## Solvability by radicals

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 ${n}_{1},...,{n}_{r} \in {ℤ}_{>0}$ such that $𝔽⊆ 𝔽 α1 ⊆ 𝔽 α1 α2 ⊆ 𝔽 α1 ... αr =𝕂,$ 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

 Proof. $⇐\right)$ Assume $G=\mathrm{Gal}\left(𝔼/𝔽\right)$ is solvable. Case 1. $𝔽$ contains a primitive ${n}^{th}$ root of unity. Let $𝔽⊆ 𝔽1⊆⋯ ⊆𝔽r =𝔼$ 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 $Gal(𝔼(ω )/𝔽(ω )) → Gal(𝔼/𝔽 ) σ ↦ σ|𝔼$ is injective. Thus, since $\mathrm{Gal}\left(𝔼/𝔽\right)$ is solvable $\mathrm{Gal}\left(𝔼\left(\omega \right)/𝔽\left(\omega \right)\right)$ 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. $⇒\right)$ Let $𝔼$ be the splitting field of ${x}^{n}-a\in 𝔽\left[x\right].$ First show that $\mathrm{Gal}\left(𝔼/𝔽\right)$ is solvable. $𝔼 {1} ⊇ 𝔽(ω) Gal(𝔼/ 𝔽(ω) ) is abelian ⊇ ⊲ 𝔽 Gal(𝔼/𝔽)$ Then $\mathrm{Gal}\left(𝔼/𝔽\left(\omega \right)\right)$ is normal in $\mathrm{Gal}\left(𝔼/𝔽\right)$ and the quotient is $\mathrm{Gal}\left(𝔽\left(\omega \right)/𝔽\right),$ a cyclic group. Since $\mathrm{Gal}\left(𝔼/𝔽\left(\omega \right)\right)$ is abelian it is solvable. So $\mathrm{Gal}\left(𝔼/𝔽\right)$ 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 $\mathrm{Gal}\left({𝔼}_{r}/𝔽\right)$ is solvable. So $Gal(𝔼/𝔽) ≅ Gal( 𝔼r/𝔽 ) Gal(𝔼/𝔽) is solvable.$ $\square$

## Notes and References

Where are these from?

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