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.

Let $I \subseteq \mathbb{Z}[i]$ be a nonzero ideal. Since $\mathbb{Z}[i]$ is a principal ideal domain, it follows that $I = (\alpha)$ for some $\alpha \in \mathbb{Z}[i].$ Let $a + I$ be a coset of $I$ with $a = k \alpha + \beta$ and $\delta(\beta) < \delta(\alpha).$ In particular, every coset in $\mathbb{Z}[i]$ is represented by an element with norm less than $\delta(\alpha).$ Now let $z =  x+iy$ have norm $q.$ Then $x^2+y^2 = q$ with $|x|,|y| \leqslant q$ meaning that there are is a finite amount of possibilities for $z.$ Thus there are finitely many possible representatives of distinct cosets in $\mathbb{Z}[i].$ Therefore $\mathbb{Z}[i]/I$ is finite. $\hspace{1mm} \Box$
