FreeAlgebra doesn't reduce using ideal relations

jacksonwalters commented 3 weeks ago

Steps To Reproduce

For the free algebra, the quotient by an ideal should result in relations in the quotient ring.

Expected Behavior

For a polynomial ring, we get the expected relation.

R.<x,y> = PolynomialRing(QQ, 2)
S.<a,b> = R.quotient(x^2 + y^2)
S.gens()
(a, b)
a^2+b^2 == 0
True

Actual Behavior

For a free algebra, the variables in the quotient don't satisfy any relations.

R.<x,y> = FreeAlgebra(QQ, 2)
S.<a,b> = R.quotient(x^2 + y^2)
S.gens()
(a, b)
a^2+b^2 == 0
False

Additional Information

No response

Environment

- **OS**: Mac M1 2020, Sonoma 14.4.1
- **Sage Version**: 10.3

Checklist

[X] I have searched the existing issues for a bug report that matches the one I want to file, without success.
[X] I have read the documentation and troubleshoot guide

tscrim commented 2 weeks ago

The free algebra would need to compute a Gröbner basis, which is generally infinite (unlike for polynomial rings!). So the free algebra quotients (well, really generic quotients) do the safe thing, and only consider the representation given as there is no canonical form. You're basically wanting a solution to an undecidable problem, and ad-hoc methods aren't such a good way to implement things.

jacksonwalters commented 2 weeks ago

Ah, I see. That makes sense. I have some more questions re: quotients of the symmetric group algebra explicitly I’ll save for another place.

sagemath / sage