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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.

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After some thinking, a counter example happened to be surprisingly simple. Let $R$ be the formal power series ring $\mathbb{F}_3[[1/t]]$, and $f(x)=x^3-x-t$. Then the discriminant of $f(x)$ is $-4A^3-27B^2=4$ since we are working over fields of characteristic three. However, the extension is totally ramified. This is an example of Artin Shreier equation. –  Jiangwei Xue May 26 '11 at 8:08

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