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 $L/K$.

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    $\begingroup$ $k_L$ is well defined only if $S$ is local. Is this in your assumotion ? $\endgroup$ – Cantlog Feb 26 '14 at 18:08
  • $\begingroup$ Well sure, the integral closure of a discrete valuation ring is still discrete, right? If necessary, I may need to add the assumption that $K$ is complete with respect to the absolute value induced by $\nu_R$. $\endgroup$ – D_S Feb 26 '14 at 18:11
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    $\begingroup$ In general $S$ is a semi-local Dedekind domain. And yes, tt is a DVR if K is complete. I would try to prove $R[x]$ is integrally closed. This implies immediately that $S=R[x]$. $\endgroup$ – Cantlog Feb 26 '14 at 18:14
  • $\begingroup$ Oh I didn't know that! I'll edit my question then. $\endgroup$ – D_S Feb 26 '14 at 18:15

I wasn't able to solve this using the hint in the comments, but I've figured out a solution nevertheless. Since $\mathfrak d = R$, we have $$ 0 = \nu_K(\mathfrak d) = \nu_K(N_{L/K}(\mathfrak D)) = \nu_L(\mathfrak D)$$ where $\mathfrak D$ is the different. Hence $S = \mathfrak D$.

Of course $g \in R[X]$ and $x \in S$. Furthermore $\overline{g}$ is the minimal polynomial of $\overline{x}$ over $k$, and since $k_L/k$ is separable we cannot have $\overline{g}'(\overline{\alpha}) = \overline{0}$, so $\nu_L(g'(\alpha)) = 0$. Hence also $g'(\alpha)S = S$.


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