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Linear Independence of Square Roots of Primes


Date: 11/07/96 at 07:17:49
From: Mats Oldin
Subject: Linear independence of square roots of primes

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.

Best regards,

Mats Oldin
Sweden


Date: 11/11/96 at 14:37:15
From: Doctor Rob
Subject: Re: Linear independence of square roots of primes

Actually, something more general is true.  You need only assume that 
the integers you are taking the square roots of are >1, squarefree and 
pairwise relatively prime.  One proof goes as follows.

We will prove that as a vector space, the dimension of the field 
extension gotten from Q, the field of rational numbers, by adjoining 
the square roots of s integers > 1 which are squarefree and pairwise 
relatively prime, is 2^s.  Proceed by induction on the number of such 
integers adjoined.

We know it works for s = 1 and s = 2, as you state in your question.  
Assume it works for 1, 2, ..., s roots.  Let n(1), n(2), ... n(s), 
n(s+1) be such a set of s+1 integers.  Let us use the letters E and F 
to represent the following fields:

  E = Q(sqrt[n(1)], ..., sqrt[n(s-1)])    
  F = E(sqrt[n(s)])

By induction, dim(E) = 2^(s-1) and dim(F) = 2^s.  We would be done 
with the proof if we could show that sqrt[n(s+1)] is not an element 
of F.

Assume otherwise, that sqrt[n(s+1)] is in F.  Then for some a, b in E, 
we can write:

  sqrt[n(s+1)] = a + b*sqrt[n(s)]

Squaring both sides gives:

  n(s+1) = a^2 + 2*a*b*sqrt[n(s)] + b^2*n(s)
or
  2*a*b*sqrt[n(s)] = n(s+1) - a^2 - b^2*n(s)

The righthand side lies in E.  Three cases are possible:

  1) a = 0.
  2) b = 0.
  3) sqrt[n(s)] is in E.

Case 1:  a = 0.  Then sqrt[n(s+1)] = b*sqrt[n(s)].  This implies that
sqrt[n(s)*n(s+1)] = b*n(s) lies in E.  Then the set of s integers 
n(1), n(2), ..., n(s-1), n(s)*n(s+1) satisfies the induction 
hypothesis, so the dimension of this field extension must be 2^s.  
On the other hand, this field extension must be exactly E, whose 
dimension is 2^(s-1), a clear contradiction.  Thus this case is 
impossible.

Case 2:  b = 0.  Then sqrt[n(s+1)] = a lies in E, and the set of 
s integers n(1), n(2), ..., n(s-1), n(s+1) satisfies the induction 
hypothesis, so the dimension of this field extension must be 2^s.  
On the other hand, this field extension must be exactly E, whose 
dimension is 2^(s-1), again a contradiction.  Thus this case is 
impossible.

Case 3:  sqrt[n(s)] is in E.  Then F = E, which is also a 
contradiction, since the dimension of F is twice that of E.  
Thus this case, too, is impossible.

The conclusion is that sqrt[n(s+1)] cannot lie in F, so we are done.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Modern Algebra

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