﻿ ﻿GALOIS THEORY: PREAMBLE - A Book of Abstract Algebra

A Book of Abstract Algebra, Second Edition (1982)

Chapter 31. GALOIS THEORY: PREAMBLE

Field extensions were used in Chapter 30 to settle some of the most puzzling questions of classical geometry. Now they will be used to solve a problem equally ancient and important: they will give us a definite and elegant theory of solutions of polynomial equations.

We will be concerned not so much with finding solutions (which is a problem of computation) as with the nature and properties of these solutions. As we shall discover, these properties turn out to depend less on the polynomials themselves than on the fields which contain their solutions. This fact should be kept in mind if we want to clearly understand the discussions in this chapter and Chapter 32. We will be speaking of field extensions, but polynomials will always be lurking in the background. Every extension will be generated by roots of a polynomial, and every theorem about these extensions will actually be saying something about the polynomials.

Let us quickly review what we already know of field extensions, filling in a gap or two as we go along. Let F be a field; an element a (in an extension of F) is algebraic over F if a is a root of some polynomial with its coefficients in F. The minimum polynomial of a over F is the monic polynomial of lowest degree in F[x] having a as a root; every other polynomial in F[x] having a as a root is a multiple of the minimum polynomial.

The basic theorem of field extensions tells us that any polynomial of degree n in F[x] has exactly n roots in a suitable extension of F. However, this does not necessarily mean n distinct roots. For example, in [x] the polynomial (x − 2)5 has five roots all equal to 2. Such roots are called multiple roots. It is perfectly obvious that we can come up with polynomials such as (x − 2)5 having multiple roots; but are there any irreducible polynomials with multiple roots? Certainly the answer is not obvious. Here it is:

Theorem 1 If F has characteristic 0, irreducible polynomials over F can never have multiple roots.

PROOF: To prove this, we must define the derivative of the polynomial a(x) = a0 + a1x + ⋯ +anxn. It is a′(x) = a1 + 2a2x + ⋯ + nanxnl. As in elementary calculus, it is easily checked that for any two polynomials f(x) and g(x),

(f + g)′= f′ + g′ and (fg)′ = fg′ + fg

Now suppose a(x) is irreducible in F[x] and has a multiple root c: then in a suitable extension we can factor a(x) as a(x) = (x − c)2q(x), and therefore a′(x) = 2(xc)q(x) + (xc)2q′ (x). So xc is a factor of a′(x), and therefore c is a root of a′(x). Let p(x) be the minimum polynomial of c over F; since both a(x) and a′(x) have casa root, they are both multiples of p(x).

But a(x) is irreducible: its only nonconstant divisor is itself; so p(x) must be a(x). However, a(x) cannot divide a′(x) unless a′(x) = 0 because a′(x) is of lower degree than a(x). So a′(x) = 0 and therefore its coefficient nan is 0. Here is where characteristic 0 comes in: if nan = 0 then an = 0, and this is impossible because an is the leading coefficient of a(x). ■

In the remaining three chapters we will confine our attention to fields of characteristic 0. Thus, by Theorem 1, any irreducible polynomial of degree n has n distinct roots.

Let us move on with our review. Let E be an extension of F. We call E a finite extension of F if E, as a vector space with scalars in F, has finite dimension. Specifically, if E has dimension n, we say that the degree of E over F is equal to n, and we symbolize this by writing [E: F] = n. If c is algebraic over F, the degree of F(c) over F turns out to be equal to the degree of p(x), the minimum polynomial of c over F.

F(c), obtained by adjoining an algebraic element c to F, is called a simple extension of F. F(cl, …, cn), obtained by adjoining n algebraic elements in succession to F, is called an iterated extension of F. Any iterated extension of F is finite, and, conversely, any finite extension of F is an iterated extension F(c1, …, cn). In fact, even more is true; let F be of characteristic 0.

Theorem 2 Every finite extension of F is a simple extension F(c).

PROOF: We already know that every finite extension is an iterated extension. We will now show that any extension F(a, b) is equal to F(c) for some c. Using this result several times in succession yields our theorem. (At each stage, we reduce by 1 the number of elements that must be adjoined to F in order to get the desired extension.)

Well, given F(a, b), let A(x) be the minimum polynomial of a over F, and let B(x) be the minimum polynomial of b over F. Let K denote any extension of F which contains all the roots a1, …, an of A(x) as well as all the roots b1, …, bm of B(x). Let a1 be a and let b1 be b.

Let t be any nonzero element of F such that

Cross multiplying and setting c = a + tb, it follows that cai + tbj, that is,

ctbjai for all i ≠ 1 and j ≠ 1

while for every ; j ≠ 1,

Thus, b is the only common root of h(x) and B(x).

We will prove that bF(c), hence also a = ctbF(c), and therefore F(a, b) ⊆ F(c). But cF(a, b), so F(c) ⊆ F(a, b). Thus F(a, b) = F(c).

So, it remains only to prove that bF(c). Let p(x) be the minimum polynomial of b over F(c). If the degree of p(x) is 1, then p(x) is xb, so bF(c), and we are done. Let us suppose p(x) 2 and get a contradiction: observe that h(x) and B(x) must both be multiples of p(x) because both have b as a root, and p(x) is the minimum polynomial of b. But if h(x) and B(x) have a common factor of degree 2, they must have two or more roots in common, contrary to the fact that b is their only common root. Our proof is complete. ■

For example, we may apply this theorem directly to (, ). Taking t = 1, we get c = + , hence (, ) = ( + ).

If a(x) is a polynomial of degree n in F[x], let its roots be c1, …, cn. Then F(c1, …, cn) is clearly the smallest extension of F containing all the roots of a(x). F(c1, …, cn) is called the root field of a(x) over F. We will have a great deal to say about root fields in this and subsequent chapters.

Isomorphisms were important when we were dealing with groups, and they are important also for fields. You will remember that if F1 and F2 are fields, an isomorphism from F1 to F2 is a bijective function h: F1F2satisfying

h(a + b) = h(a) + h(b) and h(ab) = h(a)h(b)

From these equations it follows that h(0) = 0, h(l) = 1, h(−a) = − h(a), and h(al) = (h(a))1

Suppose F1 and F2 are fields, and h: F1F2 is an isomorphism. Let K1 and K2 be extensions of F1 and F2, and let : K1K2 also be an isomorphism. We call and extension of if (x) = (x) for every x in Fl that is, if h and are the same on F1. ( is an extension of h in the plain sense that it is formed by “adding on” to h.)

As an example, given any isomorphism h: F1F2, we can extend h to an isomorphism : F1[x]→ F2[x]. (Note that F[x] is an extension of F when we think of the elements of F as constant polynomials; of course, F[x] is not a field, simply an integral domain, but in the present example this fact is unimportant.) Now we ask: What is an obvious and natural way of extending h? The answer, quite clearly, is to let send the polynomial with coefficients a0, a1, …, an to the polynomial with coefficients h(a0), h(a1), …, h(an):

(a0 + a1x + ⋯ + anxn) = h(a0) + h(a1)x + ⋯ + h(an)xn

It is child’s play to verify formally that is an isomorphism from F1[x] to F2[x]. In the sequel, the polynomial (a(x)), obtained in this fashion, will be denoted simply by ha(x). Because is an isomorphism, a(x) is irreducible iff ha(x) is irreducible.

A very similar isomorphism extension is given in the next theorem.

Theorem 3 Let h: F1F2 be an isomorphism, and let p(x) be irreducible in F1[x]. Suppose a is a root of p(x), and b a root of hp(x). Then h can be extended to an isomorphism

: F1,(a) → F2(b)

Furthermore, (a) = b.

PROOF: Remember that every element of F1(a) is of the form

c0 + c1a + ⋯ + cnan

where c0, …, cn are in F1, and every element of F2(b) is of the form d0 + d1b + ⋯ + dnbn where d0, …, dn are in F2. Imitating what we did successfully in the preceding example, we let send the expression with coefficients c0, …, cn to the expression with coefficients h(c0), …, h(cn):

(c0 + c1a + ⋯ + cnan) = h(c0) + h(c1)b + ⋯ + h(cn)bn

Again, it is routine to verify that is an isomorphism. Details are laid out in Exercise H at the end of the chapter. ■

Most often we yse Theorem 3 in the special case where F1 and F2 are the same field—let us call it F— and h is the identity function ε: FF. [Remember that the identity function is ε(x) = x.] When we apply Theorem 3 to the identity function ε: FF, we get

Theorem 4 Suppose a and b are roots of the same irreducible polynomial p(x) in F[x].Then there is an isomorphism g: F(a)→ F(b) such that g(x) = x for every x in F, and g(a) = b.

Now let us consider the following situation: K and K′ are finite extensions of F, and K and K′ have a common extension E. If h : KK′ is an isomorphism such that h(x) = x for every x in F, we say that h fixes F. Let c be an element of K; if h fixes F, and c is a root of some polynomial a(x) = a0 + ⋯ + anxn in F[x], h(c) also is a root of a(x). It is easy to see why: the coefficients of a(x) are in F and are therefore not changed by h. So if a(c) = 0, then

What we have just shown may be expressed as follows:

(*) Let a(x) be any polynomial in F[x]. Any isomorphism which fixes F sends roots of a(x) to roots of a(x).

If K happens to be the root field of a(x) over F, the situation becomes even more interesting. Say K= F(cl, c2, …, cn), where cl, c2, …, cn are the roots of a(x). If h: KK′ is any isomorphism which fixes F, then by (*), h permutes cl, c2, …, cn. Now, by the brief discussion headed “For later reference” on page 296, every element of F(c1, …, cn) is a sum of terms of the form

where the coefficient k is in F. Because h fixes F, h(k) = k. Furthermore, c1, c2, …, cn are the roots of a(x), so by (*), the product is transformed by h into another product of the same form. Thus, h sends every element of F(c1, c2, …, cn) to another element of F(c1, c2, …, cn).

The above comments are summarized in the next theorem.

Theorem 5 Let K and Kbe finite extensions of F. Assume K is the root field of some polynomial over F. If h : KKis an isomorphism which fixes F, then K = K′.

PROOF: From Theorem 2, K and K′ are simple extensions of F, say K = F(a) and K′ = F(b). Then E = F(a, b) is a common extension of K and K′. By the comments preceding this theorem, h maps every element of K to an element of K′; hence K′⊆K. Since the same argument may be carried out for h1, we also have KK′.■

Theorem 5 is often used in tandem with the following (see the figure on the next page):

Theorem 6 Let L and Lbe finite extensions of F. Let K be an extension of L such that K is a root field over F. Any isomorphism h:LLwhich fixes F can be extended to an isomorphism :KK.

PROOF: From Theorem 2, K is a simple extension of L, say K = L(c). Now we can use Theorem 3 to extend the isomorphism h : LL′ to an isomorphism

By Theorem 5 applied to , K = K′. ■

REMARK: It follows from the theorem that L′ ⊆,K since ran h ⊆ ran = K.

For later reference. The following results, which are of a somewhat technical nature, will be needed later. The first presents a surprisingly strong property of root fields.

Theorem 7 Let K be the root field of some polynomial over F. For every irreducible polynomial p(x) in F[x], if p(x) has one root in K, then p(x) must have all of its roots in K.

PROOF: Indeed, suppose p(x) has a root a in K, and let b be any other root of p(x). From Theorem 4, there is an isomorphism h : F(a)→ F(b) fixing F. But F(a) ⊆ K; so from Theorem 6 and the remark following it F(b) ⊆K; hence bK.■

Theorem 8 Suppose IEK, where E is a finite extension of I and K is a finite extension of E. If K is the root field of some polynomial over I, then K is also the root field of some polynomial over E.

PROOF: Suppose K is a root field of some polynomial over I. Then K is a root field of the same polynomial over E. ■

EXERCISES

A. Examples of Root Fields over

Example Find the root field of a(x) = (x2 − 3)(x3 − 1) over .

SOLUTION The complex roots of a(x) are ±,1, (−1 ± i), so the root field is ,1, (−1± i)). The same field can be written more simply as (, i).

1 Show that (, i) is the root field of (x2 −2x − 2)(x2 + 1) over .

Comparing part 1 with the example, we note that different polynomials may have the same root field. This is true even if the polynomials are irreducible.

2 Prove that x2 − 3 and x2 − 2x − 2 are both irreducible over . Then find their root fields over and show they are the same.

3 Find the root field of x4 − 2, first over , then over .

4 Explain: (i, ) is the root field of x4 − 2x2 + 9 over , and is the root field of x2 − 2x + 3 over ().

5 Find irreducible polynomials a(x) over , and b(x) over (i), such that (i, ) is the root field of a(x) over , and is the root field of b(x) over (i). Then do the same for (, ).

# 6 Which of the following extensions are root fields over ? Justify your answer: (i); (); (), where is the real cube root of 2; (2 + ); (i + ); (i, , ).

B. Examples of Root Fields over p

Example Find the root field of x2 + 1 over 3.

SOLUTION By the basic theorem of field extensions,

where u is a root of x2 + 1. In 3(u), x2 + 1 = (x + u)(xu), because u2 + 1 = 0. Since 3(u) contains ± u, it is the root field of x2 + 1 over 3. Note that 3(u) has nine elements, and its addition and multiplication tables are easy to construct. (See Chapter 27, Exercise C4).

1 Show that, in any extension of 3 which contains a root u of

a(x) = x3 + 2x + 1 ∈ 3[x]

it happens that u + 1 and u + 2 are the remaining two roots of a(x). Use this fact to find the root field of x3 + 2x + 1 over 3. List the elements of the root field.

2 Find the root field of x2 + x + 2 over 3, and write its addition and multiplication tables.

3 Find the root field of x3 + x2 + 1 ∈ 2[x] over 2. Write its addition and multiplication tables.

4 Find the root field over 2 of x3 + x + 1 ⊆ 2[x]. (CAUTION: This will prove to be a little more difficult than part 3.)

# 5 Find the root field of x3 + x2 + x + 2 over 3. Find a basis for this root field over 3.

C. Short Questions Relating to Root Field

Prove each of the following

1 Every extension of degree 2 is a root field.

2 If FIK and K is a root field of a(x) over F, then K is a root field of a(x) over I.

3 The root field over of any polynomial in [x] is or .

4 If c is a complex root of a cubic a(x) ∈ [x], then (c) is the root field of a(x) over .

# 5 If p(x) = x4 +ax2 + b is irreducible in F[x], then F[x]/⟨p(x)⟩ is the root field of p(x) over F.

6 If K = F(a) and K is the root field of some polynomial over F, then K is the root field of the minimum polynomial of a over F.

7 Every root field over F is the root field of some irreducible polynomial over F. (HINT: Use part 6 and Theorem 2.)

8 Suppose [K :F] = n, where K is a root field over F. Then K is the root field over F of every irreducible polynomial of degree n in F[x] having a root in K.

9 If a(x) is a polynomial of degree n in F[x], and K is the root field of a(x) over F, then [K: F] divides n!

D. Reducing Iterated Extensions to Simple Extensions

1 Find c such that (, ) = (c). Do the same for (,)

2Let a be a root of x3x + 1, and b a root of x2 − 2x − 1. Find c such that (a, b) = (c). (HINT: Use calculus to show that x3x + 1 has one real and two complex roots, and explain why no two of these may differ by a real number.)

# 3 Find c such that (, ,) = (c).

4 Find an irreducible polynomial p(x) such that (, ) is the root field of p(x) over . (HINT: Use Exercise C6.)

5 Do the same as in part 4 for (, , ).

De Moivre’s theorem provides an explicit formula to write the n complex nth roots of 1. (See Chapter 16, Exercise H.) By de Moivre’s formula, the nth roots of unity consist of ω = cos (2π/n) + i sin(2π/n) and its first n powers, namely, 1, ω,ω2, …, ωn1 We call ω a primitive nth root of unity, because all the other nth roots of unity are powers of ω. Clearly, every nth root of unity (except 1) is a root of

† E. Roots of Unity and Radical Extensions

This polynomial is irreducible if n is a prime (see Chapter 26, Exercise D3). Prove parts 1−3, where ω denotes a primitive nth root of unity.

1 (ω) is the root field of xn 1 over .

2 If n is a prime, [(ω): ] = n − 1.

3 If n is a prime, ωn1 is equal to a linear combination of 1, ω, …, ωn2 with rational coefficients.

4 Find [(ω):], where ω is a primitive nth root of unity, for n = 6, 7, and 8.

5 Prove that for any r ∈ {1, 2, …, n − 1}, ωr is an nth root of a. Conclude that ,ω, …,ωn1 are the n complex nth roots of a.

6 Prove that (ω,) is the root field of xna over .

7 Find the degree of (ω,) over , where ω is a primitive cube root of 1. Also show that (ω,) = (, iV3) (HINT: Compute ω.)

8 Prove that if K is the root field of any polynomial over , and K contains an nth root of any number a, then K contains all the nth roots of unity.

† F. Separable and Inseparable Polynomials

Let F be a field. An irreducible polynomial p(x) in F[x] is said to be separable over F if it has no multiple roots in any extension of F. If p(x) does have a multiple root in some extension, it is inseparable over F.

1 Prove that if F has characteristic 0, every irreducible polynomial in F[x] is separable.

Thus, for characteristic 0, there is no question whether an irreducible polynomial is separable or not. However, for characteristic p ≠ 0, it is different. This case is treated next. In the following problems, let F be a field of characteristic p ≠ 0.

2 If a′(x) = 0, prove that the only nonzero terms of a(x) are of the form ampxmp for some m. [In other words, a(x) is a polynomial in powers of xp.]

3 Prove that if an irreducible polynomial a(x) is inseparable over F, then a(x) is a polynomial in powers of xp. (HINT: Use part 2, and reason as in the proof of Theorem 1.)

4 Use Chapter 27, Exercise J (especially the conclusion following J6) to prove the converse of part 3.

Thus, if F is a field of characteristic p ≠ 0, an irreducible polynomial a(x) ∈ F[x] is inseparable iff a(x) is a polynomial in powers of xp. For finite fields, we can say even more:

5 Prove that if F is any field of characteristic p ≠0, then in F[x],

(HINT: See Chapter 24, Exercise D6.)

6 If F is a finite field of characteristic p ≠ 0, prove that, in F[x], every polynomial a(xp) is equal to [b(x)]p for some b(x). [HINT: Use part 5 and the fact that in a finite field of characteristic p, every element has a pth root (see Chapter 20, Exercise F).]

7 Use parts 3 and 6 to prove: In any finite field, every irreducible polynomial is separable.

Thus, fields of characteristic 0 and finite fields share the property that irreducible polynomials have no multiple roots. The only remaining case is that of infinite fields with finite characteristic. It is treated in the next exercise set.

† G. Multiple Roots over Infinite Fields of Nonzero Characteristic

If p[y] is the domain of polynomials (in the letter y) over p, let E = p(y) be the field of quotients of p[y]. Let K denote the subfield p(yp) of p(y).

1 Explain why p(y) and p(yp) are infinite fields of characteristic p.

2 Prove that a(x) = xpyp has the factorization xpyp = (xy)p in E[x], but is irreducible in K[x]. Conclude that there is an irreducible polynomial a(x) in K[x] with a root whose multiplicity is p.

Thus, over an infinite field of nonzero characteristic, an irreducible polynomial may have multiple roots. Even these fields, however, have a remarkable property: all the roots of any irreducible polynomial have the same multiplicity. The details follow: Let F be any field, p(x) irreducible in F[x], a and b two distinct roots of p(x), and K the root field of p(x) over F. Let i: Ki(K) = K′ be the isomorphism of Theorem 4, and : K[x] → K′[x] the isomorphism described immediately preceding Theorem 3.

3 Prove that leaves p(x) fixed.

4 Prove that ((xa)m) = (xb)m.

5 Prove that a and b have the same multiplicity.

† H. An Isomorphism Extension Theorem (Proof of Theorem 3)

Let F1, F2, h, p(x),a, b, and be as in the statement of Theorem 3. To prove that is an isomorphism, it must first be shown that it is properly defined: that is, if c(a) = d(a) in F1(a), then (c(a)) = (d(a)).

1 If c(a) = d(a), prove that c(x) −d(x) is a multiple of p(x). Deduce from this that hc(x) − hd(x) is a multiple of hp(x).

# 2 Use part 1 to prove that (c(a)) = (d(a)).

3 Reversing the steps of the preceding argument, show that is injective.

4 Show that is surjective.

5 Show that is a homomorphism.

† I. Uniqueness of the Root Field

Let h: F1F2 be an isomorphism. If a(x) ∈ F1[x], let Kl be the root field of a(x) over F1, and K2 the root field of ha(x) over F2.

1 Prove: If p(x) is an irreducible factor of a(x), uK1 is a root of p(x), and υ ∈ K2 is a root of hp(x), then F1(u) ≅ F2(υ).

2 F1(u) = Kl iff F2(υ) = K2.

# 3 Use parts 1 and 2 to form an inductive proof that K1K2.

4 Draw the following conclusion: The root field of a polynomial a(x) over a field F is unique up to isomorphism.

† J. Extending Isomorphism

In the following, let F be a subfield of . An injective homomorphism h: F is called a monomorphism; it is obviously an isomorphism Fh(F).

1 Let ω be a complex pth root of unity (where p is a prime), and let h: (ω)→ be a monomorphism fixing . Explain why h is completely determined by the value of h(ω). Then prove that there exist exactly p −1 monomorphisms (ω)→ which fix .

# 2 Let p(x) be irreducible in F[x], and c a complex root of p(x). Let h: F be a monomorphism. If deg p(x) = n, prove that there are exactly n monomorphisms F(c)→ which are extensions of h.

3 Let FK, with [K: F] = n. If h: F is a monomorphism, prove that there are exactly n monomorphisms K which are extensions of h.

# 4 Prove: The only possible monomorphism h: is h(x) = x. Thus, any monomorphism h: (a) → necessarily fixes .

5 Prove: There are exactly three monomorphisms ()→, and they are determined by the conditions: ;ω;ω2, where ω is a primitive cube root of unity.

K. Normal Extensions

If K is the root field of some polynomial a(x) over F, K is also called a normal extension of F. There are other possible ways of defining normal extensions, which are equivalent to the above. We consider the two most common ones here: they are precisely the properties expressed in theorems 7 and 6. Let K be a finite extension of F.

1 Suppose that for every irreducible polynomial p(x) in F[x], if p(x) has one root in K, then p(x) must have all its roots in K. Prove that K is a normal extension of F.

2 Suppose that, if h is any isomorphism with domain K which fixes F, then h(K) ⊆ K. Prove that K is a normal extension of F.

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