# Thread: Galois extension over Q

1. ## Galois extension over Q

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

2. 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.

3. 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.

4. 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.