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
