Let $K$ be the field of fractions of a strictly Henselian discrete valuation ring $A$. Following Milne ("Etale Cohomology", Chapter I, Paragraph 5, example e)), $Spec(K)$ is the algebraic analogue of a punctured disc in the plane.

The cited example is about the calculation of the tame fundamental group (in the Grothendieck sense) of that disc. Milne, citing Serre ("Corps Locaux", Chapter IV), says that every tamely ramified extension of $K$ (so every tamely ramified covering of $Spec(K)$) is a Kummer extension of $K$. Unfortunately the book of Serre seems to explain only the restrictive case: $K=\mathbb{C}((z))$. Why this is true in general? In specific the problems seem to arise when the characteristic of the residue field of $A$ is not zero anymore.

I thought also to have problems, related to this question, regarding the possible non algebraic closure of the residue field of $A$. But, since we are looking only for tamely ramified extensions, the strict Henselianity of $A$ is enough to avoid those problems.

Moreover I'm interested in the more general case of $A$ being the strict Henselianization of the stalk $\mathcal{O}_{X,x}$ of a sheaf on a point $x$ of a scheme $X$, so $A$ being strictly Henselian but not a discrete valuation ring anymore. I would like to have the same result (or something similar) als in this situation. So, how much is deep the request of $A$ being a discrete valuation ring? Or, can I restrict to that case in any situation?

Thank you for your time.


Here is an answer to your question in the context of DVRs.

The argument with ramification filtrations from Serre shows that if $L/K$ is tame, then $Gal(L/K)$ embeds into $\mathcal O_L^{\times}/(1+\mathfrak m_L)$, and hence is cyclic (being isomorphic to a finite subgroup of the multiplicative group of a field). Since $K$ contains all prime-to-$p$ power roots of $1$, by the strict Henselian assumption (here $p$ is the char. of the residue field $\mathcal O_K/\mathfrak m_K$), we see that $L/K$ is a Kummer extension, i.e. obtained by extracting the $n$th root of some element of $K$, for some $n$ prime to $p$. Again by strict Henselianess, the $n$th root of all elements of $O_K^{\times}$ already lie in $K$ (using the fact that $n$ is prime-to-$p$), and so in fact $L$ comes by extracting the $n$th root of a uniformizer of $\mathcal O_K$.

If I can get my thoughts in order, I might come back and address the non-DVR aspects of the question as well.

  • $\begingroup$ Dear Matt, thank you for your answer. I took some time in order to have a clear picture of your answer, for me seems that everything fits. If I discover something about the non-DVR part I will write here. +1 and accepted answer! =) $\endgroup$ – Giovanni De Gaetano Jun 15 '11 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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