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