possible bug: kernel of ring homomorphism

[Akos M]

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

[John Cremona]

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

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[Dima Pasechnik]

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.

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[Akos M]

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. 

[Samuel Lelievre]

For reference this is also asked on Ask Sage:

https://ask.sagemath.org/question/55618

[Markus Wageringel]

Thank you for reporting this problem. I have opened https://trac.sagemath.org/ticket/31367 for it and will provide a fix there shortly.