Can anyone give me a hint with proving the following,
Let $\alpha\in\mathbb Z[i],$ and let $P$ be a non-zero prime ideal of $\mathbb Z[\alpha].$ Show that the quotient $\mathbb Z[\alpha]/P$ is a finite ring.
My ideas have been to try use the euclidean function for Z[i] to show all coset representatives have norm bounded but I haven't been able to get this to work.
This has nothing to do with prime ideals. For every nonzero ideal $\mathfrak a$ in $\mathbf Z[\alpha]$, $\mathbf Z[\alpha]/\mathfrak a$ is a finite ring.
After all, $\mathfrak a$ contains a positive integer $n$ (pick a nonzero element $\gamma$ of $\mathfrak a$ and then the positive integer $n = {\rm N}(\gamma) = \gamma\overline{\gamma}$ is in $\mathfrak a$), so we have the containment of ideals $(n) \subset \mathfrak a \subset \mathbf Z[\alpha]$. Therefore to show $\mathfrak a$ has finite index, it suffices to show the smaller ideal $(n) = n\mathbf Z + n\alpha\mathbf Z$ has finite index in $\mathbf Z[\alpha]$, and indeed its index is $n^2$; do you see why?