# How to prove that $\sqrt{p_n} \notin \mathbb{Q}(\sqrt{p_1},\dots,\sqrt{p_{n-1}})$

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 $n>2$.

• Mar 21, 2018 at 7:05
• Kummer theory${}$? Mar 21, 2018 at 7:18
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 $2$.
• 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? Mar 22, 2018 at 4:26
• I dont think its true for general groups, maybe even it characterizes the product of $\mathbb{Z}_2$. Mar 22, 2018 at 13:22