# 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:

Apparently, it works when F is defined as a polynomial ring. The free algebra on one generator should equal the polynomial ring. But SageMath thinks differently.