Let $m(x) = x^n + a_{n-1}x^{n-1} + \dots + a_1 x+ a_0$, $a_i \in \mathbb{Z}$, be an irreducible polynomial over $\mathbb{Q}$ and $K = \mathbb{Q}[x] / {m(x)\mathbb{Q}[x]}$, so $K$ is an algebraic number field. Let $\mathcal{O}_K$ be its ring of algebraic integers and $x\mathcal{O}_K$ be the ideal in $\mathcal{O}_K$ generated by $x$.

Since $\mathcal{O}_K$ is a Dedekind domain, $x\mathcal{O}_K$ is a product of prime ideals, $P_1^{j_1} P_2^{j_2} \dots P_m^{j_m}$, where each $j_i$ is a positive integer and each $P_i$ is a prime ideal. $\mathcal{O}_K/P_i$ is a finite field. My question is:

$$ Is ~\mathcal{O}_K/P_i ~always~ isomorphic~ to~ F_{p_i}~ for~ some~ prime~ p_i~ ?$$

That is the order of $\mathcal{O}_K/P_i$ is equal to $p_i$, not some power of $p_i$. Someone pointed out to me before that this is not likely to be true. But I could not find a counter example.

  • $\begingroup$ @PeteL.Clark Shouldn't $x$ be $i$ in your example? At least that's how I interpreted the question. $\endgroup$ Jul 27 '13 at 20:02
  • $\begingroup$ @Mahdi: yes, thanks. I read the question too quickly and the notation confused me. $\endgroup$ Jul 27 '13 at 20:05

No. Let $p$ be a prime number with $p \equiv 3 \pmod{4}$, and let $m(x) = x^2+p^2$. Then $K = \mathbb{Q}[x]/(m) \cong \mathbb{Q}(i)$ and the element $x$ may be identified with $ip$. Since $i$ is a unit in $\mathcal{O}_K$, we have $(ip) = (p)$ in $\mathcal{O}_K$. Since $p \equiv 3 \pmod{4}$, $\mathcal{O}_K/(ip) \cong \mathbb{F}_{p^2}$. This also shows that $(ip)$ is a prime ideal.


If you're interested not in every prime having this property, but just "some prime" then the answer is in fact yes. You can always find infinitely many primes p (given the number field K) where the residue field will still be just size p. In Pete's example, it would be the 1 mod 4 primes.


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