Ask Your Question
3

possible bug: kernel of ring homomorphism

asked 4 years ago

makos gravatar image

updated 2 years ago

tmonteil 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?

Preview: (hide)

Comments

slelievre gravatar imageslelievre ( 4 years ago )
rburing gravatar imagerburing ( 4 years ago )

1 Answer

Sort by » oldest newest most voted
2

answered 3 years ago

rburing gravatar image

updated 3 years ago

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
Preview: (hide)
link

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: 4 years ago

Seen: 374 times

Last updated: Apr 13 '21