﻿ ﻿GALOIS THEORY: THE HEART OF THE MATTER - A Book of Abstract Algebra

## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 32. GALOIS THEORY: THE HEART OF THE MATTER

If K is a field and h is an isomorphism from K to K, we call h an automorphism of K (automorphism = “self-isomorphism”).

We begin this chapter by restating Theorems 5 and 6 of Chapter 31:

Let K be the root field of some polynomial over F; suppose aK:

(i)Any isomorphism with domain K which fixes F is an automorphism of K.

(ii)If a and b are roots of an irreducible polynomial p(x) in F[x], there is an automorphism of K fixing F and sending a to b.

Rule (i) is merely a restatement of Theorem 5 of Chapter 31, using the notion of automorphism. Rule (ii) is a result of combining Theorem 4 of Chapter 31 [which asserts that there exists an F-fixing isomorphism from L = F(a) to L′ = F(b)] with Theorem 6 of the same chapter.

Let K be the root field of a polynomial a(x) in F[x], If c1, c2, …, cn are the roots of a(x), then K = F(c1, c2, …, cn), and, by (*) on page 316, any automorphism h of K which fixes F permutes c1, c2, …, cn. On the other hand, remember that every element a in F(c1, c2, …, cn) is a sum of terms of the form

where the coefficient k of each term is in F. If h is an automorphism which fixes F, h does not change the coefficients, so h(a) is completely determined once we know h(c1), …, h(cn). Thus, every automorphism of K fixing F is completely determined by a permutation of the roots of a(x).

This is very important!

What it means is that we may identify the automorphisms of K which fix F with permutations of the roots of a(x).

It must be pointed out here that, just as the symmetries of geometric figures determine their geometric properties, so the symmetries of equations (that is, permutations of their roots) give us all the vital information needed to analyze their solutions. Thus, if K is the root field of our polynomial a(x) over F, we will now pay very close attention to the automorphisms of K which fix F.

To begin with, how many such automorphisms are there? The answer is a classic example of mathematical elegance and simplicity.

Theorem 1 Let K be the root field of some polynomial over F. The number of automorphisms of K fixing F is equal to the degree of K over F.

PROOF: Let [K : F] = n, and let us show that K has exactly n automorphisms fixing F. By Theorem 2 of Chapter 31, K = F(a) for some aK.Let p(x) be the minimum polynomial of a over F; if b is any root of p(x), then by (ii) on the previous page, there is an automorphism of K fixing F and sending a to b. Since p(x) has n roots, there are exactly n choices of b, and therefore n automorphisms of K fixing F.

[Remember that every automorphism h which fixes F permutes the roots of p(x) and therefore sends a to some root of p(x); and h is completely determined once we have chosen h(a).] ■

For example, we have already seen that () is of degree 2 over . () is the root field of x2 – 2 over because () contains both roots of x2 – 2, namely ±. By Theorem 1, there are exactly two automorphisms of () fixing : one sends to ; it is the identity function. The other sends to –, and is therefore the function a + bab.

Similarly, we saw that = (i), and is of degree 2 over . The two automorphisms of which fix are the identity function and the function a + biabi which sends every complex number to its complex conjugate.

As a final example, we have seen that (, ) is an extension of degree 4 over , so by Theorem 1, there are four automorphisms of (, ) which fix : Now, (, ) is the root field of (x2 – 2)(x2 − 3) over for it contains the roots of this polynomial, and any extension of containing the roots of (x2 – 2)(x2 − 3) certainly contains and . Thus, by (*) on page 316, each of the four automorphisms which fix sends roots of x2 – 2 to roots of x2 − 2, and roots of x2 − 3 to roots of x2 – 3. But there are only four possible ways of doing this, namely,

Since every element of (, ) is of the form , these four automorphisms (we shall call them ε, α, β, and γ) are the following:

If K is an extension of F, the automorphisms of K which fix F form a group. (The operation, of course, is composition.) This is perfectly obvious: for if g and h fix F, then for every x in F,

that is, gh fixes F. Furthermore, if

that is, if h fixes F so does h–1

This fact is perfectly obvious, but nonetheless of great importance, for it means that we can now use all of our accumulated knowledge about groups to help us analyze the solutions of polynomial equations. And that is precisely what Galois theory is all about.

If K is the root field of a polynomial a(x) in F[x], the group of all the automorphisms of K which fix F is called the Galois group of a(x). We also call it the Galois group of K over F, and designate it by the symbol

Gal(K : F)

In our last example we saw that there are four automorphisms of (, ) which fix . We called them ε, α, β, and γ. Thus, the Galois group of (, ) over is Gal((, ) : ) = {ε, α, β, γ}; the operation is composition, giving us the table

As one can see, this is an abelian group in which every element is its own inverse; almost at a glance one can verify that it is isomorphic to 2 × 2.

Let K be the root field of a(x), where a(x) is in F[x]. In our earlier discussion we saw that every automorphism of K fixing F [that is, every member of the Galois group of a(x)] may be identified with a permutation of the roots of a(x). However, it is important to note that not every permutation of the roots of a(x) need be in the Galois group of a(x), even when a(x) is irreducible. For example, we saw that (, ) = ( + ), where + is a root of the irreducible polynomial x4 – 10x2 + 1 over . Since x4 – 10x2 + 1 has four roots, there are 4! = 24 permutations of its roots, only four of which are in its Galois group. This is because only four of the permutations are genuine symmetries of x4 – 10x2 + 1, in the sense that they determine automorphisms of the root field.

In the discussion throughout the remainder of this chapter, let F and K remain fixed. F is an arbitrary field and K is the root field of some polynomial a(x) in F[x]. The thread of our reasoning will lead us to speak about fields Iwhere FIK, that is, fields “between” F and K. We will

refer to them as intermediate fields. Since K is the root field of a(x) over F, it is also the root field of a(x) over I for every intermediate field I.

The letter G will denote the Galois group of K over F. With each intermediate field I, we associate the group

I* = Gal(K : I)

that is, the group of all the automorphisms of K which fix I. It is obviously a subgroup of G. We will call I* the fixer of I.

Conversely, with each subgroup H of G we associate the subfield of K containing all the a in K which are not changed by any πH. That is,

{aK : π(a) = a for every πH}

One verifies in a trice that this is a subfield of K. It obviously contains F, and is therefore one of the intermediate fields. It is called the fixed field of H. For brevity and euphony we call it the fixfield of H.

Let us recapitulate: Every subgroup H of G fixes an intermediate field I, called the fixfield of H. Every intermediate field I is fixed by a subgroup H of G, called the fixer of I. This suggests very strongly that there is a one-to-one correspondence between the subgroups of G and the fields intermediate between F and K. Indeed, this is correct. This one-to-one correspondence is at the very heart of Galois theory, because it provides the tie-in between properties of field extensions and properties of subgroups.

Just as, in Chapter 29, we were able to use vector algebra to prove new things about field extensions, now we will be able to use group theory to explore field extensions. The vector-space connection was a relative lightweight. The connection with group theory, on the other hand, gives us a tool of tremendous power to study field extensions.

We have not yet proved that the connection between subgroups of G and intermediate fields is a one-to-one correspondence. The next two theorems will do that.

Theorem 2 If H is the fixer of I, then I is the fixfield of H.

PROOF: Let H be the fixer of I, and I′ be the fixfield of H. It follows from the definitions of fixer and fixfield that II′, so we must now show that I′ ⊆ I. We will do this by proving that aI implies aI′. Well, if a is an element of K which is not in I, the minimum polynomial p(x) of a over I must have degree ≥2 (for otherwise, aI). Thus, p(x) has another root b. By Rule (ii) given at the beginning of this chapter, there is an automorphism of Kfixing I and sending a to b. This automorphism moves a, so aI′. ■

Lemma Let H be a subgroup of G, and I the fixfield of H. The number of elements in H is equal to [K : I].

PROOF: Let H have r elements, namely, h1, …, hr. Let K = I(a). Much of our proof will revolve around the following polynomial:

b(x) = [xh1(a)][xh2(a)] ⋯ [xhr(a)]

Since one of the hi is the identity function, one factor of b(x) is (xa), and therefore a is a root of b(x). In the next paragraph we will see that all the coefficients of b(x) are in I, so b(x) ∈ I [x]. It follows that b(x) is a multiple of the minimum polynomial of a over I, whose degree is exactly [K : I]. Since b(x) is of degree r, this means that r ≥ [K : I], which is half our theorem.

Well, let us show that all the coefficients of b(x) are in I. We saw on page 314 that every isomorphism hi : KK can be extended to an isomorphism i, : K[x] → K[x]. Because i is an isomorphism of polynomials, we get

But hih1, hih2, …, hihr are r distinct elements of H, and H has exactly r elements, so they are all the elements of H (that is, they are h1, …, hr, possibly in a different order). So the factors of i (b(x)) are the same as the factors of b(x), merely in a different order, and therefore i(b(x)) = b(x). Since equal polynomials have equal coefficients, hi leaves the coefficients of b(x) invariant. Thus, every coefficient of b(x) is in the fixfield of H, that is, in I.

We have just shown that [K : I] ≤ r. For the opposite inequality, remember that by Theorem 1, [K : I] is equal to the number of I-fixing automorphisms of K. But there are at least r such automorphisms, namely h1, …, hr. Thus, [K : I] ≥ r, and we are done. ■

Theorem 3 If I is the fixfield of H, then H is the fixer of I.

PROOF: Let I be the fixfield of H, and I* the fixer of I. It follows from the definitions of fixer and fixfield that HI*. We will prove equality by showing that there are as many elements in H as in I*. By the lemma, the order of H is equal to [K : I]. By Theorem 2, I is the fixfield of I*, so by the lemma again, the order of I* is also equal to [K : I]. ■

It follows immediately from Theorems 2 and 3 that there is a one-to-one correspondence between the subgroups of Gal(K : F) and the intermediate fields between K and F. This correspondence, which matches every subgroup with its fixfield (or, equivalently, matches every intermediate field with its fixer) is called a Galois correspondence. It is worth observing that larger subfields correspond to smaller subgroups; that is,

As an example, we have seen that the Galois group of (, ) over is G = {ε, α, β, γ} with the table given on page 325. This group has exactly five subgroups—namely, {ε}, {ε, α}, {ε, β}, {ε, γ}, and the whole group G. They may be represented in the “inclusion diagram”:

On the other hand, there are exactly five fields intermediate between and (, ), which may be represented in the inclusion diagram:

If H is a subgroup of any galois group, let designate the fixfield of H. The subgroups of G in our example have the following fixfields:

(This is obvious by inspection of the way ε, α, β, and γ were defined on page 325.) The Galois correspondence, for this example, may therefore be represented as follows:

In order to effectively tie in subgroups of G with extensions of the field F, we need one more fact, to be presented next.

Suppose EIK, where K is a root field over E and I is a root field over E. (Hence by Theorem 8 of Chapter 31, K is a root field over I.) If hGal(K : E), h is an automorphism of K fixing E. Consider the restriction of h to I, that is, h restricted to the smaller domain I. It is an isomorphism with domain I fixing E, so by Rule (i) given at the beginning of this chapter, it is an automorphism of I, still fixing E. We have just shown that if hGal(K : E), then the restriction of h to I is in Gal(I : E). This permits us to define a function μ : Gal(K : E) → Gal(I : E) by the rule

μ(h) = the restriction of h to I

It is very easy to check that μ is a homomorphism. μ is surjective, because every E-fixing automorphism of I can be extended to an E-fixing automorphism of K, by Theorem 6 in Chapter 31.

Finally, if hGal(K : E), the restriction of h to I is the identity function iff h(x) = x for every xI, that is, iff h fixes I. This proves that the kernel of μ is Gal(K : I).

To recapitulate: μ is a homomorphism from Gal(K : E) onto Gal(I : E) with kernel Gal(K : I). By the FHT, we immediately conclude as follows:

Theorem 4 Suppose EIK, where I is a root field over E and K is a root field over E. Then

It follows, in particular, that Gal(K : I) is a normal subgroup of Gal(K : E).

EXERCISES

† A. Computing a Galois Group

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

# 2Find the degree of (i, ) over .

3List the elements of Gal((i, ) : ) and exhibit its table.

4Write the inclusion diagram for the subgroups of Gal((i, ) : ), and the inclusion diagram for the fields intermediate between and (i, ). Indicate the Galois correspondence.

† B. Computing a Galois Group of Eight Elements

1Show that (, , ) is the root field of (x2 – 2)(x2 – 3)(x2 − 5) over .

2Show that the degree of (, , ) over is 8.

3List the eight elements of G = Gal((, , ) : ) and write its table.

4List the subgroups of G. (By Lagrange’s theorem, any proper subgroup of G has either two or four elements.)

5For each subgroup of G, find its fixfield.

6Indicate the Galois correspondence by means of a diagram like the one on page 329.

† C. A Galois Group Equal to S3

1Show that (, i) is the root field of x3 – 2 over , where designates the real cube root of 2. (HINT: Compute the complex cube roots of unity.)

2Show that [() : ] = 3.

3Explain why x2 + 3 is irreducible over (), then show that [(, i): ()] = 2. Conclude that [(, i) : ] = 6.

4Use part 3 to explain why Gal((, ) : ) has six elements. Then use the discussion following Rule (ii) on page 323 to explain why every element of Gal((, i) : ) may be identifed with a permutation of the three cube roots of 2.

5Use part 4 to prove that Gal((, i) : ) ≅ S3.

† D. A Galois Group Equal to D4

If α = is a real fourth root of 2, then the four fourth roots of 2 are ±α and ±. Explain parts 1–6, briefly but carefully:

# 1 (α, i) is the root field of x4 − 2 over .

2 [(α) : ] = 4.

3 i(α); hence [(α, i) : (α)] = 2.

4 [(α, i) : ] = 8.

5 {1, α, α2, α3, i, , 2, 3} is a basis for (α, i) over .

6 Any -fixing automorphism h of (α, i) is determined by its effect on the elements in the basis. These, in turn, are determined by h(α) and h(i).

7 Explain: h(α) must be a fourth root of 2 and h(i) must be equal to ±i. Combining the four possibilities for h(α) with the two possibilities for h(i) gives eight possible automorphisms. List them in the format

8 Compute the table of the group Gal((α, i) : ) and show that it is isomorphic to D4, the group of symmetries of the square.

† E. A Cyclic Galois Group

# 1 Describe the root field K of x7 − 1 over . Explain why [K : ] = 6.

2 Explain: If α is a primitive seventh root of unity, any hGal(K : ) must send α to a seventh root of unity. In fact, h is determined by h(α).

3 Use part 2 to list explicitly the six elements of Gal(K : ). Then write the table of Gal(K : ) and show that it is cyclic.

4 List all the subgroups of Gal(K : ), with their fixfields. Exhibit the Galois correspondence.

5 Describe the root field L of x6 − 1 over , and show that [L : ] = 2. Explain why it follows that there are no intermediate fields between and L (except for and L themselves).

# 6 Let L be the root field of x6 – 2 over . List the elements of Gal(L : ) and write its table.

† F. A Galois Group Isomorphic to S5

Let a(x) = x5 − 4x4 + 2x + 2 ∈ [x], and let r1, …, r5 be the roots of a(x) in . Let K (r1, …, r5) be the root field of a(x) over .

Prove: parts 1–3:

1 a(x) is irreducible in x⌉.

2 a(x) has three real and two complex roots. [HINT: Use calculus to sketch the graph of y = a(x), and show that it crosses the x axis three times.]

3 If r1 denotes a real root of a(x), [(r1) : ] = 5. Use this to prove that [K : ] is a multiple of 5.

4 Use part 3 and Cauchy’s theorem (Chapter 13, Exercise E) to prove that there is an element α of order 5 in Gal(K : ). Since α may be identified with a permutation of {r1, …, r5}, explain why it must be a cycle of length 5. (HINT: Any product of disjoint cycles on {r1, …, r5} has order ≠ 5.)

5 Explain why there is a transposition in Gal (K : ). [It permutes the conjugate pair of complex roots of a(x).]

6 Prove: Any subgroup of S5 which contains a cycle of length 5 and a transposition must contain all possible transpositions in S5, hence all of S5. Thus, Gal(K : ) = S5.

G. Shorter Questions Relating to Automorphisms and Galois Groups

Let F be a field, and K a finite extension of F. Suppose a, bK. Prove parts 1–3:

1 If an automorphism h of K fixes F and a, then h fixes F(a).

2 F(a, b)* = F(a)* F(b)*.

3 Aside from the identity function, there are no -fixing automorphisms of (). [HINT: Note that () contains only real numbers.]

4 Explain why the conclusion of part 3 does not contradict Theorem 1.

In the next three parts, let ω be a primitive pth root of unity, where p is a prime.

5 Prove: If hGal((ω) : ), then h(ω) = ωk for some k where 1 ≤ kp − 1.

6 Use part 5 to prove that Gal((ω) : ) is an abelian group.

7 Use part 5 to prove that Gal((ω) : ) is a cyclic group.

† H. The Group of Automorphisms of C

1 Prove: The only automorphism of is the identity function. [HINT: If h is an automorphism, h(1) = 1; hence h(2) = 2, and so on.]

2 Prove: Any automorphism of sends squares of numbers to squares of numbers, hence positive numbers to positive numbers.

3 Using part 2, prove that if h is any automorphism of , a < b implies h(a) < h(b).

# 4 Use parts 1 and 3 to prove that the only automorphism of is the identity function.

5 List the elements of Gal( : ).

6 Prove that the identity function and the function a + biabi are the only automorphisms of which fix .

I. Further Questions Relating to Galois Groups

Throughout this set of questions, let K be a root field over F, let G = Gal(K : F), and let I be any intermediate field. Prove the following:

1 I* = Gal(K : I) is a subgroup of G.

2 If H is a subgroup of G and H° = {aK : π(a) = a for every πH}, then H° is a subfield of K, and FH°.

3 Let H be the fixer of I, and I′ the fixfield of H. Then II′. Let I be the fixfield of H, and I* the fixer of I. Then HI*.

# 4 Let I be a normal extension of F (that is, a root field of some polynomial over F). If G is abelian, then Gal(K : I) and Gal(I : F) are abelian. (HINT: Use Theorem 4.)

5 Let I be a normal extension of F. If G is a cyclic group, then Gal(K : I) and Gal(I : F) are cyclic groups.

6 If G is a cyclic group, there exists exactly one intermediate field I of degree k, for each integer k dividing [K : F].

† J. Normal Extensions and Normal Subgroups

Suppose FK, where K is a normal extension of F. (This means simply that K is the root field of some polynomial in F[x]: see Chapter 31, Exercise K.) Let I1I2 be intermediate fields.

1 Deduce from Theorem 4 that, if I2 is a normal extension of I1, then is a normal subgroup of .

2 Prove the following for any intermediate field I: Let hGal (K : F), gI*, aI, and b = h(a). Then [hgh 1](b) = b. Conclude that

hI* h 1h(I)*

3 Use part 2 to prove that hI* h 1 = h(I)*.

Two intermediate fields I1 and I2 are called conjugate iff there is an automorphism [i.e., an element iGal(K : F)] such that i(I1) = I2.

4 Use part 3 to prove that I1 and I2 are conjugate iff and are conjugate subgroups in the Galois group.

5 Use part 4 to prove that for any intermediate fields I1 and I2 : iff is a normal subgroup of , then I2 is a normal extension of I1.

Combining parts 1 and 5 we have: I2 is a normal extension of I1 iff is a normal subgroup of . (Historically, this result is the origin of the word “normal” in the term “normal subgroup.”)

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