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Section 11.1 Solvability 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}\text{.} \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 \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 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 \(f(x)\text{.}\)
The easiest polynomial to solve by radicals is one of the form \(x^n - a\text{.}\) As we discussed in Example 7.2.4, the roots of \(x^n - 1\) are called the nth roots of unity. These roots are a finite subgroup of the splitting field of \(x^n -1\text{.}\) By Corollary 7.3.5, the \(n\)th roots of unity form a cyclic group. Any generator of this group is called a primitive nth root of unity.

Example 11.1.1.

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)\text{.} \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 \}\text{,} \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 9.3.4.

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 )\text{.} \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 )\text{.} \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.

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 \(K_1 = F(\alpha_1)\text{.}\) 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 \(\beta, \omega \beta, \ldots, \omega^{n_1-1} \beta\text{,}\) 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.

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 11.1.3, 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)\text{.} \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})\text{.} \end{equation*}
By Lemma 11.1.2, \(G(F_i/F_{i - 1})\) is solvable; hence, \(G(E/F)\) is also solvable.
The converse of Theorem 11.1.4 is also true. For a proof, see any of the references at the end of this chapter.
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