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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 23.3 Applications

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Subsection Solvability by Radicals

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Throughout this section we shall assume that all fields have characteristic zero to ensure that irreducible polynomials do not have multiple roots. The immediate goal of this section is to determine when the roots of a polynomial f(x) can be computed with a finite number of operations on the coefficients of f(x). The allowable operations are addition, subtraction, multiplication, division, and the extraction of nth roots. Certainly the solution to the quadratic equation, ax2+bx+c=0, illustrates this process:
x=−b±b2−4ac2a.
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The only one of these operations that might demand a larger field is the taking of nth roots. We are led to the following definition.
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An extension field E of a field F is an extension by radicals if there exists a chain of subfields
F=F0⊂F1⊂F2⊂⋯⊂Fr=E
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such for i=1,2,…,r, we have Fi=Fi−1(αi) and αini∈Fi−1 for some positive integer ni. A polynomial f(x) is solvable by radicals over F if the splitting field K of f(x) over F is contained in an extension of F by radicals. Our goal is to arrive at criteria that will tell us whether or not a polynomial f(x) is solvable by radicals by examining the Galois group f(x).
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The easiest polynomial to solve by radicals is one of the form xn−a. As we discussed in Chapter 4, the roots of xn−1 are called the nth roots of unity. These roots are a finite subgroup of the splitting field of xn−1. By Corollary 22.11, the nth roots of unity form a cyclic group. Any generator of this group is called a primitive nth root of unity.
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Example 23.27.

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The polynomial xn−1 is solvable by radicals over Q. The roots of this polynomial are 1,ω,ω2,…,ωn−1, where
ω=cos⁡(2πn)+isin⁡(2πn).
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The splitting field of xn−1 over Q is Q(ω).
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We shall prove that a polynomial is solvable by radicals if its Galois group is solvable. Recall that a subnormal series of a group G is a finite sequence of subgroups
G=Hn⊃Hn−1⊃⋯⊃H1⊃H0={e},
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where Hi is normal in Hi+1. A group G is solvable if it has a subnormal series {Hi} such that all of the factor groups Hi+1/Hi are abelian. For example, if we examine the series {id}⊂A3⊂S3, we see that S3 is solvable. On the other hand, S5 is not solvable, by Theorem 10.11.

Proof.

The roots of xn−a are an,ωan,…,ωn−1an, where ω is a primitive nth root of unity. Suppose that F contains all of its nth roots of unity. If ζ is one of the roots of xn−a, then distinct roots of xn−a are ζ,ωζ,…,ωn−1ζ, and E=F(ζ). Since G(E/F) permutes the roots xn−a, the elements in G(E/F) must be determined by their action on these roots. Let σ and τ be in G(E/F) and suppose that σ(ζ)=ωiζ and τ(ζ)=ωjζ. If F contains the roots of unity, then
στ(ζ)=σ(ωjζ)=ωjσ(ζ)=ωi+jζ=ωiτ(ζ)=τ(ωiζ)=τσ(ζ).
Therefore, στ=τσ and G(E/F) is abelian, and G(E/F) must be solvable.
Now suppose that F does not contain a primitive nth root of unity. Let ω be a generator of the cyclic group of the nth roots of unity. Let α be a zero of xn−a. Since α and ωα are both in the splitting field of xn−a, ω=(ωα)/α is also in E. Let K=F(ω). Then F⊂K⊂E. Since K is the splitting field of xn−1, K is a normal extension of F. Therefore, any automorphism σ in G(F(ω)/F) is determined by σ(ω). It must be the case that σ(ω)=ωi for some integer i since all of the zeros of xn−1 are powers of ω. If τ(ω)=ωj is in G(F(ω)/F), then
στ(ω)=σ(ωj)=[σ(ω)]j=ωij=[τ(ω)]i=τ(ωi)=τσ(ω).
Therefore, G(F(ω)/F) is abelian. By the Fundamental Theorem of Galois Theory the series
{id}⊂G(E/F(ω))⊂G(E/F)
is a normal series. By our previous argument, G(E/F(ω)) is abelian. Since
G(E/F)/G(E/F(ω))≅G(F(ω)/F)
is also abelian, G(E/F) is solvable.

Proof.

Since E is a radical extension of F, there exists a chain of subfields
F=F0⊂F1⊂F2⊂⋯⊂Fr=E
such for i=1,2,…,r, we have Fi=Fi−1(αi) and αini∈Fi−1 for some positive integer ni. We will build a normal radical extension of F,
F=K0⊂K1⊂K2⊂⋯⊂Kr=K
such that K⊇E. Define K1 for be the splitting field of xn1−α1n1. The roots of this polynomial are α1,α1ω,α1ω2,…,α1ωn1−1, where ω is a primitive n1th root of unity. If F contains all of its n1 roots of unity, then K1=F(α1). On the other hand, suppose that F does not contain a primitive n1th root of unity. If β is a root of xn1−α1n1, then all of the roots of xn1−α1n1 must be β,ωβ,…,ωn1−1β, where ω is a primitive n1th root of unity. In this case, K1=F(ωβ). Thus, K1 is a normal radical extension of F containing F1. Continuing in this manner, we obtain
F=K0⊂K1⊂K2⊂⋯⊂Kr=K
such that Ki is a normal extension of Ki−1 and Ki⊇Fi for i=1,2,…,r.
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We will now prove the main theorem about solvability by radicals.

Proof.

Since f(x) is solvable by radicals there exists an extension E of F by radicals F=F0⊂F1⊂⋯⊂Fn=E. By Lemma 23.29, we can assume that E is a splitting field f(x) and Fi is normal over Fi−1. By the Fundamental Theorem of Galois Theory, G(E/Fi) is a normal subgroup of G(E/Fi−1). Therefore, we have a subnormal series of subgroups of G(E/F):
{id}⊂G(E/Fn−1)⊂⋯⊂G(E/F1)⊂G(E/F).
Again by the Fundamental Theorem of Galois Theory, we know that
G(E/Fi−1)/G(E/Fi)≅G(Fi/Fi−1).
By Lemma 23.28, G(Fi/Fi−1) is solvable; hence, G(E/F) is also solvable.
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The converse of Theorem 23.30 is also true. For a proof, see any of the references at the end of this chapter.
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Subsection Insolvability of the Quintic

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We are now in a position to find a fifth-degree polynomial that is not solvable by radicals. We merely need to find a polynomial whose Galois group is S5. We begin by proving a lemma.

Proof.

Let G be a subgroup of Sp that contains a transposition σ and τ a cycle of length p. We may assume that σ=(12). The order of τ is p and τn must be a cycle of length p for 1≤n<p. Therefore, we may assume that μ=τn=(1,2,i3,…,ip) for some n, where 1≤n<p (see Exercise 5.4.13 in Chapter 5). Noting that (12)(1,2,i3,…,ip)=(2,i3,…,ip) and (2,i3,…,ip)k(12)(2,i3,…,ip)−k=(1ik), we can obtain all the transpositions of the form (1n) for 1≤n<p. However, these transpositions generate all transpositions in Sp, since (1j)(1i)(1j)=(ij). The transpositions generate Sp.
The graph is one a set of rectangular axes and starts below the horizontal axis, increases to about 68 at x approximately equal to 2, while passing through the horizontal axis at about minus 3. The graph then decreases to about minus 73 at x equal to about 2.3, while passing through the horizontal axis at about minus 0.2. Finally, the graph increases to 20 at about 3.2, while passing through the horizontal axis at about 3.
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Figure 23.32. The graph of f(x)=x5−6x3−27x−3
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Example 23.33.

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We will show that f(x)=x5−6x3−27x−3∈Q[x] is not solvable. We claim that the Galois group of f(x) over Q is S5. By Eisenstein’s Criterion, f(x) is irreducible and, therefore, must be separable. The derivative of f(x) is f′(x)=5x4−18x2−27; hence, setting f′(x)=0 and solving, we find that the only real roots of f′(x) are
x=±66+95.
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Therefore, f(x) can have at most one maximum and one minimum. It is easy to show that f(x) changes sign between −3 and −2, between −2 and 0, and once again between 0 and 4 (Figure 23.32). Therefore, f(x) has exactly three distinct real roots. The remaining two roots of f(x) must be complex conjugates. Let K be the splitting field of f(x). Since f(x) has five distinct roots in K and every automorphism of K fixing Q is determined by the way it permutes the roots of f(x), we know that G(K/Q) is a subgroup of S5. Since f is irreducible, there is an element in σ∈G(K/Q) such that σ(a)=b for two roots a and b of f(x). The automorphism of C that takes a+bi↦a−bi leaves the real roots fixed and interchanges the complex roots; consequently, G(K/Q) contains a transposition. If α is one of the real roots of f(x), then [Q(α):Q]=5 by Exercise 21.5.28. Since Q(α) is a subfield of K, it must be the case that [K:Q] is divisible by 5. Since [K:Q]=|G(K/Q)| and G(K/Q)⊂S5, we know that G(K/Q) contains a cycle of length 5. By Lemma 23.31, S5 is generated by a transposition and an element of order 5; therefore, G(K/Q) must be all of S5. By Theorem 10.11, S5 is not solvable. Consequently, f(x) cannot be solved by radicals.
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Subsection The Fundamental Theorem of Algebra

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It seems fitting that the last theorem that we will state and prove is the Fundamental Theorem of Algebra. This theorem was first proven by Gauss in his doctoral thesis. Prior to Gauss’s proof, mathematicians suspected that there might exist polynomials over the real and complex numbers having no solutions. The Fundamental Theorem of Algebra states that every polynomial over the complex numbers factors into distinct linear factors.

Proof.

Suppose that E is a proper finite field extension of the complex numbers. Since any finite extension of a field of characteristic zero is a simple extension, there exists an α∈E such that E=C(α) with α the root of an irreducible polynomial f(x) in C[x]. The splitting field L of f(x) is a finite normal separable extension of C that contains E. We must show that it is impossible for L to be a proper extension of C.
Suppose that L is a proper extension of C. Since L is the splitting field of f(x)=(x2+1) over R, L is a finite normal separable extension of R. Let K be the fixed field of a Sylow 2-subgroup G of G(L/R). Then L⊃K⊃R and |G(L/K)|=[L:K]. Since [L:R]=[L:K][K:R], we know that [K:R] must be odd. Consequently, K=R(β) with β having a minimal polynomial f(x) of odd degree. Therefore, K=R.
We now know that G(L/R) must be a 2-group. It follows that G(L/C) is a 2-group. We have assumed that L≠C; therefore, |G(L/C)|≥2. By the first Sylow Theorem and the Fundamental Theorem of Galois Theory, there exists a subgroup G of G(L/C) of index 2 and a field E fixed elementwise by G. Then [E:C]=2 and there exists an element γ∈E with minimal polynomial x2+bx+c in C[x]. This polynomial has roots (−b±b2−4c)/2 that are in C, since b2−4c is in C. This is impossible; hence, L=C.
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Although our proof was strictly algebraic, we were forced to rely on results from calculus. It is necessary to assume the completeness axiom from analysis to show that every polynomial of odd degree has a real root and that every positive real number has a square root. It seems that there is no possible way to avoid this difficulty and formulate a purely algebraic argument. It is somewhat amazing that there are several elegant proofs of the Fundamental Theorem of Algebra that use complex analysis. It is also interesting to note that we can obtain a proof of such an important theorem from two very different fields of mathematics.
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