Why $\mathbb{Z}[\theta]\,/\,\mathcal{P} \simeq \mathbb{F}_{p^e}$ for any non-zero prime ideal $\mathcal{P}$ of $\mathbb{Z}[\theta]$? 
If $f(x)$ is a monic,irreducible polynomial in $\mathbb{Z}[x]$ with $\theta\in\mathbb{C}$ as root, why $\mathbb{Z}[\theta]\,/\,\mathcal{P} \simeq \mathbb{F}_{p^e}$ for any non-zero prime ideal $\mathcal{P}$ of $\mathbb{Z}[\theta]$?

My attempt:
If $\mathcal{P}$ is a prime ideal of $\mathbb{Z}[\theta]$, then $\mathbb{Z}[\theta]/\mathcal{P}$ is an integral domain which contains a subring $\mathbb{F} \simeq \mathbb{Z}/(\mathbb{Z}\cap\mathcal{P}) \simeq \mathbb{Z}/p\mathbb{Z}$, for some prime $p$.
Now, since every element in $\mathbb{Z}[\theta]$ is algebraic, then $\mathbb{Z}[\theta]/\mathcal{P}$ is an algebraic extension over $\mathbb{Z}/p\mathbb{Z}$. But how I conclude that the extension is finite?
Furthermore, I cannot use the norm to say that $N(\mathcal{P}) = p^e$ because $\mathbb{Z}[\theta]$ is not a Dedekind domain.
 A: The ring $R:=\Bbb{Z}[\theta]$ has a finite rank, say $m$, as a free (additive) Abelian group. More precisely, its rank equals the degree of the minimal polynomial of $\theta$.
To see this combine the following points:


*

*$R$ is generated as an abelian group by $1,\theta,\theta^2,\ldots,\theta^{m-1}$.

*As a subgroup of the additive group of $\Bbb{C}$ $R$ is surely torsion free.

*A torsion free finitely generated Abelian group is free of a finite rank.


As $p\in\mathcal{P}$ it follows that the quotient ring has finitely many elements because:


*

*$p R\subseteq\mathcal{P}$ because $\mathcal{P}$ is an ideal, and

*$[R:pR]=p^m$. For the purposes of calculating this index we can equate $R$ with the additive group $\Bbb{Z}^m$.


Anyway, $pR\subseteq \mathcal{P}\subseteq R$, so the index of $\mathcal{P}$ in $R$ is a factor of the index of $pR$.
A: An alternative approach to Jyrki's:
All the preliminaries are the same, noting that $[\Bbb Z[\theta]:(m)]<\infty$ and the like. Then note that as $(p)\supseteq \mathcal{P}$, the projection map
$$\Bbb Z[\theta]\to\Bbb Z[\theta]/\mathcal{P}$$
has in its kernel $(p)$ hence it is a $\Bbb Z/p$-module. From this you get even more than the order (if only a little).
