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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.

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    $\begingroup$ 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. $\endgroup$
    – WhatsUp
    May 11 at 13:25

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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?

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  • $\begingroup$ Thanks! Its strange my text said prime ideal, do you have any idea why? $\endgroup$
    – RickSmith
    May 15 at 21:28
  • $\begingroup$ 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. $\endgroup$
    – KCd
    May 15 at 21:55

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