MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
up vote 9 down vote accepted

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.

share|cite|improve this answer
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 $\kappa$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.