Subgroups of an abelian group, related ideals of the group ring, and quotients by those ideals

TL;DR

This paper explores the relationship between subgroups of an abelian group and certain ideals in its group ring, providing conditions for when these ideals produce specific quotient structures.

Contribution

It characterizes when ideals associated with subgroups of an abelian group generate quotients isomorphic to group rings of quotient groups, extending known injections.

Findings

Conditions for the distribution of subgroup-related ideals in the set of non-unit ideals.

Criteria for selecting elements in the group ring to produce desired quotient isomorphisms.

Abstract

Let $RG$ be the group ring of an abelian group $G$ over a commutative ring $R$ with identity. An injection $\Phi$ from the subgroups of $G$ to the non-unit ideals of $RG$ is well-known. It is defined by $\Phi(N)=I(R,N)RG$ where $I(R,N)$ is the augmentation ideal of $RN$, and each ideal $\Phi(N)$ has a property : $RG/\Phi(N)$ is $R$-algebra isomorphic to $R(G/N)$. Let $T$ be the set of non-unit ideals of $RG$. While the image of $\Phi$ is rather a small subset of $T$, we give conditions on $R$ and $G$ for the image of $\Phi$ to have some distribution in $T$. In the last section, we give criteria for choosing an element $x$ of $RG$ satisfying $RG/xRG$ is $R$-algebra isomorphic to $R(G/N)$ for a subgroup $N$ of $G$.

Equations

$$\Phi :\mathfrak{S}⟶\mathfrak{T},\Phi (N)=I(R,N)RG$$

$$\Psi :\mathfrak{T}⟶\mathfrak{S},\Psi (J)=G\cap (1+J).$$

$$\Psi \circ \Phi =\text{identity}$$

$$\Phi$$

$$\Phi (N)\in {\Psi }^{-1}(N)$$

$$\mathfrak{T}=\bigcup _{N\in \mathfrak{S}}{\Psi }^{-1}(N)(\text{disjoint union}).$$

$$RG/\Phi (N)\cong R(G/N),$$

$$G_p$$

$$\mathrm{supp}G$$

$$RG/\mathfrak{N}(RG)=RG/\Phi (N)\cong R(G/N),$$

$$RG/\Phi (G_p)\cong R(G/G_p)$$

$$\mathfrak{J}(RG)=\mathfrak{J}(R)G+\{r(g-1)|g\in G_p,r\in \mathfrak{J}(R){:}_{R}p\text{for some p∈suppG}\}RG$$

$$G_p\subseteq G\cap (1+\mathfrak{J}(RG))=\Psi (\mathfrak{J}(RG))=\Psi (\Phi (N))=N.$$

$$r(g-1)\in \mathfrak{J}(RG)=\Phi (G_p)$$

$$RG/\Phi (G_p)$$

$$0=\overline{r(g-1)}$$

$$\Phi (H)=I(H)RG=(g-1)RG.$$

$$\Phi (K)=(g-1{)}^{2}RG.$$

$$H=\Psi (\Phi (H))⫌\Psi (\Phi (K))=K.$$

$$(g-1{)}^3=g^3-3g^2+3g-1=g^3+g^2+g+1\ne 0$$

$$0⫋I⫋(g-1{)}^{2}RG=(g^2-1)RG=\Phi (H)$$

$$\Phi (1)⫋\Phi (K)⫋\Phi (H)$$

$$1⫋K⫋H$$

$$x\notin \mathrm{U}(RG)\iff x\in I(G)$$

$$RG\cong R[x]/(x^2-1)=R[x]/((x-1{)}^2),$$

$$A_x=\left(\begin{array}{ccccc}{r}_{m-1}& r_{m-2}& \cdots & r_1& r_0\\ r_{m-2}& r_{m-3}& \cdots & r_0& r_{m-1}\\ r_{m-3}& r_{m-4}& \cdots & r_{m-1}& r_{m-2}\\ ⋮& ⋮& & ⋮& ⋮\\ ⋮& ⋮& & ⋮& ⋮\\ r_2& r_1& \cdots & r_4& r_3\\ r_1& r_0& \cdots & r_3& r_2\\ r_0& r_{m-1}& \cdots & r_2& r_1\end{array}\right),$$

$$\tilde{A}_{x,n}=\begin{array}[]{cc|ccccccccccc|ccc}\cline{3-13}\cr\ldelim({9}{4pt}&&&&&&&&&&&&&0&\rdelim){9}{4pt}&\rdelim\}{4}{4pt}[$m-n$]\\ &&&&&&&&&&&&&\vdots&&\\ &&&&&&&&&&&&&0&&\\ &&&&&&\mbox{\smash{\Large$A_{x}$}}&&&&&&&1&&\\ &&&&&&&&&&&&&0&&\\ &&&&&&&&&&&&&\vdots&&\\ &&&&&&&&&&&&&0&&\\ &&&&&&&&&&&&&-1&&\\ \cline{3-13}\cr\end{array}.$$

$$\{\begin{array}{l}{r}_{m-1}{s}_0+r_{m-2}{s}_1+\cdots +r_1{s}_{m-2}+r_0{s}_{m-1}=0\\ r_{m-2}{s}_0+r_{m-3}{s}_1+\cdots +r_0{s}_{m-2}+r_{m-1}{s}_{m-1}=0\\ 1\\ r_n{s}_0+r_{n-1}{s}_1+\cdots +r_{n+2}{s}_{m-2}+r_{n+1}{s}_{m-1}=1\\ 1\\ r_1{s}_0+r_0{s}_1+\cdots +r_3{s}_{m-2}+r_2{s}_{m-1}=0\\ r_0{s}_0+r_{m-1}{s}_1+\cdots +r_2{s}_{m-2}+r_1{s}_{m-1}=-1\end{array}$$

$$\{\begin{array}{l}{r}_0+r_d+r_{2d}+\cdots +r_{(e-1)d}=0\\ r_1+r_{d+1}+r_{2d+1}+\cdots +r_{(e-1)d+1}=0\\ 1\\ r_{d-1}+r_{2d-1}+r_{3d-1}+\cdots +r_{m-1}=0\end{array}$$

$$\begin{array}[]{ccccccccccccccccccc}&\lx@intercol\overbrace{\hskip 150.00023pt}^{\mbox{$m-d+1$}}\hfil\lx@intercol&&&&&&&&&&&\\ \ldelim({14}{2pt}&-1&&&&&&1&&&&\rdelim){14}{2pt}&\ldelim({14}{2pt}&t_{0}&\rdelim){14}{2pt}&&\ldelim({14}{2pt}&r_{0}&\rdelim){14}{2pt}\\ &&-1&&&&&&1&\lx@intercol\hfil\text{\Huge 0}\hfil\lx@intercol&&&t_{1}&&&&r_{1}&\\ &&&\ddots&&&&&&\ddots&&&&\vdots&&&&\vdots&\\ &&&&-1&&&\lx@intercol\hfil\text{\Huge 0}\hfil\lx@intercol&&1&&&t_{d-1}&&&&r_{d-1}&\\ &1&&&&-1&&&&&&&&t_{d}&&=&&r_{d}&\\ &&\ddots&&&&\ddots&&&&&&&\vdots&&&&\vdots&\\ &&&\ddots&\lx@intercol\hfil\text{\Huge 0}\hfil\lx@intercol&&\ddots&&&&&&\vdots&&&&\vdots&\\ &&&&\ddots&&&&\ddots&&&&&\vdots&&&&\vdots&\\ &&\lx@intercol\hfil\text{\Huge 0}\hfil\lx@intercol&&\ddots&&&&\ddots&&&&\vdots&&&&\vdots&\\ &&&&&&1&&&&-1&&&t_{m-1}&&&&r_{m-1}&\\ &&&&&&\lx@intercol\underbrace{\hskip 100.00015pt}_{\mbox{$d+1$}}\hfil\lx@intercol&&&&&&&&\end{array}\enskip.$$

$$\left(\begin{array}[]{cccccc}\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$-I_{d}$}&\\ &&\\ \hline\cr\end{array}&&&&\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$\ I_{d}\ $}&\\ &&\\ \hline\cr\end{array}&\begin{array}[]{c}r_{0}\\ \vdots\\ r_{d-1}\end{array}\\ \begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$\ I_{d}\ $}&\\ &&\\ \hline\cr\end{array}&\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$-I_{d}$}&\\ &&\\ \hline\cr\end{array}&&\text{\Huge 0}&&\begin{array}[]{c}\vdots\\ \vdots\\ \vdots\end{array}\\ &\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$\ I_{d}\ $}&\\ &&\\ \hline\cr\end{array}&\begin{array}[]{ccc}\ddots&&\\ &\ddots&\\ &&\ddots\end{array}&&&\begin{array}[]{c}\vdots\\ \vdots\\ \vdots\end{array}\\ &&\begin{array}[]{ccc}\ddots&&\\ &\ddots&\\ &&\ddots\end{array}&\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$-I_{d}$}&\\ &&\\ \hline\cr\end{array}&&\begin{array}[]{c}\vdots\\ \vdots\\ \vdots\end{array}\\ \text{\Huge 0}&&&\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$\ I_{d}\ $}&\\ &&\\ \hline\cr\end{array}&\begin{array}[]{|ccc|}\hline\cr&&\\ &\text{\large$-I_{d}$}&\\ &&\\ \hline\cr\end{array}&\begin{array}[]{c}r_{m-d}\\ \vdots\\ r_{m-1}\end{array}\end{array}\right),$$

Full text

HIDEYASU KAWAI

General Education, Ishikawa College, National Institute of Technology,

Ishikawa 929-0392, Japan

kawai@ishikawa-nct.ac.jp

Abstract. Let $RG$ be the group ring of an abelian group $G$ over a commutative ring $R$ with identity. An injection $\Phi$ from the subgroups of $G$ to the non-unit ideals of $RG$ is well-known. It is defined by $\Phi(N)=I(R,N)RG$ where $I(R,N)$ is the augmentation ideal of $RN$, and each ideal $\Phi(N)$ has a property : $RG/\Phi(N)$ is $R$-algebra isomorphic to $R(G/N)$. Let $\mathfrak{T}$ be the set of non-unit ideals of $RG$. While the image of $\Phi$ is rather a small subset of $\mathfrak{T}$, we give conditions on $R$ and $G$ for the image of $\Phi$ to have some distribution in $\mathfrak{T}$. In the last section, we give criteria for choosing an element $x$ of $RG$ satisfying $RG/xRG$ is $R$-algebra isomorphic to $R(G/N)$ for a subgroup $N$ of $G$.

Keywords: abelian group, group ring, ideal, nilradical, Jacobson radical, residue class ring

1. Introduction

In this paper, all groups are abelian and written multiplicatively, all rings are commutative with identity. We denote the group ring of a group $G$ over a ring $R$ by $RG$, and the augmentation ideal by $I(R,G)$. Recall that $I(R,G)$ is the kernel of the $R$-algebra homomorphism from $RG$ to $R$ induced by collapsing $G$ to $1$, and that $I(R,G)$ is a free $R$-module with $\{g-1|1\neq g\in G\}$ as a basis. Let $\mathfrak{T}$ denote the set of non-unit ideals of $RG$, and let $\mathfrak{S}$ denote the set of subgroups of $G$. We have the following two maps between $\mathfrak{T}$ and $\mathfrak{S}$.

[TABLE]

and

[TABLE]

We know that

[TABLE]

(see [2, Chap.2, Corollary 2.11]). So,

[TABLE]

This correspondence between $\mathfrak{S}$ and $\mathfrak{T}$ is well-known as we can see in literature. Generally, $\Phi(\mathfrak{S})=\{\Phi(N)|N\in\mathfrak{S}\}$ is rather a small subset of $\mathfrak{T}$ since

[TABLE]

and

[TABLE]

However, each element $\Phi(N)$ of $\Phi(\mathfrak{S})$ is the kernel of the $R$-algebra homomorphism $RG\rightarrow R(G/N)$ induced by the canonical surjection $G\rightarrow G/N$ (see [2, Chap.2, Proposition 2.10]). So

[TABLE]

the quotient is a group ring over $R$ as well, that is why we take notice of this class of ideals. We give conditions on $R$ and $G$ for $\Phi(\mathfrak{S})$ to have some distribution in $\mathfrak{T}$.

In Section 2, first we discuss when the nilradical of $RG$ is contained in $\Phi(\mathfrak{S})$. A sufficient condition has been shown in [3]. It is proved to be necessary in this paper(Proposition 2.2). Next we also give a necessary and sufficient condition for $\Phi(\mathfrak{S})$ to contain the Jacobson radical (Proposition 2.3). In Section 3, we consider conditions for $\Phi(\mathfrak{S})$ to include some classes of ideals as prime ideals, principal ideals and so on. In particular, it is shown that the set of non-unit principal ideals of a group ring cannot be included by $\Phi(\mathfrak{S})$ except for an almost trivial case (Theorem 3.7). So, in Section 4, we give conditions for an element of the group ring of a cyclic group to generate an ideal contained in $\Phi(\mathfrak{S})$.

In the rest of this paper, we abbreviate $I(R,G)$ to $I(G)$ if it is not ambiguous, and denote the characteristic, the nilradical and the Jacobson radical of a ring $A$ by char $A$, $\mathfrak{N}(A)$ and $\mathfrak{J}(A)$, respectively.

2. The nilradical and the Jacobson radical

In this section, $R$ denotes a ring, and $G$ denotes a group. We set

[TABLE]

Here two conditions $\mathfrak{N}(RG)\in\Phi(\mathfrak{S})$ and $\mathfrak{J}(RG)\in\Phi(\mathfrak{S})$ are investigated.

By definition of augmentation ideals, $\Phi(1)=I(1)RG=0$. We know that $\mathfrak{N}(RG)=\Phi(1)$, namely $RG$ is reduced if and only if $R$ is reduced and $p$ is not a zero divisor of $R$ for all $p\in{\rm supp}\>G$ ([2, Chap.3, Corollary 4.3]). So we exclusively discuss the case $\mathfrak{N}(RG)=\Phi(N)$ for a subgroup $N\neq 1$, and use the following proposition repeatedly.

Proposition 2.1** ([2, Chap.3, Proposition 3.5],[1]).**

Let $R$ be a ring, and let $G$ be a group. Then $I(G)\subseteq\mathfrak{N}(RG)$ if and only if there exists a prime number $p$ such that $G$ is a $p$-group and $p\in\mathfrak{N}(R)$.

Then we have

Proposition 2.2**.**

*For group rings $RG$ the following conditions are equivalent.

$(1)$ $\mathfrak{N}(RG)=\Phi(N)$ for a subgroup $N\neq 1$.

$(2)$ $R$ is reduced and ${\rm char}\>R=p$ for some $p\in{\rm supp}\>G$.

Moreover, if these conditions are satisfied, $\mathfrak{N}(RG)=\Phi(G_{p})$.*

Proof.

(1)$\Rightarrow$(2). We note $I(N)\subseteq\mathfrak{N}(RN)$ since $I(N)\subseteq\Phi(N)=\mathfrak{N}(RG)$. By Proposition 2.1, there exists a prime number $p$ such that $N$ is a $p$-group and $p\in\mathfrak{N}(R)$. So $N\subseteq G_{p}$, accordingly $p\in{\rm supp}\>G$. Moreover, from the isomorphism

[TABLE]

we can say $R(G/N)$ is reduced and so is $R$. Taking notice of $p\in\mathfrak{N}(R)$, we have ${\rm char}\>R=p$. Now we can show $\mathfrak{N}(RG)=\Phi(G_{p})$ as follows : the inclusion $N\subseteq G_{p}$ implies $\mathfrak{N}(RG)=\Phi(N)\subseteq\Phi(G_{p})$, and $\Phi(G_{p})=I(G_{p})RG\subseteq\mathfrak{N}(RG)$ since $I(G_{p})\subseteq\mathfrak{N}(RG_{p})$ by Proposition 2.1.

(2)$\Rightarrow$(1).([3, Lemma 2.3]) Using the isomorphism

[TABLE]

with a necessary and sufficient condition for a group ring to be reduced, which is mentioned at the top of this section, we can say that $RG/\Phi(G_{p})$ is reduced. So, $\mathfrak{N}(RG)\subseteq\Phi(G_{p})$. By Proposition 2.1 it holds that $I(G_{p})\subseteq\mathfrak{N}(RG_{p})$. Thus $\Phi(G_{p})=I(G_{p})RG\subseteq\mathfrak{N}(RG)$, and $\mathfrak{N}(RG)=\Phi(G_{p})$. This completes the proof. ∎

Like the case of $\mathfrak{N}(RG)$, we know a necessary and sufficient condition for the equality $\mathfrak{J}(RG)=\Phi(1)$ (namely, $=0$) to hold (see [2, Chap.3, Corollary 4.6]). Hence it is enough to consider the case $\mathfrak{J}(RG)=\Phi(N)$ for a subgroup $N\neq 1$. In addition, we may assume $G$ is torsion since $\mathfrak{J}(RG)=\mathfrak{N}(RG)$ if $G$ is not torsion ([2, Chap.3, Corollary 4.7]). The following is crucial for the rest of this section : if $G$ is torsion, then

[TABLE]

([2, Chap.3, Theorem 4.5]). So we have

Proposition 2.3**.**

*If $G$ is torsion, the following conditions are equivalent.

$(1)$ $\mathfrak{J}(RG)=\Phi(N)$ for a subgroup $N\neq 1$.

$(2)$ $\mathfrak{J}(R)=0$ and ${\rm char}\>R=p$ for some $p\in{\rm supp}\>G$.

Moreover, if these conditions are satisfied, $\mathfrak{J}(RG)=\Phi(G_{p})$.*

Proof.

(1)$\Rightarrow$(2). Note that $RG$ is integral over $RN$. Hence for any maximal ideal $m$ of $RN$, there exists a maximal ideal $M$ of $RG$ such that $M\cap RN=m$. By definition of $\Phi$ and condition $(1)$, $I(N)\subseteq\Phi(N)\subseteq M$. Thus $I(N)\subseteq m$. Accordingly $I(N)\subseteq\mathfrak{J}(RN)$. It follows that $N$ is a $p$-group and $p\in\mathfrak{J}(R)$ for a prime $p$ by [1, Proposition 15 (i)], and that $p\in{\rm supp}\>G$. By virtue of (2.1), if $g\in G_{p}$, then $g-1\in\mathfrak{J}(RG)$. Hence $G_{p}\subseteq 1+\mathfrak{J}(RG)$, and

[TABLE]

So we have $G_{p}=N$ and $\mathfrak{J}(RG)=\Phi(G_{p})$. Now suppose that there exists a prime $q$ distinct from $p$ in ${\rm supp}\>G$. We know that $q$ cannot be contained in any maximal ideal of $R$ since $p\in\mathfrak{J}(R)$. Thus $\mathfrak{J}(R):_{R}q=\mathfrak{J}(R)$. Let $g$ be an element of $G$ whose order is $q$. Let $r\in\mathfrak{J}(R)$. Then by (2.1)

[TABLE]

and we have the following correspondence

[TABLE]

where $\overline{r(g-1)}=r(g-1)\>{\rm mod}\>\Phi(G_{p})$, $\bar{g}=gG_{p}$, and $\bar{1}=G_{p}$. Thus $r=0$. Therefore $\mathfrak{J}(R)=0$ and ${\rm char}\>R=p$. Next suppose ${\rm supp}\>G=\{p\}$. Since $G=G_{p}$, $\mathfrak{J}(RG)=\Phi(G)=I(G)$. We know that $\mathfrak{J}(R)=\mathfrak{J}(RG)\cap R$ ([2, Chap.3, Lemma 4.4]). Hence $\mathfrak{J}(R)=0$ and ${\rm char}\>R=p$.

(2)$\Rightarrow$(1). Use Lemma 2.4. ∎

Lemma 2.4**.**

Let $G$ be a group not necessarily torsion. If $\mathfrak{J}(R)=0$ and ${\rm char}\>R=p$ for some $p\in{\rm supp}\>G$, then $\mathfrak{J}(RG)=\Phi(N)$ for a subgroup $N\neq 1$.

Proof.

By [2, Chap.3, Corollary 4.7], we see $\mathfrak{J}(RG)=\mathfrak{N}(RG)$. Apply Proposition 2.2. ∎

The converse of Lemma 2.4 does not hold in general. Using a group $G$, which is not torsion, we can show a counterexample to the converse of Lemma 2.4 as follows. Let $F$ be a field of characteristic $p>0$ and let $F[x]$ be the polynomial ring in one variable over $F$. Denote by $F[x]_{(x)}$ the localization of $F[x]$ at a prime ideal $xF[x]$. We define $R=F[x]_{(x)}\times F[x]_{(x)}$ and $G=C_{\infty}\times C_{p}$, where $C_{\infty}$ and $C_{p}$ denote an infinite cyclic group and a cyclic group of order $p$ respectively. Then $\mathfrak{J}(RG)=\mathfrak{N}(RG)$ ([2, Chap.3, Corollary 4.7]), $p\in{\rm supp}\>G$, ${\rm char}\>R=p$ and $\mathfrak{N}(R)=0$. We see $\mathfrak{J}(RG)=\Phi(G_{p})$ by Proposition 2.2. However clearly $\mathfrak{J}(R)\neq 0$.

3. Prime ideals and principal ideals

Lemma 3.1**.**

Let $R$ be a ring, and let $G$ be a group. If the maximal ideals of $RG$ are images of $\Phi$, then there exists a prime number $p$ such that $R$ is a field of characteristic $p$ and $G$ is a $p$-group.

Proof.

Let $P$ be a maximal ideal of $RG$. Then there exists a subgroup $N$ of $G$ such that $P=\Phi(N)=I(N)RG$. We see $P\subseteq I(G)$ since $I(N)RG\subseteq I(G)$, and thus $I(G)$ is the unique maximal ideal of $RG$, and $R$ is a field since $RG/I(G)\cong R$. By using [2, Chap.3, Theorem 4.8], we can conclude that $G$ is a $p$-group and the characteristic of $R$ is $p$ with some prime number $p$. ∎

From Proposition 2.1, we get the following.

Lemma 3.2**.**

Let $R$ be a field of characteristic $p>0$, and let $G$ be a $p$-group. Then $I(G)$ is the unique prime ideal of $RG$.

Using the preceding lemmas, we have a corollary.

Corollary 3.3**.**

*Let $R$ be a ring, and let $G$ be a group. Then the following are equivalent.

$(1)$ The maximal ideals of $RG$ are contained in $\Phi(\mathfrak{S})$.

$(2)$ $R$ is a field of characteristic $p>0$ and $G$ is a $p$-group.

$(3)$ $I(G)$ is the unique prime ideal of $RG$.

$(4)$ The prime ideals of $RG$ are contained in $\Phi(\mathfrak{S})$.*

Lemma 3.4**.**

Let $R$ be a field of characteristic $p>2$, and let $G$ be a $p$-group with $G\neq 1$. Then there exists a non-unit principal ideal of $RG$ which is not contained in $\Phi(\mathfrak{S})$.

Proof.

Let $g$ be an element of $G$ whose order is $p$, and let $H$ be the subgroup of $G$ generated by $g$. Then by definition

[TABLE]

Here, if $(g-1)RG=(g-1)^{2}RG$, then $(g-1)RG=(g-1)^{p}RG=0$. So, we have $(g-1)RG\supsetneqq(g-1)^{2}RG$. Now assume that there exists a subgroup $K$ of $G$ such that

[TABLE]

Since $\Phi(H)\supsetneqq\Phi(K)$ and $\Psi\circ\Phi=\text{identity}$ (see 1. Introduction), we have

[TABLE]

Hence $K=1$, accordingly $\Phi(K)=0$. Thus $(g-1)^{2}=0$, whereas $g^{2}-2g+1\neq 0$ because $g^{2}$, $g$, and $1$ are different elements of $G$ by the assumption that the order of $g$ is $p>2$. This contradiction shows that $(g-1)^{2}RG\not\in\Phi(\mathfrak{S})$. ∎

Lemma 3.5**.**

Let $R$ be a field of characteristic $2$, and let $G$ be a $2$-group. Suppose $G$ has an element of order $4$. Then there exists a non-unit principal ideal of $RG$ which is not contained in $\Phi(\mathfrak{S})$.

Proof.

Let $g$ be an element of $G$, and suppose the order of $g$ is $4$. Define $I=(g-1)^{3}RG$. If $(g-1)^{2}RG=(g-1)^{3}RG$, then $(g-1)^{3}RG=(g-1)^{4}RG=(g^{4}-1)RG=0$. Thus $(g-1)^{3}=0$. However

[TABLE]

since the order of $g$ is $4$. Hence $(g-1)^{2}RG\supsetneqq(g-1)^{3}RG=I$. Moreover we see that $I\neq 0$ by (3.1). So, we have

[TABLE]

where $H$ denotes the subgroup of $G$ generated by $g^{2}$.

Suppose there exists a subgroup $K$ of $G$ such that $I=\Phi(K)$. Rewriting (3.2), we have

[TABLE]

and get

[TABLE]

by mapping through $\Psi$. This is contradictory to the fact that $H$ is generated by $g^{2}$ and the order of $g$ is $4$. Therefore $I\not\in\Phi(\mathfrak{S})$. ∎

Lemma 3.6**.**

Let $R$ be a field of characteristic $2$, and let $G$ be a $2$-group which has no element of order $4$. Suppose that the order of $G$ is greater than $2$. Then there exists a non-unit principal ideal of $RG$ which is not contained in $\Phi(\mathfrak{S})$.

Proof.

By the assumption on the order of $G$, we have two different elements $f_{1},f_{2}\in G\setminus\{1\}$ satisfying $f_{1}f_{2}\not\in\{1,f_{1},f_{2}\}$. Define an element $x$ of $RG$ by $x=1+f_{1}+f_{2}+f_{1}f_{2}$. Clearly, $xRG\neq RG$ since $x\in I(G)$. Now we assume $xRG\in\Phi(\mathfrak{S})$. Then $xRG$ has an element $1+g$ for some $g\in G\setminus\{1\}$ by [2, Chap.3, Lemma 1.3]. Hence there is an element $y$ of $RG$ such that $xy=1+g$. Note that $x=(1+f_{1})(1+f_{2})$, where $(1+f_{i})^{2}=0$ for $i=1,2$. So, we have $(1+f_{i})(1+g)=(1+f_{i})xy=0$ for $i=1,2$. Thus $1+f_{1},1+f_{2}\in{\rm ann}(1+g)$, where ${\rm ann}(1+g)$ denotes the annihilator of $(1+g)RG$. Using the fact that the order of $g$ is $2$ and [2, Chap.2, Proposition 2.18 (iv)], we get ${\rm ann}(1+g)=(1+g)RG$. Therefore $x=(1+f_{1})(1+f_{2})\in(1+g)^{2}RG=0$, whereas $x\neq 0$. By this contradiction, we conclude $xRG\not\in\Phi(\mathfrak{S})$. ∎

As a result, we can show

Theorem 3.7**.**

*Let $R$ be a ring, and let $G$ be a group with $G\neq 1$. Then the following are equivalent.

$(1)$ $\Phi(\mathfrak{S})=\mathfrak{T}$.

$(2)$ Every non-unit principal ideal of $RG$ is contained in $\Phi(\mathfrak{S})$.

$(3)$ $R$ is a field of characteristic $2$ and $G$ is a group of order $2$.*

Proof.

(2)$\Rightarrow$(3). We know $\Phi(N)\subseteq I(G)$ for any subgroup $N$ of $G$ by definition. So it follows from (2) that $J\subseteq I(G)$ for every non-unit principal ideal $J$ of $RG$. Hence if $x$ is an element of $RG$ then we have

[TABLE]

where ${\rm U}(RG)$ denotes the unit group of $RG$. Thus $I(G)$ is the unique maximal ideal of $RG$. By Corollary 3.3, we see that $R$ is a field of characteristic $p>0$ and $G$ is a $p$-group. By Lemma 3.4, we get $p=2$. By Lemma 3.5, we can say $G$ has no element of order 4. By Lemma 3.6, we conclude that the order of $G$ is 2.

(3)$\Rightarrow$(1). We use the fact that $RG$ is isomorphic to a quotient ring of the polynomial ring $R[x]$ in an indeterminate $x$ over $R$. So, using the isomorphism

[TABLE]

we see $\mathfrak{T}=\{0,I(G)\}$. This completes the proof. ∎

4. Conditions for an element of the group ring of a cyclic group to generate an ideal contained

in $\Phi(\mathfrak{S})$

As shown in Theorem 3.7, there exist elements in a group ring each element of which generates an ideal not contained in $\Phi(\mathfrak{S})$ except for a particular case. So, we give conditions for an element of the group ring of a cyclic group to generate an ideal contained in $\Phi(\mathfrak{S})$.

Lemma 4.1**.**

*Let $R$ be a field, and let $G$ be a finite cyclic group of order $m$ with a generator $g$. Suppose $x$ is an element of $RG$, and write $x=\sum_{i=0}^{m-1}r_{i}g^{i}\enskip(r_{i}\in R)$. Let $n$ be a positive integer with $n<m$. Then, $g^{n}-1\in xRG$ if and only if ${\rm rank}(A_{x})={\rm rank}(\tilde{A}_{x,n})$, where rank( ) denotes the rank of a matrix, and matrices $A_{x}$, $\tilde{A}_{x,n}$ are defined as follows:

[TABLE]

[TABLE]

Proof.

Suppose that $g^{n}-1=xy$ for an element $y$ of $RG$, and $y=\sum_{j=0}^{m-1}s_{j}g^{j}\enskip(s_{j}\in R)$. By calculating coefficients of $g^{k}$ ( $k=0,1,\ldots,m-1$ ) in $xy$, we have

[TABLE]

So, $g^{n}-1\in xRG$ if and only if the system (4.1) of linear equations for variables $s_{0},s_{1},\ldots,s_{m-1}$ has a solution. It is a well-known fact in linear algebra that the latter condition is equivalent to the condition ${\rm rank}(A_{x})={\rm rank}(\tilde{A}_{x,n})$. ∎

We denote the greatest common divisor of positive integers $m$ and $n$ by $(m,n)$. And we denote the subgroup generated by an element $s$ of a group by $\langle s\rangle$.

The following is easily shown.

Lemma 4.2**.**

Let $G$ be a finite cyclic group of order $m$, and let $g$ be a generator of $G$. Supposing $n$ is a positive integer with $n<m$ and $(m,n)=d$, then $\langle g^{n}\rangle=\langle g^{d}\rangle$.

So,we can show

Lemma 4.3**.**

Let $R$ be a field, and let $G$ be a finite cyclic group of order $m$ with a generator $g$. Suppose $x$ is an element of $RG$, and write $x=\sum_{i=0}^{m-1}r_{i}g^{i}\enskip(r_{i}\in R)$. Let $n$ be a positive integer with $n<m$ and $(m,n)=d$. Then, $x\in(g^{n}-1)RG$ if and only if the following holds:

[TABLE]

where $e=m/d$.

Proof.

We know that $I(R,\langle g^{n}\rangle)RG=(g^{n}-1)RG$ and $I(R,\langle g^{d}\rangle)RG=(g^{d}-1)RG$ by [2, Chap.3, Lemma 1.3]. Hence $(g^{n}-1)RG=(g^{d}-1)RG$ by Lemma 4.2, therefore it suffices to show that $x\in(g^{d}-1)RG$ if and only if the condition (4.2) holds. It is clear that we have $x=(g^{d}-1)z$ with $z=\sum_{j=0}^{m-1}t_{j}g^{j}\enskip(t_{j}\in R)$ if and only if the following system of linear equations has a solution for $t_{0},t_{1},\ldots,t_{m-1}$ :

[TABLE]

The corresponding matrix of coefficients can also be expressed as follows :

[TABLE]

where $I_{d}$ denotes the identity matrix of size $d$. Using elementary row operations, we can transform this matrix into the following

[TABLE]

Hence we see that the system (4.3) has a solution if and only if the condition (4.2) holds. This completes the proof. ∎

In the following statements, for a field $R$, ${\rm dim}_{R}(\quad)$ denotes the dimension of an $R$-vector space.

Lemma 4.4**.**

Let $R$ be a field, $G$ a finite cyclic group, and $x$ a nonzero element of $RG$. Then ${\rm dim}_{R}\thinspace xRG={\rm rank}(A_{x})$, where $A_{x}$ is a matrix defined in Lemma 4.1.

Proof.

Suppose $G=\langle g\rangle$ with $|G|=m$, and write $x=\sum_{i=0}^{m-1}r_{i}g^{i}\enskip(r_{i}\in R)$. We see that $xRG$ is generated over $R$ by $\{g^{j}x\}_{0\leq j\leq m-1}$. Choosing $G$ as a basis of $RG$, we have a canonical isomorphism of $R$-vector spaces $RG\cong R^{m}$. This isomorphism maps the elements of $\{g^{j}x\}_{0\leq j\leq m-1}$ as follows.

[TABLE]

Thus we get ${\rm dim}_{R}\thinspace xRG={\rm rank}(A_{x})$ as desired. ∎

Proposition 4.5**.**

*Let $R$ be a field, and let $G$ be a finite cyclic group of order $m$ with a generator $g$. Suppose $x$ is a nonzero element of $RG$, and set $d=m-{\rm rank}(A_{x})$, where $A_{x}$ is a matrix defined in Lemma 4.1. Then the following are equivalent.

$(1)$ $xRG=\Phi(N)$ for a subgroup $N\neq 1$.

$(2)$ $d$ is a divisor of $m$ for which (4.2) of Lemma 4.3 holds and ${\rm rank}(A_{x})={\rm rank}(\tilde{A}_{x,d})$, where $\tilde{A}_{x,d}$ is a matrix defined in Lemma 4.1.

Moreover, if these are satisfied, $xRG=\Phi(\langle g^{d}\rangle)$, $RG/xRG\cong R(G/\langle g^{d}\rangle)$.*

Proof.

(1)$\Rightarrow$(2). Assume $N=\langle g^{n}\rangle$ with $0<n<m$. Let $(m,n)=d^{\prime}$, then $\langle g^{n}\rangle=\langle g^{d^{\prime}}\rangle$ by Lemma 4.2. Since $\Phi(\langle g^{d^{\prime}}\rangle)=I(R,\langle g^{d^{\prime}}\rangle)RG=(g^{d^{\prime}}-1)RG$ by [2, Chap.3, Lemma 1.3], we have $xRG=\Phi(\langle g^{d^{\prime}}\rangle)=(g^{d^{\prime}}-1)RG$. Hence $RG/xRG=RG/\Phi(\langle g^{d^{\prime}}\rangle)\cong R(G/\langle g^{d^{\prime}}\rangle)$ (see Section 1). Therefore, ${\rm dim}_{R}\thinspace RG/xRG=|G/\langle g^{d^{\prime}}\rangle|=d^{\prime}$. On the other hand, ${\rm dim}_{R}\thinspace RG/xRG=m-{\rm dim}_{R}\thinspace xRG$. Thus $d^{\prime}=m-{\rm rank}(A_{x})$ by Lemma 4.4 and, consequently, $d^{\prime}=d$. Since $xRG=(g^{d}-1)RG$, from Lemma 4.1 and Lemma 4.3, (4.2) of Lemma 4.3 holds and ${\rm rank}(A_{x})={\rm rank}(\tilde{A}_{x,d})$ for $d$.

(2)$\Rightarrow$(1). From Lemma 4.1 and Lemma 4.3, we have $xRG=(g^{d}-1)RG$. Thus $xRG=I(R,\langle g^{d}\rangle)RG=\Phi(\langle g^{d}\rangle)$ with $\langle g^{d}\rangle\neq 1$. ∎

Example 4.6**.**

Let $\mathbb{F}_{5}=\mathbb{Z}/5\mathbb{Z}$, the prime field of order 5, and let $\alpha=2{\rm mod}5\in\mathbb{F}_{5}$. We denote by $C_{12}$ a cyclic group of order 12 with a generator $g$. Let $x$ be as follows :

[TABLE]

Then, ${\rm rank}(A_{x})=8$, $d=|C_{12}|-{\rm rank}(A_{x})=4$. Denoting $x=\sum_{i=0}^{11}r_{i}g^{i}\enskip(r_{i}\in\mathbb{F}_{5})$, we have

[TABLE]

which shows (4.2) of Lemma 4.3 holds. Since ${\rm rank}(\tilde{A}_{x,4})=8$, we get $x\thinspace\mathbb{F}_{5}C_{12}=\Phi(\langle g^{4}\rangle)$ and $\mathbb{F}_{5}C_{12}/x\thinspace\mathbb{F}_{5}C_{12}\cong\mathbb{F}_{5}(C_{12}/\langle g^{4}\rangle)$ as $\mathbb{F}_{5}$-algebras.

For infinite cyclic groups, we have

Proposition 4.7**.**

*Let $R$ be an integral domain, and let $G$ be an infinite cyclic group. Suppose $x$ is a nonzero element of $RG$. Then the following are equivalent.

$(1)$ $xRG=\Phi(N)$ for a subgroup $N\neq 1$.

$(2)$ $x=ug_{1}-ug_{2}$ for a unit $u$ of $R$ and two elements $g_{1}$, $g_{2}$ of $G$.

Moreover, if these are satisfied, $xRG=\Phi(\langle h\rangle)$, $RG/xRG\cong R(G/\langle h\rangle)$, where $h=g_{1}g_{2}^{-1}$.*

Proof.

Since $G$ is a cyclic group, we see that (1) holds if and only if there exists an element $h$ of $G$ satisfying $(h-1)RG=xRG$. So, we replace (1) with the condition $\rm{(1)}^{\prime}$: $xRG=(h-1)RG$ for an element $h$ of $G$.

\rm{(1)}^{\prime}$$\Rightarrow(2). We have $h-1=xy$ for an element $y$ of $RG$, and $x=(h-1)z$ for an element $z$ of $RG$. Hence $h=xy+1=(h-1)zy+1=hyz-yz+1$. Thus $h-hyz=1-yz$, that is, $h(1-yz)=1-yz$. For each element $w$ of $RG$, define ${\rm Supp}(w)$ as follows: if $w=\sum_{g\in G}r_{g}g(r_{g}\in R)$ then ${\rm Supp}(w)=\{~{}g\in G~{}|~{}r_{g}\neq 0~{}\}$. Using this notation, we can say that if $1-yz\neq 0$ then ${\rm Supp}(h(1-yz))\neq{\rm Supp}(1-yz)$ since $G$ is an infinite cyclic group. Thus $1-yz=0$, so $y$ is a unit of $RG$. Because $R$ is an integral domain and $G$ is torsion-free, $y$ is a trivial unit, that is, $y=rt$ for a unit $r$ of $R$ and an element $t$ of $G$ by [2, Chap.2, Proposition 2.23]. Recalling $xy=h-1$, we get $x=y^{-1}h-y^{-1}=r^{-1}(t^{-1}h)-r^{-1}t^{-1}$. If we set $u=r^{-1}$, $g_{1}=t^{-1}h$, and $g_{2}=t^{-1}$, then we have $x=ug_{1}-ug_{2}$ for a unit $u$ of $R$ and $g_{1},g_{2}\in G$.

(2)\Rightarrow$$\rm{(1)}^{\prime}. Setting $h=g_{1}g_{2}^{-1}$, we get $x=uhg_{2}-ug_{2}=(h-1)ug_{2}$. ∎