From a problem in finite field to a problem in number theory When I was studying about "Finite field", I had a excercise require find $[\mathbb{Q}(\sqrt{3}, \sqrt{5}, \sqrt{11}):\mathbb{Q}]$, and I need to prove that $\sqrt{11}\notin \mathbb{Q}(\sqrt{3}, \sqrt{5})$ to complete.
I had tried to prove it, but it is quite complex. And I think I will prove a general problem
Fix $p_1, p_2,...,p_n$ are prime numbers. Suppose there are $a_1, a_2,...,a_n\in \mathbb{Q}$ such as: $a_1\sqrt{p_1}+a_2\sqrt{p_2}+...+a_n\sqrt{p_n}=0$ if and only if $a_1=a_2=...=a_n=0$ (*)
I think that this is a familiar result, but I don't solve it. I would like to receive some feedback!
 A: $\newcommand{\Q}{\mathbb Q}$
We show that

Let $p_1,\ldots,p_{r}$ be distinct primes.
  Let $\alpha_i=\sqrt{p_i}$.
  Then
  $$\mathbb [\Q(\alpha_1,\ldots,\alpha_{r}):\Q]=2^{r}$$

We prove a stronger statement using induction: Let $b_1,\ldots,b_{r}$ be distinct, positive, pairwise relatively prime, and square-free integers.
Define $\beta_i=\sqrt{b_i}$.
Then $[\Q(\beta_1,\ldots,\beta_{r}):\Q]=2^{r}$.
We prove by induction on $r$.
The base case $r=1$ clearly holds.
Let the theorem be true for some $r\geq 1$ and all smaller values.
We prove for $r+1$.
Define $E=\Q(\beta_1,\ldots,\beta_{r-1})$ and $F=E(\beta_r)$.
By induction we know that $[E:\Q]=2^{r-1}$ and $[F:E]=2$.
We just need to prove that $[F(\beta_{r+1}):F]=2$.
Suppose not.
So $\beta_{r+1}\in F$ and we can write $\beta_{r+1}=x+y\beta_r$ for some $x,y\in E$.
Squaring we get $b_{r+1}-(x^2+y^2b_r)=2xy\beta_r$.
The LHS of above is in $E$.
But for the RHS to be in $E$ one or both of $x$ or $y$ has to be $0$.
Since $\beta_{r+1}=x+y\beta_r$, by induction, we cannot have $y=0$.
Thus $x=0$.
Now $\beta_{r+1}=y\beta_r$.
Therefore
$$\beta_{r+1}\beta_r\in E \tag{1}$$
Note that since $\beta_{r}$ and $\beta_{r+1}$ are pairwise relatively prime and square-free, $\beta_{r+1}\beta_r$ is also square-free.
Write $\beta=\beta_{r+1}\beta_r$.
It is not hard to see that $\beta_1,\ldots,\beta_{r-1},\beta$ are $r$ integers satisfying the induction hypothesis.
Thus 
$$
[\Q(\beta_1,\ldots,\beta_{r-1},\beta):\Q]=2^r=[E(\beta):E][E:\Q]=2^{r-1}[E(\beta):E]
$$
But this gives $[E(\beta):E]=2$ which means that $\beta\not\in E$.
Therefore we arrive at a contradiction and the induction is complete.
Now using the proven statement, we can show that if
$$a_1\sqrt{p_1}+ \cdots +a_n\sqrt{p_n}=0$$
then each $a_i=0$.
For if $a_n\neq 0$, then we would have $\sqrt{p_n}\in \Q(\sqrt{p_1}, \ldots, \sqrt{p_{n-1}})$, which is a contradiction.
A: Let $L=\mathbb{Q}(\sqrt{3},\sqrt{5})=\mathbb{Q}(\sqrt{3}+\sqrt{5})$. 
Then $L/\mathbb{Q}$ is Galois (the Galois conjugates of $\alpha=\sqrt{3}+\sqrt{5}$ are $\pm\alpha, \pm 2/\alpha$) and we know that $[L:\mathbb{Q}]=deg P=4$ where $P$ is the minimal polynomial of $\sqrt{3}+\sqrt{5}$.
(Check: $Gal(L/\mathbb{Q})=<\phi>\times<\psi>\cong C_2\times C_2$ where $\phi:\sqrt{5}\to-\sqrt{5}, \sqrt{3}\to\sqrt{3}$ and $\phi:\sqrt{5}\to\sqrt{5}, \sqrt{3}\to-\sqrt{3}$)
So, there are three intermediate fields of $L/\mathbb{Q}$.$F_1=\mathbb{Q}(\sqrt{3})=L^{<\phi>}$, $F_2=(\mathbb{Q}(\sqrt{5})=L^{<\psi>}$
 and $F_3=\mathbb{Q}(\sqrt{15})=L^{<\phi \psi>}$
If $\sqrt{11}\in L$ then, we have $ \sqrt{11}\in F_i$ for some $i$ since $\mathbb{Q}(\sqrt{11})/\mathbb{Q}$ is a degree 2 extension. 
Then you need to show that $a\sqrt{11}+b\sqrt{m}=0$ has no rational solutions where $m=3,5, 15$. And this I hope you agree that this is elementary number theory.
