As the title says, I am trying to prove that $\sqrt{2} \notin \mathbb{Q}(a)$, for all $ a \neq \sqrt[k]{2}$. My intuition says that this is true but I haven't made much progress.
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1Take $a = 1 + \sqrt{2}$. – anomaly Mar 23 '20 at 12:10
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This isn't true. For instance, $$ \sqrt 2\in \Bbb Q\left(\sqrt{5 + \sqrt2}\right) $$ yet $\left(\sqrt{5 + \sqrt2}\right)^k$ never becomes $2$ for any integer $k$.
Even simpler, we could have $\Bbb Q(\sqrt 2 + 1)$, or just $\Bbb Q(2\sqrt 2)$.
Arthur
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Ahh I see, is there any way then to find for which $a \in \mathbb{R}$ the proposition is true? – Kagk Mar 23 '20 at 12:03
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@AVlachos You mean a way to decide whether $\sqrt 2\in \Bbb Q(a)$? I don't know a good general way to decide this, no. There are a few theoretical ways, like "does the polynomial $x^2-2$ split over $\Bbb Q(a)$", but it eventually just boils down to the same thing. There are also a few definitive and practical ways, but they don't work all the time. For instance, if $[\Bbb Q(a):\Bbb Q]$ is odd, then $\Bbb Q(a)$ cannot contain $\sqrt 2$. Or the other way (which would work in the examples in my answer post), if you can make $\sqrt 2$, then clearly $\sqrt 2$ is in the field. – Arthur Mar 23 '20 at 12:06
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I was actually trying to prove it for when $a=\sqrt{3+\sqrt{3+\sqrt{3}}}$ but I figured it would be easier to prove the general case then move down to the specific one. – Kagk Mar 23 '20 at 12:10
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@AVlachos No, I really don't think it is. This is one of those times where details of the proof basically has to be tailored to the number in question. At least as far as I'm aware. – Arthur Mar 23 '20 at 12:23
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@AVlachos See the links here for general algorithms for testing (sub)field membership and related algorithms. – Bill Dubuque Mar 23 '20 at 15:23