Let $R$ be a discrete valuation ring with quotient field $K$, and $L/K$ a finite separable extension which is unramified over $K$. Also suppose that $K$ is complete with respect to the valuation of $R$. Let $S$ be the integral closure of $R$ in $L$, and $x \in S$ an element with minimal polynomial $g \in K[X]$ for which $L = K[x]$ and $k_L = k[\overline x]$ ($k$ is the residue field of $K$ and $k_L$ is that of $L$).
Why does it follow that $S = R[x]$? I know that this is true if and only if $\mathfrak D = g'(x)S$, where $\mathfrak D$ is the different of $L/K$. I also know that since $L/K$ is unramified, $\mathfrak d = R$ where $\mathfrak d$ is the discriminant of $L/K$.