# possible bug: kernel of ring homomorphism

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### Akos M

Feb 8, 2021, 4:20:34 AM2/8/21
to sage-devel
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.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

Feb 8, 2021, 5:06:27 AM2/8/21
to SAGE devel
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

Feb 8, 2021, 5:09:52 AM2/8/21
to sage-devel
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.

### Akos M

Feb 8, 2021, 5:42:59 AM2/8/21
to sage-devel
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

Feb 8, 2021, 9:54:42 PM2/8/21
to sage-devel