Is the following statement true?
Let $R$ be a discrete valuation ring with quotient field $K$ and valuation $\nu$. Suppose that $f(x)\in K[x]$ is an irreducible separable polynomial with discriminant $\Delta_f$. We denote the finite extension $K[x]/(f(x))$ of $K$ by $L$. If $\nu(\Delta_f)=0$, then $L$ is unramified over $K$.
Even if the statement if false in general, will it be true if we assume that $R$ is complete w.r.t. $\nu$?
Clearly, if $f(x)\in R[x]$, then the statement is true since the discriminant of the extension $\Delta_{L/K}$ divides $\Delta_f$ in this case. But I am having trouble to prove it in general.