Solvability by radicals

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 01 February 2012

Solvability by radicals

Let $𝔽$ be a field.

• A polynomial $f\left(x\right)\in 𝔽\left[x\right]$ is solvable by radicals if the splitting field $𝔼$ of $f\left(x\right)$ is contained in a radical extension of $𝔽$.
• A radical extension of $𝔽$ is an extension $𝕂$ such that there are elements ${\alpha }_1,...,{\alpha }_r \in 𝕂$ and $n_1,...,n_r \in {\mathbb{Z}}_{>0}$ such that $$𝔽\subseteq 𝔽\left({\alpha }_1\right)\subseteq 𝔽\left({\alpha }_1,{\alpha }_2\right)\subseteq 𝔽\left({\alpha }_1,...,{\alpha }_r\right)=𝕂,$$ with ${\alpha }_i^{n_i}\in 𝔽\left({\alpha }_1,...,{\alpha }_{i-1}\right),$ for each $1\le i\le r.$
• A finite group is solvable if ???

Let $𝔽$ be a field with $\mathrm{char}\left(𝔽\right)=0$ and let $𝔼$ be the splitting field of $f\left(x\right)\in 𝔽\left[x\right].$ Then $$f\left(x\right) \; is\; solvable\; by\; radicalsif\; and\; only\; if\mathrm{Gal}(𝔼/𝔽) \; is\; a\; solvable\; group.$$

		Proof.
	$\Leftarrow )$ Assume $G=\mathrm{Gal}(𝔼/𝔽)$ is solvable. Case 1. $𝔽$ contains a primitive $n^{th}$ root of unity. $$G\supseteq G_1\supseteq \cdots \supseteq G_r=\left\{1\right\}with\; G_i<G_{i-1} \; and\; G_i/G_{i-1} \; cyclic.$$ Let $$𝔽\subseteq {𝔽}_1\subseteq \cdots \subseteq {𝔽}_r=𝔼$$ be the corresponding fixed subfields. The normality of $G_i/G_{i-1}$ implies that ${𝔽}_i/{𝔽}_{i-1}$ is a normal extension and $G_i/G_{i-1}$ cyclic implies ${𝔽}_i/{𝔽}_{i-1}$ is cyclic. Thus, by the cyclotomy theorem, ${𝔽}_i$ is the splitting field of an irreducible $x^{n_i}-b_i\in {𝔽}_{i-1}\left[x\right],$ where $b_i={\alpha }_i^{n_i}$ and ${𝔽}_i={𝔽}_{i-1}\left({\alpha }_i\right).$ So $f\left(x\right)$ is solvable by radicals. Case 2. $𝔽$ does not contain a primitive $n^{th}$ root of unity. Let $\omega$ be a primitive $n^{th}$ root of unity. Then $𝔼\left(\omega \right)$ is the splitting field of $f\left(x\right)$ over $𝔽\left(\omega \right)$. The map $$\begin{array}{ccc}\mathrm{Gal}\left(𝔼\right(\omega )/𝔽(\omega \left)\right)& \to & \mathrm{Gal}(𝔼/𝔽)\\ \sigma & \mapsto & \sigma {|}_{𝔼}\end{array}$$ is injective. Thus, since $\mathrm{Gal}(𝔼/𝔽)$ is solvable $\mathrm{Gal}\left(𝔼\right(\omega )/𝔽(\omega \left)\right)$ is solvable. So $𝔼\left(\omega \right)$ is a radical extension of $𝔽$ and $𝔼\subseteq 𝔼\left(\omega \right).$ So $f\left(x\right)$ is solvable by radicals. $\Rightarrow )$ Let $𝔼$ be the splitting field of $x^n-a\in 𝔽\left[x\right].$ First show that $\mathrm{Gal}(𝔼/𝔽)$ is solvable. $$\begin{array}{ccc}𝔼& \left\{1\right\}& \\ \supseteq & & \\ 𝔽\left(\omega \right)& \mathrm{Gal}(𝔼/𝔽(\omega \left)\right)& is\; abelian\\ \supseteq & ⊲& \\ 𝔽& \mathrm{Gal}(𝔼/𝔽)& \end{array}$$ Then $\mathrm{Gal}(𝔼/𝔽(\omega \left)\right)$ is normal in $\mathrm{Gal}(𝔼/𝔽)$ and the quotient is $\mathrm{Gal}\left(𝔽\right(\omega )/𝔽),$ a cyclic group. Since $\mathrm{Gal}(𝔼/𝔽(\omega \left)\right)$ is abelian it is solvable. So $\mathrm{Gal}(𝔼/𝔽)$ is solvable. Case 2. We have PICTURE IS MISSING since each ${𝔼}_i/{𝔼}_{i-1}$ is a splitting field of $x^{n_i}-{\alpha }_i\in {𝔼}_{i-1}\left[x\right].$ This series can be refined to a solvable series. So $\mathrm{Gal}({𝔼}_r/𝔽)$ is solvable. So $$\mathrm{Gal}(𝔼/𝔽)\cong \frac{\mathrm{Gal}({𝔼}_r/𝔽)}{\mathrm{Gal}(𝔼/𝔽)}is\; solvable.$$ $\square$

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