Let $p\in\mathbb{Z}$ be a rational prime. Let $f(x)\in\mathbb{Q}[x]$ be an irreducible polynomial. If $\theta_1$ and $\theta_2$ are two roots of $f$, we know that the fields $K_1=\mathbb{Q}(\theta_1)$ and $K_2=\mathbb{Q}(\theta_2)$ are isomorphic and their discriminants are same. So $p$ ramifies in $K_1$ iff $p$ ramifies in $K_2$. Is there any other way to arrive at this fact without using the fact that $p$ ramifies in a field iff $p$ divides the discriminant ?
Also, say we know that the factorization of $\langle p\rangle$ in $K_1$: $$pO_1=P_1^{e_1}\dots P_k^{e_k}$$
I have a guess that the factorization of $\langle p\rangle$ in $K_2$ will be of the form $$pO_2=\mathfrak{P}_1^{e_1}\dots \mathfrak{P}_k^{e_k}$$
Basically, the ramification indices of $P_i$ and $\mathfrak{P}_i$ are same. I also think that the corresponding inertia degrees will be same.
Is my guess correct ? If so, how to prove it ?