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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 $[L:K]$?

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    $\begingroup$ Because $A$ is a PID (every DVR is a PID) and you can use discriminants to stick $B$ between two finite free $A$-modules of rank $n$; a module over a PID that lies between two free modules of rank $n$ is also free of rank $n$. In fact, this is the same proof as the proof that the integral closure of the integers (another PID) in a number field is a free ${\mathbf Z}$-module of rank $n$, where $n$ is the degree of the number field over ${\mathbf Q}$. $\endgroup$ – KCd Mar 16 '12 at 17:57
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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 $A$.

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