Hello,


I've been reading the excellent online book on Algebraic Number Theory by J.S.Milne. In the section described above there is a footnote maintaining that the separability of the residue field extension implies the separability of the original field extension in the unramified case (base field complete w.r.t. a discrete valuation). Unfortunately, I haven't been able to find a reference nor been able to supply a proof of this myself. Any help would be greatly appreciated ! Kind regards, Stephan.



Unramified (finite) extensions of fields complete with respect to a discrete valuation
Let $R\to R'$ be a such extension of DVR, let $k'/k$ be the residue extension. Lift a generator of $k'/k$ to $\theta\in R'$. Then $R'=R[\theta]$ by Nakayama. The minimal polynomial $P(X)\in R[X]$ of $\theta$ has degree $[k':k]$, and it is separable because it is separable in the residue field (consider the GCD of $P(X)$ and $P'(X)$ if you want). Therefore $\mathrm{Frac}(R')/\mathrm{Frac}(R)$ is separable. 


[**EDIT**] Some more explanations. Let $[k':k]=d$ and let $\pi$ be a uniformizing element of $R$. Then $N:=R+\theta R + \dots +\theta^{d-1}R$ is a finite submodule of $R'$ and we have $R'=N+\pi^n R'$ for any $n\ge 1$. 


Let us show $R'=N$. Let $b\in R'$. For all $n\ge 1$, we have $b=a\_n+\pi^n b\_n$ with $a\_n\in N$ and $b\_n\in R'$. This implies that the sequence $(a\_n)\_n$ is Cauchy. Now use the hypothesis that $R$ is complete. It implies that $N$ is complete, hence the $a\_n$ converge in $N$ and $b\in N$, thus $R'=N=R[\theta]$. 


Now use the hypothesis $R'/R$ is unramified: $k'=R'/\pi R'=R[X]/(\pi, P(X)))=k[X]/(p(X))$, where $p(X)$ is the image of $P(X)$ in $k[X]$. Therefore $p(X)$ is separable because $k'/k$ is. 


