Ticket #5477 (closed defect: fixed)
Make R.quotient_ring(I) normalize generator in the univariate case (easy to fix!)
|Reported by:||was||Owned by:||malb|
|Authors:||Alex Ghitza||Merged in:||sage-4.3.2.alpha0|
From a Sage Days 14 user (see below).
In short, in the univariate over-a-field case, R.quotient_ring(I) should normalize the generator of the ideal before forming the quotient.
In each case below "I" and "J" are defined by different choices of generators and are recognized as the same ideal. In case 1 the quotients are considered equal and in case 2 they are considered unequal. (I checked this with the latest version) Case 1: ---------- sage: R.<x> = PolynomialRing(QQ) sage: I = R.ideal([x + x^2, x]) sage: J = R.ideal([2*x + 2*x^2, x]) sage: S = R.quotient_ring(I) sage: U = R.quotient_ring(J) sage: I==J True sage: S==U True Case 2: ---------- sage: R.<x> = PolynomialRing(QQ) sage: I = R.ideal([x + x^2]) sage: J = R.ideal([2*x + 2*x^2]) sage: S = R.quotient_ring(I) sage: U = R.quotient_ring(J) sage: I==J True sage: S==U False
- Status changed from new to needs_review
- Report Upstream set to N/A
- Authors set to Alex Ghitza