# Does $\mathbb{Q}(\sqrt{m_1},…,\sqrt{m_k})$ has degree $2^k$ over $\mathbb{Q}? [duplicate] Let$m_1 ,...,m_k$be$k$distinct squarefree integers$\neq 1$which are pairwise coprime. Then is$[\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k}):\mathbb{Q}]=2^k$? Intuitively, I guess that the statement is true, by observing the result when$k=1,2$, but I have trouble actually proving this for the general case. What I know is that$\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})$is Galois over$\mathbb{Q}$, so$[\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k}):\mathbb{Q}]=|\mathrm{Gal}(\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})/\mathbb{Q})|$and every element of$G=\mathrm{Gal}(\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})/\mathbb{Q})$sends$\sqrt{m_i}$to either itself or$-\sqrt{m_i}$for each$i=1,...,k$, so$G$has at most$2^k$elements. I think the key of the proof is to prove that if$\phi(\sqrt{m_i})$is given, then there is always an automorphism$\phi$of$\mathbb{Q}(\sqrt{m_1},...,\sqrt{m_k})$having such value of$\phi(\sqrt{m_i})\$, and here is where I stuck.

Does anyone have ideas? Any advice or comments will be helpful!

If it is hard to give the answer here directly, then any bibliography for reference is also acceptable.

## marked as duplicate by anomaly, Rolf Hoyer, user99914, Claude Leibovici, Guy FsoneOct 30 '17 at 15:38

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