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.



Prime ideals in the ring of algebraic integers
**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. 


