Let $G$ be a finite group and let $p$ be a prime dividing the order of $G$. Let $\chi$ be a $\mathbb{C}\_p$-valued irreducible character of $G$ and let $e\_{\chi} = |G|^{-1}\chi(1)\sum\_{g \in G} \chi(g^{-1})g$ be the associated primitive central idempotent in $\mathbb{C}\_p[G]$. Let $\mathbb{Q}\_{p}(\chi)=\mathbb{Q}\_p(\chi(g) : g \in G)$ be the character field. Let $H=\mathrm{Gal}(\mathbb {Q}\_{p}(\chi)/\mathbb{Q}\_p)$ and let $e=\sum\_{h \in H} e\_{\chi^h}$ ($H$ acts on characters in the usual way.) Then $e$ is a central primitive idempotent of $\mathbb{Q}\_p[G]$. Let $v\_p$ denote the usual $p$-adic valuation.


Claim: $v\_p(|G|)=v\_p(\chi(1))$ if and only if
$e \in \mathbb{Z}\_p[G]$ and $e\mathbb{Z}\_p[G]$ is a maximal $\mathbb{Z}\_p$-order.


If $v\_p(|G|)=v\_p(\chi(1))$ then it is clear that $e \in \mathbb{Z}\_p[G]$. That $e\mathbb{Z}\_p[G]$ is a maximal $\mathbb{Z}\_p$-order follows from Jacobinski's formula for the central conductor of $\mathbb{Z}\_p[G]$ in a maximal order (see Curtis-Reiner, Methods of representation theory, vol 1 section 27).


For the converse, I can prove the claim for $p$ odd again using Jacobinski's formula and some calculations of the different of the extension $\mathbb{Q}\_p(\chi)/\mathbb{Q}\_p$.


Question: can anyone provide a proof of counterexample for the missing part for $p=2$?


Here is a related claim that would prove the claim and make everything much simpler if true:
If $e \in \mathbb{Z}\_p[G]$ then $\mathbb{Q}\_p(\chi)/\mathbb{Q}\_p$ is unramified (i.e. $\mathbb{Q}\_p(\chi) \subseteq \mathbb{Q}\_p(\zeta\_n)$ for some $n$ relatively prime to $p$).


Also, maybe I can drop the maximal order part of the claim altogether?


I have a reasonable knowledge of ordinary representation theory but have only really started to look at modular representation theory in the past few days. I know that this is related to "blocks of defect zero", but in the books I have looked at (Serre, Curtis-Reiner) it is assumed that the ground field is "sufficiently large", which doesn't really help me. But I suspect this is an easy problem for someone who knows the subject well. 



Modular representation theory: central idempotents in $\mathbb{Z}_p[G]$
EDIT: There is a MUCH simpler proof than my first one, which I found after looking up the proof of (90.4) in Curtis-Reiner (Rep. Th. of finite groups...) mentioned by Florian Eisele in the comments:  

The central idempotent $e\in \mathbb{Z}\_p[G]$ is supported on $p$-regular elements. (By the way, a quite elementary proof of this fact can be given using the ideas in a paper of M. Leitz ([Proc. Amer. Math. Soc. 128 (2000), no. 11, 3149–3152](http://dx.doi.org/10.1090/S0002-9939-00-05422-8), [MR1676316](http://www.ams.org/mathscinet-getitem?mr=1676316)), see also Külshammers [paper](http://dx.doi.org/10.1017/S1446788700034881) cited at the end.)
Since 
$$ e = \frac{\chi(1)}{|G|}\sum\_{h\in H}\sum\_{g\in G} \chi(g^{-1})^h g, $$
it follows that the character $\beta= \sum\_{h\in H} \chi^h$ vanishes on $p$-singular elements. Let $P$ be a Sylow $p$-subgroup of $G$. Then $\beta$ vanishes on $P\setminus 1 $, so
the multiplicity of the trivial character of $P$ in the restriction $\beta\_P$ is 
$$ (\beta\_P, 1\_P) = \frac{\beta(1)}{|P|}= \frac{|H|\chi(1)}{|P|}. $$
On the other hand, we have
$$ (\beta\_P,1\_P) = \sum\_{h\in H} (\chi^h\_P,1\_P)= |H|(\chi\_P,1\_P). $$
Thus $|P|=|G|\_p$ divides $\chi(1)= |P|(\chi\_P,1\_P)$, q.e.d. 


