In this answer, one can read:
Let $P \in \Bbb Z[X]$. Let $K$ be the splitting field of $P(x)$. A prime $p$ splits in $K$ if and only if$^1$ $P(x)$ splits into distinct linear factors modulo $p$.
1 With finitely many exceptions, related to the fact that the ring of integers in $K$ may not be $\mathbb{Z}[x]/P(x)$.
Question: I would like to know the precise statement of this claim (e.g. what are these "finitely many exceptions"), and how to prove this statement.
I am aware of Dedekind's factorization theorem, stated here :
Let $K$ be a number field and $a \in O_K$ such that $K=\Bbb Q(a)$. Let $f(T)$ be the minimal polynomial of $a$ in $\Bbb Z[T]$. For any prime $p$ not dividing $[O_K:\Bbb Z[a]]$,write $$f(T) = \pi_1(T)^{e_1}\cdots \pi_r(T)^{e_r} \pmod p,$$ where the $\pi(T)$'s are distinct monic irreducibles in $\Bbb Fp[T]$. Then $(p) =pO_K$ factors into prime ideals as $$(p) = P_1^{e_1}\cdots P_r^{e_r},$$ where there is a bijection between the $P_i$'s and $\pi(T)$'s such that $N(P_i)=p^{deg(\pi_i)}$.
In order to prove $\Longleftarrow$ in the statement above, I would like to apply Dedekind's factorization theorem. But I don't know how I could take $P(T)=f(T)$, i.e. $P$ is the minimal polynomial of an element $a \in O_K$ such that $K=\Bbb Q(a)$ (only knowing that the roots of $P$ are $r_1,\dots,r_n$ such that $K=\Bbb Q(r_1,\dots,r_n)$).
Moreover, I don't see how to get "$\implies$".
