# Galois subfields of real radical extensions are at most quadratic

###### Theorem 1.

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

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 $K^{\prime}=K(\zeta_{n})=F^{\prime}(\sqrt[n]{\alpha})$.

 $\xymatrix@R1pc@C.3pc{&K^{\prime}=K(\zeta_{n})=F^{\prime}(\sqrt[n]{\alpha})\ar@% {-}[dl]\ar@{-}[dr]&&\\ K=F(\sqrt[n]{\alpha})\ar@{-}[dr]&&L^{\prime}=L(\zeta_{n})\ar@{-}[dl]\ar@{-}[dr% ]&\\ &L\ar@{-}[dr]&&F^{\prime}=F(\zeta_{n})\ar@{-}[dl]\\ &&F&}$

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 $\operatorname{Gal}(K^{\prime}/F^{\prime})$). 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 $n\leq 2$ (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 GaloisSubfieldsOfRealRadicalExtensionsAreAtMostQuadratic 2013-03-22 17:43:05 2013-03-22 17:43:05 rm50 (10146) rm50 (10146) 7 rm50 (10146) Theorem msc 12F10 msc 12F05