Linear Independence of Square Roots of PrimesDate: 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/ |
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