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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 $p$?

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    $\begingroup$ This is a result of computing $\mathcal O_K/(p)$ in two ways. First, by writing it as $\mathbf Z[\theta]/(p,f(x)) \cong (\mathbf Z/p\mathbf Z)[x]/(\overline{f(x)})$ and using the Chinese remainder theorem we get the product of rings $(\mathbf Z/p\mathbf Z)[x]/(f_i(x)^{e_i})$. Second, by writing $(p) = P_1^{e_1}\cdots P_g^{e_g}$ and using the Chinese remainder theorem in $\mathcal O_K$, $\mathcal O_K/(p)$ is the product of rings $\mathcal O_K/P_i^{e_i}$. By counting maximal ideals and their powers in $\mathcal O_K/(p)$ you can recover the similar data from both ways of looking at the ring. $\endgroup$
    – KConrad
    Mar 6, 2018 at 17:31
  • $\begingroup$ So is the theorem also true if instead of requiring that $\mathcal{O}_{K}=\mathbb{Z}[\theta]$ and taking $f$ as the minimal polynomial of $\theta$ one just considers a polynomial $f$ such that $p$ does not divide the discriminant of $f$, without making further hypotheses on $\mathcal{O}_{K}$? $\endgroup$
    – near
    Mar 6, 2018 at 18:29
  • $\begingroup$ For a prime $p$, you can weaken the assumption $\mathcal O_K = \mathbf Z[\theta]$ to $\theta$ being chosen in $\mathcal O_K$ so that $K = \mathbf Q(\theta)$ and $p$ does not divide the index $[\mathcal O_K:\mathbf Z[\theta]]$. Since ${\rm disc}(f) = [\mathcal O_K:\mathbf Z[\theta]]^2{\rm disc}(\mathcal O_K)$, $p$ not dividing that index is weaker than $p$ not dividing ${\rm disc}(f)$. If you suppose $p$ doesn't divide ${\rm disc}(f)$ you will not be able to apply the result to ramified primes: if $p$ does not divide ${\rm disc}(f)$ then it does not divide ${\rm disc}(\mathcal O_K)$ either. $\endgroup$
    – KConrad
    Mar 7, 2018 at 0:38
  • $\begingroup$ That's true, I hadn't considered the problems with ramified primes. Thank you again! $\endgroup$
    – near
    Mar 7, 2018 at 9:06

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