Thank you for reporting this problem. I have opened
https://trac.sagemath.org/ticket/31367 for it and will provide a fix there
shortly.
Akos M schrieb am Montag, 8. Februar 2021 um 11:42:59 UTC+1:
> It seems that unfortunately the problem persists for multivariate rings as
> well:
>
> A. = QQ[
For reference this is also asked on Ask Sage:
https://ask.sagemath.org/question/55618
--
You received this message because you are subscribed to the Google Groups
"sage-devel" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to sage-devel+unsubscr...@google
It seems that unfortunately the problem persists for multivariate rings as
well:
A. = QQ[]
B. = 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
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 wrote:
>
> It looks like a bug to me. f.kernel() expands to
> f._inverse_image_ideal(f.codomai
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 wro
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. = QQ[]
B. = 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
