TITLE: Unramified (finite) extensions of fields complete with respect to a discrete valuation
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: The statement that every unramified extension is separable holds in fact for any Henselian base field $(K,v)$, and Nakayama's lemma is unneccesary for the proof. It is sufficient to prove the assertion for finite extensions. So let $L|K$ denote a finite unramified extension, and let $\lambda|\kappa$ denote the corresponding finite field extension. Since $[\lambda:\kappa] = [L:K]$ is finite and $\lambda|\kappa$ is separable, we can find some $\bar{\alpha} \in \lambda$ such that 
$\lambda = \kappa(\bar{\alpha})$ holds. Let $\alpha \in \mathcal{O}_L$ denote 
a preimage. Since $\kappa(\bar{\alpha})$ is the residue class field of $K(\alpha)$ and every subextension of an unramified extension is unramified, we obtain
$$
[L:K] = [\lambda:\kappa] = [\kappa(\bar{\alpha}):\kappa] = [K(\alpha):K]
$$
and $L = K(\alpha)$. Let $f \in \mathcal{O}_K[x]$ denote the minimum polynomial
of $\alpha$ over $K$. Then the image $\bar{f} \in \kappa[x]$ is the minimum polynomial 
of $\bar\alpha$ over $\kappa$. If $f$ were inseparable, then $f' = 0$ and 
$\bar{f}' = 0$. But the latter contradicts the separability of $\bar\alpha$
over $\kappa$.