It seems that unfortunately the problem persists for multivariate rings as
well:
A.<t,u> = QQ[]
B.<x,y,z> = QQ[]
H = B.quotient(B.ideal([B.2]))
f = A.hom([H.0, H.1], H)
f
f.kernel()
Ring morphism:
From: Multivariate Polynomial Ring in t, u over Rational Field
To: Quotient of Multivariate Polynomial Ring in x, y, z over Rational
Field by the ideal (z)
Defn: t |--> xbar
u |--> ybar
Ideal (-t, -u, 0) of Multivariate Polynomial Ring in t, u over Rational
Field
I have the impression that the fact that the ring homomorphism is to a
quotient ring introduces the error, but that's just a wild guess.
On Monday, February 8, 2021 at 11:09:52 AM UTC+1 [email protected] wrote:
> A wild guess would be that it's due to univariate and multivariate
> rings handled by different backends in Sage, one sees this kinds of
> corner cases errors.
>
> On Mon, Feb 8, 2021 at 10:06 AM John Cremona <[email protected]> wrote:
> >
> > It looks like a bug to me. f.kernel() expands to
> > f._inverse_image_ideal(f.codomain().zero_ideal()) and
> > f.codomain().zero_ideal() looks OK so the problem must be in the
> > inverse image. The author is apparently Simon King (2011). Simon,
> > can you help?
> >
> > John
> >
> > On Mon, 8 Feb 2021 at 09:20, Akos M <[email protected]> wrote:
> > >
> > > Hi,
> > >
> > > I'm not sure whether this is a bug or not, but the kernel of a ring
> homomorphism to a quotient ring gives unexpected results:
> > >
> > > A.<t> = QQ[]
> > > B.<x,y> = QQ[]
> > > H = B.quotient(B.ideal([B.1]))
> > > f = A.hom([H.0], H)
> > > f
> > > f.kernel()
> > >
> > > outputs:
> > >
> > > Ring morphism: From: Univariate Polynomial Ring in t over Rational
> Field
> > > To: Quotient of Multivariate Polynomial Ring in x, y over Rational
> Field by the ideal (y) Defn: t |--> xbar
> > > Principal ideal (t) of Univariate Polynomial Ring in t over Rational
> Field
> > >
> > > whereas the kernel of f:A[t]->B[x,y]->B[x,y]/(y), for f(t)=x should be
> (0).
> > >
> > > Is this a bug?
> > >
> > > Thanks,
> > > Akos
> > >
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