Homomorphisms from the additive groups of virtual characters into certain idele groups This is a question from Frohlich's book 'Galois Module Structure of Algebraic Integers', Ch.1. 
Let $K$ be a number field and $\Omega_K=\text{Gal}(K^c/K)$ where $K^c$ is the separable closure of $K$. Also let $\Gamma$be any finite group and $R_\Gamma$ ring of virtual characters of $\Gamma$. Next let $\mathfrak I(\mathbb Q^c)$ be the direct limit of the idele groups $\mathfrak I(E)$ as $E$ runs over the number fields in $\mathbb Q^c$. The values of the virtual characters of $\Gamma$ lie in some number field $F$ containing $K$.
The book states that $$\text{Hom}_{\Omega_K}(R_\Gamma,\mathfrak{I}(\mathbb Q^c))=\text{Hom}_{\Omega_K}(R_\Gamma,\mathfrak{I}(F))$$
How can we show this? For a start, how do we know that the LHS is a subgroup of the RHS? 
Many thanks for your help.
 A: Suppose you have a group homomorphism $f$ from $R_\Gamma$ to $\mathfrak{I}(\mathbb Q^c)$. To say that it's an $\Omega_K$-homomorphism is to say that for any $\sigma\in \Omega_K$ and any $\chi\in R_\Gamma$, $\sigma(f(\sigma^{-1}\chi))=f(\chi)$. This is the usual action of a group $G$ on homs between two $G$-modules. In particular, this should be true for $\sigma$ that lie in the subgroup $\Omega_F$. But by definition of $F$, $\sigma^{-1}\chi=\chi$ for all such $\sigma$, so you are demanding that $\sigma(f(\chi))=f(\chi)$ for all $\chi\in R_\Gamma$. In other words, you want the image of $f$ to land in the part of $\mathfrak{I}(\mathbb Q^c)$ that is invariant under $\Omega_F$. Finally, you use Galois descent for idele groups to conclude that the image of any $f$ that commutes with Galois must land in $\mathfrak{I}(\mathbb Q^c)^{\Omega_F} = \mathfrak{I}(F)$, which is precisely what Fröhlich claims.
Note that we didn't even use that $f$ was an $\Omega_K$-hom, merely that it was an $\Omega_F$-hom.
Addendum: Galois descent for idele groups says that if $L/k$ is a finite Galois extension of number fields with Galois group $\Gamma$, then $\mathfrak{I}(L)^\Gamma = \mathfrak{I}(k)$. This is very easy to convince yourself of. You can then also easily see that the same is true for a direct limit of finite extensions. In the above argument, this is applied with $L=\mathbb Q^c$, and with $k=F$. This is discussed a bit in the book by Neukirch, Wingberg, Schmidt, Cohomology of Number Fields.
