Degree of extension of $\mathbb Q(\sqrt{a_1},\cdots,\sqrt{a_n})$ over $\mathbb Q$

Asked Dec 21 '23 at 15:47

Active Dec 21 '23 at 15:47

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Assume $a_1,\cdots,a_n \in \mathbb N_+$, none of which is perfect square and each pair of them is coprime. Proof that $$ [\mathbb Q(\sqrt{a_1},\cdots,\sqrt{a_n}): \mathbb Q]=2^n $$

I have tried a lot but I don't have any progress. Can someone please give me a hint?

asked Dec 21 '23 at 15:47

QiFeng233

Another duplicate is this post, where the proof is not only for prime numbers. – Dietrich Burde Dec 21 '23 at 15:58

Degree of extension of $\mathbb Q(\sqrt{a_1},\cdots,\sqrt{a_n})$ over $\mathbb Q$

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