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I am at the moment reading a paper where they use some Algebraic Number Theory, which I know little of and I am curious about the following:

Let $L/K$ be a finite Galois extension and let v be a place of K that does not ramify L. If v is a non-archimedean place, we have a notion of an inertia group and a decomposition group. What can we, in general, say about the relation between the inertia group here and the absolute galois group of L? The Galois Group of L over K? I know this is a vague question, but I am looking for what we can say here, for example, can we say that the inertia group is a normal subgroup of the galois group?

Thanks in advance!

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I assume $L$ and $K$ are supposed to be number fields (or global fields). There are in general several decomposition (inertia) groups in $\mathrm{Gal}(L/K)$ attached to a finite prime $v$ of $K$, one for each finite prime of $L$ lying over $v$. If $w$ is such a prime of $L$, then the decomposition group $G_w$ is the subgroup of $\sigma\in\mathrm{Gal}(L/K)$ such that $\sigma w=w$ (if you think of $w$ as a prime ideal in the ring of integers in $L$, this means that $\sigma$ fixes the prime ideal, though not necessarily pointwise). Every element of $G_w$ gives rise to a well-defined automorphism of the residue field $k(w)$ of $w$, which can be thought of as $\mathscr{O}_L$ modulo the corresponding prime ideal. This gives a homomorphism from $G_w$ to $\mathrm{Gal}(k(w)/k(v))$. Its kernel is, by definition, the inertia group $I_w$ of $w$ in $\mathrm{Gal}(L/K)$. So the inertia group is a normal subgroup of the decomposition group. If $w^\prime$ is another prime of $L$ lying over $v$, then there is an element $\sigma\in\mathrm{Gal}(L/K)$ with $\sigma w=w^\prime$, and then the corresponding decomposition groups are conjugate (via $\sigma$). So, unless, e.g., there is only one prime of $L$ lying over $v$, the decomposition group $G_w$ will not be normal in $\mathrm{Gal}(L/K)$. The same goes for inertia groups.

To say that $v$ is unramified in $L/K$ in this context is to say that $I_w$ is trivial for every $w\mid v$ (or equivalently, for any $w\mid v$). So in this case the inertia group of $w$ is (trivially) a normal subgroup of $\mathrm{Gal}(L/K)$. When $G$ is abelian, conjugation by any element is the identity, so there really is a well-defined decomposition (resp. inertia) group attached to any prime $v$ of $K$, and they are of course normal in $\mathrm{Gal}(L/K)$.

I'm not really sure what you're asking about the absolute Galois group of $L$. This is a different object. Namely it is $\mathrm{Gal}(L_s/L)$ where $L_s$ is a choice of separable closure of $L$. The Galois group $\mathrm{Gal}(L/K)$ is can be regarded as a quotient of $\mathrm{Gal}(K_s/K)$ once an embedding of $L$ into $K_s$ is chosen. One can also define decomposition groups and inertia groups for infinite Galois extensions like $K_s/K$, and they have similar properties to the ones defined for finite Galois extensions. Also, if $\bar{w}$ is a prime of $K_s$ lying over the prime $w$ of $L$ which itself lies over $v$, then the image of the decomposition group $D(\bar{w})\subseteq\mathrm{Gal}(K_s/K)$ in $\mathrm{Gal}(L/K)$ is the decomposition group $D(w)$. The same goes for inertia groups.

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  • $\begingroup$ Thank you for your answer. I will wait with accepting it for just a little while, to get (maybe) more answers. However, I think your answer was very good. To you know if there is any general theory for when we can claim that the inertia group is equal to the decomposition group? $\endgroup$ – Dedalus Apr 2 '13 at 16:14
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    $\begingroup$ This happens when the extension of completions $L_w/K_v$ is totally ramified, meaning that the ramification index of $v$ in $L/K$ is equal to $[L_w:K_v]$, or equivalently, the residue extension $k(w)/k(v)$ is trivial, that is, $k(w)=k(v)$. $\endgroup$ – Keenan Kidwell Apr 2 '13 at 16:19
  • $\begingroup$ Hm, Appearently the paper deals with the case where the decomposition group is profinite (so I guess they are talking about the absolute galois group here). But the case carries over analogously? $\endgroup$ – Dedalus Apr 2 '13 at 16:33
  • $\begingroup$ For the decomposition and inertia group, yes. The decomposition (resp. inertia) groups in the Galois group of an infinite extension are naturally the inverse limits of appropriate decomposition (resp. inertia) groups in the Galois groups of the finite subextensions. $\endgroup$ – Keenan Kidwell Apr 2 '13 at 16:43
  • $\begingroup$ I am very thankful for your careful answer, it has helpen me a great deal. Thank you! $\endgroup$ – Dedalus Apr 2 '13 at 16:46

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