TITLE: The square roots of different primes are linearly independent over the field of rationals
QUESTION [153 upvotes]: I need to find a way of proving that the square roots of a finite set 
  of different primes are linearly independent over the field of 
  rationals. 

I've tried to solve the problem using elementary algebra 
and also using the theory of field extensions, without success. To 
prove linear independence of two primes is easy but then my problems 
arise. I would be very thankful for an answer to this question.

REPLY [36 votes]: Assume that there was some linear dependence relation of the form
$$ \sum_{k=1}^n c_k \sqrt{p_k} + c_0 = 0 $$
where $ c_k \in \mathbb{Q} $ and the $ p_k $ are distinct prime numbers. Let $ L $ be the smallest extension of $ \mathbb{Q} $ containing all of the $ \sqrt{p_k} $. We argue using the field trace $ T =  T_{L/\mathbb{Q}} $. First, note that if $ d \in \mathbb{N} $ is not a perfect square, we have that $ T(\sqrt{d}) = 0 $. This is because $ L/\mathbb{Q} $ is Galois, and $ \sqrt{d} $ cannot be a fixed point of the action of the Galois group as it is not rational. This means that half of the Galois group maps it to its other conjugate $ -\sqrt{d} $, and therefore the sum of all conjugates cancel out. Furthermore, note that we have $ T(q) = 0 $ iff $ q = 0 $ for rational $ q $.
Taking traces on both sides we immediately find that $ c_0 = 0 $. Let $ 1 \leq j \leq n $ and multiply both sides by $ \sqrt{p_j} $ to get
$$ c_j p_j + \sum_{1 \leq k \leq n, k\neq j} c_k \sqrt{p_k p_j} = 0$$
Now, taking traces annihilates the second term entirely and we are left with $ T(c_j p_j) = 0 $, which implies $ c_j = 0 $. Since $ j $ was arbitrary, we conclude that all coefficients are zero, proving linear independence.