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Section23.3Applications

SubsectionSolvability by Radicals

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 \(n\)th roots. Certainly the solution to the quadratic equation, \(a x^2 + b x +c = 0\), illustrates this process: \begin{equation*}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\end{equation*} The only one of these operations that might demand a larger field is the taking of \(n\)th roots. We are led to the following definition.

An extension field \(E\) of a field \(F\) is an extension by radicals if there exists a chain of subfields \begin{equation*}F = F_0 \subseteq F_1 \subseteq F_2 \subseteq \cdots \subseteq F_r = E\end{equation*} such for \(i = 1, 2, \ldots, r\), we have \(F_i = F_{i - 1}(\alpha_i)\) and \(\alpha_i^{n_i} \in F_{i-1}\) for some positive integer \(n_i\). 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 \(f(x)\).

The easiest polynomial to solve by radicals is one of the form \(x^n - a\). As we discussed in Chapter 4, the roots of \(x^n - 1\) are called the nth roots of unity. These roots are a finite subgroup of the splitting field of \(x^n -1\). By Corollary 22.11, the \(n\)th roots of unity form a cyclic group. Any generator of this group is called a primitive nth root of unity.

Example23.26

The polynomial \(x^n - 1\) is solvable by radicals over \({\mathbb Q}\). The roots of this polynomial are \(1, \omega, \omega^2, \ldots, \omega^{n - 1}\), where \begin{equation*}\omega = \cos\left( \frac{2 \pi}{n} \right) + i \sin\left( \frac{2 \pi}{n} \right).\end{equation*} The splitting field of \(x^n - 1\) over \({\mathbb Q}\) is \({\mathbb Q}(\omega)\).

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 \begin{equation*}G = H_n \supset H_{n-1} \supset \cdots \supset H_1 \supset H_0 = \{ e \},\end{equation*} where \(H_i\) is normal in \(H_{i+1}\). A group \(G\) is solvable if it has a subnormal series \(\{ H_i \}\) such that all of the factor groups \(H_{i+1} /H_i\) are abelian. For example, if we examine the series \(\{ \identity \} \subset A_3 \subset S_3\), we see that \(S_3\) is solvable. On the other hand, \(S_5\) is not solvable, by Theorem 10.11.

We will now prove the main theorem about solvability by radicals.

Proof

The converse of Theorem 23.29 is also true. For a proof, see any of the references at the end of this chapter.

SubsectionInsolvability of the Quintic

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 \(S_5\). We begin by proving a lemma.

Figure23.31The graph of \(f(x) = x^5 - 6 x^3 - 27 x - 3\)
Example23.32

We will show that \(f(x) = x^5 - 6 x^3 - 27 x - 3 \in {\mathbb Q}[x]\) is not solvable. We claim that the Galois group of \(f(x)\) over \({\mathbb Q}\) is \(S_5\). By Eisenstein's Criterion, \(f(x)\) is irreducible and, therefore, must be separable. The derivative of \(f(x)\) is \(f'(x) = 5 x^4 - 18 x^2 - 27\); hence, setting \(f'(x) = 0\) and solving, we find that the only real roots of \(f'(x)\) are \begin{equation*}x = \pm \sqrt{ \frac{6 \sqrt{6} + 9 }{5} }.\end{equation*} 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 23.31). 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 \({\mathbb Q}\) is determined by the way it permutes the roots of \(f(x)\), we know that \(G(K/{\mathbb Q})\) is a subgroup of \(S_5\). Since \(f\) is irreducible, there is an element in \(\sigma \in G(K/{\mathbb Q})\) such that \(\sigma(a) = b\) for two roots \(a\) and \(b\) of \(f(x)\). The automorphism of \({\mathbb C}\) that takes \(a + bi \mapsto a - bi\) leaves the real roots fixed and interchanges the complex roots; consequently, \(G(K/{\mathbb Q} ) \subset S_5\). By Lemma 23.30, \(S_5\) is generated by a transposition and an element of order 5; therefore, \(G(K/{\mathbb Q} )\) must be all of \(S_5\). By Theorem 10.11, \(S_5\) is not solvable. Consequently, \(f(x)\) cannot be solved by radicals.

SubsectionThe Fundamental Theorem of Algebra

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

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.