On Wed, 30 Apr 2003 06:51:52 -0500, David C. Ullrich <email@example.com> wrote:
>Are the square roots of the positive square-free integers >(including 1) linearly independent over Q? > >Hmm, in case this looks like homework I should say how >far I got: Say we have a dependence among sqrt(n_0), >... sqrt(n_N) (say n_0 = 1 to make the notation nice >below). To make things look nice, we can assume by >adding more n_j that in fact n_0, ... n_N is a list of >all the square-free products of a certain finite set of >primes. Now we have > >(*) a_0 sqrt(n_0) + ... + a_N sqrt(n_N) = 0 > >for integers a_0, ... a_N, which we can take to have >no common factor.
Oh: Say n_j "appears" here if a_j <> 0. There must be a prime p such that p is a factor of at least one of the n_j which appears but not all of them (if not then we could divide the whole thing by sqrt(p) and get a counterexample with fewer primes).
Put all the terms where p appears on the left side and the other terms on the right side. Square both sides. After squaring both sides, p will not be a factor of any of the n_j which appear on either side. So after squaring, by induction, the coefficients on both sides must be the same. This is impossible because ___(???)
(After squaring, all the coefficients on the left side are divisible by p. It's impossible for all the coefficients on the right to be divisible by p because ___???)
>By our hypothesis on the n_i's, >if we repeatedly square (*) we get other relations >that look the same (it's not clear to me whether or >not the coefficients in the new relations can have >common factors.) So we have Av = 0, where >v is the (column) vector consisting of the sqrt(n_i) >and A has as many rows as we want. _If_ we >could show that A eventually has rank N+1 then >we could solve for the sqrt's, showing that they're >rational. Does the rank of A have to increase >when we add more rows by repeated squaring >of (*)? (Or taking other powers of (*), come >to think of it...) > >****************** > >David C. Ullrich