Re: Kronecker symbol

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

On Tuesday, August 26, 2003, at 03:33 AM, Olivier Ramare wrote:

> > 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.
>
> Of course, the kernel of this symbol is a subgroup of order 2

Are you sure about this?  I think the kernel has *index* 2, but not 
order 2.

> while the subgroup of squares is of order 2^{omega(q)}
> (false if 2|q but but ...).

In the case of (Z/(15))^*, {1,4} is the group of squares.  That doesn't 
seem to match 2^omega(q); am I missing something?

Regards,

Justin

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