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.


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 $pR$.

  • $\begingroup$ Thank you (altough i miss some algebraic definitions and concepts). It is not clear to me why we can say $p$ is in $R$. What I can see is that there is an isomorphic copy of $\mathbb{Z}/p\mathbb{Z}$ in $R$. $\endgroup$ – ilmarchese May 26 '17 at 16:55
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    $\begingroup$ Hmm. $R=\Bbb{Z}[\theta]$ is a subring of $\Bbb{C}$. I had not passed to the quotient ring yet at that point. I thought you had proved yourself that $\Bbb{Z}\cap\mathcal{P}$ contains a prime number $p$. How else did you conclude that $\Bbb{Z}/(\Bbb{Z}\cap\mathcal{P})$ is isomorphic to $\Bbb{Z}/p$? $\endgroup$ – Jyrki Lahtonen May 26 '17 at 16:58
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    $\begingroup$ @ilmarchese $p\in\Bbb Z\subseteq \Bbb Z[\theta]=R$ $\endgroup$ – Adam Hughes May 26 '17 at 16:59
  • $\begingroup$ Of course!!! Maybe I need a break... :) Thank you all! $\endgroup$ – ilmarchese May 26 '17 at 17:16

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).


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