﻿ ﻿EXTENSIONS OF FIELDS - A Book of Abstract Algebra

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

### Chapter 27. EXTENSIONS OF FIELDS

In the first 26 chapters of this book we introduced the cast and set the scene on a vast and complex stage. Now it is time for the action to begin. We will be surprised to discover that none of our effort has been wasted; for every notion which was defined with such meticulous care, every subtlety, every fine distinction will have its use and play its prescribed role in the story which is about to unfold.

We will see modern algebra reaching out and merging with other disciplines of mathematics; we will see its machinery put to use for solving a wide range of problems which, on the surface, have nothing whatever to do with modern algebra. Some of these problems—ancient problems of geometry, riddles about numbers, questions concerning the solutions of equations—reach back to the very beginnings of mathematics. Great masters of the art of mathematics puzzled over them in every age and left them unsolved, for the machinery to solve them was not there. Now, with a light touch modern algebra uncovers the answers.

Modern algebra was not built in an ivory tower but was created part and parcel with the rest of mathematics—tied to it, drawing from it, and offering it solutions. Clearly it did not develop as methodically as it has been presented here. It would be pointless, in a first course in abstract algebra, to replicate all the currents and crosscurrents, all the hits and misses and false starts. Instead, we are provided with a finished product in which the agonies and efforts that went into creating it cannot be discerned. There is a disadvantage to this: without knowing the origin of a given concept, without knowing the specific problems which gave it birth, the student often wonders what it means and why it was ever invented.

We hope, beginning now, to shed light on that kind of question, to justify what we have already done, and to demonstrate that the concepts introduced in earlier chapters are correctly designed for their intended purposes.

Most of classical mathematics is set in a framework consisting of fields, especially , , and . The theory of equations deals with polynomials over and , calculus is concerned with functions over , and plane geometry is set in × . It is not surprising, therefore, that modern efforts to generalize and unify these subjects should also center around the study of fields. It turns out that a great variety of problems, ranging from geometry to practical computation, can be translated into the language of fields and formulated entirely in terms of the theory of fields. The study of fields will therefore be our central concern in the remaining chapters, though we will see other themes merging and flowing into it like the tributaries of a great river.

If F is a field, then a subfield of F is any nonempty subset of F which is closed with respect to addition and subtraction, multiplication and division. (It would be equivalent to say: closed with respect to addition and negatives, multiplication and multiplicative inverses.) As we already know, if K is a subfield of F, then K is a field in its own right.

If K is a subfield of F, we say also that F is an extension field of K. When it is clear in context that both Fand K are fields, we say simply that F is an extension of K.

Given a field F, we may look inward from F at all the subfields of F. On the other hand, we may look outward from F at all the extensions of F. Just as there are relationships between F and its subfields, there are also interesting relationships between F and its extensions. One of these relationships, as we shall see later, is highly reminiscent of Lagrange’s theorem—an inside-out version of it.

Why should we be interested in looking at the extensions of fields? There are several reasons, but one is very special. If F is an arbitrary field, there are, in general, polynomials over F which have no roots in F. For example, x2+ 1 has no roots in . This situation is unfortunate but, it turns out, not hopeless. For, as we shall soon see, every polynomial over any field F has roots. If these roots are not already in F, they are in a suitable extension of F. For example, x2 + 1 = 0 has solutions in .

In the matter of factoring polynomials and extracting their roots, is utopia! In every polynomial a(x) of degree n has exactly n roots c1, …, cn and can therefore be factored as a(x) = k(xc1)(xc2) ⋯ (xcn). This ideal situation is not enjoyed by all fields—far from it! In an arbitrary field F, a polynomial of degree n may have any number of roots, from no roots to n roots, and there may be irreducible polynomials of any degree whatever. This is a messy situation, which does not hold the promise of an elegant theory of solutions to polynomial equations. However, it turns out that F always has a suitable extension E such that any polynomial a{x) of degree n over F has exactly n solutions in E. Therefore, a(x) can be factored in E[x] as

a(x) = k(xc1)(xc2) ⋯ (xcn)

Thus, paradise is regained by the expedient of enlarging the field F. This is one of the strongest reasons for our interest in field extensions. They will give us a trim and elegant theory of solutions to polynomial equations.

Now, let us get to work! Let E be a field, F a subfield of E, and c any

element of E. We define the substitution function σc as follows:

For every polynomial a(x) in F[x],

σc (a(x)) = a(c)

Thus, σc is the function “substitute c for x.” It is a function from F[x] into E. In fact, σc is a homomorphism. This is true because

and

The kernel of the homomorphism σc is the set of all the polynomials a(x) such that a(c) = σc(a(x)) = 0. That is, the kernel of σc consists of all the polynomials a(x) in F[x] such that c is a root of a(x).

Let Jc denote the kernel of ac ; since the kernel of any homomorphism is an ideal, Jc is an ideal of F[x).

An element c in E is called algebraic over F if it is the root of some nonzero polynomial a(x) in F[x]. Otherwise, c is called transcendental over F. Obviously c is algebraic over F iff Jc contains nonzero polynomials, and transcendental over F iff Jc = {0}.

We will confine our attention now to the case where c is algebraic. The transcendental case will be examined in Exercise G at the end of this chapter.

Thus, let c be algebraic over F, and let Jc be the kernel of σc (where σc is the function “substitute c for x”). Remember that in F[x] every ideal is a principal ideal; hence Jc = ⟨p(x)⟩ = the set of all multiples of p(x), for some polynomial p(x). Since every polynomial in Jc is a multiple of p(x), p(x) is a polynomial of lowest degree among all the nonzero polynomials in Jc. It is easy to see that p(x) is irreducible; otherwise we could factor it into polynomials of lower degree, say p(x) = f(x)g(x). But then 0 = p(c) = f(c)g(c), so f(c) = 0 or g(c) = 0, and therefore either f(x) or g(x) is in Jc. This is impossible, because we have just seen thatp{x) has the lowest degree among all the polynomials in Jc, whereas f(x) and g(x) both have lower degree than p(x).

Since every constant multiple of p(x) is in Jc, we may take p(x) to be monic, that is, to have leading coefficient 1. Then p(x) is the unique monic polynomial of lowest degree in Jc. (Also, it is the only monic irreducible polynomial in Jc.) This polynomial p(x) is called the minimum polynomial of c over F, and will be of considerable importance in our discussions in a later chapter.

Let us look at an example: is an extension field of , and contains the irrational number . The function is the function “substitute for x”; for example (x4 − 3x2 + 1) = − 3 + 1 = −1. By our discussion above, : [x] → . is a homomorphism and its kernel consists of all the polynomials in [x] which have as one of their roots. The monic polynomial of least degree in [x] having as a root is p(x) = x2 − 2; hence x2 − 2 is the minimum polynomial of over .

Now, let us turn our attention to the range of σc. Since σc is a homomorphism, its range is obviously closed with respect to addition, multiplication, and negatives, but it is not obviously closed with respect to multiplicative inverses. Not obviously, but in fact it is closed for multiplicative inverses, which is far from self-evident, and quite a remarkable fact. In order to prove this, letf(c) be any nonzero element in the range of σc. Since f(c) ≠ 0, f(x) is not in the kernel of σc. Thus, f(x) is not a multiple of p(x), and since p(x) is irreducible, it follows that f(x) and p(x) are relatively prime. Therefore there are polynomials s(x) and t(x) such that s(x)f(x) + t(x)p(x) = 1. But then

and therefore s(c) is the multiplicative inverse of f(c).

We have just shown that the range of σc is a subfield of E. Now, the range of σc is the set of all the elements a(c), for all a(x) in F[x]:

Range σc = {a(c): a(x) ∈ F[x]}

We have just seen that range σc is a field. In fact, it is the smallest field containing F and c: indeed, any other field containing F and c would inevitably contain every element of the form

a0 + a1c + ⋯ + ancn(a0, …, anF)

in other words, would contain every element in the range of σc.

By the smallest field containing F and c we mean the field which contains F and c and is contained in any other field containing F and c. It is called the field generated by F and c, and is denoted by the important symbol

F(c)

Now, here is what we have, in a nutshell: σc is a homomorphism with domain F[x], range F(c), and kernel Jc = ⟨p(x)⟩. Thus, by the fundamental homomorphism theorem,

Finally, here is an interesting sidelight: if c and d are both roots of p(x), where c and d are in E, then, by what we have just proved, F(c) and F(d) are both isomorphic to F[x]/⟨p(x)⟩, and therefore isomorphic to each other:

If c and d are roots of the same irreducible polynomial p(x) in F[x], then F(c) ≅ F(d)

In particular, this shows that, given F and c, F(c) is unique up to isomorphism.

It is time now to recall our main objective: if a(x) is a polynomial in F[x] which has no roots in F, we wish to enlarge F to a field E which contains a root of a(x). How can we manage this?

An observation is in order: finding extensions of F is not as easy as finding subfields of F. A subfield of F is a subset of an existing set: it is therel But an extension of F is not yet there. We must somehow build it around F.

Let p(x) be an irreducible polynomial in F[x]. We have just seen that if F can be enlarged to a field E containing a root c of p(x), then F(c) is already what we are looking for: it is an extension of F containing a root of p(x). Furthermore, F(c) is isomorphic to F[x]/⟨p(x)⟩. Thus, the field extension we are searching for is precisely F[x]/⟨p(x)⟩. Our result is summarized in the next theorem.

Basic theorem of field extensions Let F be a field and a(x) a nonconstant polynomial in F[x]. There exists an extension field E of F and an element c in E such that c is a root of a(x).

PROOF: To begin with, a(x) can be factored into irreducible polynomials in F[x]. If p(x) is any nonconstant irreducible factor of a(x), it is clearly sufficient to find an extension of F containing a root of p(x), since such a root will also be a root of a(x).

In Exercise D4 of Chapter 25, the reader was asked to supply the simple proof that, if p(x) is irreducible in F[x], then ⟨p(x)⟩ is a maximal ideal of F[x]. Furthermore, by the argument at the end of Chapter 19, if ⟨p(x)⟩ is a maximal ideal of F[x], then the quotient ring F[x]/⟨p(x)⟩ is a field.

It remains only to prove that F[x]/ ⟨p(x)⟩ is the desired field extension of F. When we write J = ⟨p(x)⟩, let us remember that every element of F[x]/J is a coset of J. We will prove that F[x]/J is an extension of F by identifying each element a in F with its coset J + a.

To be precise, define h: FF[x]/J by h(a) = J + a. Note that h is the function which matches every a in F with its coset J + a in F[x]/J. We will now show that h is an isomorphism.

By the familiar rules of coset addition and multiplication, h is a homomorphism. Now, every homomorphism between fields is injective. (This is true because the kernel of a homomorphism is an ideal, and a field has no nontrivial ideals.) Thus, h is an isomorphism between its domain and its range.

What is the range of h? It consists of all the cosets J + a where aF, that is, all the cosets of constant polynomials. (If a is in F, then a is a constant polynomial.) Thus, F is isomorphic to the subfield of F[x]/J containing all the cosets of constant polynomials. This subfield is therefore an isomorphic copy of F, which may be identified with F, so F[x]/J is an extension of F.

Finally, if p(x) = a0 + a1x + ⋯ + anxn, let us show that the coset J + x is a root of p(x) in F[x]/J. Of course, in F[x]/J, the coefficients are not actually a0, a1, …, an, but their cosets J + a0, J + a1, …, J + an. Writing

J + a0 = ā0, …, J + an = ānandJ + x =

we must prove that

ā0 + ā1 + ⋯ + ānn = J(J is the zero coset)

Well,

This completes the proof of the basic theorem of field extensions. Observe that we may use this theorem several times in succession to get the following:

Let a(x) be a polynomial of degree n in F[x], There is an extension field ∈ of F which contains all n roots of a(x).

EXERCISES

A. Recognizing Algebraic Elements

Example To show that is algebraic over , one must find a polynomial p(x) ∈ [x] such that is a root of p(x).

Let a = ; then a2 = 1 + , a2 − 1 = , and finally, (a2 − l)2 = 2. Thus, a satisfies p(x) = x4 − 2x2 −1 = 0.

1 Prove that each of the following numbers is algebraic over :

(a)i

(b)

(c)2 + 3i

(d)

# (e)

(f) +

(g)

2 Prove that each of the following numbers is algebraic over the given field:

(a) over (π)

(b) over (π2)

(c)π2 − 1 over (π3)

NOTE: Recognizing a transcendental element is much more difficult, since it requires proving that the element cannot be a root of any polynomial over the given field. In recent times it has been proved, using sophisticated mathematical machinery, that π and e are transcendental over .

B. Finding the Minimum Polynomial

1 Find the minimum polynomial of each of the following numbers over . (Where appropriate, use the methods of Chapter 26, Exercises D, E, and F to ensure that your polynomial is irreducible.)

(a)1 + 2i

(b)1 +

(c)1 +

#(d)

(e) +

(f)

2 Show that the minimum polynomial of + i is

(a)x2 − 2x + 3 over

(b)x4 − 2x2 + 9 over

(c)x2 − 2ix − 3 over (i)

3 Find the minimum polynomial of the following numbers over the indicated fields:

 + i over ; over : over (i); over () over ; over (i); over (); over

4 For each of the following polynomials p(x), find a number a such that p(x) is the minimum polynomial of a over :

(a)x2 + 2x − 1

(b)x4 + 2x2 − 1

(c)x4 − 10x2 + 1

5 Find a monic irreducible polynomial p(x) such that [x]/⟨p(x)⟩ is isomorphic to

(a)()

(b)(1 + )

(c)

C. The Structure of Fields F[x]/⟨p (x)⟩

Let p(x) be an irreducible polynomial of degree n over F. Let c denote a root of p(x) in some extension of F (as in the basic theorem on field extensions).

1 Prove: Every element in F(c) can be written as r(c), for some r(x) of degree < n in F[x]. [HINT: Given any element t(c) ∈. F(c), use the division algorithm to divide t(x) by p(x).]

2 If s(c) = t(c) in F(c), where s(x) and t(x) have degree < n, prove that s(x) = t(x).

3 Conclude from parts 1 and 2 that every element in F(c) can be written uniquely as r(c), with deg r(x) < n.

# 4 Using part 3, explain why there are exactly four elements in 2[x]/⟨x2 + x + 1⟩. List these four elements, and give their addition and multiplication tables. {HINT: Identify 2[x]/⟨x2 + x + 1⟩ with 2(c), where c is a root of x2 + x + 1. Write the elements of 2(c) as in part 3. When computing the multiplication table, use the fact that c2 + c + 1 = 0.}

5 Describe 2[x]/⟨x3 + x + 1⟩, as in part 4.

6 Describe 3[x]/⟨x3 + x2 + 2⟩, as in part 4.

D. Short Questions Relating of Field Extensions

Let F be any field.

Prove parts 1–5:

# 1 If c is algebraic over F, so are c + 1 and kc (where kF).

2 If c ≠ 0 and c is algebraic over F, so is 1/c.

3 If cd is algebraic over F, then c is algebraic over F(d). If c + d is algebraic over F, then c is algebraic over F(d) (Assume c ≠ 0 and d ≠ 0.)

4 If the minimum polynomial of a over F is of degree 1, then aF, and conversely.

5 Suppose FK and aK. If p(x) is a monic irreducible polynomial in F[x], and p(a) = 0, then p(x) is the minimum polynomial of a over F.

6 Name a field (≠ or ) which contains a root of x5 + 2x3 + 4x2 + 6.

# 7 Prove: (1 + i) ≅ (1 − i). However, () ≅ ().

8 If p(x) is irreducible and has degree 2, prove that F[x]/⟨p(x)⟩ contains both roots of p(x).

E. Simple Extensions

Recall the definition of F(a). It is a field such that (i) FF(a); (ii) aF(a); (iii) any field containing F and a contains F(a).

Use this definition to prove parts 1–5, where FK, cF, and aK:

1 F(a) = F(a + c) and F(a) = F(ca). (Assume c ≠ 0.)

2 F(a2) ⊆ F(a) and F(a + b) ⊆ F(a, b). [F(a, b) is the field containing F, a, and b, and contained in any other field containing F, a and b.] Why are the reverse inclusions not necessarily true?

3 a + c is a root of p(x) iff a is a root of p(x + c); ca is a root of p(x) iff a is a root of p(cx).

4 Let p(x) be irreducible, and let a be a root of p{x + c). Then

F[x]/⟨p(x + c)⟩ ≅ F(a)andF[x]/⟨p(x)⟩ ≅ F(a + c)

Conclude that F[x]/⟨p(x + c)⟩ ≅ F[x]/⟨p(x)⟩.

5 Let p(x) be irreducible, and let a be a root of p(cx). Then F[x]/⟨p(cx)⟩ ≅ F(a) and F[x]/⟨p(x)⟩ ≅ F(ca). Conclude that F[x]/⟨p(cx)⟩ ≅ F[x]/⟨p(x)⟩.

6 Use parts 4 and 5 to prove the following:

(a)11[x]/⟨x2 + 1⟩ ≅ 11[x]/⟨x2 + x + 4⟩.

(b)If a is a root of x2 − 2 and b is a root of x2 − 4x + 2, then (a) ≅ (b).

(c)If a is a root of x2 − 2 and b is a root of x2, then (a) ≅ (b).

If the minimum polynomial of a over F has degree 2, we call F(a) a quadratic extension of F.

1 Prove that, if F is a field whose characteristic is ≠2, any quadratic extension of F is of the form F(), for some aF (HINT: Complete the square, and use Exercise E4.)

Let F be a finite field, and F* the multiplicative group of nonzero elements of F. Obviously H = {x2: xF*} is a subgroup of F*; since every square x2 in F* is the square of only two different elements, namely ±x, exactly half the elements of F* are in H. Thus, H has exactly two cosets: H itself, containing all the squares, and aH (where aH), containing all the nonsquares. If a and b are nonsquares, then by Chapter 15, Theorem 5(i),

Thus: if a and b are nonsquares, a/b is a square. Use these remarks in the following:

2 Let F be a finite field. If a, bF, let p(x) = x2a and q{x) = x2b be irreducible in F[x], and let and denote roots of p(x) and q(x) in an extension of F. Explain why a/b is a square, say a/b = c2 for some cF. Prove that is a root of p(cx).

3 Use part 2 to prove that F[x]/⟨p(cx)⟩ ≅ F(); then use Exercise E5 to conclude that F() ≅ F() .

4 Use part 3 to prove: Any two quadratic extensions of a finite field are isomorphic.

5 If a and b are nonsquares in , a/b is a square (why?). Use the same argument as in part 4 to prove that any two simple extensions of are isomorphic (hence isomorphic to ).

G. Questions Relating to Transcendental Elements

Let F be a field, and let c be transcendental over F. Prove the following:

1 {a(c):a(x) ∈ F[x]} is an integral domain isomorphic to F[x].

# 2 F(c) is the field of quotients of {a(c): a(x) ∈ F[x]}, and is isomorphic to F(x), the field of quotients of F[x].

3 If c is transcendental over F, so are c + 1, kc (where kF and k ≠ 0), c2.

4 If c is transcendental over F, every element in F(c) but not in F is transcendental over F.

† H. Common Factors of Two Polynomials: Over F and over Extensions of F

Let F be a field, and let a(x), b(x) ∈ F[x]. Prove the following:

1 If a{x) and b{x) have a common root c in some extension of F, they have a common factor of positive degree in F[x]. [Use the fact that a(x), b(x) ∈ ker σc.]

2 If a(x) and b(x) are relatively prime in F[x], they are relatively prime in K[x], for any extension K of F. Conversely, if they are relatively prime in K[x], then they are relatively prime in F[x].

† I. Derivatives and Their Properties

Let a(x) = a0 + a1x + ⋯ + anxnF[x]. The derivative of a(x) is the following polynomial a′(x) ∈ F[x]:

a′(x) = a1 + 2a2x + … + nanxn 1

(This is the same as the derivative of a polynomial in calculus.) We now prove the analogs of the formal rules of differentiation, familiar from calculus.

Let a(x), b(x) ∈ F[x], and let kF.

Prove parts 1–4:

1 [a (x) + b (x)]′ = a′(x) + b′(x)

2 [a(x)b(x)]′ = a′(x)b(x) + a(x)b′(x)

3 [ka(x)]′ = ka′(x)

4 If F has characteristic 0 and a′(x) = 0, then a(x) is a constant polynomial. Why is this conclusion not necessarily true if F has characteristic p ≠ 0?

5 Find the derivative of the following polynomials in 5[x]:

x6 + 2x3 + x + 1x5 + 3x2 + 1x15 + 3x10 + 4x5 + 1

6 If F has characteristic p ≠ 0, and a′(x) = 0, prove that the only nonzero terms of a(x) are of the form ampxmp for some m. [That is, a(x) is a polynomial in powers of xp.]

† J. Multiple Roots

Suppose a(x) ≅ F[x], and K is an extension of F. An element cK is called a multiple root of a(x) if (xc)m|a(x) for some m > 1. It is often important to know if all the roots of a polynomial are different, or not. We now consider a method for determining whether an arbitrary polynomial a(x) ≅ F[x] has multiple roots in any extension of F.

Let K be any field containing all the roots of a(x) .Suppose a(x) has a multiple root c.

1 Prove that a(x) = (xc)2q(x) ∈ K[x].

2 Compute a′(x), using part 1.

3 Show that xc is a common factor of a(x) and a′(x).Use Exercise hi to conclude that a(x) and a′(x) have a common factor of degree >1 in F[x].

Thus, if a(x) has a multiple root, then a(x) and a′(x) have a common factor in F[x]. To prove the converse, suppose a(x) has no multiple roots. Then a(x) can be factored as a(x) = (xc1) ⋯ (xcn) where c1, …, cn are all different.

4 Explain why a′(x) is a sum of terms of the form

(xc1)⋯(xci 1)(xci + 1)⋯(xcn)

5 Using part 4, explain why none of the roots c1, …, cn of a(x) are roots of a′(x).

6 Conclude that a(x) and a′(x) have no common factor of degree >1 in F[x].

This important result is stated as follows: A polynomial a(x) in F[x] has a multiple root iff a(x) and a′ (x) have a common factor of degree >1 in F[x].

7 Show that each of the following polynomials has no multiple roots in any extension of its field of coefficients:

x3 − 7x2 + 8 ∈ [x]x2 + x + 1 ∈ 5[x]x100 − 1 ∈ 7[x]

The preceding example is most interesting: it shows that there are 100 different hundredth roots of 1 over 7. (The roots ±1 are in 7, while the remaining 98 roots are in extensions of 7.) Corresponding results hold for most other fields.

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