Another (in)dependence over the nonzero rationals question About one hour ago I asked a question which at first sight looked non-trivial to me but it is really trivial. Shame on me, whether I want it or not.
Now I have, solely for fun, another question which is slightly modified and maybe little less trivial.
Let $S_2$ be the set of square roots of all prime numbers, $S_3$ the set of cube roots of all prime numbers, ... , $S_n$ the set of $n$-th roots of all prime numbers, ...
Now let us define set $S$ as $S=\bigcup_{n=2}^\infty S_n$.
The question is:
Does there exist natural number $k$ and rational numbers $r_1,r_2,...,r_k$ (all different from zero) such that there are k numbers from the set $S$, let us denote them as $s_1,s_2,...,s_k$ such that we have
$\sum_{i=1}^{k}r_is_i=0$
(All roots in this question are unique real roots.)
 A: Oh well, I will answer the question just because it seems to me that it is better to have as many as possible questions on the site that are answered. I would like to thank Erick Wong who pointed me in the right direction with his comment.
In this paper Louis Joel Mordell proves much more general result in a theorem which he states in this way:

A polynomial $P(x_1,x_2,...,x_s)$ with coefficients in $K$ and of degrees in $x_1,x_2,...,x_s$ less than $n_1,n_2,...,n_s$, respectively, can vanish only if all its coefficients vanish provided that the algebraic number field $K$ is such that there exists no relation of the form $x_1^{m_1}x_2^{m_2}\cdot \cdot \cdot x_s^{m_s}=a$, where $a$ is a number in $K$, unless $m_1 \equiv 0 \pmod {n_1}, m_2 \equiv 0 \pmod {n_2},..., m_s \equiv 0 \pmod {n_s}$

A: This is not really an answer, only a hint,  but may be too long to fit as a comment.
Assume such a relation exists. 
Let us assume $m$th roots are the highest appearing in this relation (i.e. no $n$th roots with $n>m$ appear in this relation).
So this means we have primes numbers $p_1,p_2,\ldots, p_r$ and integers $a_1,a_2,\ldots,a_r$ such that  the algebraic number $\alpha$ defined by
$$\alpha=a_1\sqrt[m]{p_1}+a_2\sqrt[m]{p_2}+\cdots+a_r\sqrt[m]{p_r}$$
belongs to the field generated by lower radicals.
My guess is $\alpha$ is of degree $m^r$.
And this relation will lead to a conclusion that $m^r$ divides a number
factorizable as a product of numbers less than $m$. I am stuck there.
