# possible bug: kernel of ring homomorphism

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?

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Sort by » oldest newest most voted 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

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