ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 04 Jan 2021 15:12:56 +0100inverse image under ring homomorphismhttps://ask.sagemath.org/question/55099/inverse-image-under-ring-homomorphism/Hi there!
Just starting to learn Sage and can't get why both `inverse_image` calls
below give an error saying the given element does not have a preimage,
while `y^2` clearly has `xbar` as preimage.
Referring to the generators of `R1` and `R2` explicitly doesn't help either.
sage: R.<x, y> = QQ[]
sage: R1 = R.quotient(R.ideal(y^2 - x^3))
sage: R2 = R.quotient(R.ideal(x))
sage: h = R1.hom([y^2, y^3], R2)
sage: h
Ring morphism:
From: Quotient of Multivariate Polynomial Ring in x, y over Rational Field by the ideal (-x^3 + y^2)
To: Quotient of Multivariate Polynomial Ring in x, y over Rational Field by the ideal (x)
Defn: xbar |--> y^2
ybar |--> y^3
sage: h.inverse_image(y)
Traceback (most recent call last)
...
ValueError: element y does not have preimage
sage: h.inverse_image(y^2)
Traceback (most recent call last)
...
ValueError: element y^2 does not have preimage
Thanks!KarinaMon, 04 Jan 2021 15:12:56 +0100https://ask.sagemath.org/question/55099/How do I define a homomorphism of a graded commutative algebra?https://ask.sagemath.org/question/42202/how-do-i-define-a-homomorphism-of-a-graded-commutative-algebra/Why does the following throw a `TypeError: images do not define a valid homomorphism`?
E = GradedCommutativeAlgebra(QQ,'x,y',degrees=(1,1))
E.inject_variables()
f = E.hom([x,y])
I expected it to define $f$ to be the identity homomorphism of $E$. What is the right way to define a homomorphism of $E$? I'm more interested in the one that switches $x$ and $y$ than the identity homomorphism, but this seemed a more obvious version of the question.ronnoSat, 28 Apr 2018 19:14:46 +0200https://ask.sagemath.org/question/42202/