I citate Serge Lang's Algebra (Second edition, page 23).
If $A$ is a discrete valuation ring, then in particular, $A$ is a principal ideal ring, and any finitely generated torsion-free module $M$ over $A$ is free. If its rank is $n$, and if $\mathfrak{p}$ is the maximal ideal of $A$, then $M/\mathfrak{p}M$ is a free module of rank $n$.
Proposition 19. Let $A$ be a local ring and $M$ a free module of rank $n$ over $A$. Let $\mathfrak{p}$ be the maximal ideal of $A$. Then $M/\mathfrak{p}M$ is a vector space of dimension $n$ over $A/\mathfrak{p}$.
Proof. [...]
Let $A$ be a Dedekind ring, $K$ its quotient field, $L$ a finite separable extension of $K$, and $B$ the integral closure of $A$ in $L$. If $\mathfrak{p}$ is a prime ideal of $A$, then $\mathfrak{p}B$ is an ideal of $B$ and has a factorization $$ \mathfrak{p}B = \mathfrak{B}_1^{e_1} \cdots \mathfrak{B}_r^{e_r} \quad (e_i \geq 1)$$ into primes of $B$. It is clear that a prime $\mathfrak{B}$ of $B$ occurs in this factorization if and only if $\mathfrak{B}$ lies above $\mathfrak{p}$.
I'm having difficulties understanding this last section. I can see why a prime $\mathfrak{B}$ of $B$ occurs in this factorization if and only if $\mathfrak{B}$ lies above $\mathfrak{p}$ given that there exists a factorization. I don't know why there must exist such a factorization since $B$ is not a Dedekind ring (is it? I didn't come across a proof yet). The 'only' thing I know about $B$ from previous theorems and propositions is that $B$ is a free module of rank $n$ over $A$. Any help understanding this section?
Added note: apparently $B$ is Dedekind. However Lang does not use this fact to formulate this statement. Do I have to find a proof by reading between the lines or don't you need this fact?
Yes, $B$ is Dedekind. There is a proof, for example, here (and probably in Lang somewhere, but I don't have the book handy).