ring R. But K[x]/(m) is a field, since m is irreducible (Proposition 4.6). Therefore R is a subfield of K(α) containing K ∪ {α}, and hence R = K(α).

Let z ∈ K(α). Then z = g(α) for some g ∈ K[x]. But then there exist polynomials l and f belonging to K[x] such that g = lm+f and either f = 0 or deg f < deg m (Lemma 4.1). But then z = f(α) since m(α) = 0.

Suppose that z = h(α) for some polynomial h ∈ K[x], where either h = 0 or deg h < deg m. Then m divides h−f, since α is a zero of h−f. But if h−f were non-zero then its degree would be less than that of m, and thus h − f would not be divisible by m. We therefore conclude that h = f. Thus any element z of K(α) can be expressed in the form z = f(α) for some uniquely determined polynomial f ∈ K[x] satisfying either f = 0 or deg f < deg m. Thus if n = deg m then 1, α, α^{2} …, α^{n−1} is a basis of K(α) over K. It follows that the extension K(α): K is finite and [K(α): K] = deg m, as required.

Corollary 4.14 A field extension L: K is finite if and only if there exists a finite subset {α_{1}, α_{2}, …, α_{k}} of L such that α_{i} is algebraic over K for i = 1, 2, …, k and L = K(α_{1}, α_{2}, …, α_{k}).

Proof Suppose that the field extension L: K is a finite. Then it is algebraic (Lemma 4.11). Thus if {α_{1}, α_{2}, …, α_{k}} is a basis for L, considered as a vector space over K, then each α_{i} is algebraic and L = K(α_{1}, α_{2}, …, α_{k}).

Conversely suppose that L = K(α_{1}, α_{2}, …, α_{k}), where α_{i} is algebraic over K for i = 1, 2, …, k. Let K_{i} = K(α_{1}, α_{2}, …, α_{i}) for i = 1, 2, …, k. Clearly K_{i−1}(α_{i}) ⊂ K_{i} for all i > 1, since K_{i−1} ⊂ K_{i} and α_{i} ∈ K_{i}. Also K_{i} ⊂ K_{i−1}(α_{i}), since K_{i−1}(α_{i}) is a subfield of L containing K ∪ {α_{1}, α_{2}, …, α_{i}} We deduce that K_{i} = K_{i−1}(α_{i}) for i = 2, 3, …, k. Moreover α_{i} is clearly algebraic over K_{i−1} since it is algebraic over K, and K ⊂ K_{i−1}. It follows from Theorem 4.13 that the field extension K_{i}: K_{i−1} is finite for each i. Using the Tower Law (Proposition 4.10), we deduce that L: K is a finite extension, as required.

Corollary 4.15 Let M: L and L: K be algebraic field extensions. Then M: K is an algebraic field extension.

Proof Let α be an element of M. We must show that α is algebraic over K. Now there exists some non-zero polynomial f ∈ L[x] with coefficients in L such that f(α) = 0, since M: L is algebraic. Let β_{1}, β_{2}, …, β_{k} be the coefficients of f(x), and let L_{0} = K(β_{1}, β_{2}, …, β_{k}). Now each β_{i} is algebraic over K (since L: K is algebraic). Thus L_{0}: K is finite. Moreover α is algebraic over L_{0}, since the coefficients of the polynomial f belong to L_{0}, and thus L_{0}(α): L_{0} is finite (Theorem 4.13). It follows from the Tower Law (Proposition 4.10) that L_{0}(α): K is finite. But then K(α): K is finite, and hence α is algebraic over K, as required.

4.6 Algebraically Closed Fields

Definition A field K is said to be algebraically closed if, given any nonconstant polynomial f ∈ K[x] with coefficients in K, there exists some α ∈ K satisfying f(α) = 0.

The field ℂ of complex numbers is algebraically closed. This result is the Fundamental Theorem of Algebra.

Lemma 4.16 Let K be an algebraically closed field, and let L: K be an algebraic extension of K. Then L = K.

Proof Let α ∈ L, and let m_{α} ∈ K[x] be the minimal polynomial of α over K. Then the polynomial m_{α}(x) has a root a in K, and is therefore divisible by the polynomial x − a. It follows that m_{α}(x) = x − a, since m_{α}(x) is an irreducible monic polynomial. But then α = a, and therefore α ∈ K. This shows that every element of L belongs to K, and thus L = K, as required.

4.7 Ruler and Compass Constructions

One can make use of the Tower Law in order to prove the impossibility of performing a number of geometric constructions in a finite number of steps using straightedge and and compasses alone. These impossible constructions include the following:

• the trisection of an arbitrary angle;

• the construction of the edge of a cube having twice the volume of some given cube;

• the construction of a square having the same area as a given circle.

Definition Let P_{0} and P_{1} be the points of the Euclidean plane given by P_{0} = (0, 0) and P_{1} = (1, 0). We say that a point P of the plane is constructible using straightedge and compasses alone if P = P_{n} for some finite sequence P_{0}, P_{1}, …, P_{n} of points of the plane, where P_{0} = (0, 0), P_{1} = (1, 0) and, for each j > 1, the point P_{j} is one of the following:— • the intersection of two distinct straight lines, each passing through at

least two points belonging to the set {P_{0}, P_{1}, …, P_{j−1}};

• the point at which a straight line joining two points belonging to the set {P_{0}, P_{1}, …, P_{j−1}} intersects a circle which is centred on a point of this set and passes through another point of the set;

• the point of intersection of two distinct circles, where each circle is centred on a point of the set {P_{0}, P_{1}, …, P_{j−1}} and passes through another point of the set.

Constructible points of the plane are those that can be constructed from the given points P_{0} and P_{1} using straightedge (i.e., unmarked ruler) and compasses alone.

Theorem 4.17 Let (x, y) be a constructible point of the Euclidean plane. Then [ℚ(x, y): ℚ] = 2^{r} for some non-negative integer r.

Proof Let P = (x, y) and let P_{0}, P_{1}, …, P_{n} be a finite sequence of points of the plane with the properties listed above. Let K_{0} = K_{1} = ℚ and K_{j} = K_{j−1}(x_{j}, y_{j}) for j = 2, 3, …, n, where P_{j} = (x_{j}, y_{j}). Straightforward coordinate geometry shows that, for each j, the real numbers x_{j} and y_{j} are both roots of linear or quadratic polynomials with coefficients in K_{j−1}. It follows that [K_{j−1}(x_{j}): K_{j−1}] = 1 or 2 and [K_{j−1}(x_{j}, y_{j}): K_{j−1}(x_{j})] = 1 or 2 for each j. It follows from the Tower Law (Proposition 4.10) that [K_{n}: ℚ] = 2^{s} for some non-negative integer s. But [K_{n}: ℚ] = [K_{n}: ℚ(x, y)][ℚ(x, y): ℚ]. We deduce that [ℚ(x, y): ℚ] divides 2^{s}, and therefore [ℚ(x, y): ℚ] = 2^{r} for some non-negative integer r.

One can apply this criterion to show that there is no geometrical construction that enables one to trisect an arbitrary angle using straightedge and compasses alone. The same method can be used to show the impossibility of ‘duplicating a cube’ or ‘squaring a circle’ using straightedge and compasses alone.

Example We show that there is no geometrical construction for the trisection of an angle of \frac{π}{3} radians (i.e., 60^{∘}) using straightedge and compasses alone. Let a = cos \frac{π}{9} and b = sin \frac{π}{9}. Now the point (cos \frac{π}{3}, sin \frac{π}{3}) (i.e, the point (\frac{1}{2}, \frac{1}{2} \sqrt{3})) is constructible. Thus if an angle of \frac{π}{3} radians could be trisected using straightedge and compasses alone, then the point (a, b) would be constructible. Now

cos 3θ = cos θ cos 2θ − sin θ sin 2θ = cos θ(cos^{2} θ − sin^{2} θ) − 2 sin^{2} θ cos θ
= 4 cos^{3} θ − 3 cos θ

for any angle θ. On setting θ = \frac{π}{9} we deduce that 4a^{3} − 3a = \frac{1}{2} and thus 8a^{3} − 6a − 1 = 0. Now 8a^{3} − 6a − 1 = f(2a − 1), where f(x) = x^{3} + 3x^{2} − 3. An immediate application of Eisenstein’s criterion for irreducibility shows that the polynomial f is irreducible over the field ℚ of rational numbers, and thus [ℚ(a): ℚ] = [ℚ(2a − 1): ℚ] = 3. It now follows from Theorem 4.17 that the point (cos \frac{π}{9}, sin \frac{π}{9}) is not constructible using straightedge and compasses alone. Therefore it is not possible to trisect an angle of \frac{π}{3} radians using straightedge and compasses alone. It follows that there is no geometrical construction for the trisection of an arbitrary angle using straightedge and compasses alone.

Example It is not difficult to see that if it were possible to construct two points in the plane a distance ^{3} \sqrt{2} apart, then the point ( ^{3} \sqrt{2}, 0) would be constructible. But it follows from Theorem 4.17 that this is impossible, since ^{3} \sqrt{2} is a root of the irreducible monic polynomial x^{3} − 2, and therefore [ℚ( ^{3} \sqrt{2}), ℚ] = 3. We conclude that there is no geometric construction using straightedge and compasses alone that will construct from a line segment in the plane a second line segment such that a cube with the second line segment as an edge will have twice the volume of a cube with the first line segment as an edge.

Example It can be shown that π is not algebraic over the field ℚ of rational numbers. Therefore \sqrt{π} is not algebraic over ℚ. It then follows from Theorem 4.17 it is not possible to give a geometrical construction for obtaining a square with the same area as a given circle, using straightedge and compasses alone. (Thus it is not possible to ‘square the circle’ using straightedge and compasses alone.)

Lemma 4.18 If the endpoints of any line segment in the plane are constructible, then so is the midpoint.

Proof Let P and Q be constructible points in the plane. Let S and T be the points where the circle centred on P and passing through Q intersects the circle centred on Q and passing through P. Then S and T are constructible points in the plane, and the point R at which the line ST intersects the line P Q is the midpoint of the line segment P Q. Thus this midpoint is a constructible point.

Lemma 4.19 If any three vertices of a parallelogram in the plane are constructible, then so is the fourth vertex.

Proof Let the vertices of the parallelogram listed in anticlockwise (or in clockwise) order be A, B, C and D, where A, B and D are constructible points. We must show that C is also constructible. Now the midpoint E of the line segment BD is a constructible point, and the circle centred on E and passing though A will intersect the line AE in the point C. Thus C is a constructible point, as required.

Theorem 4.20 Let 𝕂 denote the set of all real numbers x for which the point (x, 0) is constructible using straightedge and compasses alone. Then 𝕂 is a subfield of the field of real numbers, and a point (x, y) of the plane is constructible using straightedge and compass alone if and only if x ∈ 𝕂 and y ∈ 𝕂. Moreover if x ∈ 𝕂 and x > 0 then \sqrt{x} ∈ 𝕂.

Proof Clearly 0 ∈ 𝕂 and 1 ∈ 𝕂. Let x and y be real numbers belonging to 𝕂. Then (x, 0) and (y, 0) are constructible points of the plane. Let M be the midpoint of the line segment whose endpoints are (x, 0) and (y, 0). Then M is constructible (Lemma 4.18), and M = (\frac{1}{2}(x + y), 0). The circle centred on M and passing through the origin intersects the x-axis at the origin and at the point (x + y, 0). Therefore (x + y, 0) is a constructible point, and thus x + y ∈ 𝕂. Also the circle centred on the origin and passing through (x, 0) intersects the x-axis at (−x, 0). Thus (−x, 0) is a constructible point, and thus −x ∈ 𝕂.

We claim that if x ∈ 𝕂 then the point (0, x) is constructible. Now if x ∈ 𝕂 and x ≠ 0 then (x, 0) and (−x, 0) are constructible points, and the circle centred on (x, 0) and passing through (−x, 0) intersects the circle centred on (−x, 0) and passing through (x, 0) in two points that lie on the y-axis. These two points (namely (0, \sqrt{3}x) and (0, −\sqrt{3}x)) are constructible, and therefore the circle centred on the origin and passing though (x, 0) intersects the y-axis in two constructible points which are (0, x) and (0, −x). Thus if x ∈ 𝕂 then the point (0, x) is constructible.

Let x and y be real numbers belonging to 𝕂. Then the points (x, 0), (0, y) and (0, 1) are constructible. The point (x, y − 1) is then constructible, since it is the fourth vertex of a parallelogram which has three vertices at the constructible points (x, 0), (0, y) and (0, 1) (Lemma 4.19). But the line which passes through the two constructible points (0, y) and (x, y − 1) intersects the x-axis at the point (xy, 0). Therefore the point (xy, 0) is constructible, and thus xy ∈ 𝕂.

Now suppose that x ∈ 𝕂, y ∈ 𝕂 and y ≠ 0. The point (x, 1 − y) is constructible, since it is the fourth vertex of a parallelogram with vertices at the constructible points (x, 0), (0, y) and (0, 1). The line segment joining the constructible points (0, 1) and (x, 1−y) intersects the x-axis at the point (xy^{−1}, 0). Thus xy^{−1} ∈ 𝕂.

The above results show that 𝕂 is a subfield of the field of real numbers. Moreover if x ∈ 𝕂 and y ∈ 𝕂 then the point (x, y) is constructible, since it is the fourth vertex of a rectangle with vertices at the constructible points (0, 0), (x, 0) and (0, y). Conversely, suppose that the point (x, y) is constructible. We claim that the point (x, 0) is constructible and thus x ∈ 𝕂. This result is obviously true if y = 0. If y ≠ 0 then the circles centred on the points (0, 0) and (1, 0) and passing through (x, y) intersect in the two points (x, y) and (x, −y). The point (x, 0) is thus the point at which the line passing through the constructible points (x, y) and (x, −y) intersects the x-axis, and is thus itself constructible. The point (0, y) is then the fourth vertex of a rectangle with vertices at the constructible points (0, 0), (x, 0) and (x, y), and thus is itself constructible. The circle centred on the origin and passing though (0, y) intersects the x-axis at (y, 0). Thus (y, 0) is constructible, and thus y ∈ 𝕂. We have thus shown that a point (x, y) is constructible using straightedge and compasses alone if and only if x ∈ 𝕂 and y ∈ 𝕂.

Suppose that x ∈ 𝕂 and that x > 0. Then \frac{1}{2}(1 − x) ∈ 𝕂. Thus if C = (0, \frac{1}{2}(1 − x)) then C is a constructible point. Let (u, 0) be the point at which the circle centred on C and passing through the constructible point (0, 1) intersects the x-axis. (The circle does intersect the x-axis since it passes through (0, 1) and (0, −x), and x > 0.) The radius of this circle is \frac{1}{2}(1 + x)), and therefore \frac{1}{4}(1 − x)^{2} + u^{2} = \frac{1}{4}(1 + x)^{2} (Pythagoras’ Theorem.) But then u^{2} = x. But (u, 0) is a constructible point. Thus if x ∈ 𝕂 and x > 0 then \sqrt{x} ∈ 𝕂, as required.

The above theorems can be applied to the problem of determining whether or not it is possible to construct a regular n-sided polygon with a straightedge and compass, given its centre and one of its vertices. The impossibility of trisecting an angle of 60^{∘} shows that a regular 18-sided polygon is not constructible using straightedge and compass. Now if one can construct a regular n-sided polygon then one can easily construct a regular 2n-sided polygon by bisecting the angles of the n-sided polygon. Thus the problem reduces to that of determining which regular polygons with an odd number of sides are constructible. Moreover it is not difficult to reduce down to the case where n is a power of some odd prime number.

Gauss discovered that a regular 17-sided polygon was constructible in 1796, when he was 19 years old. Techniques of Galois Theory show that the regular n-sided polygon is constructible using straightedge and compass if and only if n = 2^{s}p_{1}p_{2} ⋯ p_{t}, where p_{1}, p_{2}, …, p_{t} are distinct Fermat primes: a Fermat prime is a prime number that is of the form 2^{k}+1 for some integer k.

If k = uv, where u and v are positive integers and v is odd, then 2^{k} + 1 = w^{v} + 1 = (w + 1)(w^{v−1} − w^{v−2} + ⋯ − w + 1), where w = 2^{u}, and hence 2^{k} + 1 is not prime. Thus any Fermat prime is of the form 2^{2^{m}} + 1 for some non-negative integer m. Fermat observed in 1640 that F_{m} is prime when m ≤ 4. These Fermat primes have the values F_{0} = 3, F_{1} = 5, F_{2} = 17, F_{3} = 257 and F_{4} = 65537. Fermat conjectured that all the numbers F_{m} were prime. However it has been shown that F_{m} is not prime for any integer m between 5 and 16. Moreover F_{16} = 2^{65536} + 1 ≈ 10^{20000}. Note that the five Fermat primes 3, 5, 17, 257 and 65537 provide only 32 constructible regular polygons with an odd number of sides.

It is not difficult to see that the geometric problem of constructing a regular n-sided polygon using straightedge and compasses is equivalent to the algebraic problem of finding a formula to express the nth roots of unity in the complex plane in terms of integers or rational numbers by means of algebraic formulae which involve finite addition, subtraction, multiplication, division and the successive extraction of square roots. Thus the problem is closely related to that of expressing the roots of a given polynomial in terms of its coefficients by means of algebraic formulae which involve only finite addition, subtraction, multiplication, division and the successive extraction of pth roots for appropriate prime numbers p.

4.8 Splitting Fields

Definition Let L: K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. The polynomial f is said to split over L if f is a constant polynomial or if there exist elements α_{1}, α_{2}, …, α_{n} of L such that

f(x) = c(x − α_{1})(x − α_{2}) ⋯ (x − α_{n}),

where c ∈ K is the leading coefficient of f.

We see therefore that a polynomial f ∈ K[x] splits over an extension field L of K if and only if f factors in L[x] as a product of constant or linear factors.

Definition Let L: K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. The field L is said to be a splitting field for f over K if the following conditions are satisfied:—

• the polynomial f splits over L;

• the polynomial f does not split over any proper subfield of L that contains the field K.

Lemma 4.21 Let M: K be a field extension, and let f ∈ K[x] be a polynomial with coefficients in K. Suppose that the polynomial f splits over M. Then there exists a unique subfield L of M which is a splitting field for f over K.

Proof Let L be the intersection of all subfields M′ of M containing K with the property that the polynomial f splits over M′. One can readily verify that L is the unique splitting field for f over K contained in M.

The Fundamental Theorem of Algebra ensures that a polynomial f ∈ ℚ[x] with rational coefficients always splits over the field ℂ of complex numbers. Thus some unique subfield L of ℂ is a splitting field for f over ℚ.

Note that if the polynomial f ∈ K[x] splits over an extension field M of K, and if α_{1}, α_{2}, …, α_{n} are the roots of the polynomial f in M, then the unique splitting field of f over K contained in M is the field K(α_{1}, α_{2}, …, α_{n}) obtaining on adjoining the roots of f to K.

Example The field ℚ(\sqrt{2}) is a splitting field for the polynomial x^{2} − 2 over ℚ.

We shall prove below that splitting fields always exist and that any two splitting field extensions for a given polynomial over a field K are isomorphic.

Given any homomorphism σ: K → M of fields, we define

σ_{∗}(a_{0} + a_{1}x + ⋯ + a_{n}x^{n}) = σ(a_{0}) + σ(a_{1})x + ⋯ + σ(a_{n})x^{n}

for all polynomials a_{0} + a_{1}x + ⋯ + a_{n}x^{n} with coefficients in K. Note that σ_{∗}(f + g) = σ_{∗}(f) + σ_{∗}(g) and σ_{∗}(fg) = σ_{∗}(f)σ_{∗}(g) for all f, g ∈ K[x].

Theorem 4.22 (Kronecker) Let K be a field, and let f ∈ K[x] be a nonconstant polynomial with coefficients in K. Then there exists an extension field L of K and an element α of L for which f(α) = 0.

Proof Let g be an irreducible factor of f, and let L = K[x]/(g), where (g) is the ideal of K[x] generated by g. For each a ∈ K let i(a) = a + (g). Then i: K → L is a monomorphism. We embed K in L on identifying a ∈ K with i(a).

Now L is a field, since g is irreducible (Proposition 4.6). Let α = x + (g). Then g(α) is the image of the polynomial g under the quotient homomorphism from K[x] to L, and therefore g(α) = 0. But g is a factor of the polynomial f. Therefore f(α) = 0, as required.

Corollary 4.23 Let K be a field and let f ∈ K[x]. Then there exists a splitting field for f over K.

Proof We use induction on the degree deg f of f. The result is trivially true when deg f = 1 (since f then splits over K itself). Suppose that the result holds for all fields and for all polynomials of degree less than deg f. Now it follows from Theorem 4.22 that there exists a field extension K_{1}: K of K and an element α of K_{1} satisfying f(α) = 0. Moreover f(x) = (x − α)g(x) for some polynomial g with coefficients in K(α). Now deg g < deg f. It follows from the induction hypothesis that there exists a splitting field L for g over K(α). Then f splits over L.

Suppose that f splits over some field M, where K ⊂ M ⊂ L. Then α ∈ M and hence K(α) ⊂ M. But M must also contain the roots of g, since these are roots of f. It follows from the definition of splitting fields that M = L. Thus L is the required splitting field for the polynomial f over K.

Any two splitting fields for a given polynomial with coefficients in a field K are K-isomorphic. This result is a special case of the following theorem.

Theorem 4.24 Let K_{1} and K_{2} be fields, and let σ: K_{1} → K_{2} be an isomorphism between K_{1} and K_{2}. Let f ∈ K_{1}[x] be a polynomial with coefficients in K_{1}, and let L_{1} and L_{2} be splitting fields for f and σ_{∗}(f) over K_{1} and K_{2} respectively. Then there exists an isomorphism τ: L_{1} → L_{2} which extends σ: K_{1} → K_{2}.

Proof We prove the result by induction on [L_{1}: K_{1}]. The result is trivially true when [L_{1}: K_{1}] = 1. Suppose that [L_{1}: K_{1}] > 1 and the result holds for splitting field extensions of lower degree. Choose a root α of f in L_{1}\K_{1}, and let m be the minimum polynomial of α over K_{1}. Then m divides f and σ_{∗}(m) divides σ_{∗}(f), and therefore σ_{∗}(m) splits over L_{2}. Moreover the polynomial σ_{∗}(m) is irreducible over K_{2}, since σ: K_{1} → K_{2} induces an isomorphism between the polynomial rings K_{1}[x] and K_{2}[x]. Choose a root β of σ_{∗}(m).

Let g and h be polynomials with coefficients in K_{1}. Now g(α) = h(α) if and only if m divides g − h. Similarly σ_{∗}(g)(β) = σ_{∗}(h)(β) if and only if σ_{∗}(m) divides σ_{∗}(g) − σ_{∗}(h). Therefore σ_{∗}(g)(β) = σ_{∗}(h)(β) if and only if g(α) = h(α), and thus there is a well-defined isomorphism φ: K_{1}(α) → K_{2}(β) which sends g(α) to σ_{∗}(g)(β) for any polynomial g with coefficients in K.

Now L_{1} and L_{2} are splitting fields for the polynomials f and σ_{∗}(f) over the fields K_{1}(α) and K_{2}(β) respectively, and [L_{1}: K_{1}(α)] < [L_{1}: K_{1}]. The induction hypothesis therefore ensures the existence of an isomorphism τ: L_{1} → L_{2} extending φ: K_{1}(α) → K_{2}(β). Then τ: L_{1} → L_{2} is the required extension of σ: K_{1} → K_{2}.

Corollary 4.25 Let L: K be a splitting field extension, and let α and β be elements of L. Then there exists a K-automorphism of L sending α to β if and only if α and β have the same minimum polynomial over K.

Proof Suppose that there exists a K-automorphism σ of L which sends α to β. Then h(β) = σ(h(α)) for all polynomials h ∈ K[x] with coefficients in K. Therefore h(α) = 0 if and only if h(β) = 0. It follows that α and β must have the same minimum polynomial over K.

Conversely suppose that α and β are elements of L that have the same minimum polynomial m over K. Let h_{1} and h_{2} be polynomials with coefficients in K. Now h_{1}(α) = h_{2}(α) if and only if h_{1} − h_{2} is divisible by the minimum polynomial m. It follows that h_{1}(α) = h_{2}(α) if and only if h_{1}(β) = h_{2}(β). Therefore there is a well-defined K-isomorphism φ: K(α) → K(β) that sends h(α) to h(β) for all polynomials h with coefficients in K. Then φ(α) = β.

Now L is the splitting field over K for some polynomial f with coefficients in K. The field L is then a splitting field for f over both K(α) and K(β). It follows from Theorem 4.24 that the K-isomorphism φ: K(α) → K(β) extends to a K-automorphism τ of L that sends α to β, as required.

4.9 Normal Extensions

Definition A field extension L: K is said to be normal if every irreducible polynomial in K[x] with at least one root in L splits over L.

Note that a field extension L: K is normal if and only if, given any element α of L, the minimum polynomial of α over K splits over L.

Theorem 4.26 Let K be a field, and let L be an extension field of K. Then L is a splitting field over K for some polynomial with coefficients in K if and only if the field extension L: K is both finite and normal.

Proof Suppose that L: K is both finite and normal. Then there exist algebraic elements α_{1}, α_{2}, …, α_{n} of L such that L = K(α_{1}, α_{2}, …, α_{n}) (Corollary 4.14). Let f(x) = m_{1}(x)m_{2}(x) ⋯ m_{n}(x), where m_{j} ∈ K[x] is the minimum polynomial of α_{j} over K for j = 1, 2, …, n. Then m_{j} splits over L since m_{j} is irreducible and L: K is normal. Thus f splits over L. It follows that L is a splitting field for f over K, since L is obtained from K by adjoining roots of f.

Conversely suppose that L is a splitting field over K for some polynomial f ∈ K[x]. Then L is obtained from K by adjoining the roots of f, and therefore the extension L: K is finite. (Corollary 4.14).

Let g ∈ K[x] be irreducible, and let M be a splitting field for the polynomial fg over L. Then L ⊂ M and the polynomials f and g both split over M. Let β and γ be roots of g in M. Now the polynomial f splits over the fields L(β) and L(γ). Moreover if f splits over any subfield of M containing K(β) then that subfield must contain L (since L is a splitting field for f over K) and thus must contain L(β). We deduce that L(β) is a splitting field for f over K(β). Similarly L(γ) is a splitting field for f over K(γ).

Now there is a well-defined K-isomorphism σ: K(β) → K(γ) which sends h(β) to h(γ) for all polynomials h with coefficients in K, since two such polynomials h_{1} and h_{2} take the same value at a root of the irreducible polynomial g if and only if their difference h_{1}−h_{2} is divisible by g. This isomorphism σ: K(β) → K(γ) extends to an K-isomorphism τ: L(β) → L(γ) between L(β) and L(γ), since L(β) and L(β) are splitting fields for f over the field K(β) and K(γ) respectively (Theorem 4.24). Thus the extensions L(β): K and L(γ): K are isomorphic, and [L(β): K] = [L(γ): K]. But [L(β): K] = [L(β): L][L: K] and [L(γ): K] = [L(γ): L][L: K] by the Tower Law (Theorem 4.10). It follows that [L(β): L] = [L(γ): L]. In particular β ∈ L if and only if γ ∈ L. This shows that that any irreducible polynomial with a root in L must split over L, and thus L: K is normal, as required.

4.10 Separability

Let K be a field. We recall that nk is defined inductively for all integers n and for all elements k of K so that 0k = 0 and (n + 1)k = nk + k for all n ∈ ℤ and k ∈ K. Thus 1k = k, 2k = k + k, 3k = k + k + k etc., and (−n)k = −(nk) for all n ∈ ℤ.

Definition Let K be a field, and let f ∈ K[x] be a polynomial with coeffi-

cients c_{0}, c_{1}, …, c_{n} in K, where f(x) = ∑_{j=0}^{n} c_{j}x^{j}. The formal derivative Df

of f is defined by the formula (Df)(x) = ∑_{j=1}^{n} jc_{j}x^{j−1}.

(The definition of formal derivative given above is a purely algebraic definition, applying to polynomials with coefficients in any field whatsoever, which corresponds to the formula for the derivative of a polynomial with real coefficients obtained by elementary calculus.)

Let K be a field. One can readily verify by straightforward calculation that D(f + g) = Df + Dg and D(fg) = (Df)g + f(Dg) for all f ∈ K[x]. If f is a constant polynomial then Df = 0.

Let K be a field, and let f ∈ K[x]. An element α of an extension field L of K is said to be a repeated zero if (x − α)^{2} divides f(x).

Proposition 4.27 Let K be a field, and let f ∈ K[x]. The polynomial f has a repeated zero in a splitting field for f over K if and only if there exists a non-constant polynomial with coefficients in K that divides both f and its formal derivative Df in K[x].

Proof Suppose that f ∈ K[x] has a repeated root α in a splitting field L. Then f(x) = (x − α)^{2}h(x) for some polynomial h ∈ L[x]. But then

(Df)(x) = 2(x − α)h(x) + (x − α)^{2}(Dh)(x)

and hence (Df)(α) = 0. It follows that the minimum polynomial of α over K is a non-constant polynomial with coefficients in K which divides both f and Df.

Conversely let f ∈ K[x] be a polynomial with the property that f and Df are both divisible by some non-constant polynomial g ∈ K[x]. Let L be a splitting field for f over K. Then g splits over L (since g is a factor of f). Let α ∈ L be a root of g. Then f(α) = 0, and hence f(x) = (x − α)e(x) for some polynomial e ∈ L[x]. On differentiating, we find that (Df)(x) = e(x) + (x − α)De(x). But (Df)(α) = 0, since g(α) = 0 and g divides Df in K[x]. It follows that e(α) = (Df)(α) = 0, and thus e(x) = (x − α)h(x) for some polynomial h ∈ L[x]. But then f(x) = (x − α)^{2}h(x), and thus the polynomial f has a repeated root in the splitting field L, as required.

Definition Let K be a field. An irreducible polynomial in K[x] is said to be separable over K if it does not have repeated roots in a splitting field. A polynomial in K[x] is said to separable over K if all its irreducible factors are separable over K. A polynomial is said to be inseparable if it is not separable.

Corollary 4.28 Let K be a field. An irreducible polynomial f is inseparable if and only if Df = 0.

Proof Let f ∈ K[x] be an irreducible polynomial. Suppose that f is inseparable. Then f has a repeated root in a splitting field, and it follows from Proposition 4.27 that there exists a non-constant polynomial g in K[x] dividing both f and its formal derivative Df. But then g = cf for some non-zero element c of K, since f is irreducible, and thus f divides Df. But if Df were non-zero then deg Df < deg f, and thus f would not divide Df. Thus Df = 0.

Conversely if Df = 0 then f divides both f and Df. It follows from Proposition 4.27 that f has a repeated root in a splitting field, and is thus inseparable.

Definition An algebraic field extension L: K is said to be separable over K if the minimum polynomial of each element of L is separable over K.

Suppose that K is a field of characteristic zero. Then n.k ≠ 0 for all n ∈ ℤ and k ∈ K satisfying n ≠ 0 and k ≠ 0. It follows from the definition of the formal derivative that Df = 0 if and only if f ∈ K[x] is a constant polynomial. The following result therefore follows immediately from Corollary 4.28.

Corollary 4.29 Suppose that K is a field of characteristic zero. Then every polynomial with coefficients in K is separable over K, and thus every field extension L: K of K is separable.

4.11 Finite Fields

Lemma 4.30 Let K be a field of characteristic p, where p > 0. Then (x + y)^{p} = x^{p} + y^{p} and (xy)^{p} = x^{p}y^{p} for all x, y ∈ K. Thus the function x ↦ x^{p} is a monomorphism mapping the field K into itself.

Proof The Binomial Theorem tells us that (x+y)^{p} = ∑_{j=0}^{p} \binom{p}{j}x^{j}y^{p−j}, where

\binom{p}{0} = 1 and \binom{p}{j} = \frac{p(p − 1) ⋯ (p − j + 1)}{j!} for j = 1, 2, …, p. The de-

nominator of each binomial coefficient must divide the numerator, since this coefficient is an integer. Now the characteristic p of K is a prime number. Moreover if 0 < j < p then p is a factor of the numerator but is not a factor of the denominator. It follows from the Fundamental Theorem of Arithmetic that p divides \binom{p}{j} for all j satisfying 0 < j < p. But px = 0 for all x ∈ K,

since charK = p. Therefore (x + y)^{p} = x^{p} + y^{p} for all x, y ∈ K. The identity (xy)^{p} = x^{p}y^{p} is immediate from the commutativity of K.

Let K be a field of characteristic p, where p > 0. The monomorphism x ↦ x^{p} is referred to as the Frobenius monomorphism of K. If K is finite then this monomorphism is an automorphism of K, since any injection mapping a finite set into itself must be a bijection.

Theorem 4.31 A field K has p^{n} elements if and only if it is a splitting field for the polynomial x^{p^{n}} − x over its prime subfield 𝔽_{p}, where 𝔽_{p} ≅ ℤ/pℤ.

Proof Suppose that K has q elements, where q = p^{n}. If α ∈ K \ {0} then α^{q−1} = 1, since the set of non-zero elements of K is a group of order q − 1 with respect to multiplication. It follows that α^{q} = α for all α ∈ K. Thus all elements of K are roots of the polynomial x^{q} − x. This polynomial must therefore split over K, since its degree is q and K has q elements. Moreover the polynomial cannot split over any proper subfield of K. Thus K is a splitting field for this polynomial.

Conversely suppose that K is a splitting field for the polynomial f over 𝔽_{p}, where f(x) = x^{q} − x and q = p^{n}. Let σ(α) = α^{q} for all α ∈ K. Then σ: K → K is a monomorphism, being the composition of n successive applications of the Frobenius monomorphism of K. Moreover an element α of K is a root of f if and only if σ(α) = α. It follows from this that the roots of f constitute a subfield of K. This subfield is the whole of K, since K is a splitting field. Thus K consists of the roots of f. Now Df(x) = qx^{q−1} − 1 = −1, since q is divisible by the characteristic p of 𝔽_{p}. It follows from Proposition 4.27 that the roots of f are distinct. Therefore f has q roots, and thus K has q elements, as required.

Let K be a finite field of characteristic p. Then K has p^{n} elements, where n = [K: 𝔽_{p}], since any vector space of dimension n over a field of order p must have exactly p^{n} elements. The following result is now a consequence of the existence of splitting fields (Corollary 4.23) and the uniqueness of splitting fields up to isomorphism (Theorem 4.24)

Corollary 4.32 There exists a finite field GF(p^{n}) of order p^{n} for each prime number p and positive integer n. Two finite fields are isomorphic if and only if they have the same number of elements.

The field GF(p^{n}) is referred to as the Galois field of order p^{n}.

The non-zero elements of a field constitute a group under multiplication. We shall prove that all finite subgroups of the group of non-zero elements of a field are cyclic. It follows immediately from this that the group of non-zero elements of a finite field is cyclic.

For each positive integer n, we denote by φ(n) the number of integers x satisfying 0 ≤ x < n that are coprime to n. We show that the sum ∑_{d|n} φ(d)

of φ(d) taken over all divisors of a positive integer n is equal to n.

Lemma 4.33 Let n be a positive integer. Then ∑_{d|n} φ(d) = n.

Proof If x is an integer satisfying 0 ≤ x < n then (x, n) = n/d for some divisor d of n. It follows that n = ∑_{d|n} n_{d}, where n_{d} is the number of integers x

satisfying 0 ≤ x < n for which (x, n) = n/d. Thus it suffices to show that n_{d} = φ(d) for each divisor d of n.

Let d be a divisor of n, and let a = n/d. Given any integer x satisfying 0 ≤ x < n that is divisible by a, there exists an integer y satisfying 0 ≤ y < d such that x = ay. Then (x, n) = (ay, ad) = a(y, d). It follows that the integers x satisfying 0 ≤ x < n for which (x, n) = a are those of the form ay, where y is an integer, 0 ≤ y < d and (y, d) = 1. It follows that there are exactly φ(d) integers x satisfying 0 ≤ x < n for which (x, n) = n/d, and thus n_{d} = φ(d) and n = ∑_{d|n} φ(d), as required.

The set of all non-zero elements of a field is a group with respect to the operation of multiplication.

Theorem 4.34 Let G be a finite subgroup of the group of non-zero elements of a field. Then the group G is cyclic.

Proof Let n be the order of the group G. It follows from Lagrange’s Theorem that the order of every element of G divides n. For each divisor d of n, let ψ(d) denote the number of elements of G that are of order d. Clearly ∑_{d|n} ψ(d) = n.

Let g be an element of G of order d, where d is a divisor of n. The elements 1, g, g^{2}, …, g^{d−1} are distinct elements of G and are roots of the polynomial x^{d} − 1. But a polynomial of degree d with coefficients in a field has at most d roots in that field. Therefore every element x of G satisfying x^{d} = 1 is g^{k} for some uniquely determined integer k satisfying 0 ≤ k < d. If k is coprime to d then g^{k} has order d, for if (g^{k})^{n} = 1 then d divides kn and hence d divides n. Conversely if g^{k} has order d then d and k are coprime, for if e is a common divisor of k and d then (g^{k})^{d/e} = g^{d(k/e)} = 1, and hence e = 1. Thus if there exists at least one element g of G that is of order d then the elements of G that are of order d are the elements g^{k} for those integers k satisfying 0 ≤ k < d that are coprime to d. It follows that if ψ(d) > 0 then ψ(d) = φ(d), where φ(d) is the number of integers k satisfying 0 ≤ k < d that are coprime to d.

Now 0 ≤ ψ(d) ≤ φ(d) for each divisor d of n. But ∑_{d|n} ψ(d) = n and

∑_{d|n} φ(d) = n. It follows that ψ(d) = ϕ(d) for each divisor d of n. In particular ψ(n) = φ(n) ≥ 1. Thus there exists an element of G whose order is the order n of G. This element generates G, and thus G is cyclic, as required.

Corollary 4.35 The group of non-zero elements of a finite field is cyclic.

4.12 The Primitive Element Theorem

Theorem 4.36 (Primitive Element Theorem) Every finite separable field extension is simple.

Proof Let L: K be a finite separable field extension. Suppose that K is a finite field. Then L is also a finite field, since it is a finite-dimensional vector space over K. The group of non-zero elements of L is therefore generated by a single non-zero element θ of L (Corollary 4.35). But then L = K(θ) and thus L: K is simple. This proves the Primitive Element Theorem in the case where the field K is finite.

Next suppose that L = K(β, γ), where K is infinite, β and γ are algebraic over K and L: K is separable. Let N be a splitting field for the polynomial fg, where f and g are the minimum polynomials of β and γ respectively over K. Then f and g both split over N. Let β_{1}, β_{2}, …, β_{q} be the roots of f in N, and let γ_{1}, γ_{2}, …, γ_{r} be the roots of g in N, where β_{1} = β and γ_{1} = γ. The separability of L: K ensures that γ_{k} ≠ γ_{j} when k ≠ j.

Now K is infinite. We can therefore choose c ∈ K so that c ≠ (β_{i} − β)/(γ − γ_{j}) for any i and j with j ≠ 1. Let h(x) = f(θ − cx), where θ = β + cγ. Then h is a polynomial in the indeterminate x with coefficients in K(θ) which satisfies h(γ) = f(β) = 0. Moreover h(γ_{j}) ≠ 0 whenever j ≠ 1, since θ − cγ_{j} ≠ β_{i} for all i and j with j ≠ 1. Thus γ is the only common root of g and h. It follows that x − γ is a highest common factor of g and h in the polynomial ring K(θ)[x], and therefore γ ∈ K(θ). But then β ∈ K(θ), since β = θ − cγ and c ∈ K. It follows that L = K(θ).

It now follows by induction on m that if L = K(α_{1}, α_{2}, …, α_{m}), where K is infinite, α_{1}, α_{2}, …, α_{m} are algebraic over K, and L: K is separable, then the extension L: K is simple. Thus all finite separable field extensions are simple, as required.

4.13 The Galois Group of a Field Extension

Definition The Galois group Γ(L: K) of a field extension L: K is the group of all automorphisms of the field L that fix all elements of the subfield K.

Lemma 4.37 If L: K is a finite separable field extension then |Γ(L: K)| ≤ [L: K].

Proof It follows from the Primitive Element Theorem (Theorem 4.36) that there exists some element α of L such that L = K(α). Let λ be an element of L. Then λ = g(α) for some polynomial g with coefficients in K. But then σ(λ) = g(σ(α)) for all σ ∈ Γ(L: K), since the coefficients of g are fixed by σ. It follows that each automorphism σ in Γ(L: K) is uniquely determined once σ(α) is known.

Let f be the minimum polynomial of α over K. Then

f(σ(α)) = σ(f(α)) = 0

for all σ ∈ Γ(L: K) since the coefficients of f are in K and are therefore fixed by σ. Thus σ(α) is a root of f. It follows that the order |Γ(L: K)| of the Galois group is bounded above by the number of roots of f that belong to L, and is thus bounded above by the degree deg f of f. But deg f = [L: K] (Theorem 4.13). Thus |Γ(L: K)| ≤ [L: K], as required.

Definition Let L be a field, and let G be a group of automorphisms of L. The fixed field of G is the subfield K of L defined by

K = {a ∈ L : σ(a) = a for all σ ∈ G}.

Proposition 4.38 Let L be a field, let G be a finite group of automorphisms of L, and let K be the fixed field of G. Then each element α of L is algebraic over K, and the minimum polynomial of α over K is the polynomial

(x − α_{1})(x − α_{2}) ⋯ (x − α_{k}),

where α_{1}, α_{2}, …, α_{k} are distinct and are the elements of the orbit of α under the action of G on L.

Proof Let f(x) = (x − α_{1})(x − α_{2}) ⋯ (x − α_{k}). Then the polynomial f is invariant under the action of G, since each automorphism in the group G permutes the elements α_{1}, α_{2}, …, α_{k} and therefore permutes the factors of f amongst themselves. It follows that the coefficients of the polynomial f belong to the fixed field K of G. Thus α is algebraic over K, as it is a root of the polynomial f.

Now, given any root α_{i} of f, there exists some σ ∈ G such that α_{i} = σ(α). Thus if g ∈ K[x] is a polynomial with coefficients in K which satisfies g(α) = 0 then g(α_{i}) = σ(g(α)) = 0, since the coefficients of g are fixed by σ. But then f divides g. Thus f is the minimum polynomial of α over K, as required.

Definition A field extension is said to be a Galois extension if it is finite, normal and separable.

Theorem 4.39 Let L be a field, let G be a finite subgroup of the group of automorphisms of L, and let K be the fixed field of G. Then the field extension L: K is a Galois extension. Moreover G is the Galois group Γ(L: K) of L: K and |G| = [L: K].

Proof It follows from Proposition 4.38 that, for each α ∈ L, the minimum polynomial of α over K splits over L and has no multiple roots. Thus the extension L: K is both normal and separable.

Let M be any field satisfying K ⊂ M ⊂ L for which the extension M: K is finite. The extension M: K is separable, since L: K is separable. It follows from the Primitive Element Theorem (Theorem 4.36) that the extension M: K is simple. Thus M = K(α) for some α ∈ L. But then [M: K] is equal to the degree of the minimum polynomial of α over K (Theorem 4.13). It follows from Proposition 4.38 that [M: K] is equal to the number of elements in the orbit of α under the action of G on L. Therefore [M: K] divides |G| for any intermediate field M for which the extension M: K is finite.

Now let the intermediate field M be chosen so as to maximize [M: K]. If λ ∈ L then λ is algebraic over K, and therefore [M(λ): M] is finite. It follows from the Tower Law (Theorem 4.10) that [M(λ): K] is finite, and [M(λ): K] = [M(λ): M][M: K]. But M has been chosen so as to maximize [M: K]. Therefore [M(λ): K] = [M: K], and [M(λ): M] = 1. Thus λ ∈ M. We conclude that M = L. Thus L: K is finite and [L: K] divides |G|.

The field extension L: K is a Galois extension, since it has been shown to be finite, normal and separable. Now G ⊂ Γ(L: K) and |Γ(L: K)| ≤ [L: K] (Lemma 4.37). Therefore |Γ(L: K)| ≤ [L: K] ≤ |G| ≤ |Γ(L: K)|, and thus G = Γ(L: K) and |G| = [L: K], as required.

Theorem 4.40 Let Γ(L: K) be the Galois group of a finite field extension L: K. Then |Γ(L: K)| divides [L: K]. Moreover |Γ(L: K)| = [L: K] if and only if L: K is a Galois extension, in which case K is the fixed field of Γ(L: K).

Proof Let M be the fixed field of Γ(L: K). It follows from Theorem 4.39 that L: M is a Galois extension and |Γ(L: K)| = [L: M]. Now [L: K] = [L: M][M: K] by the Tower Law (Theorem 4.10). Thus |Γ(L: K)| divides [L: K]. If |Γ(L: K)| = [L: K] then M = K. But then L: K is a Galois extension and K is the fixed field of Γ(L: K).

Conversely suppose that L: K is a Galois extension. We must show that |Γ(L: K)| = [L: K]. Now the extension L: K is both finite and separable. It follows from the Primitive Element Theorem (Theorem 4.36) that there exists some element θ of L such that L = K(θ). Let f be the minimum polynomial of θ over K. Then f splits over L, since f is irreducible and the extension L: K is normal. Let θ_{1}, θ_{2}, …, θ_{n} be the roots of f in L, where θ_{1} = θ and n = deg f. If σ is a K-automorphism of L then f(σ(θ)) = σ(f(θ)) = 0, since the coefficients of the polynomial f belong to K and are therefore fixed by σ. Thus σ(θ) = θ_{j} for some j. We claim that, for each root θ_{j} of f, there is exactly one K-automorphism σ_{j} of L satisfying σ_{j}(θ) = θ_{j}.

Let g(x) and h(x) be polynomials with coefficients in K. Suppose that g(θ) = h(θ). Then g − h is divisible by the minimum polynomial f of θ. It follows that g(θ_{j}) = h(θ_{j}) for any root θ_{j} of f. Now every element of L is of the form g(θ) for some g ∈ K[x], since L = K(θ). We deduce therefore that there is a well-defined function σ_{j}: L → L with the property that σ_{j}(g(θ)) = g(θ_{j}) for all g ∈ K[x]. The definition of this function ensures that it is the unique automorphism of the field L that fixes each element of K and sends θ to θ_{j}.

Now the roots of the polynomial f in L are distinct, since f is irreducible and L: K is separable. Moreover the order of the Galois group Γ(L: K) is equal to the number of roots of f, since each root determines a unique element of the Galois group. Therefore |Γ(L: K)| = deg f. But deg f = [L: K] since L = K(θ) and f is the minimum polynomial of θ over K (Theorem 4.13). Thus |Γ(L: K)| = [L: K], as required.

4.14 The Galois correspondence

Proposition 4.41 Let K, L and M be fields satisfying K ⊂ M ⊂ L. Suppose that L: K is a Galois extension. Then so is L: M. If in addition M: K is normal, then M: K is a Galois extension.

Proof Let α ∈ L and let f_{K} ∈ K[x] and f_{M} ∈ M[x] be the minimum polyomials of α over K and M respectively. Then f_{K} splits over L, since f_{K} is irreducible over K and L: K is a normal extension. Also the roots of f_{K} in L are distinct, since L: K is a separable extension. But f_{M} divides f_{K}, since f_{K}(α) = 0 and the coefficients of f_{K} belong to M. It follows that f_{M} also splits over L, and its roots are distinct. We deduce that the finite extension L: M is both normal and separable, and is therefore a Galois extension.

The finite extension M: K is clearly separable, since L: K is separable. Thus if M: K is a normal extension then it is a Galois extension.

Proposition 4.42 Let L: K be a Galois extension, and let M be a field satisfying K ⊂ M ⊂ L. Then the extension M: K is normal if and only if σ(M) = M for all σ ∈ Γ(L: K).

Proof Let α be an element of M, and let f ∈ K[x] be the minimum polynomial of α over K. Now K is the fixed field of the Galois group Γ(L: K), since the field extension L: K is a Galois extension (Theorem 4.40). It follows that the polynomial f splits over L, and the roots of f are the elements of the orbit of α under the action of Γ(L: K) on L (Proposition 4.38). Therefore f splits over M if and only if σ(α) ∈ M for all σ ∈ Γ(L: K). Now the extension M: K is normal if and only if the minimum polynomial of any element of M over K splits over M. It follows that the extension M: K is normal if and only if σ(M) ⊂ M for all σ ∈ Γ(L: K). But if σ(M) ⊂ M for all σ ∈ Γ(L: K) then σ^{−1}(M) ⊂ M and M = σ(σ^{−1}(M)) ⊂ σ(M) and thus σ(M) = M for all σ ∈ Γ(L: K). Therefore the extension M: K is normal if and only if σ(M) = M for all σ ∈ Γ(L: K).

Corollary 4.43 Let L: K be a Galois extension, and let M be a field satisfying K ⊂ M ⊂ L. Suppose that the extension M: K is normal. Then the restriction σ|_{M} to M of any K-automorphism σ of L is a K-automorphism of M.

Proof Let σ ∈ Γ(L: K) be a K-automorphism of L. We see from Proposition 4.42 that σ(M) = M. Similarly σ^{−1}(M) = M. It follows that the restrictions σ|_{M}: M → M and σ^{−1}|_{M}: M → M of σ and σ^{−1} to M are Khomomorphisms mapping M into itself. Moreover σ^{−1}|_{M}: M → M is the inverse of σ|_{M}: M → M. Thus σ|_{M}: M → M is an isomorphism, and is thus a K-automorphism of M, as required.

Theorem 4.44 (The Galois Correspondence) Let L: K be a Galois extension of a field K. Then there is a natural bijective correspondence between fields M satisfying K ⊂ M ⊂ L and subgroups of the Galois group Γ(L: K) of the extension L: K. If M is a field satisfying K ⊂ M ⊂ L then the subgroup of Γ(L: K) corresponding to M is the Galois group Γ(L: M) of the extension L: M. If G is a subgroup of Γ(L: K) then the subfield of L corresponding to G is the fixed field of G. Moreover the extension M: K is normal if and only if Γ(L: M) is a normal subgroup of the Galois group Γ(L: K), in which case Γ(M: K) ≅ Γ(L: K)/Γ(L: M).

Proof Let M be a subfield of L containing K. Then L: M is a Galois extension (Proposition 4.41). The existence of the required bijective correspondence between fields M satisfying K ⊂ M ⊂ L and subgroups of the Galois group Γ(L: K) follows immediately from Theorem 4.39 and Theorem 4.40.

Let M be a field satisfying K ⊂ M ⊂ L. Now the extension M: K is normal if and only if σ(M) = M for all σ ∈ Γ(L: K). (Proposition 4.42).

Let H = Γ(L: M). Then M = σ(M) if and only if H = σHσ^{−1}, since M and σ(M) are the fixed fields of H and σHσ^{−1} respectively, and there is a bijective correspondence between subgroups of the Galois group Γ(L: K) and their fixed fields. Thus the extension M: K is normal if and only if Γ(L: M) is a normal subgroup of Γ(L: K).

Finally suppose that M: K is a normal extension. For each σ ∈ Γ(L: K), let ρ(σ) be the restriction σ|_{M} of σ to M. Then ρ: Γ(L: K) → Γ(M: K) is a group homomorphism whose kernel is Γ(L: M). We can apply Theorem 4.39 to the extension M: K to deduce that ρ(Γ(L: K)) = Γ(M: K), since the fixed field of ρ(Γ(L: K)) is K. Therefore the homomorphism ρ: Γ(L: K) → Γ(M: K) induces the required isomorphism between Γ(L: K)/Γ(L: M) and Γ(M: K).

4.15 Quadratic Polynomials

We consider the problem of expressing the roots of a polynomial of low degree in terms of its coefficients. Then the well-known procedure for locating the roots of a quadratic polynomial with real or complex coefficients generalizes to quadratic polynomials with coefficients in a field K whose characteristic does not equal 2. Given a quadratic polynomial ax^{2} + bx + c with coefficients a and b belonging to some such field K, let us adjoin to K an element δ satisfying δ^{2} = b^{2} −4ac. Then the polynomial splits over K(δ), and its roots are (−b ± δ)/(2a). We shall describe below analogous procedures for expressing the roots of cubic and quartic polynomials in terms of their coefficients.

4.16 Cubic Polynomials

Consider a cubic polynomial x^{3} + ax^{2} + bx + c, where the coefficients a, b and c belong to some field K of characteristic zero. If f(x) = x^{3} + ax^{2} + bx + c then f(x − \frac{1}{3}a) = x^{3} − px − q, where p = \frac{1}{3}a^{2} − b and q = \frac{1}{3}ba − \frac{2}{27}a^{3} − c. It therefore suffices to restrict our attention to cubic polynomials of the form x^{3} − px − q, where p and q belong to K.

Let f(x) = x^{3} − px − q, and let u and v be elements of some splitting field for f over K. Then

f(u + v) = u^{3} + v^{3} + (3uv − p)(u + v) − q.

Suppose that 3uv = p. Then f(u + v) = u^{3} + p^{3}/(27u^{3}) − q. Thus f(u + p/(3u)) = 0 if and only if u^{3} is a root of the quadratic polynomial x^{2} − xq + p^{3}/27. Now the roots of this quadratic polynomial are

\frac{q}{2} ± \sqrt{\frac{q^{2}}{4} − \frac{p^{3}}{27}},

and the product of these roots is p^{3}/27. Thus if one of these roots is equal to u^{3} then the other is equal to v^{3}, where v = p/(3u). It follows that the roots of the cubic polynomial f are

3 \sqrt{\frac{q}{2} + \sqrt{\frac{q^{2}}{4} − \frac{p^{3}}{27}}} + ^{3} \sqrt{\frac{q}{2} − \sqrt{\frac{q^{2}}{4} − \frac{p^{3}}{27}}}

where the two cube roots must be chosen so as to ensure that their product is equal to \frac{1}{3}p. It follows that the cubic polynomial x^{3} −px−q splits over the field K(ϵ, ξ, ω), where ϵ^{2} = \frac{1}{4}q^{2} − \frac{1}{27}p^{3} and ξ^{3} = \frac{1}{2}q + ϵ and where ω satisfies ω^{3} = 1 and ω ≠ 1. The roots of the polynomial in this extension field are α, β and γ, where

α = ξ + \frac{p}{3ξ} ,  β = ωξ + ω^{2} \frac{p}{3ξ} ,  γ = ω^{2}ξ + ω^{3} \frac{p}{3ξ} .

Now let us consider the possibilities for the Galois group Γ(L: K), where L is a splitting field for f over K. Now L = K(α, β, γ), where α, β and γ are the roots of f. Also a K-automorphism of L must permute the roots of f amongst themselves, and it is determined by its action on these roots. Therefore Γ(L: K) is isomorphic to a subgroup of the symmetric group Σ_{3} (i.e., the group of permutations of a set of 3 objects), and thus the possibilities for the order of Γ(L: K) are 1, 2, 3 and 6. It follows from Corollary 4.25 that f is irreducible over K if and only if the roots of f are distinct and the Galois group acts transitively on the roots of f. By considering all possible subgroups of Σ_{3} it is not difficult to see that f is irreducible over K if and only if |Γ(L: K)| = 3 or 6. If f splits over K then |Γ(L: K)| = 1. If f factors in K[x] as the product of a linear factor and an irreducible quadratic factor then |Γ(L: K)| = 2.

Let δ = (α−β)(α−γ)(β−γ). Then δ^{2} is invariant under any permutation of α β and γ, and therefore δ^{2} is fixed by all automorphisms in the Galois group Γ(L: K). Therefore δ^{2} ∈ K. The element δ^{2} of K is referred to as the discriminant of the polynomial f. A straightforward calculation shows that if f(x) = x^{3} − px − q then δ^{2} = 4p^{3} − 27q^{2}. Now δ changes sign under any permutation of the roots α, β and γ that transposes two of the roots whilst leaving the third root fixed. But δ ∈ K if and only if δ is fixed by all elements of the Galois group Γ(L: K), in which case the Galois group must induce only cyclic permutations of the roots α, β and γ. Therefore Γ(L: K) is isomorphic to the cyclic group of order 3 if and only if f is irreducible and the discriminant 4p^{3} − 27q^{2} of f has a square root in the field K. If f is irreducible but the discriminant does not have a square root in K then Γ(L: K) is isomorphic to the symmetric group Σ_{3}, and |Γ(L: K)| = 6.

4.17 Quartic Polynomials

We now consider how to locate the roots of a quartic polynomial with coefficients in a field K of characteristic zero. A substitution of the form x ↦ x−c, where c ∈ K, will reduce the problem to that of locating the roots α, β, γ and δ of a quartic polynomial f of the form f(x) = x^{4} − px^{2} − qx − r in some splitting field L.

Now the roots α, β, γ and δ of the quartic polynomial

x^{4} − px^{2} − qx − r,

must satisfy the equation

(x − α)(x − β)(x − γ)(x − δ) = x^{4} − px^{2} − qx − r.

Equating coefficients of x, we find that

α + β + γ + δ = 0,

and

p = −(αβ + αγ + αδ + βγ + βδ + γδ),
q = βγδ + αγδ + αβδ + αβγ,
r = −αβγδ.

Let

λ = (α + β)(γ + δ) = −(α + β)^{2} = −(γ + δ)^{2},
μ = (α + γ)(β + δ) = −(α + γ)^{2} = −(β + δ)^{2},
ν = (α + δ)(β + γ) = −(α + δ)^{2} = −(β + γ)^{2}.

We shall show that λ + μ + ν, μν + λν + λμ and λμν can all be expressed in terms of p, q and r.

To do this we eliminate α from the above expressions using the identity α + β + γ + δ = 0. We find

p = (β + γ + δ)(β + γ + δ) − γδ − βδ − βγ
= β^{2} + γ^{2} + δ^{2} + γδ + βδ + βγ,
q = βγδ − (β + γ + δ)(γδ + βδ + βγ)
= −(β^{2}γ + β^{2}δ + γ^{2}β + γ^{2}δ + δ^{2}β + δ^{2}γ) − 2βγδ,
r = β^{2}γδ + γ^{2}βδ + δ^{2}βγ.

Then

λ + μ + ν = −((γ + δ)^{2} + (β + δ)^{2} + (β + γ)^{2})
= −2(β^{2} + γ^{2} + δ^{2} + γδ + βδ + βγ)
= −2p,
λ^{2} + μ^{2} + ν^{2} = (γ + δ)^{4} + (β + δ)^{4} + (β + γ)^{4}
= γ^{4} + 4γ^{3}δ + 6γ^{2}δ^{2} + 4γδ^{3} + δ^{4}
+ β^{4} + 4β^{3}δ + 6β^{2}δ^{2} + 4βδ^{3} + δ^{4}
+ β^{4} + 4β^{3}γ + 6β^{2}γ^{2} + 4βγ^{3} + γ^{4}
= 2(β^{4} + γ^{4} + δ^{4}) + 4(β^{3}γ + β^{3}δ + γ^{3}β + γ^{3}δ + δ^{3}β + δ^{3}γ)
+ 6(γ^{2}δ^{2} + β^{2}δ^{2} + β^{2}γ^{2}),
p^{2} = β^{4} + γ^{4} + δ^{4} + 3(γ^{2}δ^{2} + β^{2}δ^{2} + β^{2}γ^{2})
+ 4(β^{2}γδ + γ^{2}βδ + δ^{2}βγ)
+ 2(β^{3}γ + β^{3}δ + γ^{3}β + γ^{3}δ + δ^{3}β + δ^{3}γ).

Therefore

λ^{2} + μ^{2} + ν^{2} = 2p^{2} − 8(β^{2}γδ + γ^{2}βδ + δ^{2}βγ)
= 2p^{2} − 8r.

But

4p^{2} = (λ + μ + ν)^{2} = λ^{2} + μ^{2} + ν^{2} + 2(μν + λν + λμ)

Therefore

μν + λν + λμ = 2p^{2} − \frac{1}{2}(λ^{2} + μ^{2} + ν^{2})
= p^{2} + 4r.

Finally, we note that

λμν = −((γ + δ)(β + δ)(β + γ))^{2}.

Now

(γ + δ)(β + δ)(β + γ) = β^{2}γ + β^{2}δ + γ^{2}β + γ^{2}δ + δ^{2}β + δ^{2}γ + 2βγδ
= −q.
(α + β)(α + γ)(α + δ) = −(γ + δ)(β + δ)(β + γ) = q.

Therefore

λμν = −(−q)^{2} = −q^{2}.

Thus λ, μ and ν are the roots of the resolvent cubic

x^{3} + 2px^{2} + (p^{2} + 4r)x + q^{2}.

One can then verify that the roots of f take the form \frac{1}{2}(\sqrt{−λ} + \sqrt{−μ} + \sqrt{−ν}), where these square roots are chosen to ensure that \sqrt{−λ}\sqrt{−μ}\sqrt{−ν} = q. (It should be noted that there are four possible ways in which the square roots can be chosen to satisfy this condition; these yield all four roots of the polynomial f.) We can therefore determine the roots of f in an appropriate splitting field once we have expressed the quantities λ, μ and ν in terms of the coefficients of the polynomial.

Remark Any permutation of the roots of the quartic

x^{4} − px^{2} − qx − r,

will permute the roots λ, μ and ν of the resolvent cubic

g(x) = (x − λ)(x − μ)(x − ν)

amongst themselves, and will therefore permute the factors of g. Therefore the coefficients of g are fixed by all elements of the Galois group Γ(L: K) and therefore must belong to the ground field K. As we have seen from the calculations above, these coefficients can be expressed in terms of p, q, r.

4.18 The Galois group of the polynomial x^{4} − 2

We shall apply the Galois correspondence to investigate the structure of the splitting field for the polynomial x^{4} − 2 over the field ℚ of rational numbers. A straightforward application of Eisenstein’s Irreducibility Criterion (Proposition 4.9) shows that the polynomial x^{4} − 2 is irreducible over ℚ. Let ξ be the unique positive real number satisfying ξ^{4} = 2. Then the roots of x^{4} − 2 in the field ℂ of complex numbers are ξ, iξ, −ξ and −iξ, where i = \sqrt{−1}. Thus if L = ℚ(ξ, i) then L is a splitting field for the polynomial x^{4} − 2 over ℚ.

Now the polynomial x^{4} − 2 is the minimum polynomial of ξ over ℚ, since this polynomial is irreducible. We can therefore apply Theorem 4.13 to deduce that [ℚ(ξ): ℚ] = 4. Now i does not belong to ℚ(ξ), since ℚ(ξ) ⊂ ℝ. Therefore the polynomial x^{2} + 1 is the minimum polynomial of i over ℚ(ξ). Another application of Theorem 4.13 now shows that [L: ℚ(ξ)] = [ℚ(ξ, i): ℚ(ξ)] = 2. It follows from the Tower Law (Theorem 4.10) that [L: ℚ] = [L: ℚ(ξ)][ℚ(ξ): ℚ] = 8. Moreover the extension L: ℚ is a Galois extension, and therefore its Galois group Γ(L: ℚ) is a group of order 8 (Theorem 4.40).

Another application of the Tower Law now shows that [L: ℚ(i)] = 4, since [L: ℚ] = [L: ℚ(i)][ℚ(i): ℚ] and [ℚ(i): ℚ] = 2. Therefore the minimum polynomial of ξ over ℚ(i) is a polynomial of degree 4 (Theorem 4.13). But ξ is a root of x^{4}−2. Therefore x^{4}−2 is irreducible over ℚ(i), and is the minimum polynomial of ξ over ℚ(i). Corollary 4.25 then ensures the existence of an automorphism σ of L that sends ξ ∈ L to iξ and fixes each element of ℚ(i). Similarly there exists an automorphism τ of L that sends i to −i and fixes each element of ℚ(ξ). (The automorphism τ is in fact the restriction to L of the automorphism of ℂ that sends each complex number to its complex conjugate.)

Now the automorphisms σ, σ^{2}, σ^{3} and σ^{4} fix i and therefore send ξ to iξ, −ξ, −iξ and ξ respectively. Therefore σ^{4} = ι, where ι is the identity automorphism of L. Similarly τ^{2} = ι. Straightforward calculations show that τσ = σ^{3}τ, and (στ)^{2} = (σ^{2}τ)^{2} = (σ^{3}τ)^{2} = ι. It follows easily from this that Γ(L: ℚ) = {ι, σ, σ^{2}, σ^{3}, τ, στ, σ^{2}τ, σ^{3}τ}, and Γ(L: ℚ) is isomorphic to the dihedral group of order 8 (i.e., the group of symmetries of a square in the plane).

The Galois correspondence is a bijective correspondence between the subgroups of Γ(L: ℚ) and subfields of L that contain ℚ. The subfield of L corresponding to a given subgroup of Γ(L: ℚ) is set of all elements of L that are fixed by all the automorphisms in the subgroup. One can verify that the correspondence between subgroups of Γ(L: ℚ) and their fixed fields is as follows:—

Subgroup of Γ(L: ℚ) Fixed field
Γ(L: K)  ℚ
{ι, σ, σ^{2}, σ^{3}}  ℚ(i)
\{ι, σ^{2}, τ, σ^{2}τ\}  ℚ(\sqrt{2})
\{ι, σ^{2}, στ, σ^{3}τ\}  ℚ(i\sqrt{2})
\{ι, σ^{2}\}  ℚ(\sqrt{2}, i)
{ι, τ}  ℚ(ξ)
{ι, σ^{2}τ}  ℚ(iξ)
{ι, στ}  ℚ((1 − i)/ξ)
{ι, σ^{3}τ}  ℚ((1 + i)/ξ)
{ι}  ℚ(ξ, i)

4.19 The Galois group of a polynomial

Definition Let f be a polynomial with coefficients in some field K. The Galois group Γ_{K}(f) of f over K is defined to be the Galois group Γ(L: K) of the extension L: K, where L is some splitting field for the polynomial f over K.

We recall that all splitting fields for a given polynomial over a field K are K-isomorphic (see Theorem 4.24), and thus the Galois groups of these splitting field extensions are isomorphic. The Galois group of the given polynomial over K is therefore well-defined (up to isomorphism of groups) and does not depend on the choice of splitting field.

Lemma 4.45 Let f be a polynomial with coefficients in some field K and let M be an extension field of K. Then Γ_{M}(f) is isomorphic to a subgroup of Γ_{K}(f).

Proof Let N be a splitting field for f over M. Then N contains a splitting field L for f over K. An element σ of Γ(N: M) is an automorphism of N that fixes every element of M and therefore fixes every element of K. Its restriction σ|_{L} to L is then a K-automorphism of L (Corollary 4.43). Moreover (σ ∘ τ)|_{L} = (σ|_{L}) ∘ (τ|_{L}) for all σ, τ ∈ Γ(N: M). Therefore there is a group homomorphism from Γ(N: M) to Γ(L: K) which sends an automorphism σ ∈ Γ(N: M) to its restriction σ|_{L} to L.

Now if σ ∈ Γ(N: M) is in the kernel of this group homomorphism from Γ(N: M) to Γ(L: K) then σ|_{L} must be the identity automorphism of L. But f splits over L, and therefore all the roots of f are elements of L. It follows that σ(α) = α for each root α of f. The fixed field of σ must therefore be the whole of N, since M is contained in the fixed field of σ, and N is a splitting field for f over M. Thus σ must be the identity automorphism of N. We conclude therefore that the group homomorphism from Γ(N: M) to Γ(L: K) sending σ ∈ Γ(N: M) to σ|_{L} is injective, and therefore maps Γ(N: M) isomorphically onto a subgroup of Γ(L: K). The result therefore follows from the definition of the Galois group of a polynomial.

Let f be a polynomial with coefficients in some field K and let the roots of f is some splitting field L be α_{1}, α_{2}, …, α_{n}. An element σ of Γ(L: K) is a K-automorphism of L, and therefore σ permutes the roots of f. Moreover two automorphism σ and τ in the Galois group Γ(L: K) are equal if and only if σ(α_{j}) = τ(α_{j}) for j = 1, 2, …, n, since L = K(α_{1}, α_{2}, …, α_{n}). Thus the Galois group of a polynomial can be represented as a subgroup of the group of permutations of its roots. We deduce immediately the following result.

Lemma 4.46 Let f be a polynomial with coefficients in some field K. Then the Galois group of f over K is isomorphic to a subgroup of the symmetric group Σ_{n}, where n is the degree of f.

4.20 Solvable Groups

The concept of a solvable group was introduced into mathematics by Evariste Galois, in order to state and prove his fundamental general theorems concerning the solvability of polynomial equations. We now investigate the basic properties of such solvable groups.

Definition A group G is said to be solvable (or soluble) if there exists a finite sequence G_{0}, G_{1}, …, G_{n} of subgroups of G, where G_{0} = {1} and G_{n} = G, such that G_{i−1} is normal in G_{i} and G_{i}/G_{i−1} is Abelian for i = 1, 2, …, n.

Example The symmetric group Σ_{4} is solvable. Indeed let V_{4} be the Kleinsche Viergruppe consisting of the identity permutation ι and the permutations (12)(34), (13)(24) and (14)(23), and let A_{4} be the alternating group consisting of all even permutations of {1, 2, 3, 4}. Then {ι} ◁ V_{4} ◁ A_{4} ◁ Σ_{4}, V_{4} is Abelian, A_{4}/V_{4} is cyclic of order 3, and Σ_{4}/A_{4} is cyclic of order 2.

In order to prove certain basic results concerning solvable groups, we need to make use of the Isomorphism Theorems for groups, which may be described as follows.

Let G be a group, let H be a subgroup of G, and let N be a normal subgroup of G. Then

\frac{HN}{N} ≅ \frac{H}{N ∩ H} .

This result is often referred to as the First Isomorphism Theorem. Also if M and N are normal subgroups of a group G, and if M ⊂ N, then

\frac{G}{N} ≅ \frac{G/M}{N/M} .

This result is often referred to as the Second Isomorphism Theorem.

Lemma 4.47 Let G be a group, let H_{1} and H_{2} be subgroups of G, where H_{1} ◁ H_{2}, and let J_{1} = H_{1} ∩ N, J_{2} = H_{2} ∩ N, K_{1} = H_{1}N/N and K_{2} = H_{2}N/N, where N is some normal subgroup of G. Then J_{1} ◁ J_{2} and K_{1} ◁ K_{2}. Moreover there exists a normal subgroup of H_{2}/H_{1} isomorphic to J_{2}/J_{1}, and the quotient of H_{2}/H_{1} by this normal subgroup is isomorphic to K_{2}/K_{1}.

Proof It is a straightforward exercise to verify that J_{1} ◁ J_{2} and K_{1} ◁ K_{2}. Let θ: H_{2} → K_{2} be the surjective homomorphism sending h ∈ H_{2} to the coset hN. Now θ induces a well-defined surjective homomorphism ψ: H_{2}/H_{1} → K_{2}/K_{1}, since θ(H_{1}) ⊂ K_{1}. Also θ^{−1}(K_{1}) = H_{2} ∩ (H_{1}N). But H_{2} ∩ (H_{1}N) = H_{1}(H_{2} ∩ N), for if a ∈ H_{1}, b ∈ N and ab ∈ H_{2} then b ∈ H_{2} ∩ N. Therefore

ker ψ = θ^{−1}(K_{1})/H_{1} = H_{1}(H_{2} ∩ N)/H_{1} ≅ H_{2} ∩ N/H_{1} ∩ N = J_{2}/J_{1}

by the First Isomorphism Theorem. Moreover the quotient of H_{2}/H_{1} by the normal subgroup ker ψ is isomorphic to the image K_{2}/K_{1} of ψ. Thus ker ψ is the required normal subgroup of H_{2}/H_{1}.

Proposition 4.48 Let G be a group, and let H be a subgroup of G. Then

(i) if G is solvable then any subgroup H of G is solvable;

(ii) if G is solvable then G/N is solvable for any normal subgroup N of G;

(iii) if N is a normal subgroup of G and if both N and G/N are solvable then G is solvable.

Proof Suppose that G is solvable. Let G_{0}, G_{1}, …, G_{m} be a finite sequence of subgroups of G, where G_{0} = {1}, G_{n} = G, and G_{i−1} ◁ G_{i} and G_{i}/G_{i−1} is Abelian for i = 1, 2, …, m.

We first show that the subgroup H is solvable. Let H_{i} = H ∩ G_{i} for i = 0, 1, …, m. Then H_{0} = {1} and H_{m} = H. If u ∈ H_{i} and v ∈ H_{i−1} then uvu^{−1} ∈ H, since H is a subgroup of G. Also uvu^{−1} ∈ G_{i−1}, since u ∈ G_{i−1}, v ∈ G_{i} and G_{i−1} is normal in G_{i}. Therefore uvu^{−1} ∈ H_{i−1}. Thus H_{i−1} is a normal subgroup of H_{i} for i = 1, 2, …, m. Moreover

\frac{H_{i}}{H_{i−1}} = \frac{G_{i} ∩ H}{G_{i−1} ∩ (G_{i} ∩ H)} ≅ \frac{G_{i−1}(G_{i} ∩ H)}{G_{i−1}}

by the First Isomorphism Theorem, and thus H_{i}/H_{i−1} is isomorphic to a subgroup of the Abelian group G_{i}/G_{i−1}. It follows that H_{i}/H_{i−1} must itself be an Abelian group. We conclude therefore that the subgroup H of G is solvable.

Now let N be a normal subgroup of G, and let K_{i} = G_{i}N/N for all i. Then K_{0} is the trivial subgroup of G/N and K_{m} = G/N. It follows from Lemma 4.47 that K_{i−1} ◁ K_{i} and K_{i}/K_{i−1} is isomorphic to the quotient of G_{i}/G_{i−1} by some normal subgroup. But a quotient of any Abelian group must itself be Abelian. Thus each quotient group K_{i}/K_{i−1} is Abelian, and thus G/N is solvable.

Finally suppose that G is a group, N is a normal subgroup of G and both N and G/N are solvable. We must prove that G is solvable. Now the solvability of N ensures the existence of a finite sequence G_{0}, G_{1}, …, G_{m} of subgroups of N, where G_{0} = {1}, G_{m} = N, and G_{i−1} ◁ G_{i} and G_{i}/G_{i−1} is Abelian for i = 1, 2, …, m. Also the solvability of G/N ensures the existence of a finite sequence K_{0}, K_{1}, …, K_{n} of subgroups of G/N, where K_{0} = N/N, K_{n} = G/N, and K_{i−1} ◁ K_{i} and K_{i}/K_{i−1} is Abelian for i = 1, 2, …, n. Let G_{m+i} be the preimage of K_{i} under the the quotient homomorphism ν: G → G/N, for i = 1, 2, …, n. The Second Isomorphism Theorem ensures that G_{m+i}/G_{m+i−1} ≅ K_{i}/K_{i−1} for all i > 0. Therefore G_{0}, G_{1}, …, G_{m+n} is a finite sequence of subgroups of G, where G_{0} = {1}, G_{n} = G, and G_{i−1} ◁ G_{i} and G_{i}/G_{i−1} is Abelian for i = 1, 2, …, m + n. Thus the group G is solvable, as required.

Example The alternating group A_{5} is simple. It follows that A_{5} is not solvable, since the definition of solvable groups ensures that that any simple solvable group is cyclic, and A_{5} is not cyclic. Now if n ≥ 5 the symmetric group Σ_{n} of all permutations of a set of n elements contains a subgroup isomorphic to A_{5}. (Take as this subgroup the set of all even permutations of five of the elements permuted by the elements of Σ_{n}.) Moreover any subgroup of a solvable group is solvable (Proposition 4.48.) It follows therefore that the symmetric group Σ_{n} is not solvable when n ≥ 5.

4.21 Solvable polynomials and their Galois groups

Definition We say that a polynomial with coefficients in a given field is solvable by radicals if the roots of the polynomial in a splitting field can be constructed from its coefficients in a finite number of steps involving only the operations of addition, subtraction, multiplication, division and extraction of nth roots for appropriate natural numbers n.

It follows from the definition above that a polynomial with coefficients in a field K is solvable by radicals if and only if there exist fields K_{0}, K_{1}, …, K_{m} such that K_{0} = K, the polynomial f splits over K_{m}, and, for each integer i between 1 and m, the field K_{i} is obtained on adjoining to K_{i−1} an element α_{i} with the property that α_{i}^{p_{i}} ∈ K_{i−1} for some positive integer p_{i}. Moreover we can assume, without loss of generality that p_{1}, p_{2}, …, p_{m} are prime numbers, since an nth root α of an element of a given field can be adjoined that field by successively adjoining powers α^{n_{1}}, α^{n_{2}}, …, α^{n_{k}} of α chosen such that n/n_{1} is prime, n_{i}/n_{i−1} is prime for i = 2, 3, …, k, and n_{k} = 1.

We shall prove that a polynomial with coefficients in a field K of characteristic zero is solvable by radicals if and only if its Galois group Γ_{K}(f) over K is a solvable group.

Let L be a field, and let p be a prime number that is not equal to the characteristic of L. Suppose that the polynomial x^{p} − 1 splits over L. Then the polynomial x^{p} − 1 has distinct roots, since its formal derivative px^{p−1} is non-zero at each root of x^{p} − 1. An element ω of L is said to be a primitive pth root of unity if ω^{p} = 1 and ω ≠ 1. The primitive pth roots of unity are the roots of the polynomial x^{p−1} +x^{p−2} +⋯+1, since x^{p} −1 = (x−1)(x^{p−1} + x^{p−2} + ⋯ + 1). Also the group of pth roots of unity in L is a cyclic group over order p which is generated by any primitive pth root of unity.

Lemma 4.49 Let K be a field, and let p be a prime number that is not equal to the characteristic of K. If ω is a primitive pth root of unity in some extension field of K then the Galois group of the extension K(ω): K is Abelian.

Proof Let L = K(ω). Then L is a splitting field for the polynomial x^{p} − 1. Let σ and τ be K-automorphisms of L. Then σ(ω) and τ(ω) are roots of x^{p}−1 (since the automorphisms σ and τ permute the roots of this polynomial) and therefore there exist non-negative integers q and r such that σ(ω) = ω^{q} and τ(ω) = ω^{r}. Then σ(τ(ω)) = ω^{qr} = τ(σ(ω)). But there is at most one K-automorphism of L sending ω to ω^{qr}. It follows that σ ∘ τ = τ ∘ σ. Thus the Galois group Γ(L: K) is Abelian, as required.

Lemma 4.50 Let K be a field of characteristic zero and let M be a splitting field for the polynomial x^{p} − c over K, where p is some prime number and c ∈ K. Then the Galois group Γ(M: K) of the extension M: K is solvable.

Proof The result is trivial when c = 0, since M = K in this case.

Suppose c ≠ 0. The roots of the polynomial x^{p} − c are distinct, and each pth root of unity is the ratio of two roots of x^{p} − c. Therefore M = K(α, ω), where α^{p} = c and ω is some primitive pth root of unity. Now K(ω): K is a normal extension, since K(ω) is a splitting field for the polynomial x^{p} − 1 over K (Theorem 4.26). On applying the Galois correspondence (Theorem 4.44), we see that Γ(M: K(ω)) is a normal subgroup of Γ(M: K), and Γ(M: K)/Γ(M: K(ω)) is isomorphic to Γ(K(ω): K). But Γ(K(ω): K) is Abelian (Lemma 4.49). It therefore suffices to show that Γ(M: K(ω)) is also Abelian.

Now the field M is obtained from K(ω) by adjoining an element α satisfying α^{p} = c. Therefore each automorphism σ in Γ(M: K(ω)) is uniquely determined by the value of σ(α). Moreover σ(α) is also a root of x^{p} − c, and therefore σ(α) = αω^{j} for some integer j. Thus if σ and τ are automorphisms of M belonging to Γ(M: K(ω)), and if σ(α) = αω^{j} and τ(α) = αω^{k}, then σ(τ(α)) = τ(σ(α)) = αω^{j+k}, since σ(ω) = τ(ω) = ω. Therefore σ ∘ τ = τ ∘ σ. We deduce that Γ(M: K(ω)) is Abelian, and thus Γ(M: K) is solvable, as required.

Lemma 4.51 Let f be a polynomial with coefficients in a field K of characteristic zero, and let K′ = K(α), where α ∈ K′ satisfies α^{p} ∈ K for some prime number p. Then Γ_{K}(f) is solvable if and only if Γ_{K′}(f) is solvable.

Proof Let N be a splitting field for the polynomial f(x)(x^{p} − c) over K, where c = α^{p}. Then N contains a splitting field L for f over K and a splitting field M for x^{p} − c over K. Then N: K, L: K and M: K are Galois extensions. The Galois correspondence (Theorem 4.44) ensures that Γ(N: L) and Γ(N: M) are normal subgroups of Γ(N: K). Moreover Γ(L: K) is isomorphic to Γ(N: K)/Γ(N: L), and Γ(M: K) is isomorphic to Γ(N: K)/Γ(N: M). Now M and N are splitting fields for the polynomial x^{p} − c over the fields K and L respectively. It follows from Lemma 4.50 that Γ(M: K) and Γ(N: L) are solvable. But if H is a normal subgroup of a finite group G then G is solvable if and only both H and G/H are solvable (Proposition 4.48). Therefore Γ(N: K) is solvable if and only if Γ(N: M) is solvable. Also Γ(N: K) is solvable if and only if Γ(L: K) is solvable. It follows that Γ(N: M) is solvable if and only if Γ(L: K) is solvable. But Γ(N: M) ≅ Γ_{M}(f) and Γ(L: K) ≅ Γ_{K}(f), since L and N are splitting fields for f over K and M respectively. Thus Γ_{M}(f) is solvable if and only if Γ_{K}(f) is solvable.

Now M is also a splitting field for the polynomial x^{p} − c over K′, since K′ = K(α), where α is a root of the polynomial x^{p} − c. The above argument therefore shows that Γ_{M}(f) is solvable if and only if Γ_{K′}(f) is solvable. Therefore Γ_{K}(f) is solvable if and only if Γ_{K′}(f) is solvable, as required.

Theorem 4.52 Let f be a polynomial with coefficients in a field K of characteristic zero. Suppose that f is solvable by radicals. Then the Galois group Γ_{K}(f) of f is a solvable group.

Proof The polynomial f is solvable by radicals. Therefore there exist fields K_{0}, K_{1}, …, K_{m} such that K_{0} = K, the polynomial f splits over K_{m}, and, for each integer i between 1 and m, the field K_{i} is obtained on adjoining to K_{i−1} an element α_{i} with the property that α_{i}^{p_{i}} ∈ K_{i−1} for some prime number p_{i}. Now Γ_{K_{m}}(f) is solvable, since it is the trivial group consisting of the identity automorphism of K_{m} only. Also Lemma 4.51 ensures that, for each i > 0, Γ_{K_{i}}(f) is solvable if and only if Γ_{K_{i−1}}(f) is solvable. It follows that Γ_{K}(f) is solvable, as required.

Lemma 4.53 Let p be a prime number, let K be a field whose characteristic is not equal to p, and let L: K be a Galois extension of K of degree p. Suppose that the polynomial x^{p} − 1 splits over K. Then there exists α ∈ L such that L = K(α) and α^{p} ∈ K.

Proof The Galois group Γ(L: K) is a cyclic group of order p, since its order is equal to the degree p of the extension L: K. Let σ be a generator of Γ(L: K), let β be an element of L \ K, and let

α_{j} = β_{0} + ω^{j}β_{1} + ω^{2j}β_{2} + ⋯ + ω^{(p−1)j}β_{p−1}

for j = 0, 1, …, p − 1, where β_{0} = β, β_{i} = σ(β_{i−1}) for i = 1, 2, …, p − 1, and ω is a primitive pth root of unity contained in K. Now σ(α_{j}) = ω^{−j}α_{j} for j = 0, 1, …, p − 1, since σ(ω) = ω, σ(β_{p−1}) = β_{0} and ω^{p} = 1. Therefore σ(α_{j}^{p}) = α_{j}^{p} and hence α_{j}^{p} ∈ K for j = 0, 1, 2, …, p − 1. But

α_{0} + α_{1} + α_{2} + ⋯ + α_{p−1} = pβ,

since ω^{j} is a root of the polynomial 1 + x + x^{2} + ⋯ + x^{p−1} for all integers j that are not divisible by p. Moreover pβ ∈ L \ K, since β ∈ L \ K and p ≠ 0 in K. Therefore at least one of the elements α_{0}, α_{1}, …, α_{p−1} belongs to L \ K. Let α = α_{j}, where α_{j} ∈ L \ K. It follows from the Tower Law (Theorem 4.10) that [K(α), K] divides [L: K]. But [L: K] = p and p is prime. It follows that L = K(α). Moreover α^{p} ∈ K, as required.

Theorem 4.54 Let f be a polynomial with coefficients in a field K of characteristic zero. Suppose that the Galois group Γ_{K}(f) of f over K is solvable. Then f is solvable by radicals.

Proof Let ω be a primitive pth root of unity. Then Γ_{K(ω)}(f) is isomorphic to a subgroup of Γ_{K}(f) (Lemma 4.45) and is therefore solvable (Proposition 4.48). Moreover f is solvable by radicals over K if and only if f is solvable by radicals over K(ω), since K(ω) is obtained from K by adjoining an element ω whose pth power belongs to K. We may therefore assume, without loss of generality, that K contains a primitive pth root of unity for each prime p that divides |Γ_{K}(f)|.

The result is trivial when |Γ_{K}(f)| = 1, since in that case the polynomial f splits over K. We prove the result by induction on the degree |Γ_{K}(f)| of the Galois group. Thus suppose that the result holds when the order of the Galois group is less than |Γ_{K}(f)|. Let L be a splitting field for f over K. Then L: K is a Galois extension and Γ(L: K) ≅ Γ_{K}(f). Now the solvable group Γ(L: K) contains a normal subgroup H for which the corresponding quotient group Γ(L: K)/H is a cyclic group of order p for some prime number p dividing |Γ(L: K)|. Let M be the fixed field of H. Then Γ(L: M) = H and Γ(M: K) ≅ Γ(L: K)/H. (Theorem 4.44), and therefore [M: K] = |Γ(L: K)/H| = p. It follows from Lemma 4.53 that M = K(α) for some element α ∈ M satisfying α^{p} ∈ K. Moreover Γ_{M}(f) ≅ H, and H is solvable, since any subgroup of a solvable group is solvable (Proposition 4.48). The induction hypothesis ensures that f is solvable by radicals when considered as a polynomial with coefficients in M, and therefore the roots of f lie in some extension field of M obtained by successively adjoining radicals. But M is obtained from K by adjoining the radical α. Therefore f is solvable by radicals, when considered as a polynomial with coefficients in K, as required.

On combining Theorem 4.52 and Theorem 4.54, we see that a polynomial with coefficients in a field K of characteristic zero is solvable by radicals if and only if its Galois group Γ_{K}(f) over K is a solvable group.

4.22 A quintic polynomial that is not solvable by radicals

Lemma 4.55 Let p be a prime number and let f be a polynomial of order p with rational coefficients. Suppose that f has exactly p − 2 real roots and is irreducible over the field ℚ of rational numbers. Then the Galois group of f over ℚ is isomorphic to the symmetric group Σ_{p}.

Proof If α is a root of f then [ℚ(α): ℚ] = p since f is irreducible and deg f = p (Theorem 4.13). Thus if L is a splitting field extension for f over ℚ then [L: ℚ] = [L: ℚ(α)][ℚ(α): ℚ] by the Tower Law (Proposition 4.10) and therefore [L: ℚ] is divisible by p. But [L: ℚ] is the order of the Galois group G of f, and therefore |G| is divisible by p. It follows from a basic theorem of Cauchy that G must contain at least one element of order p. Moreover an element of G is determined by its action on the roots of f. Thus an element of G is of order p if and only if it cyclically permutes the roots of f.

The irreducibility of f ensures that f has distinct roots (Corollary 4.29). Let α_{1} and α_{2} be the two roots of f that are not real. Then α_{1} and α_{2} are complex conjugates of one another, since f has real coefficients. We have already seen that G contains an element of order p which cyclically permutes the roots of f. On taking an appropriate power of this element, we obtain an element σ of G that cyclically permutes the roots of f and sends α_{1} to α_{2}. We label the real roots α_{3}, α_{4}, …, α_{p} of f so that α_{j} = σ(α_{j−1}) for j = 2, 3, 4, …, p. Then σ(α_{p}) = α_{1}. Now complex conjugation restricts to a ℚ-automorphism τ of L that interchanges α_{1} and α_{2} but fixes α_{j} for j > 2. But if 2 ≤ j ≤ p then σ^{1−j}τσ^{j−1} transposes the roots α_{j−1} and α_{j} and fixes the remaining roots. But transpositions of this form generate the whole of the group of permutations of the roots. Therefore every permutation of the roots of f is realised by some element of the Galois group G of f, and thus G ≅ Σ_{p}, as required.

Example Consider the quintic polynomial f where f(x) = x^{5} − 6x + 3. Eisenstein’s Irreducibility Criterion (Proposition 4.9) can be used to show that f is irreducible over ℚ. Now f(−2) = −17, f(−1) = 8, f(1) = −2 and f(2) = 23. The Intermediate Value Theorem ensures that f has at least 3 distinct real roots. If f had at least 4 distinct real roots then Rolle’s Theorem would ensure that the number of distinct real roots of f′ and f′′ would be at least 3 and 2 respectively. But zero is the only root of f′′ since f′′(x) = 20x^{3}. Therefore f must have exactly 3 distinct real roots. It follows from Lemma 4.55 that the Galois group of f is isomorphic to the symmetric group Σ_{5}. This group is not solvable. Theorem 4.52 then ensures that the polynomial f is not solvable by radicals over the field of rational numbers.

The above example demonstrates that there cannot exist any general formula for obtaining the roots of a quintic polynomial from its coefficients in a finite number of steps involving only addition, subtraction, multiplication, division and the extraction of nth roots. For if such a general formula were to exist then every quintic polynomial with rational coefficients would be solvable by radicals.