prime ideal of integral closure on the decomposition iff lies above the prime ideal of the ring

I'm having troubles proving the following proposition. In every reference I read, they mark this proposition as "clear" or "trivial", but I am unable to prove it. Some help?

Let $A$ be a Dedekind ring, $K$ its field of fractions, $L$ a finite separable extension of $K$, $B$ the integral closure of $A$ in $L$ and $\mathfrak{p}$ a prime ideal of $A$. Let $\mathfrak{p}B=\mathfrak{B}_1^{e_1}\dots\mathfrak{B_k}^{e_k}$ be the (unique) decomposition of the ideal $\mathfrak{p}B$ into prime ideals of $B$ (as the integral closure of a Dedekind ring is still a Dedekind ring). Then a prime ideal $\mathfrak{B}$ of $B$ appears in the decomposition if and only if $\mathfrak{B}$ lies above $\mathfrak{p}$ (that is, $\mathfrak{B} \cap A=\mathfrak{p}$).

Thanks.

If $\mathfrak{P}$ is one of the $\mathfrak{P}_i$, then clearly $\mathfrak{P}\cap A \supseteq \mathfrak{p}$ since $\mathfrak{p} \subseteq A$ and $\mathfrak{p} \subset \mathfrak{p}B \subset \mathfrak{P}$. But $\mathfrak{p}$ is maximal in $A$, so $\mathfrak{P}\cap A = \mathfrak{p}$ (it is not equal to $A$ since it does not contain $1$).
Conversely, if $\mathfrak{P}\cap A = \mathfrak{p}$, then $\mathfrak{P}\supset \mathfrak{p}B$. But since $B$ is Dedekind, this implies that $\mathfrak{P}$ is a factor of $\mathfrak{p}$ so that by unique factorization $\mathfrak{P}$ is one of the $\mathfrak{P}_i$.