idempotents of a character

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.

• 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 $1$.) – LSpice Nov 1 '18 at 0:47
• 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? – LSpice Nov 1 '18 at 1:34
• The assumptions are: $[K:\mathbb{Q}]$ is not divisible by $p$ and $K\subseteq \mathbb{Q}(\mu_p)\cap \mathbb{R}$. – Anwesh Ray Nov 1 '18 at 3:01
• 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 $P$? – LSpice Nov 1 '18 at 14:39

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, $$\sum_{\chi} e_{\chi}=1$$.
• 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. – LSpice Nov 2 '18 at 17:19