PlanetMath: Galois subfields of real radical extensions are at most quadratic

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Galois subfields of real radical extensions are at most quadratic

(Theorem)

Theorem 1 Suppose $F\subset L\subset K=F(\sqrt[n]\alpha)\subset\mathbb{R}$ are fields with $\alpha\in F$ and Galois over . Then $[L:F]\leq 2$ .

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

$\displaystyle \xymatrix @R1pc@C.3pc{ & K'=K(\zeta_n)=F'(\sqrt[n]\alpha) \ar@{-}... ...]\ar@{-}[dr] & \ & L \ar@{-}[dr] & & F'=F(\zeta_n) \ar@{-}[dl] \ & & F & }$

Now,

is Galois since

is. But

is a Kummer extension of

, so has cyclic Galois group and thus

has cyclic Galois group as well (being a quotient of ${\mathrm{Gal}}(K'/F')$ ). Thus

is a Kummer extension of

, so that $L'=F'(\sqrt[n]{\beta})$ for some $\beta\in L$ . It follows that $L=F(\sqrt[n]{\beta})$ . But since

is Galois over

, it follows that $n\leq 2$ (since otherwise in order to be Galois,

would have to contain the non-real $n^{\mathrm{th}}$ roots of unity).

"Galois subfields of real radical extensions are at most quadratic" is owned by rm50.

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Cross-references: contain, order, quotient, Galois group, cyclic, Kummer extension, root of unity, primitive, fields
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This is version 3 of Galois subfields of real radical extensions are at most quadratic, born on 2007-12-30, modified 2007-12-30.
Object id is 10163, canonical name is GaloisSubfieldsOfRealRootExtensionsAreAtMostQuadratic.
Accessed 196 times total.

Classification:

AMS MSC:	12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)
	12F10 (Field theory and polynomials :: Field extensions :: Separable extensions, Galois theory)

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