It's multivariate (sorry I didn't specify), so being irreducible and having no non-trivial idempotents aren't equivalent.
On Wednesday, April 18, 2012 4:04:04 PM UTC-4, David Loeffler wrote: > > On 18 April 2012 15:50, diophan <[email protected]> wrote: > > I'll check out the reference. If it makes the situation any better the > ring > > is a quotient of a polynomial ring over a finite field. > > If it's a quotient of a one-variable polynomial ring, you can just > factor the defining polynomial, can't you? The quotient will have > non-trivial idempotents iff the polynomial has more than one > irreducible factor. > > David > > -- To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
