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I'm a bit confused concerning tamely ramified covers of arithmetic schemes. I guess they would reduce to tamely ramified extensions of number fields, but they don't seem to do so. Let me elaborate:

First of all, let me recall the standard definition of tamely ramified covers of arithmetic schemes (by which we mean connected, flat, regular schemes of finite type over a Dedekind domain, or $\mathbb{Z}$ if someone prefer this). Let $D$ be a divisor on $S$ of an arithmetic cover $X\to S=\mathrm{Spec}(A)$, where $A$ is a Dedekind domain. Then $X\to S$ is said to be tamely ramified along $D$ if $X\times_S (S\setminus D)\to (S\setminus D)$ is finite and étale and tamely ramified along $D$.

Now, $X$ being connected and étale over $S\setminus D$ means that it is actually a $G:=\mathrm{Aut}_S(X)$-torsor (see Milne, Étale cohomology, page 40), that is, a Galois cover. Since being a $G$-torsor implies that $G$ acts freely and transitively on the fibers, the number of closed points in the fiber is equal to the order of the group $G$. But this means that the fibers all have the same number of points over $S\setminus D$.

Ok, so far so good. Restricting to rings of integers in Galois number fields is where my problem begins. So, let $X=\mathrm{Spec}(\mathfrak{o}_L)\to S=\mathrm{Spec}(\mathfrak{o}_K)$ be a cover of the associated arithmetic schemes to $L/K$.

In this case, a divisor $D$ is a $\mathit{finite}$ sum $\sum_\mathfrak{p} a_\mathfrak{p} \mathfrak{p}$, $a_i\in \mathbb{Z}$. So $X\to S$ being tamely ramified along $D$ means that $L/K$ is at most tamely ramified at the primes in $D$. Outside $D$, the primes can be unramified, inert and completely split (in $L$). $\mathit{However}$, the only possibility, comparing with tamely ramified coverings of arithmetic schemes is that $\mathfrak{p}\in S\setminus D$ is completely split, but there are infinitely many unramified and inert primes by the Cebotarev density theorem.

What the heck am I missing?

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I'm not completely clear on what you're trying to say in your last sentence. –  stankewicz Apr 8 '11 at 13:47
    
A $G$-torsor has the number of points as $G$ if it is split, i.e, if it has at least rational point. I think this is not true for your fibers. Non-trivial residue field extensions should lead to non-trivial torsor. –  Holger Partsch Apr 8 '11 at 14:02
    
@stankewicz: My point is that the points outside $D$ is completely split, but $D$ cannot possibly contain the unramified (i.e., unramified and NOT completely split) and inert primes since $D$ is finite and the number of unramified and inert primes is infinite. @Holger: Hm, could you elaborate. I'm not sure I understand you point. Thanks! –  Daniel Larsson Apr 8 '11 at 14:40
    
Aha, I think I see. If I would only have read the assumptions in Milne on page 39 where he defines the fiber functor, I would possibly have saved myself some embaressment. Milne observes that the definition implies that there is a morphism of fields $k(y)\to k(\bar{x})$ and this doesn't seem to be the case in the number field case above. –  Daniel Larsson Apr 8 '11 at 14:54
    
However, the point concerning G-torsors and fibers still stands. I'm totally confused right now. Mustn't $G$ act transitively and freely on the fibers; isn't this exactly what the condition $G\times_S X\approx X\times_S X$ in the definition of a G-torsor over S says? –  Daniel Larsson Apr 8 '11 at 18:27

1 Answer 1

I think the confusion concerns the definition of free action on a scheme. We do not demand that free actions act freely on the underlying topological space. Instead, we ask that for every scheme $S$, the induced action on the set of maps from $S$ to the scheme is free. It suffices to restrict the set of schemes $S$ to the spectra of fields, and even algebraically closed fields whose cardinality isn't too big.

For example, the action of $GL_n$ on its underlying variety by translation is free, even though the generic point is fixed. Another example: If you have a finite Galois extension $L/K$ of fields, then the action of the Galois group on $\operatorname{Spec} L$ is free, even though it may be a nontrivial group acting on a point.

In your case, each inert place is fixed by a nontrivial subgroup of Galois (the decomposition group), but the induced scheme-theoretic action of decomposition on a given point is free (i.e., there is trivial inertia).

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Thanks! I think you might be right (in fact, there's a squirrel in a tree outside my window mocking me as I write this). –  Daniel Larsson Apr 13 '11 at 11:19

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