Abstract Algebra: Theory and Applications, Remixed!

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

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

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

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

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

Proof.

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

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 \(F\text{,}\)

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

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

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 10.36, 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 \(G(E/F)\text{:}\)

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

By Lemma 10.35, \(G(F_i/F_{i - 1})\) is solvable; hence, \(G(E/F)\) is also solvable.

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 2.7.94). 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 \(S_p\text{.}\)

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, \(L = {\mathbb C}\text{.}\)