#5477: Make R.quotient_ring(I) normalize generator in the univariate case (easy
to
fix!)
---------------------------------+------------------------------------------
Reporter: was | Owner: malb
Type: defect | Status: new
Priority: major | Milestone: sage-3.4
Component: commutative algebra | Keywords:
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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
}}}
--
Ticket URL: <http://trac.sagemath.org/sage_trac/ticket/5477>
Sage <http://sagemath.org/>
Sage - Open Source Mathematical Software: Building the Car Instead of
Reinventing the Wheel
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