# Revision history [back]

### Working with finitely presented algebras

I am trying to work with finitely presented algebras in SageMath. But apparently, I am doing something wrong.

For a simple example, I want to construct A = Q[x]/<x^2>. The image a of x in that algebra should satisfy a^2 = 0. But SageMath tells me that this is not the case.

sage: F = FreeAlgebra(QQ,1,'x')
sage: F
Free Algebra on 1 generators (x,) over Rational Field
sage: x in F
True
sage: I = F.ideal([x^2])
sage: A = F.quotient(I)
sage: A
Quotient of Free Algebra on 1 generators (x,) over Rational Field by the ideal (x^2)
sage: a = A.gen()
sage: a^2 == 0
False
sage: a^2 == A.zero()
False


What am I doing wrong here?

### Working with finitely presented algebras

I am trying to work with finitely presented algebras in SageMath. But apparently, I am doing something wrong.

For a simple example, I want to construct A = Q[x]/<x^2>. The image a of x in that algebra should satisfy a^2 = 0. But SageMath tells me that this is not the case.

sage: F = FreeAlgebra(QQ,1,'x')
sage: F
Free Algebra on 1 generators (x,) over Rational Field
sage: x in F
True
sage: I = F.ideal([x^2])
sage: A = F.quotient(I)
sage: A
Quotient of Free Algebra on 1 generators (x,) over Rational Field by the ideal (x^2)
sage: a = A.gen()
sage: a^2 == 0
False
sage: a^2 == A.zero()
False


What am I doing wrong here?

When the ideal is <1>, the quotient should be trivial. But again, SageMath does not believe this.

### Working with finitely presented algebras

I am trying to work with finitely presented algebras in SageMath. But apparently, I am doing something wrong.

For a simple example, I want to construct A = Q[x]/<x^2>. The image a of x in that algebra should satisfy a^2 = 0. But SageMath tells me that this is not the case.case. (EDIT: the code has been updated below.)

sage: F F.<x> = FreeAlgebra(QQ,1,'x')
FreeAlgebra(QQ,1)
sage: F
Free Algebra on 1 generators (x,) over Rational Field
sage: x in F
True
sage: I = F.ideal([x^2])
sage: A = F.quotient(I)
sage: A
Quotient of Free Algebra on 1 generators (x,) over Rational Field by the ideal (x^2)
sage: a = A.gen()
sage: a^2 == 0
False
sage: a^2 == A.zero()
False


What am I doing wrong here?

When the ideal is <1>, the quotient should be trivial. But again, SageMath does not believe this.

### Working with finitely presented algebras

I am trying to work with finitely presented algebras in SageMath. But apparently, I am doing something wrong.

For a simple example, I want to construct A = Q[x]/<x^2>. The image a of x in that algebra should satisfy a^2 = 0. But SageMath tells me that this is not the case. (EDIT: the code has been updated below.)

sage: F.<x> = FreeAlgebra(QQ,1)
sage: F
Free Algebra on 1 generators (x,) over Rational Field
sage: x in F
True
sage: I = F.ideal([x^2])
sage: A = F.quotient(I)
sage: A
Quotient of Free Algebra on 1 generators (x,) over Rational Field by the ideal (x^2)
sage: a = A.gen()
sage: a^2 == 0
False
sage: a^2 == A.zero()
False


What am I doing wrong here?

When the ideal is <1>, the quotient should be trivial. But again, SageMath does not believe this.

It seems that this bug has been reported here before: