How would one show that $[\mathbb{Q}(\sqrt2, \sqrt3,...,\sqrt{p_n},...)]=\infty$? I know that we want to show there is no finite basis over the rationals, but I'm not sure how one would determine that such a basis does not exist.
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Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question. – David R. Nov 07 '18 at 22:27
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1See e.g. https://math.stackexchange.com/a/1609061/300700 – nguyen quang do Nov 09 '18 at 16:35
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Kummer theory tells us that as $ 2$, $3,\ldots,{p_n}$ generate a subgroup of order $2^n$ in $\Bbb Q^*/(\Bbb Q^*)^2$ then the extension field $K_n=\Bbb Q(\sqrt2,\ldots,\sqrt {p_n})$ has degree $2^n$ over $\Bbb Q$. So your field contains subfields of arbitrarily large degree over $\Bbb Q$.

Angina Seng
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