How do I find a basis for $\mathbb{Z}_5[x]/\langle x^3-x^2-1 \rangle$?

Question

I wish to find a basis for the field $\mathbb{Z}_5[x]/\langle x^3-x^2-1 \rangle$. Treating the polynomial as the additive identity, my intuition tells me that it should be $\{1,x,x^2\}$. Thus, finding different forms of $\alpha = x + \langle x^3-x^2-1\rangle$ such as $\alpha^2$ is simply done by computing and continuing to treat $x^3-x^2-1$ as $0$. Is this correctly laid out?

Answer 1

(Since comment box is getting too small)

@Andrew: So let's compute $\alpha^4$. As you said, $\alpha^4 = x^4+\langle x^3-x^2-1\rangle$. Now since $x^3-x^2-1$ becomes zero in the quotient ring, we get that $x^3+\langle x^3-x^2-1\rangle = (x^2+1)+\langle x^3-x^2-1\rangle$. Now multiply both sides by $x$ to get $x^4+\langle x^3-x^2-1\rangle = (x^3+x)+\langle x^3-x^2-1\rangle$. You can rewrite this as: $$ x^4+\langle x^3-x^2-1\rangle = [x^3+\langle x^3-x^2-1\rangle] + [x+\langle x^3-x^2-1\rangle]$$ Therefore, $\alpha^4=\alpha^3+\alpha$.