From: Robin Chapman
Subject: Re: Quadratic units problem
Date: Fri, 03 Dec 1999 10:04:30 GMT
Newsgroups: sci.math
Keywords: identification of units in a real quadratic number field
In article <3aA14.31$6c6.6744@den-news1.rmi.net>,
Kurt Foster wrote:
> I'm sure I've seen this - somewhere - but I can't remember where, and
> can't seem to club it into submission easily. It's a matter of showing
> that two different expressions, each giving a unit in the same real
> quadratic field, both give the SAME unit.
>
> I know that, if m is odd and squarefree, and m == 3 (mod 4) then the
> discriminant D of R = Z(sqrt(m)) is D = 4m. Let X() be the primitive
> quadratic character (mod D). Then I know that, if u > 1 is the
> fundamental unit of R, and
>
> U = [PRODUCT, sin(pi*b/D)]/[PRODUCT, sin(pi*a/D)]
>
> the products being over integers 1 =< a, b < D/2, X(a) = +1, X(b) = -1,
> then
>
> U = u^h
>
> where h is the class number of R.
>
> But I *also* know that
>
> U' = PRODUCT, (1 - 2*pi*i*b/D); 1 =< b < D, X(b) = -1
You probably want 1 - exp(2 pi i b/D) here?
> is a unit of R; this is VERY easy to prove. But it doesn't seem so easy
> to tell when U' = U. It does when m = 3 and m = 7. Does U' = U always?
> I wind up having to decide whether an expression like
>
> PRODUCT, 2*sin(pi*2*a/D), 1 =< a < D/2, X(a) = +1
>
> is equal to 1 (or perhaps -1, I was too lazy to bother with that detail).
All the terms in the product are positive, so it can't equal -1 :-)
> Anyone have a reference that treats this? Thanks.
I don't know about a reference, but let P denote the product. Then
sin(2 pi a/D) = sin(2 pi (2m - a)/D) and X(2m - a) = X(a). Thus
P^2 is the product of all 2 sin(2pi a/D) over a between 0 and D/2.
coprime to D. If we extend the range to between 0 and D we get P^4. Now
for any N it's not too hard to compute the product of sin(2 pi j/N)
over all 0 < j < N. Then one can use inclusion/exclusion to get at
P^4.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'"
Greg Egan, _Teranesia_
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