Newsgroups: sci.math From: rusin@mp.cs.niu.edu (David Rusin) Subject: Re: A tarnished Gaussian sum? Date: Thu, 10 Sep 1992 19:36:34 GMT In article <1992Sep10.151807.25693@tdb.uu.se> matsa@tdb.uu.se (Mats Andersson) writes: > >Let p be a prime and w a pth root of unity. Does there always exist a k >such that > > (1-w)(1-w^2)...(1-w^((p-1)/2)) = w^k * Gaussian sum ? Let Z be the left side. Then if Z* is its complex conjugate, Z*=(1-w^(-1))...(1-w^((p+1)/2))=(1-w^((p+1)/2)...(1-w^(p-1)), so Z.Z* is the product of all conjugates of (1-w); that is, you let x=1 in the polynomial \Prod (x-w^j) = 1+x+...+x^(p-1): Z.Z*=p. Thus certainly Z is \sqrt(p) times a complex number of norm 1. On the other hand, (1-w^(2j)) = w^j . (w^(-j)-w^j) and similarly (1-w^(2j+1))=(w^p-w^(2j+1))=w^p.(1-w^(2j+1-p)) can be factored in the form w^something.(w^a-w^(-a)). So if you collect all the factors in Z you get a big power of w times a product of terms (w^a-w^(-a)), all of which are purely imaginary. So Z is of the form (power of w ) . (i ^ (p-1)/2)) . real. So up to powers of w this product Z is just i^((p-1)/2) . sqrt(p). dave rusin@math.niu.edu PS - yes, I think there is a way to decide on the signs in the square root but I forget what it is. ============================================================================== To: rusin@mp.cs.niu.edu (David Rusin) Subject: Re: A tarnished Gaussian sum? Date: Mon, 14 Sep 92 16:16:47 EDT From: "Victor S. Miller" >>>>> On Thu, 10 Sep 1992 19:36:34 GMT, rusin@mp.cs.niu.edu (David Rusin) said: David> In article <1992Sep10.151807.25693@tdb.uu.se> matsa@tdb.uu.se (Mats Andersson) writes: > >Let p be a prime and w a pth root of unity. Does there always exist a k >such that > > (1-w)(1-w^2)...(1-w^((p-1)/2)) = w^k * Gaussian sum ? David> Let Z be the left side. Then if Z* is its complex conjugate, David> Z*=(1-w^(-1))...(1-w^((p+1)/2))=(1-w^((p+1)/2)...(1-w^(p-1)), David> so Z.Z* is the product of all conjugates of (1-w); that is, you let David> x=1 in the polynomial \Prod (x-w^j) = 1+x+...+x^(p-1): Z.Z*=p. David> Thus certainly Z is \sqrt(p) times a complex number of norm 1. David> On the other hand, (1-w^(2j)) = w^j . (w^(-j)-w^j) and similarly David> (1-w^(2j+1))=(w^p-w^(2j+1))=w^p.(1-w^(2j+1-p)) can be factored in the David> form w^something.(w^a-w^(-a)). So if you collect all the factors in David> Z you get a big power of w times a product of terms (w^a-w^(-a)), David> all of which are purely imaginary. So Z is of the form David> (power of w ) . (i ^ (p-1)/2)) . real. David> So up to powers of w this product Z is just i^((p-1)/2) . sqrt(p). David> dave rusin@math.niu.edu David> PS - yes, I think there is a way to decide on the signs in the square root David> but I forget what it is. David> David, the sign in this case is easy (a lot easier than determining the sign of the quadratic Gauss sum). As you observed (1-w^k) = -2 i w^{k/2} (w^{k/2} - w^{-k/2}) = -2 i w^{k/2} sin( \pi k /p). Since all k's in the original sum are between 0 and p/2, we have 0< k/p < 1/2, so that all the sin's are positive. Victor ============================================================================== To: matsa@tdb.uu.se (Mats Andersson) Cc: rusin@mp.cs.niu.edu (David Rusin) Subject: Re: A tarnished Gaussian sum? Date: Mon, 14 Sep 92 16:33:16 EDT From: "Victor S. Miller" Your product is one of the ways of determining the sign of the quadratic Gauss Sum. Look at Ireland and Rosen "A Classical Introduction to Modern Number Theory", Springer Graduate Texts in Math #84, pages 70-76. Victor