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The square roots of the primes are linearly independent over the field of rationals

I am trying to classify the Galois group of the field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})/\mathbb{Q}$ and am getting stuck on trying to show that the extension is degree 8. I understand that you can look at intermediate fields in the tower to get

$[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) : \mathbb{Q(\sqrt{2}, \sqrt{3})}][\mathbb{Q}(\sqrt{2}, \sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}) : \mathbb{Q}]$

and each of these has degree 2. But is there an easy way to show $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) : \mathbb{Q}(\sqrt{2}, \sqrt{3})] = 2$? I tried showing that $\sqrt{5}$ cannot be written as a linear combination of $\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}\}$ but it's a long mess, and I'm wondering if there's a more clever way to do this.

More generally, is it true that $\mathbb{Q}(\sqrt{p_1}, ..., \sqrt{p_n})/\mathbb{Q}$ where $p_i$ are distinct primes has degree $2^n$?

Community
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Brian
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1 Answers1

22

An easy way to show that $\sqrt{5}$ is not in $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is to note that $[\mathbb{Q}(\sqrt{2},\sqrt{3})\colon\mathbb{Q}]=4$, so by the Fundamental Theorem of Galois Theory it has either one or three intermediate fields of degree $2$ over $\mathbb{Q}$ (since a group of order $4$ has either a single subgroup of index $2$, when the group is cyclic, or exactly three, when it is the Klein $4$-group). Since this field contains the three distinct intermediate field $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, and $\mathbb{Q}(\sqrt{6})$, it cannot also contain the field $\mathbb{Q}(\sqrt{5})$, which is distinct from those three.

For your second question, the answer is likewise "yes". The only intermediate fields of degree $2$ are $\mathbb{Q}(\sqrt{d})$, where $d$ is the of some (but at least one) of the $p_i$. You can prove it by induction on $n$.

Arturo Magidin
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