Solvability by radicals

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 01 February 2012

Solvability by radicals

Let 𝔽 be a field.

Let 𝔽 be a field with char𝔽=0 and let 𝔼 be the splitting field of fx𝔽 [x]. Then fx   is solvable by radicals if and only if Gal(𝔼/𝔽)   is a solvable group.

) Assume G=Gal(𝔼/𝔽) is solvable.
Case 1. 𝔽 contains a primitive nth root of unity. G G1 Gr= {1} with   Gi< Gi-1   and   Gi/ Gi-1   cyclic. Let 𝔽 𝔽1 𝔽r =𝔼 be the corresponding fixed subfields. The normality of Gi/ Gi-1 implies that 𝔽i/ 𝔽i-1 is a normal extension and Gi/ Gi-1 cyclic implies 𝔽i/ 𝔽i-1 is cyclic. Thus, by the cyclotomy theorem, 𝔽i is the splitting field of an irreducible xni -bi 𝔽i-1[x], where bi= αini and 𝔽i= 𝔽i-1 αi. So fx is solvable by radicals.

Case 2. 𝔽 does not contain a primitive nth root of unity. Let ω be a primitive nth root of unity. Then 𝔼ω is the splitting field of fx over 𝔽ω. The map Gal(𝔼(ω )/𝔽(ω )) Gal(𝔼/𝔽 ) σ σ|𝔼 is injective. Thus, since Gal(𝔼/𝔽) is solvable Gal(𝔼(ω) /𝔽(ω)) is solvable. So 𝔼(ω) is a radical extension of 𝔽 and 𝔼𝔼(ω). So f(x) is solvable by radicals.
) Let 𝔼 be the splitting field of xn-a 𝔽[x].
First show that Gal(𝔼/𝔽) is solvable. 𝔼 {1} 𝔽(ω) Gal(𝔼/ 𝔽(ω) ) is abelian 𝔽 Gal(𝔼/𝔽) Then Gal(𝔼/𝔽(ω)) is normal in Gal(𝔼/𝔽) and the quotient is Gal(𝔽(ω)/𝔽), a cyclic group. Since Gal(𝔼/𝔽(ω)) is abelian it is solvable. So Gal(𝔼/𝔽) is solvable.
Case 2. We have
since each 𝔼i/ 𝔼i-1 is a splitting field of xni- αi 𝔼i-1 [x]. This series can be refined to a solvable series. So Gal( 𝔼r/𝔽) is solvable. So Gal(𝔼/𝔽) Gal( 𝔼r/𝔽 ) Gal(𝔼/𝔽) is solvable.

Notes and References

Where are these from?



page history