Justin C. Walker on Mon, 25 Aug 2003 20:11:06 -0700 |
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Re: Kronecker symbol |
I'm not sure that your assumptions are correct.Starting with Legendre, the symbol is defined for a, p, with p prime (and (a|p)=+/- 1); Jacobi extends this to (a|n) for general odd n, making it multiplicative in n. Kronecker then extends Jacobi to n=2. I'd think that you'd end up with something like (3|4)=(3|2)^2=1.
Once you extend Legendre, I think you are out of the "a is a square mod b" realm; for example, (-7|15)=1, but -7 (== 8) is not a square mod 15.
I can't answer the second question, and a check of the source offers no hints. FWIW, I believe that quadratic (== real?) characters all have the form (a|b) (due to Dirichlet, I think).
Regards, Justin On Monday, August 25, 2003, at 12:06 PM, Jack Fearnley wrote:
I have been having some trouble with kronecker(x,y) which is described in thedocumentation as a generalization of the Legendre symbol (x|y). I wasexpecting kronecker to behave like a Dirichlet character. In particular, Iexpected kronecker(3,4) to equal -1 whereas it computes as 1. I have two questions: 1) What is the precise definition of kronecker in Pari?2) Is there a quick and easy way to generate quadratic Dirichlet characters inPari? 2a) What about higher order Dirichlet characters? Best Regards, Jack Fearnley
-- Justin C. Walker, Curmudgeon-At-Large *Institute for General Semantics | It's not whether you win or lose...
| It's whether *I* win or lose. *--------------------------------------*-------------------------------*