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