Re: Kronecker symbol

["Justin C. Walker" <justin@mac.com>]

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 the
> documentation as a generalization of the Legendre symbol (x|y).  I was
> expecting kronecker to behave like a Dirichlet character.  In 
> particular, I
> expected 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 in
> Pari?
>
> 2a) What about higher order Dirichlet characters?
>
> Best Regards,
> 		Jack Fearnley
--
Justin C. Walker, Curmudgeon-At-Large  *
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