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 "$\implies$".

Here are some related questions: (1), (2), (3), (4).


Apply the Dedekind theorem to $\mathbb{Q}(r_i)$ and use the fact that if a prime splits in a several fields then it splits in their composite.

  • $\begingroup$ Thank you. But what is the precise statement of the claim? And how to prove $\implies$ ? $\endgroup$ – Watson Dec 29 '16 at 16:32
  • $\begingroup$ You can't finish the proof on your own ? It might be a long wait. $\endgroup$ – Rene Schipperus Jan 2 '17 at 21:29

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