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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.

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up vote 6 down vote accepted

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, MR1676316), see also Külshammers paper 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.

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You reduce the claim of the OP to the claim that whenever you have a block $B$ such that all characters in $\textrm{Irr}(B)$ are Galois conjugates of one another, then $B$ has defect zero. In literature (specifically: Curtis-Reiner: "Rep. Theory of fin. grps. and ass. alg." Theorem (90.4)) it is claimed that a block $B$ such that all characters in $B$ are $p$-conjugate is necessarily of defect zero. Here $p$-conjugate means "conjugate by an element of $Gal(\mathbb Q (\zeta_{|G|})/\mathbb Q(\zeta_{|G|_{p'}}))$". –  Florian Eisele Aug 8 '12 at 20:40
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(continued) Is it maybe possible to reduce your claim to this theorem? I believe it would suffice to show that (in your notation) $L \subseteq K\cdot \mathbb Q_p(\zeta_{|G|_p})$. This seems plausible, but I don't quite see how to show it at the moment. –  Florian Eisele Aug 8 '12 at 20:40
    
@F. Landish - Thanks! I guess I first need to read a lot more on modular representation theory before I fully understand this; I'll probably ask some more questions on specific points tomorrow. –  Henri Johnston Aug 8 '12 at 20:51
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There is a purely character theoretic proof that if chi is an irreducible character of G and with defect zero (which means that chi(1) is divisible by the full p-part of |G|), then chi(x) = 0 whenever x has order divisible by p. (This proof depends on Brauer's characterization of characters.) By an amazing result of Knorr, there is a very strong converse. If chi vanishes on every element of order p exactly, then chi has defect zero. –  Marty Isaacs Aug 17 '12 at 23:15
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@Marty Isaacs: Actually, Knörr's result is even stronger: if $\sum \chi(x) =0$, where the sum runs over elements of order $p$ exactly, then $\chi$ has $p$-defect zero. Even more, it is enough to assume that the $p$-part of $\chi(1)$ is strictly smaller than the $p$-part of the above sum. –  Frieder Ladisch Aug 20 '12 at 11:48

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