TITLE: Question concerning the coefficients of block idempotents
QUESTION [6 upvotes]: Let $G$ be a finite group. Let $p$ be a prime number such that $p \mid |G|$.
Let Irr$(G)$ denote the set of ordinary irreducible characters of $G$.
For $\chi\in$ Irr$(G)$ define $e_{\chi} := \frac{\chi(1)}{|G|}\sum_{g\in G} {\chi(g^{-1})\cdot g}$.
Let $B$ denote a $p$-block of $G$. Let Irr$(B)$ be the set of those ordinary irreducible characters of $G$ which lie in $B$.
Define $e_B:=\sum_{\chi\in \text{Irr}(B)} {e_{\chi}}$. This can be rewritten as
$$e_B = \sum_{g\in G} {\bigg(\underbrace{\frac{1}{|G|}\sum_{\chi\in \text{Irr}(B)} {\chi(1)\chi(g^{-1})}}_{=:a_g}\bigg)} \cdot g.$$
Question:

Is it always true that $a_g$ lies in $\mathbb{Q}_p[\zeta]$ for $\zeta = \exp(\frac{2\pi i}{p^n-1})$ for some positive integer $n$?

Example:

Set $G:=A_5$ and $p:=5$. Then there are exactly two $p$-blocks of $G$.
In the coefficients of some of the $e_{\chi}$ there is the number $z:=\exp(\frac{2\pi i}{5})$ involved. The number $z$ cannot be expressed via the $\zeta$ above and $e_{\chi}$ does not lie in $\mathbb{Q}_p[\zeta]$ for $\zeta$ as above.
But the two block idempotents $e_{B_0}$ and $e_{B_1}$ do not involve $z$ anymore.

In other examples, the ${e_{\chi}}'s$ add up nicely to ${e_{B_j}}'s$, or such $z$'s don't appear at all.
If this is always the case, I would be fond of a reference.
Thank you very much for the help.

REPLY [9 votes]: Yes, this is true. By block orthogonality relations due to R. Brauer, it is true that $a_{g}=0$ whenever the element $g$ has order divisible by $p$.  But when $g$ has order prime to  $p$, it is clear that $a_{g}$ lies in $\mathbb{Q}[\omega]$ for some $t$-th root of unity $\omega$, where $t$ is not divisible by $p$.
Furthermore, the same block orthogonality relations imply that when the order of $g$ is prime to $p$, then the class function $\theta = \sum_{\chi \in B} \chi(g^{-1} )\chi$ vanishes of all non-identity elements of $P$, a Sylow $p$-subgroup of $G$. Taking the inner product of $\theta$
(restricted to $P$) with the trivial character of $P$ then shows that $[G:P] a_{g}$ is an algebraic integer.
These two facts give what you want. The necessary block orthogonality relations can be found in almost any text which deals with blocks and characters, eg those by G.Navarro  or by M. Isaacs.