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I want to prove $\mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})=\mathbb{Q}(\sum\limits_{i =1}^n\sqrt{p_i})$, where $p_i$ are different prime integers.

But I must solve this problem firstly: $\sqrt{p_n} \notin \mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})$.

It's trivial when $n=2$, but I don't have any idea for $n>2$.

user26857
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user469246
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1 Answers1

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Generalise the assumption from primes to assuming that no product is a perfect square. The field $\mathbb{Q}(\sqrt{p_1},\ldots ,\sqrt{p_n}) $ has $2^n-1$ many subfields of order $2$ namely for $X\subseteq n$ and $X\neq \emptyset$, there is the subfield, $$\mathbb{Q}(\sqrt{\prod\limits_{i\in X}p_i}) $$ and these are distinct by the better assumption used in the $n=2$ case. If $\sqrt{p_n}\in \mathbb{Q}(\sqrt{p_1},\ldots ,\sqrt{p_{n-1}}) $ then the Galois group of $\mathbb{Q}(\sqrt{p_1},\ldots ,\sqrt{p_n}) $ would have size at most $2^{n-1}$ and thus there could not be enough subfields of degree $2$.

Rene Schipperus
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  • It is really cool! A nice proof. Thank you for your help. – user469246 Mar 21 '18 at 15:12
  • My pleasure, for your main problem you might want to look into Cogalois theory, I think it has a lot to say about that sort of question in general. (Of course your case it can be solved directly). – Rene Schipperus Mar 21 '18 at 15:19
  • Thank you for your advices.BTW,I have an another question: actually I can understand that $(\mathbb{Z}/2\mathbb{Z})^{n-1}$which exactly has $2^{n-1}-1$ subgroups with index 2.But for any group G with order $2^{n-1}$, it’s true or not that G has at most $2^{n-1}-1$ subgroups with index 2? Can you give me any suggestions on this question? – user469246 Mar 22 '18 at 04:26
  • I dont think its true for general groups, maybe even it characterizes the product of $\mathbb{Z}_2$. – Rene Schipperus Mar 22 '18 at 13:22