# Unramified (finite) extensions of fields complete with respect to a discrete valuation

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.

-

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.

-
First of all thank you very much for your answer (this is my first time here, and I'm not sure whether I'm posting my reply in the correct spot !). I don't quite understand, however, why the image of $P(X)$ in $k[X]$ is separable (I only know it is divisible by the minimal polynomial of the image of $\theta$ in $k'$ over $k$ ?! (I guess I'm just a bit slow tonight !) Kind regards, Stephan. – Stephan F. Kroneck Mar 6 '11 at 22:45
Or is this due to the degrees of the polynomials involved, forcing the image of $P(X)$ in $k[X]$ to coincide with the minimal polynomial of the image of $\theta$ in $k′$ over $k$ ? Stephan. – Stephan F. Kroneck Mar 6 '11 at 23:18
I added some details. – Qing Liu Mar 6 '11 at 23:38
Excellent ! Thank you again ! Stephan. – Stephan F. Kroneck Mar 8 '11 at 14:32

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 $\kappa$.

-