$$R = \mathbb{Z}[\sqrt{-5}]$$
$I = (3, 2 + \sqrt{-5})$ is an ideal of $R$.
Show that $R/(3)$ has $9$ elements and $I/(3)$ has $3$ elements.
I have tried $(3) = \lbrace 3a + 3b\sqrt{-5} \rbrace$, but I can only guess some elements in $R/(3)$. How shall I prove that there are exactly 9 elements in $R/(3)$?
The first part follows from the Chinese Remainder Theorem.
$$\mathbb{Z}[\sqrt{-5}]\cong\mathbb{Z}[x]/(x^2+5)$$ So $R∕(3)\cong\mathbb{Z}_3[x]/(x^2-1)\cong\mathbb{Z}_3\times\mathbb{Z}_3$ which has $9$ elements.