Let $A$ be discrete valuation domain, and $K$ be quotient field of $A$. Let $L$ be a finite extension of $K$ and $B$ be the integral closure of $A$ in $L$.Does separability of residue fields implies separability of $L/K$
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3$\begingroup$ No, since the residue field degree can be $1$. Let $F$ be a field of characteristic $p$, $K$ be the Laurent series field $F((X))$, and $A = F[[X]]$. The residue field of $A$ is $A/(X) = F$. Let $L = K(\sqrt[p]{X}) = F((\sqrt[p]{X}))$, so $L/K$ is an inseparable extension. The integral closure of $A$ in $L$ is $B = A[\sqrt[p]{X}] = F[[\sqrt[p]{X}]]$ and the residue field of $B$ is $B/(\sqrt[p]{X}) = F$ so that the residue field extension is just $F$ extended to $F$. $\endgroup$– KConradCommented Apr 23, 2019 at 7:31
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3$\begingroup$ And conversely, separability of $L/K$ does not imply separability of the residue field extension. See kconrad.math.uconn.edu/blurbs/gradnumthy/…. $\endgroup$– KConradCommented Apr 23, 2019 at 7:36
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$\begingroup$ could you please help me to understand why B is the integral closure of A in L $\endgroup$– SUNIL PASUPULATICommented Apr 24, 2019 at 6:14
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$\begingroup$ Easily $B$ is integral over $A$. To show $B$ is integrally closed in its fraction field $L$ there are a few methods: (1) $B$ is the valuation ring of the complete valued field $L$ (with respect to the $\sqrt[p]{X}$-adic topology), and valuation rings are integrally closed (2) $B$ is a DVR (it's a power series ring in $\sqrt[p]{X}$ over the field $F$), so it's a PID in a simple way and PIDs are integrally closed, (3) $L/K$ is totally ramified (you are adjoining a root of $T^p-X$ to $K$) and the integral closure in that case is generated over $A$ as a ring by a prime element ($\sqrt[p]{X}$). $\endgroup$– KConradCommented Apr 24, 2019 at 19:43
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$\begingroup$ Check that you can prove similar things over the $p$-adics first: in a finite extension $L/\mathbf Q_p$ the integral closure $\mathcal O_L$ of $\mathbf Z_p$ is the unit disc of $L$ and if $L/\mathbf Q_p$ is totally ramified with prime element $\pi$ in $\mathcal O_L$ then $\mathcal O_L = \mathbf Z_p[\pi]$. $\endgroup$– KConradCommented Apr 24, 2019 at 19:45
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