Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

The $\mathbb{Z}_p$ rank of a finitely generated $\mathbb{Z}_p$-module simply ignores the torsion and computes the rank of the other (free) piece.

By Dirichlet's Unit theorem, $\mathcal{O}_K^{\times} \otimes \mathbb{Z}_p$ is a direct sum of a torsion piece (isomorphic to the $p$-power roots of unity in $K$, which here are just the $p$th roots of unity) with a free $\mathbb{Z}_p$-module. Thus, the $\chi$ component will be a direct sum of a free part (say of rank $r_{\chi}$) with a torsion part as well. Now the $\mathbb{Z}_p$-rank will simply ignore the torsion part and produce $r_{\chi}$.

$(\mathcal{O}_K^{\times} \otimes \mathbb{F}_p)^{\chi}$ is obtained from $\mathcal{O}_K^{\times} \otimes \mathbb{Z}_p$ by tensoring over $\mathbb{Z}_p$ with $\mathbb{F}_p$, so the free part will produce an $\mathbb{F}_p$ vector space of rank $r_{\chi}$ along with the same torsion part. This torsion part is either isomorphic to the $p$th roots of unity or is trivial, depending on whether the $p$th roots of unity live in the $\chi$-isotypic component or not. The dimension of this $\mathbb{F}_p$-vector space is thus $r_{\chi}$ plus the dimension of this torsion part.

The roots of unity happen to live in the $\omega$-isotypic component! (You can tell because "G acts on them through $\omega$".) So when $\chi = \omega$, the dimension of the $\mathbb{F}_p$-vector space includes a contribution from the roots of unity giving $r_{\chi}+1$, while otherwise, the dimension is just $r_{\chi}$. Since the $\mathbb{Z}_p$ rank just ignores torsion, you get $r_{\chi}$ in either case.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.