I found this theorem in "Problems in Algebraic Number Theory" by Jody Esmonde and M.Ram Murty, and I feel like I'm not fully understanding it. It describes factorization in $\mathcal{O}_{K}$, where $K$ is a number field:
Suppose that there is a $\theta\in K$ such that $\mathcal{O}_{K}=\mathbb{Z}[\theta]$. Let $f(x)$ be the minimal polynomial of $\theta$ over $\mathbb{Z}[x]$. Let $p$ be a rational prime, and suppose $f(x)\equiv f_{1}(x)^{e_{1}}\cdots f_{g}(x)^{e_{g}}$ (mod $p$), where each $f_{i}(x)$ is irreducible in $\mathbb{F}_{p}[x]$. Then $p\mathcal{O}_{K}=P_{1}^{e_{1}}\cdots P_{g}^{e_{g}}$, where $P_{i}=(p_{i},f_{i}(\theta))$ are prime ideals, with $N(P_{i})=p^{deg(f_{i})}$.
This theorem links the exponents in the factorization of $f$ with the ones in the factorization of $p\mathcal{O}_{K}$, and it seems also to link the degrees of the irreducible factors of $f$ with the inertial degrees of primes in the factorization of $p\mathcal{O}_{K}$, is that correct? Specifically, is it true that the degrees of the irreducible factors of $f$ are exactly the inertial degrees of primes in $\mathcal{O}_{K}$ lying above $p$?