TITLE: How to prove that $\mathbb{Q}[\sqrt{p_1}, \sqrt{p_2}, \ldots,\sqrt{p_n} ] = \mathbb{Q}[\sqrt{p_1}+ \sqrt{p_2}+\cdots + \sqrt{p_n}]$, for $p_i$ prime?
QUESTION [38 upvotes]: This is Exercise 18.14 from Algebra, Isaacs.
$p_{1}\ ,\ p_{2}\ ,\ ... p_{n}$ are different prime numbers. How to show that $$\mathbb{Q}[\sqrt{p_{1}}, \sqrt{p_{2}}, \ldots, \sqrt{p_{n}} ] = \mathbb{Q}[\sqrt{p_{1}}+ \sqrt{p_{2}}+\cdots + \sqrt{p_{n}}] \quad ?$$  First we can note that the Galois group of the first extension over $\mathbb{Q}$ ( which I call $E$ ) is elementary abelian of order $2^{n}$, so we can prove that the orbit of $\sqrt{p_{1}}+\sqrt{p_{2}}+ ... + \sqrt{p_{n}}$ under $\operatorname{Gal}(E/\mathbb{Q})$ contains $2^{n}$ elements, but how to do so?

REPLY [11 votes]: $\newcommand{\Q}{\mathbf Q}$This is a simplification of Geoff Robinson's answer.
For each $i$ write $a_i=\sqrt{p_i}$. Also write $K=\Q(a_1, \ldots, a_n)$. Define $s=a_1+\cdots+a_n$. As Geoff has shown, the only element of $G:=\text{Aut}(K:\Q)$ which fixes $s$ is the identity.
Therefore $\text{Aut}(K:\Q(s))$ is a singleton. But since $K:\Q$ is Galois, so is $K:\Q(s)$, and therefore $[K:\Q(s)]=|\text{Aut}(K:\Q(s))|$ showing that $[K:\Q(s)]=1$ and hence $\Q(s)=K$.
Note that we do not require to know that $\text{Aut}(K:\Q)$ is abelian.
A neat proof of a much more general result has here been given by @orangeskid here.