Why is $\mathbb F_3[x]/(x^2 + 1) \cong \mathbb F_9$ I want to show that $\mathbb Z[i]/(3) \cong \mathbb F_9$, and I know that $\mathbb Z[i]/(3) \cong \mathbb Z[x]/(3, x^2 + 1)  \cong \mathbb F_3[x]/(x^2 + 1)$. How do we know that $\mathbb F_3[x]/(x^2 + 1) \cong \mathbb F_9$? I know that $x^2 + 1$ is irreducible, is this important?

$F_3[X]/(x^2+1)$ is an extension of degree 2 of $F_3$ so its cardinal is $3^2$. Two finite field which have the same order are isomorphic.
https://en.wikipedia.org/wiki/Finite_field
