I am trying to find some ring isomorphic to $\mathbb Z[i] / \langle 5 \rangle$ . I know that
$\langle 5 \rangle=\langle (2+i)(2-i)\rangle=\langle 2+i\rangle \langle2-i\rangle$ , now if $d$ is the gcd of $2+i , 2-i$ , then
$d|2+i-(2-i)=2i$ , so $d|2$ , then $d|2+i-2=i$ , so the gcd is a unit , hence
$\langle 2+i\rangle+ \langle2-i\rangle=\mathbb Z[i]$ , so by Chinese remainder theorem ,
$\mathbb Z[i] / \langle 5 \rangle \cong \mathbb Z[i] /\langle 2+i\rangle \times \mathbb Z[i] / \langle 2-i\rangle $ . Now if gcd$(a,b)=1$ , then $\mathbb Z[i] / \langle a+ib \rangle \cong \mathbb Z_{a^2+b^2}$ , so
$\mathbb Z[i] / \langle 2-i\rangle \cong \mathbb Z_5$ and $\mathbb Z[i] /\langle 2+i\rangle \cong \mathbb Z_5$ , thus $\mathbb Z[i] / \langle 5 \rangle \cong \mathbb Z_5 \times \mathbb Z_5$ . Am I correct ? If so , then
I want to know , what is the explicit surjective ring homomorphism from $\mathbb Z[i]$ to $\mathbb Z_5 \times \mathbb Z_5$ with
kernel $\langle 5 \rangle$ ? I have already checked that the map $f(a+ib)=([a]_5 , [b]_5)$ is a group
homomorphism but not a ring homomorphism . Please help . Thanks in advance