TITLE: Linear independence of roots over Q
QUESTION [5 upvotes]: Let $p_1,\ldots,p_k$ be $k$ distinct primes (in $\mathbb{N}$) and $n>1$. Is it true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$? (all the roots are in $\mathbb{R}^+$) 
Iurie Boreico proved here that a linear combination $\sum q_i\sqrt[n]{a_i}$ with positive rational coefficients $q_i$ (and no $\sqrt[n]{a_i}\in\mathbb{Q}$) can't be rational, but this question seems to be more difficult..

REPLY [10 votes]: Below are links to classical proofs. Nowadays such results are usually derived as special cases of results in Kummer Galois theory. See my post here for a very simple proof of the quadratic case.

Besicovitch, A. S. $\ $ On the linear independence of fractional powers of integers.
J. London Math. Soc. 15 (1940). 3-6.  MR 2,33f 10.0X
Let  $\ a_i = b_i\ p_i,\ i=1,\ldots s\:,\:$ where the  $p_i$  are  $s$  different primes and 
the  $b_i$  positive integers not divisible by any of them. The author proves 
by an inductive argument that, if  $x_j$  are positive real roots of 
 $x^{n_j} - a_j = 0,\  j=1,...,s ,$ and  $P(x_1,...,x_s)$  is a polynomial with 
rational coefficients and of degree not greater than  $n_j - 1$  with respect 
to  $x_j,$ then  $P(x_1,...,x_s)$  can vanish only if all its coefficients vanish. $\quad$ Reviewed by W. Feller.   

Mordell, L. J. $\ $ On the linear independence of algebraic numbers.
Pacific J. Math. 3 (1953). 625-630.   MR 15,404e  10.0X   
Let $K$ be an algebraic number field and  $x_1,\ldots,x_s$  roots of the equations 
$\ x_i^{n_i} = a_i\ (i=1,2,...,s)$ and suppose that (1) $K$ and all $x_i$ are real, or 
(2) $K$ includes all the $n_i$ th roots of unity, i.e. $ K(x_i)$ is a Kummer field. 
The following theorem is proved. A polynomial $P(x_1,...,x_s)$ with coefficients 
in $K$ and of degrees in $x_i$ , less than $n_i$ for $i=1,2,\ldots s$ , 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  $\ x_1^{m_1}\ x_2^{m_2}\:\cdots\: x_s^{m_s} = a$, where $a$ is a number in $K$ unless  $\ m_i = 0\ mod\ n_i\ (i=1,2,...,s)$ . When $K$ is of the second type, the theorem was proved earlier by Hasse [Klassenkorpertheorie, 
Marburg, 1933, pp. 187--195] by help of Galois groups. When K is of the first 
type and K also the rational number field and the $a_i$ integers, the theorem was proved by Besicovitch in an elementary way. The author here uses a proof analogous to that used by Besicovitch [J. London Math. Soc. 15b, 3--6 (1940) these Rev. 2, 33].  $\quad$  Reviewed by H. Bergstrom.

Siegel, Carl Ludwig. $\ $ Algebraische Abhaengigkeit von Wurzeln.
Acta Arith. 21 (1972), 59-64.   MR 46 #1760  12A99  
Two nonzero real numbers are said to be equivalent with respect to a real 
field  $R$  if their ratio belongs to  $R$ . Each real number  $r \ne 0$  determines 
a class  $[r]$  under this equivalence relation, and these classes form a 
multiplicative abelian group  $G$  with identity element $[1]$. If  $r_1,\cdots,r_h$ 
are nonzero real numbers such that $r_i^{n_i}\in R$ for some positive integers $n_i\  
(i=1,...,h)$ , denote by $G(r_1,...,r_h) = G_h$ the subgroup of $G$ generated by 
[r_1],...,[r_h] and by R(r_1,...,r_h) = R_h the algebraic extension field of 
$R = R_0$ obtained by the adjunction of $r_1,...,r_h$ . The central problem 
considered in this paper is to determine the degree and find a basis of $R_h$ 
over $R$ . Special cases of this problem have been considered earlier by A. S. 
Besicovitch [J. London Math. Soc. 15 (1940), 3-6; MR 2, 33] and by L. J. 
Mordell [Pacific J. Math. 3 (1953), 625-630; MR 15, 404]. The principal 
result of this paper is the following theorem: the degree of $R_h$ with respect 
to $R_{h-1}$ is equal to the index $j$ of $G_{h-1}$ in $G_h$ , and the powers $r_i^t\ 
(t=0,1,...,j-1)$ form a basis of $R_h$ over $R_{h-1}$ . Several interesting 
applications and examples of this result are discussed. $\quad$ Reviewed by H. S. Butts

REPLY [6 votes]: Yes, it is true that $[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$ and it was proved by Besicovitch ( a student of A.A. Markov) in 1940.  
Although there are (almost) infinitely many books on field and Galois theory, the only book I know which  proves Besicovich's  theorem (but only for odd $n$) is Roman's Field Theory (Theorem 14.3.2, page 305 of the second edition).