# The Field of Values of the Height Zero Characters

Gabriel Navarro  Departament de Matemàtiques, Universitat de València, 46100
Burjassot, València, Spain<EMAIL_ADDRESS>,  Lucas Ruhstorfer  School of
Mathematics and Natural Sciences University of Wuppertal, 42119 Wuppertal,
Germany<EMAIL_ADDRESS>,  Pham Huu Tiep  Department of
Mathematics, Rutgers University, Piscataway, NJ 08854, USA
<EMAIL_ADDRESS>and  Carolina Vallejo  Dipartimento di Matematica e
Informatica ‘Ulisse Dini’, 50134 Firenze (Italy)
<EMAIL_ADDRESS>
###### Abstract.

We determine what are the fields of values of the irreducible $p$-height zero
characters of all finite groups for $p=2$; we conjecture what they should be
for odd primes, and reduce this statement to a problem on blocks of quasi-
simple groups.

###### 2010 Mathematics Subject Classification:

Primary 20D20; Secondary 20C15

The research of the first author is supported by Ministerio de Ciencia e
Innovación PID2019-103854GB-I00. The third author gratefully acknowledges the
support of the NSF (grants DMS-1840702 and DMS-2200850), the Simons
Foundation, and the Joshua Barlaz Chair in Mathematics. The fourth author
acknowledges support from the Rita Levi Montalcini Program (bando 2019) and
from the INdAM-GNSAGA

Part of this work was done when the third author visited Princeton University
and MIT. It is a pleasure to thank Princeton University and MIT for generous
hospitality and stimulating environment.

##  1\. Introduction

Every Abelian number field can be realized as the field of values of a complex
irreducible character of a finite group (see, for instance, Theorem 2.2 of
[NT]). Motivated by the McKay conjecture and the McKay-Navarro conjecture
[Nav1, Conjecture A], it is of great interest to characterize the field of
values of the irreducible characters of degree not divisible by a fixed prime
$p$. This was established in [NT] for $p=2$ following seminal work on the
fields of values of odd-degree characters in [ILNT]. For instance, the
quadratic field of values of the odd-degree irreducible characters of all
finite groups are exactly the fields ${\mathbb{Q}}$, ${\mathbb{Q}}(i)$ and
${\mathbb{Q}}(\sqrt{d})$, where $d\equiv 1$ mod 4 is a square-free number.

The characters of degree not divisible by $p$ constitute only a part of the
Brauer’s $p$-height zero characters, namely those that lie in $p$-blocks of
maximal defect. The McKay conjecture admits a version for Brauer’s $p$-blocks
(the celebrated Alperin–McKay conjecture) where characters of degree not
divisible by $p$ are replaced by $p$-height zero characters, and this
conjecture also admits a version taking into account the action of Galois
automorphisms [Nav1, Conjecture B] (which is sometimes referred to as the
Alperin-McKay-Navarro conjecture).

The main question now is: What are then the field of values of the 2-height
zero characters? As we have mentioned, ${\mathbb{Q}}(\sqrt{3})$, say, cannot
be the field of values of an irreducible character of odd degree, but it is
easy to find many 2-height zero characters having this field of values. (For
instance, in a double cover of $\sf S_{5}$.) However, ${\mathbb{Q}}(\sqrt{2})$
or ${\mathbb{Q}}(\sqrt{-2})$, say, do not appear to be the field of values of
any 2-height zero character.

When studying fields of values of characters, character conductors are a
fundamental invariant. If $\chi$ is a character of a group $G$ and
${\mathbb{Q}}(\chi)$ is the smallest field extension of ${\mathbb{Q}}$
containing the values of $\chi$, then we define $c(\chi)$, the conductor of
$\chi$, to be the smallest integer $n$ such that ${\mathbb{Q}}(\chi)$ is
contained in the $n$-th cyclotomic field
${\mathbb{Q}}_{n}={\mathbb{Q}}(e^{2\pi i/n})$. If $F$ is any subfield of
${\mathbb{C}}$, then we write $F(\chi)=\langle F,{\mathbb{Q}}(\chi)\rangle$.

The following is the main result of this paper. Its proof uses the
Classification of Finite Simple Groups, together with the work of [BoDR], and
its refinement by [KL].

######  THEOREM A.

Let $G$ be a finite group, and let $\chi\in{\rm Irr}(G)$ of 2-height zero.
Write $c(\chi)=2^{a}m$, where $m$ is odd and $a\geq 0$. Then
${\mathbb{Q}}_{2^{a}}\subseteq{\mathbb{Q}}_{m}(\chi)$.

In fact, the fields of values of the 2-height zero irreducible characters can
be characterized in the following way. Let $\mathcal{F}_{2}$ be the set of
Abelian number fields $F$ such that
${\mathbb{Q}}_{n}=\langle{\mathbb{Q}}_{m},F\rangle$, where here $n$ is the
conductor of the field $F$, and $n=2^{a}m$, for some odd number $m$.

######  THEOREM B.

The set consisting of the fields of values of the 2-height zero characters of
finite groups is exactly $\mathcal{F}_{2}$.

As a consequence of Theorem B we obtain that the following are the quadratic
number fields that appear as fields of values of $2$-height zero characters.

######  COROLLARY C.

Let $F$ be a quadratic number field. Then $F={\mathbb{Q}}(\chi)$ for some
2-height zero character $\chi$ if and only if
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\sqrt{d})$ where $d\neq 1$ is an odd square-
free integer.

In this context, it is natural to wonder what happens for odd primes. The
fields of values of the characters of degree not divisible by $p$ are
conjectured to be precisely the Abelian number fields $F$ such that
$[{\mathbb{Q}}_{p^{a}}:{\mathbb{Q}}_{p^{a}}\cap F]$ is not divisible by $p$ in
[NT, Conjecture C]. This conjecture does not seem to follow from the McKay-
Navarro conjecture [Nav1, Conjecture A]. As happens in the case of characters
of degree not divisible by $p$, we can only conjecture what the fields of
values of the $p$-height zero characters should be for odd primes. This is
Conjecture D below. The novelty is that we can show that the statement of
Conjecture D follows from the statement of the Alperin-McKay-Navarro
conjecture.

For any prime $p$, let $\mathcal{F}_{p}$ be the set of Abelian number fields
$F$ with conductor $n=p^{a}m$, where $p$ does not divide $m$, such that the
degree $|{\mathbb{Q}}_{n}:\langle{\mathbb{Q}}_{m},F\rangle|$ is not divisible
by $p$. Notice that the fields $F$ with $p$-part of the conductor $p^{a}$ such
that $[{\mathbb{Q}}_{p^{a}}:{\mathbb{Q}}_{p^{a}}\cap F]$ is not divisible by
$p$ are a subclass contained in $\mathcal{F}_{p}$.

######  CONJECTURE D.

The set of fields of values of the $p$-height zero characters of finite groups
is exactly $\mathcal{F}_{p}$.

We show that any field in $\mathcal{F}_{p}$ is the field of values of a
$p$-height zero character in Theorem 5.1. Hence settling one of the
containments in Conjecture D (and also reducing the proof of Theorem B to
proving Theorem A). We reduce the verification of the other containment to a
problem on quasi-simple groups in Theorem 6.3.

We show that the statement of Conjecture D follows from the statement of the
Alperin-McKay-Navarro conjecture [Nav1, Conjecture B] in Theorem 5.3. A
fundamental part of our work is devoted to show that Theorem A is true for
quasi-simple groups. We believe that this will be useful in the final
verification of the Alperin-McKay-Navarro conjecture.

##  2\. Conductors

Let us start by recording some elementary results on characters and conductors
that we will frequently use. Recall that if $\psi$ is a character of a finite
group, then $\psi(g)$ is a sum of $o(g)$-th roots of unity for $g\in G$, and
therefore the field of values ${\mathbb{Q}}(\psi)$, which is the smallest
field containing $\psi(g)$ for all $g\in G$, is contained in
${\mathbb{Q}}_{|G|}={\mathbb{Q}}(e^{2\pi i/|G|})$. The conductor $c(\psi)$ is
the smallest $n$ such that
${\mathbb{Q}}(\psi)\subseteq{\mathbb{Q}}_{n}={\mathbb{Q}}(e^{2\pi i/n})$.
Therefore $c(\psi)$ divides $|G|$. Moreover,
${\mathbb{Q}}(\psi)\subseteq{\mathbb{Q}}_{m}$ if and only if $c(\psi)$ divides
$m$. If $F$ is an Abelian number field, that is $F\subseteq{\mathbb{C}}$ and
$F/{\mathbb{Q}}$ is a Galois extension with ${\rm Gal}(F/{\mathbb{Q}})$
abelian, then the Kronecker-Weber theorem implies that
$F\subseteq{\mathbb{Q}}_{n}$ for some $n$ and $c(F)$, the conductor of $F$, is
the smallest such $n$. By elementary Galois theory, recall that $c(\langle
F_{1},F_{2}\rangle)$ is the least common multiple of $c(F_{1})$ and
$c(F_{2})$.

In this paper, if $p$ is a prime and $n\geq 1$ is an integer, then $n_{p}$ is
the largest power of $p$ dividing $n$, and $n_{p^{\prime}}=n/n_{p}$. We call
$n_{p}$ the $p$-part of $n$ and $n_{p^{\prime}}$ the $p^{\prime}$-part of $n$.
For a fixed prime $p$, we are interested in the $p$-parts of conductors. If
$\psi$ is a character and $c(\psi)_{p}=1$, then $\psi$ is called $p$-rational.
If $p=2$, $\psi$ is either 2-rational or $c(\psi)_{2}\geq 4$. Notice that if
$\chi$ is linear character then $c(\chi)=o(\lambda)$ unless $o(\lambda)_{2}=2$
in which case $c(\chi)=o(\lambda)/2$.

######  Lemma 2.1.

Let $p$ be a prime. Suppose that $\chi\in{\rm Irr}(G)$, and write
$c(\chi)=p^{a}m$, where $n$ is not divisible by $p$. If $n$ is a natural
number not divisible by $p$ with
${\mathbb{Q}}_{p^{f}}\subseteq{\mathbb{Q}}_{pn}(\chi)$, then
${\mathbb{Q}}_{p^{f}}\subseteq{\mathbb{Q}}_{pm}(\chi)$. Moreover $f\leq a$
unless possibly when $f=1$ and $a=0$.

###### Proof.

By replacing $n$ by $mn$, we we may assume that $m$ divides $n$. If $a=0$ and
$f=1$ then ${\mathbb{Q}}_{p}\subseteq{\mathbb{Q}}_{pm}(\chi)$.

Hence we may assume that $a\geq 1$. Then $a\geq 2$ if $p=2$. In either case
${\mathbb{Q}}_{p^{f}}\subseteq{\mathbb{Q}}_{pn}(\chi)\subseteq{\mathbb{Q}}_{p^{a}n}(\chi)={\mathbb{Q}}_{p^{a}n}$
because $m$ divides $n$, so $f\leq a$. If $p=2$, we may also assume that
$f\geq 2$, because otherwise the result is trivial.

Write $F={\mathbb{Q}}_{n}$, $K={\mathbb{Q}}_{m}$, $L={\mathbb{Q}}_{p^{a}m}$
and $E=\langle F,L\rangle={\mathbb{Q}}_{p^{a}n}$. We have that $F\cap L=K$.
Let $J={\mathbb{Q}}_{pm}(\chi)$, so that $K\subseteq J\subseteq L$. Let
$M={\mathbb{Q}}_{pn}(\chi)=\langle F,J\rangle$. Since
${\mathbb{Q}}_{p^{f}}\subseteq M\subseteq{\mathbb{Q}}_{p^{a}n}$, we have that
$f\leq a$. Now, ${\mathbb{Q}}_{p^{f}}\subseteq M\cap L=J$, by Lemma 2.6(i) of
[NT], for instance. ∎

######  Lemma 2.2.

Let $p$ be a prime. Suppose that $\chi$ and $\psi$ are characters of groups
$G$ and $H$. Suppose that ${\mathbb{Q}}_{pn}(\chi)={\mathbb{Q}}_{pn}(\psi)$
for some $n$ not divisible by $p$.

  1. (i)

If $n$ divides $m$, then ${\mathbb{Q}}_{pm}(\chi)={\mathbb{Q}}_{pm}(\psi)$.

  2. (ii)

If $p=2$, or $p$ is odd and $c(\chi)_{p},c(\psi)_{p}\geq p$, then
$c(\chi)_{p}=c(\psi)_{p}$.

###### Proof.

To prove part (i) just notice that

${\mathbb{Q}}_{pm}(\chi)={\mathbb{Q}}(\zeta_{p},\zeta_{m},\chi)={\mathbb{Q}}(\zeta_{p},\zeta_{n},\zeta_{m},\chi)={\mathbb{Q}}_{np}(\chi)(\zeta_{m})={\mathbb{Q}}_{np}(\psi)(\zeta_{m})={\mathbb{Q}}_{pm}(\psi).$

To prove part (ii) notice that
${\mathbb{Q}}(\psi)\subseteq{\mathbb{Q}}_{pn}(\chi)\subseteq{\mathbb{Q}}_{c(\chi)_{p}m}$
with $m=nc(\chi)_{p^{\prime}}$. In particular $c(\psi)_{p}$ divides
$c(\chi)_{p}$. By reversing the roles played by $\chi$ and $\psi$ we obtain
the result. ∎

##  3\. Fields and Height Zero Characters

Our notation for blocks follows [Nav2]. We will frequently use the following
facts on height zero characters.

######  Theorem 3.1.

Let $B$ be a $p$-block of a finite group $G$, and let $\chi\in{\rm Irr}(B)$
with height zero.

  1. (i)

If $\psi^{G}=\chi$, where $\psi\in{\rm Irr}(H)$ of some subgroup $H$ of $G$,
then $\psi$ has height zero in its $p$-block, and any defect group of the
block of $\psi$ is a defect group of $B$.

  2. (ii)

If $N\triangleleft\,G$ and $\theta\in{\rm Irr}(N)$ is under $\chi$, then
$\theta$ has height zero.

###### Proof.

Part (i) is Proposition 2.5(e) of [NS]. Part (ii) is due to M. Murai, and is
Proposition 2.5(a) of [NS]. ∎

######  Theorem 3.2.

Let $p$ be a prime, and suppose that $\chi\in{\rm Irr}(G)$ has height zero in
its $p$-block. Let $N\triangleleft\,G$, let $\theta\in{\rm Irr}(N)$ be under
$\chi$, and let $\psi\in{\rm Irr}(T|\theta)$ be the Clifford correspondent of
$\chi$ over $\theta$. Then $\psi$ and $\theta$ have height zero. Also,
${\mathbb{Q}}_{pn}(\chi)={\mathbb{Q}}_{pn}(\psi)$, where $n=|G|_{p^{\prime}}$.
Therefore, if $p=2$ or $p$ odd and $c(\chi)_{p}\geq p$, then
$c(\chi)_{p}=c(\psi)_{p}$.

###### Proof.

We have that $\psi$ and $\theta$ have height zero by Theorem 3.1. We argue by
induction on $|G:N|$. Since $\psi^{G}=\chi$, we have that
${\mathbb{Q}}(\chi)\subseteq{\mathbb{Q}}(\psi)$. Let $T^{*}$ be the semi-
inertia group of $\theta$ in $G$ consisting of the elements $g\in G$ for which
there is some $\sigma\in{\rm Gal}({\mathbb{Q}}(\theta)/{\mathbb{Q}})$ such
that $\theta^{g}=\theta^{\sigma}$, as in Problem 3.9 of [Nav3]. Let
$\eta=\psi^{T^{*}}$. Since $\eta^{G}=\chi$, then we have that $\eta$ has
height zero. Also, ${\mathbb{Q}}(\chi)={\mathbb{Q}}(\eta)$,
$T\triangleleft\,T^{*}$ and $T^{*}/T$ is abelian.

If $T^{*}<G$, by induction, we have that
${\mathbb{Q}}_{p|T^{*}|_{p^{\prime}}}(\eta)={\mathbb{Q}}_{p|T^{*}|_{p^{\prime}}}(\psi)$.
By Lemma 2.2(i), we have that
${\mathbb{Q}}_{pn}(\psi)={\mathbb{Q}}_{pn}(\eta)={\mathbb{Q}}_{pn}(\chi)$, and
we are done. Thus we may assume that $T^{*}=G$. Then $T\triangleleft\,G$, and
that $G/T$ is abelian. By induction, we may assume that $T=N$, and
$\psi=\theta$. If $M$ is a maximal normal subgroup of $G$ with $N<M<G$, then
again by induction (and using Lemma 2.2(i)),
${\mathbb{Q}}_{pn}(\theta^{M})={\mathbb{Q}}_{pn}(\theta)$. Hence it is enough
to prove the statement in the case where $G/N$ has prime order.

Now $G/N$ has prime order. Let $\sigma\in{\rm
Gal}({\mathbb{Q}}_{pn}(\theta)/{\mathbb{Q}}_{pn}(\chi))$. We want to show
$\sigma$ is trivial. Assume that $\sigma\neq 1$. Notice that $\sigma$ is a
$p$-element, since ${\rm Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{pn})\leq{\rm
Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{pn})\cong{\rm
Gal}({\mathbb{Q}}_{|G|_{p}}/{\mathbb{Q}}_{p})$ is a $p$-group. By Clifford’s
theorem, we have that $\theta^{\sigma}=\theta^{g}$ for some $g\in G$. Also
$\theta^{\sigma}\neq\theta$ because $\sigma$ is not trivial. In particular
$\langle gN\rangle=G/N$ is a group of order $p$. Let $b$ be the block of
$\theta$. Since $\sigma$ fixes $p^{\prime}$-roots of unity, it follows that
$b^{\sigma}=b$. (Use, for instance, Theorem 3.19 of [Nav2].) Then $b^{g}=b$
and $b$ is $G$-invariant. Then we apply Corollary 9.6 and Corollary 9.18 of
[Nav2], and conclude that $\theta$ is $G$-invariant, a contradiction.

The second part of the statement follows from Lemma 2.2. Notice if $p$ is odd
then $c(\chi)_{p}\geq p$ implies that $c(\psi)_{p}\geq 1$ because if
$c(\psi)_{p}=1$, then $\psi$ is $p$-rational, and $\chi=\psi^{G}$ is also
$p$-rational. ∎

Notice that the hypothesis on the odd case of the second statement of the
above theorem is necessary: if $p=3$, $\chi\in{\rm Irr}({\sf S}_{3})$ has
degree 2 and $\psi\in{\rm Irr}(N)$ is under $\chi$ with $|N|=3$, then
$c(\chi)_{3}=1$ but $c(\psi)_{3}=3$.

######  Corollary 3.3.

Let $N\triangleleft\,G$, let $\chi\in{\rm Irr}(G)$ of height zero in its
$p$-block, and let $\theta$ be an irreducible constituent of $\chi_{N}$. If
$p$ is odd, assume that $c(\chi)_{p}\geq p$. Then $c(\theta)_{p}\leq
c(\chi)_{p}$. In particular, if $p=2$ and $\chi$ is 2-rational, then $\theta$
is $2$-rational.

###### Proof.

By Theorem 3.2, we may assume that $\theta$ is $G$-invariant. Then
$\chi_{N}=e\theta$, ${\mathbb{Q}}(\theta)\subseteq{\mathbb{Q}}(\chi)$, and the
statement is clear. ∎

Of course, Corollary 3.3 is about height zero characters. (Consider, for
instance, $G={\sf D}_{8}$, $\chi\in{\rm Irr}(G)$ of degree 2, and $N$ a cyclic
subgroup of $G$ of order 4.)

Next we prove the normal defect case of Theorem A. We first need a lemma.
Suppose that $\chi\in{\rm Irr}(G)$ lies in a block $B$ with defect group
$D\triangleleft\,G$. Let $C={\bf C}_{G}(D)$. Let $b$ a block of $CD$ covered
by $B$. By Corollary 9.21 of [Nav2], we have that $B=b^{G}$ is the only block
of $G$ covering $b$. By Theorem 9.26 of [Nav2], we have that $b$ has defect
group $D$. By Theorem 9.12 of [Nav2], there is a unique irreducible character
$\theta\in{\rm Irr}(b)$ such that $D\subseteq{\rm ker}(\theta)$. This
character has defect zero, viewed as a character of $CD/D$, and it is called
the canonical character of $b$, which is uniquely defined up to $G$-conjugacy.
The irreducible characters of $b$ are described in Theorem 9.12 of [Nav2].

######  Lemma 3.4.

Suppose that $\chi\in{\rm Irr}(G)$ has height zero and belongs to a $p$-block
$B$ with a normal defect group $D$. Let $C={\bf C}_{G}(D)$ and $Z={\bf Z}(D)$.
If $\chi_{CD}$ is homogeneous, then $\chi_{D}$ is homogeneous, the canonical
character $\theta\in{\rm Irr}(CD/D)$ of $B$ is $G$-invariant, and $G/CD$ is a
$p^{\prime}$-group. If $\lambda$ the irreducible constituent of $\chi_{D}$,
then $\lambda$ is linear and $c(\chi)_{p}=c(\lambda)$.

###### Proof.

Let $\eta\in{\rm Irr}(CD)$ be under $\chi$. By Theorem 3.1, we have that
$\eta$ has height zero. By Theorem 9.12 of [Nav2], we know that we can write
$\eta=\theta_{\lambda}$, where $\lambda\in{\rm Irr}(D)$ is linear and
$\theta\in{\rm Irr}(CD/D)$ has defect zero. Moreover, $\eta(x)=0$ if
$x_{p}\not\in D$, and $\eta(x)=\theta(x_{p^{\prime}})\lambda(x_{p})$ if
$x_{p}\in D$. Thus $\eta_{D}=\theta(1)\lambda$. Since $\chi_{CD}$ is
homogeneous, it follows that $\chi_{D}$ is homogeneous then $\lambda$ and
$\eta$ are $G$-invariant. We claim that $\theta\in{\rm Irr}(CD/D)$ is
$G$-invariant. View $\theta$ as a character of $C/Z$. Let $x\in C$ and $g\in
G$. Since $\theta\in{\rm Irr}(C/Z)$ has $p$-defect zero, if $xZ$ is
$p$-singular, then $\theta(x)=0=\theta(x^{g})$ because $x^{g}Z$ is also
$p$-singular. If $xZ$ is $p$-regular, then $x^{g}Z$ is also $p$-regular,
$xZ=x_{p^{\prime}}Z$ and $x^{g}Z=(x_{p^{\prime}})^{g}Z$. Since $\eta$ is
$G$-invariant, we have that
$\theta(x_{p^{\prime}})\lambda(x_{p})=\eta(x)=\eta(x^{g})=\theta((x_{p^{\prime}})^{g})\lambda((x_{p^{\prime}})^{g})$.
Since $\lambda$ is $G$-invariant and linear, we deduce that
$\theta(x_{p^{\prime}})=\theta((x_{p^{\prime}})^{g})$. Now
$\theta(x)=\theta(x_{p^{\prime}})=\theta((x_{p^{\prime}})^{g})=\theta(x^{g})$
because $Z$ is contained in the kernel of $\theta$, and we deduce that
$\theta$ is $G$-invariant. Therefore, we have that $G/CD$ has order coprime to
$p$ by Theorem 9.22 of [Nav2].

Since $\chi_{CD}=e\eta$ and $\eta_{D}=\theta(1)\lambda$, we have that
${\mathbb{Q}}_{c(\lambda)}\subseteq{\mathbb{Q}}(\eta)$. Now, $\theta$ is
$p$-rational, because it is a defect zero character, and therefore,
${\mathbb{Q}}(\theta)\subseteq{\mathbb{Q}}_{m}$, where $m$ is a
$p^{\prime}$-number. Then, using the formula for the values of $\eta$, we have
that
${\mathbb{Q}}_{c(\lambda)}\subseteq{\mathbb{Q}}(\eta)\subseteq{\mathbb{Q}}_{c(\lambda)m}$.
Therefore, $c(\lambda)$ divides $c(\eta)$ which divides $c(\lambda)m$,
implying that $c(\eta)_{p}=c(\lambda)$. We apply Lemma 4.2(ii) of [NT], and we
get that $c(\chi)_{2}=c(\eta)_{2}$ if $p=2$ and $c(\chi)_{p}=c(\eta)_{p}$ if
$p$ is odd and $c(\lambda)=c(\eta)_{p}>1$. If $p$ is odd and
$c(\lambda)=c(\eta)_{p}=1$, then $\lambda=1_{D}$. Therefore $\eta=\theta$, and
$\chi$ has $p$-defect zero, so $\chi$ is $p$-rational and $c(\chi)_{p}=1$. In
any case, we conclude that $c(\chi)_{p}=c(\lambda)$. ∎

Next we prove Theorem A, and one of the containments of Conjecture D, in the
case of blocks with a normal defect group.

######  Lemma 3.5.

Let $\chi\in{\rm Irr}(B)$ of height zero, where $B$ is a $p$-block with a
normal defect group $D$. Write $c(\chi)=p^{a}m$, where $m$ is not divisible by
$p$. Then ${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pn}(\chi)$.

###### Proof.

First notice that we may assume that $a\geq 2$ as otherwise the result
trivially holds. We argue by induction on $|G|$.

Write $C={\bf C}_{G}(D)$. Let $\eta\in{\rm Irr}(CD)$ be an irreducible
constituent of $\chi_{CD}$. Let $T=G_{\eta}$ be the stabilizer of $\eta$ in
$G$ and $\psi\in{\rm Irr}(T|\eta)$ be the Clifford correspondent of $\chi$
over $\eta$. By Theorem 3.1, we know that $\psi$ has height zero and that $D$
is a defect group of its block. By Theorem 3.2, we have that
${\mathbb{Q}}_{pn}(\chi)={\mathbb{Q}}_{pn}(\psi)$ where $n=|G|_{p^{\prime}}$
and $c(\chi)_{p}=c(\psi)_{p}$. Assume that $T<G$. By induction
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pc(\psi)_{p^{\prime}}}(\psi)\subseteq{\mathbb{Q}}_{pn}(\psi)={\mathbb{Q}}_{pn}(\chi)$.
By Lemma 2.1 we conclude that
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm}(\chi)$.

Hence $T=G$ and we are under the hypotheses of Lemma 3.4. If $\lambda\in{\rm
Irr}(D)$ lies under $\chi$, then $p^{a}=c(\chi)_{p}=c(\lambda)$. Notice that
$\chi_{D}=f\lambda$ and hence
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}(\chi)\subseteq{\mathbb{Q}}_{pm}(\chi)$.
∎

The next results are key to understand the statement of Conjecture D when the
group possesses a normal subgroup of index $p$ (a fundamental step in our
reduction theorem).

######  Lemma 3.6.

Suppose that $G/N$ is a $p$-group, and $b$ is a $G$-invariant $p$-block of $N$
covered by a block $B$ of $G$ with defect group $D$. Suppose that $D_{0}=D\cap
N\triangleleft\,G$. Then $G=DN$ and $b$ has a $G$-invariant height zero
$p$-rational irreducible character.

###### Proof.

By Corollary 9.6 of [Nav2], $B$ is the unique $p$-block covering $b$. Then
Theorem 9.17 of [Nav2] implies that $G=ND$ and $D_{0}$ is the unique defect
group of $b$. Let $C={\bf C}_{N}(D_{0})$. Notice that
$CD_{0}\triangleleft\,G$. By the Fong-Reynolds correspondence, Theorem 9.14 of
[Nav2], we can find $e$ a block of $CD_{0}$ covered by $b$ such that the block
$b_{T}$ of $T=G_{e}$, the stabilizer of $e$ in $G$, inducing $B$ and covering
$e$ has defect group $D$. Notice that $e$ has defect group $D_{0}$. Since $b$
is $G$-invariant, notice that $TN=G$, $b_{T}$ covers a unique block $f$ of
$N_{e}=N\cap T$ that induces $b$ and covers $e$. By induction and the Fong-
Reynolds correspondence, we may assume that $e$ is $G$-invariant. Then we have
that $N/CD_{0}$ is a $p^{\prime}$-group, by Theorem 9.22 of [Nav2]. Since $e$
is $G$-invariant, we have that the canonical character $\theta\in{\rm
Irr}(CD_{0}/D_{0})$ of $e$ is $G$-invariant. By Theorem 13.31 of [Isa], some
irreducible constituent $\xi$ of $\theta^{N}$ is $D$-invariant. Thus $\xi$ is
$G$-invariant. Since $b$ is the only block of $N$ that covers $e$, we have
that $\xi\in{\rm Irr}(b)$. Also, $\xi$ is $p$-rational, because it has defect
zero considered as a character of $N/D_{0}$. It also has height zero because
$\theta$ has height zero and $N/CD_{0}$ is a $p^{\prime}$-group. ∎

######  Lemma 3.7.

Suppose that $G/N$ is a cyclic $p$-group, and $\theta\in{\rm Irr}(N)$ is
$G$-invariant of $p$-height zero. Then every $\chi\in{\rm Irr}(G|\theta)$ has
$p$-height zero. Also, if $D$ is a defect group of the block of $\chi$, then
$DN=G$ and $D\cap N$ is a defect group of the block of $\theta$.

###### Proof.

Let $b$ be the block of $\theta$. Let $B$ be the unique $p$-block of $G$
covering $b$ by Corollary 9.6 of [Nav2]. Let $\chi\in{\rm Irr}(G|\theta)$, so
that $\chi\in{\rm Irr}(B)$. Since $b$ is $G$-invariant, we have that $G=DN$,
where $D$ is a defect group of $B$, and $D_{0}=D\cap N$ is a defect group of
$b$ by Theorem 9.17 of [Nav2]. We have that $\chi_{N}=\theta$ because $G/N$ is
cyclic and $\theta$ is $G$-invariant (using Theorem 5.1 of [Nav3] and the
Gallagher correspondence Corollary 1.23 of [Nav3]). Then
$\chi(1)_{p}=\theta(1)_{p}=|N:D_{0}|=|G:D|$. Thus $\chi$ has height zero. ∎

######  Lemma 3.8.

Suppose that $G/N$ is a $p$-group. Let $\theta\in{\rm Irr}(N)$ be of
$p$-height zero and $G$-invariant. Let $n=|G|_{p^{\prime}}$. Let $D_{0}$ be a
defect group of the block of $\theta$, let $H={\bf N}_{G}(D_{0})$. Then there
exists an $H$-invariant $\varphi\in{\rm Irr}(N\cap H)$ of $p$-height zero such
that $[\theta_{H\cap N},\varphi]\not\equiv 0\bmod\,p$, and
${\mathbb{Q}}_{pn}(\varphi)\subseteq{\mathbb{Q}}_{pn}(\theta)$.

###### Proof.

Let $b$ be the $p$-block of $\theta$ and let $B$ be the only $p$-block of $G$
covering $b$. Since $b$ is $G$-invariant, we have that $G=DN$, where $D$ is a
defect group of $B$, and $D_{0}=D\cap N$ is a defect group of $B_{0}$, by
Theorem 9.17 of [Nav2].

Since $D\subseteq H$, note that $G=HN$. Let $M=H\cap N={\bf N}_{N}(D_{0})$.
Then $H=MD$. Let $e$ be the Brauer correspondent block of $M$ (with defect
group $D_{0}$) inducing $b$. By the Harris–Knörr Theorem [Nav2, Theorem 9.28],
there is a unique block $E$ of $H$ covering $e$ that induces $B$. This block
$E$ has defect group $D$.

Let $\mathcal{U}={\rm Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{pn}(\theta))$.
Notice that $\mathcal{U}\leq{\rm Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{pn})$
which is a $p$-group. We must then work to show that $e$ has a
$D\times\mathcal{U}$-invariant height zero character $\varphi$ with
$[\theta_{M},\varphi]\not\equiv 0$ mod $p$. We will use the $\tilde{}$
construction as in [Nav2, page 27] and the fact that $e$ possesses an
irreducible character $\psi$ which is $H$-invariant and $p$-rational by Lemma
3.6.

Let $\delta=\theta_{M}$. Then $\tilde{\delta}$ defined as
$\tilde{\delta}(x)=|M|_{p}\delta(x)$, if $x\in M$ is $p$-regular, and 0,
otherwise, is a generalized character by Lemma 2.15 of [Nav2]. We can write

$\tilde{\delta}=\sum_{\xi\in{\rm Irr}(M)}[\delta,\xi]\tilde{\xi}\,.$

By Lemma 6.5.(b) of [Nav2], we have that

${[\tilde{\delta},\psi]\over\psi(1)}\not\equiv 0\,{\rm mod}\,\mathcal{P}\,,$

where $\mathcal{P}$ is the maximal ideal of ${\bf R}_{M}$ the localization of
the ring of algebraic integers $\bf R$ at a prime $M$ containing $p$ (see
[Nav2, page 16]). Therefore

$\Lambda=\sum_{\xi\in{\rm
Irr}(M)}[\delta,\xi]{[\tilde{\xi},\psi]\over\psi(1)}\not\equiv 0\,{\rm
mod}\,\mathcal{P}\,.$

Note that $[\tilde{\xi},\psi]=[\xi,\tilde{\psi}]$ whenever $\xi\in{\rm
Irr}(M)$. By Lemma 3.20 of [Nav2], recall that $[\xi,\tilde{\psi}]=0$ if $\xi$
is not in $e$, so that

$\Lambda=\sum_{\xi\in{\rm
Irr}(e)}[\delta,\xi]{[\tilde{\xi},\psi]\over\psi(1)}\not\equiv 0\,{\rm
mod}\,\mathcal{P}\,.$

By Lemma 3.22(a) of [Nav2] ${[\tilde{\xi},\psi]\over\psi(1)}\in{\bf
R}_{M}\cap{\mathbb{Q}}$ for every $\xi\in{\rm Irr}(e)$, so
$\nu({[\tilde{\xi},\psi]\over\psi(1)})\geq 0$, where $\nu$ is the valuation
function defined in [Nav2, page 64]. By Theorem 3.24 of [Nav2], we have that
$\nu({[\tilde{\xi},\psi]\over\psi(1)})=0$ if, and only if, $\xi$ has height
zero in the $p$-block $e$ ($\xi\in{\rm Irr}_{0}(e)$). By Lemma 3.21 of [Nav2]
we have that ${[\tilde{\xi},\psi]\over\psi(1)}\in\mathcal{P}$ whenever $\xi$
does not have height zero in $e$. Hence

$\Lambda\equiv\sum_{\xi\in{\rm
Irr}_{0}(e)}[\delta,\xi]{[\tilde{\xi},\psi]\over\psi(1)}\not\equiv 0\,{\rm
mod}\,\mathcal{P}\,.$

Consider $\Omega=\\{\xi\in{\rm Irr}_{0}(e)\ |\ [\delta,\xi]\not\equiv
0\bmod\,p\\}$. We have that

$\Lambda\equiv\sum_{\xi\in\Omega}[\delta,\xi]{[\tilde{\xi},\psi]\over\psi(1)}\not\equiv
0\,{\rm mod}\,\mathcal{P}\,.$

The $p$-group $D\times\mathcal{U}$ acts on $\Omega$. Let
$\Omega=\Omega_{\xi_{1}}\cup\ldots\cup\Omega_{\xi_{r}}$ be the orbit
decomposition of $\Omega$. Then, given that $\delta$ and $\psi$ are
$D\times\mathcal{U}$-invariant, we have that

$\Lambda\equiv\sum_{i=1}^{r}|\Omega_{\xi_{i}}|[\delta,\xi_{i}]{[\tilde{\xi_{i}},\psi]\over\psi(1)}\not\equiv
0\,{\rm mod}\,\mathcal{P}\,.$

In particular there is some $D\times\mathcal{U}$-invariant $\varphi\in\Omega$.
The $D$-invariance of $\varphi$ implies that $\varphi$ is $H$-invariant and
the $\mathcal{U}$-invariance of $\varphi$ implies that
${\mathbb{Q}}_{pn}(\varphi)\subseteq{\mathbb{Q}}_{pn}(\theta)$. ∎

The following is a consequence of an argument of J. Thompson, we refer the
reader to Theorem 6.9 of [Nav3].

######  Lemma 3.9.

Suppose that $G/N$ is a $p$-group, and let $H\leq G$ such that $G=NH$. Write
$M=N\cap H$. Let $\theta\in{\rm Irr}(N)$ be $G$-invariant, and let
$\varphi\in{\rm Irr}(M)$ be $H$-invariant such that
$[\theta_{M},\varphi]\not\equiv 0$ mod $p$.

  1. (a)

Suppose that $\xi\in{\rm Irr}(H)$ extends $\varphi$. Then there is an
extension $\chi\in{\rm Irr}(G)$ of $\theta$ such that
${\mathbb{Q}}(\chi)\subseteq{\mathbb{Q}}(\xi,\theta)$.

  2. (b)

Suppose that $\chi\in{\rm Irr}(G)$ extends $\theta$. Then there is an
extension $\xi\in{\rm Irr}(H)$ of $\varphi$ such that
${\mathbb{Q}}(\xi)\subseteq{\mathbb{Q}}(\chi,\varphi)$.

  3. (c)

Suppose that $G/N$ has order $p$. Then
${\mathbb{Q}}_{p}(\chi,\varphi)={\mathbb{Q}}_{p}(\theta,\xi)$ for every
$\chi\in{\rm Irr}(G|\theta)$ and $\xi\in{\rm Irr}(H|\varphi)$.

###### Proof.

Write $m=|G|$. In order to prove part (a), let $\sigma\in{\rm
Gal}({\mathbb{Q}}_{m}/{\mathbb{Q}}(\xi,\theta))$. Since $\xi^{\sigma}=\xi$, in
particular $\varphi^{\sigma}=\varphi$. By Theorem 6.9 of [Nav3], there is a
unique extension $\chi\in{\rm Irr}(G)$ of $\theta$ such that
$\chi_{H}=\Psi\xi+\Delta$, where $\Delta$ is a character of $H$ or zero, with
$[\Delta_{M},\varphi]=0$ and $\Psi$ is a character of $H/M$ with trivial
determinant. Now, $\chi^{\sigma}$ is an extension of $\theta=\theta^{\sigma}$
and $(\chi^{\sigma})_{H}=\Psi^{\sigma}\xi+\Delta^{\sigma}$, where
$[(\Delta^{\sigma})_{M},\varphi]=0$ and $\Psi^{\sigma}$ is a character of
$H/M$ with trivial determinant. By the uniqueness of $\chi$, we get that
$\chi=\chi^{\sigma}$. Part (b) is proved in the same way.

We prove part (c). We have that $\theta$ extends to $G$ by Theorem 5.1 of
[Nav3]. In fact, every character in ${\rm Irr}(G|\theta)$ is an extension of
$\theta$ by the Gallagher correspondence. Let $\chi\in{\rm Irr}(G|\theta)$. By
part (b), there is an extension $\xi\in{\rm Irr}(H)$ of $\varphi$ such that
${\mathbb{Q}}(\xi)\subseteq{\mathbb{Q}}(\chi,\varphi)$. By part (a), there is
an extension $\chi^{\prime}\in{\rm Irr}(G)$ of $\theta$ such that
${\mathbb{Q}}(\chi^{\prime})\subseteq{\mathbb{Q}}(\xi,\theta)$. Since
$\chi^{\prime}=\lambda\chi$ for some $\lambda\in{\rm Irr}(G/N)$, we have that
${\mathbb{Q}}_{p}(\chi)={\mathbb{Q}}_{p}(\chi^{\prime})$. Since
${\mathbb{Q}}(\theta)\subseteq{\mathbb{Q}}(\chi)$ and
${\mathbb{Q}}(\varphi)\subseteq{\mathbb{Q}}(\xi)$, part (c) follows. ∎

In order to treat later the case where $N$ is a normal subgroup of $G$ of
index $p$, we need to extend Lemma 3.5, and prove the statement of Conjecture
D in a slightly more general case than the normal defect group case.

######  Lemma 3.10.

Suppose that $G/N$ has order $p$. Let $n=|G|_{p^{\prime}}$. Let $\chi\in{\rm
Irr}(G)$ have $p$-height zero. Suppose that $\chi_{N}=\theta\in{\rm Irr}(N)$
and the defect group $D_{0}$ of the block of $\theta$ is normal in $G$. Then
${\mathbb{Q}}_{c(\chi)_{p}}\subseteq{\mathbb{Q}}_{pn}(\chi)$.

###### Proof.

Write $p^{a}=c(\chi)_{p}$. We may assume that $a\geq 2$ as the statement is
trivially satisfied otherwise. We argue by induction on $|G|$.

Write $K={\bf C}_{N}(D_{0})D_{0}\triangleleft\,G$. Let $\eta$ be an
irreducible constituent of $\chi_{K}$ and $\psi\in{\rm Irr}(G_{\eta}|\eta)$ be
the Clifford correspondence of $\chi$. Using Theorem 3.2 and induction, we may
assume that $\eta$ is $G$-invariant. In particular, $\theta_{K}=\chi_{K}$ is
homogeneous. By Lemma 3.4, $\chi_{D_{0}}=\theta_{D_{0}}$ is homogeneous. Write
$\theta_{D_{0}}=\theta(1)\lambda$, where $\lambda\in{\rm Irr}(D_{0})$ is
linear and $G$-invariant.

Let $D$ be a defect group of the block of $\chi$. We have that $G=ND$ and
$D_{0}=N\cap D$, using for instance Lemma 3.7. Let $H={\bf N}_{G}(D)$ and
$C={\bf N}_{N}(D)=N\cap H$. If $H=G$ then $\chi$ lies in a block of normal
defect, and we are done by Lemma 3.5. Hence we may assume that $H<G$. By
Theorem A of [NS], there are a block $b^{\prime}$ of $C$ with defect $D_{0}$
and a $D$-invariant character $\theta^{\prime}$ in $b^{\prime}$ satisfying
that $[\theta_{C},\theta^{\prime}]\equiv\pm 1\,{\rm mod}\,p$. Since $\lambda$
is $G$-invariant $\theta^{\prime}$ is the unique irreducible constituent of
$\theta_{C}$ such that $\theta^{\prime}(1)_{p}=|C:D_{0}|_{p}$, and
$[\theta_{C},\theta^{\prime}]\not\equiv 0\bmod\,p$. The block $B^{\prime}$ of
$H=CD$ covering $\theta^{\prime}$ has defect group $D$, the block of
$\theta^{\prime}$ has defect group $D_{0}$, and $\theta^{\prime}$ has height
zero.

Given $\sigma\in{\rm Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{n})$, we have that
$b^{\sigma}=b$ and $(b^{\prime})^{\sigma}=b^{\prime}$ because $\sigma$ fixes
$p^{\prime}$-roots of unity. Moreover
$(\theta^{\sigma})^{\prime}=(\theta^{\prime})^{\sigma}$ under the canonical
correspondence given by Theorem A of [NS] (because of part (c) in that
statement, noticing that $\lambda^{\sigma}$ is also $G$-invariant). This
implies that ${\mathbb{Q}}_{n}(\theta^{\prime})={\mathbb{Q}}_{n}(\theta)$ by
elementary Galois theory.

Let $\xi\in{\rm Irr}(H|\theta^{\prime})$. Then $\xi$ extends $\theta^{\prime}$
and, by Lemma 3.7, notice that $\xi$ has height zero. By Lemma 3.9(c), we have
that ${\mathbb{Q}}_{p}(\chi,\theta^{\prime})={\mathbb{Q}}_{p}(\theta,\xi)$.
Since ${\mathbb{Q}}_{n}(\theta)={\mathbb{Q}}_{n}(\theta^{\prime})$ we have
that
${\mathbb{Q}}_{pn}(\chi,\theta)={\mathbb{Q}}_{pn}(\chi,\theta^{\prime})={\mathbb{Q}}_{pn}(\xi,\theta)={\mathbb{Q}}_{pn}(\xi,\theta^{\prime})$.
We easily deduce that ${\mathbb{Q}}_{pn}(\chi)={\mathbb{Q}}_{pn}(\xi)$ using
that ${\mathbb{Q}}(\theta)\subseteq{\mathbb{Q}}(\chi)$ and
${\mathbb{Q}}(\theta^{\prime})\subseteq{\mathbb{Q}}(\xi)$. We want to apply
Lemma 2.2(ii). If $c(\xi)_{p}=1$ then
${\mathbb{Q}}(\xi)\subseteq{\mathbb{Q}}_{n}$ and consequently
${\mathbb{Q}}(\chi)\subseteq{\mathbb{Q}}_{pn}(\chi)\subseteq{\mathbb{Q}}_{pn}$,
but this is impossible as $c(\chi)_{p}\geq p^{2}$. Hence $c(\chi)_{p}\geq p$
and by Lemma 2.2(ii) $c(\chi)_{p}=c(\xi)_{p}$. Recall that $H<G$. Write
$k=|H|_{p^{\prime}}.$ By induction,
${\mathbb{Q}}_{c(\chi)_{p}}={\mathbb{Q}}_{c(\xi)_{p}}\subseteq{\mathbb{Q}}_{pk}(\xi)\subseteq{\mathbb{Q}}_{pn}(\xi)={\mathbb{Q}}_{pn}(\chi)$,
then we are done. ∎

##  4\. Character Triples and Fields

If $\chi\in{\rm Irr}(G)$ lies in a $p$-block $B$, we denote $h(\chi)$ the
$p$-height of $\chi$ (we will sometimes just refer to $h(\chi)$ as the height
of $\chi$). We remind the reader that if $N\subseteq{\rm ker}(\chi)$, then the
height of $\chi$ as a character of $G$ and as a character of $G/N$ can be
different. For instance, the character of degree 2 of $\sf S_{3}$ has 2-height
zero, but as a character of $\sf S_{4}$ has 2-height 1. The next result
clarifies this situation.

######  Lemma 4.1.

Suppose that $\chi\in{\rm Irr}(B)$, where $B$ is a $p$-block of $G$ with
defect group $P$. Suppose that $K\subseteq{\rm ker}(\chi)$ and let
$\bar{\chi}\in{\rm Irr}(G/K)$ the character $\chi$ viewed as a character in
$G/K$. Let $\bar{B}$ be the $p$-block of $G/K$ containing $\bar{\chi}$.

  1. (i)

There is a defect group $\bar{D}$ of $\bar{B}$ such that $\bar{D}\leq PK/K$.

  2. (ii)

We have that

$p^{h(\chi)}={|PK/K|\over|\bar{D}|}p^{h(\bar{\chi})}\,.$

In particular, $h(\chi)\geq h(\bar{\chi})$ and if $h(\chi)=0$ then
$PK/K=\bar{D}$ and $h(\bar{\chi})=0$.

  3. (iii)

If $K\subseteq{\bf Z}(G)$, then $PK/K=\bar{D}$ and $h(\chi)=h(\bar{\chi})$.

###### Proof.

The first part is Theorem 9.9 of [Nav2]. Since $\chi$ lies over $1_{K}$, it
follows that $B$ covers the principal block of $K$, by Theorem 9.2 of [Nav2].
By Theorem 9.26 of [Nav2], we have that $P\cap K\in{\rm Syl}_{p}(K)$, so
$|K:K\cap P|_{p}=1$. Now,

$\chi(1)_{p}=\bar{\chi}(1)_{p}={|G/K|_{p}\over|PK/K|}{|PK/K|\over|\bar{D}|}p^{h(\bar{\chi})}={|G|_{p}\over|P|}{|PK/K|\over|\bar{D}|}p^{h(\bar{\chi})}\,,$

and we use the definition of $h(\chi)$. The third part follows from Lemma 2.2
of [Riz]. ∎

######  Lemma 4.2.

Suppose that $G^{*}$ is a finite group, $N,Z\triangleleft\,G^{*}$ such that
$N\cap Z=1$, where $Z\subseteq{\bf Z}(G^{*})$. Let $N^{*}=N\times Z$. Let
$\theta\in{\rm Irr}(N)$ be $G$-invariant, and $\lambda\in{\rm Irr}(Z)$. Let
$\theta^{*}=\theta\times 1_{Z}$, $\lambda^{*}=1_{N}\times\lambda$, and assume
that $(\lambda^{*})^{-1}\theta^{*}$ extends to some $\tau\in{\rm Irr}(G^{*})$.
Then the map $\chi^{*}\mapsto\chi^{*}\tau$ defines a character triple
isomorphism $(G^{*},N^{*},\lambda^{*})\rightarrow(G^{*},N^{*},\theta^{*})$.
Let $\chi^{*}\in{\rm Irr}(G^{*}|\lambda^{*})$. If $\chi=\chi^{*}\tau$ has
height zero in $G^{*}$, then $\chi^{*}$ has height zero in $G^{*}/N$.

###### Proof.

We have that $\tau_{N}=\theta$. The fact that $\chi^{*}\mapsto\chi^{*}\tau$
defines a character triple isomorphism follows from Lemma 11.27 of [Isa].

Let $\chi^{*}\in{\rm Irr}(G^{*}|\lambda^{*})$. Then $N\subseteq{\rm
ker}(\chi^{*})$ and we can see $\chi^{*}$ has a character of $G^{*}/N$ lying
over $\lambda$ (identified with a character of $N^{*}/N$). Assume that
$\chi=\chi^{*}\tau\in{\rm Irr}(G^{*}|\theta^{*})$ has height zero in $G^{*}$,
we want to prove that $\chi^{*}$ has height zero in $G^{*}/N$. Let $B$ be the
$p$-block of $G^{*}/N$ that contains $\chi^{*}$, and let $D^{*}/N$ be a defect
group of $B$. By Proposition 2.5(b) of [NS], $D^{*}/N$ is contained in $DN/N$,
where $D$ is a defect group of $\chi$. Since $\theta$ is an irreducible
constituent of $\chi_{N}$ and $\chi$ has height zero, by Theorem 3.1, we know
that $\theta$ has height zero. By Theorem 9.26 of [Nav2], we know that $D\cap
N$ is a defect group of the block of $\theta$. Therefore:

$\chi^{*}(1)_{p}=(\chi(1)/\theta(1))_{p}=|G^{*}:DN|_{p}\,.$

By definition,

$\chi^{*}(1)_{p}=|G^{*}/N:D^{*}/N|_{p}p^{h(\chi^{*})}\geq|G^{*}/N:DN/N|_{p}p^{h(\chi^{*})}=|G^{*}:DN|_{p}p^{h(\chi^{*})}\,.$

We conclude that $p^{h(\chi^{*})}=1$, as wanted. ∎

Next we use the theory of character triples, as developed in [Isa, Chapter
11]. Recall that if $G/N$ is a group, by [Isa, Theorem 11.17] there exists a
finite central extension $(\Gamma,\pi)$ of $G/N$ such that $A={\rm
ker}(\pi)\cong M(G/N)$ and the standard map $\eta\colon{\rm Irr}(A)\to M(G/N)$
is an isomorphism. In particular, by [Isa, Theorem 11.19], if $G/N$ is perfect
then $\Gamma$ is perfect. We will also make use of the results contained in
[GP, Section 3].

######  Theorem 4.3.

Suppose that $(G,N,\theta)$ is a character triple, where $\theta\in{\rm
Irr}(N)$ and $G/N$ is perfect. Then there exists a character triple
isomorphism

$(G,N,\theta)\rightarrow(\Gamma,A,\lambda),$

where $\Gamma$ is perfect, $A={\bf Z}(\Gamma)$,
${\mathbb{Q}}(\lambda)\subseteq{\mathbb{Q}}(\theta)$ and
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\chi^{*},\theta)$ for every $\chi\in{\rm
Irr}(G|\theta)$, where $\chi^{*}$ corresponds to $\chi$ under the character
triple isomorphism. Furthermore, if $\chi$ has height zero in $G$, then
$\chi^{*}$ has height zero in $\Gamma$.

###### Proof.

We consider a canonically constructed character triple $(\Gamma,A,\lambda)$
isomorphic to $(G,N,\theta)$ in the sense of [GP]. Notice that the values of
any $\psi\in{\rm Irr}(\Gamma|\lambda)$ are in ${\mathbb{Q}}_{|G|}$. (See the
paragraph before Corollary 3.3 of [GP].) By Theorem 3.6 of [GP], we have that
whenever $\sigma\in{\rm Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}(\theta))$, then
$(\chi^{*})^{\sigma}=(\chi^{\sigma})^{*}$. Hence
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\chi,\theta)={\mathbb{Q}}(\chi^{*},\theta)$,
as wanted. We do notice that $\Gamma$ is perfect using Theorem 11.19 of [Isa].
For the second part, we notice that the construction of the character triple
isomorphism in [GP] follows the construction in Theorem 11.28 of [Isa]. We
have that $(G,N,\theta)$ isomorphic to $(G^{*},N^{*},\theta^{*})$ where
$G^{*}\subseteq G\times\Gamma$, $N^{*}=N\times A$, $A\subseteq{\bf Z}(G^{*})$
and $\theta^{*}=\theta\times 1_{A}$; in fact $G\cong G^{*}/A$. Also
$(G^{*},N^{*},\theta^{*})$ is isomorphic to $(G^{*},N^{*},\lambda^{*})$ where
$\lambda^{*}=1_{N}\times\lambda$ (here $\theta^{*}(\lambda^{*})^{-1}$ extends
to $\tau\in{\rm Irr}(G^{*})$). Finally $(G^{*},N^{*},\lambda^{*})$ is
isomorphic to $(\Gamma,A,\lambda)$ using that $\Gamma\cong G^{*}/N$. Given
$\chi\in{\rm Irr}(G|\theta)$ of height zero in $G\cong G^{*}/A$, the first
character triple isomorphism just sends $\chi$ to $\chi$ viewed as a character
of $G^{*}$. By Lemma 4.1(iii), we have that $\chi$ has height zero as a
character of $G^{*}$. By Lemma 4.2 we have that $\chi=\chi^{*}\tau$, and
$\chi^{*}\in{\rm Irr}(G^{*}|\lambda^{*})$ has height zero in $G^{*}/N$. Since
the last character triple isomorphism just sends $\chi^{*}$ to $\chi^{*}$ seen
as a character of $\Gamma\cong G^{*}/N$, we have that $\chi^{*}$ has height
zero as a character of $\Gamma$, and the second part of the statement follows.
∎

We believe that the following result might be useful in the future. (It can be
used, together with the ideas of the proof of Theorem 5.1 of [NTT], as an
alternative to Theorem 4.3 in the proof of our main result.)

######  Theorem 4.4.

Suppose that $\chi\in{\rm Irr}(G)$ has 2-height zero. Let
$n=|G|_{2^{\prime}}$, $F={\mathbb{Q}}_{n}(\chi)$. Then $\chi$ can be afforded
by an absolutely irreducible $F$-representation.

###### Proof.

We want to show that $m_{F}(\chi)=1$. We know by Corollary 10.13 of [Isa] that
$m_{F}(\chi)\leq 2$. Suppose that $D$ is a defect group of the block of
$\chi$, and let $H={\bf N}_{G}(D)$. Let $C={\bf C}_{G}(D)$ and $Z={\bf Z}(D)$.
By Lemma 3.8, we have that $\chi_{H}$ contains some $\psi\in{\rm Irr}(H)$ with
2-height zero, ${\mathbb{Q}}(\psi)\subseteq F$ and $[\chi_{H},\psi]$ is odd.
If we can show that $\psi$ is afforded by an $F$-representation, then so it is
$\psi^{G}$ and by Corollary 10.2(c) of [Isa] we have that $m_{F}(\chi)$
divides $[\psi^{G},\chi]=[\psi,\chi_{H}]$ which is odd.

We work thus to show that $\psi$ can be afforded by an $F$-representation. By
Theorem 3.2, and arguing by induction on $|G|$, we may assume that $\psi$ is
quasiprimitive. In particular, as $DC\triangleleft\,H$, we have that
$\psi_{DC}$ is homogeneous. By Lemma 3.4 we have that $H/CD$ is an odd-order
group, and the canonical character $\theta\in{\rm Irr}(C/Z)$ of the block of
$\psi$ is $H$-invariant. Then $\psi_{DC}=e\nu$ is an $F$-valued character and
$\eta\in{\rm Irr}(CD)$. Notice that by Corollary 11.29 of [Isa] $e$ is odd.
Using again Lemma 3.4, we have that $\eta=\theta_{\lambda}$, where
$\lambda\in{\rm Irr}(D)$ is linear. On the other hand, as $CD$ is a central
product of $C$ and $D$, we have that $\eta=\nu\cdot\lambda$, where $\nu\in{\rm
Irr}(C|\mu)$ and $\lambda\in{\rm Irr}(D|\mu)$ being $\mu=\lambda_{Z}$. In
particular, $\eta_{C}=\nu$.

We claim that $\eta$ can be afforded by an $F$-representation. Let $P$ be a
Sylow $p$-subgroup of $C$. Let $x\in P-Z$. As
$\nu=\eta_{C}=(\theta_{\lambda})_{C}$, using the values of $\theta_{\lambda}$
we have that $\nu(x)=0$. If $x\in Z$, then $\nu(x)=\theta(1)\lambda(x)$.
Therefore

$\nu_{P}={\theta(1)\over|P:D|}\mu^{P}\,.$

Notice that ${\theta(1)\over|P:D|}$ is not divisible by 2 because
$\theta\in{\rm Irr}(C/Z)$ has 2-defect zero. Finally, since $\psi_{DC}=e\eta$,
where $e$ is odd, then $\psi_{C}=e\nu$ and thus $[\psi_{P},\mu^{P}]$ is odd.
Hence $[\psi,\mu^{H}]$ is odd. Since $\mu$ is afforded by an
$F$-representation, we have that $m_{F}(\psi)$ is odd, by Corollary 10.2(c) of
[Isa]. But Corollary 10.13 of [Isa] implies that $m_{F}(\chi)\leq 2$.
Therefore $m_{F}(\psi)=1$, and this completes the proof. ∎

According to M. Geline, Theorem 4.4 can also be proved by using the main
theorem of [GG] and some number theoretical arguments. We notice that the case
where $\chi$ has odd degree of Theorem 4.4 follows from a theorem of Fong
[Isa, Corollary 10.13] using [Isa, Corollary 10.2(h)].

##  5\. Fields of Characters in $\mathcal{F}_{p}$

We briefly pause in our journey to the proof of Theorem A and the reduction
theorem for Conjecture D to show the easy containments in Theorem B and
Conjecture D. In other words, we show that every number field in
$\mathcal{F}_{p}$ is the field of values of some irreducible character of
$p$-height zero. We will also show that the statement of Conjecture D is
implied by the statement of [Nav1, Conjecture B].

######  Theorem 5.1.

Suppose that ${\mathbb{Q}}\subseteq F\subseteq{\mathbb{Q}}_{n}$, where $n$ is
the conductor of $F$. Write $n=p^{a}m$, where $m$ is not divisible by $p$. If
$|{\mathbb{Q}}_{n}:\langle F,{\mathbb{Q}}_{m}\rangle|$ is not divisible by
$p$, then there is a solvable group $G$ and $\chi\in{\rm Irr}(G)$ of
$p$-height zero such that ${\mathbb{Q}}(\chi)=F$.

###### Proof.

Let $\zeta_{n}$ be a primitive $n$-th root of unity, and let
$C_{n}=\langle\zeta_{n}\rangle$ be the cyclic group of order $n$, which is
acted on faithfully by $\mathcal{G}={\rm Gal}({\mathbb{Q}}_{n}/{\mathbb{Q}})$.
Let $G$ be the semidirect product of $C_{n}$ with $H={\rm
Gal}({\mathbb{Q}}_{n}/F)\leq\mathcal{G}$. If $\lambda\in{\rm Irr}(C_{n})$ has
order $n$, then $\lambda^{G}=\chi\in{\rm Irr}(G)$ has field of values $F$. Let
$\nu=\lambda_{C_{m}}$, where $C_{m}\leq C_{n}$ has order m. Then $\nu$ has
order $m$. Notice that $H_{\nu}={\rm Gal}({\mathbb{Q}}_{n}/\langle
F,{\mathbb{Q}}_{m}\rangle)$ has order not divisible by $p$ by hypothesis. Thus
$G_{\nu}=C_{n}H_{\nu}$. By the Fong-Reynolds Theorem 9.14 of [Nav2], it
follows that $\chi$ has height zero if and only if $\lambda^{G_{\nu}}$ has
height zero, which it has, because it has degree not divisible by $p$. ∎

We can also prove one of the implications in Corollary C.

######  Corollary 5.2.

Suppose that $d\neq 1$ is an odd square-free integer. Then there is a group
$G$ and a $2$-height zero character $\chi\in{\rm Irr}(G)$ such that
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\sqrt{d})$.

###### Proof.

By considering the cyclic group of order 4, we may assume that $d\neq\pm 1$ is
an odd square-free integer. Let $F={\mathbb{Q}}(\sqrt{d})$. If $d\equiv 1$ mod
4, then $F\subseteq{\mathbb{Q}}_{|d|}$, $|d|$ is the conductor of $F$, and we
are done by Theorem 5.1. Suppose that $d\equiv 3$ mod 4. Let $n=2^{2}|d|$. By
Theorem 5.1, we only need to show that
$\langle{\mathbb{Q}}_{|d|},{\mathbb{Q}}(\sqrt{d})\rangle={\mathbb{Q}}_{n}$.
Since $|{\mathbb{Q}}_{n}:{\mathbb{Q}}_{|d|}|=2$, this can only fail if
${\mathbb{Q}}(\sqrt{d})\subseteq{\mathbb{Q}}_{|d|}$, which is not possible
because the conductor of ${\mathbb{Q}}(\sqrt{d})$ is $n$. ∎

We finish this section by showing that Conjecture D follows from the Alperin-
McKay-Navarro conjecture [Nav1, Conjecture B]. We recall that the Alperin-
McKay-Navarro conjecture predicts, for a $p$-block $B$ of a finite group $G$
and its Brauer first main correspondent $b$, that there is a bijection between
the sets of height zero characters of $B$ and $b$ such that, if $\chi$
corresponds to $\chi^{*}$ then
${\mathbb{Q}}_{n}(\chi)={\mathbb{Q}}_{n}(\chi^{*})$ where
$n=|G|_{p^{\prime}}$. We care to mention that the currently accepted and most
studied form of [NT, Conjecture C] is more general, predicting the existence
of an $\mathcal{H}$-equivariant bijection between height zero characters of
$B$ and $b$, where the Galois group $\mathcal{H}$, which contains ${\rm
Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{n})$, is defined in [Nav1, Section 2].
When we write the Alperin-McKay-Navarro conjecture we refer to the statement
predicting $\mathcal{H}$-equivariant character bijections.

######  Theorem 5.3.

Conjecture D follows from the Alperin-McKay-Navarro conjecture.

###### Proof.

Let $\chi\in{\rm Irr}(B)$ be of $p$-height zero, where $B$ has defect group
$D$. By the Alperin-McKay-Navarro conjecture, there is $\tau\in{\rm Irr}({\bf
N}_{G}(D))$ of heigh zero, in the Brauer correspondent block $b$ of ${\bf
N}_{G}(D)$ such that ${\mathbb{Q}}_{n}(\chi)={\mathbb{Q}}_{n}(\tau)$, where
$n=|G|_{p^{\prime}}$. In particular $c(\chi)_{p}=c(\tau)_{p}$ reasoning as in
Lemma 2.2. Hence, it is enough to prove that
$Q_{c(\tau)_{p}}\subseteq{\mathbb{Q}}_{pm}(\tau)$ with $m=|{\bf
N}_{G}(D)|_{p^{\prime}}$. We may then assume that $D\triangleleft\,G$. In this
case, we can apply Lemma 3.5. ∎

##  6\. The Reduction

There is one more issue that we have to solve before proving a reduction to
quasi-simple groups of Conjecture D. If $G/N$ is a $p^{\prime}$-group,
$\theta\in{\rm Irr}(N)$ is $G$-invariant and $p$-rational, it is not
necessarily true that the characters of $G$ over $\theta$ are $p$-rational
(even if they extend $\theta$ and $p$ is odd), as shown by $p=3$ and ${\tt
SmallGroup}(24,4)$. We need the following.

######  Theorem 6.1.

Suppose that $G/N$ is a simple group of order coprime to $p$. Let
$\theta\in{\rm Irr}(N)$ be $G$-invariant and $p$-rational. If $\chi\in{\rm
Irr}(G|\theta)$, then $c(\chi)_{p}\leq p$.

###### Proof.

Suppose first that $G/N$ has order $q$. Then this follows from Theorem B of
[Val0]. Suppose now that $G/N$ is perfect. By Theorem 4.3, there is an
isomorphic character triple $(G^{*},N^{*},\theta^{*})$ such that $N^{*}={\bf
Z}(G^{*})$, ${\mathbb{Q}}(\theta^{*})\subseteq{\mathbb{Q}}(\theta)$ and
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\chi^{*},\theta^{*})$ whenever $\chi\in{\rm
Irr}(G|\theta)$ and $\chi^{*}$ corresponds to $\chi$ under the character
triple isomorphism. In particular, we have that $G^{*}/N^{*}$ is a simple non
abelian $p^{\prime}$-group, and $\theta^{*}$ is $p$-rational. Let $x\in
G^{*}$, and let $\delta\in{\rm Irr}(N^{*}\langle x\rangle)$ be over
$\theta^{*}$. Since $N^{*}\langle x\rangle/N^{*}$ is cyclic and $\theta^{*}$
is $G^{*}$-invariant, $\delta_{N^{*}}=\theta^{*}$. Since $x_{p}\in N^{*}$ and
$\delta$ is linear, we have that
$\delta(x)=\delta(x_{p})\delta(x_{p^{\prime}})=\theta^{*}(x_{p})\delta(x_{p^{\prime}})\in{\mathbb{Q}}_{|G^{*}|_{p^{\prime}}}$.
Then every $\delta\in{\rm Irr}(N^{*}\langle x\rangle|\theta^{*})$ is
$p$-rational. Since $\chi_{N^{*}\langle x\rangle}$ is a sum of irreducible
characters lying over $\theta^{*}$ and the choice of $x\in G^{*}$ was
arbitrary, we are done. ∎

######  Conjecture 6.2.

Let $\chi\in{\rm Irr}(G)$ of $p$-height zero, where $G$ is a quasi-simple
group. Assume in addition that the $p$-block $B$ containing $\chi$ is not
(virtual) Morita equivalent over an absolutely unramified complete discrete
valuation ring to a $p$-block of any group $H$ with $|H:{\bf Z}(H)|<|G:{\bf
Z}(G)|$. Write $c(\chi)=p^{a}m$. Then
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm}(\chi)$.

Notice that if the $p$-block $B$ of $\chi$ is virtual Morita equivalent over
an absolutely unramified complete discrete valuation ring to a $p$-block
$\tilde{B}$, by Theorem 1.6 of [KL], there is a height zero character
$\tilde{\chi}$ in $\tilde{B}$ and an integer $l$ not divisible by $p$ such
that ${\mathbb{Q}}_{l}(\chi)={\mathbb{Q}}_{l}(\tilde{\chi})$. By Lemma 2.2, we
have that $c(\chi)_{p}=c(\tilde{\chi})_{p}=p^{a}$, and therefore
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pl}(\chi)$ if and only if
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pl}(\tilde{\chi})$. Hence, if
$c(\chi)_{p^{\prime}}=m$ and $c(\tilde{\chi})_{p^{\prime}}=m_{1}$, then
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm}(\chi)$ if, and only if,
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm_{1}}(\tilde{\chi})$, by Lemma
2.1. We will use this argument below.

######  Theorem 6.3.

Let $G$ be a finite group, and let $p$ be a prime. Let $\chi\in{\rm Irr}(G)$
be of $p$-height zero, and write $c(\chi)=p^{a}m$, where $a\geq 0$ and $p$
does not divide $m$. If Conjecture 6.2 is true, then
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm}(\chi)$.

###### Proof.

We argue by induction on $|G:{\bf Z}(G)|$. We may assume that $a\geq 2$,
because otherwise the statement is trivially satisfied. Let
$n=|G|_{p^{\prime}}$. By Lemma 2.1, it is enough to show that
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pn}(\chi)$.

Step 1. If $N\triangleleft\,G$ is a proper normal subgroup, then we may assume
that $\chi_{N}=e\theta$ for some $\theta\in{\rm Irr}(N)$, with
$c(\theta)_{p}<c(\chi)_{p}$.

Let $T$ be the stabilizer of $\theta$ in $G$, let $\psi\in{\rm Irr}(T|\theta)$
be the Clifford correspondent of $\chi$ over $\theta$. By Theorem 3.2, we have
that ${\mathbb{Q}}_{pn}(\psi)={\mathbb{Q}}_{pn}(\chi)$ and
$c(\chi)_{p}=p^{a}=c(\psi)_{p}$. Assume that $T<G$. Then $|T:{\bf
Z}(T)|<|G:{\bf Z}(G)|$, and by induction, if $m_{1}$ is the $p^{\prime}$-part
of the conductor of $\psi$, we have that
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm_{1}}(\psi)\subseteq{\mathbb{Q}}_{pn}(\psi)={\mathbb{Q}}_{pn}(\chi)$.
By Lemma 2.1, we deduce that
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm}(\chi)$. Hence, we may assume
that $\chi_{N}=e\theta$. In particular,
${\mathbb{Q}}(\theta)\subseteq{\mathbb{Q}}(\chi)$, and $c(\theta)$ divides
$c(\chi)$. If $c(\theta)_{p}=c(\chi)_{p}$, and $m_{2}$ is the
$p^{\prime}$-part of the conductor of $\theta$, then by induction
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pm_{2}}(\theta)\subseteq{\mathbb{Q}}_{pm_{2}}(\chi)$.
Again by Lemma 2.1, we deduce what we want.

Step 2. $G$ does not have a normal subgroup of index $p$.

Otherwise, let $N$ be a normal subgroup of $G$ with $G/N$ of order $p$. By
Step 1, $\chi_{N}$ is homogeneous. Since $G/N$ is cyclic we can write
$\chi_{N}=\theta\in{\rm Irr}(N)$.

Suppose that $G=NH$, $M=N\cap H$ and there exists some $H$-invariant
$\varphi\in{\rm Irr}(M)$ of $p$-height zero such that $[\theta_{M},\varphi]$
is not divisible by $p$ and
${\mathbb{Q}}_{pn}(\varphi)\subseteq{\mathbb{Q}}_{pn}(\theta)$. Let
$\xi\in{\rm Irr}(H)$ be an extension of $\varphi$. By Lemma 3.7, we have that
$\xi$ has $p$-height zero. By Lemma 3.9(c), we have that
${\mathbb{Q}}_{p}(\xi,\theta)={\mathbb{Q}}_{p}(\chi,\varphi)$. Notice that
${\mathbb{Q}}_{pn}(\chi)\subseteq{\mathbb{Q}}_{pn}(\chi,\varphi)\subseteq{\mathbb{Q}}_{pn}(\chi,\theta)={\mathbb{Q}}_{pn}(\chi)$.
Therefore
${\mathbb{Q}}_{pn}(\chi)={\mathbb{Q}}_{pn}(\chi,\varphi)={\mathbb{Q}}_{pn}(\theta,\xi)$.
Also, ${\mathbb{Q}}_{pn}(\xi)\subseteq{\mathbb{Q}}_{pn}(\chi)$. Write
$c(\theta)_{p}=p^{b}$ and $c(\xi)_{p}=p^{c}$. We have that
${\mathbb{Q}}(\xi)\subseteq{\mathbb{Q}}_{pn}(\xi)\subseteq{\mathbb{Q}}_{pn}(\chi)\subseteq{\mathbb{Q}}_{p^{a}n}$.
Therefore $c\leq a$. Notice that if $\theta$ and $\xi$ are $p$-rational, then
${\mathbb{Q}}(\chi)\subseteq{\mathbb{Q}}_{pn}(\chi)={\mathbb{Q}}_{pn}(\chi,\varphi)={\mathbb{Q}}_{pn}(\theta,\xi)={\mathbb{Q}}_{pn}$,
but we are assuming that $a\geq 2$. Hence $b,c\geq 1$. Now
${\mathbb{Q}}_{pn}(\theta)\subseteq\langle{\mathbb{Q}}_{p^{b}n},{\mathbb{Q}}_{p}\rangle$,
and
${\mathbb{Q}}_{pn}(\xi)\subseteq\langle{\mathbb{Q}}_{p^{c}n},{\mathbb{Q}}_{p}\rangle$.
Thus

${\mathbb{Q}}(\chi)\subseteq{\mathbb{Q}}_{pn}(\chi)=\langle{\mathbb{Q}}_{pn}(\theta),{\mathbb{Q}}_{pn}(\xi)\rangle\subseteq\langle{\mathbb{Q}}_{p^{b}n},{\mathbb{Q}}_{p^{c}n},{\mathbb{Q}}_{p}\rangle\subseteq{\mathbb{Q}}_{p^{d}n}\,,$

where $d={\rm max}(b,c)$. Hence $p^{a}\leq p^{d}$. Since $b<a$ by Step 1 and
$c\leq a$, we conclude that $d=c=a$. If $|H:{\bf Z}(H)|<|G:{\bf Z}(G)|$ then,
by induction, we have that
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pn}(\xi)\subseteq{\mathbb{Q}}_{pn}(\chi)$,
and we are done (using Lemma 2.1).

By Lemma 3.8, we may assume that the defect group $D_{0}$ of the block of
$\theta$ is normal in $G$. By Lemma 3.10, we are done in this case.

Step 3. $G$ does not have a proper normal subgroup of index not divisible by
$p$.

Otherwise, let $N\triangleleft\,G$ such that $G/N$ is simple of order not
divisible by $p$. By Step 1, $\chi_{N}=e\theta$. Recall that $c(\chi)_{p}\geq
p^{2}$. Hence, by Theorem 6.1, we have that $c(\theta)_{p}\geq p$, in this
case. By Lemma 4.2.(ii) of [NT], we conclude that $c(\chi)_{p}=c(\theta)_{p}$,
contradicting Step 1.

Final Step. Let $N$ be a maximal normal subgroup of $G$. By Steps 2 and 3,
$G/N$ is simple non-abelian of order divisible by $p$. By Step 1, write
$\chi_{N}=e\theta$, where $c(\theta)_{p}=p^{b}$ and $b<a$. By Theorem 4.3,
there is a quasi-simple group $G^{*}$ and a $p$-height zero character
$\chi^{*}$ of $G^{*}$ such that
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\chi^{*},\theta)$. Write
$c(\chi^{*})=p^{c}k$, where $k$ is not divisible by $p$. Then the conductor of
the field
${\mathbb{Q}}(\chi^{*},\theta)=\langle{\mathbb{Q}}(\chi^{*}),{\mathbb{Q}}(\theta)\rangle$
is the least common multiple of the conductors of $\chi^{*}$ and $\theta$. In
particular, its $p$-part is $p^{{\rm max}(c,b)}$. As
$c(\chi)=c({\mathbb{Q}}(\chi^{*},\theta))$, we have that $a={\rm max}(c,b)$.
Since $b<a$, we have that $a=c$. Thus $c(\chi)_{p}=c(\chi^{*})_{p}$. If the
$p$-block $B^{*}$ of $\chi^{*}$ is virtual Morita equivalent over an
absolutely unramified complete discrete valuation ring to a $p$-block
$\tilde{B}$ of a group $H$ with $|H:{\bf Z}(H)|<|G:{\bf Z}(G)|$, then
$c(\chi^{*})_{p}=c(\tilde{\chi})_{p}$ for some $\tilde{\chi}\in\tilde{B}$. As
explained before the statement of this theorem, we would be done in this case,
using induction. Therefore, by Conjecture 6.2, we have that
${\mathbb{Q}}_{p^{a}}\subseteq{\mathbb{Q}}_{pk}(\chi^{*})\subseteq{\mathbb{Q}}_{pk}(\chi)$.
By using Lemma 2.1, the proof of the theorem follows. ∎

##  7\. Theorem A for quasisimple groups

The aim of this section is to prove Conjecture 6.2 in the case that $p=2$, see
Theorem 7.2 below. In the light of Theorem 6.3 (and Theorem 5.1), this will
complete the proof of Theorem A (and Theorem B).

The following is useful when working on extensions of quasi-simple groups.

######  Theorem 7.1.

Suppose that $G/N$ is abelian. Write $n=|G|_{2^{\prime}}$. Suppose that
$\chi\in{\rm Irr}(G)$ has 2-height zero and suppose that
${\mathbb{Q}}_{2^{a}}\subseteq{\mathbb{Q}}_{n}(\chi)$, where
$c(\chi)_{2}=2^{a}$. Let $\theta\in{\rm Irr}(N)$ be an irreducible constituent
of $\chi_{N}$, and write $c(\theta)_{2}=2^{b}$. Then
${\mathbb{Q}}_{2^{b}}\subseteq{\mathbb{Q}}_{n}(\theta)$.

###### Proof.

We argue by induction on $|G:N|$. We may assume that $G/N$ has prime index. By
Theorem 3.2, we may assume that $\chi_{N}=\theta$ is irreducible. By Lemma
4.2.(ii) of [NT], we may assume that $G/N$ has order 2. Let $D$ be a defect
group of the block of $\chi$, such that $DN=G$ and $D_{0}=D\cap N$ is a defect
group of the block of $\theta$. Let $H={\bf N}_{G}(D_{0})$ and $M=H\cap N$. By
Lemma 3.8, there is $\varphi\in{\rm Irr}(M)$ of height zero,
${\mathbb{Q}}_{n}(\varphi)\subseteq{\mathbb{Q}}_{n}(\theta)$ with an extension
$\xi\in{\rm Irr}(H)$ such that
${\mathbb{Q}}(\chi,\varphi)={\mathbb{Q}}(\theta,\xi)$. Thus
${\mathbb{Q}}_{n}(\chi,\varphi)={\mathbb{Q}}_{n}(\theta,\xi)$. Hence
${\mathbb{Q}}_{2^{a}}\subseteq{\mathbb{Q}}_{n}(\theta,\xi)$. We may assume
that $b\geq 2$. Suppose that $c(\xi)_{2}=2^{c}$. We know that
${\mathbb{Q}}_{2^{c}}\subseteq{\mathbb{Q}}_{n}(\xi)$ by Lemma 3.10. If $\xi$
is 2-rational, then we are done. So we may assume that $c\geq 2$, and thus
$i\in{\mathbb{Q}}_{n}(\xi)$. Then
${\mathbb{Q}}_{n}(i)\subseteq{\mathbb{Q}}_{n}(\xi)\cap{\mathbb{Q}}_{n}(\theta)$.
Since
${\mathbb{Q}}_{n}(\xi),{\mathbb{Q}}_{n}(\theta)\subseteq{\mathbb{Q}}_{|G|}$,
and ${\rm Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{n}(i))$ is a cyclic 2-group,
we then have that ${\mathbb{Q}}_{n}(\xi)\subseteq{\mathbb{Q}}_{n}(\theta)$ or
${\mathbb{Q}}_{n}(\theta)\subseteq{\mathbb{Q}}_{n}(\xi)$. Since
${\mathbb{Q}}_{2^{a}}\subseteq{\mathbb{Q}}_{n}(\theta,\xi)$, we may assume the
second. Then ${\mathbb{Q}}_{2^{a}}\subseteq{\mathbb{Q}}_{n}(\xi)$. Thus $a\leq
c$. If $H<G$, then by induction
${\mathbb{Q}}_{2^{c}}\subseteq{\mathbb{Q}}_{n}(\varphi)\subseteq{\mathbb{Q}}_{n}(\theta)$.
Thus we may assume that $D_{0}\triangleleft\,G$. But in this case, we are done
by Lemma 3.5. ∎

######  Theorem 7.2.

Let $\chi\in{\rm Irr}(G)$ of 2-height zero, where $G$ is a quasi-simple group.
Assume in addition that the $2$-block $B$ containing $\chi$ is not (virtual)
Morita equivalent over an absolutely unramified complete discrete valuation
ring to a $2$-block of any group $H$ with $|H:{\bf Z}(H)|<|G:{\bf Z}(G)|$. If
$c(\chi)=2^{a}m$, where $m$ is odd, then
${\mathbb{Q}}_{2^{a}}\subseteq{\mathbb{Q}}_{m}(\chi)$.

######  Theorem 7.3.

Theorem 7.2 is true in the case $G/{\mathbf{Z}}(G)$ is a simple group of Lie
type in characteristic $2$.

###### Proof.

In the case $S:=G/{\mathbf{Z}}(G)$ is isomorphic to ${\sf A}_{5}$, ${\sf
A}_{6}$, ${\sf A}_{8}$, $\operatorname{SL}_{3}(2)$,
$\operatorname{SU}_{4}(2)$, $\operatorname{Sp}_{6}(2)$,
$\operatorname{PSL}_{3}(4)$, $\operatorname{PSU}_{6}(2)$, $\Omega^{+}_{8}(2)$,
${}^{2}\\!B_{2}(8)$, $G_{2}(4)$, $F_{4}(2)$, ${}^{2}\\!F_{4}(2)^{\prime}$, or
${}^{2}\\!E_{6}(2)$, the statement is checked using [GAP]. Hence we may assume
that $S$ is not isomorphic to any of these simple groups. This implies that
$G$ is a quotient (by a central subgroup) of ${\mathcal{G}}^{F}$, where
${\mathcal{G}}$ is a simple, simply connected, algebraic group in
characteristic $2$ and $F:{\mathcal{G}}\to{\mathcal{G}}$ a Steinberg
endomorphism. It follows from the main result of [Hum] that any $2$-block $B$
of $G$ is either of defect $0$, or of maximal defect. Moreover, in the former
case $\chi\in\operatorname{Irr}(B)$ is just the Steinberg character; in
particular it is rational and so we are done. In the latter case, $\chi(1)$ is
odd, and the statement follows from [NT, Theorem A1]. ∎

######  Theorem 7.4.

Theorem 7.2 is true in the case $G/{\mathbf{Z}}(G)$ is an alternating or
sporadic simple group.

###### Proof.

In the case $S:=G/{\mathbf{Z}}(G)$ is isomorphic to ${\sf A}_{n}$ with $5\leq
n\leq 8$ or one of $26$ sporadic simple groups, the statement is checked using
[GAP]. Hence we may assume that $S={\sf A}_{n}$ with $n\geq 9$.

(a) First we consider the case $G=S$. If $\chi$ extends to ${\sf S}_{n}$ then
$\chi$ is rational. Otherwise, [JK, Theorem 2.5.13] shows that the ${\sf
S}_{n}$-character lying above $\chi$ is labeled by a self-associated partition
of $n$, with hook lengths along the main diagonal of the Young diagram being
the $k\geq 1$ odd integers $2h_{1}+1>2h_{2}+1>\ldots>2h_{k}+1$, in which case
the only possible irrational values of $\chi$ are

$\frac{(-1)^{(n-k)/2}\pm\sqrt{(-1)^{(n-k)/2}\prod^{k}_{i=1}(2h_{i}+1)}}{2}.$

Since $n=\sum^{k}_{i=1}(2h_{i}+1)$, we see that $(-1)^{(n-k)/2}=1$ if and only
if $\prod^{k}_{i=1}(2h_{i}+1)\equiv 1\pmod{4}$. It follows that $\chi$ is
$2$-rational, and we are done in this case.

(b) It remains to handle the case $G=2{\sf A}_{n}$. We change the notation,
and let $\tilde{B}$ the $2$-block of $G$ containing $\chi$. We also embed $G$
in a double cover $\tilde{G}=2{\sf S}_{n}$ of ${\sf S}_{n}$. By [Den, Lemma
2.2], $\tilde{B}$ contains a unique block $B$ of $G/{\mathbf{Z}}(G)\cong{\sf
A}_{n}$. Similarly, by [Den, Lemma 2.1], $\tilde{B}$ is covered by a unique
block $\tilde{B}_{s}$ of $\tilde{G}$, and $B$ is covered by a unique block
$B_{S}$ of $(\tilde{G})/{\mathbf{Z}}(G)\cong{\sf S}_{n}$. All these blocks
$\tilde{B}$, $\tilde{B}_{S}$, $B$, and $B_{S}$ have the same weight $w\geq 0$.

If $\chi$ is trivial on ${\mathbf{Z}}(G)$, then we are done by (a). Hence we
may assume that $\chi$ is a spin character of $G$. Since $\tilde{B}$ contains
a spin character of height zero, by [Den, Proposition 3.1] we must have that
$w\in\\{0,1\\}$ (and $\chi$ is the unique spin character of height zero in
$\tilde{B}$). The defect groups $D$ of $B$ are the Sylow $2$-subgroups of
${\sf A}_{2w}$, and hence $D=1$. This implies by [Den, Lemma 2.2] that the
defect groups $\tilde{D}$ of $\tilde{B}$ are precisely ${\mathbf{Z}}(G)$. Thus
we are in the case of central defect, and $\chi$ has relative defect zero,
with respect to the faithful linear character $\mu$ of ${\mathbf{Z}}(G)$. Now
[Nav0, Theorem 2.1] gives an explicit bijection between defect zero characters
of $G/{\mathbf{Z}}(G)$ and relative defect zero characters of $G$. In
particular, this bijection implies that $\chi(g)=0$ whenever the $2$-part
$g_{2}$ of $g$ is not in ${\mathbf{Z}}(G)$.

We now show that $\chi$ is $2$-rational. Indeed, write $g=g_{2}g_{2^{\prime}}$
as the product of the $2$-part and the $2^{\prime}$-part of $g$. By the above,
$\chi(g)=0$ if $g_{2}\notin{\mathbf{Z}}(G)$. If $g_{2}\in{\mathbf{Z}}(G)$,
then $\chi(g)=\mu(g_{2})\chi(g_{2^{\prime}})=\pm\chi(g_{2^{\prime}})$ is
$2$-rational. Hence the statement follows in this case as well. ∎

######  Theorem 7.5.

Theorem 7.2 is true in the case $G/{\mathbf{Z}}(G)$ is a simple classical
group in odd characteristic.

###### Proof.

In the case $S:=G/{\mathbf{Z}}(G)$ is isomorphic to
$\operatorname{PSU}_{4}(3)$, $\operatorname{PSU}_{6}(2)$, $\Omega_{7}(3)$, or
$G_{2}(3)$, the statement is checked using [GAP]. (Note that in Theorem 7.2 we
are dealing with a $2$-block $B$ of $G$, and so we may assume ${\bf
O}_{2^{\prime}}({\bf Z}(G))$ is cyclic. Hence in the case of covers of
$\operatorname{PSU}_{4}(3)$, it suffices to handle the two covers
$12_{1}\cdot\operatorname{PSU}_{4}(3)$ and
$12_{2}\cdot\operatorname{PSU}_{4}(3)$, which are given in [GAP].)

Hence we may assume that $S$ is not isomorphic to any of these simple groups,
as well as any Lie type group in characteristic $2$. This implies that $G$ is
a quotient (by a central subgroup) of ${\mathcal{G}}^{F}$, where
${\mathcal{G}}$ is a simple, simply connected, algebraic group in odd
characteristic $r\neq 2$ and $F:{\mathcal{G}}\to{\mathcal{G}}$ a Steinberg
endomorphism. Without any loss, we may replace $G$ by ${\mathcal{G}}^{F}$. Let
$({\mathcal{G}}^{*},F^{*})$ be dual to $({\mathcal{G}},F)$; in particular,
${\mathcal{G}}^{*}$ is of adjoint type, and let
$G^{*}:=({\mathcal{G}}^{*})^{F^{*}}$.

By the main result of [BrM], the set $\operatorname{Irr}(B)$ of complex
characters in the $2$-block $B$ containing $\chi$ is contained in
${\mathcal{E}}_{2}(G,s)$ for some semisimple element $s\in G^{*}$ of odd
order. Suppose that $s$ is not quasi-isolated (in the sense of [Bon]). Then,
by the main result of [BoR], $B$ is Morita equivalent to a $2$-block of a
group $H$ with $|H:{\mathbf{Z}}(H)|<|G:{\mathbf{Z}}(G)|$. Moreover, by [FK,
Proposition 4.2] this Morita equivalence descends to an absolutely unramified
discrete valuation ring contrary to our assumption. Hence we may assume that
$s$ is quasi-isolated.

Assume in addition that ${\mathcal{G}}$ is not of type $A$. By the
classification result of Bonnafé [Bon, Table 2], the odd-order assumption on
$s$ implies that $s=1$. In this case, by [CE, Theorem 21.14],
${\mathcal{E}}_{2}(G,s)$ is just the set of irreducible characters in the
principal $2$-block $B_{0}$ of $G$. In such a case, $\chi(1)$ is odd, and the
statement follows from [NT, Theorem A1].

It remains to consider the case ${\mathcal{G}}$ is of type $A$. The same
arguments as in the preceding paragraph allow us to assume that $s\neq 1$, and
so $s$ is not isolated, see [Bon, Table 2]. The main result of [BoDR] together
with [FK, Proposition 4.2] now shows that $B$ is again Morita equivalent over
an absolutely unramified complete discrete valuation ring to a $2$-block of a
group $H$ with $|H:{\mathbf{Z}}(H)|<|G:{\mathbf{Z}}(G)|$, contrary to our
assumption. ∎

######  Theorem 7.6.

Theorem 7.2 is true in the case $G/{\mathbf{Z}}(G)$ is a simple exceptional
group in odd characteristic.

###### Proof.

We keep the same notation from Theorem 7.5. That is $G$ is a (quotient by a
central subgroup) of ${\mathcal{G}}^{F}$, where ${\mathcal{G}}$ is a simple,
simply connected, algebraic group in odd characteristic $r\neq 2$ and
$F:{\mathcal{G}}\to{\mathcal{G}}$ is a Steinberg endomorphism. Arguing as in
Theorem 7.5, we can also assume that $s$ is isolated. By Lemma 2.1, we observe
that is suffices to prove that
${\mathbb{Q}}_{c(\psi)_{2}}\subset{\mathbb{Q}}_{|G|_{2^{\prime}}}(\psi)$ for
every height zero character $\psi\in\operatorname{Irr}_{0}(B)$. We may also
assume $c(\psi)_{2}\geq 4$ since otherwise
${\mathbb{Q}}_{c(\psi)_{2}}={\mathbb{Q}}$ and the statement is trivally true.

(a) Let us first assume that the defect group of $B$ has order
$|G^{\ast}:{\mathbf{C}}_{G^{\ast}}(s)|_{2}$ and
${\mathbf{C}}_{{\mathcal{G}}^{\ast}}^{\circ}(s)$ has only components of
classical type. Then $\psi\in\mathcal{E}(G,st)$ for some element $t\in
G^{\ast}$ which is $2$-central in the group
$H:={\mathbf{C}}^{\circ}_{{\mathcal{G}}^{\ast}}(s)^{F^{\ast}}$, see [Mal,
(2.1)].

If ${\mathbf{Z}}({\mathcal{G}})\neq 1$ then we let
${\mathcal{G}}\lhd\tilde{{\mathcal{G}}}$ a regular embedding with dual
surjective morphism
$\iota^{\ast}:\tilde{{\mathcal{G}}}^{\ast}\to{\mathcal{G}}^{\ast}$. Otherwise
set $\tilde{{\mathcal{G}}}:={\mathcal{G}}$. There exists a semisimple element
$\tilde{s}\in\tilde{G}^{\ast}:=(\tilde{{\mathcal{G}}}^{\ast})^{F^{\ast}}$ of
$2^{\prime}$-order such that $\iota^{\ast}(\tilde{s})=s$ and
$\tilde{t}\in{\mathbf{C}}_{\tilde{G}^{\ast}}(\tilde{s})_{2}$ with
$\iota^{\ast}(\tilde{t})=t$. We let
$\chi\in\mathcal{E}(\tilde{G},\tilde{s}\tilde{t})$ be a character covering
$\psi$. By [GM, Theorem 4.7.9], [GM, Proposition 4.5.5] and using that
${\mathbf{C}}_{\tilde{G}^{\ast}}(\tilde{s}\tilde{t})$ has only components of
classical type we have
${\mathbb{Q}}(\chi)\subset{\mathbb{Q}}_{\mathrm{o}(\tilde{s}\tilde{t})}$ and
so $c(\chi)_{2}\leq\mathrm{o}(\tilde{t})$.

(a1) Assume now first that $|\chi(1):\psi(1)|_{2}>1$. Then $G=E_{7}(q)$ and
$\chi=(\psi^{\prime})^{\tilde{G}}$ for some
$\psi^{\prime}\in\operatorname{Irr}(G{\mathbf{Z}}(\tilde{G})\mid\psi)$. In
this case, $c(\chi)_{2}=c(\psi^{\prime})_{2}$ by Theorem 3.2 and
${\mathbb{Q}}(\chi)\subset{\mathbb{Q}}(\psi^{\prime})$. Since
$\psi\in\operatorname{Irr}(G)$ has height zero, it follows that
$\psi_{{\mathbf{Z}}(G)}$ is trivial by [Ruh, Lemma 8.7]. Hence, we can choose
$\chi\in\operatorname{Irr}(\tilde{G}\mid\psi)$ with the additional property
that $\chi$ is trivial on ${\mathbf{Z}}(\tilde{G})$. A consequence of this
choice is that $c(\psi^{\prime})=c(\psi)$ and
${\mathbb{Q}}(\psi^{\prime})={\mathbb{Q}}(\psi)$.

Since $\chi$ is the unique character in its
$\operatorname{Irr}(\tilde{G}/G)$-orbit which is trivial on
${\mathbf{Z}}(\tilde{G})$, it follows that any Galois automorphism that
stabilizes the $\operatorname{Irr}(\tilde{G}/G)$-orbit of $\chi$ also
stabilizes $\chi$. Hence, [GM, Theorem 4.7.9] and [GM, Proposition 4.5.5] show
therefore that
${\mathbb{Q}}(\chi)\subset{\mathbb{Q}}_{\mathrm{o}(\tilde{s})\mathrm{o}(t)}$.

Recall that $\psi\in\mathcal{E}(G,st)$ has height zero and the defect group of
$B$ has order $|G^{\ast}:{\mathbf{C}}_{G^{\ast}}(s)|_{2}$. Assume that $e$ is
an integer coprime to $\mathrm{o}(t)$ such that $t$ is $H$-conjugate to
$t^{e}$. We claim that $t=t^{e}$. Arguing as in [Mal, Theorem 5.9], we see
that $t$ lies in the centralizer of a Sylow $d$-torus $\mathcal{S}_{d}$ of
${\mathbf{C}}^{\circ}_{{\mathcal{G}}^{\ast}}(s)$ for $d$ the order of $q$
modulo $4$. By the proof of [BrR, Corollary 2.4], the Sylow $2$-subgroup
$W_{2}$ of the Weyl group of
$W:=\mathbf{N}_{H}(\mathcal{S}_{d})/{\mathbf{C}}_{H}(\mathcal{S}_{d})$ is
self-normalizing. Moreover, since $t$ is $2$-central its centralizer $W(t)$ in
$W$ contains a Sylow $2$-subgroup $W_{2}$ of $W$.

Since $\mathbf{N}_{H}(\mathcal{S}_{d})$ controlls $H$-fusion in
${\mathbf{C}}_{H}(\mathcal{S}_{d})$ by [Mal, Proposition 5.11] it follows that
$t$ and $t^{e}$ are conjugate by an element $w\in\mathbf{N}_{W}(W(t))$. In
particular, by conjugacy of Sylow subgroups in $W(t)$ we can assume that
$w\in\mathbf{N}_{W}(W_{2})=W_{2}$ and so $t=t^{e}$ as claimed. This implies
that ${\mathbb{Q}}_{\mathrm{o}(t)}\subset{\mathbb{Q}}_{o(s)}(\psi)$ by [GM,
Proposition 3.3.15] and thus $c(\psi)_{2}\geq\mathrm{o}(t)_{2}$ by Lemma 2.1.
On the other hand,

$c(\chi)_{2}=c(\psi)_{2}\geq\mathrm{o}(t)\geq c(\chi)_{2},$

and so $\mathrm{o}(t)_{2}=c(\psi)_{2}$. Hence,
${\mathbb{Q}}_{c(\psi)_{2}}\subset{\mathbb{Q}}_{o(s)}(\psi)$.

(a2) Assume now that $|\chi(1):\psi(1)|_{2}=1$. In this case, $\tilde{t}$ is
$2$-central in ${\mathbf{C}}_{\tilde{G}^{\ast}}(\tilde{s})$. The argument from
the first case now show that
${\mathbb{Q}}_{c(\chi)_{2}}\subset{\mathbb{Q}}_{\mathrm{o}(\tilde{t})_{2}}\subset{\mathbb{Q}}_{o(\tilde{s})}(\chi)$.
Hence, the claim follows in this case from Theorem 7.1.

(b) Suppose now that $s=1$, i.e. that $B$ is a unipotent block. We can assume
that $B$ has non-maximal defect since otherwise the statement follows from
[NT, Theorem A.1]. By Lemma 3.5 we can also assume that $B$ has non-central
defect. In this case, $B$ is one of the blocks considered in [Ruh, Lemma 7.1].
In case (i) of [Ruh, Lemma 7.1], the defect group of $B$ is dihedral, so every
character in $\operatorname{Irr}_{0}(B)$ is $2$-rational by [Sam, Theorem
8.1]. Hence, the claim holds. In case (ii), $G=E_{8}(q)$ and the height zero
characters where explicitly described in [Ruh, Lemma 7.4]. It follows from
this description that
$\operatorname{Irr}_{0}(B)\subset\cup_{t}\mathcal{E}(G,t)$, where $t\in
G^{\ast}$ runs over elements with $t^{2}=1$, and all height zero characters of
$\operatorname{Irr}_{0}(B)$ have ${\mathbf{Z}}(G)$ in their kernel. Now [GM,
Theorem 4.7.9] and [GM, Proposition 4.5.5] show that ${\mathbb{Q}}(\chi)$ is a
cyclotomic field or ${\mathbb{Q}}(\chi)\subset{\mathbb{Q}}(\sqrt{r})$ which in
both cases implies the statement.

(c) An analysis of the tables in [KM1] shows that in the remaining casese
$G=E_{8}(q)$ and ${\mathbf{C}}_{{\mathcal{G}}^{\ast}}(s)$ is of type
$E_{6}A_{2}$. Let $G(s)$ be the $F$-fixed points of the connected reductive
group in duality with ${\mathbf{C}}_{{\mathcal{G}}^{\ast}}(s)$. For
$t\in{\mathbf{C}}_{G^{\ast}}(s)_{2}$ let
$\psi_{G,st}:\mathcal{E}(G,st)\to\mathcal{E}({\mathbf{C}}_{G^{\ast}}(st),1)$
be Digne–Michel’s unique Jordan decomposition for groups with connected center
as in in [GM, Theorem 4.7.1]. Moreover, note that the centralizer of $st$ in
${\mathbf{C}}_{{\mathcal{G}}^{\ast}}(s)$ is connected. Hence, there exists a
bijection
$\psi_{G(s),st}:\mathcal{E}(G(s),st)\to\mathcal{E}({\mathbf{C}}_{G^{\ast}}(st),1)$
which can be uniquely determined from Digne–Michel’s unique Jordan
decomposition in a regular embedding of $G(s)$. We have a bijection
$\mathcal{J}:\mathcal{E}_{2}(G,s)\to\mathcal{E}_{2}(G(s),s)$ which is the
union of the bijections $\psi_{G(s),st}^{-1}\circ\psi_{G,st}$ with
$t\in{\mathbf{C}}_{G^{\ast}}(s)_{2}$, see [Ruh, Lemma 2.3]. By [GM, Theorem
4.7.9] and the construction of $\mathcal{J}$ it follows that $\mathcal{J}$ is
$\mathrm{Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{\mathrm{o}(s)})$-equivariant.
Moreover, by [Ruh, Proposition D] there exists a bijection $c\mapsto b$
between blocks contained in $\mathcal{E}_{2}(G(s),s)$ and the blocks contained
in $\mathcal{E}_{2}(G,s)$ such that
$\mathcal{J}(\operatorname{Irr}_{0}(b))=\operatorname{Irr}_{0}(c)$. From this
it follows that
${\mathbb{Q}}_{o(s)}(\mathcal{J}(\chi))={\mathbb{Q}}_{o(s)}(\chi)$ and
$c(\chi)_{2}=c(\mathcal{J}(\chi))_{2}$. Hence, it suffices to consider the
height zero characters of the unipotent blocks of $G(s)$. However, by [CE,
Theorem 17.7] the unipotent blocks of $G(s)$ are isomorphic in a natural way
to the blocks of $G(s)_{\mathrm{sc}}$, the $F$-fixed points of the simply
connected covering of $G(s)$. By what we have established about unipotent
blocks,
${\mathbb{Q}}_{c(\mathcal{J}(\chi))_{2}}\subset{\mathbb{Q}}_{|G|_{2^{\prime}}}(\mathcal{J}(\chi)_{2})$.
Therefore,
${\mathbb{Q}}_{c(\chi)_{2}}\subset{\mathbb{Q}}_{|G|_{2^{\prime}}}(\psi)$ for
all height zero characters $\psi$. ∎

We finish this paper proving another corollary of Theorem A. For an integer
$e\geq 1$, let $\sigma_{e}$ be the Galois automorphism in ${\rm
Gal}({\mathbb{Q}}^{\rm ab}/{\mathbb{Q}})$ fixing $2^{\prime}$-roots of unity
and sending $\xi$ to $\xi^{1+2^{e}}$, where $e\geq 1$. If $G$ is any finite
group, then

$\mathcal{G}={\rm
Gal}({\mathbb{Q}}_{|G|}/{\mathbb{Q}}_{|G|_{2^{\prime}}})=\langle\tau_{1},\tau_{2}\rangle,$

where $\tau_{i}$ is the restriction of $\sigma_{i}$ to ${\mathbb{Q}}_{|G|}$.
Notice that a character $\chi\in{\rm Irr}(G)$ is 2-rational if, and only if,
$\chi$ is $\mathcal{G}$-fixed. The set of 2-height zero characters fixed under
the action of $\langle\sigma_{1}\rangle$ has been recently studied in
connection with the number of generators of 2-defect groups (see [RSV, NRSV,
Val1]).

We have the following.

######  Theorem 7.7.

Let $\chi\in{\rm Irr}(G)$ of 2-height zero. Then $\chi$ is 2-rational if, and
only if, $\chi$ is $\sigma_{1}$-fixed.

###### Proof.

Let $m=|G|_{2^{\prime}}$. Suppose that $\chi$ is $\sigma_{1}$-fixed. Then
${\mathbb{Q}}_{m}(\chi)$ is also fixed by $\sigma_{1}$. If $\chi$ is not
2-rational, then $i\in{\mathbb{Q}}_{m}(\chi)$ by Theorem A. However,
$\sigma_{1}(i)=i^{3}\neq i$, a contradiction. ∎

Theorem 7.7 is not true for characters which do not have height zero. The
smallest example is an irreducible character $\chi$ of degree 2 of a
semidihedral group of order 16 with field of values
${\mathbb{Q}}(\chi)={\mathbb{Q}}(\sqrt{-2})$.

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