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.

Since I'm not sure what you wrote, I'll try to sketch the argument. Let $a+bi\in I$, with $a+bi\neq 0$. Then $b(a+bi)+ai(a+bi)\in I$, but this is equal to $(b^2+a^2)i\in I$. Thus also $b^2+a^2\in I$ by multiplication by $i^3$. Now we have that $\mathbb{Z}[i]/I$ contains less than $(b^2+a^2)^2$ elements (can you show how?). This gives the result.
