# On the existence of free sublattices of
bounded index and arithmetic applications

Henri Johnston  Department of Mathematics
University of Exeter
Exeter
EX4 4QF
U.K<EMAIL_ADDRESS>http://emps.exeter.ac.uk/mathematics/staff/hj241
and  Alex Torzewski  Department of Mathematics
King’s College London
Strand
London
WC2R 2LS
U.K<EMAIL_ADDRESS>https://nms.kcl.ac.uk/alex.torzewski/

(Date: Version of 30th June 2023)

###### Abstract.

Let $\mathcal{O}$ be a Dedekind domain whose field of fractions $K$ is a
global field. Let $A$ be a finite-dimensional separable $K$-algebra and let
$\Lambda$ be an $\mathcal{O}$-order in $A$. Let $n$ be a positive integer and
suppose that $X$ is a $\Lambda$-lattice such that $K\otimes_{\mathcal{O}}X$ is
free of rank $n$ over $A$. Then $X$ contains a (non-unique) free
$\Lambda$-sublattice of rank $n$. The main result of the present article is to
show there exists such a sublattice $Y$ such that the generalised module index
$[X:Y]_{\mathcal{O}}$ has an explicit bound that is independent of $X$ and can
be chosen to satisfy certain conditions. We give examples of applications to
the approximation of normal integral bases and strong Minkowski units, and to
the Galois module structure of rational points over abelian varieties.

###### Key words and phrases:

Lattices, orders, normal integral bases, Minkowski units, abelian varieties

###### 2020 Mathematics Subject Classification:

16H20, 11R33, 11R27, 11G05, 11G10

##  1\. Introduction

Let $A$ be a finite-dimensional semisimple $\mathbb{Q}$-algebra and let
$\Lambda$ be an order in $A$. For example, if $G$ is a finite group then the
group ring $\mathbb{Z}[G]$ is an order in the group algebra $\mathbb{Q}[G]$. A
$\Lambda$-lattice is a (left) $\Lambda$-module that is finitely generated and
torsion-free over $\mathbb{Z}$. A special case of the Jordan-Zassenhaus
theorem says that for each positive integer $t$, there are only finitely many
isomorphism classes of $\Lambda$-lattices of $\mathbb{Z}$-rank at most $t$.

Now fix a positive integer $n$. Then there exists a positive integer $m$ with
the following property: given any $\Lambda$-lattice $X$ such that
$\mathbb{Q}\otimes_{\mathbb{Z}}X$ is free of rank $n$ as an $A$-module, there
exists a free $\Lambda$-sublattice $Y$ of $X$ such that the index $[X:Y]$ is
finite and divides $m$. To see this, first note that by clearing denominators
of a free basis of $\mathbb{Q}\otimes_{\mathbb{Z}}X$ over $A$, any such $X$
must contain a (non-unique) free $\Lambda$-sublattice of rank $n$, necessarily
of finite index $m_{X}$ in $X$. Then note that the Jordan-Zassenhaus theorem
implies that there are only finitely many choices for $X$ up to isomorphism
and take $m$ to be any common multiple of the $m_{X}$ as $X$ ranges over all
such choices.

The main goal of the present article is to make $m$ explicit and to give
arithmetic applications. In fact, the setting generalises to the case in which
$\Lambda$ is an $\mathcal{O}$-order where $\mathcal{O}$ is a Dedekind domain
whose field of fractions $K$ is a global field and $A$ is a finite-dimensional
separable $K$-algebra. In this situation, the group index $[X:Y]$ is replaced
by the generalised module index $[X:Y]_{\mathcal{O}}$ and the main theorem is
Theorem 4.5.

We now give examples of our algebraic results and arithmetic applications. The
following result is a weaker version of Theorem 5.3 obtained via
specialisation and Remark 7.2.

######  Theorem 1.1.

Let $G$ be a finite group. Let $k$ and $n$ be positive integers. Then there
exists a positive integer $i$, which can be chosen to be coprime to $k$, with
the following property: given any $\mathbb{Z}[G]$-lattice $X$ such that
$\mathbb{Q}\otimes_{\mathbb{Z}}X$ is free of rank $n$ over $\mathbb{Q}[G]$,
there exists a free $\mathbb{Z}[G]$-sublattice $Z$ of $X$ such that the index
$[X:Z]$ divides $i\cdot|G|^{\lceil 3|G|/2\rceil n}$.

Note that when $G$ satisfies certain conditions we can take $i=1$. Moreover,
$|G|^{\lceil 3|G|/2\rceil n}$ is a crude but neat upper bound for a more
precise term that will be made explicit. The following result is Theorem 5.15,
which is just one example of the stronger results that can be obtained in
special cases.

######  Theorem 1.2.

Let $G$ be a finite group and suppose that there exist positive integers
$n_{1},\ldots,n_{t}$ such that
$\mathbb{Q}[G]\cong\prod_{i=1}^{t}\mathrm{Mat}_{n_{i}}(\mathbb{Q})$. Let $n$
be a positive integer. If $X$ is an $\mathbb{Z}[G]$-lattice such that
$\mathbb{Q}\otimes_{\mathbb{Z}}X$ is free of rank $n$ over $\mathbb{Q}[G]$
then there exists a free $\mathbb{Z}[G]$-sublattice $Z$ of $X$ such that
$[X:Z]$ divides

$\left(|G|^{|G|}\textstyle{\prod_{i=1}^{t}}n_{i}^{-n_{i}^{2}}\right)^{\frac{3n}{2}}.$

Before sketching the ideas used in the proof of our main result, we discuss
how a variant of Theorem 1.1 can be applied in the following arithmetic
situation. Let $L/K$ be a finite Galois extension such that $K$ is equal to
either $\mathbb{Q}$ or an imaginary quadratic field. Let $G=\mathrm{Gal}(L/K)$
and let $\mu_{L}$ denote the roots of unity of $L$. In this setting,
$\mathcal{O}_{L}^{\times}/\mu_{L}$ is a $\mathbb{Z}[G]$-lattice and one can
show that $L/K$ has a so-called Minkowski unit, that is, an element
$\varepsilon\in\mathcal{O}_{L}^{\times}/\mu_{L}$ such that
$\mathbb{Q}\otimes_{\mathbb{Z}}(\mathcal{O}_{L}^{\times}/\mu_{L})=\mathbb{Q}[G]\cdot\varepsilon$.
Such an $\varepsilon$ is said to be a strong Minkowski unit if
$\mathcal{O}_{L}^{\times}/\mu_{L}=\mathbb{Z}[G]\cdot\varepsilon$. The
existence of strong Minkowski units (which some authors refer to as Minkowski
units) has been studied in numerous articles; see Remark 8.3. In §8, we give
several results on the approximation of strong Minkowski units. The following
result is a weakening of Theorem 8.5 obtained via Remark 8.6.

######  Theorem 1.3.

Let $G$ be a finite group and let $k$ be a positive integer. Then there exists
a positive integer $i$, which can be chosen to be coprime to $k$, with the
following property: given any finite Galois extension $L/K$ with
$\mathrm{Gal}(L/K)\cong G$ and $K$ equal to either $\mathbb{Q}$ or an
imaginary quadratic field, there exists a Minkowski unit
$\varepsilon\in\mathcal{O}_{L}^{\times}/\mu_{L}$ such that the index
$[\mathcal{O}_{L}^{\times}/\mu_{L}:\mathbb{Z}[\mathrm{Gal}(L/K)]\cdot\varepsilon]$
divides $i\cdot|G|^{\lceil 3|G|/2\rceil-2}$.

Again, stronger results can be obtained in special cases. Analogous
applications to the approximation of normal integral bases are given in §7 and
to the Galois module structure of rational points on abelian varieties are
given in §9.

We now outline the ideas used in the proof of the main result Theorem 4.5. Let
$\mathcal{O}$ be a Dedekind domain whose field of fractions $K$ is a global
field and assume that $\mathcal{O}\neq K$. Let $A$ be a finite-dimensional
separable $K$-algebra and let $\Lambda$ be an $\mathcal{O}$-order in $A$. Let
$\mathcal{M}$ be a maximal $\mathcal{O}$-order in $A$ containing $\Lambda$ and
let $X$ be a $\Lambda$-lattice such that $K\otimes_{\mathcal{O}}X$ is free of
rank $1$ over $A$ (the higher rank case is similar). We consider the unique
$\mathcal{M}$-lattice $\leftindex^{\mathcal{M}}{X}$ contained in $X$ that is
maximal with respect to inclusion. Then $\leftindex^{\mathcal{M}}{X}$ is
locally free over $\mathcal{M}$, and as explained in Corollary 4.2,
$\leftindex^{\mathcal{M}}{X}$ contains a free $\mathcal{M}$-sublattice
$\mathcal{M}\cdot\varepsilon$ with an index that can be controlled (the key
ingredients here are the Jordan-Zassenhaus theorem and Roiter’s lemma).
Hypotheses on $\mathcal{M}$ can also be given to ensure that this index is
trivial (see Lemma 2.2). We then obtain a bound on the index
$[X:\Lambda\cdot\varepsilon]_{\mathcal{O}}$ by taking the product of bounds on
the indices corresponding to each of the three inclusions

$\Lambda\cdot\varepsilon\subseteq\mathcal{M}\cdot\varepsilon\subseteq\leftindex^{\mathcal{M}}{X}\subseteq
X.$

Note that
$[\mathcal{M}\cdot\varepsilon:\Lambda\cdot\varepsilon]_{\mathcal{O}}=[\mathcal{M}:\Lambda]_{\mathcal{O}}$,
which can be computed by localisation. Moreover,
$[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$ divides
$[\mathcal{M}X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$, where
$\mathcal{M}X$ is the unique $\mathcal{M}$-lattice containing $X$ that is
minimal with respect to inclusion. In Corollary 3.3, we show that
$[\mathcal{M}X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$ divides
$[\mathcal{M}:J]_{\mathcal{O}}$ where $J$ is any full two-sided ideal of
$\mathcal{M}$ contained in $\Lambda$. Again, $[\mathcal{M}:J]_{\mathcal{O}}$
can be computed by localisation. Crucially, the product of bounds of indices
obtained is independent of the choice of $\Lambda$-lattice $X$.

If $G$ is a finite group such that $|G|$ is invertible in $K$ and
$\Lambda=\mathcal{O}[G]$, then $J$ can be taken to be the (left) conductor of
$\mathcal{M}$ into $\Lambda$ (the left and right conductors are equal in this
case) and $[\mathcal{M}:J]_{\mathcal{O}}$ can be computed explicitly using
Jacobinski’s conductor formula [Jac66]. We also obtain an explicit formula for
$[\mathcal{M}:\mathcal{O}[G]]_{\mathcal{O}}$, which may be of independent
interest. Note that in the setting of Theorem 1.1 with $n=1$, the term
$|G|^{\lceil 3|G|/2\rceil}$ is a crude but neat upper bound for
$[\mathcal{M}:\mathbb{Z}[G]]\cdot[\mathcal{M}:J]=[\mathcal{M}:\mathbb{Z}[G]]^{3}$
and the term $i$ is the upper bound for
$[\leftindex^{\mathcal{M}}{X}:\mathcal{M}\cdot\varepsilon]$ given by Corollary
4.2. Moreover, we can take $i=1$ when $G$ satisfies the hypotheses of
Proposition 5.11 or Corollary 5.13.

### Acknowledgements

The authors are grateful to Werner Bley, Nigel Byott, Frank Calegari, Hebert
Gangl, Tommy Hofmann, Donghyeok Lim, Daniel Macias Castillo, Alexandre Maksoud
and John Nicholson for helpful comments and discussions. For the purpose of
open access, the authors have applied a Creative Commons Attribution (CC BY)
licence to any author accepted manuscript version arising.

##  2\. Preliminaries on lattices and orders

For further background, we refer the reader to [CR81, Rei03, FT93]. Let
$\mathcal{O}$ be a Dedekind domain with field of fractions $K$. To avoid
trivialities, we assume that $\mathcal{O}\neq K$.

###  2.1. Lattices over Dedekind domains

An $\mathcal{O}$-lattice $M$ is a finitely generated torsion-free
$\mathcal{O}$-module, or equivalently, a finitely generated projective
$\mathcal{O}$-module. Using the former definition and the fact that
$\mathcal{O}$ is noetherian, we see that any $\mathcal{O}$-submodule of an
$\mathcal{O}$-lattice is again an $\mathcal{O}$-lattice.

For any finite-dimensional $K$-vector space $V$, an $\mathcal{O}$-lattice in
$V$ is a finitely generated $\mathcal{O}$-submodule $M$ in $V$. We define a
$K$-vector subspace of $V$ by

$KM:=\\{\alpha_{1}m_{1}+\alpha_{2}m_{2}+\cdots+\alpha_{r}m_{r}\mid
r\in\mathbb{Z}_{\geq 0},\alpha_{i}\in K,m_{i}\in M\\}$

and say that $M$ is a full $\mathcal{O}$-lattice in $V$ if $KM=V$. Each
$\mathcal{O}$-lattice $M$ may be viewed as an $\mathcal{O}$-lattice in the
finite-dimensional $K$-vector space $K\otimes_{\mathcal{O}}M$ by identifying
$M$ with its image $1\otimes M$. We may identify $K\otimes_{\mathcal{O}}M$
with $KM$.

Let $M$ and $N$ be a pair of full $\mathcal{O}$-lattices in a finite-
dimensional $K$-vector space $V$. Since $N$ contains a $K$-basis for $V$, for
each $m\in M$ there is a nonzero $r\in\mathcal{O}$ such that $rm\in N$.
Therefore there exists a nonzero $r\in\mathcal{O}$ such that $rM\subseteq N$
since $M$ is finitely generated over $\mathcal{O}$.

For a maximal ideal $\mathfrak{p}$ of $\mathcal{O}$, let
$\mathcal{O}_{\mathfrak{p}}$ denote the localisation of $\mathcal{O}$ at
$\mathfrak{p}$. Let $\widehat{\mathcal{O}}_{\mathfrak{p}}$ denote the
completion of $\mathcal{O}$ at $\mathfrak{p}$ and let
$\widehat{K}_{\mathfrak{p}}$ denote its field of fractions. For an
$\mathcal{O}$-lattice $M$, we define the localisation $M$ at $\mathfrak{p}$ to
be the $\mathcal{O}_{\mathfrak{p}}$-lattice
$M_{\mathfrak{p}}:=\mathcal{O}_{\mathfrak{p}}\otimes_{\mathcal{O}}M$ and the
completion of $M$ at $\mathfrak{p}$ to be the
$\widehat{\mathcal{O}}_{\mathfrak{p}}$-lattice
$\widehat{M}_{\mathfrak{p}}:=\widehat{\mathcal{O}}_{\mathfrak{p}}\otimes_{\mathcal{O}}M$.
By identifying $M$ with its image $1\otimes M$, we may view $M$ as embedded in
$M_{\mathfrak{p}}$. Viewing $M$ and each $M_{\mathfrak{p}}$ as embedded in
$KM$, we have $M=\bigcap_{\mathfrak{p}}M_{\mathfrak{p}}$, where $\mathfrak{p}$
ranges over all maximal ideals of $\mathcal{O}$ (see [Rei03, (4.21)]).

###  2.2. Generalised module indices

Let $M,N$ be full $\mathcal{O}$-lattices in a finite-dimensional $K$-vector
space $V$. First consider the case in which $\mathcal{O}$ is a discrete
valuation ring. Then $M$ and $N$ are both free and of equal rank over
$\mathcal{O}$, and so there exists an $\alpha\in\mathrm{Aut}_{K}(V)$ with
$\alpha(M)=N$. Moreover, $\alpha$ is unique modulo
$\mathrm{Aut}_{\mathcal{O}}(N)$; hence its determinant is unique modulo
$\mathcal{O}^{\times}$, and so the ideal
$[M:N]_{\mathcal{O}}:=\mathcal{O}\det(\alpha)$ is a uniquely defined
fractional ideal of $\mathcal{O}$. Now consider the case in which
$\mathcal{O}$ is an arbitrary Dedekind domain. For almost all maximal ideals
$\mathfrak{p}$ of $\mathcal{O}$ we have $M_{\mathfrak{p}}=N_{\mathfrak{p}}$
and hence
$[M_{\mathfrak{p}}:N_{\mathfrak{p}}]_{\mathcal{O}_{\mathfrak{p}}}=\mathcal{O}_{\mathfrak{p}}$
(see [CR81, Exercise 4.7]). Therefore there is a unique fractional ideal
$[M:N]_{\mathcal{O}}$ of $\mathcal{O}$ such that
$([M:N]_{\mathcal{O}})_{\mathfrak{p}}=[M_{\mathfrak{p}}:N_{\mathfrak{p}}]_{\mathcal{O}_{\mathfrak{p}}}$
for all $\mathfrak{p}$. Note that if $M_{1},M_{2},M_{3}$ are full
$\mathcal{O}$-lattices in $V$ then
$[M_{1}:M_{3}]_{\mathcal{O}}=[M_{1}:M_{2}]_{\mathcal{O}}\cdot[M_{2}:M_{3}]_{\mathcal{O}}$.
Moreover, if $\mathcal{O}^{\prime}$ is a Dedekind domain containing
$\mathcal{O}$ then
$\mathcal{O}^{\prime}\otimes_{\mathcal{O}}[M:N]_{\mathcal{O}}=[\mathcal{O}^{\prime}\otimes_{\mathcal{O}}M:\mathcal{O}^{\prime}\otimes_{\mathcal{O}}N]_{\mathcal{O}^{\prime}}$.
If $M\subseteq N$ are $\mathbb{Z}$-lattices of equal rank then we abbreviate
$[N:M]_{\mathbb{Z}}$ to $[N:M]$, which is consistent with the fact that
$[N:M]_{\mathbb{Z}}$ is the $\mathbb{Z}$-ideal generated by the usual group
index of $M$ in $N$.

######  Lemma 2.1.

Suppose we have a diagram of $\mathcal{O}$-lattices with exact rows

${0}$${N_{1}}$${N_{2}}$${N_{3}}$${0}$${0}$${M_{1}}$${M_{2}}$${M_{3}}$${0,}$

such that $KN_{i}=KM_{i}$ for $i=1,2,3$. Then
$[M_{2}:N_{2}]_{\mathcal{O}}=[M_{1}:N_{1}]_{\mathcal{O}}\cdot[M_{3}:N_{3}]_{\mathcal{O}}$.

###### Proof.

For $i=1,3$, fix $K$-linear maps $\alpha_{i}\colon KM_{i}\rightarrow KM_{i}$
such that $\alpha_{i}(M_{i})=N_{i}$. Let $\pi\colon N_{2}\rightarrow N_{3}$
denote the map in the above diagram. Since $N_{3}$ is
$\mathcal{O}$-projective, there exists an $\mathcal{O}$-section $s\colon
N_{3}\rightarrow N_{2}$ of $\pi$. Then define $\tilde{\alpha}_{3}\colon
KM_{3}\rightarrow KM_{2}$ by
$\tilde{\alpha}_{3}=(K\otimes_{\mathcal{O}}s)\circ\alpha_{3}$. Fixing an
$\mathcal{O}$-linear splitting $M_{2}\cong M_{1}\oplus M_{3}$ (which exists
since $M_{3}$ is $\mathcal{O}$-projective) and thus a $K$-linear splitting
$KM_{2}\cong KM_{1}\oplus KM_{3}$, we then obtain a $K$-linear map
$\alpha_{2}:=(\alpha_{1}+\tilde{\alpha}_{3})\colon KM_{2}\rightarrow KM_{2}$
such that $\alpha_{2}(M_{2})=N_{2}$ and $\alpha_{1}(M_{1})=N_{1}$. Hence, with
respect to a $K$-basis of $KM_{2}$ extending a $K$-basis of $KM_{1}$, the
matrix representing $\alpha_{2}$ is block upper triangular. Consequently,
$\det(\alpha_{2})=\det(\alpha_{1})\det(\alpha_{3})$, and thus we obtain the
desired result. ∎

###  2.3. Duals of lattices

Let $M$ be an $\mathcal{O}$-lattice. The linear dual
$M^{\vee}:=\mathrm{Hom}_{\mathcal{O}}(X,\mathcal{O})$ is also an
$\mathcal{O}$-lattice and there is a canonical identification
$(M^{\vee})^{\vee}=M$. Moreover, $(-)^{\vee}$ is inclusion-reversing. For a
maximal ideal $\mathfrak{p}$ of $\mathcal{O}$, we have
$(M_{\mathfrak{p}})^{\vee}=(M^{\vee})_{\mathfrak{p}}$. Together the fact that
determinants are invariant under transposes, this implies that if $M$ and $N$
are full $\mathcal{O}$-lattices in a finite-dimensional $K$-vector space $V$
then $[M:N]_{\mathcal{O}}=[N^{\vee}:M^{\vee}]_{\mathcal{O}}$.

###  2.4. Lattices over orders

Let $A$ be a finite-dimensional $K$-algebra and let $\Lambda$ be an
$\mathcal{O}$-order in $A$, that is, a subring of $A$ that is also a full
$\mathcal{O}$-lattice in $A$. Note that $\Lambda$ is both left and right
noetherian since $\Lambda$ is finitely generated over $\mathcal{O}$. A left
$\Lambda$-lattice $X$ is a left $\Lambda$-module that when considered as an
$\mathcal{O}$-module is also an $\mathcal{O}$-lattice; in this case, $KX$ may
be viewed as a left $A$-module.

Henceforth all modules (resp. lattices) shall be assumed to be left modules
(resp. lattices) unless otherwise stated. Two $\Lambda$-lattices are said to
be isomorphic if they are isomorphic as $\Lambda$-modules.

For a maximal ideal $\mathfrak{p}$ of $\mathcal{O}$, the localisation
$\Lambda_{\mathfrak{p}}$ is an $\mathcal{O}_{\mathfrak{p}}$-order in $A$, and
the completion $\widehat{\Lambda}_{\mathfrak{p}}$ is a
$\widehat{\mathcal{O}}_{\mathfrak{p}}$-order in
$\widehat{K}_{\mathfrak{p}}\otimes_{K}A$. Localising a $\Lambda$-lattice $X$
at $\mathfrak{p}$ yields a $\Lambda_{\mathfrak{p}}$-lattice
$X_{\mathfrak{p}}$, and completing $X$ at $\mathfrak{p}$ yields a
$\widehat{\Lambda}_{\mathfrak{p}}$-lattice $\widehat{X}_{\mathfrak{p}}$. Given
$\Lambda$-lattices $X$ and $Y$, we have that $X_{\mathfrak{p}}\cong
Y_{\mathfrak{p}}$ as $\Lambda_{\mathfrak{p}}$-lattices if and only if
$\widehat{X}_{\mathfrak{p}}\cong\widehat{Y}_{\mathfrak{p}}$ as
$\widehat{\Lambda}_{\mathfrak{p}}$-lattices (see [Rei03, (18.2)]). For a
positive integer $n$, a $\Lambda$-lattice $X$ is said to be locally free of
rank $n$, if for each maximal ideal $\mathfrak{p}$ of $\mathcal{O}$, the
$\Lambda_{\mathfrak{p}}$-lattice $X_{\mathfrak{p}}$ is free of rank $n$, or
equivalently, the $\widehat{\Lambda}_{\mathfrak{p}}$-lattice
$\widehat{X}_{\mathfrak{p}}$ is free of rank $n$. Note that every locally free
$\Lambda$-lattice is projective by [CR81, (8.19)].

###  2.5. Maximal orders

Suppose that $A$ is a separable finite-dimensional $K$-algebra (see [Rei03,
§7c]). A maximal $\mathcal{O}$-order in $A$ is an $\mathcal{O}$-order that is
not properly contained in any other $\mathcal{O}$-order in $A$. For any
$\mathcal{O}$-order $\Lambda$ in $A$, there exists a (not necessarily unique)
maximal $\mathcal{O}$-order $\mathcal{M}$ in $A$ containing $\Lambda$ by
[Rei03, (10.4)]. If $\mathcal{M}$ is a maximal $\mathcal{O}$-order, $X$ is an
$\mathcal{M}$-lattice, and $n$ is a positive integer, then by [CR81,
(31.2)(iii)] we have that $KX$ is free of rank $n$ over $A$ if and only if $X$
is locally free of rank $n$.

###  2.6. Locally free class groups and cancellation properties

Suppose that $K$ is a global field and that $A$ is separable finite-
dimensional $K$-algebra. Let $\Lambda$ be an $\mathcal{O}$-order $A$. Let
$P(\Lambda)$ be the free abelian group generated by symbols $[X]$, one for
each isomorphism class of locally free $\Lambda$-lattices $X$, modulo
relations $[X]=[X_{1}]+[X_{2}]$ whenever $X\cong X_{1}\oplus X_{2}$. We define
the locally free class group $\operatorname{Cl}(\Lambda)$ of $\Lambda$ to be
the subgroup of $P(\Lambda)$ consisting of all expressions $[X]-[Y]$, with
$X,Y$ locally free and $KX\cong KY$.

We remark that $[X]-[Y]=0$ in $\operatorname{Cl}(\Lambda)$ if and only if $X$
is stably isomorphic to $Y$, that is, $X\oplus\Lambda^{(k)}\cong
Y\oplus\Lambda^{(k)}$ for some positive integer $k$ (here $\Lambda^{(k)}$
denotes the direct sum of $k$ copies of $\Lambda$). The order $\Lambda$ is
said to have the locally free cancellation property if given any pair of
locally free $\Lambda$-lattices $X$ and $Y$,

$X\oplus\Lambda^{(k)}\cong Y\oplus\Lambda^{(k)}\textrm{ for some
}k\in\mathbb{Z}_{\geq 0}\implies X\cong Y.$

Moreover, $\Lambda$ is said to have the stably free cancellation property if
this holds in the special case that $Y$ is free. If $A$ satisfies the so-
called Eichler condition relative to $\mathcal{O}$ then $\Lambda$ has the
locally free cancellation property; this condition is satisfied if $A$ is
commutative (see [CR87, §51] for further details).

If $n$ is a positive integer and $Y$ is any locally free $\Lambda$-lattice of
rank $n$, then by [CR81, (31.14)] there exists a locally free
$\Lambda$-lattice $X$ in $A$ such that $Y\cong\Lambda^{(n-1)}\oplus X$. Hence
every element of $\operatorname{Cl}(\Lambda)$ is expressible in the form
$[X_{1}]-[X_{2}]$, where $X_{1}$ and $X_{2}$ are locally free
$\Lambda$-lattices in $A$. Moreover, for each such pair $X_{1},X_{2}$, there
exists another such lattice $X_{3}$ such that $X_{2}\oplus
X_{3}\cong\Lambda\oplus X_{1}$ by [CR81, (31.7)]. Therefore every element of
$\operatorname{Cl}(\Lambda)$ is in fact expressible in the form
$[X]-[\Lambda]$ for some locally free $\Lambda$-lattice $X$ in $A$. This
together with the Jordan-Zassenhaus theorem [Rei03, (26.4)] implies that
$\operatorname{Cl}(\Lambda)$ is finite.

The following result is easily deduced from the above discussion.

######  Lemma 2.2.

The following statements are equivalent:

  1. (i)

every locally free $\Lambda$-lattice is in fact free;

  2. (ii)

every locally free $\Lambda$-lattice of rank $1$ is in fact free;

  3. (iii)

$\operatorname{Cl}(\Lambda)$ is trivial and $\Lambda$ has the stably free
cancellation property;

  4. (iv)

$\operatorname{Cl}(\Lambda)$ is trivial and $\Lambda$ has the locally free
cancellation property.

##  3\. Overlattices and sublattices for overorders

Let $\mathcal{O}$ be a Dedekind domain with field of fractions $K$ and assume
that $\mathcal{O}\neq K$.

###  3.1. Setup and definitions

Let $A$ be a finite-dimensional $K$-algebra. Let $\Lambda\subseteq\Gamma$ be
$\mathcal{O}$-orders in $A$ and let $X$ be a $\Lambda$-lattice. Define

$\Gamma X:=\\{\gamma_{1}m_{1}+\gamma_{2}m_{2}+\cdots+\gamma_{r}m_{r}\mid
r\in\mathbb{Z}_{\geq 0},m_{i}\in X,\gamma_{i}\in\Gamma\\}\subseteq KX.$

This is the unique $\Gamma$-lattice in $KX$ containing $X$ that is minimal
with respect to inclusion.

There exists a nonzero $r\in\mathcal{O}$ such that $r\Gamma\subseteq\Lambda$
(see §2.1) and so $r\Gamma X$ is a $\Gamma$-lattice contained in $X$ of finite
index. Since the sum of any two $\Gamma$-lattices contained in $X$ is also a
$\Gamma$-lattice contained in $X$, we see that there exists a unique
$\Gamma$-lattice contained in $X$ that is maximal with respect to inclusion,
which we shall denote by $\leftindex^{\Gamma}{X}$. For a right
$\Lambda$-lattice $X$, we define $X^{\Gamma}$ similarly. Note that
$\leftindex^{\Gamma}{\Lambda}$ (resp. $\Lambda^{\Gamma}$) coincides with the
right (resp. left) conductor of $\Gamma$ into $\Lambda$ (see [CR81, (27.2)]).

###  3.2. Bounds on indices

The following result gives a bound on

$[\Gamma X:\leftindex^{\Gamma}{X}]_{\mathcal{O}}=[\Gamma
X:X]_{\mathcal{O}}\cdot[X:\leftindex^{\Gamma}{X}]_{\mathcal{O}}$

that only depends on $\Gamma$ and $\Lambda$, and not on the particular choice
of lattice $X$.

######  Proposition 3.1.

Let $A$ be a finite-dimensional $K$-algebra and let $\Lambda\subseteq\Gamma$
be $\mathcal{O}$-orders in $A$. Let $J$ be any full two-sided ideal of
$\Gamma$ contained in $\Lambda$. Let $n$ be a positive integer and let $X$ be
a $\Lambda$-lattice such that $\Gamma X$ is locally free of rank $n$ over
$\Gamma$. Then $[\Gamma X:\leftindex^{\Gamma}{X}]_{\mathcal{O}}$ divides
$[\Gamma:J]_{\mathcal{O}}^{n}$.

######  Remark 3.2.

There are many possible choices of $J$, and the best choice will be context
specific. For example, a weak but general choice is
$J=[\Gamma:\Lambda]_{\mathcal{O}}\cdot\Gamma$. Moreover, $J$ can always be
taken to be the two-sided ideal of $\Gamma$ generated by the central conductor
of $\Gamma$ into $\Lambda$, that is, by $\\{x\in C\mid
x\Gamma\subseteq\Lambda\\}$, where $C$ denotes the centre of $A$.

###### Proof of Proposition 3.1.

Since $J$ is a left $\Gamma$-lattice contained in $\Lambda$, we have that $JX$
is a left $\Gamma$-lattice contained in $X$. Hence $JX$ is contained in
$\leftindex^{\Gamma}{X}$. The chain of containments

$JX\subseteq\leftindex^{\Gamma}{X}\subseteq X\subseteq\Gamma X$

implies that $[\Gamma X:\leftindex^{\Gamma}{X}]_{\mathcal{O}}$ divides
$[\Gamma X:JX]_{\mathcal{O}}$. Thus it remains to show that $[\Gamma
X:JX]_{\mathcal{O}}=[\Gamma:J]_{\mathcal{O}}^{n}$. Since indices are defined
locally and $([\Gamma
X:JX]_{\mathcal{O}})_{\mathfrak{p}}=[\Gamma_{\mathfrak{p}}X_{\mathfrak{p}}:J_{\mathfrak{p}}X_{\mathfrak{p}}]_{\mathcal{O}_{\mathfrak{p}}}$
for every maximal ideal $\mathfrak{p}$ of $\mathcal{O}$, we can and do assume
without loss of generality that $\mathcal{O}$ is a discrete valuation ring.
Then by hypothesis there exist
$\varepsilon_{1},\ldots,\varepsilon_{n}\in\Gamma X$ such that $\Gamma
X=\Gamma\varepsilon_{1}\oplus\cdots\oplus\Gamma\varepsilon_{n}$. Since $J$ is
a right $\Gamma$-module, we have

$JX=J\Gamma
X=J(\Gamma\varepsilon_{1}\oplus\cdots\oplus\Gamma\varepsilon_{n})=J\varepsilon_{1}\oplus\cdots\oplus
J\varepsilon_{n}.$

Therefore $[\Gamma
X:JX]_{\mathcal{O}}=[\Gamma\varepsilon_{1}\oplus\cdots\oplus\Gamma\varepsilon_{n}:J\varepsilon_{1}\oplus\cdots\oplus
J\varepsilon_{n}]_{\mathcal{O}}=[\Gamma:J]_{\mathcal{O}}^{n}$. ∎

######  Corollary 3.3.

Let $A$ be a separable finite-dimensional $K$-algebra and let $\Lambda$ be an
$\mathcal{O}$-order in $A$. Let $\mathcal{M}$ be a maximal $\mathcal{O}$-order
in $A$ containing $\Lambda$ and let $J$ be any full two-sided ideal of
$\mathcal{M}$ contained in $\Lambda$. Let $n$ be a positive integer and let
$X$ be a $\Lambda$-lattice such that $KX$ is free of rank $n$ over $A$. Then
$[\mathcal{M}X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$ divides
$[\mathcal{M}:J]_{\mathcal{O}}^{n}$.

###### Proof.

Since $KX$ is free of rank $n$ over $A$, we have that $\mathcal{M}X$ is
locally free of rank $n$ over $\mathcal{M}$ (see §2.5), and so the desired
result follows directly from Proposition 3.1. ∎

######  Remark 3.4.

In Proposition 3.1 and Corollary 3.3, the order $\Lambda$ can be replaced by
the so-called associated order $\mathcal{A}(X)=\\{\alpha\in A\mid\alpha
X\subseteq X\\}$. Thus if the containment $\Lambda\subseteq\mathcal{A}(X)$ is
strict, then it may be possible to make a better choice of $J$. For example,
if $\mathcal{M}$ is a maximal order containing $\Lambda$ and we take
$X=\mathcal{M}$, then $\mathcal{A}(X)=\mathcal{M}$ and so we can take
$J=\mathcal{M}$, which is consistent with the fact that
$\mathcal{M}X=X=\leftindex^{\mathcal{M}}{X}$ in this situation. Of course, the
disadvantage of this approach is that $\mathcal{A}(X)$ depends on $X$.

###  3.3. Duals and overorders

For an $\mathcal{O}$-order $\Lambda$ in a finite-dimensional $K$-algebra and
any left (resp. right) $\Lambda$-lattice $X$, the dual
$X^{\vee}=\mathrm{Hom}_{\mathcal{O}}(X,\mathcal{O})$ has the structure of a
right (resp. left) $\Lambda$-lattice, and there is a canonical identification
$(X^{\vee})^{\vee}=X$.

######  Lemma 3.5.

Let $\Lambda\subseteq\Gamma$ be $\mathcal{O}$-orders in a finite-dimensional
$K$-algebra.

  1. (i)

If $X$ is a left $\Lambda$-lattice then we have an equality of right
$\Gamma$-lattices $(\Gamma X)^{\vee}=(X^{\vee})^{\Gamma}$ and an equality of
indices $[\Gamma
X:X]_{\mathcal{O}}=[X^{\vee}:(X^{\vee})^{\Gamma}]_{\mathcal{O}}$.

  2. (ii)

If $X$ is a right $\Lambda$-lattice then we have an equality of left
$\Gamma$-lattices $(X\Gamma)^{\vee}=\leftindex^{\Gamma}{(X^{\vee})}$ and an
equality of indices
$[X\Gamma:X]_{\mathcal{O}}=[X^{\vee}:\leftindex^{\Gamma}{(X^{\vee})}]_{\mathcal{O}}$.

###### Proof.

We only prove part (i). Since $(-)^{\vee}$ reverses inclusions, $(\Gamma
X)^{\vee}$ is a right $\Gamma$-lattice contained in $X^{\vee}$. Hence $(\Gamma
X)^{\vee}$ is contained in $(X^{\vee})^{\Gamma}$ by definition of the latter.
Dualising, we also have that

(3.1) $\Gamma X=((\Gamma
X)^{\vee})^{\vee}\supseteq((X^{\vee})^{\Gamma})^{\vee}.$

Since $((X^{\vee})^{\Gamma})^{\vee}$ is itself a left $\Gamma$-lattice
containing $X$, this forces equality in (3.1) and hence $(\Gamma
X)^{\vee}=(X^{\vee})^{\Gamma}$ as desired. Finally, since $(-)^{\vee}$
preserves indices (see §2.3) we have that $[\Gamma
X:X]_{\mathcal{O}}=[X^{\vee}:(\Gamma
X)^{\vee}]_{\mathcal{O}}=[X^{\vee}:(X^{\vee})^{\Gamma}]_{\mathcal{O}}$. ∎

###  3.4. The commutative separable setting

In the setting of commutative separable algebras, the following result of
Fröhlich is a refinement of Corollary 3.3.

######  Theorem 3.6 (Fröhlich).

Let $A$ be a commutative separable finite-dimensional $K$-algebra and let
$\Lambda$ be an $\mathcal{O}$-order in $A$. Let $\mathcal{M}$ be the unique
maximal $\mathcal{O}$-order in $A$. Let $n$ be a positive integer and let $X$
be a $\Lambda$-lattice such that $KX$ is free of rank $n$ over $A$. Then both
$[\mathcal{M}X:X]_{\mathcal{O}}$ and
$[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$ divide
$[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$.

###### Proof.

The claim that $[\mathcal{M}X:X]_{\mathcal{O}}$ divides
$[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$ is contained in [Frö65, Theorem 4].

It remains to show that $[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$
divides $[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$. Since $A$ is separable
there is an isomorphism of (right) $A$-modules $A\cong\mathrm{Hom}_{K}(A,K)$
induced by the pairing of [CR81, (7.41)]. Thus there are $A$-module
isomorphisms

$K(X^{\vee})\cong\mathrm{Hom}_{K}(KX,K)\cong\mathrm{Hom}_{K}(A^{(n)},K)\cong\mathrm{Hom}_{K}(A,K)^{(n)}\cong
A^{(n)}.$

Lemma 3.5(ii) implies that
$[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}=[X^{\vee}\mathcal{M}:X^{\vee}]_{\mathcal{O}}=[\mathcal{M}X^{\vee}:X^{\vee}]_{\mathcal{O}}$,
where in the last equality, we consider $X^{\vee}$ as a left
$\mathcal{M}$-lattice, as we may since $\mathcal{M}$ is commutative. Moreover,
since $K(X^{\vee})$ is free of rank $n$ over $A$, the first claim and the
appropriate substitution imply that
$[\mathcal{M}X^{\vee}:X^{\vee}]_{\mathcal{O}}$ divides
$[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$. ∎

######  Remark 3.7.

[Bal71, §7, Example 1] shows that the result analogous to Theorem 3.6 does not
always hold in the noncommutative separable setting.

##  4\. Free sublattices of bounded index

Let $\mathcal{O}$ be a Dedekind domain with field of fractions $K$. Assume
that $K$ is a global field and that $\mathcal{O}\neq K$. Let $A$ be a
separable finite-dimensional $K$-algebra.

###  4.1. Free sublattices of locally free lattices

The following result gives a bound on the index of a free sublattice in a
locally free lattice.

######  Proposition 4.1.

Let $\Gamma$ be an $\mathcal{O}$-order in $A$ and let $\mathcal{K}$ be any
non-zero ideal of $\mathcal{O}$. Then there exists a non-zero ideal
$\mathcal{I}$ of $\mathcal{O}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: for every locally free
$\Gamma$-lattice $X$, there exists a free $\Gamma$-sublattice $Y$ of $X$ such
that $[X:Y]_{\mathcal{O}}$ divides $\mathcal{I}$.

###### Proof.

By [CR81, (31.14)], for a positive integer $n$ and a locally free
$\Lambda$-lattice $X$ of rank $n$, there exists a locally free
$\Lambda$-lattice $W$ of rank $1$ such that $X\cong\Lambda^{(n-1)}\oplus W$.
Thus the problem reduces to the case of locally free $\Lambda$-lattices $W$ of
rank $1$. The number of isomorphism classes of such lattices is finite by the
Jordan–Zassenhaus theorem [Rei03, (26.4)]. For each such class, choose a
representative $W$ and note that by Roiter’s lemma [CR81, (31.6)] there exists
an inclusion $\iota_{W}:W\hookrightarrow\Gamma$ such that
$[\Gamma:\iota_{W}(W)]_{\mathcal{O}}$ is coprime to $\mathcal{K}$. Now take
$\mathcal{I}$ to be any common multiple of the (finite number of) ideals
$[\Gamma:\iota_{W}(W)]_{\mathcal{O}}$ as $W$ varies. ∎

######  Corollary 4.2.

Let $\mathcal{M}$ be a maximal $\mathcal{O}$-order in $A$ and let
$\mathcal{K}$ be any non-zero ideal of $\mathcal{O}$. Then there exists a non-
zero ideal $\mathcal{I}$ of $\mathcal{O}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any $\mathcal{M}$-lattice
$X$ such that $KX$ is free as an $A$-module, there exists a free
$\mathcal{M}$-sublattice $Y$ of $X$ such that $[X:Y]_{\mathcal{O}}$ divides
$\mathcal{I}$.

###### Proof.

Let $X$ be an $\mathcal{M}$-lattice. Then $KX$ is free as an $A$-module if and
only if $\mathcal{M}X$ is locally free over $\mathcal{M}$ (see §2.5). Hence
the result follows from Proposition 4.1. ∎

######  Remark 4.3.

If $\Gamma$ (resp. $\mathcal{M}$) satisfies the equivalent conditions of Lemma
2.2 then it is clear that we can take $\mathcal{I}=\mathcal{O}$ in Proposition
4.1 (resp. Corollary 4.2).

Given a finite set $S$ of maximal ideals of $\mathcal{O}$, let
$\mathcal{O}_{S}=\bigcap_{\mathfrak{p}\notin S}\mathcal{O}_{\mathfrak{p}}$,
where $\mathfrak{p}$ ranges over all maximal ideals of $\mathcal{O}$ not in
$S$. We include the following neat result for general interest.

######  Corollary 4.4.

Let $\Gamma$ be an $\mathcal{O}$-order in $A$ and let $T$ be a finite set of
maximal ideals of $\mathcal{O}$. Then there exists a finite set $S$ of maximal
ideals of $\mathcal{O}$ such that $S\cap T=\emptyset$ and
$\mathcal{O}_{S}\otimes_{\mathcal{O}}\Gamma$ satisfies the equivalent
conditions of Lemma 2.2.

###### Proof.

If $\mathcal{I}$ is the ideal given by Proposition 4.1 and $\mathcal{K}$ the
product of the maximal ideals in $T$, then we can take $S$ to be the set of
maximal ideals dividing $\mathcal{I}$. ∎

###  4.2. The main theorem

The main theorem of the present article is as follows.

######  Theorem 4.5.

Let $\mathcal{O}$ be a Dedekind domain with field of fractions $K$. Assume
that $K$ is a global field and that $\mathcal{O}\neq K$. Let $A$ be a
separable finite-dimensional $K$-algebra and let $\Lambda$ be an
$\mathcal{O}$-order in $A$. Let $\mathcal{M}$ be a maximal $\mathcal{O}$-order
in $A$ containing $\Lambda$ and let $J$ be any full two-sided ideal of
$\mathcal{M}$ contained in $\Lambda$. Let $n$ be a positive integer and let
$\mathcal{K}$ be any non-zero ideal of $\mathcal{O}$. Then there exists a non-
zero ideal $\mathcal{I}$ of $\mathcal{O}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any $\Lambda$-lattice $X$
such that $KX$ is free of rank $n$ over $A$, there exists a free
$\Lambda$-sublattice $Z$ of $X$ such that $[X:Z]_{\mathcal{O}}$ divides
$\mathcal{I}\cdot[\mathcal{M}:\Lambda]_{\mathcal{O}}^{2n}$ if $A$ is
commutative or
$\mathcal{I}\cdot[\mathcal{M}:J]_{\mathcal{O}}^{n}\cdot[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$
otherwise. Moreover, if $\mathcal{M}$ satisfies the equivalent conditions of
Lemma 2.2 then we can take $\mathcal{I}=\mathcal{O}$.

###### Proof.

Let $\mathcal{I}$ be the ideal of $\mathcal{O}$ given by Corollary 4.2. If
$\mathcal{M}$ satisfies the equivalent conditions of Lemma 2.2 then we can
take $\mathcal{I}=\mathcal{O}$ by Remark 4.3. Then there exists a free
$\mathcal{M}$-sublattice $Y$ of $\leftindex^{\mathcal{M}}{X}$ such that
$[\leftindex^{\mathcal{M}}{X}:Y]_{\mathcal{O}}$ divides $\mathcal{I}$. Let
$\varepsilon_{1},\ldots,\varepsilon_{n}$ be a free $\mathcal{M}$-basis of $Y$,
so that
$Y=\mathcal{M}\varepsilon_{1}\oplus\cdots\oplus\mathcal{M}\varepsilon_{n}$,
and let $Z=\Lambda\varepsilon_{1}\oplus\cdots\oplus\Lambda\varepsilon_{n}$.
Then $Z\subseteq Y\subseteq\leftindex^{\mathcal{M}}{X}\subseteq X$ and
$[X:Z]_{\mathcal{O}}=[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}\cdot[\leftindex^{\mathcal{M}}{X}:Y]_{\mathcal{O}}\cdot[Y:Z]_{\mathcal{O}}$.
Note that $[Y:Z]_{\mathcal{O}}=[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$.
Moreover, Corollary 3.3 implies that
$[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$ divides
$[\mathcal{M}:J]_{\mathcal{O}}^{n}$, and under the assumption that $A$ is
commutative, Theorem 3.6 implies that in fact
$[X:\leftindex^{\mathcal{M}}{X}]_{\mathcal{O}}$ divides
$[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$. Therefore we obtain the desired
result. ∎

######  Remark 4.6.

The statement of Theorem 4.5 extends to $\Lambda$-lattices $X$ admitting a
surjection $A^{(n)}\twoheadrightarrow KX$ of $A$-modules. More specifically,
the ideal $\mathcal{I}$ has the following property: given any
$\Lambda$-lattice $X$ admitting a surjection $A^{(n)}\twoheadrightarrow KX$ of
$A$-modules, there exists a $\Lambda$-sublattice $Z$ of $X$ generated by at
most $n$ elements such that $[X:Z]_{\mathcal{O}}$ divides
$\mathcal{I}\cdot[\mathcal{M}:J]_{\mathcal{O}}^{n}\cdot[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$.
This can be seen as follows. There exists an $A$-module $B$ such that
$KX\oplus B\cong A^{(n)}$. Thus given any full $\Lambda$-lattice $W$ in $B$,
the $\Lambda$-lattice $X\oplus W$ satisfies the conditions of Theorem 4.5 and
so admits a free $\Lambda$-sublattice $Z^{\prime}$ of index dividing
$\mathcal{I}\cdot[\mathcal{M}:J]_{\mathcal{O}}^{n}\cdot[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}$,
and the image of $Z^{\prime}$ under the projection $X\oplus
W\twoheadrightarrow X$ is the desired sublattice $Z$. Note that one should
expect stronger bounds if one specifies the isomorphism class of $KX$; one
such situation is considered in §6.

##  5\. Group rings

###  5.1. Conductors of group rings

The extra structure of group rings is exploited in the following result, which
will allow us to make an optimal choice of the two-sided ideal $J$ that
appears in the statement of Theorem 4.5.

######  Proposition 5.1.

Let $\mathcal{O}$ be a Dedekind domain with field of fractions
$K\neq\mathcal{O}$. Let $G$ be a finite group and let $\Gamma$ be an
$\mathcal{O}$-order in $K[G]$ containing $\mathcal{O}[G]$. Then
${\mathcal{O}[G]}^{\Gamma}=\leftindex^{\Gamma}{\mathcal{O}[G]}$ and we have

(5.1)
$[\Gamma:\mathcal{O}[G]]_{\mathcal{O}}=[\mathcal{O}[G]:\leftindex^{\Gamma}{\mathcal{O}[G]}]_{\mathcal{O}}=[\Gamma:\leftindex^{\Gamma}{\mathcal{O}[G]}]_{\mathcal{O}}^{\frac{1}{2}}.$

Moreover, if $|G|$ is invertible in $K$ and $\Gamma=\mathcal{M}$ is a maximal
$\mathcal{O}$-order, this index is independent of the choice of $\mathcal{M}$.

######  Remark 5.2.

In the case that $|G|$ is invertible in $K$ and $\Gamma=\mathcal{M}$ is a
maximal $\mathcal{O}$-order, Jacobinski has given an explicit description of
${\mathcal{O}[G]}^{\Gamma}=\leftindex^{\Gamma}{\mathcal{O}[G]}$ (see Theorem
5.5) and this leads to an explicit formula for the index of (5.1) (see
Corollary 5.6).

###### Proof of Proposition 5.1.

Given an $\mathcal{O}$-order $\Lambda$ in $K[G]$, let $\Lambda^{\mathrm{op}}$
denote the $\mathcal{O}$-order defined by the image of $\Lambda$ under the
involution on $K[G]$ induced by $g\mapsto g^{-1}$. Any left (resp. right)
$\Lambda$-lattice carries a canonical structure of a right (resp. left)
$\Lambda^{\mathrm{op}}$-lattice with $g^{-1}$ acting as $g$ did previously.
Given a left (resp. right) $\Lambda$-lattice $X$, we denote by $X^{*}$ the
dual lattice $X^{\vee}=\mathrm{Hom}_{\mathcal{O}}(X,\mathcal{O})$ considered
as a left (resp. right) $\Lambda^{\mathrm{op}}$-lattice. Note that for a left
$\Lambda$-lattice $X$, we have
$[(X^{\vee})\Lambda:X^{\vee}]_{\mathcal{O}}=[\Lambda
X^{*}:X^{*}]_{\mathcal{O}}$, etc.

Now observe that $\Gamma^{\mathrm{op}}$ is an $\mathcal{O}$-order containing
$\mathcal{O}[G]=\mathcal{O}[G]^{\mathrm{op}}$. Hence
$\Gamma^{\mathrm{op}}\mathcal{O}[G]=\Gamma^{\mathrm{op}}=\mathcal{O}[G]\Gamma^{\mathrm{op}}$.
We also have that

$\displaystyle(\Gamma^{\mathrm{op}}\mathcal{O}[G])^{\vee}$
$\displaystyle=(\mathcal{O}[G]^{\vee})^{\Gamma^{\mathrm{op}}}=\leftindex^{\Gamma}{(\mathcal{O}[G]}^{*}),$
$\displaystyle(\mathcal{O}[G]\Gamma^{\mathrm{op}})^{\vee}$
$\displaystyle=\leftindex^{\Gamma^{{\mathrm{op}}}}{(\mathcal{O}[G]^{\vee})}=(\mathcal{O}[G]^{*})^{\Gamma},$

where in each case the first equality follows from Lemma 3.5 and the second
equality follows from the definition of $(-)^{*}$. Therefore
$\leftindex^{\Gamma}{(\mathcal{O}[G]}^{*})=(\mathcal{O}[G]^{*})^{\Gamma}$.
Furthermore, there is an
$\mathcal{O}[G]=\mathcal{O}[G]^{\mathrm{op}}$-isomorphism
$\mathcal{O}[G]^{*}\overset{\sim}{\to}\mathcal{O}[G]$ given by
$\mathbbm{1}_{g}\mapsto g$, where $\mathbbm{1}_{g}$ denotes the element of
$\mathrm{Hom}_{\mathcal{O}}(\mathcal{O}[G],\mathcal{O})$ defined by $h\mapsto
0$ for $h\neq g$ and $g\mapsto 1$. Hence we conclude that
$\leftindex^{\Gamma}{\mathcal{O}[G]}=\mathcal{O}[G]^{\Gamma}$.

We have that
$[\Gamma:\mathcal{O}[G]]_{\mathcal{O}}=[\Gamma^{\mathrm{op}}:\mathcal{O}[G]^{\mathrm{op}}]_{\mathcal{O}}$
since $(-)^{\mathrm{op}}$ is an $\mathcal{O}$-linear isomorphism. As
$\mathcal{O}[G]=\mathcal{O}[G]^{\mathrm{op}}$, we then have

$\displaystyle[\Gamma:\mathcal{O}[G]]_{\mathcal{O}}$
$\displaystyle=[\Gamma^{\mathrm{op}}:\mathcal{O}[G]]_{\mathcal{O}}$
$\displaystyle=[(\mathcal{O}[G])^{\vee}:((\mathcal{O}[G])^{\vee})^{\Gamma^{\mathrm{op}}}]_{\mathcal{O}}$
$\displaystyle=[\mathcal{O}[G]^{*}:\leftindex^{\Gamma}{((\mathcal{O}[G])^{*})}]_{\mathcal{O}}$
$\displaystyle=[\mathcal{O}[G]:\leftindex^{\Gamma}{\mathcal{O}[G]}]_{\mathcal{O}},$

where the second equality follows from Lemma 3.5(i). Since

$[\Gamma:\leftindex^{\Gamma}{\mathcal{O}[G]}]_{\mathcal{O}}=[\Gamma:\mathcal{O}[G]]_{\mathcal{O}}\cdot[\mathcal{O}[G]:\leftindex^{\Gamma}{\mathcal{O}[G]}]_{\mathcal{O}},$

the second equality of (5.1) follows.

For the last statement, note that the hypotheses ensure that $K[G]$ is
separable and hence maximal orders exist (see §2.5). For any
$\mathcal{O}$-order $\Lambda$ in $K[G]$, let $\mathrm{Disc}(\Lambda)$ denote
the discriminant of $\Lambda$ with respect to the reduced trace map
$\mathrm{tr}:K[G]\rightarrow K$. Then $\mathrm{Disc}(\mathcal{M})$ is
independent of the choice of maximal $\mathcal{O}$-order $\mathcal{M}$ of
$K[G]$ by [Rei03, (25.3)]. Moreover, by [CR81, (26.3)(iii)] we have
$\mathrm{Disc}(\mathcal{O}[G])=[\mathcal{M}:\mathcal{O}[G]]_{\mathcal{O}}^{2}\cdot\mathrm{Disc}(\mathcal{M})$,
and so $[\mathcal{M}:\mathcal{O}[G]]_{\mathcal{O}}$ is independent of the
choice of $\mathcal{M}$. ∎

###  5.2. The main theorem for group rings

We now obtain a more precise version of Theorem 4.5 for lattices over group
rings.

######  Theorem 5.3.

Let $\mathcal{O}$ be a Dedekind domain with field of fractions $K$. Assume
that $K$ is a global field and that $\mathcal{O}\neq K$. Let $G$ be a finite
group such that $|G|$ is invertible in $K$. Set $s=2$ if $G$ is abelian and
$s=3$ otherwise. Let $\mathcal{M}$ be a maximal $\mathcal{O}$-order in $K[G]$
containing $\mathcal{O}[G]$. Let $n$ be a positive integer and let
$\mathcal{K}$ be any non-zero ideal of $\mathcal{O}$. Then there exists a non-
zero ideal $\mathcal{I}$ of $\mathcal{O}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any $\mathcal{O}[G]$-lattice
$X$ such that $KX$ is free of rank $n$ over $K[G]$, there exists a free
$\mathcal{O}[G]$-sublattice $Z$ of $X$ such that $[X:Z]_{\mathcal{O}}$ divides
$\mathcal{I}\cdot[\mathcal{M}:\mathcal{O}[G]]_{\mathcal{O}}^{sn}$. Moreover,
if $\mathcal{M}$ satisfies the equivalent conditions of Lemma 2.2 then we can
take $\mathcal{I}=\mathcal{O}$.

######  Remark 5.4.

In the case $\mathcal{O}=\mathbb{Z}$, explicit conditions on $G$ under which
$\mathcal{M}$ satisfies the equivalent conditions of Lemma 2.2 are given in
Proposition 5.11 and Corollary 5.13.

###### Proof of Theorem 5.3.

We apply Theorem 4.5 with $\Lambda=\mathcal{O}[G]$. If $G$ is abelian then the
desired result follows directly. Otherwise, by Proposition 5.1 we can and do
take
$J={\mathcal{O}[G]}^{\mathcal{M}}=\leftindex^{\mathcal{M}}{\mathcal{O}[G]}$,
and we have
$[\mathcal{M}:J]_{\mathcal{O}}^{n}\cdot[\mathcal{M}:\Lambda]_{\mathcal{O}}^{n}=[\mathcal{M}:\Lambda]_{\mathcal{O}}^{3n}$.
∎

###  5.3. Jacobinski’s formula and the index of a group ring in a maximal
order

For further details on the following setup and notation, we refer the reader
to [CR81, §27] and the references therein.

Let $\mathcal{O}$ be a Dedekind domain with field of fractions
$K\neq\mathcal{O}$. Let $G$ be a finite group such that $|G|$ is invertible in
$K$. Then $K[G]$ is a separable finite-dimensional $K$-algebra. We may write
$K[G]=A_{1}\oplus\cdots\oplus A_{t}$, where each $A_{i}$ is a simple
$K$-algebra. For each $i$, let $K_{i}$ denote the centre of $A_{i}$. Then each
$K_{i}$ is a finite separable field extension of $K$, and there exist integers
$n_{1},\ldots,n_{t}$ such that $\dim_{K_{i}}A_{i}=n_{i}^{2}$ for each $i$. Let
$\mathrm{tr}_{i}$ denote the reduced trace from $A_{i}$ to $K$ (see [CR81,
§7D]). Then $\mathrm{tr}_{i}=T_{K_{i}/K}\circ\mathrm{tr}_{A_{i}/K_{i}}$, where
$T_{K_{i}/K}$ is the ordinary trace from $K_{i}$ to $K$, and
$\mathrm{tr}_{A_{i}/K_{i}}$ is the reduced trace from $A_{i}$ to $K_{i}$.

Let $\mathcal{M}$ be a maximal $\mathcal{O}$-order such that
$\mathcal{O}[G]\subseteq\mathcal{M}\subseteq K[G]$. For each $i$, let
$\mathcal{M}_{i}=\mathcal{M}\cap A_{i}$, let $\mathcal{O}_{i}$ denote the
integral closure of $\mathcal{O}$ in $K_{i}$, and define the _inverse
different_ of $\mathcal{M}_{i}$ with respect to $\mathrm{tr}_{i}$ to be
$\mathcal{D}_{i}^{-1}=\\{x\in
A_{i}:\mathrm{tr}_{i}(x\mathcal{M}_{i})\subseteq\mathcal{O}\\}$. Then
$\mathcal{M}=\mathcal{M}_{1}\oplus\cdots\oplus\mathcal{M}_{t}$ and each
$\mathcal{D}_{i}^{-1}$ is a two-sided $\mathcal{M}_{i}$-lattice containing
$\mathcal{M}_{i}$.

######  Theorem 5.5 (Jacobinski [Jac66]).

In the notation above, we have

$\leftindex^{\mathcal{M}}{\mathcal{O}[G]}={\mathcal{O}[G]}^{\mathcal{M}}=\bigoplus_{i=1}^{t}|G|n_{i}^{-1}\mathcal{D}_{i}^{-1}.$

######  Corollary 5.6.

In the notation above, we have

$[\mathcal{M}:{\mathcal{O}[G]}]_{\mathcal{O}}=[\mathcal{O}[G]:\leftindex^{\mathcal{M}}{\mathcal{O}[G]}]_{\mathcal{O}}=\left(|G|^{|G|}\prod_{i=1}^{t}\left(n_{i}^{[K_{i}:K]n_{i}^{2}}[\mathcal{D}_{i}^{-1}:\mathcal{M}_{i}]_{\mathcal{O}}\right)^{-1}\right)^{\frac{1}{2}},$

and this index is independent of the choice of $\mathcal{M}$.

###### Proof.

By Theorem 5.5 we have

$\displaystyle[\mathcal{M}:\leftindex^{\mathcal{M}}{\mathcal{O}[G]}]_{\mathcal{O}}$
$\displaystyle=\prod_{i=1}^{t}[\mathcal{M}_{i}:(|G|n_{i}^{-1}\mathcal{D}_{i}^{-1})]_{\mathcal{O}}$
$\displaystyle=\prod_{i=1}^{t}(|G|n_{i}^{-1})^{\dim_{K}A_{i}}[\mathcal{M}_{i}:\mathcal{D}_{i}^{-1}]_{\mathcal{O}}$
$\displaystyle=|G|^{|G|}\prod_{i=1}^{t}(n_{i}^{[K_{i}:K]n_{i}^{2}}[\mathcal{D}_{i}^{-1}:\mathcal{M}_{i}]_{\mathcal{O}})^{-1},$

where in the last equality we have used that
$\dim_{K}A_{i}=[K_{i}:K]n_{i}^{2}$ for each $i$ and that
$\prod_{i=1}^{t}\dim_{K}A_{i}=|G|$. The desired result now follows from
Proposition 5.1. ∎

######  Corollary 5.7.

In the notation above, if $A_{i}\cong\mathrm{Mat}_{n_{i}}(K)$ for
$i=1,\ldots,t$, then

$[\mathcal{M}:{\mathcal{O}[G]}]_{\mathcal{O}}=[\mathcal{O}[G]:\leftindex^{\mathcal{M}}{\mathcal{O}[G]}]_{\mathcal{O}}=\left(|G|^{|G|}\prod_{i=1}^{t}n_{i}^{-n_{i}^{2}}\right)^{\frac{1}{2}}.$

###### Proof.

The hypotheses imply that $K_{i}=K$ and $\mathcal{D}_{i}^{-1}=\mathcal{M}_{i}$
for $i=1,\ldots,t$. Thus the desired result follows from Corollary 5.6. ∎

######  Corollary 5.8.

In the notation above, if $G$ is abelian then

$[\mathcal{M}:{\mathcal{O}[G]}]_{\mathcal{O}}=[\mathcal{O}[G]:\leftindex^{\mathcal{M}}{\mathcal{O}[G]}]_{\mathcal{O}}=\left(|G|^{|G|}\prod_{i=1}^{t}(\Delta_{K_{i}/K})^{-1}\right)^{\frac{1}{2}},$

where $\Delta_{K_{i}/K}$ denotes the discriminant of $\mathcal{O}_{i}$ with
respect to $\mathcal{O}$.

###### Proof.

Since $A$ is commutative, for every $i$ we have $n_{i}=1$, $A_{i}=K_{i}$, and
$\mathcal{M}_{i}=\mathcal{O}_{i}$. Thus the reduced trace $\mathrm{tr}_{i}$
coincides with the ordinary trace $T_{K_{i}/K}$ and so

$\mathcal{D}_{i}^{-1}=\\{x\in
K_{i}:T_{K_{i}/K}(x\mathcal{O}_{i})\subseteq\mathcal{O}\\}$

is the usual inverse different of $\mathcal{O}_{i}$ with respect to
$\mathcal{O}$. Moreover,

$[\mathcal{D}_{i}^{-1}:\mathcal{M}_{i}]_{\mathcal{O}}=[\mathcal{D}_{i}^{-1}:\mathcal{O}_{i}]_{\mathcal{O}}=[\mathcal{O}_{i}:\mathcal{D}_{i}]_{\mathcal{O}}=\mathrm{Norm}_{K_{i}/K}(\mathcal{D}_{i})=\Delta_{K_{i}/K},$

where for the third equality, it suffices to first localise and then consider
the determinant of the $K$-linear endomorphism of $K_{i}$ given by
multiplication by a generator of $\mathcal{D}_{i}$. Therefore the desired
result now follows from Corollary 5.6. ∎

We now make the last result completely explicit in the case $K=\mathbb{Q}$.

######  Proposition 5.9.

Let $G$ be a finite abelian group and let $e$ denote its exponent. Let
$\mathcal{M}$ be the unique maximal $\mathbb{Z}$-order in $\mathbb{Q}[G]$.
Then

$\textstyle{[\mathcal{M}:\mathbb{Z}[G]]=\left(|G|^{|G|}\prod_{d\mid
e}\left(d^{-\phi(d)}\prod_{p\mid
d}p^{\frac{\phi(d)}{p-1}}\right)^{t_{d}}\right)^{\frac{1}{2}}},$

where $t_{d}$ denotes the number of cyclic subgroups of $G$ of order $d$ and
$\phi(-)$ denotes the Euler totient function.

###### Proof.

By [AA69, Theorem 2], we have $\mathbb{Q}[G]\cong\prod_{d\mid
e}\mathbb{Q}(\zeta_{d})^{(t_{d})}$, where $\mathbb{Q}(\zeta_{d})^{(t_{d})}$
denotes the direct product of $t_{d}$ copies of $\mathbb{Q}(\zeta_{d})$ (see
also [PW50]). Moreover,

$\textstyle{\Delta_{\mathbb{Q}(\zeta_{d})/\mathbb{Q}}^{-1}=d^{-\phi(d)}\prod_{p\mid
d}p^{\frac{\phi(d)}{p-1}}}$

by [Was97, Proposition 2.7]. Therefore the desired result now follows from a
straightforward calculation and Corollary 5.8 in the case $K=\mathbb{Q}$. ∎

The following special case is equivalent to [Wit04, Lemma 5.2], which was
proven using different methods.

######  Corollary 5.10.

Let $p$ be a prime, let $k$ be a positive integer, and let $G$ be the cyclic
group of order $p^{k}$. Let $\mathcal{M}$ be the unique maximal
$\mathbb{Z}$-order in $\mathbb{Q}[G]$. Then

$[\mathcal{M}:\mathbb{Z}[G]]=p^{1+p+\cdots+p^{k-1}}.$

###  5.4. The case of group rings over the integers

For a finite group $G$, let $\mathrm{Irr}_{\mathbb{C}}(G)$ denote the set of
complex irreducible characters of $G$. For each
$\chi\in\mathrm{Irr}_{\mathbb{C}}(G)$, let $\mathbb{Q}(\chi)$ denote the field
generated by the values of $\chi$, and let $C(\chi)$ be the ideal class group
of its ring of integers.

The following result is well-known to experts, but the authors were unable to
locate it in this precise form in the literature.

######  Proposition 5.11.

Let $G$ be a finite group and let $\mathcal{M}$ be a maximal
$\mathbb{Z}$-order in $\mathbb{Q}[G]$. Suppose that no factor of
$\mathbb{R}[G]$ is isomorphic to the quaternions $\mathbb{H}$. Then
$\mathcal{M}$ satisfies the equivalent conditions of Lemma 2.2 if and only if
$C(\chi)$ is trivial for each $\chi\in\mathrm{Irr}_{\mathbb{C}}(G)$.

###### Proof.

The hypothesis on $\mathbb{R}[G]$ ensures that $\mathbb{Q}[G]$ satisfies the
Eichler condition relative to $\mathbb{Z}$ (see [CR87, §51A]). Hence
Jacobinski’s cancellation theorem [CR87, (51.24)] implies that every
$\mathbb{Z}$-order in $\mathbb{Q}[G]$, in particular $\mathcal{M}$, has the
locally free cancellation property. Now write
$\mathbb{Q}[G]=A_{1}\oplus\cdots\oplus A_{t}$, where each $A_{i}$ is a central
simple $\mathbb{Q}$-algebra. For each $i$, let $K_{i}$ denote the centre of
$A_{i}$ and let $\mathcal{M}_{i}=A_{i}\cap\mathcal{M}$. Then
$\mathcal{M}=\mathcal{M}_{1}\oplus\cdots\oplus\mathcal{M}_{t}$ and
$\operatorname{Cl}(\mathcal{M})\cong\operatorname{Cl}(\mathcal{M}_{1})\oplus\cdots\oplus\operatorname{Cl}(\mathcal{M}_{t})$.
Each $K_{i}$ is isomorphic to $\mathbb{Q}(\chi)$ for some
$\chi\in\mathrm{Irr}_{\mathbb{C}}(G)$ and by [CR87, (49.32)]
$\operatorname{Cl}(\mathcal{M}_{i})$ is isomorphic to $C(\chi)$. Therefore we
obtain the ‘if’ direction of the desired equivalence. The ‘only if’ direction
now follows from the fact that $\\{K_{i}:1\leq i\leq
t\\}=\\{\mathbb{Q}(\chi):\chi\in\mathrm{Irr}_{\mathbb{C}}(G)\\}$. ∎

######  Remark 5.12.

The hypothesis in Proposition 5.11 that no factor of $\mathbb{R}[G]$ is
isomorphic to the quaternions $\mathbb{H}$ can be weakened to the requirement
that $\mathcal{M}$ has the stably free cancellation property, once $C(\chi)$
is taken to be the narrow class group in the case that $\chi$ is an
irreducible symplectic character of $G$. Maximal $\mathbb{Z}$-orders in
finite-dimensional semisimple $\mathbb{Q}$-algebras with the stably free
cancellation property have been classified in [SV19]; see also [Swa83, Theorem
II] for the case of maximal $\mathbb{Z}$-orders in $\mathbb{Q}[G]$ for $G$ a
binary polyhedral group. For example, the unique maximal $\mathbb{Z}$-order in
$\mathbb{Q}[Q_{8}]$ satisfies the equivalent conditions of Lemma 2.2, where
$Q_{8}$ denotes the quaternion group of order $8$.

######  Corollary 5.13.

Let $G$ be a finite abelian group and let $e$ denote its exponent. Define

$\Sigma=\\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24,25\\\
26,27,28,30,32,33,34,35,36,38,40,42,44,45,48,50,54,60,66,70,84,90\\}$

and let $\mathcal{M}$ be the unique maximal $\mathbb{Z}$-order in
$\mathbb{Q}[G]$. Then $\mathcal{M}$ satisfies the equivalent conditions of
Lemma 2.2 if and only if $e\in\Sigma$.

###### Proof.

First write $G\cong C_{n_{1}}\times C_{n_{2}}\times\cdots\times C_{n_{k}}$ for
positive integers $k,n_{1},\ldots,n_{k}$ such that $n_{i}\mid n_{i+1}$ for
$1\leq i\leq k-1$. Then $e=n_{k}$ and
$\\{\mathbb{Q}(\chi):\chi\in\mathrm{Irr}_{\mathbb{C}}(G)\\}=\\{\mathbb{Q}(\zeta_{d}):d\mid
e\\}$. Since $\mathbb{Z}[\zeta_{d}]$ is the ring of integers of
$\mathbb{Q}(\zeta_{d})$, this implies that
$\\{C(\chi):\chi\in\mathrm{Irr}_{\mathbb{C}}(G)\\}=\\{\operatorname{Cl}(\mathbb{Z}[\zeta_{d}]):d\mid
e\\}$. The set $\Sigma$ is precisely the set positive integers $n$ for which
$\operatorname{Cl}(\mathbb{Z}[\zeta_{n}])=0$ by [MM76]; see also [Was97,
Theorem 11.1] (note that we include choices of $n$ such that $n\equiv
2\bmod{4}$). It can easily be checked that $\Sigma$ is also precisely the set
of choices of $e$ for which $\operatorname{Cl}(\mathbb{Z}[\zeta_{d}])=0$ for
all $d\mid e$. Since no factor of $\mathbb{Q}[G]$ is isomorphic to the
quaternions $\mathbb{H}$, the desired result now follows from Proposition
5.11. ∎

######  Remark 5.14.

The hypotheses of Proposition 5.11 and Corollary 5.13 are much weaker than
$\operatorname{Cl}(\mathbb{Z}[G])$ itself being trivial. If $G$ is a finite
abelian group, then a result of Ph. Cassou-Noguès [CN72] shows that
$\text{Cl}(\mathbb{Z}[G])$ is trivial if and only if either $G\cong
C_{2}\times C_{2}$ or $G\cong C_{n}$ where $1\leq n\leq 11$ or
$n\in\\{13,14,17,19\\}$. If $G$ is a finite non-abelian non-dihedral group
then a result of Endô and Hironaka [EH79] shows that
$\text{Cl}(\mathbb{Z}[G])$ is trivial if and only if $G\cong A_{4}$, $S_{4}$
or $A_{5}$ (the if direction was already shown by Reiner and Ullom [RU74]).
For partial results in the dihedral case, see [EM80].

Many specialisations of Theorem 5.3 can now be obtained by applying the
results of §5.3, Proposition 5.11 and/or Corollary 5.13; the following is just
one such example.

######  Theorem 5.15.

Let $G$ be a finite group and suppose that there exist positive integers
$n_{1},\ldots,n_{t}$ such that
$\mathbb{Q}[G]\cong\prod_{i=1}^{t}\mathrm{Mat}_{n_{i}}(\mathbb{Q})$. Let $n$
be a positive integer. If $X$ is an $\mathbb{Z}[G]$-lattice such that
$\mathbb{Q}X$ is free of rank $n$ over $\mathbb{Q}[G]$ then there exists a
free $\mathbb{Z}[G]$-sublattice $Z$ of $X$ such that $[X:Z]$ divides

$\left(|G|^{|G|}\textstyle{\prod_{i=1}^{t}}n_{i}^{-n_{i}^{2}}\right)^{\frac{3n}{2}}.$

###### Proof.

Proposition 5.11 implies that any maximal $\mathbb{Z}$-order in
$\mathbb{Q}[G]$ satisfies the equivalent conditions of Lemma 2.2. Thus the
result follows from Theorem 5.3 and Corollary 5.7. ∎

######  Remark 5.16.

The collection of finite groups $G$ satisfying the hypothesis of Theorem 5.15
is closed under direct products and includes symmetric groups and
hyperoctahedral groups (e.g. the dihedral group of order $8$). See [Sol74] and
[Kle84] for more on this topic.

##  6\. Group rings modulo trace

Let $\mathcal{O}$ be a Dedekind domain with field of fractions
$K\neq\mathcal{O}$. Let $G$ be a finite group such that $|G|$ is invertible in
$K$ and let $\mathrm{Tr}_{G}=\sum_{g\in G}g$. Then both $K[G]$ and its
quotient $K[G]/(\mathrm{Tr}_{G})$ are separable finite-dimensional
$K$-algebras. The purpose of this section is to consider lattices over the
$\mathcal{O}$-order $\mathcal{O}[G]/(\mathrm{Tr}_{G})$.

Let $e=1-|G|^{-1}\mathrm{Tr}_{G}$, which is a central idempotent of $K[G]$.
Let $\pi_{e}:K[G]\rightarrow eK[G]$ be the projection map associated to $e$.
Given a subset $X\subseteq K[G]$, let $X_{e}=\pi_{e}(X)$ and let
$X^{{1-e}}=X\cap\ker(\pi_{e})$. In particular, $K[G]_{e}=eK[G]\cong
K[G]/(\mathrm{Tr}_{G})$ and $K[G]^{{1-e}}=\mathrm{Tr}_{G}K$.

Let $\Lambda=\mathcal{O}[G]$ and let $\mathcal{M}$ be a maximal
$\mathcal{O}$-order in $K[G]$ containing $\Lambda$. Then
$\mathcal{M}_{e}=e\mathcal{M}$ is a maximal $\mathcal{O}$-order of $K[G]_{e}$
containing $\Lambda_{e}$. By Proposition 5.1,
$\leftindex^{\mathcal{M}}{\Lambda}$ is a two-sided ideal of $\mathcal{M}$
contained in $\Lambda$. Hence $(\leftindex^{{\mathcal{M}}}{\Lambda})_{e}$ is a
choice of two-sided ideal of $\mathcal{M}_{e}$ contained in $\Lambda_{e}$.
This is not necessarily the largest such choice, but its form allows us to
make use of our previous computations.

######  Lemma 6.1.

We have
$[\mathcal{M}_{e}:\Lambda_{e}]_{\mathcal{O}}=|G|^{-1}[\mathcal{M}:\Lambda]_{\mathcal{O}}$
and
$[\Lambda_{e}:(\leftindex^{{\mathcal{M}}}{\Lambda})_{e}]_{\mathcal{O}}=[\mathcal{M}:\Lambda]_{\mathcal{O}}$.

###### Proof.

Consider the following diagram of $\mathcal{O}$-lattices with exact rows

${0}$${\Lambda^{1-e}}$${\Lambda}$${\Lambda_{e}}$${0}$${0}$${\mathcal{M}^{1-e}}$${\mathcal{M}}$${\mathcal{M}_{e}}$${0.}$

Then by Lemma 2.1 we have
$[\mathcal{M}:\Lambda]_{\mathcal{O}}=[\mathcal{M}^{1-e}:\Lambda^{1-e}]_{\mathcal{O}}\cdot[\mathcal{M}_{e}:\Lambda_{e}]_{\mathcal{O}}$.
Note that
$\mathcal{M}^{1-e}=\left(|G|^{-1}\mathrm{Tr}_{G}\right)\cdot\mathcal{O}$ and
$\Lambda^{1-e}=\mathcal{M}^{1-e}\cap\Lambda=\mathrm{Tr}_{G}\cdot\mathcal{O}$.
Hence $[\mathcal{M}^{1-e}:\Lambda^{1-e}]_{\mathcal{O}}=|G|$, and so we obtain
the first equality.

Similarly, we also have the following diagram of $\mathcal{O}$-lattices with
exact rows

(6.1)
${0}$${(\leftindex^{\mathcal{M}}{\Lambda})^{1-e}}$${\leftindex^{\mathcal{M}}{\Lambda}}$${(\leftindex^{\mathcal{M}}{\Lambda})_{e}}$${0}$${0}$${\Lambda^{1-e}}$${\Lambda}$${\Lambda_{e}}$${0.}$

Then by Lemma 2.1 we have
$[\Lambda:\leftindex^{\mathcal{M}}{\Lambda}]_{\mathcal{O}}=[\Lambda^{1-e}:(\leftindex^{\mathcal{M}}{\Lambda})^{1-e}]_{\mathcal{O}}\cdot[\Lambda_{e}:(\leftindex^{\mathcal{M}}{\Lambda})_{e}]_{\mathcal{O}}$.
By maximality of $\leftindex^{\mathcal{M}}{\Lambda}$, the subset
$(\leftindex^{\mathcal{M}}{\Lambda})^{1-e}$ is the largest
$\mathcal{M}^{1-e}$-sublattice contained in $\Lambda^{1-e}$. Since
$\mathcal{M}^{1-e}\cong\mathcal{O}$, we find that $\Lambda^{1-e}$, an
$\mathcal{O}$-lattice, is already a $\mathcal{M}^{1-e}$-sublattice so that the
left vertical map of (6.1) is an equality. Hence we have
$[\Lambda_{e}:(\leftindex^{{\mathcal{M}}}{\Lambda})_{e}]_{\mathcal{O}}=[\Lambda:\leftindex^{\mathcal{M}}{\Lambda}]_{\mathcal{O}}$.
But
$[\Lambda:\leftindex^{\mathcal{M}}{\Lambda}]_{\mathcal{O}}=[\mathcal{M}:\Lambda]_{\mathcal{O}}$
by Proposition 5.1, and thus we obtain the desired result. ∎

######  Theorem 6.2.

Let $\mathcal{O}$ be a Dedekind domain with field of fractions $K$. Assume
that $K$ is a global field and that $\mathcal{O}\neq K$. Let $G$ be a finite
group such that $|G|$ is invertible in $K$. Set $s=2$ if $G$ is abelian and
$s=3$ otherwise. Let $\mathcal{M}$ be a maximal $\mathcal{O}$-order in $K[G]$
containing $\mathcal{O}[G]$. Let $n$ be a positive integer and let
$\mathcal{K}$ be any non-zero ideal of $\mathcal{O}$. Then there exists a non-
zero ideal $\mathcal{I}$ of $\mathcal{O}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any
$\mathcal{O}[G]/(\mathrm{Tr}_{G})$-lattice $X$ such that $KX$ is free of rank
$n$ over $K[G]/(\mathrm{Tr}_{G})$, there exists a free
$\mathcal{O}[G]/(\mathrm{Tr}_{G})$-sublattice $Z$ of $X$ such that
$[X:Z]_{\mathcal{O}}$ divides
$\mathcal{I}\cdot|G|^{-2n}\cdot[\mathcal{M}:\mathcal{O}[G]]_{\mathcal{O}}^{sn}$.
Moreover, if $\mathcal{M}$ satisfies the equivalent conditions of Lemma 2.2
then we can take $\mathcal{I}=\mathcal{O}$.

###### Proof.

Let $\Lambda=\mathcal{O}[G]$. Then $\mathcal{M}_{e}$ is a maximal
$\mathcal{O}$-order of $K[G]_{e}$ containing
$\Lambda_{e}=\mathcal{O}[G]/(\mathrm{Tr}_{G})$. Note that if $\mathcal{M}$
satisfies the equivalent conditions of Lemma 2.2 then so does
$\mathcal{M}_{e}$. By Lemma 6.1 we have
$[\mathcal{M}_{e}:\Lambda_{e}]_{\mathcal{O}}=|G|^{-1}[\mathcal{M}:\Lambda]_{\mathcal{O}}$.
By Proposition 5.1, $J:=(\leftindex^{\mathcal{M}}{\Lambda})_{e}$ is a two-
sided ideal of $\mathcal{M}_{e}$ contained in $\Lambda_{e}$. Then we have

$[\mathcal{M}_{e}:\Lambda_{e}]_{\mathcal{O}}\cdot[\mathcal{M}_{e}:J]_{\mathcal{O}}=[\mathcal{M}_{e}:\Lambda_{e}]_{\mathcal{O}}^{2}\cdot[\Lambda_{e}:J]_{\mathcal{O}}=|G|^{-2}\cdot[\mathcal{M}:\Lambda]_{\mathcal{O}}^{3}.$

Therefore we obtain the desired result by applying Theorem 4.5 for the
$\mathcal{O}$-order $\Lambda_{e}$, the maximal $\mathcal{O}$-order
$\mathcal{M}_{e}$ and the ideal $J$. ∎

##  7\. Application: approximation of normal integral bases

We refer the reader to [Frö83] for an overview of normal integral bases, on
which there is a vast literature. In this section, we consider examples of
applications of the algebraic machinery of previous sections to the
approximation of normal integral bases.

Beyond the base field and the isomorphism type of the Galois group, the
following result does not use any arithmetic information about the Galois
extensions concerned.

######  Theorem 7.1.

Let $K$ be a number field and let $\mathcal{K}$ be any non-zero ideal of
$\mathcal{O}_{K}$. Let $G$ be a finite group and let $\mathcal{M}$ be a
maximal $\mathcal{O}$-order in $K[G]$ containing $\mathcal{O}_{K}[G]$. Set
$s=2$ if $G$ is abelian and $s=3$ otherwise. Then there exists a non-zero
ideal $\mathcal{I}$ of $\mathcal{O}_{K}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any Galois extension $L/K$
with $\mathrm{Gal}(L/K)\cong G$, there exists $\alpha\in\mathcal{O}_{L}$ such
that
$[\mathcal{O}_{L}:\mathcal{O}_{K}[\mathrm{Gal}(L/K)]\cdot\alpha]_{\mathcal{O}}$
divides
$\mathcal{I}\cdot[\mathcal{M}:\mathcal{O}_{K}[G]]_{\mathcal{O}_{K}}^{s}$.
Moreover, if $\mathcal{M}$ satisfies the equivalent conditions of Lemma 2.2
then we can take $\mathcal{I}=\mathcal{O}_{K}$.

###### Proof.

The normal basis theorem says that for a finite Galois extension of fields
$L/K$ we have $L\cong K[\mathrm{Gal}(L/K)]$ as $K[\mathrm{Gal}(L/K)]$-modules.
Therefore the desired result now follows easily from Theorem 5.3 with $n=1$
and $\mathcal{O}=\mathcal{O}_{K}$. ∎

######  Remark 7.2.

An explicit formula for $[\mathcal{M}:\mathcal{O}_{K}[G]]_{\mathcal{O}_{K}}$
is given in Corollary 5.6. In particular, a weak but general bound is that
$[\mathcal{M}:\mathcal{O}_{K}[G]]_{\mathcal{O}_{K}}^{s}$ divides $|G|^{\lceil
s|G|/2\rceil}$.

By making the further assumption that the extensions concerned are at most
tamely ramified, we obtain the following result with a stronger conclusion.

######  Theorem 7.3.

Let $K$ be a number field, let $\mathcal{K}$ be any non-zero ideal of
$\mathcal{O}_{K}$, and $G$ be a finite group. Then there exists a non-zero
ideal $\mathcal{I}$ of $\mathcal{O}_{K}$, that can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any at most tamely ramified
Galois extension $L/K$ with $\mathrm{Gal}(L/K)\cong G$, there exists
$\alpha\in\mathcal{O}_{L}$ such that
$[\mathcal{O}_{L}:\mathcal{O}_{K}[\mathrm{Gal}(L/K)]\cdot\alpha]_{\mathcal{O}_{K}}$
divides $\mathcal{I}$. Moreover, if $\mathcal{O}_{K}[G]$ satisfies the
equivalent conditions of Lemma 2.2 then we can take
$\mathcal{I}=\mathcal{O}_{K}$.

###### Proof.

For an at most tamely ramified Galois extension $L/K$ with
$\mathrm{Gal}(L/K)\cong G$, we have that $\mathcal{O}_{L}$ is a locally free
$\mathcal{O}_{K}[G]$-lattice of rank $1$ by [Frö83, Chapter I, §3, Corollary
2], for example. Therefore the desired result now follows easily from
Proposition 4.1 with $\mathcal{O}=\mathcal{O}_{K}$ and
$\Gamma=\mathcal{O}_{K}[G]$. ∎

######  Remark 7.4.

Improved bounds can be obtained in special cases. For example, if $G$ is a
finite group with no irreducible symplectic characters (e.g. $G$ is abelian or
of odd order), then every (at most) tamely ramified Galois extension
$L/\mathbb{Q}$ with $\mathrm{Gal}(L/\mathbb{Q})\cong G$ has a normal integral
basis by a special case of Taylor’s proof [Tay81] of a conjecture of Fröhlich
(see [Frö83, §I] for an overview). Moreover, if $G$ is a finite abelian group
then Leopoldt’s theorem [Leo59] (see also [Let90]) implies that for every
Galois extension $L/\mathbb{Q}$ with $\mathrm{Gal}(L/\mathbb{Q})\cong G$,
there exists $\alpha\in\mathcal{O}_{L}$ such that
$[\mathcal{O}_{L}:\mathbb{Z}[\mathrm{Gal}(L/\mathbb{Q})]\cdot\alpha]$ divides
$[\mathcal{M}:\mathbb{Z}[G]]$, where $\mathcal{M}$ is the unique maximal
$\mathbb{Z}$-order in $\mathbb{Q}[G]$. By contrast, Theorems 7.1 and 7.3 are
very general and their short proofs use little or no arithmetic information
about the particular field extensions concerned.

##  8\. Application: approximation of strong Minkowski units

In this section, we consider examples of applications of the algebraic
machinery of previous sections to the approximation of strong Minkowski units.

######  Definition 8.1.

Let $L/K$ be a Galois extension of number fields and let
$G=\mathrm{Gal}(L/K)$. Let $\mu_{L}$ denote the roots of unity of $L$. An
element $\varepsilon\in\mathcal{O}_{L}^{\times}/\mu_{L}$ is said to be

  1. (i)

a _Minkowski unit_ of $L/K$ if
$\mathbb{Q}\otimes_{\mathbb{Z}}(\mathcal{O}_{L}^{\times}/\mu_{L})=\mathbb{Q}[G]\cdot\varepsilon$,

  2. (ii)

a _strong Minkowski unit_ of $L/K$ if
$\mathcal{O}_{L}^{\times}/\mu_{L}=\mathbb{Z}[G]\cdot\varepsilon$.

######  Lemma 8.2.

Let $L/K$ be a Galois extension of number fields and let
$G=\mathrm{Gal}(L/K)$. Then $L/K$ has a Minkowski unit if and only $K$ is
equal to either $\mathbb{Q}$ or an imaginary quadratic field. Moreover, if
either $L$ is totally real or $K$ is imaginary quadratic, then
$\mathbb{Q}\otimes_{\mathbb{Z}}(\mathcal{O}_{L}^{\times}/\mu_{L})\cong\mathbb{Q}[G]/(\mathrm{Tr}_{G})$
as $\mathbb{Q}[G]/(\mathrm{Tr}_{G})$-modules (and as $\mathbb{Q}[G]$-modules).

###### Proof.

By definition, $L/K$ has a Minkowski unit if and only if
$\mathbb{Q}\otimes_{\mathbb{Z}}\mathcal{O}_{L}^{\times}$ is cyclic as a
$\mathbb{Q}[G]$-module. By a theorem of Herbrand (see [Tat84, Chapter I,
§4.3], for example) there is an isomorphism
$(\mathbb{Q}\otimes_{\mathbb{Z}}\mathcal{O}_{L}^{\times})\oplus\mathbb{Q}\cong\mathbb{Q}[S_{\infty}]$
of $\mathbb{Q}[G]$-modules, where $S_{\infty}$ denotes the set of infinite
places of $L$. So the existence of a Minkowski unit is equivalent to
$\mathbb{Q}[S_{\infty}]$ being cyclic as a $\mathbb{Q}[G]$-module, which in
turn is equivalent to $S_{\infty}$ being transitive as a $G$-set. This occurs
precisely when $K$ has a unique infinite place. If either $L$ is totally real
or $K$ is imaginary quadratic, then the unique infinite place of $K$ splits
completely in $L/K$ and thus $\mathbb{Q}[S_{\infty}]\cong\mathbb{Q}[G]$ as
$\mathbb{Q}[G]$-modules. ∎

######  Remark 8.3.

The existence of strong Minkowski units (which some authors refer to as
Minkowski units) in special cases has been studied in numerous articles,
including [Pay75, Mos75, BP79, Duv81, Mos79a, Mos79b, Mos83, BP75, Mar88,
Mar91, Frö92, Bur93, MU04, Mar05, Mar11a, Mar11b], as well as [Nar04, §3.3 &
§3.5.1] and the references therein. Also see the articles cited below.

######  Remark 8.4.

If $L/\mathbb{Q}$ is a Galois extension with $[L:\mathbb{Q}]$ odd then $L$ is
totally real and
$\mathcal{O}_{L}^{\times}/\mu_{L}\cong\\{u\in\mathcal{O}_{L}^{\times}:\mathrm{Norm}_{L/\mathbb{Q}}(u)=1\\}$
as $\mathbb{Z}[\mathrm{Gal}(L/\mathbb{Q})]$-lattices.

The following result is a strong refinement and generalisation of [Mar96,
Theorem 1], [All13, Proposition 5.2] and [All14, Theorem 3.3], which only
consider finite cyclic or abelian totally real extensions of $\mathbb{Q}$ and
do not actually bound the index in question.

######  Theorem 8.5.

Let $G$ be a finite group and let $k$ be a positive integer. Set $s=2$ if $G$
is abelian and $s=3$ otherwise. Then there exists a positive integer $i$,
which can be chosen to be coprime to $k$, with the following property: given
any finite Galois extension $L/K$ with $\mathrm{Gal}(L/K)\cong G$ and $K$
equal to either $\mathbb{Q}$ or an imaginary quadratic field, there exists a
Minkowski unit $\varepsilon\in\mathcal{O}_{L}^{\times}/\mu_{L}$ such that
$[\mathcal{O}_{L}^{\times}/\mu_{L}:\mathbb{Z}[\mathrm{Gal}(L/K)]\cdot\varepsilon]$
divides $i\cdot|G|^{-2}[\mathcal{M}:\mathbb{Z}[G]]^{s}$, where $\mathcal{M}$
is a maximal $\mathbb{Z}$-order in $\mathbb{Q}[G]$ containing $\mathbb{Z}[G]$.

######  Remark 8.6.

An explicit formula for $[\mathcal{M}:\mathbb{Z}[G]]$ is given in Corollary
5.6. In particular, a weak but general bound is that
$|G|^{-2}[\mathcal{M}:\mathbb{Z}[G]]^{s}$ divides $|G|^{\lceil
s|G|/2\rceil-2}$.

######  Remark 8.7.

It is interesting to compare Theorem 8.5 with [AV17, Theorem 1.1], which in
the case that $L/\mathbb{Q}$ is a finite Galois extension asserts the
existence of a Minkowski unit $\varepsilon$ such that the index
$[\mathcal{O}_{L}^{\times}/\mu_{L}:\mathbb{Z}[\mathrm{Gal}(L/\mathbb{Q})]\cdot\varepsilon]$
is bounded by an expression involving the Weil height of $\varepsilon$, the
regulator of $L$, the degree $[L:\mathbb{Q}]$ and
$\mathrm{rank}_{\mathbb{Z}}\mathcal{O}_{L}^{\times}$.

###### Proof of Theorem 8.5.

Let $\Gamma=\mathbb{Z}[G]/(\mathrm{Tr}_{G})$. By Theorem 6.2 there exists a
positive integer $i$, which can be chosen to be coprime to $k$, with the
following property: given any $\Gamma$-lattice $X$ such that
$\mathbb{Q}X\cong\mathbb{Q}[G]/(\mathrm{Tr}_{G})$ as
$\mathbb{Q}[G]/(\mathrm{Tr}_{G})$-modules, there exists a free
$\Gamma$-sublattice $Y$ of $X$ such that $[X:Y]$ divides
$i\cdot|G|^{-2}[\mathcal{M}:\mathbb{Z}[G]]^{s}$. Note that if $\varepsilon\in
X$ is a free $\Gamma$-generator of $Y$, then
$Y=\Gamma\cdot\varepsilon=\mathbb{Z}[G]\cdot\varepsilon$. For $L/K$ with
either $L$ totally real or $K$ imaginary quadratic, the desired result now
follows from Lemma 8.2 after fixing an isomorphism $\mathrm{Gal}(L/K)\cong G$.
For $L/K$ with $L$ totally imaginary and $K=\mathbb{Q}$, the result follows
from Lemma 8.2 and Remark 4.6. ∎

######  Corollary 8.8.

Let $G$ be a finite group satisfying the conditions of Proposition 5.11 or
Corollary 5.13. Set $s=2$ if $G$ is abelian and $s=3$ otherwise. Then given
any finite Galois extension $L/K$ with $\mathrm{Gal}(L/K)\cong G$ and $K$
equal to either $\mathbb{Q}$ or an imaginary quadratic field, there exists a
Minkowski unit $\varepsilon\in\mathcal{O}_{L}^{\times}/\mu_{L}$ such that
$[\mathcal{O}_{L}^{\times}/\mu_{L}:\mathbb{Z}[\mathrm{Gal}(L/K)]\cdot\varepsilon]$
divides $|G|^{-2}[\mathcal{M}:\mathbb{Z}[G]]^{s}$, where $\mathcal{M}$ is a
maximal $\mathbb{Z}$-order in $\mathbb{Q}[G]$ containing $\mathbb{Z}[G]$.

###### Proof.

The hypotheses on $G$ imply that $\mathcal{M}$ satisfies the equivalent
conditions of Lemma 2.2. Thus the desired result follows as in the proof of
Theorem 8.5 after noting that we can take $i=1$ in the application of Theorem
6.2. ∎

######  Remark 8.9.

Let $L/\mathbb{Q}$ be a finite Galois extension such that $L$ is CM and let
$L^{+}$ denote its maximal totally real subfield. Let
$\varepsilon\in\mathcal{O}_{L^{+}}^{\times}/\\{\pm 1\\}$ be a Minkowski unit
of $L^{+}/\mathbb{Q}$ and by abuse of notation let this also denote its image
in $\mathcal{O}_{L}^{\times}/\mu_{L}$. By [Was97, Theorem 4.12] we have that
$[\mathcal{O}_{L}^{\times}:\mu_{L}\mathcal{O}_{L^{+}}^{\times}]=1$ or $2$, and
so
$[\mathcal{O}_{L}^{\times}/\mu_{L}:\mathbb{Z}[\mathrm{Gal}(L/\mathbb{Q})]\cdot\varepsilon]$
divides $2[\mathcal{O}_{L^{+}}^{\times}/\\{\pm
1\\}:\mathbb{Z}[\mathrm{Gal}(L^{+}/\mathbb{Q})]\cdot\varepsilon]$. Thus
$\varepsilon$ is also a Minkowski unit of $L/\mathbb{Q}$, and we obtain
stronger analogues of Theorem 8.5 and Corollary 8.8 in this situation.

The above results can be strengthened for extensions of prime degree.

######  Theorem 8.10.

Let $p$ be an odd prime and let $k$ be a positive integer. Then there exists a
positive integer $i$, which can be chosen to be coprime to $k$, with the
following property: given any cyclic field extension $L/K$ with $[L:K]=p$ and
$K$ equal to either $\mathbb{Q}$ or an imaginary quadratic field, there exists
a Minkowski unit $\varepsilon\in\mathcal{O}_{L}^{\times}/\mu_{L}$ such that
$[\mathcal{O}_{L}^{\times}/\mu_{L}:\mathbb{Z}[\mathrm{Gal}(L/K)]\cdot\varepsilon]$
divides $i$.

###### Proof.

Let $G$ be the cyclic group of order $p$ and let
$\mathcal{M}=\mathbb{Z}[G]/(\mathrm{Tr}_{G})$. Then
$\mathcal{M}\cong\mathbb{Z}[\zeta_{p}]$, which is a maximal
$\mathbb{Z}$-order. By Corollary 4.2 there exists a positive integer $i$,
which can be chosen to be coprime to $k$, with the following property: given
any $\mathcal{M}$-lattice $X$ such that $\mathbb{Q}X$ is free of rank $1$ as a
$\mathbb{Q}[G]/(\mathrm{Tr}_{G})$-module, there exists a free
$\mathcal{M}$-sublattice $Y$ of $X$ such that $[X:Y]$ divides $i$. Note that
if $\varepsilon\in X$ is a free $\mathcal{M}$-generator of $Y$, then
$Y=\mathcal{M}\cdot\varepsilon=\mathbb{Z}[G]\cdot\varepsilon$. The desired
result now follows from Lemma 8.2 since the hypotheses ensure that either $L$
is totally real or $K$ is imaginary quadratic. ∎

The following result is not new, as it is the combination of [Zeĭ65] (see also
[Bru69, Corollary]) and an easy consequence of [Gil80, Théorèm]; we include it
for completeness.

######  Corollary 8.11.

Let $p$ be a prime such that $3\leq p\leq 19$. Then every cyclic field
extension $L/K$ with $[L:K]=p$ and $K$ equal to either $\mathbb{Q}$ or an
imaginary quadratic field has a strong Minkowski unit.

###### Proof.

In the proof of Theorem 8.10, adding the hypothesis that $p\leq 19$ implies
that $\mathcal{M}\cong\mathbb{Z}[\zeta_{p}]$ has trivial class group and so
satisfies the equivalent conditions of Lemma 2.2. Hence we can take $i=1$ by
Remark 4.3, and this implies the desired result. ∎

######  Remark 8.12.

It is interesting to compare Theorem 8.10 to (i) [Bru69, Theorem] when
$K=\mathbb{Q}$ and (ii) [Gil80, Théorèm] when $K$ is imaginary quadratic.
Result (i) considers cyclic extensions $L/\mathbb{Q}$ of odd prime degree $p$
and gives sufficient conditions on ideals of $\mathbb{Z}[\zeta_{p}]$ of norm
equal to the class number $h_{L}$ of $\mathcal{O}_{L}$ for both the existence
and non-existence of a strong Minkowski unit of $L/\mathbb{Q}$. The proof uses
the fact that $\mathcal{O}_{L}^{\times}/\\{\pm 1\\}$ contains a free
$\mathbb{Z}[\zeta_{p}]$-submodule of index $h_{L}$ generated by a cyclotomic
unit. Result (ii) is analogous and uses elliptic units. By contrast, the proof
and statement of Theorem 8.10 do not depend on the particular extension $L/K$.

##  9\. Application: rational points on abelian varieties

In this section, we consider examples of applications of the algebraic
machinery of previous sections to the Galois module structure of rational
points of abelian varieties. By the Mordell–Weil theorem, for every abelian
variety $A$ over a number field $K$, the group $A(K)/A(K)_{\mathrm{tors}}$ is
a free $\mathbb{Z}$-module of finite rank. If $L/K$ is a Galois extension of
number fields, then $A(L)/A(L)_{\mathrm{tors}}$ is a
$\mathbb{Z}[\mathrm{Gal}(L/K)]$-lattice, so is amenable to study via our
methods.

######  Theorem 9.1.

Let $G$ be a finite group and let $k$ be a positive integer. Set $s=2$ if $G$
is abelian and $s=3$ otherwise. Then there exists a positive integer $i$,
which can be chosen to be coprime to $k$, with the following property: given
any Galois extension of number fields $L/K$ with $\mathrm{Gal}(L/K)\cong G$,
and any abelian variety $A/K$ such that $\mathbb{Q}\otimes_{\mathbb{Z}}A(L)$
is cyclic as a $\mathbb{Q}[G]$-module, there exists $\varepsilon\in
A(L)/A(L)_{\mathrm{tors}}$ such that
$[A(L)/A(L)_{\mathrm{tors}}:\mathbb{Z}[G]\cdot\varepsilon]$ is finite and
divides $i\cdot[\mathcal{M}:\mathbb{Z}[G]]^{s}$, where $\mathcal{M}$ is any
maximal $\mathbb{Z}$-order in $\mathbb{Q}[G]$ containing $\mathbb{Z}[G]$.

######  Remark 9.2.

An explicit formula for $[\mathcal{M}:\mathbb{Z}[G]]$ is given in Corollary
5.6. In particular, a weak but general bound is that
$[\mathcal{M}:\mathbb{Z}[G]]^{s}$ divides $|G|^{\lceil s|G|/2\rceil}$.

######  Remark 9.3.

Let $G$ be a finite group. The isomorphism class of a finite-dimensional
$\mathbb{Q}[G]$-module $X$ is entirely determined by the values of
$\dim_{\mathbb{Q}}X^{H}$ as $H$ runs over a set of representatives of the set
of cyclic subgroups of $G$ up to conjugacy (see [Ser77, §13.1, Corollary to
Theorem 30´]). In particular, $X$ is free of rank $1$ if and only if
$\dim_{\mathbb{Q}}X^{H}=[G:H]$ for all cyclic subgroups $H$ of $G$ up to
conjugacy.

###### Proof of Theorem 9.1.

By Theorem 5.3, there exists a positive integer $i$, which can be chosen to be
coprime to $n$, with the following property: given any $\mathbb{Z}[G]$-lattice
$X$ such that $\mathbb{Q}\otimes_{\mathbb{Z}}X$ is free of rank $1$ as a
$\mathbb{Q}[G]$-module, there exists a free $\mathbb{Z}[G]$-sublattice $Y$ of
$X$ such that $[X:Y]$ divides $i\cdot[\mathcal{M}:\mathbb{Z}[G]]^{s}$. By
Remark 4.6, $i$ also has the property that given any $\mathbb{Z}[G]$-lattice
$X$ such that $\mathbb{Q}\otimes_{\mathbb{Z}}X$ is cyclic as a
$\mathbb{Q}[G]$-module, there exists a cyclic $\mathbb{Z}[G]$-sublattice $Y$
of $X$ such that $[X:Y]$ divides $i\cdot[\mathcal{M}:\mathbb{Z}[G]]^{s}$. In
particular, this holds for $X=A(L)/A(L)_{\mathrm{tors}}$ after fixing an
isomorphism $G\cong\mathrm{Gal}(L/K)$. ∎

######  Theorem 9.4.

Let $p$ be an odd prime and let $k$ be a positive integer. Then there exists a
positive integer $i$, which can be chosen to be coprime to $k$, with the
following property: given any cyclic extension $L/K$ of number fields with
$[L:K]=p$ and any abelian variety $A/K$ such that
$\mathrm{rank}_{\mathbb{Z}}A(K)=0$ and $\mathrm{rank}_{\mathbb{Z}}A(L)=p-1$,
there exists $\varepsilon\in A(L)/A(L)_{\mathrm{tors}}$ such that
$[A(L)/A(L)_{\mathrm{tors}}:\mathbb{Z}[\mathrm{Gal}(L/K)]\cdot\varepsilon]$ is
finite and divides $i$.

###### Proof.

Let $G$ be the cyclic group of order $p$ and let
$\mathcal{M}=\mathbb{Z}[G]/(\mathrm{Tr}_{G})$. Then
$\mathcal{M}\cong\mathbb{Z}[\zeta_{p}]$, which is a maximal
$\mathbb{Z}$-order. By Corollary 4.2 there exists a positive integer $i$,
which can be chosen to be coprime to $n$, with the following property: given
any $\mathcal{M}$-lattice $X$ such that
$\mathbb{Q}X\cong\mathbb{Q}[G]/(\mathrm{Tr}_{G})$ as
$\mathbb{Q}[G]/(\mathrm{Tr}_{G})$-modules, there exists a free
$\mathcal{M}$-sublattice $Y$ of $X$ such that $[X:Y]$ divides $i$. After
fixing an isomorphism $G\cong\mathrm{Gal}(L/K)$, the desired result now
follows since the rank hypotheses ensure that
$\mathbb{Q}\otimes_{\mathbb{Z}}(A(L)/A(L)_{\mathrm{tors}})\cong\mathbb{Q}[G]/(\mathrm{Tr}_{G})$
as $\mathbb{Q}[G]$-modules (see Remark 9.3) and hence as
$\mathbb{Q}[G]/(\mathrm{Tr}_{G})$-modules. ∎

######  Remark 9.5.

Note that for $\mathbb{Q}\otimes_{\mathbb{Z}}(A(L)/A(L)_{\mathrm{tors}})$ to
be cyclic as a ${\mathbb{Q}}[G]/(\mathrm{Tr}_{G})$-module, it is necessary
that $\mathrm{rank}_{\mathbb{Z}}A(L)=0$ or $p-1$.

######  Corollary 9.6.

Let $p$ be a prime such that $3\leq p\leq 19$. Then given any cyclic extension
of number fields $L/K$ with $[L:K]=p$ and any abelian variety $A/K$ such that
$\mathrm{rank}_{\mathbb{Z}}A(K)=0$ and $\mathrm{rank}_{\mathbb{Z}}A(L)=p-1$,
there exists $\varepsilon\in A(L)/A(L)_{\mathrm{tors}}$ such that
$A(L)/A(L)_{\mathrm{tors}}=\mathbb{Z}[\mathrm{Gal}(L/K)]\cdot\varepsilon$.

###### Proof.

In the proof of Theorem 9.4, the additional hypothesis that $p\leq 19$ implies
that $\mathcal{M}\cong\mathbb{Z}[\zeta_{p}]$ has trivial class group and so
satisfies the equivalent conditions of Lemma 2.2. Hence we can take $i=1$ by
Remark 4.3, and this implies the desired result. ∎

######  Theorem 9.7.

Let $p$ be a prime, let $F$ be an imaginary quadratic field with discriminant
coprime to $p$, and let $\mathcal{K}$ be a non-zero ideal of
$\mathcal{O}_{F}$. Then there exists an ideal $\mathcal{I}$ of
$\mathcal{O}_{F}$, which can be chosen to be coprime to $\mathcal{K}$, with
the following property: given any cyclic extension of number fields $L/K$ with
$[L:K]=p$ and such that $K$ contains $F$, and any elliptic curve $E/K$ with
complex multiplication by $\mathcal{O}_{F}$ and with
$\mathrm{rank}_{\mathbb{Z}}E(K)=0$ and
$\mathrm{rank}_{\mathbb{Z}}E(L)=2(p-1)$, there exists $\varepsilon\in
E(L)/E(L)_{\mathrm{tors}}$ such that
$[E(L)/E(L)_{\mathrm{tors}}:\mathcal{O}_{F}[\mathrm{Gal}(L/K)]\cdot\varepsilon]_{\mathcal{O}_{F}}$
divides $\mathcal{I}$.

###### Proof.

Let $G$ be the cyclic group of order $p$. The discriminant of
$\mathbb{Z}[\zeta_{p}]$ is a power of $p$ and in particular is coprime to the
discriminant of $\mathcal{O}_{F}$. Hence $\mathbb{Q}(\zeta_{p})$ and $F$ are
linearly disjoint over $\mathbb{Q}$. Moreover, by [Lan94, III, §3, Proposition
17] we have $\mathcal{O}_{F}[\zeta_{p}]=\mathcal{O}_{F(\zeta_{p})}$. Thus
$\mathcal{O}_{F}[G]/(\mathrm{Tr}_{G})\cong\mathcal{O}_{F}[\zeta_{p}]$ is a
maximal $\mathcal{O}_{F}$-order. By Corollary 4.2 there exists a non-zero
ideal $\mathcal{I}$ of $\mathcal{O}_{F}$, which can be chosen to be coprime to
$\mathcal{K}$, with the following property: given any
$\mathcal{O}_{F}[G]/(\mathrm{Tr}_{G})$-lattice $X$ such that
$F\otimes_{\mathcal{O}_{F}}X\cong F[G]/(\mathrm{Tr}_{G})$ as
$F[G]/(\mathrm{Tr}_{G})$-modules, there exists a free
$\mathcal{O}_{F}[G]/(\mathrm{Tr}_{G})$-sublattice $Y$ of $X$ such that
$[X:Y]_{\mathcal{O}_{F}}$ divides $\mathcal{I}$.

Let $E$, $L$ and $K$ be as in the theorem and fix an isomorphism
$G\cong\mathrm{Gal}(L/K)$. By assumption, $E$ has CM by $\mathcal{O}_{F}$
defined over $K$. The commuting Galois action and action by endomorphisms then
give $E(L)/E(L)_{\mathrm{tors}}$ the structure of an
$\mathcal{O}_{F}[G]$-lattice. Moreover, as $\mathrm{rk}_{\mathbb{Z}}E(K)=0$,
it is in fact a $\mathcal{O}_{F}[G]/(\mathrm{Tr}_{G})$-lattice and since
$\mathrm{rank}_{\mathbb{Z}}E(L)=2(p-1)$, we have that
$\dim_{\mathbb{Q}}F\otimes_{\mathcal{O}_{F}}(E(L)/E(L)_{\mathrm{tors}})=2(p-1)$.
Since $F$ and $\mathbb{Q}(\zeta_{p})$ are linearly disjoint, the unique
$F[G]/(\mathrm{Tr}_{G})$-module with these properties is
$F[G]/(\mathrm{Tr}_{G})$ itself. Therefore $E(L)/E(L)_{\mathrm{tors}}$ is an
example of an $\mathcal{O}_{F}[G]/(\mathrm{Tr}_{G})$-lattice such that
$F\otimes_{\mathcal{O}_{F}}X\cong F[G]/(\mathrm{Tr}_{G})$. ∎

######  Remark 9.8.

Since $\mathbb{Q}(\zeta_{p})$ and $F$ are linearly disjoint over $\mathbb{Q}$,
the ${F}[G]/(\mathrm{Tr}_{G})$-module
$F\otimes_{\mathcal{O}_{F}}(E(L)/E(L)_{\mathrm{tors}})$ is cyclic if and only
if either $\mathrm{rank}_{\mathbb{Z}}E(L)=0$ or $2(p-1)$.

######  Corollary 9.9.

Let $p$ be a prime, let $F$ be an imaginary quadratic field with discriminant
coprime to $p$ such that $\mathcal{O}_{F(\zeta_{p})}$ has trivial class group.
Then for every cyclic extension of number fields $L/K$ such that $K$ contains
$F$ and $[L:K]=p$, and for every elliptic curve $E/K$ with complex
multiplication by $\mathcal{O}$ and with $\mathrm{rank}_{\mathbb{Z}}E(K)=0$
and $\mathrm{rank}_{\mathbb{Z}}E(L)=2(p-1)$, we have that
$E(L)/E(L)_{\mathrm{tors}}$ is free as an
$\mathcal{O}_{F}[G]/(\mathrm{Tr}_{G})$-module.

###### Proof.

In the proof of Theorem 9.7, the additional hypothesis that
$\mathcal{O}_{F(\zeta_{p})}$ has trivial class group ensures that
$\mathcal{O}_{F}/(\mathrm{Tr}_{G})\cong\mathcal{O}_{F(\zeta_{p})}$ satisfies
the equivalent conditions of Lemma 2.2. Hence we can take
$\mathcal{I}=\mathcal{O}_{F}$ by Remark 4.3, and this implies the desired
result. ∎

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