Let $ I $ be an ideal in $\mathbb Z [i]$. Show that $\mathbb Z[i] /I $ is finite. 
Let $I$ be an ideal in $\mathbb Z[i]$. I want to show that $\mathbb Z[i]/I$ is finite.

I start with $Z[i]/I$ is isomorphic to $Z$.
$Z$ is ID then $I$ is prime .Here i get stuck.
Thanks for help.

First of all you need $I \neq 0$ of course. Then you have some $0 \neq a+bi \in I$, in particular we have $\mathbb Z \ni D := a^2+b^2 = (a+bi)(a-bi) \in I$.
We obtain $(D) \subset I$, hence a surjection $\mathbb Z[i]/(D) \twoheadrightarrow \mathbb Z[i]/I$.
Thus it suffices to show that $\mathbb Z[i]/(D)$ is finite. $\mathbb Z[i]/(D)$ is an abelian group, generated by $1$ and $i$, which are both of finite order at most $D$.
But a finitely generated abelian group is finite if and only if all its generators have finite order.
