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Consider $K=\mathbb Q(\sqrt{2},\sqrt{3})$ - I think $K$ is Galois since it's the splitting field of $(x-\sqrt 2)(x+\sqrt 2)(x-\sqrt 3)(x+\sqrt 3)$. I feel like $G(K/F)$ is isomorphic to the Klein 4 group since you can swap $\pm\sqrt{2}$, $\pm\sqrt{3}$ or both (since the roots have no F-relation). But since $K$ is Galois, $|G(K/F)|=[K:F]$ which means $[K:F]=4$, which clashes with my intuition that it should have degree three (the basis units being 1, $\sqrt{2}$, $\sqrt 3$).

What am I not understanding?

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What about $\sqrt{6}$? – Zhen Lin Sep 22 '11 at 2:20
up vote 10 down vote accepted

As a field it also has to be closed under multiplication so $\mathbb{Q}(\sqrt{2},\sqrt{3})$ also has to include $\sqrt{6}=\sqrt{2}\sqrt{3}$. Since $\sqrt{6}$ isn't a rational linear combination of $\sqrt{2}$ and $\sqrt{3}$, the set $ \{1,\sqrt{2},\sqrt{3},\sqrt{6} \}$ is linearly independent (over $\mathbb{Q}$). Since this set has $4=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]$ elements it in fact forms a basis.

Another way you can see this is by observing that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2})(\sqrt{3})$ (adjoin $\sqrt{2}$ and then adjoin $\sqrt{3}$), for then $$ \mathbb{Q}(\sqrt{2})(\sqrt{3}) = \{ \alpha + \beta\sqrt{3} : \alpha,\beta\in\mathbb{Q}(\sqrt{2}) \}$$ $$ = \{ (a+b\sqrt{2}) + (c+d\sqrt{2})\sqrt{3} : a,b,c,d\in\mathbb{Q} \} $$ $$ = \{ a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} : a,b,c,d\in\mathbb{Q} \} $$

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A basis includes $\sqrt{6}$; perhaps better to note that $\sqrt{6}$ lies in the field, but is not a rational linear combination of $1$, $\sqrt{2}$, and $\sqrt{3}$... – Arturo Magidin Sep 22 '11 at 2:44
@ArturoMagidin: Thanks for noticing that. It's been fixed now. – Riley E Sep 22 '11 at 3:01

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