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.
 A: 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. $\hspace{1mm} \Box$
A: 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.
A: 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.
