Let $O_K$ be a dvr with fraction field $K$. Let $L/K$ be a tamely ramified finite Galois extension. Then, Abhyankar's Lemma implies that there exists a finite Galois extension $K^\prime/K$ such that the compositum $L^\prime$ of $K^\prime$ and $L$ is unramified over $K^\prime$. (This statement is wrong: QiL explains that one must normalize)

In other words, we're starting with a finite flat surjective morphism $Spec \ O_L\to Spec \ O_K$. Then, we make a base change along the morphism $Spec \ O_{K^\prime}\to Spec \ O_K$ and obtain an etale morphism $Spec \ O_{L^\prime} \to Spec \ O_{K^\prime}$. But doesn't faithful flat descent imply then that the morphism $Spec \ O_L\to Spec \ O_K$ was already etale?

Certainly not, but what am I misunderstanding here?


You are right and you are wrong.

First you are right because a morphism which becomes étale after a faithfully flat base change is certainly étale.

Second, your interpretation of Abhyankar's lemma is wrong. It says that the normalization of the tensor product is unramified over $O_{K'}$. In the finite flat Galois case, it is trivial, just take $O_{K'}=O_{L}$ and normalize the tensor product: you will get a finite disjoint union of copies of $O_L$. Before normalization, theses copies have intersections points in the closed fiber.

There is a more general result (See Epp, Inventiones Math. in the 70's) which says that Abhyankar's lemma holds also for wilde ramifications (and for more general type of extensions of DVR's, not necessarily finite).

  • $\begingroup$ By "some maximal ideal" do you mean "any maximal ideal", or does Abhyankar really mean that there is maybe just "one" maximal ideal? $\endgroup$ – Harry Jan 25 '13 at 8:48
  • $\begingroup$ I think I understand what I'm doing wrong now. Starting with $Y\to X$ as in the question, the base change gives $Y\times_X X^\prime \to X^\prime$. This is not necessarily etale, but once you normalize you obtain $Y^\prime\to X^\prime$ and this morphism is etale by Abhyankar. Now, if by some chance $Y^\prime = Y\times_X X^\prime$ (i.e., the base change is normal) then faithfully flat descent implies that the morphism we started out with was already etale. Thank you very much QiL!! $\endgroup$ – Harry Jan 25 '13 at 8:57
  • $\begingroup$ @Harry, yes I wrote it in a hurry, it is "any maximal ideal". $\endgroup$ – user18119 Jan 25 '13 at 16:20
  • $\begingroup$ Yes, I just wanted to be sure. Thank you for this. It completely cleared up everything. $\endgroup$ – Harry Jan 25 '13 at 17:00

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