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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.

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

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