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? 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?  
 A: I think it's not so trivial to show that $\sqrt{p_n}$ is not already in the field $\mathbb Q(\sqrt{p_1},\ldots,\sqrt{p_{n-1}})$. Bare-hands elementary algebra and unique factorization in $\mathbb Z$ do suffice for $n=1,2,3$, and perhaps a little beyond, but things quickly turn ugly. I do not know what the context is at that point in Isaacs' text, but I can offer two reasonably-structured approaches. The first is easy to explain in words, but is uses some algebraic number theory, which I suspect is not what is intended: in the corresponding rings of algebraic integers, adjoining $\sqrt{p_n}$ introduces ramification at $p_n$. Ignoring the prime $2$, the earlier adjoinings did not introduce ramification at $p_n$... done. The other approach first observes that $\sqrt{p_n}$ is a value of a quadratic Gauss sum, which exhibits that square root as lying in the cyclotomic field obtained by adjoining a $p_n$-th or $4p_n$-th root of unity. The "independence" comes from the "fact" that the degree of the $N$th cyclotomic field over $\mathbb Q$ is Euler totient-function of $N$, and that we have an explicit description of the Galois groups, if needed. Thus, there is a Galois automorphism fixing all those square roots but one, etc. The usual proofs about independence of cyclotomic fields are perhaps not so transparent... and I confess that the only proof I easily remember uses Dirichlet's theorem on primes in arithmetic progressions to obtain a prime $p$ so that all but $\mathbb Q(\zeta_{p_n})$ collapse to give trivial extensions of $\mathbb Q_p$. 
EDIT: in response to comment/question... my in-quotes use of "independence" is meant to refer, albeit imprecisely, to the general problem of showing that a given polynomial in $\mathbb Q[x]$ has no root in some field extension of $\mathbb Q$. The exercise as posed asks about "independence" of square roots of primes: $\sqrt{p_n}$ ought not lie in the field extension of $\mathbb Q$ obtained by adjoining square roots of other primes. Similarly, the Gauss-sum argument still does depend on the "independence" of roots of unity of relatively-prime orders. The ramification argument from algebraic number theory is a different device to prove such things.
LATER EDIT: As in Geoff Robinson's answer, there are more elementary arguments! A day in which one learns something so basic is a good day!
A: OK, in view of what is written above, I will write down what I had in mind more carefully. Let us take the positive choice of $\sqrt{p_i}$ in each case, so clearly the sum (and each non-empty subsum) of the square roots is non-zero. Let $E$ be as stated in the question, the Galois group of the left-hand extension. The extension is certainly a Galois extension of $\mathbb{Q}.$ Let $\alpha$ be an element of $E$. Then $\alpha(\sqrt{p_i}) = \pm \sqrt{p_i}$ for each $i.$ Let $I$ be the subset of $\{i: 1 \leq i \leq n \}$ such that $\alpha$ fixes $\sqrt{p_i}$ for each $i \in I$, but no other $i.$
Let $s = \sum_{i=1}^{n} \sqrt{p_i}$. Then $s- \alpha(s) = 2\sum_{i \not \in I} \sqrt{p_i},$ which is strictly positive unless $\alpha$ is trivial. Since $E$ is elementary Abelian (I do not think we need to know that it has order $2^n$), we know that every intermediate extension is a Galois extension. Hence $s$ lies in no intermediate extension, and must generate the whole of the left-hand extension. 
A: $\newcommand{\Q}{\mathbf Q}$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.
