# $\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)=\mathbb Q(\sqrt 2+\sqrt 3+\sqrt 5)$ - Generalisation?

Problem:

We know that $\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)=\mathbb Q(\sqrt 2+\sqrt 3+\sqrt 5)$. Generalise this fact.

My idea is to use the (proof of the) primitive element theorem. Looking at the equations given, I think I would phrase a generalisation as follows:

Let $a_i,...,a_n\in\mathbb N$ be distinct integers such that gcd$(a_i,a_j)=1$ for all $i\ne j$. Then $$\mathbb Q(\sqrt{a_1},\ldots,\sqrt{a_n})=\mathbb Q(\sqrt{a_1}+\ldots+\sqrt{a_n})$$

Is this generalisation in fact correct, and if so, is it possible to generalise this notion of field extensions even more?

• Why don't you just write down your proof? – Martin Brandenburg Apr 9 '13 at 17:26
• Also this should answer your question: math.stackexchange.com/questions/15074 – Martin Brandenburg Apr 9 '13 at 17:31
• @Martin: the link you give doesn't look like it's related to this question at all. Did you mean some other link? – KCd Apr 20 '13 at 12:24