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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 21.1 Extension Fields

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A field E is an extension field of a field F if F is a subfield of E. The field F is called the base field. We write F⊂E.
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Example 21.1.

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For example, let
F=Q(2)={a+b2:a,b∈Q}
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and let E=Q(2+3) be the smallest field containing both Q and 2+3. Both E and F are extension fields of the rational numbers. We claim that E is an extension field of F. To see this, we need only show that 2 is in E. Since 2+3 is in E, 1/(2+3)=3−2 must also be in E. Taking linear combinations of 2+3 and 3−2, we find that 2 and 3 must both be in E.
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Example 21.2.

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Let p(x)=x2+x+1∈Z2[x]. Since neither 0 nor 1 is a root of this polynomial, we know that p(x) is irreducible over Z2. We will construct a field extension of Z2 containing an element α such that p(α)=0. By Theorem 17.22, the ideal ⟨p(x)⟩ generated by p(x) is maximal; hence, Z2[x]/⟨p(x)⟩ is a field. Let f(x)+⟨p(x)⟩ be an arbitrary element of Z2[x]/⟨p(x)⟩. By the division algorithm,
f(x)=(x2+x+1)q(x)+r(x),
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where the degree of r(x) is less than the degree of x2+x+1. Therefore,
f(x)+⟨x2+x+1⟩=r(x)+⟨x2+x+1⟩.
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The only possibilities for r(x) are then 0, 1, x, and 1+x. Consequently, E=Z2[x]/⟨x2+x+1⟩ is a field with four elements and must be a field extension of Z2, containing a zero α of p(x). The field Z2(α) consists of elements
0+0α=01+0α=10+1α=α1+1α=1+α.
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Notice that α2+α+1=0; hence, if we compute (1+α)2,
(1+α)(1+α)=1+α+α+(α)2=α.
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Other calculations are accomplished in a similar manner. We summarize these computations in the following tables, which tell us how to add and multiply elements in E.
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+01α1+α001α1+α1101+αααα1+α011+α1+αα10
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Figure 21.3. Addition Table for Z2(α)
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⋅01α1+α00000101α1+αα0α1+α11+α01+α1α
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Figure 21.4. Multiplication Table for Z2(α)
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The following theorem, due to Kronecker, is so important and so basic to our understanding of fields that it is often known as the Fundamental Theorem of Field Theory.

Proof.

To prove this theorem, we will employ the method that we used to construct Example 21.2. Clearly, we can assume that p(x) is an irreducible polynomial. We wish to find an extension field E of F containing an element α such that p(α)=0. The ideal ⟨p(x)⟩ generated by p(x) is a maximal ideal in F[x] by Theorem 17.22; hence, F[x]/⟨p(x)⟩ is a field. We claim that E=F[x]/⟨p(x)⟩ is the desired field.
We first show that E is a field extension of F. We can define a homomorphism of commutative rings by the map ψ:F→F[x]/⟨p(x)⟩, where ψ(a)=a+⟨p(x)⟩ for a∈F. It is easy to check that ψ is indeed a ring homomorphism. Observe that
ψ(a)+ψ(b)=(a+⟨p(x)⟩)+(b+⟨p(x)⟩)=(a+b)+⟨p(x)⟩=ψ(a+b)
and
ψ(a)ψ(b)=(a+⟨p(x)⟩)(b+⟨p(x)⟩)=ab+⟨p(x)⟩=ψ(ab).
To prove that ψ is one-to-one, assume that
a+⟨p(x)⟩=ψ(a)=ψ(b)=b+⟨p(x)⟩.
Then a−b is a multiple of p(x), since it lives in the ideal ⟨p(x)⟩. Since p(x) is a nonconstant polynomial, the only possibility is that a−b=0. Consequently, a=b and ψ is injective. Since ψ is one-to-one, we can identify F with the subfield {a+⟨p(x)⟩:a∈F} of E and view E as an extension field of F.
It remains for us to prove that p(x) has a zero α∈E. Set α=x+⟨p(x)⟩. Then α is in E. If p(x)=a0+a1x+⋯+anxn, then
p(α)=a0+a1(x+⟨p(x)⟩)+⋯+an(x+⟨p(x)⟩)n=a0+(a1x+⟨p(x)⟩)+⋯+(anxn+⟨p(x)⟩)=a0+a1x+⋯+anxn+⟨p(x)⟩=0+⟨p(x)⟩.
Therefore, we have found an element α∈E=F[x]/⟨p(x)⟩ such that α is a zero of p(x).
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Example 21.6.

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Let p(x)=x5+x4+1∈Z2[x]. Then p(x) has irreducible factors x2+x+1 and x3+x+1. For a field extension E of Z2 such that p(x) has a root in E, we can let E be either Z2[x]/⟨x2+x+1⟩ or Z2[x]/⟨x3+x+1⟩. We will leave it as an exercise to show that Z2[x]/⟨x3+x+1⟩ is a field with 23=8 elements.
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Subsection Algebraic Elements

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An element α in an extension field E over F is algebraic over F if f(α)=0 for some nonzero polynomial f(x)∈F[x]. An element in E that is not algebraic over F is transcendental over F. An extension field E of a field F is an algebraic extension of F if every element in E is algebraic over F. If E is a field extension of F and α1,…,αn are contained in E, we denote the smallest field containing F and α1,…,αn by F(α1,…,αn). If E=F(α) for some α∈E, then E is a simple extension of F.
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Example 21.7.

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Both 2 and i are algebraic over Q since they are zeros of the polynomials x2−2 and x2+1, respectively. Clearly π and e are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over Q. Numbers in R that are algebraic over Q are in fact quite rare. Almost all real numbers are transcendental over Q.
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The probability that a real number chosen at random from the interval [0,1] will be transcendental over the rational numbers is one.
(In many cases we do not know whether or not a particular number is transcendental; for example, it is still not known whether π+e is transcendental or algebraic.)
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A complex number that is algebraic over Q is an algebraic number. A transcendental number is an element of C that is transcendental over Q.
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Example 21.8.

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We will show that 2+3 is algebraic over Q. If α=2+3, then α2=2+3. Hence, α2−2=3 and (α2−2)2=3. Since α4−4α2+1=0, it must be true that α is a zero of the polynomial x4−4x2+1∈Q[x].
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It is very easy to give an example of an extension field E over a field F, where E contains an element transcendental over F. The following theorem characterizes transcendental extensions.

Proof.

Let ϕα:F[x]→E be the evaluation homomorphism for α. Then α is transcendental over F if and only if ϕα(p(x))=p(α)≠0 for all nonconstant polynomials p(x)∈F[x]. This is true if and only if ker⁡ϕα={0}; that is, it is true exactly when ϕα is one-to-one. Hence, E must contain a copy of F[x]. The smallest field containing F[x] is the field of fractions F(x). By Theorem 18.4, E must contain a copy of this field.
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We have a more interesting situation in the case of algebraic extensions.

Proof.

Let ϕα:F[x]→E be the evaluation homomorphism. The kernel of ϕα is a principal ideal generated by some p(x)∈F[x] with deg⁡p(x)≥1. We know that such a polynomial exists, since F[x] is a principal ideal domain and α is algebraic. The ideal ⟨p(x)⟩ consists exactly of those elements of F[x] having α as a zero. If f(α)=0 and f(x) is not the zero polynomial, then f(x)∈⟨p(x)⟩ and p(x) divides f(x). So p(x) is a polynomial of minimal degree having α as a zero. Any other polynomial of the same degree having α as a zero must have the form βp(x) for some β∈F.
Suppose now that p(x)=r(x)s(x) is a factorization of p(x) into polynomials of lower degree. Since p(α)=0, r(α)s(α)=0; consequently, either r(α)=0 or s(α)=0, which contradicts the fact that p is of minimal degree. Therefore, p(x) must be irreducible.
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Let E be an extension field of F and α∈E be algebraic over F. The unique monic polynomial p(x) of the last theorem is called the minimal polynomial for α over F. The degree of p(x) is the degree of α over F.
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Example 21.11.

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Let f(x)=x2−2 and g(x)=x4−4x2+1. These polynomials are the minimal polynomials of 2 and 2+3, respectively.

Proof.

Let ϕα:F[x]→E be the evaluation homomorphism. The kernel of this map is ⟨p(x)⟩, where p(x) is the minimal polynomial of α. By the First Isomorphism Theorem for rings, the image of ϕα in E is isomorphic to F(α) since it contains both F and α.

Proof.

Since ϕα(F[x])≅F(α), every element in E=F(α) must be of the form ϕα(f(x))=f(α), where f(α) is a polynomial in α with coefficients in F. Let
p(x)=xn+an−1xn−1+⋯+a0
be the minimal polynomial of α. Then p(α)=0; hence,
αn=−an−1αn−1−⋯−a0.
Similarly,
αn+1=ααn=−an−1αn−an−2αn−1−⋯−a0α=−an−1(−an−1αn−1−⋯−a0)−an−2αn−1−⋯−a0α.
Continuing in this manner, we can express every monomial αm, m≥n, as a linear combination of powers of α that are less than n. Hence, any β∈F(α) can be written as
β=b0+b1α+⋯+bn−1αn−1.
To show uniqueness, suppose that
β=b0+b1α+⋯+bn−1αn−1=c0+c1α+⋯+cn−1αn−1
for bi and ci in F. Then
g(x)=(b0−c0)+(b1−c1)x+⋯+(bn−1−cn−1)xn−1
is in F[x] and g(α)=0. Since the degree of g(x) is less than the degree of p(x), the irreducible polynomial of α, g(x) must be the zero polynomial. Consequently,
b0−c0=b1−c1=⋯=bn−1−cn−1=0,
or bi=ci for i=0,1,…,n−1. Therefore, we have shown uniqueness.
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Example 21.14.

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Since x2+1 is irreducible over R, ⟨x2+1⟩ is a maximal ideal in R[x]. So E=R[x]/⟨x2+1⟩ is a field extension of R that contains a root of x2+1. Let α=x+⟨x2+1⟩. We can identify E with the complex numbers. By Proposition 21.12, E is isomorphic to R(α)={a+bα:a,b∈R}. We know that α2=−1 in E, since
α2+1=(x+⟨x2+1⟩)2+(1+⟨x2+1⟩)=(x2+1)+⟨x2+1⟩=0.
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Hence, we have an isomorphism of R(α) with C defined by the map that takes a+bα to a+bi.
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Let E be a field extension of a field F. If we regard E as a vector space over F, then we can bring the machinery of linear algebra to bear on the problems that we will encounter in our study of fields. The elements in the field E are vectors; the elements in the field F are scalars. We can think of addition in E as adding vectors. When we multiply an element in E by an element of F, we are multiplying a vector by a scalar. This view of field extensions is especially fruitful if a field extension E of F is a finite dimensional vector space over F, and Theorem 21.13 states that E=F(α) is finite dimensional vector space over F with basis {1,α,α2,…,αn−1}.
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If an extension field E of a field F is a finite dimensional vector space over F of dimension n, then we say that E is a finite extension of degree n over F. We write
[E:F]=n.
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to indicate the dimension of E over F.

Proof.

Let α∈E. Since [E:F]=n, the elements
1,α,…,αn
cannot be linearly independent. Hence, there exist ai∈F, not all zero, such that
anαn+an−1αn−1+⋯+a1α+a0=0.
Therefore,
p(x)=anxn+⋯+a0∈F[x]
is a nonzero polynomial with p(α)=0.
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Remark 21.16.

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Theorem 21.15 says that every finite extension of a field F is an algebraic extension. The converse is false, however. We will leave it as an exercise to show that the set of all elements in R that are algebraic over Q forms an infinite field extension of Q.
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The next theorem is a counting theorem, similar to Lagrange’s Theorem in group theory. Theorem 21.17 will prove to be an extremely useful tool in our investigation of finite field extensions.

Proof.

Let {α1,…,αn} be a basis for E as a vector space over F and {β1,…,βm} be a basis for K as a vector space over E. We claim that {αiβj} is a basis for K over F. We will first show that these vectors span K. Let u∈K. Then u=∑j=1mbjβj and bj=∑i=1naijαi, where bj∈E and aij∈F. Then
u=∑j=1m(∑i=1naijαi)βj=∑i,jaij(αiβj).
So the mn vectors αiβj must span K over F.
We must show that {αiβj} are linearly independent. Recall that a set of vectors v1,v2,…,vn in a vector space V are linearly independent if
c1v1+c2v2+⋯+cnvn=0
implies that
c1=c2=⋯=cn=0.
Let
u=∑i,jcij(αiβj)=0
for cij∈F. We need to prove that all of the cij’s are zero. We can rewrite u as
∑j=1m(∑i=1ncijαi)βj=0,
where ∑icijαi∈E. Since the βj’s are linearly independent over E, it must be the case that
∑i=1ncijαi=0
for all j. However, the αj are also linearly independent over F. Therefore, cij=0 for all i and j, which completes the proof.
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The following corollary is easily proved using mathematical induction.

Proof.

We know that deg⁡p(x)=[F(α):F] and deg⁡q(x)=[F(β):F]. Since F⊂F(β)⊂F(α),
[F(α):F]=[F(α):F(β)][F(β):F].
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Example 21.20.

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Let us determine an extension field of Q containing 3+5. It is easy to determine that the minimal polynomial of 3+5 is x4−16x2+4. It follows that
[Q(3+5):Q]=4.
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We know that {1,3} is a basis for Q(3) over Q. Hence, 3+5 cannot be in Q(3). It follows that 5 cannot be in Q(3) either. Therefore, {1,5} is a basis for Q(3,5)=(Q(3))(5) over Q(3) and {1,3,5,35=15} is a basis for Q(3,5)=Q(3+5) over Q. This example shows that it is possible that some extension F(α1,…,αn) is actually a simple extension of F even though n>1.
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Example 21.21.

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Let us compute a basis for Q(53,5i), where 5 is the positive square root of 5 and 53 is the real cube root of 5. We know that 5i∉Q(53), so
[Q(53,5i):Q(53)]=2.
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It is easy to determine that {1,5i} is a basis for Q(53,5i) over Q(53). We also know that {1,53,(53)2} is a basis for Q(53) over Q. Hence, a basis for Q(53,5i) over Q is
{1,5i,53,(53)2,(56)5i,(56)7i=556i or 56i}.
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Notice that 56i is a zero of x6+5. We can show that this polynomial is irreducible over Q using Eisenstein’s Criterion, where we let p=5. Consequently,
Q⊂Q(56i)⊂Q(53,5i).
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But it must be the case that Q(56i)=Q(53,5i), since the degree of both of these extensions is 6.

Proof.

(1) ⇒ (2). Let E be a finite algebraic extension of F. Then E is a finite dimensional vector space over F and there exists a basis consisting of elements α1,…,αn in E such that E=F(α1,…,αn). Each αi is algebraic over F by Theorem 21.15.
(2) ⇒ (3). Suppose that E=F(α1,…,αn), where every αi is algebraic over F. Then
E=F(α1,…,αn)⊃F(α1,…,αn−1)⊃⋯⊃F(α1)⊃F,
where each field F(α1,…,αi) is algebraic over F(α1,…,αi−1).
(3) ⇒ (1). Let
E=F(α1,…,αn)⊃F(α1,…,αn−1)⊃⋯⊃F(α1)⊃F,
where each field F(α1,…,αi) is algebraic over F(α1,…,αi−1). Since
F(α1,…,αi)=F(α1,…,αi−1)(αi)
is simple extension and αi is algebraic over F(α1,…,αi−1), it follows that
[F(α1,…,αi):F(α1,…,αi−1)]
is finite for each i. Therefore, [E:F] is finite.
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Subsection Algebraic Closure

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Given a field F, the question arises as to whether or not we can find a field E such that every polynomial p(x) has a root in E. This leads us to the following theorem.

Proof.

Let α,β∈E be algebraic over F. Then F(α,β) is a finite extension of F. Since every element of F(α,β) is algebraic over F, α±β, αβ, and α/β (β≠0) are all algebraic over F. Consequently, the set of elements in E that are algebraic over F form a field.
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Let E be a field extension of a field F. We define the algebraic closure of a field F in E to be the field consisting of all elements in E that are algebraic over F. A field F is algebraically closed if every nonconstant polynomial in F[x] has a root in F.

Proof.

Let F be an algebraically closed field. If p(x)∈F[x] is a nonconstant polynomial, then p(x) has a zero in F, say α. Therefore, x−α must be a factor of p(x) and so p(x)=(x−α)q1(x), where deg⁡q1(x)=deg⁡p(x)−1. Continue this process with q1(x) to find a factorization
p(x)=(x−α)(x−β)q2(x),
where deg⁡q2(x)=deg⁡p(x)−2. The process must eventually stop since the degree of p(x) is finite.
Conversely, suppose that every nonconstant polynomial p(x) in F[x] factors into linear factors. Let ax−b be such a factor. Then p(b/a)=0. Consequently, F is algebraically closed.

Proof.

Let E be an algebraic extension of F; then F⊂E. For α∈E, the minimal polynomial of α is x−α. Therefore, α∈F and F=E.
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It is a nontrivial fact that every field has a unique algebraic closure. The proof is not extremely difficult, but requires some rather sophisticated set theory. We refer the reader to [3], [4], or [8] for a proof of this result.
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We now state the Fundamental Theorem of Algebra, first proven by Gauss at the age of 22 in his doctoral thesis. This theorem states that every polynomial with coefficients in the complex numbers has a root in the complex numbers. The proof of this theorem will be given in Chapter 23.
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