Let $L/K$ be a finite separable extension of fields. Let $A$ be a discrete valuation ring in $K$. Let $B$ be the integral closure of $A$ in $L$. Why is $B$ a free $A$-module of rank equal to $[L:K]$?
As KCd mentioned in a comment, every DVR is a PID. Now you can look at Lang's "Algebraic Number Theory", Theorem 1 in page 7, where he shows the following theorem:
Theorem 1. Let $A$ be a principal ideal ring, and $L$ a finite separable extension of its quotient field, of degree $n$. Let $B$ be the integral closure of $A$ in $L$. Then $B$ is a free module of rank $n$ over $A$.