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Let $K|k$ be a finite Galois extension of number fields, inside a given "maximal" (infinite) Galois extension $k_S$ of $k$. Let $G = Gal(k_S | K)$ denote the Galois group and let $A$ be a $Gal(k_S|k)$-module. In a paper the following is stated

Let $K|k$ be large enough so that $G$ acts trivially on $A$.

I now don't understand how can $A$ be trivialized by a bigger extension? I know that $G$ acts trivial on $A$ if and only if

$$g a = a \text{ for every } g \in G, a\in A.$$ Also, if the extension $K$ over $k$ gets bigger, the group $G$ gets smaller. Furthermore $A$ is a $G$ module via the continuous inclusion $G \rightarrow Gal(k_S | K)$.

But why does such a finite extension as mention above exist? For infinite extensions the problem is clear to me: Just take $K = k_S$.

Every hint is appreciated. Thank you a lot!

Tom

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It's definitely not true in general that if $G$ acts continuously on a topological abelian group $A$ that $A$ is point-wise fixed by an open subgroup of $G$ (note that every open subgroup of $G$ has the form $\mathrm{Gal}(k_S/K)$ for $K/k$ finite). Are there conventions implicit in the statement that the author is making? Is $A$ discrete? Often a ``Galois module" is implicitly taken to be discrete, meaning that each element in $A$ is fixed by an open subgroup, but if $A$ is not finite, or, say, finitely generated as a $\mathbf{Z}$-module, there's no reason that every element of $A$ should be fixed by a single open subgroup (if you intersect infinitely many open subgroups of $G$ you're going to get a closed, but not open subgroup, in general. If however $A$ is finite, or more generally is finitely generated over $\mathbf{Z}$, then you can intersect the stabilizers of the finitely many generators to get an open subgroup (finite intersection of open is open) which will fix every generator, and hence every element of $A$ by linearity of the $G$-action.

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  • $\begingroup$ Thanks for your answer :-) You were right. $A$ is a discrete, finitely generated (as $\mathbb{Z}$-module) abelian group, though in general not finite. It was implicitly assumed because of the category, $A$ is taken from. $\endgroup$ – BIS HD Feb 15 '14 at 16:35
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    $\begingroup$ Yes, so then indeed it will be point-wise fixed by an open subgroup of the Galois group. $\endgroup$ – Keenan Kidwell Feb 15 '14 at 16:56

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