From: John Rickard Subject: Re: The degree of this field Date: 14 Dec 2000 19:13:48 +0000 (GMT) Newsgroups: sci.math Summary: [missing] Charles Matthews wrote: : It is an old question, I believe of Besicovitch, asking you to prove : that square roots of integers are linearly independent over the : rationals if the obvious necessary condition is fulfilled. So this : can presumably be verified by the sufficiently patient. It is : certainly a consequence of standard algebraic number theory. I came up with this elementary proof a while ago; it basically just uses linear algebra and very elementary number theory. The statement is that the square roots of the square-free positive integers are linearly independent over Q, which follows from the following proposition: Define S_n (n >= 0) to be the set of square roots of products of subsets of the first n primes. (So S_n has 2^n elements, S_0 = {1}, S_1 = {1, sqrt(2)}, S_2 = {1, sqrt(2), sqrt(3), sqrt(6)}, ...) Define F_n to be the subset of R generated as a vector space over Q by the elements of S_n. Then: (i) The elements of S_n are linearly independent over Q. (ii) Every element of F_n whose square is rational is a rational multiple of an element of S_n. (iii) F_n is a field. I prove this proposition by induction over n. It is clearly true for n = 0 (S_0 = {1}, F_0 = Q). Suppose it is true for n = k. Let p be the (k+1)-st prime, r = sqrt(p). Then every element of F_(k+1) is of the form a + b r, where a and b are in F_k. If S_(k+1) is not linearly independent, then 0 can be expressed in this form with a and b linear combinations of elements of S_k, not both a and b being the trivial linear combination: since S_k is linearly independent, this implies that a and b are not both zero. Since a + b r = 0, it follows that neither a nor b is zero; we have r = -a/b (using the fact that F_k is a field), and p = (a/b)^2; this contradicts (ii) for n = k, using unique factorization in the integers and counting powers of p. Hence we have (i) for n = k + 1. Now (a + b r)^2 = a^2 + p b^2 + 2ab r; if this is rational, then since S_(k+1) is linearly independent, 2ab = 0, so either a = 0 or b = 0. If a = 0, then (a + b r)^2 = p b^2, so if this is rational then b^2 is rational, so b is a rational multiple of an element of S_k, so a + b r = b r is a rational multiple of an element of S_(k+1); similarly if b = 0 then a + b r is a rational multiple of an element of S_(k+1) (indeed, of an element of S_k). This shows (ii) for n = k + 1. F_(k+1) is clearly a ring; to show that it is a field, it suffices to show that every non-zero element is invertible. This is true because (a + b r)^(-1) = (a - b r)/(a^2 - p b^2); a^2 - p b^2 is non-zero because it equals (a + b r)(a - b r), so it is invertible because it is in F_k, which is a field. This shows (iii) for n = k + 1. -- John Rickard