Galois subfields of real radical extensions are at most quadratic

Theorem 1.

Suppose $F\mathrm{\subset }L\mathrm{\subset }K\mathrm{=}F\mathit{}\mathrm{(}\sqrt[n]{\alpha }\mathrm{)}\mathrm{\subset }\mathrm{R}$ are fields with $\alpha \mathrm{\in }F$ and $L$ Galois over $F$. Then $\mathrm{[}L\mathrm{:}F\mathrm{]}\mathrm{\le }\mathrm{2}$.

Proof. Let ${\zeta }_n$ be a primitive $n^{\mathrm{th}}$ root of unity, and define $F^{\prime }=F({\zeta }_n)$, $L^{\prime }=L({\zeta }_n)$, and $K^{\prime }=K({\zeta }_n)=F^{\prime }(\sqrt[n]{\alpha })$.

Now, $L^{\prime }/F^{\prime }$ is Galois since $L/F$ is. But $K^{\prime }$ is a Kummer extension of $F^{\prime }$, so has cyclic Galois group and thus $L^{\prime }/F^{\prime }$ has cyclic Galois group as well (being a quotient of $\mathrm{Gal}(K^{\prime }/F^{\prime })$). Thus $L^{\prime }$ is a Kummer extension of $F^{\prime }$, so that $L^{\prime }=F^{\prime }(\sqrt[n]{\beta })$ for some $\beta \in F^{\prime }$. It follows that $L=F(\sqrt[n]{\beta })$. But since $L$ is Galois over $F$, it follows that $n\le 2$ (since otherwise in order to be Galois, $L$ would have to contain the non-real $n^{\mathrm{th}}$ roots of unity).

Title	Galois subfields of real radical extensions are at most quadratic
Canonical name	GaloisSubfieldsOfRealRadicalExtensionsAreAtMostQuadratic
Date of creation	2013-03-22 17:43:05
Last modified on	2013-03-22 17:43:05
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 12F10
Classification	msc 12F05