Galois extension over Q

**Stiger** · December 21st 2008, 02:23 AM

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

**ThePerfectHacker** · December 21st 2008, 08:46 AM

Originally Posted by Stiger

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

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

**NonCommAlg** · December 21st 2008, 01:41 PM

in general if $\{K_i \}_{i \in I}$ is a family of (finite) Galois extensions of $F,$ then $\bigcap_{i \in I}K_i$ is also Galois over $F.$ 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.

**ThePerfectHacker** · December 21st 2008, 04:08 PM

Originally Posted by NonCommAlg

in general if $\{K_i \}_{i \in I}$ is a family of (finite) Galois extensions of $F,$ then $\bigcap_{i \in I}K_i$ is also Galois over $F.$ just recall that a (finite) extension is Galois iff it's separable and normal.

But $\mathbb{Q}(\sqrt[n]{a})/\mathbb{Q}$ is not Galois in general.

Thread: Galois extension over Q

LinkBack

Thread Tools

Display

Galois extension over Q

Similar Math Help Forum Discussions

Galois Extension

finite Extension of Fp Galois

Galois extension

Galois extension

Galois Extension

Search Tags