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Suppose that $F$ is the infinite extension of $\mathbb{Q}$ obtained by adjoining the square root of every integer (positive or negative). I'm trying to show that $\sqrt[3]{2}\notin F$. I have no idea how to prove it. Anyone can help me?

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  • $\begingroup$ I guess you have to say that beside $F$ being an infinite extension of $\mathbb{Q}$, if $a \in F$ then there is some $n$ such that $a \in \mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{3}, \ldots,\sqrt{n})$ $\endgroup$
    – reuns
    Apr 8 '16 at 10:13
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$F$ is the union of adjoining of finitely many quadratic extension which have degree over $\mathbb{Q} $ a power of 2. Then your element must be in one of those finite adjoining, but it has degree 3 over $\mathbb{Q} $ and that's impossible

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  • $\begingroup$ Is "grade" means "degree"? I never see this word in my textbook. Can you define it? $\endgroup$
    – Kelan
    Apr 8 '16 at 10:13
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    $\begingroup$ Yes, I'm sorry but in italian they have the same translation, I'll correct $\endgroup$
    – karmalu
    Apr 8 '16 at 10:13

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