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So I tried by claiming each extension $ \mathbb{Q}(a_i) $ was of degree 2 because of the $a_i^2 \in \mathbb{Q}.$ Apparently that wasn't necessarily true. I said that $ \mathbb{Q}(\sqrt[3]2) $ was a degree 3 extension and $F$ had degree $2^n$ so, it wasn't possible. But apparently that isn't necessarily true? SO I'm not too sure how to do it.

Fyi, this was a homework problem but no solutions were given and so, just wanna know how to do it, or get some help.

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  • $\begingroup$ The degree theorem implies it because $3\nmid 2^m$, you are right. You just have to consider the right tower of field extensions. $\endgroup$ – Dietrich Burde Apr 17 at 20:02
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    $\begingroup$ $F$ doesn't necessarily have degree $2^n$—consider for example $a_1=a_2=\cdots=a_n$. But you should be able to show (using a tower of extensions) that its degree is some power of $2$. $\endgroup$ – Greg Martin Apr 17 at 20:08
  • $\begingroup$ Probably the objection is: for example if $a_1 = a_2 = \sqrt{2}$ then it satisfies the conditions but $[F : \mathbb{Q}] = 2$ not $2^2$. For a slightly more complex counterexample, consider $a_1 = \sqrt{2}$, $a_2 = \sqrt{3}$, $a_3 = \sqrt{6}$. $\endgroup$ – Daniel Schepler Apr 17 at 20:09
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Hint: $[\mathbb{Q}(a_1,\dots,a_{k+1}:\mathbb{Q}(a_1,\dots,a_k)]=1$ or $2$.

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