Vincent Delecroix wrote:
> Does polynomial over non-commutative ring make sense ? Because in that
> context axbx is not abx^2. Depending on what you call polynomial, they
> may or may not form a ring.
Just to clarify, I'm asking about polynomials over a non-commutative
rings but with an indeterminate that commutes with the coefficients. And
yes, this makes sense--you only have to be careful that p \mapsto p(a)
is not necessarily a ring homomorphism.
("Noncommutative polynomials", where the coefficients do not commute
with the indeterminate make sense, see for instance
sage.rings.polynomial.plural in Sage.)
--
Marc
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