Finding inverse in a quotient ring

Question

Say we define $E$ to be the field $Q[\alpha]$ where $\alpha$ is a root of $x^3-3x-1$. So, $E=\{a+b\alpha+c\alpha^2:a,b,c\in\Bbb{Q}\}$. I want to find the linear combination s.t $\sum_{i=0}^2a_i\alpha^i=\alpha^{-1}$.

I know that there exists an isomorphism $\phi:\Bbb{Q[x]}/I\rightarrow \Bbb{Q[\alpha]}$ (evaluation at $\alpha$). So the question is equivalent to finding inverse of $x+I$ in my quotient ring (which is a field).

Using Euclid's algorithm, one can easily note that $(x^2-3)x=(x^3-3x-1)+1$, so $(x+I)^{-1}=(x^2-3)+I$ and so $\alpha^{-1}=\phi(x^2-3+I)=\alpha^2-3$, but looking online for general inverses in such fields, my answer doesn't match.

Is my approach correct or am I missing something? any help would be appreciated.

Answer 1

Your work is correct. You can also check it using SageMath:

sage: E.<a> = NumberField(x^3 - 3*x - 1)
sage: 1/a
a^2 - 3

You can also write the linear map multiplication-by-$\alpha$ as a matrix $M_\alpha$ with respect to the basis $1,\alpha,\alpha^2$ and find a preimage of the vector $(1,0,0)$ by solving the linear system:

sage: a.matrix().transpose()
[0 0 1]
[1 0 3]
[0 1 0]
sage: E.one().vector()
(1, 0, 0)
sage: a.matrix().transpose() \ E.one().vector()
(-3, 0, 1)
sage: E(a.matrix().transpose() \ E.one().vector())
a^2 - 3