﻿ ﻿DEGREES OF FIELD EXTENSIONS - A Book of Abstract Algebra

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

### Chapter 29. DEGREES OF FIELD EXTENSIONS

In this chapter we will see how the machinery of vector spaces can be applied to the study of field extensions.

Let F and K be fields. If K is an extension of F, we may regard K as being a vector space over F. We may treat the elements in K as “vectors”

and the elements in F as “scalars.” That is, when we add elements in K, we think of it as vector addition; when we add and multiply elements in F, we think of this as addition and multiplication of scalars; and finally, when we multiply an element of F by an element of K, we think of it as scalar multiplication.

We will be especially interested in the case where the resulting vector space is of finite dimension. If K, as a vector space over F, is of finite dimension, we call K a finite extension of F. If the dimension of the vector space K is n, we say that K is an extension of degree n over F. This is symbolized by writing

[K : F] = n

which should be read, “the degree of K over F is equal to n.”

Let us recall that F(c) denotes the smallest field which contains F and c. This means that F(c) contains F and c, and that any other field K containing F and c must contain F(c). We saw in Chapter 27 that if c is algebraic over F, then F(c) consists of all the elements of the form a(c), for all a(x) in F[x]. Since F(c) is an extension of F, we may regard it as a vector space over F. Is F(c) a finite extension of F?

Well, let c be algebraic over F, and let p(x) be the minimum polynomial of c over F. [That is, p(x) is the monic polynomial of lowest degree having casa root.] Let the degree of the polynomial p(x) be equal to n. It turns out, then, that the n elements

1, c,c2, …, cn 1

are linearly independent and span F(c). We will prove this fact in a moment, but meanwhile let us record what it means. It means that the set of n “vectors” {1,c,c2, …, cn 1} is a basis of F(c); hence F(c) is a vector space of dimension n over the field F. This may be summed up concisely as follows:

Theorem 1 The degree of F(c) over F is equal to the degree of the minimum polynomial of c over F.

PROOF: It remains only to show that the n elements 1, c, …, cn l span F(c) and are linearly independent. Well, if a(c) is any element of F(c), use the division algorithm to divide a(x) by p(x):

a(x) = p(x)q(x) + r(x) where deg r(x)≤n − 1

Therefore,

This shows that every element of F(c) is of the form r(c) where r(x) has degree n − l or less. Thus, every element of F(c) can be written in the form

a0 + a1c + ⋯ + an 1cn 1

which is a linear combination of 1,c,c2, …, cn 1.

Finally, to prove that 1,c,c2, …, cn l are linearly independent, suppose that a0 + a1c + ⋯ + an lcn 1 = 0. If the coefficients a0,a1, …, an 1 were not all zero, c would be the root of a nonzero polynomial of degree n − 1 or less, which is impossible because the minimum polynomial of c over F has degree n. Thus, a0 = a1 = ⋯ = an 1 = 0.■

For example, let us look at (): the number is not a root of any monic polynomial of degree 1 over . For such a polynomial would

have to be ,and the latter is not in [x] because is irrational. However, is a root of x2 − 2, which is therefore the minimum polynomial of over , and which has degree 2. Thus,

In particular, every element in () is therefore a linear combination of 1 and ,that is, a number of the form where a,b.

As another example, i is a root of the irreducible polynomial of over x2 + 1 in [x].Therefore x2 + 1 is the minimum polynomial of i over x2 + 1 has degree 2, so [(i) : ] = 2. Thus, (i) consists of all the linear combinations of 1 and i with real coefficients, that is, all the a + bi where a,b.Clearly then, (i) = , so the degree of over is equal to 2.

In the sequel we will often encounter the following situation: E is a finite extension of K, where K is a finite extension of F. If we know the

degree of E over K and the degree of K over F, can we determine the degree of E over F This is a question of major importance! Fortunately, it has an easy answer, based on the following lemma:

Lemma Let a1a2, …, am be a basis of the vector space K over F, and let b1,b2, …, bn be a basis of the vector space E over K. Then the set of mn products {aibj} is a basis of the vector space E over the field F.

PROOF: To prove that the set {aibj} spans E, note that each element c in E can be written as a linear combination c = k1b1 + ⋯ + knbn with coefficients ki in K. But each k,i because it is in K, is a linear combination

ki = li1a1 + ⋯ + lim am

with coefficients lij in F. Substituting,

and this is a linear combination of the products aibj with coefficient lij in F.

To prove that {aibj} is linearly independent, suppose ∑ lijaibj = 0. This can be written as

(l11a1 + ⋯ + l1m am) b1 + ⋯ + (ln1a1 + ⋯ + lnmam)bn = 0

and since b1, …, bn are independent, lila1 + · · · + limam = 0 for each i. But a1, …, am are also independent, so every lij = 0. ■

With this result we can now conclude the following:

Theorem 2 Suppose FKE where E is a finite extension of K and K is a finite extension of F.Then ∈ is a finite extension of F, and

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

This theorem is a powerful tool in our study of fields. It plays a role in field theory analogous to the role of Lagrange’s theorem in group theory. See what it says about any two extensions, K and E of a fixed “base field” F : If Kis a subfield of E9 then the degree of K (over F) divides the degree of (over F).

If c is algebraic over F, we say that F(c) is obtained by adjoining c to F. If c and d are algebraic over F, we may find adjoin c to F, thereby obtaining F(c), and then adjoin d to F(c). The resulting field is denoted F(c, d) and is the smallest field containing F, c and d. [Indeed, any field containing F, c and d must contain F(c), hence also F(c,d).] It does not matter whether we first adjoin c and then d or vice versa.

If c1, …, cn are algebraic over F, we let F(c1, …, cn) be the smallest field containing F and c1, …, cn. We call it the field obtained by adjoining c1,…,cn to F. We may form F(c1, …, cn) step by step, adjoining one ci at a time, and the order of adjoining the ci is irrelevant.

An extension F(c) formed by adjoining a single element to F is called a simple extension of F. An extension F(c1, …, cn)formed by adjoining a finite number of elements c1, …, cnis called an iterated extension. It is called “iterated” because it can be formed step by step, one simple extension at a time:

F ⊆ F(c1) ⊆ F(c1,c2) ⊆ F(cl,c2,c3) ⊆ … ⊆ F(c1, …, cn (1)

If c1, …, cn are algebraic over F, then by Theorem 1, each extension in Condition (1) is a finite extension. By Theorem 2, F(c1 c2) is a finite extension of F; applying Theorem 2 again, F(c1,c2,c3) is a finite extension of F; and so on. So finally, if c1, …, cn are algebraic over F, then F(c1, …, cn) is a finite extension of F.

Actually, the converse is true too: every finite extension is an iterated extension. This is obvious: for if K is a finite extension of F, say an extension of degree n, then K has a basis {a1, …,an} over F. This means that every element in K is a linear combination of a1, …, an with coefficients in F; but any field containing F and a1, …, an obviously contains all the linear combinations of a1, …, an; hence K is the smallest field containing F and a1, …, an. That is, K = F(a1, …, an).

In fact, if K is a finite extension of F and K = F(a1, …, an), then a1, …, an have to be algebraic over F. This is a consequence of a simple but important little theorem:

Theorem 3 If K is a finite extension of F, every element of K is algebraic over F

PROOF: Indeed, suppose K is of degree n; over F, and let c be any element of K. Then the set {1,c,c2, …, cn) is linearly dependent, because it has n + 1 elements in a vector space K of dimension n. Consequently, there are scalars a0,…,anF, not all zero, such that a0 + a1c + ⋯ + ancn = 0. Therefore c is a root of the polynomial a(x) = a0 + a1x + ⋯ + anxn in F[x].■

Let us sum up: Every iterated extension F(c1, …, cn), where c1, …, cn are algebraic over F, is finite extension of extension of F. Conversely, every finite extension of F is an iterated extension F(c1, …, cn), where c1, …, cn are algebraic over F.

Here is an example of the concepts presented in this chapter. We have already seen that () is of degree 2 over , and therefore () consists of all the numbers a + b where a,b. Observe that cannot be in(); for if it were, we would have = a + b for rational a and b; squaring both sides and solving for would give us = a rational number, which is impossible.

Since is not in (), cannot be a root of a polynomial of degree 1 over () (such a polynomial would have to be x). But is a root of x2 − 3, which is therefore the minimum polynomial of over ().Thus, (,) is of degree 2 over (), and therefore by Theorem 2, (,) is of degree 4 over .

By the comments preceding Theorem 1, {1,} is a basis of () over ,and {1,} is a basis of (,) over (). Thus, by the lemma of this chapter, {1,,,} is a basis of (,) over . This means that (,) consists of all the numbers a + b+ c + d , for all a b,c, and d in .

For later reference. The technical observation which follows will be needed later.

By the comments immediately preceding Theorem 1, every element of F(c1) is a linear combination of powers of c1, with coefficients in F. That is, every element of F(c1) is of the form

where the ki are in F. For the same reason, every element of F(c1 c2) is of the form

where the coefficients lj are in F(c1).Thus, each coefficient lj is equal to a sum of the form (2). But then, clearing brackets, it follows that every element of F(c1c2) is of the form

where the coefficients kij are in F.

If we continue this process, it is easy to see that every element of F(c1,c2,…, cn) is a sum of terms of the form

where the coefficient k of each term is in F.

EXERCISES

A. Examples of Finite Extensions

1 Find a basis for (i) over , and describe the elements of (i).(See the two examples immediately following Theorem 1.)

2 Show that every element of (2 + 3i) can be written as a + bi, where a,b Conclude that (2 + 3i) = .

# 3 If , show that {1, 21/3, 22/3, a,21/3a, 22/3a} is a basis of (a) over . Describe the elements of (a).

# 4 Find a basis of () over , and describe the elements of .

5 Find a basis of over and describe the elements of ().(See the example at the end of this chapter.)

6 Find a basis of (,,) over , and describe the elements of (,,.

7 Name an extension of over which π is algebraic of degree 3.

† B. Further Examples of Finite Extensions

Let F be a field of characteristic ≠ 2. Let ab be in F.

1 Prove that any field F containing + also contains and .[HINT: Compute ( + )2 and show that F. Then compute ( + ), which is also in F] Conclude that F( + ) = F(,).

2 Prove that if bx2a for any xF, then F(). Conclude that F(,) is of degree 4 over F.

3 Show that x = + satisfies x4 − 2(a + b)x2 + (a − b)2 = 0. Show that x = also satisfies this equation. Conclude that

4 Using parts 1 to 3, find an uncomplicated basis for (d) over , where d is a root of x4 − 14x2 + 9. Then find a basis for over .

C. Finite Extensions of Finite Fields

By the proof of the basic theorem of field extensions, if p(x) is an irreducible polynomial of degree n in F[x], then F[x]/p(x)⟩ ≅ F(c) where c is a root of p(x). By Theorem 1 in this chapter, F(c) is of degree η over F. Using the paragraph preceding Theorem 1:

1 Prove that every element of F(c) can be written uniquely as a0 + a1c + ⋯ +an lcn 1, for some a0,…, an 1F.

# 2 Construct a field of four elements. (It is to be an extension of 2.) Describe its elements, and supply its addition and multiplication tables.

3 Construct a field of eight elements. (It is to be an extension of 2).

4 Prove that if F has q elements, and a is algebraic over F of degree n, then F(a) has qn elements.

5 Prove that for every prime number p, there is an irreducible quadratic in p[x]. Conclude that for every prime p, there is a field with p2 elements.

D. Degrees of Extensions (Applications of Theorem 2)

Let F be a field, and K a field extension of F. Prove the following:

1 [K:F] = 1 iff K = F.

# 2 If [K : F] is a prime number, there is no field properly between F and K (that is, there is no field L such that F L K).

3 If [K:F] is a prime, then K = F(a) for every aKF.

4 Suppose a,bK are algebraic over F with degrees m and η, where m and « are relatively prime. Then:

(a)F(a,b)is degree mn over F.

(b) F(a) F(b)= F.

5 If the degree of F(a) over F is a prime, then F(a) = F(an) for any n (on the condition that anF).

6 If an irreducible polynomial p(x) ∈ F[x] has a root in K, then deg p(x)|[K:F].

E. Short Questions Relating to Degrees of Extensions

Let F be a field.

Prove parts 1−3:

1 The degree of a over F is the same as the degree of 1/a over F. It is also the same as the degrees of a + c and ac over F, for any cF.

2 a is of degree 1 over F iff aF.

3 If a real number c is a root of an irreducible polynomial of degree >1 in [x], then c is irrational.

4 Use part 3 and Eisentein ’ s irreducibility criterion to prove that (where ra, m n) is irrational if there is a prime number which divides ra but not n,and whose square does not divide ra.

5 Show that part 4 remains true for where q >1.

6 If a and b are algebraic over F, prove that F(a, b) is a finite extension of F.

† F. Further Properties of Degrees of Extensions

Let F be a field, and K a finite extension of F. Prove each of the following:

1 Any element algebraic over K is algebraic over F, and conversely.

2 If b is algebraic over K, then [F(b :F] | [K(b) :F].

3 If fe is algebraic over K, then [K(b) :K] ≤ [F(b) : F]. (HINT: The minimum polynomial of b over F may factor in K[x], and 6 will then be a root of one of its irreducible factors.)

# 4 If b is algebraic over K, then [K(b):F(b)] ≤ [K :F]. [HINT: Note that FKK(b) and FF(b) ⊆ K(b). Relate the degrees of the four extensions involved here, using part 3.]

# 5 Let p(x) be irreducible in F[x]. If [K : F] and deg p(x) are relatively prime, then p(x) is irreducible in K[x].

† G. Fields of Algebraic Elements: Algebraic Numbers

Let FK and a,bK. We have seen on page 295 that if a and b are algebraic over F, then F(a, b) is a finite extension of F.

Use the above to prove parts 1 and 2.

1 If a and b are algebraic over F, then a + b, ab, ab, and a / b are algebraic over F. (In the last case, assume b ≠ 0.)

2 The set {xK : x is algebraic over F} is a subfield of K, containing F.

Any complex number which is algebraic over is called an algebraic number. By part 2, the set of all the algebraic numbers is a field, which we shall designate by .

Let a(x) = a0 + a1x + ⋯ + anxn be in [x], and let c be any root of a(x). We will prove that c. To begin with, all the coefficients of a(x) are in (a0,a1, …,an).

3 Prove: (a0, a1, …, an) is a finite extension of .

Let (a0, …, an) = 1 Since a(x) ∈ 1[x], c is algebraic over 1 Prove parts 4 and 5:

4 1(c) is a finite extension of 1 hence a finite extension of . (Why?)

5 c.

Conclusion: The roots of any polynomial whose coefficients are algebraic numbers are themselves algebraic numbers.

A field F is called algebraically closed if the roots of every polynomial in F[x] are in F. We have thus proved that is algebraically closed.

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