Ask Your Question
3

possible bug: kernel of ring homomorphism

asked 2021-02-07 17:41:51 +0200

makos gravatar image

updated 2021-04-13 13:41:39 +0200

vdelecroix gravatar image

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).

Why?

edit retag flag offensive close merge delete

Comments

slelievre gravatar imageslelievre ( 2021-02-09 03:55:39 +0200 )edit
rburing gravatar imagerburing ( 2021-02-09 15:28:59 +0200 )edit

1 Answer

Sort by ยป oldest newest most voted
2

answered 2021-04-13 10:53:41 +0200

rburing gravatar image

updated 2021-04-13 11:58:55 +0200

slelievre gravatar image

This was a bug, fixed in the meantime at

(merged in Sage 9.3.beta9). In Sage 9.3.rc2 for instance:

sage: A.<t> = QQ[]
sage: B.<x, y> = QQ[]
sage: H = B.quotient(B.ideal([B.1]))
sage: f = A.hom([H.0], H)

sage: f
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

sage: f.kernel()
Principal ideal (0) of Univariate Polynomial Ring in t over Rational Field
edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2021-02-07 17:41:51 +0200

Seen: 82 times

Last updated: Apr 13