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 $K = \mathbb{Q}(x) / {m(x)\mathbb{Q}(x)}$. 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.