Justin C. Walker on Tue, 26 Aug 2003 14:10:34 -0700


[Date Prev] [Date Next] [Thread Prev] [Thread Next] [Date Index] [Thread Index]

Re: Kronecker symbol



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

--
/~\ The ASCII           Justin C. Walker, Curmudgeon-at-Large
\ / Ribbon Campaign
 X  Help cure HTML Email
/ \