> > For the record, my ring is currently S[t], where S is the ring of
> > symmetric functions in the e basis. You probably are not that
> > surprised about that :-)
>
> Is it easy to factor such polynomials? How do you do it?
Well, it's a free algebra, so you could always coerce to
QQ['e1,e2,...,], factor, and back (or better, the factorization
implementation should be generic to apply on any free algebra). But in
the short run, I actually don't care about factoring the small
denominators I start with.
Cheers,
Nicolas
--
Nicolas M. Thiéry "Isil" <nthi...@users.sf.net>
http://Nicolas.Thiery.name/
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