# Let $F = \mathbb{Q}(a_1, a_2, . . . , a_n)$ with $a_i^2 \in \mathbb{Q}.$ Prove that $\sqrt[3]2 \notin F$.

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.

• The degree theorem implies it because $3\nmid 2^m$, you are right. You just have to consider the right tower of field extensions. – Dietrich Burde Apr 17 at 20:02
• $F$ 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 $2$. – Greg Martin Apr 17 at 20:08
• 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 $a_1 = \sqrt{2}$, $a_2 = \sqrt{3}$, $a_3 = \sqrt{6}$. – Daniel Schepler Apr 17 at 20:09

Hint: $$[\mathbb{Q}(a_1,\dots,a_{k+1}:\mathbb{Q}(a_1,\dots,a_k)]=1$$ or $$2$$.