ramification of a rational prime in conjugate fields

Let p∈Zp\in\mathbb{Z} be a rational prime. Let f(x)∈Q[x]f(x)\in\mathbb{Q}[x] be an irreducible polynomial. If θ1\theta_1 and θ2\theta_2 are two roots of ff, we know that the fields K1=Q(θ1)K_1=\mathbb{Q}(\theta_1) and K2=Q(θ2)K_2=\mathbb{Q}(\theta_2) are isomorphic and their discriminants are same. So pp ramifies in K1K_1 iff pp ramifies in K2K_2. Is there any other way to arrive at this fact without using the fact that pp ramifies in a field iff pp divides the discriminant ?

Also, say we know that the factorization of ⟨p⟩\langle p\rangle in K1K_1: pO1=Pe11…PekkpO_1=P_1^{e_1}\dots P_k^{e_k}

I have a guess that the factorization of ⟨p⟩\langle p\rangle in K2K_2 will be of the form
pO2=Pe11…PekkpO_2=\mathfrak{P}_1^{e_1}\dots \mathfrak{P}_k^{e_k}

Basically, the ramification indices of PiP_i and Pi\mathfrak{P}_i are same. I also think that the corresponding inertia degrees will be same.

Is my guess correct ? If so, how to prove it ?




if f∈K[x]f \in K[x] is irreducible and f(θi)=0f(\theta_i) = 0 then K(θi)≃K[x]/(f)K(\theta_i) \simeq K[x]/(f), so your two fields are isomorphic, and the splitting behavior of pp has to be the same in both
– user1952009
Oct 20 at 19:34



Yes, I agree. But what does it imply in the context of this question ?
– pink butterfly
Oct 20 at 19:35



? the isomorphism means that K1K_1 and K2K_2 are the same field up to a renaming of some elements, so that pO1=Pe11…Pekk⟹pO2=σ(pO1)=σ(P1)e1…σ(Pk)ekpO_1 = P_1^{e_1}\ldots P_k^{e_k} \implies p O_2 = \sigma(pO_1) = \sigma(P_1)^{e_1}\ldots \sigma(P_k)^{e_k} where σ\sigma is the renaming function K1→K2K_1 \to K_2
– user1952009
Oct 20 at 19:38