A Method that Usually Works

Suppose $\alpha\in\O _K$ is such that $K=\mathbb{Q}(\alpha)$, and let $g(x)$ be the minimal polynomial of $\alpha$. Then $\mathbb{Z}[\alpha]\subset \O _K$, and we have a diagram of schemes

$$\xymatrix{ {(??)\, }\ar@{^(->}[r]\ar[d] &{\Spec(\O_K)}\ar[d] \\ ... ...]/g(x))\\ {\Spec(\mathbb{Z}/p\mathbb{Z})\,}\ar@{^(->}[r]&{\Spec(\mathbb{Z})} }$$

where $\overline{g} = \prod_i \overline{g}_i^{e_i}$ is the factorization of the image of $g$ in $(\mathbb{Z}/p\mathbb{Z})[x]$.

The cover $\Spec(\mathbb{Z}[\alpha])\rightarrow \Spec(\mathbb{Z})$ is easy to understand because it is defined by the single equation $g(x)$. To give a maximal ideal $\mathfrak{p}$ of $\mathbb{Z}[\alpha]$ such that $f(\mathfrak{p}) = p\mathbb{Z}$ is the same as giving a homomorphism $\mathbb{Z}[x]/(g) \rightarrow \overline{\mathbb{F}}_p$, which is in turn the same as giving a root of $g$ in $\overline{\mathbb{F}}_p$ (an allowed place where $x$ can go). If the index of $\mathbb{Z}[\alpha]$ in $\O _K$ is coprime to $p$, then the primes $\mathfrak{p}_i$ in the factorization of $p\O _K$ don't decompose further going from $\mathbb{Z}[\alpha]$ to $\O _K$, so we are done (the homomorphisms $\mathbb{Z}[\alpha]\rightarrow \overline{\mathbb{F}}_p$ are in bijection with the homomorphisms $\O _K\rightarrow \overline{\mathbb{F}}_p$). We formalize this in the following theorem:

Theorem 2.1   Let $g(x)$ denote the minimal polynomial of $\alpha$ over  $\mathbb{Q}$. Let $p$ be a prime number that does not divide $[\O _K:\mathbb{Z}[\alpha]]$. Suppose that

$$\overline{g} = \prod_{i=1}^t \overline{g}_i^{e_i} \in (\mathbb{Z}/p\mathbb{Z})[x]$$

with the $\overline{g}_i$ distinct monic irreducible polynomials. Let $\mathfrak{p}_i = (p,g_i(\alpha))$ with $g_i\in\mathbb{Z}[x]$ any polynomial whose image is $\overline{g}_i$ in $(\mathbb{Z}/p\mathbb{Z})[X]$. Then

$$p\O _K = \prod_{i=1}^t \mathfrak{p}_i^{e_i}.$$

Geometrically,

$$f^{-1}(p\mathbb{Z}) = \{\mathfrak{p}_1,\mathfrak{p}_2,\ldots, \mathfrak{p}_t\},$$

(with multiplicities $e_i$).