2
$\begingroup$

Let $K$ be a number field and $\Delta = Gal (K/\mathbb{Q})$ and $\chi: \Delta \rightarrow \mathbb{Z}_p^*$ be a non-trivial Dirichlet character, $e_{\chi} = (1/\mid \Delta \mid) \sum_{\sigma \in \Delta} \chi (\sigma) \sigma$ be the corresponding idempotent.

I am trying to understand why the following statement is true:

Let $p$ be a prime such that $p \not \mid [K:\mathbb{Q}]$. Since $K \subseteq \mathbb{Q}(\zeta_p) \cap \mathbb{R}$, we have that $\sum_{\chi} e_{\chi} = 1$, where $\chi$ runs over all $p$-adic Dirichlet characters of $\Delta$.

PS it's on page 15 of the paper of Thaine on Ideal Class Groups of real abelian number fields in Annals of Math, 128.

$\endgroup$
  • 1
    $\begingroup$ This is just a statement about the representation theory of compact groups (specifically, the decomposition of the regular representation); it has nothing to do with Galois theory. (Also I think that the right-hand side of the equality should be $1$ at the identity and $0$ elsewhere, not identically $1$.) $\endgroup$ – LSpice Nov 1 '18 at 0:47
  • 1
    $\begingroup$ Ah, I see that these aren't just complex representations, so I guess that there's some more to it. I don't think I understand the assumptions. You say that $K$ is a number field, and then speak of $[K : \mathbb Q_p]$; do you mean $[K : \mathbb Q]$? Then you also say $K \subseteq \mathbb Q_p$; is that really what you mean? $\endgroup$ – LSpice Nov 1 '18 at 1:34
  • $\begingroup$ The assumptions are: $[K:\mathbb{Q}]$ is not divisible by $p$ and $K\subseteq \mathbb{Q}(\mu_p)\cap \mathbb{R}$. $\endgroup$ – Anwesh Ray Nov 1 '18 at 3:01
  • $\begingroup$ Anent the recent edit, I assume that it should be (as @AnweshRay says) $K \subseteq \mathbb Q(\mu_p) \cap \mathbb R$, with a lowercase $p$, not $K \subseteq \mathbb Q(\zeta_P)$, with a capital $P$? $\endgroup$ – LSpice Nov 1 '18 at 14:39
1
$\begingroup$

The assumption $p\nmid [K:\mathbb{Q}]$ means that $p$ does not divide the order of $\Delta$, in particular, the assumption in the paper is that $K\subseteq \mathbb{Q}(\mu_p)$. By the decomposition $\mathbb{Z}_p^{\times}=\mu_{p-1}\times (1+p\mathbb{Z}_p)$ all $p$-adic characters $\chi$ have target in $\mu_{p-1}$. Expanding $\sum_{\chi} e_{\chi}$ you get $\sum_{\sigma\in \Delta} \{\frac{1}{\lvert\Delta\rvert}\sum_{\chi} \chi(\sigma)\}\sigma$. Since $K\subseteq \mathbb{Q}(\mu_p)$, the number of characters $\chi$ is equal to the order of $\Delta$. Observe that $\frac{1}{\lvert\Delta\rvert}\sum_{\chi} \chi(\sigma)=0$ if $\sigma\neq 1$ (once again use the assumption here, if $\sigma\neq 1$, there is character $\psi$ such that $\psi(\sigma)\neq 1$, multiply the expression by $\psi(\sigma)$ to see that it does not change) and is $1$ otherwise by the observation we just made. Therefore, $\sum_{\chi} e_{\chi}=1$.

$\endgroup$
  • 1
    $\begingroup$ A TeX note: $p \not| [K : \mathbb Q]$ ($p \not| [K : \mathbb Q]$) should be $p \nmid [K : \mathbb Q]$ ($p \nmid [K : \mathbb Q]$), and $\mid\Delta\mid$ (as in $a \mid\Delta\mid b$) should be $\lvert\Delta\rvert$ (as in $a \lvert\Delta\rvert b$), or at worst $|\Delta|$ (as in $a |\Delta| b$). I have edited accordingly. $\endgroup$ – LSpice Nov 2 '18 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.