Why is $\displaystyle \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a})$ a Galois extension over $\displaystyle \mathbb{Q}$??

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- Dec 21st 2008, 02:23 AMStigerGalois extension over Q
Why is $\displaystyle \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a})$ a Galois extension over $\displaystyle \mathbb{Q}$??

- Dec 21st 2008, 08:46 AMThePerfectHacker
Intersection of subfields (in a larger field) is a subfield. Therefore, $\displaystyle \mathbb{Q}(\zeta_n)\cap \mathbb{Q}(\sqrt[n]{a})$ is a subfield (of the algebraic numbers, for example). Now $\displaystyle \mathbb{Q}(\zeta)/\mathbb{Q}$ is a cyclotomic extension and therefore $\displaystyle \text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ is an abelian group. Since all subgroups of this group are normal it follows that if $\displaystyle \mathbb{Q}\subseteq L\subseteq \mathbb{Q}(\zeta_n)$ then $\displaystyle L/\mathbb{Q}$ is a Galois extension by the fundamental theorem. Since $\displaystyle \mathbb{Q}\subseteq \mathbb{Q}(\zeta)\cap \mathbb{Q}(\sqrt[n]{a})\subseteq \mathbb{Q}(\zeta)$ the rest follows.

- Dec 21st 2008, 01:41 PMNonCommAlg
in general if $\displaystyle \{K_i \}_{i \in I}$ is a family of (finite) Galois extensions of $\displaystyle F,$ then $\displaystyle \bigcap_{i \in I}K_i$ is also Galois over $\displaystyle F.$ just recall that a (finite) extension is Galois iff it's separable and normal.

__Edit__: what was i thinking?!! lol ... this, although true, but doesn't apply to Stiger's problem! thanks ThePerfectHacker for pointing that out. - Dec 21st 2008, 04:08 PMThePerfectHacker