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 Fsone Oct 30 '17 at 15:38

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