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, …, c^{n − 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, …, c^{n − 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, …, c^{n − 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 + ⋯ + a_{n − 1}c^{n − 1}

which is a linear combination of 1,c,c², …, c^{n − 1}.

Finally, to prove that 1,c,c², …, c^{n − l} are linearly independent, suppose that a₀ + a₁c + ⋯ + a_{n − l}c^{n − 1} = 0. If the coefficients a₀,a₁, …, a_{n − 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 = ⋯ = a_{n − 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, …, a_m be a basis of the vector space K over F, and let b₁,b₂, …, b_n be a basis of the vector space E over K. Then the set of mn products {a_ib_j} is a basis of the vector space E over the field F.

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

k_i = l_i1a₁ + ⋯ + l_im a_m

with coefficients l_ij in F. Substituting,

and this is a linear combination of the products a_ib_j with coefficient l_ij in F.

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

(l₁₁a₁ + ⋯ + l_1m a_m) b₁ + ⋯ + (l_n1a₁ + ⋯ + l_nma_m)b_n = 0

and since b₁, …, b_n are independent, l_ila₁ + · · · + l_ima_m = 0 for each i. But a₁, …, a_m are also independent, so every l_ij = 0. ■

With this result we can now conclude the following:

Theorem 2 Suppose F ⊆ K ⊆ E 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, …, c_n are algebraic over F, we let F(c₁, …, c_n) be the smallest field containing F and c₁, …, c_n. We call it the field obtained by adjoining c₁,…,c_n to F. We may form F(c₁, …, c_n) step by step, adjoining one c_i at a time, and the order of adjoining the c_i 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, …, c_n)formed by adjoining a finite number of elements c₁, …, c_nis 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(c_l,c₂,c₃) ⊆ … ⊆ F(c₁, …, c_n (1)

If c₁, …, c_n 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, …, c_n are algebraic over F, then F(c₁, …, c_n) 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, …,a_n} over F. This means that every element in K is a linear combination of a₁, …, a_n with coefficients in F; but any field containing F and a₁, …, a_n obviously contains all the linear combinations of a₁, …, a_n; hence K is the smallest field containing F and a₁, …, a_n. That is, K = F(a₁, …, a_n).

In fact, if K is a finite extension of F and K = F(a₁, …, a_n), then a₁, …, a_n 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,…,a_n ∈ F, not all zero, such that a₀ + a₁c + ⋯ + a_ncⁿ = 0. Therefore c is a root of the polynomial a(x) = a₀ + a₁x + ⋯ + a_nxⁿ in F[x].■

Let us sum up: Every iterated extension F(c₁, …, c_n), where c₁, …, c_n are algebraic over F, is finite extension of extension of F. Conversely, every finite extension of F is an iterated extension F(c₁, …, c_n), where c₁, …, c_n 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 k_i are in F. For the same reason, every element of F(c₁ c₂) is of the form

where the coefficients l_j are in F(c₁).Thus, each coefficient l_j 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 k_ij are in F.

If we continue this process, it is easy to see that every element of F(c₁,c₂,…, c_n) 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, 2^1/3, 2^2/3, a,2^1/3a, 2^2/3a} 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 a ≠ b 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 b ≠ x²a for any x ∈ F, 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 + ⋯ +a_{n − l}c^{n − 1}, for some a₀,…, a_{n − 1} ∈ F.

# 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 a ∈ K − F.

4 Suppose a,b ∈ K 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 aⁿ ∉ F).

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 c ∈ F.

2 a is of degree 1 over F iff a ∈ F.

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 F ⊆ K ⊆ K(b) and F ⊆ F(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 F ⊆ K and a,b ∈ K. 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, a − b, ab, and a / b are algebraic over F. (In the last case, assume b ≠ 0.)

2 The set {x ∈ K : 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 + ⋯ + a_nxⁿ 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, …,a_n).

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

Let (a₀, …, a_n) = 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.