Why is $\mathbb{Z}_3[x]/\langle x^2+1\rangle$ a field with order $9$? And does order $9$ mean it has $9$ elements?

Question

So I get that $\mathbb{Z}_3[x]/\langle x^2+1\rangle$ is a field since $x^2+1$ is irreducible. I don't get how to show it only has $9$ elements. And is that the same thing as saying it has order $9$?

Answer 1

Yes, it has $9$ elements, namely $$ \{ 0,1,2,x,x+1,x+2,2x,2x+1,2x+2\} $$ considered as equivalence classes modulo $(x^2+1)$. Because of $x^2=-1$, there are no higher-degree polynomials in it.

Answer 2

Hint $:$ $\Bbb Z_3 [x] / \langle x^2 + 1 \rangle \cong \left (\Bbb Z [x] / \langle x^2 + 1 \rangle \right ) / \langle 3 \rangle \cong \Bbb Z[i] / \langle 3 \rangle.$