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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 $f(x)\text{.}$ The allowable operations are addition, subtraction, multiplication, division, and the extraction of $n$th roots. Certainly the solution to the quadratic equation, $a x^2 + b x +c = 0\text{,}$ 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 if there exists a chain of subfields

$$\begin{equation*} F = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_r = E \end{equation*}$$

such for $i = 1, 2, \ldots, r\text{,}$ we have $F_i = F_{i - 1}(\alpha_i)$ and $\alpha_i^{n_i} \in F_{i-1}$ for some positive integer $n_i\text{.}$ A polynomial $f(x)$ is 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 $f(x)\text{.}$

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

Example23.26

The polynomial $x^n - 1$ is solvable by radicals over ${\mathbb Q}\text{.}$ The roots of this polynomial are $1, \omega, \omega^2, \ldots, \omega^{n - 1}\text{,}$ 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 ${\mathbb Q}(\omega)\text{.}$

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 $H_{i+1}\text{.}$ 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 $\{ \identity \} \subset A_3 \subset S_3\text{,}$ we see that $S_3$ is solvable. On the other hand, $S_5$ is not solvable, by Theorem 10.11.

Lemma23.27

Let $F$ be a field of characteristic zero and $E$ be the splitting field of $x^n - a$ over $F$ with $a \in F\text{.}$ Then $G(E/F)$ is a solvable group.

Proof

The roots of $x^n - a$ are $\sqrt[n]{a}, \omega \sqrt[n]{a}, \ldots, \omega^{n-1} \sqrt[n]{a}\text{,}$ where $\omega$ is a primitive $n$th root of unity. Suppose that $F$ contains all of its $n$th roots of unity. If $\zeta$ is one of the roots of $x^n - a\text{,}$ then distinct roots of $x^n - a$ are $\zeta, \omega \zeta, \ldots, \omega^{n - 1} \zeta\text{,}$ and $E = F(\zeta)\text{.}$ Since $G(E/F)$ permutes the roots $x^n - a\text{,}$ the elements in $G(E/F)$ must be determined by their action on these roots. Let $\sigma$ and $\tau$ be in $G(E/F)$ and suppose that $\sigma( \zeta ) = \omega^i \zeta$ and $\tau( \zeta ) = \omega^j \zeta\text{.}$ If $F$ contains the roots of unity, then

$$\begin{equation*} \sigma \tau( \zeta ) = \sigma( \omega^j \zeta) = \omega^j \sigma( \zeta ) = \omega^{i+j} \zeta = \omega^i \tau( \zeta ) = \tau( \omega^i \zeta ) = \tau \sigma( \zeta ). \end{equation*}$$

Therefore, $\sigma \tau = \tau \sigma$ and $G(E/F)$ is abelian, and $G(E/F)$ must be solvable.

Now suppose that $F$ does not contain a primitive $n$th root of unity. Let $\omega$ be a generator of the cyclic group of the $n$th roots of unity. Let $\alpha$ be a zero of $x^n - a\text{.}$ Since $\alpha$ and $\omega \alpha$ are both in the splitting field of $x^n - a\text{,}$ $\omega = (\omega \alpha)/ \alpha$ is also in $E\text{.}$ Let $K = F( \omega)\text{.}$ Then $F \subset K \subset E\text{.}$ Since $K$ is the splitting field of $x^n - 1\text{,}$ $K$ is a normal extension of $F\text{.}$ Therefore, any automorphism $\sigma$ in $G(F( \omega)/ F)$ is determined by $\sigma( \omega)\text{.}$ It must be the case that $\sigma( \omega ) = \omega^i$ for some integer $i$ since all of the zeros of $x^n - 1$ are powers of $\omega\text{.}$ If $\tau( \omega ) = \omega^j$ is in $G(F(\omega)/F)\text{,}$ then

$$\begin{equation*} \sigma \tau( \omega ) = \sigma( \omega^j ) = [ \sigma( \omega )]^j = \omega^{ij} = [\tau( \omega ) ]^i = \tau( \omega^i ) = \tau \sigma( \omega ). \end{equation*}$$

Therefore, $G(F( \omega ) / F)$ is abelian. By the Fundamental Theorem of Galois Theory the series

$$\begin{equation*} \{ \identity \} \subset G(E/ F(\omega)) \subset G(E/F) \end{equation*}$$

is a normal series. By our previous argument, $G(E/F(\omega))$ is abelian. Since

$$\begin{equation*} G(E/F) /G(E/F( \omega)) \cong G(F(\omega)/F) \end{equation*}$$

is also abelian, $G(E/F)$ is solvable.

Lemma23.28

Let $F$ be a field of characteristic zero and let

$$\begin{equation*} F = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_r = E \end{equation*}$$

a radical extension of $F\text{.}$ Then there exists a normal radical extension

$$\begin{equation*} F = K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_r = K \end{equation*}$$

such that $K$ that contains $E$ and $K_i$ is a normal extension of $K_{i-1}\text{.}$

Proof

Since $E$ is a radical extension of $F\text{,}$ there exists a chain of subfields

$$\begin{equation*} F = F_0 \subset F_1 \subset F_2 \subset \cdots \subset F_r = E \end{equation*}$$

such for $i = 1, 2, \ldots, r\text{,}$ we have $F_i = F_{i - 1}(\alpha_i)$ and $\alpha_i^{n_i} \in F_{i-1}$ for some positive integer $n_i\text{.}$ We will build a normal radical extension of $F\text{,}$

$$\begin{equation*} F = K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_r = K \end{equation*}$$

such that $K \supseteq E\text{.}$ Define $K_1$ for be the splitting field of $x^{n_1} - \alpha_1^{n_1}\text{.}$ The roots of this polynomial are $\alpha_1, \alpha_1 \omega, \alpha_1 \omega^2, \ldots, \alpha_1 \omega^{n_1 - 1}\text{,}$ where $\omega$ is a primitive $n_1$th root of unity. If $F$ contains all of its $n_1$ roots of unity, then $K_1 = F(\alpha_!)\text{.}$ On the other hand, suppose that $F$ does not contain a primitive $n_1$th root of unity. If $\beta$ is a root of $x^{n_1} - \alpha_1^{n_1}\text{,}$ then all of the roots of $x^{n_1} - \alpha_1^{n_1}$ must be $\beta, \omega \beta, \ldots, \omega^{n_1-1}\text{,}$ where $\omega$ is a primitive $n_1$th root of unity. In this case, $K_1 = F(\omega \beta)\text{.}$ Thus, $K_1$ is a normal radical extension of $F$ containing $F_1\text{.}$ Continuing in this manner, we obtain

$$\begin{equation*} F = K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_r = K \end{equation*}$$

such that $K_i$ is a normal extension of $K_{i-1}$ and $K_i \supseteq F_i$ for $i = 1, 2, \ldots, r\text{.}$

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

Theorem23.29

Let $f(x)$ be in $F[x]\text{,}$ where $\chr F = 0\text{.}$ If $f(x)$ is solvable by radicals, then the Galois group of $f(x)$ over $F$ is solvable.

Proof

Since $f(x)$ is solvable by radicals there exists an extension $E$ of $F$ by radicals $F = F_0 \subset F_1 \subset \cdots \subset F_n = E\text{.}$ By Lemma 23.28, we can assume that $E$ is a splitting field $f(x)$ and $F_i$ is normal over $F_{i - 1}\text{.}$ By the Fundamental Theorem of Galois Theory, $G(E/F_i)$ is a normal subgroup of $G(E/F_{i - 1})\text{.}$ Therefore, we have a subnormal series of subgroups of $G(E/F)\text{:}$

$$\begin{equation*} \{ \identity \} \subset G(E/F_{n - 1}) \subset \cdots \subset G(E/F_1) \subset G(E/F). \end{equation*}$$

Again by the Fundamental Theorem of Galois Theory, we know that

$$\begin{equation*} G(E/F_{i - 1})/G(E/F_i) \cong G(F_i/F_{i - 1}). \end{equation*}$$

By Lemma 23.27, $G(F_i/F_{i - 1})$ is solvable; hence, $G(E/F)$ is also solvable.

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 $S_5\text{.}$ We begin by proving a lemma.

Lemma23.30

If $p$ is prime, then any subgroup of $S_p$ that contains a transposition and a cycle of length $p$ must be all of $S_p\text{.}$

Proof

Let $G$ be a subgroup of $S_p$ that contains a transposition $\sigma$ and $\tau$ a cycle of length $p\text{.}$ We may assume that $\sigma = (1 2)\text{.}$ The order of $\tau$ is $p$ and $\tau^n$ must be a cycle of length $p$ for $1 \leq n \lt p\text{.}$ Therefore, we may assume that $\mu = \tau^n = (1 2 i_3 \ldots i_p)$ for some $n\text{,}$ where $1 \leq n \lt p$ (see Exercise 5.3.13 in Chapter 5). Noting that $(1 2)(12 i_3\ldots i_p) = (2 i_3\ldots i_p)$ and $(2i_3 \ldots i_p)^k(12)(2i_3 \ldots i_p)^{-k} = (1i_k)\text{,}$ we can obtain all the transpositions of the form $(1n)$ for $1 \leq n \lt p\text{.}$ However, these transpositions generate all transpositions in $S_p\text{,}$ since $(1j)(1 i)(1 j) = (i j)\text{.}$ The transpositions generate $S_p\text{.}$

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 $S_5\text{.}$ By Eisenstein's Criterion, $f(x)$ is irreducible and, therefore, must be separable. The derivative of $f(x)$ is $f'(x) = 5 x^4 - 18 x^2 - 27\text{;}$ 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 $-2\text{,}$ between $-2$ and $0\text{,}$ 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 $f(x)\text{.}$ 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 $f(x)\text{,}$ we know that $G(K/{\mathbb Q})$ is a subgroup of $S_5\text{.}$ 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 $f(x)\text{.}$ The automorphism of ${\mathbb C}$ that takes $a + bi \mapsto a - bi$ leaves the real roots fixed and interchanges the complex roots; consequently, $G(K/{\mathbb Q} )$ contains a transpostion. If $\alpha$ is one of the real roots of $f(x)\text{,}$ then $[\mathbb Q(\alpha) : \mathbb Q] = 5$ by Exercise 21.4.28. Since $\mathbb Q(\alpha)$ is a subfield of $K\text{,}$ it must be the case the $[K : \mathbb Q]$ is divisible by 5. Since $[K : \mathbb Q] = |G(K/{\mathbb Q})|$ and $G(K/{\mathbb Q}) \subset S_5\text{,}$ we know that $G(K/{\mathbb Q})$ contains a cycle of length 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 $S_5\text{.}$ 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.

Theorem23.33Fundamental Theorem of Algebra

The field of complex numbers is algebraically closed; that is, every polynomial in ${\mathbb C}[x]$ has a root in ${\mathbb C}\text{.}$

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 $\alpha \in E$ such that $E = {\mathbb C}( \alpha )$ with $\alpha$ the root of an irreducible polynomial $f(x)$ in ${\mathbb C}[x]\text{.}$ The splitting field $L$ of $f(x)$ is a finite normal separable extension of ${\mathbb C}$ that contains $E\text{.}$ We must show that it is impossible for $L$ to be a proper extension of ${\mathbb C}\text{.}$

Suppose that $L$ is a proper extension of ${\mathbb C}\text{.}$ Since $L$ is the splitting field of $f(x)(x^2 + 1)$ over ${\mathbb R}\text{,}$ $L$ is a finite normal separable extension of ${\mathbb R}\text{.}$ Let $K$ be the fixed field of a Sylow 2-subgroup $G$ of $G(L/{\mathbb R})\text{.}$ Then $L \supset K \supset {\mathbb R}$ and $|G( L / K )| =[L:K]\text{.}$ Since $[L : {\mathbb R}] = [L:K][K:{\mathbb R}]\text{,}$ we know that $[K:{\mathbb R}]$ must be odd. Consequently, $K = {\mathbb R}(\beta)$ with $\beta$ having a minimal polynomial $f(x)$ of odd degree. Therefore, $K = {\mathbb R}\text{.}$

We now know that $G(L/{\mathbb R})$ must be a 2-group. It follows that $G(L / {\mathbb C})$ is a 2-group. We have assumed that $L \neq {\mathbb C}\text{;}$ therefore, $|G(L / {\mathbb C})| \geq 2\text{.}$ By the first Sylow Theorem and the Fundamental Theorem of Galois Theory, there exists a subgroup $G$ of $G(L/{\mathbb C})$ of index 2 and a field $E$ fixed elementwise by $G\text{.}$ Then $[E:{\mathbb C}] = 2$ and there exists an element $\gamma \in E$ with minimal polynomial $x^2 + b x + c$ in ${\mathbb C}[x]\text{.}$ This polynomial has roots $( - b \pm \sqrt{b^2 - 4c}\, ) / 2$ that are in ${\mathbb C}\text{,}$ since $b^2 - 4 c$ is in ${\mathbb C}\text{.}$ This is impossible; hence, $L = {\mathbb C}\text{.}$

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.