In a paper I'm currently reading, they have the following situation:
$k$ is some number field that doesn't have a primitive $p^{th}$ root of unity, and $k(\zeta_p)$ a field above it with Galois group $\Delta (=(\mathbb{Z}/p\mathbb{Z})^{\times})$. For any $\chi \in \hat{\Delta}:=Hom(\Delta, \mathbb{Z}_p^{\times})$ they define $e_{\chi}:= \frac{1}{|\Delta|} \sum_{\delta \in \Delta} \chi(\delta)\delta^{-1}$ (I recognize this from normal representation theory) and $M^{\chi}:=e_{\chi}M$ (so $M=\oplus_{\chi \in \hat{\Delta}} M^{\chi}$).
They define the "cyclotomic character" as $\delta(\zeta_p)=\zeta_p^{\omega(\delta)}$, which doens't make a whole lot of sense since this only determines $\omega(\delta)$ mod $p$. But since it would have to be a $p^{th}$ root of unity in $\mathbb{Z}_p^{\times}$, it does determine it, and therefore does make sense.
Now they say that if $K$ is the compositum of $k(\zeta_p)$ with a $G$-Galois extension of $k$ that is linearly disjoint from $k(\zeta_p)$ then (is all of this necessary? probably not):
$dim_{\mathbb{F}_p}(O_K^{\times} \otimes \mathbb{F}_p)^{\chi}$ is $rank_{\mathbb{Z}_p}(O_K^{\times} \otimes \mathbb{Z}_p)^{\chi}$ if $\chi \neq \omega$ and $rank_{\mathbb{Z}_p}(O_K^{\times} \otimes \mathbb{Z}_p)^{\chi} +1$ if $\chi=\omega$.
I don't know what the intuition is for this fact! Would this be true for any $\mathbb{Z}_p[\Delta]$-module, $M$? It doesn't seem so...
What makes $\omega$ special? What makes $M^{\omega}$ special? If there's anything you can contribute that would give me better intuition, it would be much appreciated!