# Doesn't Abhyankar's lemma contradict faithful flat descent (no, but I'm confused)

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?

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.
• 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!! – Harry Jan 25 '13 at 8:57