# Some ranks of modules over group rings

V.A. Bovdi  and  L.A. Kurdachenko  Dedicated to Professor Igor Subbotin on the
occasion of his $70^{th}$ birthday  UAEU, Al Ain, United Arab Emirates
<EMAIL_ADDRESS>Department of Algebra and Geometry, School of Mathematics and
Mechanics, National University of Dnipro, Ukraine<EMAIL_ADDRESS>
###### Abstract.

A commutative ring $R$ has finite rank $r$, if each ideal of $R$ is generated
at most by $r$ elements. A commutative ring $R$ has the $r$-generator
property, if each finitely generated ideal of $R$ can be generated by $r$
elements. Such rings are closely related to Prüfer domains. In the present
paper we investigate some analogs of these concepts for modules over group
rings.

###### Key words and phrases:

module over ring; special rank; Prüfer domain; Dedekind domain

###### 2010 Mathematics Subject Classification:

Primary: 20C05; 20F50 Secondary: 20D25; 16D10; 16D70; 16D80

##  1\. Introduction and results

The concept of dimension of a vector space is one of the most important
concepts of mathematics. It is the source of many important numerical
invariants in many areas of mathematics.

Modules are natural generalizations of vector spaces, therefore, it is natural
to extend the concept of dimension to them. One possibility is the concept of
the $R$-rank of an $R$-module $A$, where $R$ is a ring. It is defined as the
cardinality of a maximal $R$-free subset of $A$. This definition turned out to
be effective only for the case when $A$ is an $R$-torsion free module over an
integral domain $R$. Other extensions of the concept of dimension are related
to the number of generators. Unlike in the case of vector spaces, not each
submodule of a finitely generated module is finitely generated. Moreover, when
a submodule is f.g. (finitely generated), it can have a minimal generating set
whose cardinality is greater than the number of elements in the generating
system of the entire module. The reasons of this phenomena are related to the
structure of the ring $R$, over which the module is considered. Not evry left
(or right) ideal of a ring is finitely generated. Rings having such property
are called Noetherian. However, even in notherian rings the number of
generators of left ideals are not always bounded. Following Cohen [5], a
commutative ring $R$ has finite rank $r$, if each ideal of $R$ has $r$
generators. Almost at the same time, the concept of special rank arises in
group theory (Maltsev [15]). Dedekind and principal ideal domains are examples
of such rings. Commutative rings, having finite rank, were studied in [5, 10,
11, 12]. The notion of rings of finite rank was generalized by Gilmer [10].
Following Gilmer, we say that a commutative ring $R$ has an $r$-generator
property, if each finitely generated ideal of $R$ can be generated by $r$
elements. Examples of such rings are the Bezout domains with the $1$-generator
property. Commutative rings having the $r$-generator property have been
actively studied in [10, 11, 12]. In particular, such rings proved to be
closely related to Prüfer domains.In our investigation we consider analogs of
these concepts for modules over group rings. Since the notion of ”$R$-rank” is
already used in representation theory, we adapt the terminology from group
theory as it was proposed by Maltsev in [15]. A module $A$ over a ring $R$ has
special rank $r$ if each f.g. (finitely generated) $R$-submodule of $A$ can be
generated by $r$ elements and there exists at least one f.g. $R$-submodule $D$
of $A$, which has a minimal generating subset, consisting exactly of $r$
elements. The special $R$-rank and the classical $R$-rank of $A$ are denoted
by $\mathrm{sr}_{R}(A)$ and $\mathrm{rank}_{R}(A)$, respectively. If
$\mathrm{sr}_{R}(A)$ is finite, then we write
$\mathrm{sr}_{R}(A)\in\mathbb{N}$, where $\mathbb{N}$ is the set of positive
integers.

Note that in the theory of groups, the topic related to groups of finite
special $FG$-rank have played and continue to play a significant role. A huge
array of articles has been devoted to this subject, in which many interesting
and profound results have been obtained. An excellent overview of the basic
results obtained in this topic is contained in the article of Dixon [8]. The
book [9] is devoted to the presentation of many results connected to groups of
finite special rank.

The first natural step in our investigation is the case when $RG$ is the group
algebra of a finite group $G$ over a field $R$. Our main result is the
following.

######  Theorem 1.

Let $FG$ be the group algebra of a finite group $G$ over a field $F$ of
characteristic 0. A right $FG$-module $A$ has a finite special $R$-rank
$\mathrm{sr}_{FG}(A)=r\in\mathbb{N}$ if and only if the following conditions
hold:

  * (i)

$A=\oplus_{1\leq j\leq n}H_{j}$ with $H_{j}=\oplus_{1\leq k\leq t(j)}D_{k}$,
where each $D_{k}$ is a simple right $FG$-submodule of $A$ and
$D_{1}\cong_{FG}D_{2}\cong_{FG}\cdots\cong_{FG}D_{t(j)}$;

  * (ii)

$t(j)\leq r$ for all $1\leq j\leq n$ and there exist $s\in\mathbb{N}$ such
that $t(s)=r$;

  * (iii)

$n$ is bounded by the number $\mathrm{nns}_{FG}(G)$ of pairwise non-isomorphic
simple $FG$-modules.

In particular, if $\mathrm{sr}_{FG}(A)=r\in\mathbb{N}$, then $\dim_{F}(A)\leq
r\cdot|G|\cdot\mathrm{nns}_{F}(G)$.

If $G$ is a finite group and $R$ is an integral domain of characteristic $0$,
then a description of $R$-torsion-free $RG$-modules, having finite special
rank, is given by the following.

######  Theorem 2.

Let $RG$ be the group ring of a finite group $G$ over an integral domain $R$
of characteristic 0 and let $F$ be the field of fractions of $R$. Let $A$ be a
right $RG$-module, which is $R$-torsion free. The $RG$-module $A$ has special
rank $\mathrm{sr}_{RG}(A)=r\in\mathbb{N}$ if and only if the right $FG$-module
$A\otimes_{R}F$ has the same special rank $\mathrm{sr}_{FG}(A\otimes_{R}F)=r$.
In particular, $\mathrm{rank}_{RG}(A)\leq\mathrm{nns}_{F}(G)$, where
$\mathrm{nns}_{FG}(G)$ is the number of pairwise non-isomorphic simple
$FG$-modules.

Let $A$ be a right module over an integral domain $R$. If $A$ has a finite
$R$-rank, then the section $A/\omega(RG)A$ also has a finite $R$-rank, where
$\omega(RG)=\langle g-1\mid g\in G\rangle_{R}$ is the augmentation ideal of
$RG$. Since the action of $G$ on the right module $A/\omega(RG)A$ is trivial,
$A/\omega(RG)A$ is an $R$-module. Therefore, it is natural to consider modules
over $RG$ having finite special ranks in which $R$ is a commutative ring with
the $r$-generator property. An important example is the case when $R$ is a
Dedekind domain. Our next result is the following.

######  Theorem 3.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$,
all whose capitals have characteristic $0$. Let $F$ be the field of fractions
of the ring $D$. Let $A$ be a right $DG$-module which is $D$-periodic. The
module $A$ has special rank $\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$ if and only
if the following conditions hold:

  * (i)

$A=\oplus_{P\in\pi}A_{P}$ where $\pi=\mathrm{Ass}_{R}(A)$ and $A_{P}$ is the
$P$-component of $A$;

  * (ii)

$A_{P}=\oplus_{1\leq j\leq n(P)}H_{j,P}$ where $n(P)\leq\mathrm{nns}_{F}(G)$
and each $H_{j,P}$ is a $G$-homogeneous $DG$-submodule of $A$ and it is
Artinian as a $D$-module;

  * (iii)

$\Omega_{[P,1]}(H_{j,P})$ is a direct sum of at most $r$ simple
$DG$-submodules of $A$;

  * (iv)

there exist $t\in\mathbb{N}$ and a maximal ideal $P$ such that
$\Omega_{[P,1]}(H_{t,P})$ is a direct sum of $r$ copies of simple
$DG$-submodules of $A$.

All necessary definitions and notations are given in the next section.

##  2\. Preliminaries and Lemmas

In the sequel, each module means a right module, unless otherwise specified.
Let $FG$ be the group algebra of a finite group $G$ over a field $F$ of
characteristic $0$. Each $FG$-module $A$ is semisimple (see [13, Corollary
5.15]), so the algebra $FG$ is semisimple too. The number of pairwise non-
isomorphic simple $FG$-modules is denoted by $\mathrm{nns}_{F}(G)$. When $FG$
is a finite dimensional $F$-algebra, $\mathrm{nns}_{F}(G)$ is a finite number
(see [1, 2]) which was calculated in the following way: Let $\xi$ be a
primitive $n$-th root of unity, where $n$ is the greatest divisor of $\exp(G)$
which is not divisible by $char(F)$. Two elements $a,b\in G$ are called
$F$-conjugate if there exist $x\in G$ and $m\in\mathbb{N}$ such that
$x^{-1}ax=b^{m}$ and the map $\xi\mapsto\xi^{m}$ is extensible to an
automorphism of the field $F(\xi)$ fixing the subfield $F$ elementwise.

Berman [2] and Witt [16] proved that $\mathrm{nns}_{F}(G)$ is equal to the
number of $F$-conjugate classes of $F$-regular elements of $G$ (see [7,
Theorem 42.8 p. 306]). If $F$ is algebraically closed, then Brauer [4, 7]
shows that $\mathrm{nns}_{F}(G)$ is equal to the number of the conjugate
classes of elements of $G$ (see [7, Theorem 40.1 p. 283]). In both cases
$\mathrm{nns}_{F}(G)$ is bounded by the number of conjugacy classes of $G$.

Let $A$ be a module over an integral domain $R$ and let $G$ be a finite group.
The set

$\mathrm{Ass}_{R}(A)=\\{P\;\mid P\quad\text{ is a prime ideal of }R\text{ such
that }\quad\mathrm{Ann}_{A}(P)\not=\langle 0\rangle\\}$

denotes the $R$-assassinator of $A$. The $R$-submodule

$A_{U}=\\{a\in A\;\mid\quad aU^{n}=\langle 0\rangle\quad\text{ for some }\quad
n\in\mathbb{N}\\}$

of $A$ is called the $U$-component of $A$. If $A=A_{U}$, then $A$ is called
the $U$-module.

For each $U$-module $A$ and each $k\in\mathbb{N}$ we define the following
$R$-submodules of $A$:

$\Omega_{[U,k]}(A)=\\{a\in A\mid aU^{k}=\langle
0\rangle\\}\qquad\text{and}\qquad
A_{U}=\cup_{s\in\mathbb{N}}\Omega_{[U,s]}(A).$ (1)

Obviously, $\Omega_{[U,1]}(A)\leq\Omega_{[U,2]}(A)\leq\cdots\leq A_{U}$.

Let $A$ be a $DG$-module such that $A$ is a $P$-module for some maximal ideal
$P$ of a Dedekind domain $D$. The module $A$ is called $(G,P)$-homogeneous if
$A$ has an ascending series of $DG$-submodules whose factors are isomorphic as
$DG$-modules. A capital of a Dedekind domain $D$ is a factor-ring $D/P$ in
which $P$ is a maximal ideal of $D$.

Let $A$ and $B$ be $R$-modules over an integral domain $R$. The notation
$A_{R}$ (or $(A)_{R}$) denotes the fact that $A$ is an $R$-module. Define the
following $R$-submodule:

$\mathrm{Tor}_{R}(A)=\\{a\in A\mid\mathrm{Ann}_{R}(a)\not=\langle
0\rangle\\}\leq A_{R}.$

We denoted the annihilator and the left annihilator of a module $A$ by
$\mathrm{Ann}_{R}(A)$ and $\mathrm{Ann}_{R}^{left}(A)$, respectively. The
$R$-submodule of $A$ generated by the elements $a_{1},\ldots,a_{n}\in A$ is
denoted by $\langle a_{1},\ldots,a_{n}\rangle_{R}$. The $R$-isomorphism
between $R$-modules $A$ and $B$ is denoted by $A\cong_{R}B$.

The module $A$ is called periodic as an $R$-module (or simply, $R$-periodic),
if $\mathrm{Tor}_{R}(A)=A$. In other words, $A$ is $R$-periodic if
$\mathrm{Ann}_{R}(a)\not=\langle 0\rangle$ for each $a\in A$. If
$\mathrm{Tor}_{R}(A)=\langle 0\rangle$, then we say that $A$ is $R$-torsion-
free. The intersection $\Phi(A)$ of all maximal $R$-submodules of $A$ is
called the Frattini submodule of $A$. Of course, if $A$ does not include
proper maximal submodules, then $\Phi(A)=A$.

In the sequel, we use the notions and results from the books [3, 7, 9].

We start our proof with the following.

######  Lemma 1.

Let $A$ be an $R$-module over a ring $R$ such that $\mathrm{sr}_{R}(A)$ is
finite. Let $B$ and $C$ be $R$-submodules of $A$ such that $B\subseteq C$. The
following conditions hold:

  * (i)

$\mathrm{sr}_{R}(B)\leq\mathrm{sr}_{R}(C)\leq\mathrm{sr}_{R}(A)$;

  * (ii)

$\mathrm{sr}_{R}(C/B)\leq\mathrm{sr}_{R}(A/B)\leq\mathrm{sr}_{R}(A)$.

These assertions are obvious.

######  Lemma 2.

Let $B$ be an $R$-submodule of an $R$-module $A$ over a Noetherian ring $R$.
If $\mathrm{sr}_{R}(B)$ and $\mathrm{sr}_{R}(A/B)$ are finite numbers, then
$\mathrm{sr}_{R}(A)\leq\mathrm{sr}_{R}(B)+\mathrm{sr}_{R}(A/B)$.

###### Proof.

Let $D$ be a f.g. $R$-submodule of $A$. Clearly, $D/(D\cap B)\cong_{R}(D+B)/B$
has a finite generating subset $\\{d_{j}+(D\cap B)\mid 1\leq j\leq m\\}$ with
$m\leq\mathrm{sr}_{R}(A/B)$. Since $D$ is a f.g. module over a Noetherian
ring, $D$ is a Noetherian $R$-module (see [13, Lemma 1.1]), so $D\cap B$ is a
f.g. $R$-submodule of $B$. Since $\mathrm{sr}_{R}(B)$ is finite, there exists
a finite set of generators $\\{b_{j}\mid 1\leq j\leq k\\}$ of $D\cap B$ with
$k\leq\mathrm{sr}_{R}(B)$. The subset $\\{d_{j}\mid 1\leq j\leq
m\\}\cup\\{b_{j}\mid 1\leq j\leq k\\}$ generates $D$ as an $R$-submodule and
$m+k\leq\mathrm{sr}_{R}(B)+\mathrm{sr}_{R}(A/B)$. ∎

######  Lemma 3.

Let $A=\oplus_{1\leq j\leq n}A_{j}$ be an $R$-module over a ring $R$ in which
each $A_{j}$ is a simple $R$-submodule. If
$\mathrm{Ann}_{R}^{left}(A_{j})\not=\mathrm{Ann}_{R}^{left}(A_{i})$ for all
$i\not=j$, then $A$ is a cyclic $R$-module.

###### Proof.

Since each $A_{j}$ is a simple $R$-submodule,
$U_{j}=\mathrm{Ann}_{R}^{left}(A_{j})$ is a maximal left ideal of $R$. From
$U_{i}\not=U_{j}$ follows that $U_{i}+U_{j}=R$ for all $i\not=j$.

Let us prove that $A=\langle a_{1}+\cdots+a_{n}\mid 0\not=a_{i}\in
A_{i}\rangle_{R}$. Indeed, $U_{1}+U_{2}=R$ shows that we can choose
$\alpha_{1}\in U_{2}$ such that $\alpha_{1}\not\in U_{1}$. Clearly
$\alpha_{1}a_{1}\not=0$, $\alpha_{1}a_{2}=0$ and

$\alpha_{1}(a_{1}+a_{2}+\cdots+a_{n})=\alpha_{1}a_{1}+\alpha_{1}a_{3}+\cdots+\alpha_{1}a_{n}.$

If $\alpha_{1}a_{3}+\cdots+\alpha_{1}a_{n}=0$, then $\alpha_{1}a_{1}\in
R(a_{1}+\cdots+a_{n})$. Since $A_{1}$ is a simple $R$-submodule, it is
generated by any non-zero element, so $A_{1}=Ra_{1}\leq
R(a_{1}+\cdots+a_{n})$.

Assume now that $\alpha_{1}a_{3}+\cdots+\alpha_{1}a_{n}\not=0$. Let $j_{1}$ be
a smallest positive integer (say, $j_{1}=3$), such that $j_{1}>2$ and
$\alpha_{1}a_{j_{1}}\not=0$. The equality $U_{1}+U_{3}=R$ shows that we can
choose $\alpha_{2}\in U_{3}$ with $\alpha_{2}\not\in U_{1}$. Hence
$\alpha_{2}\alpha_{1}a_{1}\not=0$, $\alpha_{2}\alpha_{1}a_{3}=0$ and

$\alpha_{2}(\alpha_{1}a_{1}+\alpha_{1}a_{3}+\cdots+\alpha_{1}a_{n})=\alpha_{2}\alpha_{1}a_{1}+\alpha_{2}\alpha_{1}a_{4}+\cdots+\alpha_{2}\alpha_{1}a_{n}.$

If $\alpha_{2}\alpha_{1}a_{4}+\cdots+\alpha_{2}\alpha_{1}a_{n}=0$ then, as
above, we have $A_{1}=Ra_{1}\leq R(a_{1}+\cdots+a_{n})$. Otherwise, we can
repeat finitely many times the mentioned argument above. Finally we obtain
that $A_{1}\leq R(a_{1}+\cdots+a_{n})$. Similarly, using the same argument,
$A_{j}=Ra_{j}\leq R(a_{1}+\cdots+a_{n})$ for each $1\leq j\leq n$, so
$A=A_{1}\oplus\cdots\oplus A_{n}=R(a_{1}+\cdots+a_{n})$. ∎

######  Lemma 4.

Let $A=\oplus_{1\leq j\leq n}A_{j}$ be an $R$-module over a ring $R$, in which
each $A_{j}$ is a simple $R$-submodule. If

$\mathrm{Ann}_{R}^{left}(A_{1})=\mathrm{Ann}_{R}^{left}(A_{2})=\cdots=\mathrm{Ann}_{R}^{left}(A_{n}),$

then $A_{j}\cong_{R}Ra$ for each $0\not=a\in A$ and $1\leq j\leq n$.

###### Proof.

Let $a=a_{1}+\cdots+a_{n}\in A$ where each $a_{j}\in A_{j}$ for $1\leq j\leq
n$. Without loss of generality, we can assume $a_{1}\not=0$. Put $B=Ra$ and
$D=A_{2}\oplus\cdots\oplus A_{n}$. Then $(B+D)/D$ is a non-zero submodule of
$A/D$. The isomorphism $A/D\cong_{R}A_{1}$ shows that $A/D$ is a simple $R$\-
module. It follows that either $A/D=(B+D)/D$ or $A=B+D$. Using the isomorphism
$(B+D)/D\cong_{R}B/(B\cap D)$, we obtain that $B/(B\cap D)\cong_{R}A_{1}$.
Suppose that $B\cap D\not=0$ and choose $0\not=c\in B\cap D$. Since $c\in
B=Ra$ and $c=ya$ for some $y\in R$, $c=ya_{1}+\cdots+ya_{n}$. On the other
hand, $c\in D$, so that $pr_{1}(c)=ya_{1}=0$ in which $pr_{j}(A)=A_{j}$. It
follows that $y\in\mathrm{Ann}_{R}^{left}(a_{1})$. Since $A_{1}$ is a simple
$R$-submodule and $a_{1}\not=0$,
$\mathrm{Ann}_{R}^{left}(a_{1})=\mathrm{Ann}_{R}^{left}(A_{1})$. This yields
$y\in\mathrm{Ann}_{R}^{left}(A_{j})$ for each $1\leq j\leq n$. Consequently,
$ya_{j}=0$ for all $j$, so $c=0$ is a contradiction. It proves that $B\cap
D=\langle 0\rangle$ and $B\cong_{R}A_{1}$. ∎

######  Corollary 1.

Let $A=\oplus_{\lambda\in\Lambda}A_{\lambda}$ be an $R$-module over a ring $R$
in which each $A_{\lambda}$ is a simple $R$-submodule, where $\Lambda$ is a
linearly ordered set. If
$\mathrm{Ann}_{R}^{left}(A_{\lambda})=\mathrm{Ann}_{R}^{left}(A_{\mu})$ for
all $\lambda,\mu\in\Lambda$, then $Ra\cong_{R}A_{\delta}$, for each
$0\not=a\in A$ and $\delta\in\Lambda$.

######  Corollary 2.

Let $A=\oplus_{1\leq j\leq n}A_{j}$ be an $R$-module over a ring $R$ in which
each $A_{j}$ is a simple $R$-submodule. If
$A_{1}\cong_{R}A_{2}\cong_{R}\cdots\cong_{R}A_{n}$, then
$\mathrm{sr}_{R}(A)=r\in\mathbb{N}$ if and only if $n=r$.

###### Proof.

Let $B=\langle S\rangle_{R}$ be a f.g. $R$-submodule of $A$ with a minimal
generating subset $S=\\{b_{1},\ldots,b_{k}\\}$. The $R$-submodule
$B_{1}=\langle b_{1}\rangle_{R}\cong_{R}A_{1}$ by Lemma 4. Since $B$ is a
semisimple $R$-module (see [13, Corollary 4.4]), there exists an $R$-submodule
$D_{1}$ such that $B=B_{1}\oplus D_{1}$. Clearly $b_{2}\in D_{1}$ and
$b_{2}\not\in B_{1}$, by the minimality of $S$. The $R$-submodule
$B_{2}=\langle b_{2}\rangle_{R}$ is isomorphic to $A_{1}$ by Lemma 4.
Repeating this argument finitely many times, we obtain a direct decomposition
$B=B_{1}\oplus B_{2}\oplus\cdots\oplus B_{k}$, where each
$B_{j}\cong_{R}A_{1}$.

Choosing a different minimal generating subset $\\{d_{1},\ldots,d_{m}\\}$ of
$B$ and repeating the same argument, we construct a direct decomposition
$B=D_{1}\oplus\cdots\oplus D_{m}$ in which each $D_{j}\cong_{R}A_{1}$. Clearly
$k=m$ by the Krull-Remak-Schmidt theorem (see [6, Chapter 6, Theorem 1.6]). In
particular, if $B=A$, then $n\leq r$ and the above argument shows that $n=r$.
∎

Let $A$ be an $R$-module over a ring $R$, such that $A=\oplus_{1\leq j\leq
n}A_{j}$ in which each $A_{j}$ is a simple $R$-submodule. The relation
”$A_{i}\cong_{R}A_{j}$” is an equivalence relation on the set $\\{A_{j}\mid
1\leq j\leq n\\}$ with equivalence classes $E_{1},\ldots,E_{k}$. The direct
sum of all submodules from the set $E_{j}$ is denoted by $S_{j}$ and it is
called the homogeneous component of $A$.

######  Corollary 3.

Let $A=\oplus_{1\leq j\leq n}A_{j}$ be an $R$-module over a ring $R$, in which
each $A_{j}$ is a simple $R$-submodule and let $E_{1},\ldots,E_{k}$ be the
homogeneous components of $A$. The number $r=\mathrm{sr}_{R}(A)$ is finite if
and only if the following conditions hold:

  * (i)

$\mathrm{sr}_{R}(E_{j})\leq r$ for each $j\in\\{1,\ldots,k\\}$;

  * (ii)

there exist at least one $j\in\\{1,\ldots,k\\}$ such that
$\mathrm{sr}_{R}(E_{j})=r$.

###### Proof.

Clearly $A=E_{1}\oplus\cdots\oplus E_{k}$, where each
$E_{j}=A_{j,1}\oplus\cdots\oplus A_{j,t(j)}$ has a special rank
$\mathrm{sr}_{R}(E_{j})=t(j)\leq r$ by Corollary 2. Set
$s=\max\\{t(1),\ldots,t(k)\\}$. It is easy to see that each
$B_{i}=\oplus_{1\leq m\leq k}A_{m,i}$ is cyclic by Lemma 3. It follows that
$A$ has a minimal generating subset, consisting exactly of $s$ elements.
Moreover, any $R$-submodule $D\leq A$ is also semisimple, so $D=\oplus_{1\leq
l\leq q}D_{l}$ in which $q\leq n$. Furthermore, each simple $R$-submodule
$D_{l}$ is isomorphic to a submodule from the set $\\{A_{1},\ldots,A_{n}\\}$
(see [13, Corollary 4.4]). Repeating the above arguments for $D$, we obtain
that the minimal generating subset of $D$ consists at most of $s$ elements, so
$s=r$. ∎

##  3\. Proofs

###### Proof of Theorem 1.

The fact $char(F)=0$ implies that $A$ is a semisimple $FG$-module (see [13,
Corollary 5.15]), so $A=\oplus_{1\leq j\leq n}H_{j}$ in which each $H_{j}$ is
a homogeneous component of $A$ and $H_{j}$ is a direct sum of at most $r$
simple submodules by Corollary 2. As we noted above,
$n\leq\mathrm{nns}_{F}(G)$. Obviously, the dimension of a simple
$FG$-submodule of $A$ is bounded by $|G|$, so $dim_{F}(H_{j})\leq r|G|$ and
$dim_{F}(A)\leq r\cdot|G|\cdot\mathrm{nns}_{F}(G)$. The sufficiency of the
conditions of our Theorem follows from Corollary 3. ∎

###### Proof of Theorem 2.

Put $B=A\otimes_{R}F$. Let $C=\langle b_{1},\ldots,b_{n}\rangle_{FG}$ be an
$FG$-submodule of $B$ generated by the subset
$\\{b_{1},\ldots,b_{n}\\}\subseteq A$. For each $b_{j}$ there exists a non-
zero $y_{j}\in R$ such that $a_{j}=y_{j}b_{j}\in A$. This choice ensures that
the $FG$-submodule $\langle a_{1},\ldots,a_{n}\rangle_{FG}$ of $B$ coincides
with $C$. Let $E=\langle a_{1},\ldots,a_{n}\rangle_{RG}$ be an $RG$-submodule.
Clearly, $C/E$ is an $R$-periodic $R$-module. The fact that $A$ has a finite
special rank $r$ implies that $E$ has an $RG$-generating subset
${e_{1},\ldots,e_{r}}$ in which $r\leq n$. The construction of $E$ implies
that $C=\langle e_{1},\ldots,e_{r}\rangle_{FG}$. Thus each finitely generated
$FG$-submodule of $B$ can be generated by at most $r$ elements. It follows
that the $FG$-module $B$ has a finite special rank at most $r$. Moreover,
using similar arguments and Theorem 1, we can prove that $B$ has a special
rank $r$ and $B$ has a finite dimension $d$ at most $r|G|\mathrm{nns}_{F}(G)$.
Let $\\{v_{1},\ldots,v_{d}\\}$ be a basis of B. Again, for each $v_{j}$ we can
choose a non-zero $z_{j}\in R$ such that $u_{j}=z_{j}v_{j}\in A$. Since
$\\{v_{1},\ldots,v_{d}\\}$ is a basis of $B$, $\\{u_{1},\ldots,u_{d}\\}$ is a
maximal independent subset of $A$. Consequently, the $R$-rank of $A$ is
exactly $d$. ∎

######  Lemma 5.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$ of
characteristic $0$. Let $A$ be a $DG$-module which is also a $P$-module for
some maximal ideal $P$ of $D$ with the property that $char(D/P)=0$. If $B$ is
a $DG$-submodule of $A$ such that $A=B\oplus C$ for a $D$-submodule $C$ of
$A$, then there exists a $DG$-submodule $K$ with the property that $A=B\oplus
K$.

###### Proof.

Put $D_{P}=D/P$. Consider the $D$-submodule $\Omega_{[P,1]}(A)$ (see (1)). It
is easy to see that $\mathrm{Ann}_{D}(a)=P$ for each $a\in\Omega_{[P,1]}(A)$,
so $\Omega_{[P,1]}(A)$ can be considered as a $(D/P)G$-module. In particular,
the additive group of $\Omega_{[P,1]}(A)$ is a direct sum of isomorphic copies
of the additive group $D/P$. Since $char(D/P)=0$, the additive group of $D/P$
is divisible, so the additive group of $\Omega_{[P,1]}(A)$ is divisible too.
Also the additive group of $\Omega_{[P,j+1]}(A)/\Omega_{[P,j]}(A)$ is
divisible for each $j$ by the same reason, so the additive group of
$A=\cup_{j\in\mathbb{N}}\Omega_{[P,j]}(A)$ has an ascending series whose
factors are divisible. It follows that the additive group of $A$ is divisible.
There exists a $DG$-submodule $K$ such that $nA\leq B+K$ and $n(B\cap
K)=\langle 0\rangle$ by [13, Theorem 5.9]. The fact that the additive group of
$A$ is divisible implies that $nA=A$, so $A=B+K$. The fact that $char(D/P)=0$
implies that an additive group of $A$ is torsion-free, so that $B\cap
K=\langle 0\rangle$ and $A=B\oplus K$. ∎

######  Lemma 6.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$ of
characteristic 0. Let $A$ be a $DG$-module such that $A$ is also a $P$-module
for some maximal ideal $P$ of $D$. If $char(D/P)=0$ and
$\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$, then $A$ is an Artinian $D$-module.

###### Proof.

Put $D_{P}=D/P$. The $D$-submodule $\Omega_{[P,1]}(A)$ is a $DG$-submodule of
$A$ and $\mathrm{Ann}_{D}(a)=P$ for each $a\in\Omega_{[P,1]}(A)$. Obviously,
$\Omega_{[P,1]}(A)$ is a $D_{P}[G]$-module and
$dim_{D_{P}}(\Omega_{[P,1]}(A))$ is finite by Lemma 1 and Theorem 1. Thus $A$
is Artinian as a $D$-module (see [14, Corollary 7.12]). ∎

Let $A$ be a simple $R$-module over a Dedekind domain $R$ (in particular,
principal ideal domain). Clearly, $A\cong R/P$ for some maximal ideal $P$. The
factors $R/P^{k}$ and $P/P^{k+1}$ are isomorphic as $R$-modules for each
$k\in\mathbb{N}$ (see [14, Corollary 1.28]). In particular, the $R$-module
$R/P^{k}$ is embedded in the $R$-module $R/P^{k+1}$ for each $k\in\mathbb{N}$
and $\\{R/P^{k}\mid k\in\mathbb{N}\\}$ is an injective family of $R$-modules.
Consider the $R$-module $C_{P^{\infty}}=\lim_{inj}\\{R/P^{k}\mid
k\in\mathbb{N}\\}$ which is called the Prüfer $P$-module. It is easy to see
that $C_{P^{\infty}}$ is a $P$-module,
$\Omega_{[P,k]}(C_{P^{\infty}})\cong_{R}R/P^{k}$ and

$\Omega_{[P,k+1]}(C_{P^{\infty}})/\Omega_{[P,k]}(C_{P^{\infty}})\cong_{R}(R/P^{k+1})/(P/P^{k+1})\cong_{R}R/P.$

Hence, if $C$ is a proper $R$-submodule of $C_{P^{\infty}}$ (i.e.
$C\not=C_{P^{\infty}}$), then $C=\Omega_{[P,k]}(C_{P^{\infty}})$ for some
$k\in\mathbb{N}$. Similarly, if $b\not\in\Omega_{[P,k-1]}(C_{P^{\infty}})$,
then $C=Rb$. Observe also that a Prüfer $P$-module is monolithic and its
monolith coincides with $\Omega_{[P,1]}(C_{P^{\infty}})$. Indeed, the
intersection $C\cap\Omega_{[P,1]}(C_{P^{\infty}})$ is non-zero for each
$R$-submodule $C$ of a Prüfer $P$-module. Since
$\Omega_{[P,1]}(C_{P^{\infty}})$ is a simple $R$-module
($\Omega_{[P,1]}(C_{P^{\infty}})$ is isomorphic to $R/P$),

$C\cap\Omega_{[P,1]}(C_{P^{\infty}})=\Omega_{[P,1]}(C_{P^{\infty}}).$

It follows that the $R$-monolith of a Prüfer $P$-module includes
$\Omega_{[P,1]}(C_{P^{\infty}})$ and, hence coincides with it, because
$\Omega_{[P,1]}(C_{P^{\infty}})$ is a simple $R$-submodule.

Indeed, let $C$ be a finitely generated $R$-submodule of a Prüfer $P$-module.
Then C is a proper $R$-submodule. As we have seen above, $C$ coincides with
$\Omega_{[P,n]}(C_{P^{\infty}})$ for some $n\in\mathbb{N}$. The last
$R$-submodule is cyclic, it is isomorphic to $R/P^{n}$. Hence each f.g.
$R$-submodule of a Prüfer $P$-module is cyclic. It follows that a Prüfer
$P$-module has $R$-rank 1.

######  Corollary 4.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $A$ be a $DG$-module of $\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$ such that it
is also a $P$-module for some maximal ideal $P$ of $D$. If $char(D/P)=0$, then
$A=K\oplus B$ in which $K,B$ are $DG$-submodules. Moreover, $K$ is a f.g.
$D$-submodule and $B$ is a direct sum of finitely many Prüfer $P$-submodules.

###### Proof.

The module $A$ is Artinian as a $D$-module by Lemma 6, so $A=B\oplus C$, where
$B$ is a direct sum of finitely many Prüfer $P$-submodules and $C$ is a f.g.
$D$-submodule. Clearly $B$ is a $DG$-submodule, so there exists a
$DG$-submodule $K$ with $A=B\oplus K$ by Lemma 5.

Since $K\cong_{D}C$, the module $K$ is finitely generated as a $D$-submodule.
∎

######  Lemma 7.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $B$ be a $DG$-submodule of a $DG$-module $A$ which satisfies the following
conditions:

  * (i)

$B=\oplus_{\lambda\in\Lambda}B_{\lambda}$ where $B_{\lambda}$ is a simple
$DG$-submodule for all $\lambda\in\Lambda$;

  * (ii)

$A/B$ is a simple $DG$-module and $A/B\not\cong_{DG}B_{\lambda}$ for all
$\lambda\in\Lambda$;

  * (iii)

$\mathrm{Ann}_{D}(B)=\mathrm{Ann}_{D}(A/B)=P$ is a maximal ideal of $D$;

  * (iv)

$char(D/P)=0$.

Then there exists a $DG$-submodule $K$ such that $A=B\oplus K$.

###### Proof.

Let us fix $y\in P\setminus P^{2}$. Consider a map $f:A\to A$ such that
$f(a)=ya$ for $a\in A$. The map $f$ is a $DG$-endomorphism and $yA=Im(f)\leq
B$. Indeed, $Ann_{D}(A/B)=P$ by (iii), $y\in P\setminus P^{2}\subset P$ and
$y\in Ann_{D}(A/B)$, so $yA\leq B$.

If $yA\not=\langle 0\rangle$, then $Ker(f)=B$ by condition (ii) and
$yA=Im(f)\cong_{DG}A/Ker(f)=A/B$, so that $yA$ is a simple $DG$-submodule of
$B$. Thus $yA\cong_{DG}B_{\lambda}$ for some $\lambda\in\Lambda$ (see [13,
Corollary 4.4]), which contradicts condition (ii). Consequently, $yA=\langle
0\rangle$ and $PA=\langle 0\rangle$. It follows that $A$ is a vector space
over $D/P$ and $A=B\oplus C$ for a $D$-submodule $C$ by Lemma 5. ∎

######  Corollary 5.

Let $DG$ be a group ring of a finite group $G$ over a Dedekind domain $D$. Let
$A$ be a $DG$-module such that the following conditions hold:

  * (i)

$\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$;

  * (ii)

$A$ is a $P$-module for some maximal ideal $P$ of $D$;

  * (iii)

$A$ is a f.d. $D$-module.

If $char(D/P)=0$, then $A=\oplus_{1\leq j\leq n}L_{j}$ in which each $L_{j}$
is a $(G,P)$-homogeneous $DG$-submodule of $A$.

###### Proof.

As before, the $(D/P)G$-module $\Omega_{[P,1]}(A)$ is a semisimple
$DG$-submodule by Lemma 1 and Theorem 1. Consequently,
$\Omega_{[P,1]}(A)=\oplus_{1\leq j\leq n}H_{j}$ in which each $H_{j}$ is a
homogeneous component of $\Omega_{[P,1]}(A)$ and $A=\oplus_{1\leq j\leq
n}L_{j}$ in which each $L_{j}$ is a $(G,P)$-homogeneous $DG$-submodule of $A$
such that $\Omega_{[P,1]}(L_{j})=H_{j}$. ∎

######  Corollary 6.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $A$ be a $DG$-module such that the following conditions hold:

  * (i)

$\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$;

  * (ii)

$A$ is a $P$-module for some maximal ideal $P$ of $D$;

If $char(D/P)=0$, then $A=\oplus_{1\leq j\leq n}L_{j}$ in which each $L_{j}$
is a $(G,P)$-homogeneous $DG$-submodule of $A$.

###### Proof.

Since $A$ is a $P$-module, $A=\cup_{j\in\mathbb{N}}\Omega_{[P,j]}(A)$. Using
induction and Corollary 5 we get the result. ∎

Let $A$ be a $D$-module over a Dedekind domain $D$. The intersection $\Phi(A)$
of all maximal $D$ \- submodules of $A$ is called the Frattini submodule of
$A$. Of course, if $A$ does not include proper maximal submodules, then
$\Phi(A)=A$.

If $A$ is a f.g. periodic module and $\\{a_{j}+\Phi(A)\mid 1\leq j\leq n\\}$
is a generating set for $A/\Phi(A)$, then $\\{a_{j}\mid 1\leq j\leq n\\}$ is a
generating set for $A$. Moreover, if $\\{a_{j}+\Phi(A)\mid 1\leq j\leq n\\}$
is a minimal generating set for $A/\Phi(A)$, then $\\{a_{j}\mid 1\leq j\leq
n\\}$ is a minimal generating set for $A$.

######  Lemma 8.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $A$ be a $DG$-module such that the following conditions hold:

  * (i)

$A$ is a $P$-module for some maximal ideal $P$ of $D$;

  * (ii)

$A$ is finitely generated as a $D$-module;

  * (iii)

$A$ is $(G,P)$-homogeneous;

  * (iv)

char(D/P) = 0.

The number $\mathrm{sr}_{DG}(A)$ is equal to $r\in\mathbb{N}$ if and only if
$\Omega_{[P,1]}(A)$ is a direct sum of $r$ copies of simple $DG$-submodules.

###### Proof.

If $\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$, then
$\Omega_{[P,1]}(A)=M_{1}\oplus\cdots\oplus M_{k}$, in which each $M_{j}$ is a
simple $DG$-submodule and $M_{j}\cong_{DG}M_{m}$ for all $j,m$, and $k\leq r$
by Lemma 1 and Theorem 1. Note that $\Omega_{[P,1]}(A)\cong_{DG}A/\Phi(A)$, so
$k=r$. Conversely, if $k=r$, then $\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$.∎

######  Corollary 7.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $A$ be a $DG$-module such that the following conditions hold:

  * (i)

$A$ is a $P$-module for some maximal ideal $P$ of $D$;

  * (ii)

$A$ is finitely generated as a $D$-module;

  * (iii)

$char(D/P)=0$.

Then $\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$ if and only if the following
conditions hold:

  * (a)

$A=\oplus_{1\leq j\leq n}H_{j}$ in which each $H_{j}$ is a $(G,P)$-homogeneous
$DG$-submodule of $A$;

  * (b)

$\Omega_{[P,1]}(H_{j})$ is a direct sum of at most $r$ simple $DG$-submodules
of $A$;

  * (c)

there exists $t\in\mathbb{N}$ such that $\Omega_{[P,1]}(H_{t})$ is a direct
sum of $r$ simple $DG$-submodules of $A$.

###### Proof.

For the proof it is sufficient to use the arguments from the proof of Theorem
1 and the isomorphism $\Omega_{[P,1]}(A)\cong_{DG}A/\Phi(A)$. ∎

######  Corollary 8.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $A$ be a $DG$-module such that the following conditions hold:

  * (i)

$A$ is a $P$-module for some maximal ideal $P$ of $D$;

  * (ii)

$char(D/P)=0$.

The number $\mathrm{sr}_{DG}(A)$ is finite and equals to $r\in\mathbb{N}$ if
and only if the following conditions hold:

  * (a)

$A=\oplus_{1\leq j\leq n}H_{j}$ in which each $H_{j}$ is a $(G,P)$-homogeneous
$DG$-submodule of $A$;

  * (b)

$\Omega_{[P,1]}(H_{j})$ is a direct sum of at most $r$ simple $DG$-submodules
of $A$;

  * (c)

there exists $t\in\mathbb{N}$ such that $\Omega_{[P,1]}(H_{t})$ is a direct
sum of $r$ simple $DG$-submodules of $A$.

In particular, if $A$ has finite special rank, then $A$ is Artinian as a
$D$-module.

######  Corollary 9.

Let $DG$ be the group ring of a finite group $G$ over a Dedekind domain $D$.
Let $A$ be a $DG$-module which satisfies the following conditions:

  * (i)

$A$ is a $D$-periodic module;

  * (ii)

$A$ is finitely generated as a $D$-module;

  * (iii)

$char(D/P)=0$ for all maximal ideal $P\in\mathrm{Ass}_{R}(A)$.

The number $\mathrm{sr}_{DG}(A)=r\in\mathbb{N}$ if and only if the following
conditions hold:

  * (a)

$A=\oplus_{1\leq j\leq n;P\in\pi}H_{j,P}$ in which each $H_{j,P}$ is a
$(G,P)$-homogeneous $DG$-submodule of $A$ and $P\in\pi=\mathrm{Ass}_{R}(A)$;

  * (b)

$\Omega_{[P,1]}(H_{j,P})$ is a direct sum of at most $r$ simple
$DG$-submodules of $A$;

  * (c)

there exists $t\in\mathbb{N}$ and a maximal ideal $P$ such that
$\Omega_{[P,1]}(H_{t,P})$ is a direct sum of $r$ simple $DG$-submodules of
$A$.

###### Proof.

Since $A$ is $D$-periodic, $A=\oplus_{P\in\pi}A_{P}$ in which
$\pi=\mathrm{Ass}_{R}(A)$ and $A_{P}$ is a $P$-component of $A$. Since $A$ is
a f.g. module, the set $\pi$ is finite (see [14, Theorem 7.8]). Now we can
apply Corollary 7. ∎

###### Proof of the Theorem 3.

Since $A$ is $D$-periodic, $A=\oplus_{P\in\pi}A_{P}$ where
$\pi=\mathrm{Ass}_{R}(A)$ and $A_{P}$ is a $P$-component of $A$ (see [14,
Corollary 3.8]). Now using Corollary 8 we obtain that $A_{P}$ satisfies (ii),
(iii) and (iv). Conversely, suppose that $A$ satisfies conditions (i)- (iv).
If $B$ is a f.g. $DG$-submodule of $A$, then $B$ has special rank at most $r$
by Corollary 8. Consequently, $A$ has special rank $r$ by (iv) and Corollary
2. ∎

##  4\. Funding

This research was supported by the UAEU UPAR Grant G00002160.

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