Splitting an ideal lying in a prime ideal An obviously naive question question I can't answer. Let $A$ be a Dedekind domain, $\mathfrak{p}$ a prime ideal of $A$ and $I$ a non-zero ideal lying in $\mathfrak{p}$. The ideal $I$ can be splitted into a product of prime ideals
$$I=\prod_i^n \mathfrak{\beta_i}^{e_i}.$$
Is then $\mathfrak{p}$ necessarily one of the  $\beta_i$?
It is the case when $A=\mathbb{Z}$.
Initially, I wanted to understand why "$B/A$ is ramified at a finite number of prime ideals" follows immediately from "$B/A$ is ramified at $\mathfrak{p}$ iff $\mathfrak{D}_{B/A} \neq(0)$", where $A$ and $B$ are the rings of integers of number fields $K$ and $L$ with $K\subset L$. 
Thanks.

We have ${\frak{p}} \supset I$ iff $p \supset \beta_i$ for some $i$ iff $p = \beta_i$ for some $i$.  
The first iff comes from the definition of a prime ideal; the second iff comes from the fact that in a Dedekind domain, every non-zero prime ideal is maximal.
