Let $A$ be a Dedekind domain. Let $K$ be the field of fractions of $A$ and $L$ is some finite field extension of $K$. Then let $B$ be the integral closure of $A$ in $L$. (Sorry I don't know how to make that diagram with the arrows but I think these are the standard letters for this situation.)
In general, if $X$ and $Y$ are ideals of $B$ which are coprime, when are $X \cap A$ and $Y\cap A$ coprime as ideals of $A$? Are there some easy hypotheses which make this true?
In particular, in Neukirch Algebraic Number Theory, Proposition 8.3, he uses the following:
Let $F$ be the conductor in $B$ of $A[\theta]$ where $L = K[\theta]$ and $\theta \in B$. Then if $p \subset A$ is prime and $pB$ is coprime to $F$, then $p$ is coprime to $F\cap A$.
Basically you need to know that when you contract $X$ and $Y$ to $A$, they will still share no common prime factors (so there are no ideals in each lying over the same prime).
In the case you mention, all primes in $B$ lying over $p$ appear in the prime factorization of $pB$, so $F$ is not divided by any of these (being coprime). Therefore, when you contract $F$, the prime ideal $p$ cannot divide it; that is, $F \cap A$ and $p$ are still coprime.