# Prime ideals of subrings of the ring of Gaussian integers

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.

• First show that there is a nonzero rational integer $n$ in $P$. Then we have a surjective map $\Bbb Z[\alpha]/n \rightarrow \Bbb Z[\alpha]/P$ and for any finitely generated $\Bbb Z$-module $\Lambda$, the quotient $\Lambda / n\Lambda$ is finite. May 11 at 13:25

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?
• I have no idea since you did not show us the surrounding text or tell us what the reference is. If the ideal $P$ is a nonzero prime ideal then $\mathbf Z[\alpha]/P$ is a field and maybe something is done with that that we can't see.