$\textbf{Problem } \mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n}]=\mathbb{Q}[\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}]$ for all $n\geq 2$.
I knew that $\mathbb{Q}[\sqrt{2},\sqrt{3}]=\mathbb{Q}[\sqrt{2}+\sqrt{3}]$.
Thus, I'll prove the problem by using induction.
Suppose $\mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n}]=\mathbb{Q}[\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}]$
It suffices to show that $\mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n}]\subset \mathbb{Q}[\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n+1}]$ since $$ \mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n}]\subset \mathbb{Q}[\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n+1}]\subset \mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n},\sqrt{n+1}] $$ and $$[ \mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n},\sqrt{n+1}]:Q[\sqrt{2},\sqrt{3},\dots,\sqrt{n}]]\leq2$$ imply $$ \mathbb{Q}[\sqrt{2}+\sqrt{3}+\dots+\sqrt{n}+\sqrt{n+1}]=\mathbb{Q}[\sqrt{2},\sqrt{3},\dots,\sqrt{n},\sqrt{n+1}]$$
Any help is appreciated..
Thank you!
The solution to this is straightforward, but in general in any sort of such problems which have to do with simple extensions, you can use the Primitive element theorem and especially its proof. What the theorem tells you is that any finite extension $\mathbb Q(a,b)$ can be made to be simple and be written as $\mathbb Q(c)$. This is well known. What is perhaps less familiar and which you can see from the proof in the link is what that value c is. To be more precise, c can take the form $c=a+\lambda b$ for almost all $\lambda$. Looking even more carefully you can point out exactly for which $\lambda$ this is not applicable. To put it simply in most of the cases (not all!) you have that $\mathbb Q(a,b)=\mathbb Q(a+b)$.
Returning to your example you have by induction that $ K=\mathbb Q(\sqrt{2},....,\sqrt{n},\sqrt{n+1})=\mathbb Q(\sqrt{2}+...+\sqrt{n},\sqrt{n+1})$. If you name $a=\sqrt{2}+....+\sqrt{n}$ and $b=\sqrt{n+1}$ you have that $K=\mathbb Q(a,b)$ If you spend some time to understand the proof you will realize whether $\lambda=1$ is applicable to your example or not in your case.