# Exploring the Unfairness of DP-SGD Across Settings

Frederik Noe, Rasmus Herskind and Anders Søgaard
University of Copenhagen

###### Abstract

End users and regulators require private and fair artificial intelligence
models, but previous work suggests these objectives may be at odds. We use the
CivilComments to evaluate the impact of applying the de facto standard
approach to privacy, DP-SGD, across several fairness metrics. We evaluate
three implementations of DP-SGD: for dimensionality reduction (PCA), linear
classification (logistic regression), and robust deep learning (Group-DRO). We
establish a negative, logarithmic correlation between privacy and fairness in
the case of linear classification and robust deep learning. DP-SGD had no
significant impact on fairness for PCA, but upon inspection, also did not seem
to lead to private representations.

## Introduction

The vast majority of artificial intelligence research projects focus on
learning accurate models, but in practice, real-life application of machine
learning models require balancing performance with the need to prevent
discrimination against protected demographic subgroups and satisfying privacy
principles. Both needs are prerequisites for employing machine learning models
at scale in many domains, and the lack of fairness and privacy in many widely
used models threatens the public perception of artificial intelligence.
Unfortunately, the needs are seemingly at odds (Ekstrand, Joshaghani, and
Mehrpouyan 2018; Cummings et al. 2019; Bagdasaryan, Poursaeed, and Shmatikov
2019; Farrand et al. 2020; Chang and Shokri 2021; Agarwal 2021). Common
fairness and privacy objectives are in conflict, because privacy-preserving
algorithms often disproportionally affect members of minority classes (Farrand
et al. 2020).111Note this is a different trade-off than the fairness-privacy
trade-off which results from the need for collecting sensitive data to learn
fair models (Veale and Binns 2017). The exact nature of this trade-off has not
been analyzed, however, and perhaps more surprisingly, there has been no
evaluations of whether learning algorithms that were developed with group
fairness in mind, such as Sagawa et al. (2020), are less sensitive to
optimization with DP-SGD.

Figure 1: A few datapoints from CivilComments including demographic identities
(Koh et al. 2021)

### Contributions

We evaluate the impact on fairness (Unequal Risk) of using DP-SGD
optimization, across three settings: PCA, logistic regression and Group-DRO
(Sagawa et al. 2020). For the baseline condition, we confirm past results and
establish a negative correlation between fairness and privacy. We further show
that this negative correlation is logarithmic. Moreover, we show that DP-SGD
has little impact on fairness in the PCA setting. This, however, seems to
because differentially private dimensionality reduction is insufficient to
guarantee the privacy of downstream classifiers.

## Fairness and Privacy

We represent each data point as
$z=(x,g,y)\in\mathcal{X}\times\mathcal{G}\times\mathcal{Y}$, with
$g\in\mathcal{G}$ encoding its protected attributes (e.g., race, gender). Let
$\mathcal{D}^{g}_{y}$ denote the distribution of data with protected attribute
$g$ and label $y$. Several definitions of fairness exist (Williamson and Menon
2019), but we use (generalized) approximately constant conditional (equal)
risk (Donini et al. 2018):

######  Definition 0.1 ($\delta$-Unequal Risk).

Let $\ell^{g_{i}}(\theta)=\mathbb{E}[\ell(\theta(x),y)|g=g_{i}]$ be the risk
of the samples in the group defined by $g_{i}$, and $\delta\in[0,1]$. We say
that a model $\theta$ exhibits $\delta$-Unequal Risk if for any two values of
$g$, say $g_{i}$ and $g_{j}$,
$|\ell^{g_{i}}(\theta)-\ell^{g_{j}}(\theta)|<\delta$.

Note that if $\ell$ coincides with the performance metric of a task, and
$\delta=0$, this is identical to performance or classification parity (Yuan et
al. 2021).222Such a notion of fairness can be derived from John Rawls’ theory
on distributive justice and stability, treating model performance as a
resource to be allocated. Rawls’ difference principle, maximizing the welfare
of the worst-off group, is argued to lead to stability and mobility in society
at large (Rawls 1971). Performance or classification parity has, however, been
argued to suffer from statistical limitations in (Corbett-Davies and Goel
2018), which remind us that when risk distributions differ, standard error
metrics are poor proxies of individual equity. This is known as the problem of
infra-marginality. Note, however, that this argument does not apply to binary
classification problems. $\delta$ directly measures what is sometimes called
Rawlsian min-max fairness (Bertsimas, Farias, and Trichakis 2011). Recall the
standard definition of $(\varepsilon,\phi)$-privacy333We talk about
$(\varepsilon,\phi)$-privacy rather than $(\varepsilon,\delta)$-privacy, to
avoid confusion with $\delta$-fairness. is as follows:

######  Definition 0.2 ($(\varepsilon,\phi)$-privacy).

$\theta$ is $(\varepsilon,\phi)$-private iff
$\mbox{Pr}[\theta(\mathcal{X})]\leq\exp(\varepsilon)\times\mbox{Pr}[\theta(\mathcal{X}^{\prime})]+\phi$
for any two distributions, $\mathcal{X}$ and $\mathcal{X^{\prime}}$, different
at most in one row.

Differential privacy ensures that an algorithm will generate similar outputs
for similar training data sets. Note the multiplicative bound
$\exp(\varepsilon)$ and the additive bound $\phi$ serve different roles: The
$\phi$ term represents the possibility that a few data points are not governed
by the multiplicative bound, which controls the level of privacy (rather than
its scope). Note that it also follows directly that if $\varepsilon=0$ and
$\phi=0$, absolute privacy is required, leading $\theta$ to be independent of
the data. In our experiments, we assume $\phi=0$ and look at the impact of DP-
SGD training across different values of $\epsilon$.

Agarwal (2021) shows that a $(\varepsilon,0)$-private and fully fair model –
using equalized odds as the definition of fairness – will be unable to learn
anything. To see this, remember that a fully private model is independent of
the data and unable to learn from correlations between input and output. If
$\theta$ is, in addition, required to be fair, it is thereby required to be
fair for all distributions, which prevents $\theta$ from encoding any prior
beliefs about the output distribution. Equal odds is often less interesting
for NLP applications than equal risk, but note this finding generalizes
straight-forwardly to equalized risk, and to approximate fairness (since even
for finite distributions, we can define a $\delta>0$, such that preserving
absolute privacy leads to a constant $\theta$).

######  Theorem 1.

For sufficiently small values of $\varepsilon$, a fully
$(\varepsilon,0)$-private model $\theta$ that is also $\delta$-fair, will have
trivial performance.

###### Proof.

This follows directly from the above. ∎

We focus on $\delta$-Unequal Risk below, but for completeness include
alternative definitions. While $\delta$-Unequal Risk only considers the
difference between extremes, $\delta$-Variance considers the overall F1-score
variance across the groups, with $\delta=\sigma^{2}$ defined as the mean of
the squared distances of $\ell^{g_{i}}$ from their mean $\bar{\ell^{g_{i}}}$,
i.e.., $\frac{\sum_{i=1}^{n}(\ell^{g_{i}}-\bar{\ell^{g_{i}}})^{2}}{n}$. Often
Unequal Risk is contrasted with Equal Odds, i.e., the idea that fairness
requires similar positive rates across groups. Zafar et al. (2017) introduce
the so-called p%-rule, a variant of Equal Odds, which we will also adopt
below. In the binary case, we say a model $\theta$ exhibits $p$-Equal Odds if

$p=\min_{i,j}\min(\frac{\mathbb{E}[\theta(\mathcal{X}^{g_{i}})]}{\mathbb{E}[\theta(\mathcal{X}^{g_{j}})]},\frac{\mathbb{E}[\theta(\mathcal{X}^{g_{j}})]}{\mathbb{E}[\theta(\mathcal{X}^{g_{i}})]})$

Note that if the p%-rule returns 1.0, $\theta$ exhibits 1.0-Equal Odds, which
amounts to identical positive rates across groups. It is easy to show that
$p\in[0,1]$, with higher values suggesting fairness under Equal Odds. We
modify the p%-rule, which serves as a compromise between Unequal Risk and
Equal Odds.

######  Definition 0.3 (Modified p%-rule).

We say a model $\theta$ exhibits $p$-Equally Correct Priors if

$p=\min_{i,j}\min(\frac{\frac{\mathbb{E}[\theta(\mathcal{X}^{g_{i}})]}{\mathbb{E}[\mathcal{Y}^{g_{i}}]}}{\frac{\mathbb{E}[\theta(\mathcal{X}^{g_{j}})]}{\mathbb{E}[\mathcal{Y}^{g_{j}}]}},\frac{\frac{\mathbb{E}[\theta(\mathcal{X}^{g_{j}})]}{\mathbb{E}[\mathcal{Y}^{g_{j}}]}}{\frac{\mathbb{E}[\theta(\mathcal{X}^{g_{i}})]}{\mathbb{E}[\mathcal{Y}^{g_{i}}]}})$

This definition says the divergence from the positive class prior should be no
different across any two groups. Note how this, unlike $\delta$-Unequal Risk,
says nothing about the actual risk of misclassification. It is merely a
weighted version of $p$-Equal Odds, which takes the true priors into account.
We report results for $\delta$-Variance and the p%-rule in the Appendix, and
focus on $\delta$-Unequal Risk and the Modified p%-rule here.

Figure 2: Baseline performance (F1) across groups

## Experiments

Our experiments evaluate the impact of differential privacy (Dwork et al.
2006) on the fairness of different machine learning architectures.
Differentially private stochastic gradient descent (DP-SGD) is the de facto
baseline approach to learning private models, and we limit ourselves to
variants of DP-SGD. Vanilla DP-SGD limits the influence of training samples by
(i) clipping the per-batch gradient where its norm exceeds a pre-determined
clipping bound $C$, and by (ii) adding Gaussian noise $\mathcal{N}$
characterized by a noise scale $\sigma$ to the aggregated per-sample
gradients. We control this influence with a privacy budget $\varepsilon$,
where lower values for $\varepsilon$ indicates a more strict level of privacy.
DP-SGD has remained popular, among other things because it generalizes to
iterative training procedures (McMahan et al. 2018), and supports tighter
bounds using the Rényi method (Mironov 2017a). We evaluate the impact of
differentially private training on fairness across dimensionality reduction,
linear classification, and deep learning.

### Bertweet and DistilBert

Transformer architectures are often trained with masked language modeling
objectives to learn representations that are useful for downstream NLP tasks
(Devlin et al. 2019a; Liu et al. 2019). We use two pretrained, Transformer-
based language models: (a) Bertweet (Nguyen, Vu, and Nguyen 2020) is trained
on English tweets with the same hyper-parameters as the original BERT-base
model (Devlin et al. 2019b). DistilBert (Sanh et al. 2020) is induced by
distillation of BERT (Devlin et al. 2019a) to obtain a more efficient and
smaller version of it.

### Principal Component Analysis

Principal components are linear combinations or mixtures of the initial
variables and internally uncorrelated to minimize the information loss. We
rely on the IBM Differential Privacy Library
(Diffprivlib)444https://github.com/IBM/differential-privacy-library (Holohan
et al. 2019) for DP-PCA, which relies on Imtiaz and Sarwate (2016), who add
Wishart noise to this reconstruction to obtain $\varepsilon$-privacy. Their
algorithm is shown to approximate the underlying data well, especially for
$\varepsilon\geq 1$. Alternatives are discussed in shang21. We apply DP-PCA to
Bertweet representations of CivilComments and pass these to a three-layer
feed-forward network. We varied $\varepsilon$-values in [0.1,10] with a step-
size of 0.5, and performed grid search for the hyper-parameters of the feed-
forward network. Its dimensionality was given (768, 64, and 32 in the
outermost layer), but all other parameters were optimized for validation
performance: We trained for 2979 epochs, with learning rate $1e^{-06}$, MSE
loss, AdamW optimization, and a batch size of 32.

### Logistic Regression

We also evaluate the impact of DP-SGD on linear classification. In logistic
regression, the weighted sum of model parameters and input is transformed to
probabilities by the logistic function, with parameters fit to minimize
logistic loss with DP-SGD. We train 400 models with
$\varepsilon\in\\{0.1,0.2,0.3\textperiodcentered\textperiodcentered\textperiodcentered,39.9,40\\}$.
Again, we rely on Bertweet and Diffprivlib.

### Group Distributionally Robust Optimization

We also consider a deep architecture optimized for group fairness, namely
GroupDRO (Sagawa et al. 2020), which relies on distributionally robust
optimization (Hashimoto et al. 2018) to minimize the worst-case loss over the
groups in the training data. We use DP-SGD in
Opacus555https://github.com/pytorch/opacus (AI 2020) to implement
differentially private GroupDRO. DP-SGD was generalized to deep learning in
Abadi et al. (2016); Opacus uses Renyi Differential Privacy (RDP) (Mironov
2017b).

### Data

We use Civil Comments (Koh et al. 2021), a toxicity detection dataset in which
comments are annotated for toxicity and the demographics of the target of
toxicity (see Fig. 1). The task is a binary classification task. There are
eight target groups:666We cite the names used in the publicly available
dataset. It is unclear to what extent the names impacted annotations. Some
seem to represent communities that can in part be referred to by other, more
general or more specific, terms, e.g., LGBT, LGBTQ+, the Rainbow Community. By
using this data, we do not mean to imply that these terms are useful for
referring to groups of citizens. We use CivilComments simply because it is the
most widely distributed NLP dataset for fairness studies. LGBTQ, male, female,
Christian, Muslim, other religions, Black, and White. For simplicity, we only
use data from these groups: Prior work has shown that toxicity classifiers
pick up on biases in the training data and spuriously associate toxicity with
the mention of certain demographics (Borkan et al. 2019). Thus, for
classifying the toxicity of online comments, the setup shown in Figure 1
ensures that we are able to keep track of fairness by analysing scores across
different groups.

Figure 3: Top (Linear Classification): Logarithmic correlations between
$\varepsilon$ ($x$) and accuracy ($y$) (first row), $\varepsilon$ and Unequal
Risk (second, left), and $\varepsilon$ and the modified p%-rule (second,
right). Bottom (Deep GroupDRO): Logarithmic correlation between $\varepsilon$
($x$) and accuracy ($y$) (third), $\varepsilon$ and Unequal Risk (fourth,
left), and the modified p%-rule (fourth, right).

## Results

The baseline F1 scores across the eight groups are plotted in Figure 2. Scores
significantly across the eight groups, with lowest toxicity detection scores
for minority religious groups and Christians, and highest toxicity detection
scores for blacks and whites, reportedly being the prototypical targets for
American and European racist slur (Croom 2015). Generally, PCA reduces
performance slightly, while our GroupDRO architecture performs slightly better
than our baseline with fixed Bertweet embeddings. We are mainly interested in
how overall performance and fairness is affected by DP-SGD.

Figure 3 (second row) shows the correlation between the degree of approximate
privacy and Unequal Risk for linear classification: The more private a model,
the less fair. This correlation contrasts with the positive correlation
between $\varepsilon$ and accuracy (first): The more private a model, the
worse it performs. Note that both of these correlations are significant
($p<0.001$) and logarithmic and have similar coefficients.

GroupDRO with DistilBert exhibits similar correlations; see Figure 3. While
odds become logarithmically more equal with higher $\varepsilon$ values for
linear models, odds exhibit high variance for GroupDRO and are not
significantly correlated with $\varepsilon$. The fact that GroupDRO exhibit a
similar, negative, logarithmic correlation between $\epsilon$ and Unequal
Risk, is particularly interesting, since GroupDRO assumes access to group
information during training and uses this information to learn a model with
Equal Risk. Under sufficient privacy guarantees, the approach is ineffective,
however.

In contrast, with PCA, we see a weak, linear correlation between $\varepsilon$
and performance: The less privacy, the better performance, again, but
differences are minimal (absolute drops in accuracy within a 0.5 margin). We
also see a weak, negative, linear correlation between $\varepsilon$ and
Unequal Risk: The less privacy, the more fairness, but this correlation is not
significant. At first sight, it may seem promising that differentially private
dimensionality reduction does not lead to Unequal Risk. The same pattern is
seen across all the fairness measures. However, this is unfortunately not a
case of getting the best of both worlds: Our feed-forward network seems to be
able to recover private information from our dimensionality-reduced
representations. To see why this may happen, consider that the algorithm
presented in Imtiaz and Sarwate (2016) is only approximate and not providing
strict guarantees. It adds stochastic noise to each dimension independently,
and only provides expected guarantees. The more dimensions that carry signal,
the higher the chance that private information is recoverable. The sensitivity
of differentially private dimensionality reduction has been observed before
(Aggarwal 2005). We hypothesize that our feed-forward network in a way similar
to spectral filtering (Kargupta et al. 2003) can filter off the random noise
by implicitly analyzing eigenstates.

## Discussion and Conclusion

The trade-off between fairness and privacy is of great concern to the machine
learning community, but previous work has not suggested promising research
directions for how to best solve this dillemma. Our results go beyond previous
work on the trade-offs between fairness and privacy (Bagdasaryan, Poursaeed,
and Shmatikov 2019; Agarwal 2021) in evaluating this trade-off across multiple
machine learning settings and in establishing logarithmic correlations between
fairness and privacy across two different settings: linear classification and
robust deep learning. Finally, we also provide an example of a successful
attack against differentially private dimensionality reduction by training a
three-layer feed-forward network on the learned representations of DP-PCA.

Uniyal et al. (2021) presented a proposal, which exhibited a better trade-off,
empirically. Liu et al. (2021) presented an approach to fairness and privacy
that relies on a self-adaptive mechanism and dynamically adjusts instance
influence in each class depending on the theoretical bias-variance
bounds.777The paper was rejected, though, because it did not prove reasonable
privacy guarantees, and because it relied on non-standard definitions of
fairness. See https://openreview.net/forum?id=IqVB8e0DlUd Fioretto, Tran, and
Hentenryck (2021) proposes various solutions, including output perturbations,
linearizations, learning piece-wise linear proxy-functions, or fairness
payment. These methods have several drawbacks, and most are not generally
applicable. Piece-wise linear proxy-functions and GroupDRO assume that
demographics are available during training, which limits their application
severely. In summary, finding robustly applicable approaches to simultaneously
optimizing for fairness and privacy remains an open problem. Our contributions
are (a) showing the logarithmic nature of this correlation, suggesting the
usefulness of maximizing the second order gradient of this trade-off; (b)
showing this holds even for group-robust deep learning; and (c) showing, by
example, that seemingly good results for both fairness and privacy may be the
result of implicit privacy attacks, when deep classifiers exploit privacy
leaks.

Based on a sequence of experiments with toxicity detection, relying on large-
scale pretrained language models, this paper established the logarithmic
nature of the relationship between fairness and privacy, in the form of DP-
SGD, and showed that this logarithmic nature generalizes to models trained to
optimize for fairness (with GroupDRO). Moreover, we saw that when applied to
dimensionality reduction, DP-SGD did not seem to violate fairness. This
anomaly was explained by leakage of private information, however. We discussed
possible research directions for jointly learning private and fair machine
learning models, a topic we believe to be of great importance to the future of
artificial intelligence.

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# On Two Classes of Modules Related to CS Trivial Extensions

Farid Kourki
Centre Régional des Métiers de l’Education
et de la Formation (CRMEF)-Tanger,
Annexe de Larache, B.P. 4063, Larache, Morocco
<EMAIL_ADDRESS>
Rachid Tribak
Centre Régional des Métiers de l’Education
et de la Formation (CRMEF)-Tanger,
Avenue My Abdelaziz, Souani, B.P. 3117,
Tanger, Morocco
<EMAIL_ADDRESS>
###### Abstract

††Mathematics Subject Classification 2020: 16D10, 16D70, 16D80
Key words and phrases: CS-ring; strongly CS module (ring); trivial extension;
weakly IN module (ring)

All rings considered are commutative. In this article we introduce and study
two notions of modules which are stronger than CS modules, namely weakly IN
modules and strongly CS modules. Our main aim is to characterize when a
trivial extension $A=R\propto M$ (of a ring $R$ by an $R$-module $M$) is a CS
ring.

##  1 Introduction

Throughout this article, all rings considered are assumed to be commutative
rings with an identity and $R$ denotes such a ring. All modules are unital. We
denote respectively by $Spec(R)$, $Max(R)$ and $Min(R)$ the set of all prime
ideals of $R$, the set of all maximal ideals of $R$ and the set of all minimal
prime ideals of $R$. The nilradical of $R$ and the Jacobson radical of $R$ are
denoted by $Nil(R)$ and $J(R)$, respectively. Let $M$ be an $R$-module and let
$x\in M$. By $Ann_{R}(x)$ and $Ann_{R}(M)$ we denote the annihilator of $x$
and $M$, respectively; i.e. $Ann_{R}(x)=\\{r\in R\mid rx=0\\}$ and
$Ann_{R}(M)=\\{r\in R\mid rM=0\\}$. The notation $N\subseteq M$ means that $N$
is a subset of $M$; $N\leq M$ means that $N$ is a submodule of $M$; and
$N\subseteq^{ess}M$ means that $N$ is an essential submodule of $M$. By
$\mathbb{Q}$ and $\mathbb{Z}$ we denote the ring of rational and integer
numbers, respectively.

An $R$-module $M$ is called CS (or extending) if every submodule of $M$ is
essential in a direct summand of $M$ and a ring $R$ is called CS if the
$R$-module $R$ is CS. By [11, Theorem 6], $R$ is CS if and only if for any
ideals $I$ and $J$ of $R$ with $I\cap J=0$, $Ann_{R}(I)+Ann_{R}(J)=R$. In our
attempt to characterize CS trivial extensions, we were lead to introduce two
types of CS modules. The first one comes form the above characterization of CS
rings. For consider $A=R\propto M$, the trivial extension of $R$ by an
$R$-module $M$, and let $\phi:M\rightarrow A$ be the natural monomorphism.
Assume that $A$ is a CS ring. Then given two submodules $N$ and $L$ of $M$
such that $N\cap L=0$, it is clear that $\phi(N)$ and $\phi(L)$ are two ideals
of $A$ such that $\phi(N)\cap\phi(L)=0$. Thus
$Ann_{A}\phi(N)+Ann_{A}\phi(L)=A$ which implies that
$Ann_{R}(N)+Ann_{R}(L)=R$. Recall that a ring $R$ is called an Ikeda-Nakayama
ring (or IN ring) if for any two ideals $T$ and $T^{\prime}$ of $R$, we have
$Ann_{R}(T\cap T^{\prime})=Ann_{R}(T)+Ann_{R}(T^{\prime})$ (see [13, p. 148]).
Call an $R$-module $M$ weakly IN if $Ann_{R}(N)+Ann_{R}(L)=R$ whenever $N$ and
$L$ are submodules of $M$ such that $N\cap L=0$. The second type is due to the
definition itself of CS rings. Suppose that $A$ is a CS ring and let $N$ be a
submodule of $M$. Then $\phi(N)$ is an ideal of $A$ and hence $\phi(N)$ is
essential in $fA$ for some idempotent $f$ of $A$. But $f$ is of the form
$(e,0)$ for some idempotent $e$ of $R$, so $N$ is essential in $eM$. Call an
$R$-module $M$ strongly CS if for every submodule $N$ of $M$, there exists an
idempotent $e$ of $R$ such that $N$ is an essential submodule of $eM$. The
above discussion justifies our choice of the title and motivates our study in
this paper.

In Section 2, we study some properties of weakly IN modules and strongly CS
modules. We investigate the connection between these two classes of modules.
It is shown that an $R$-module $M$ is strongly CS if and only if $M$ is weakly
IN and idempotents lift modulo $Ann_{R}(M)$ (Theorem 2.8). Then we
characterize the class of rings $R$ over which every cyclic $R$-module is
weakly IN (strongly CS) (Theorems 2.11 and 2.12).

In section 3, we focus on when a finite direct sum of weakly IN (strongly CS)
modules is also weakly IN (strongly CS). Let an $R$-module $M=M_{1}\oplus
M_{2}\oplus\cdots\oplus M_{n}$ be a direct sum of submodules
$M_{1},M_{2},\dots,M_{n}$. We show that $M$ is weakly IN if and only if each
$M_{i}$ is weakly IN and $Ann_{R}(M_{i})+Ann_{R}(M_{j})=R$ for all distinct
$i$, $j$ in $\\{1,\ldots,n\\}$ (Theorem 3.1). It is also shown that $M$ is a
strongly CS $R$-module if and only if $R=R_{1}\times R_{2}\times\cdots\times
R_{n}$ such that $M_{i}$ is a strongly CS $R_{i}$-module for all
$i\in\\{1,\dots,n\\}$ (Theorem 3.3). As an application, we prove that $R$ is a
clean ring (i.e. every element of $R$ is a sum of a unit and an idempotent) if
and only if every weakly IN $R$-module is strongly CS if and only if
$R/\mathfrak{m}\oplus R/\mathfrak{m}^{\prime}$ is a strongly CS $R$-module for
every distinct maximal ideals $\mathfrak{m}$ and $\mathfrak{m}^{\prime}$ of
$R$ (Theorem 3.5). Also, we characterize the class of rings $R$ for which
$R/\mathfrak{p}\oplus R/\mathfrak{p}^{\prime}$ is a strongly CS $R$-module for
every distinct minimal prime ideals $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$
of $R$ as that of the purified rings (Theorem 3.9). Recall that a ring $R$ is
said to be a purified (or coclean) ring if for every distinct minimal prime
ideals $\mathfrak{p}$ and $\mathfrak{q}$ of $R$, there exists an idempotent
$e$ of $R$ such that $e\in\mathfrak{p}$ and $1-e\in\mathfrak{q}$.

In section 4, we characterize weakly IN and strongly CS modules over Dedekind
domains. In particular, we show that if $R$ is a Dedekind domain and $M$ is an
$R$-module, then $M$ is weakly IN if and only if $M$ is cyclic, or $M\cong
E(R/\mathfrak{m})$ for some maximal ideal $\mathfrak{m}$ of $R$, or $M$ is
isomorphic to an $R$-submodule of the quotient field $K$ of $R$ (Theorem 4.3).

The last section is devoted to the study of CS trivial extensions. Let
$A=R\propto M$ denotes the trivial extension of a ring $R$ by an $R$-module
$M$. In the main result of this section, we show that $A$ is a CS ring if and
only if $M$ is weakly IN (or strongly CS) and there exists an idempotent $e$
of $R$ such that $Ann_{R}(M)=eR$ and $eR$ is a CS ring if and only if $M\oplus
Ann_{R}(M)$ is a weakly IN (or strongly CS) $R$-module (Theorem 5.4). When $M$
is a flat $R$-module, we prove that $A$ is a CS ring if and only if $R$ and
$M$ are weakly IN (or strongly CS) $R$-modules and $Ann_{R}(M)$ is a direct
summand of $R$ (Corollary 5.6).

##  2 Weakly IN and Strongly CS Modules

Recall that an $R$-module $M$ is called CS if every submodule of $M$ is
essential in a direct summand of $M$. A ring $R$ is called CS if $R$ is a CS
$R$-module. Using [11, Theorem 6], we obtain the following proposition.

###### Proposition 2.1

The following are equivalent for a ring $R$:

(1) $R$ is a CS ring;

(2) For any ideal $I$ of $R$, there exists $e=e^{2}\in R$ such that
$I\subseteq^{ess}eR$;

(3) For any ideals $I$ and $J$ of $R$ with $I\cap J=0$,
$Ann_{R}(I)+Ann_{R}(J)=R$.

A ring $R$ is called an Ikeda-Nakayama ring (or an IN-ring) if $Ann_{R}(I\cap
J)=Ann_{R}(I)+Ann_{R}(J)$ for all ideals $I$ and $J$ of $R$ (see [8]). Using
the endomorphism ring, Wisbauer, Yousif and Zhou generalized this notion to a
module theoretic version in 2002 [17]. Here, we will consider another
generalization. We will call an $R$-module $M$ an s.IN-module (scalar IN-
module) if $Ann_{R}(N\cap L)=Ann_{R}(N)+Ann_{R}(L)$ for all submodules $N$ an
$L$ of $M$. Next, we introduce two notions. The first one is weaker than that
of s.IN-modules and the latter one is stronger than that of CS modules.

###### Definition 2.2

Let $M$ be an $R$-module.

(1) $M$ is called weakly IN if for any submodules $N$ and $L$ of $M$ with
$N\cap L=0$, $Ann_{R}(N)+Ann_{R}(L)=R$.

(2) $M$ is called strongly CS if for any submodule $N$ of $M$, there exists
$e=e^{2}\in R$ such that $N\subseteq^{ess}eM$.

(3) A ring $R$ is called weakly IN (strongly CS) if $R$, as an $R$-module, has
the corresponding property.

###### Remark 2.3

(1) From Proposition 2.1, it follows easily that a ring $R$ is CS if and only
if $R$ is weakly IN if and only if $R$ is strongly CS.

(2) Consider the ring $\mathbb{Z}_{2}[x_{1},x_{2},\ldots]$ where $x_{i}^{3}=0$
for all $i$, $x_{i}x_{j}=0$ for all $i\neq j$ and $x_{i}^{2}=x_{j}^{2}\neq 0$
for all $i$ and $j$. It was shown in [8, Example 6] that $R$ is a CS ring and
hence $R$ is weakly IN, but $R$ is not an IN ring. It follows that the
$R$-module $R$ is weakly IN, but it is not an s.IN-module.

(3) If $M$ is a uniform $R$-module, then clearly $M$ is strongly CS. If $R$ is
indecomposable or $M$ is indecomposable, then the converse also holds.

###### Proposition 2.4

Any submodule of a strongly CS $($weakly IN$)$ module is strongly CS $($weakly
IN$)$.

Proof. Let $M$ be a strongly CS module and let $N$ be a submodule of $M$. Let
$L$ be a submodule of $N$. Then $L\subseteq^{ess}eM$ for some idempotent $e$
of $R$. Hence $L\subseteq^{ess}eM\cap N=eN$. The other assertion is evident.
$\Box$

###### Proposition 2.5

A free $R$-module $F$ is weakly IN if and only if $rank(F)=1$ and $R$ is a CS
ring.

Proof. Let $F$ be a weakly IN free $R$-module. Suppose that $rank(F)\geq 2$.
Then $F$ contains a submodule isomorphic to $R\oplus R$. Therefore
$Ann_{R}(R)+Ann_{R}(R)=R$, a contradiction. So $rank(F)=1$ and hence $R$ is a
CS ring (Remark 2.3(1)). The converse is clear. $\Box$

From the preceding proposition, one can directly infer that over a field $K$,
a module $M$ is weakly IN if and only if $M\cong K$.

Recall that a module $M$ is called quasi-continuous if $M$ is CS and, for any
two direct summands $M_{1}$, $M_{2}$ with $M_{1}\cap M_{2}=0$, $M_{1}\oplus
M_{2}$ is also a direct summand (see [12]). From [17, Corollary 4], it follows
that a faithful $R$-module $M$ is weakly IN if and only if $M$ is a quasi-
continuous $R$-module and for any $f^{2}=f\in End_{R}(M)$, there exists $r\in
R$ such that $f(x)=rx$ for any $x\in M$. This fact can be extended to the
general case as shown in the following characterization of weakly IN modules.

###### Proposition 2.6

Let $R$ be a ring and let $M$ be a nonzero $R$-module. Then the following are
equivalent:

(1) $M$ is a weakly IN $R$-module;

(2) $M$ is a weakly IN $R/Ann_{R}(M)$-module;

(3) $M$ is a quasi-continuous $R$-module and for any $f^{2}=f\in End_{R}(M)$,
there exists $r\in R$ such that $f(x)=rx$ for any $x\in M$;

(4) $M$ is a CS $R$-module and for any direct summand $N$ of $M$, there exists
$r\in R$ such that $N=rM$ and $r-r^{2}\in Ann_{R}(M)$;

(5) For any submodule $N$ of $M$, there exists $r\in R$ such that
$N\subseteq^{ess}rM$ and $r-r^{2}\in Ann_{R}(M)$;

(6) $M$ is strongly CS as an $R/Ann_{R}(M)$-module.

Proof. (1) $\Leftrightarrow$ (2) This is clear.

(2) $\Rightarrow$ (3) Let $\overline{R}=R/Ann_{R}M$. It is easy to see that
$M$ has a natural structure of ($\overline{R}$,$R$)-bimodule and $M$ is a
faithful left $\overline{R}$-module. Moreover, the $R$-submodules and the
$\overline{R}$-submodules of $M$ are the same. Using [17, Corollary 4], we see
that $M$ is a quasi-continuous $R$-module and for any $f^{2}=f\in End_{R}(M)$,
there exists $\overline{r}=r+Ann_{R}(M)\in\overline{R}$ such that
$f(x)=\overline{r}x=rx$ for all $x\in M$.

(3) $\Rightarrow$ (4) It is clear that $M$ is a CS $R$-module. Let $N$ and $K$
be two submodules of $M$ such that $M=N\oplus K$. Let $f$ be the projection on
$N$ along $K$. Then $f^{2}=f$ and so there exists $r\in R$ such that $f(x)=rx$
for any $x\in M$. Hence, $N=f(M)=rM$. But $f^{2}=f$, so $r-r^{2}\in
Ann_{R}(M)$.

(4) $\Rightarrow$ (5) Let $N$ be a submodule of $M$. Since $M$ is CS, there
exists a direct summand $K$ of $M$ such that $N\subseteq^{ess}K$. Moreover, by
assumption, $K=rM$ for some $r\in R$ with $r-r^{2}\in Ann_{R}(M)$.

(5) $\Rightarrow$ (1) Let $N$ and $L$ be two submodules of $M$ with $N\cap
L=0$. Then there exist $r$ and $s$ in $R$ such that $N\subseteq^{ess}rM$,
$L\subseteq^{ess}sM$, $r-r^{2}\in Ann_{R}(M)$ and $s-s^{2}\in Ann_{R}(M)$.
Thus $N\cap L\subseteq^{ess}rM\cap sM$. Since $N\cap L=0$, we have $rM\cap
sM=0$. This implies that $rsM\subseteq rM\cap sM=0$ and hence $rs\in
Ann_{R}(M)$. Moreover, since $r-r^{2}\in Ann_{R}(M)$ and $s-s^{2}\in
Ann_{R}(M)$, we have $(1-r)\in Ann_{R}(rM)\subseteq Ann_{R}(N)$ and $(1-s)\in
Ann_{R}(sM)\subseteq Ann_{R}(L)$. It follows that

$1=(1-r)+(1-s)r+sr\in Ann_{R}(N)+Ann_{R}(L).$

Therefore, $R=Ann_{R}(N)+Ann_{R}(L)$. Thus, $M$ is weakly IN.

(5) $\Leftrightarrow$ (6) This is immediate. $\Box$

In the next example, we present a CS module which is not weakly IN, and some
CS modules which are not strongly CS. More examples are provided in the next
two sections.

###### Example 2.7

(1) Let $R$ be a self injective ring. Then clearly the $R$-module $M=R\oplus
R$ is CS. However, $M$ is not weakly IN by Proposition 2.5.

(2) Let $R$ be an indecomposable ring (e.g., $R$ is a local ring or a domain).

(a) Consider the $R$-module $M=S_{1}\oplus S_{2}$ where $S_{1}$ and $S_{2}$
are simple modules. It is clear that $M$ is a CS module. On the other hand,
$M$ is not strongly CS (see Remark 2.3(3)).

(b) Let $M$ be an injective $R$-module which is not indecomposable (e.g., we
can take the $\mathbb{Z}$-module $M=\mathbb{Q}\oplus\mathbb{Q}$ or
$M=\mathbb{Z}(p_{1}^{\infty})\oplus\mathbb{Z}(p_{2}^{\infty})$ for some prime
numbers $p_{1}$ and $p_{2}$). Then $M$ is CS, but $M$ is not strongly CS by
Remark 2.3(3).

Let $R$ be a ring and let $I$ be a proper ideal of $R$. We say that
idempotents lift modulo $I$ if whenever $r$ is an element of $R$ such that
$r-r^{2}\in I$, then there exists $e=e^{2}\in R$ such that $r-e\in I$.

###### Theorem 2.8

Let $R$ be a ring and let $M$ be a nonzero $R$-module. Then the following are
equivalent:

(1) $M$ is a strongly CS $R$-module;

(2) For any submodules $N$ and $L$ of $M$ with $N\cap L=0$, there exists an
idempotent $e\in R$ such that $e\in Ann_{R}(N)$ and $1-e\in Ann_{R}(L)$;

(3) $M$ is a weakly IN $R$-module and idempotents lift modulo $Ann_{R}(M)$;

(4) $M$ is a CS module and for every direct summand $K$ of $M$, there exists
an idempotent $e$ of $R$ such that $K=eM$.

Proof. (1) $\Rightarrow$ (2) Let $N$ and $L$ be two submodules of $M$ with
$N\cap L=0$. Then there exist $e=e^{2}$ and $f=f^{2}$ in $R$ such that
$N\subseteq^{ess}fM$ and $L\subseteq^{ess}eM$. Using the fact that $N\cap
L=0$, we get $fM\cap eM=0$. Hence, $efM\subseteq fM\cap eM=0$. So $e\in
Ann_{R}(fM)\subseteq Ann_{R}(N)$ and $1-e\in Ann_{R}(eM)\subseteq Ann_{R}(L)$.

(2) $\Rightarrow$ (3) Let $N$ and $L$ be two submodules of $M$ with $N\cap
L=0$. By hypothesis, there exists an idempotent $e\in R$ such that $e\in
Ann_{R}(N)$ and $1-e\in Ann_{R}(L)$. But $e+1-e=1$, so
$Ann_{R}(N)+Ann_{R}(L)=R$. It follows that $M$ is weakly IN. Now let $r\in R$
such that $r-r^{2}\in Ann_{R}(M)$ and let $rx=(1-r)y\in rM\cap(1-r)M$ where
$x,y\in M$. Then $r^{2}x=(r-r^{2})y=0$ since $r-r^{2}\in Ann_{R}(M)$. Using
once again the fact that $r-r^{2}\in Ann_{R}(M)$, we get $rx=0$. Therefore
$rM\cap(1-r)M=0$. Hence, by (2), there exists $f=f^{2}\in R$ such that $1-f\in
Ann_{R}(rM)$ and $f\in Ann_{R}((1-r)M)$. This means that $r-fr\in Ann_{R}(M)$
and $f-fr\in Ann_{R}(M)$. Consequently,

$f-r=(f-fr)-(r-fr)\in Ann_{R}(M).$

This shows that idempotents lift modulo $Ann_{R}(M)$.

(3) $\Rightarrow$ (1) Let $N$ be a submodule of $M$. Since $M$ is a weakly IN
$R$-module, there exists $r\in R$ such that $N\subseteq^{ess}rM$ and
$r-r^{2}\in Ann_{R}(M)$ (see Proposition 2.6). But idempotents lift modulo
$Ann_{R}(M)$, so there exits an idempotent $e$ of $R$ such that $r-e\in
Ann_{R}(M)$. Therefore $rM=eM$ and consequently $N\subseteq^{ess}eM$.

(1) $\Leftrightarrow$ (4) This is clear. $\Box$

The preceding theorem shows that the class of weakly IN modules contains that
of strongly CS modules. Next, we present an example illustrating that this
inclusion is proper, in general.

###### Example 2.9

Let $p$ and $q$ be two different prime numbers and consider the ring
$R=\\{m/n\in\mathbb{Q}\mid p\nmid n$ and $q\nmid n\ (m/n$ in lowest
terms$)\\}$. It is well known that $pR$ and $qR$ are the only maximal ideals
in $R$. Moreover, idempotents do not lift modulo $J(R)=pR\cap qR$. Let
$M=R/pR\oplus R/qR$. Clearly, $Ann_{R}(M)=J(R)$. From Corollary 3.4, we infer
that $M$ is a weakly IN $R$-module which not strongly CS.

Let $R$ be a ring. Recall that the socle of $R$, denoted by $Soc(R)$, is the
sum of all its minimal ideals. In the following corollary, we provide
sufficient conditions for a weakly IN module to be strongly CS.

###### Corollary 2.10

Let $M$ be a nonzero $R$-module such that $Ann_{R}(M)$ satisfies any one of
the following conditions:

(1) $Ann_{R}(M)$ is a nil ideal of $R$ $($i.e., $Ann_{R}(M)\subseteq
Nil(R)$$)$;

(2) $Ann_{R}(M)$ is a direct summand of $R$ $($for instance, $M$ is
faithful$)$;

(3) $Ann_{R}(M)=Soc(R)$.

Then $M$ is a weakly IN $R$-module if and only if $M$ is strongly CS.

Proof. The sufficiency follows from Theorem 2.8. Conversely, suppose that $M$
is a weakly IN $R$-module. Applying again Theorem 2.8, we only need to show
that idempotents lift modulo $Ann_{R}(M)$.

(1) This follows from [2, Proposition 27.1].

(2) We will show that idempotent lift modulo every direct summand of $R$. Let
$e=e^{2}\in R$ and let $r\in R$ such that $r-r^{2}\in eR$. Then
$(1-e)(r-r^{2})=0$ and hence $((1-e)r)^{2}=(1-e)r$. Thus $(1-e)r$ is an
idempotent of $R$. Moreover, we have $r-((1-e)r)=er\in eR$.

(3) This follows from [18, Lemma 1.2]. $\Box$

Let $n\geq 2$. By Proposition 2.5, there exists no ring $R$ for which every
$n$-generated $R$-module is weakly IN. This fact should be contrasted with [9,
Corollary 13.8]. On the other hand, take a valuation ring $R$. It is easily
seen that every cyclic $R$-module is uniform. Therefore every cyclic
$R$-module is strongly CS and weakly IN (see Remark 2.3(3) and Theorem 2.8).
Next, we will characterize the class of rings $R$ for which every cyclic
$R$-module is weakly IN (strongly CS).

A ring $R$ is called a CF-ring if every $R$–module $M$ which is a direct sum
of finitely many cyclic $R$–modules has a canonical form decomposition, i.e.
$M\cong R/I_{1}\oplus\cdots\oplus R/I_{n}$ where the ideals $I_{j}$ $(1\leq
j\leq n)$ of $R$ satisfy $I_{1}\subseteq I_{2}\subseteq\cdots\subseteq
I_{n}\varsubsetneq R$ (see [15]). For the definitions of the other types of
rings used in the following two results, we refer the reader to [7].

###### Theorem 2.11

The following are equivalent for a ring $R$:

(1) Any cyclic $R$-module is weakly IN;

(2) $R/I$ is a CS ring for every proper ideal $I$ of $R$;

(3) $R$ is a CF-ring;

(4) $R$ is a finite direct product of valuation rings, h-local Prüfer domains
and torch rings.

Proof. Let $I$ be a proper ideal of $R$ and consider the $R$-module $M=R/I$.
By Proposition 2.6, $M$ is a weakly IN $R$-module if and only if $M$ is a
strongly CS $R/Ann_{R}(M)$-module. This is equivalent to the condition that
$R/I$ is a CS ring (Remark 2.3(1)). Now use [11, Theorem 9]. $\Box$

Recall that a ring $R$ is called clean if every element of $R$ is a sum of a
unit and an idempotent.

###### Theorem 2.12

The following are equivalent for a ring $R$:

(1) Any cyclic $R$-module is strongly CS;

(2) $R$ is a clean CF-ring;

(3) $R$ is a finite direct product of valuation rings.

Proof. (1) $\Leftrightarrow$ (2) From Theorem 2.8, it follows that the
assertion (1) is equivalent to the condition that for any ideal $I$ of $R$,
$R/I$ is a weakly IN $R$-module and idempotents lift modulo $Ann_{R}(R/I)=I$.
Now using Theorem 2.11 and the fact that $R$ is a clean ring if and only if
idempotents lift modulo every ideal of $R$ (see [1, Theorem 5.1]), we obtain
the desired equivalence.

(2) $\Rightarrow$ (3) Since $R$ is a CF-ring, we have $R=R_{1}\times
R_{2}\times\cdots\times R_{n}$ where each $R_{i}$ is an indecomposable ring
which is either a valuation ring or an h-local Prüfer domain or a torch ring
([15, Theorems 3.10 and 3.12]). Let $i\in\\{1,\dots,n\\}$. Since $R$ is clean,
it follows that $R_{i}$ is also clean by [3, Proposition 2]. Moreover, $R_{i}$
is a local ring by [3, Theorem 3]. Hence $R_{i}$ could not be a torch ring
since every torch ring has at least two maximal ideals ([7, page 38]). In
addition, note that any local Prüfer domain is a valuation ring.

(3) $\Rightarrow$ (2) By Theorem 2.11, $R$ is a CF-ring. Moreover, note that
any valuation ring is local. Thus, using [3, Proposition 2(1)-(3)], we
conclude that $R$ is a clean ring. $\Box$

##  3 Finite Direct Sums of Weakly IN (Strongly CS) Modules

We begin by providing necessary and sufficient conditions for a finite direct
sum of modules to be weakly IN.

###### Theorem 3.1

Let $M=M_{1}\oplus M_{2}\oplus\cdots\oplus M_{n}$ be a direct sum of
submodules $M_{i}(1\leq i\leq n)$. Then the following are equivalent:

(1) $M$ is weakly IN;

(2) (a) $M_{i}$ is a weakly IN $R$-module for all $i\in\\{1,\dots,n\\}$, and

(b) $Ann_{R}(M_{j})+Ann_{R}(M_{k})=R$ for all $j\neq k\in\\{1,\dots,n\\}$.

Proof. (1) $\Rightarrow$ (2) (a) follows from Proposition 2.4 and (b) follows
from the definition of a weakly IN module.

(2) $\Rightarrow$ (1) Let $N$ and $L$ be two submodules of $M$ such that
$N\cap L=0$. Using (b) and [6, Lemma 2.6], we get $N=\oplus_{i=1}^{n}(N\cap
M_{i})$ and $L=\oplus_{i=1}^{n}(L\cap M_{i})$. Fix $i\in\\{1,\dots,n\\}$. As
$N\cap L=0$, we have $(N\cap M_{i})\cap(L\cap M_{i})=0$. Since $M_{i}$ is
weakly IN, we have $Ann_{R}(N\cap M_{i})+Ann_{R}(L\cap M_{i})=R$. Moreover,
using (b), it follows that for any $i\neq j\in\\{1,\dots,n\\}$, we have
$Ann_{R}(N\cap M_{i})+Ann_{R}(L\cap M_{j})=R$ as $Ann_{R}(M_{i})\subseteq
Ann_{R}(N\cap M_{i})$ and $Ann_{R}(M_{j})\subseteq Ann_{R}(L\cap M_{j})$. So
$(\cap_{i=1}^{n}Ann_{R}(N\cap M_{i}))+(\cap_{i=1}^{n}Ann_{R}(L\cap M_{i}))=R$
by [19, Theorem 31]. Consequently, $Ann_{R}(N)+Ann_{R}(L)=R$. This completes
the proof. $\Box$

In the next theorem, we provide a characterization of when a finite direct sum
of modules is strongly CS. We need the following lemma.

###### Lemma 3.2

Let $R_{1}$ and $R_{2}$ be two rings and let $M_{i}$ be an $R_{i}$-module
$(i=1,2)$. Then the following hold true:

(1) $M_{1}\times M_{2}$ is a strongly CS $R_{1}\times R_{2}$-module if and
only if $M_{i}$ is a strongly CS $R_{i}$-module for each $i=1,2$.

(2) $R=R_{1}\times R_{2}$ is a CS ring if and only if so are $R_{1}$ and
$R_{2}$.

Proof. (1) We will use the following elementary property: $(\ast)$ given an
$R_{1}$-submodule $N_{1}$ of $M_{1}$ and an $R_{2}$-submodule $N_{2}$ of
$M_{2}$, $N_{1}\times N_{2}\subseteq^{ess}M_{1}\times M_{2}$ if and only if
$N_{1}\subseteq^{ess}M_{1}$ and $N_{2}\subseteq^{ess}M_{2}$.

Now suppose that $M_{1}\times M_{2}$ is a strongly CS $R_{1}\times
R_{2}$-module and let $N_{1}$ be an $R_{1}$-submodule of $M_{1}$ and $N_{2}$
an $R_{2}$-submodule of $M_{2}$. Then there exists
$(e_{1},e_{2})=(e_{1},e_{2})^{2}=(e_{1}^{2},e_{2}^{2})\in R_{1}\times R_{2}$
such that $N_{1}\times N_{2}\subseteq^{ess}(e_{1},e_{1})(M_{1}\times
M_{2})=e_{1}M_{1}\times e_{2}M_{2}$. By $(\ast)$, it follows that
$N_{1}\subseteq^{ess}e_{1}M_{1}$ and $N_{2}\subseteq^{ess}e_{2}M_{2}$. This
clearly implies that each $M_{i}$ ($i=1,2$) is a strongly CS $R_{i}$-module.
Conversely, let $N$ be an $R_{1}\times R_{2}$-submodule of $M_{1}\times
M_{2}$. Then $N=N_{1}\times N_{2}$, where $N_{1}$ is an $R_{1}$-submodule of
$M_{1}$ and $N_{2}$ is an $R_{2}$-submodule of $M_{2}$. Let
$e_{1}^{2}=e_{1}\in R_{1}$ and $e_{2}^{2}=e_{2}\in R_{2}$ such that
$N_{1}\subseteq^{ess}e_{1}M_{1}$ and $N_{2}\subseteq^{ess}e_{2}M_{2}$. Again
by $(\ast)$, we have $N_{1}\times N_{2}\subseteq^{ess}e_{1}M_{1}\times
e_{2}M_{2}=(e_{1},e_{1})(M_{1}\times M_{2})$. Since
$(e_{1},e_{2})^{2}=(e_{1},e_{2})$, we conclude that $M_{1}\times M_{2}$ is a
strongly CS $R_{1}\times R_{2}$-module.

(2) Apply (1) for $M_{1}=R_{1}$ and $M_{2}=R_{2}$ (see Remark 2.3(1)). $\Box$

###### Theorem 3.3

Let $M=M_{1}\oplus M_{2}\oplus\cdots\oplus M_{n}$ be a direct sum of
submodules $M_{i}(1\leq i\leq n)$. Then the following are equivalent:

(1) $M$ is strongly CS;

(2) $M$ satisfies the following two conditions:

(a) $M_{i}$ is a strongly CS $R$-module for every $i\in\\{1,\dots,n\\}$, and

(b) There exists a complete set of orthogonal idempotents
$\\{e_{1},\dots,e_{n}\\}$ of $R$ such that $e_{i}M_{i}=M_{i}$ for all
$i\in\\{1,\dots,n\\}$;

(3) $R=R_{1}\times R_{2}\times\cdots\times R_{n}$ such that $M_{i}$ is a
strongly CS $R_{i}$-module for all $i\in\\{1,\dots,n\\}$.

Proof. (1) $\Rightarrow$ (2) (a) follows by using Proposition 2.4. Let us show
(b) by induction on $n$. Suppose that $M=M_{1}\oplus M_{2}$ is strongly CS. By
Theorem 2.8, there exists an idempotent $e$ of $R$ such that $1-e\in
Ann_{R}(M_{1})$ and $e\in\mbox{Ann\/}_{R}(M_{2})$. This implies that
$M_{1}=eM$ and $M_{2}=(1-e)M$. Therefore (b) is true for $n=2$. Now assume (b)
holds for $n$; we will prove it for $n+1$. Let $M=M_{1}\oplus\cdots\oplus
M_{n}\oplus M_{n+1}$ be strongly CS. From the case $n=2$, we infer that there
exists an idempotent $e$ of $R$ such that $e(M_{1}\oplus\cdots\oplus
M_{n})=M_{1}\oplus\cdots\oplus M_{n}$ and $(1-e)M_{n+1}=M_{n+1}$. But
$M_{1}\oplus\cdots\oplus M_{n}$ is a strongly CS $R$-module as it is a
submodule of $M$, so, by induction hypothesis there exists a complete set of
orthogonal idempotents $\\{f_{1},\dots,f_{n}\\}$ of $R$ such that
$M_{i}=f_{i}M_{i}$ for all $i\in\\{1,\dots,n\\}$. It is easy to see that
$\\{ef_{1},\dots,ef_{n},1-e\\}$ is a complete set of orthogonal idempotents of
$R$ such that $ef_{i}M_{i}=eM_{i}=M_{i}$ for all $i\in\\{1,\dots,n\\}$ and
$(1-e)M_{n+1}=M_{n+1}$.

(2) $\Rightarrow$ (3) By (b), $R=R_{1}\times R_{2}\times\cdots\times R_{n}$
where $R_{i}=e_{i}R$ for all $i\in\\{1,\dots,n\\}$. Fix $j\in\\{1,\dots,n\\}$.
Since $e_{j}M_{j}=M_{j}$, $M_{j}$ has a natural structure of an
$R_{j}$-module. Let $N$ be an $R_{j}$-submodule of $M_{j}$. By (a), $M_{j}$ is
a strongly CS $R$-module. So there exists $e^{2}=e\in R$ such that
$N\subseteq^{ess}eM_{j}=ee_{j}M_{j}$ as $R$-modules and also as
$e_{j}R$-modules. Note that $ee_{j}$ is an idempotent of $R_{j}$. Thus $M_{j}$
is a strongly CS $R_{j}$-module.

(3) $\Rightarrow$ (1) This follows from Lemma 3.2(1). $\Box$

As an application of the preceding two theorems, we have the following
corollary.

###### Corollary 3.4

Let $\mathfrak{p}$ and $\mathfrak{q}$ be two prime ideals of a ring $R$. Then
the following hold true:

(1) The $R$-module $R/\mathfrak{p}\oplus R/\mathfrak{q}$ is weakly IN if and
only if $\mathfrak{p}+\mathfrak{q}=R$.

(2) The following are equivalent:

(a) The $R$-module $R/\mathfrak{p}\oplus R/\mathfrak{q}$ is strongly CS;

(b) There exists an idempotent $e$ of $R$ such that $e\in\mathfrak{p}$ and
$1-e\in\mathfrak{q}$;

(c) $\mathfrak{p}+\mathfrak{q}=R$ and idempotents lift modulo
$\mathfrak{p}\cap\mathfrak{q}$.

Proof. Since the $R$-modules $R/\mathfrak{p}$ and $R/\mathfrak{q}$ are
uniform, they are strongly CS (and hence also weakly IN).

(1) Use Theorem 3.1 and the fact that $R/\mathfrak{p}$ and $R/\mathfrak{q}$
are weakly IN.

(2) (a) $\Leftrightarrow$ (c) Use (1), Theorem 2.8 and the fact that
$Ann_{R}(R/\mathfrak{p}\oplus R/\mathfrak{q})=\mathfrak{p}\cap\mathfrak{q}$.

(a) $\Rightarrow$ (b) By Theorem 2.8, there exists an idempotent $e$ of $R$
such that $e\in Ann_{R}(R/\mathfrak{p})=\mathfrak{p}$ and $1-e\in
Ann_{R}(R/\mathfrak{q})=\mathfrak{q}$.

(b) $\Rightarrow$ (a) Let $e$ be an idempotent of $R$ such that
$e\in\mathfrak{p}$ and $1-e\in\mathfrak{q}$. Then
$(1-e)(R/\mathfrak{p})=R/\mathfrak{p}$ and $e(R/\mathfrak{q})=R/\mathfrak{q}$.
Now using Theorem 3.3((2) $\Rightarrow$ (1)) and the fact that
$R/\mathfrak{p}$ and $R/\mathfrak{q}$ are strongly CS $R$-modules, we deduce
that $R/\mathfrak{p}\oplus R/\mathfrak{q}$ is a strongly CS $R$-module. $\Box$

In contrast to Example 2.9, we characterize in the next theorem the class of
rings $R$ for which every weakly IN $R$-module is strongly CS.

###### Theorem 3.5

The following are equivalent for a ring $R$:

(1) Any weakly IN $R$-module is strongly CS;

(2) $R/\mathfrak{m}\oplus R/\mathfrak{m}^{\prime}$ is a strongly CS $R$-module
for every distinct maximal ideals $\mathfrak{m}$ and $\mathfrak{m}^{\prime}$
of $R$;

(3) Idempotents of $R$ lift modulo $\mathfrak{m}\cap\mathfrak{m}^{\prime}$ for
every distinct maximal ideals $\mathfrak{m}$ and $\mathfrak{m}^{\prime}$ of
$R$;

(4) Idempotents lift modulo every ideal of $R$;

(5) For every distinct maximal ideals $\mathfrak{m}$ and
$\mathfrak{m}^{\prime}$ of $R$, there exists an idempotent $e$ of $R$ such
that $e\in\mathfrak{m}$ and $1-e\in\mathfrak{m}^{\prime}$;

(6) $R$ is a clean ring.

Proof. The equivalences (4) $\Leftrightarrow$ (5) $\Leftrightarrow$ (6) follow
from [1, Theorem 5.1].

The equivalences (2) $\Leftrightarrow$ (3) $\Leftrightarrow$ (5) follow from
Corollary 3.4(2).

(4) $\Rightarrow$ (1) Use the equivalence (1) $\Leftrightarrow$ (3) in Theorem
2.8.

(1) $\Rightarrow$ (2) Let $M=R/\mathfrak{m}\oplus R/\mathfrak{m}^{\prime}$
where $\mathfrak{m}$ and $\mathfrak{m}^{\prime}$ are two distinct maximal
ideals of $R$. By Corollary 3.4(1), $M$ is a weakly IN $R$-module and so it is
strongly CS by (1). $\Box$

In the same vein of Example 2.9, we exhibit the following examples.

###### Example 3.6

(1) Let $R$ be a ring which is not clean. By Theorem 3.5, there exists an
$R$-module $M$ such that $M$ is weakly IN but not strongly CS. To construct an
explicit example of such a ring $R$ and such a module $M$, consider the ring
$R={\mathbb{Z}}$ and the $\mathbb{Z}$-module
$M={\mathbb{Z}}/2{\mathbb{Z}}\oplus{\mathbb{Z}}/3{\mathbb{Z}}$. Since
$2{\mathbb{Z}}+3{\mathbb{Z}}={\mathbb{Z}}$, $M$ is a weakly IN
${\mathbb{Z}}$-module (Corollary 3.4(1)). However, note that $M$ is not
uniform. Then $M$ can not be a strongly CS $R$-module since $R$ is
indecomposable (Remark 2.3(3)).

(2) Let $R=\mathbb{Z}$ and consider the $R$-module
$N=\mathbb{Q}/\mathbb{Z}_{(2\mathbb{Z})}$. By [15, Example 3.13], $A=R\propto
N$ is a Torch ring. Thus, every cyclic $A$-module is weakly IN by Theorem
2.11. On the other hand, since $R$ is not a clean ring, $A$ is not clean by
[4, Theorem 6.4]. Therefore the ring $A$ has a cyclic $A$-module which is not
strongly CS by Theorem 2.12.

Recall that a ring $R$ is called zero-dimensional if every prime ideal of $R$
is maximal.

###### Proposition 3.7

For a ring $R$ the following statements are equivalent:

(1) $R/\mathfrak{p}\oplus R/\mathfrak{p}^{\prime}$ is a weakly IN $R$-module
for every distinct prime ideals $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ of
$R$;

(2) $R/\mathfrak{p}\oplus R/\mathfrak{p}^{\prime}$ is a strongly CS $R$-module
for every distinct prime ideals $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ of
$R$;

(3) $R$ is a zero-dimensional ring.

Proof. (1) $\Rightarrow$ (3) Suppose that $R$ has a nonmaximal prime ideal
$\mathfrak{p}$ and let $\mathfrak{m}$ be a maximal ideal containing
$\mathfrak{p}$. By hypothesis, $R/\mathfrak{p}\oplus R/\mathfrak{m}$ is a
weakly IN $R$-module and hence $\mathfrak{p}+\mathfrak{m}=R$ (see Corollary
3.4(1)). Thus $\mathfrak{m}=R$, which is a contradiction. Therefore $R$ is a
zero-dimensional ring.

(3) $\Rightarrow$ (2) Let $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ be two
distinct prime ideals of $R$. Since $R$ is zero-dimensional, $\mathfrak{p}$
and $\mathfrak{p}^{\prime}$ are maximal. Moreover, note that $R$ is a clean
ring by [3, Corollary 11]. So $R/\mathfrak{p}\oplus R/\mathfrak{p}^{\prime}$
is a strongly CS $R$-module by Theorem 3.5.

(2) $\Rightarrow$ (1) This follows from the fact that any strongly CS
$R$-module is weakly IN (see Theorem 2.8). $\Box$

Recall that a ring $R$ is called an mp-ring if every prime ideal contains a
unique minimal prime ideal; equivalently, every maximal ideal of $R$ contains
a unique minimal prime ideal.

Replacing the term “prime” in Proposition 3.7 by “minimal prime”, we obtain
the following characterizations.

###### Proposition 3.8

The following are equivalent for a ring $R$:

  1. (1)

$R/\mathfrak{p}\oplus R/\mathfrak{p}^{\prime}$ is a weakly IN $R$-module for
every distinct minimal prime ideals $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$
of $R$;

  2. (2)

$\mathfrak{p}+\mathfrak{p}^{\prime}=R$ for every distinct minimal prime ideals
$\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ of $R$;

  3. (3)

$R$ is an mp-ring.

Proof. (1) $\Leftrightarrow$ (2) Use Corollary 3.4(1).

(2) $\Leftrightarrow$ (3) see [1, Theorem 6.2]. $\Box$

Following [1, Definition 8.1], a ring $R$ is said to be a purified ring if for
every distinct minimal prime ideals $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$
of $R$, there exists an idempotent $e$ of $R$ such that $e\in\mathfrak{p}$ and
$1-e\in\mathfrak{p}^{\prime}$. Note that every purified ring is an mp-ring.

###### Theorem 3.9

For a ring $R$ the following are equivalent:

(1) $R/\mathfrak{p}\oplus R/\mathfrak{p}^{\prime}$ is a strongly CS $R$-module
for every distinct minimal prime ideals $\mathfrak{p}$ and
$\mathfrak{p}^{\prime}$ of $R$;

(2) $R$ is an mp-ring and idempotents lift modulo
$\mathfrak{p}\cap\mathfrak{p}^{\prime}$ for every distinct minimal prime
ideals $\mathfrak{p}$ and $\mathfrak{p}^{\prime}$ of $R$;

(3) $R$ is a purified ring.

Proof. This follows by combining Corollary 3.4(2) with Proposition 3.8. $\Box$

##  4 Modules over Dedekind Domains

This short section is devoted to the description of the structure of both
weakly IN and strongly CS modules over Dedekind domains. Recall that for an
$R$-module $M$, $Ass(M)$ denotes the set of prime ideals of $R$ associated to
$M$, that is, $Ass(M)=\\{\mathfrak{p}\in Spec(R)$ $|$
$\mathfrak{p}=Ann_{R}(x)$ for some $0\neq x\in M\\}$.

###### Theorem 4.1

Let $R$ be a Dedekind domain with field of fractions $K$ and let $M$ be a
nonzero $R$-module. Then the following are equivalent:

(1) $M$ is a strongly CS $R$-module;

(2) $M$ is a uniform $R$-module;

(3) $M$ is isomorphic to an $R$-submodule of $K$ or there exists a maximal
ideal $\mathfrak{p}$ of $R$ such that $M\cong E(R/\mathfrak{p})$ or $M\cong
R/\mathfrak{p}^{n}$ for some positive integer $n$.

Proof. (1) $\Leftrightarrow$ (2) This follows from the fact that $R$ is
indecomposable (see Remark 2.3(3)).

(2) $\Rightarrow$ (3) Let $M$ be a nonzero uniform $R$-module. Since $R$ is
noetherian, $E(M)\cong E(R/\mathfrak{p})$ for some prime ideal $\mathfrak{p}$
of $R$ (see [14, Corollary of Theorem 2.32]). Hence $M$ is isomorphic to a
submodule of $E(R/\mathfrak{p})$ and $Ass(M)=\\{\mathfrak{p}\\}$. If
$\mathfrak{p}=0$, then $M$ is isomorphic to an $R$-submodule of $E(R)\cong K$.
Now assume that $\mathfrak{p}\neq 0$. Then $\mathfrak{p}$ is a maximal ideal
of $R$ as $R$ is a Dedekind domain. Clearly, $E=E(R/\mathfrak{p})$ is a
torsion $R$-module. Moreover, $M$ is indecomposable since $E$ is uniform.
Using [10, Theorem 10], we infer that $M\cong E(R/\mathfrak{p})$ or $M\cong
R/\mathfrak{q}^{n}$ for some maximal ideal $\mathfrak{q}$ of $R$ and some
positive integer $n$. In the latter case we have $\mathfrak{q}\in Ass(M)$ and
consequently $\mathfrak{q}=\mathfrak{p}$.

(3) $\Rightarrow$ (2) Let $\mathfrak{p}$ be a nonzero prime ideal of $R$. It
is clear that $K$ and $E(R/\mathfrak{p})$ are uniform $R$-modules. Also,
$R/\mathfrak{p}^{n}$ is uniform since it is a uniserial $R$-module by [14,
Lemma 6.8]. $\Box$

Let $R$ be a Dedekind domain with field of fractions $K$. Recall that an
$R$-submodule $F$ of $K$ is called a fractional ideal of $R$ if there is a
nonzero element $r$ of $R$ such that $rF\subseteq R$. Note that every finitely
generated $R$-submodule of $K$ is a fractional ideal of $R$ (see [14, Lemma
6.15]). It is well known that every injective $R$-module has no maximal
submodules. The following corollary is an immediate consequence of the
preceding theorem.

###### Corollary 4.2

Let $R$ be a Dedekind domain and let $M$ be a nonzero finitely generated
$R$-module. Then the following are equivalent:

(1) $M$ is strongly CS;

(2) $M\cong I$ where $I$ is a nonzero fractional ideal of $R$ or there exists
a maximal ideal $\mathfrak{p}$ of $R$ such that $M\cong R/\mathfrak{p}^{n}$
for some positive integer $n$.

###### Theorem 4.3

Let $R$ be a Dedekind domain with field of fractions $K$ and let $M$ be a
nonzero $R$-module. Then the following are equivalent:

(1) $M$ is a weakly IN $R$-module;

(2) $M$ is isomorphic to an $R$-submodule of $K$ or $M\cong E(R/\mathfrak{p})$
for some maximal ideal $\mathfrak{p}$ of $R$ or $M\cong
R/\mathfrak{p}_{1}^{n_{1}}\oplus\cdots\oplus R/\mathfrak{p}_{k}^{n_{k}}$ for
some positive integers $n_{i}$ $(1\leq i\leq k)$ and distinct maximal ideals
$\mathfrak{p}_{1},\dots,\mathfrak{p}_{k}$ of $R$.

Proof. (1) $\Rightarrow$ (2) Let $M$ be a nonzero weakly IN $R$-module. If
$Ann_{R}(M)=0$, then $M$ is strongly CS by Corollary 2.10. From Theorem 4.1,
it follows that $M$ is isomorphic to an $R$-submodule of $K$ or $M\cong
E(R/\mathfrak{p})$ for some maximal ideal $\mathfrak{p}$ of $R$. Now suppose
that $Ann_{R}(M)\neq 0$. By [14, Theorem 6.11], there exists a family
$\\{\mathfrak{p}_{i},n_{i}\\}_{i\in I}$ such that $\mathfrak{p}_{i}$ are
maximal ideals of $R$ and there are only finitely many distinct ones, $n_{i}$
$(i\in I)$ are positive integers and $M=\oplus_{i\in I}M_{i}$ where
$M_{i}\cong R/\mathfrak{p}_{i}^{n_{i}}$ for every $i\in I$. Let $j\neq k\in
I$. Since $M_{j}\cap M_{k}=0$ and $M$ is weakly IN, we have
$Ann_{R}(M_{j})+Ann_{R}(M_{k})=R$. Thus,
$\mathfrak{p}_{j}^{n_{j}}+\mathfrak{p}_{k}^{n_{k}}=R$. So
$\mathfrak{p}_{j}\neq\mathfrak{p}_{k}$ for all $j\neq k$ in $I$. Consequently,
$I$ is a finite set.

(2) $\Rightarrow$ (1) This follows from Theorems 3.1 and 4.1. $\Box$

###### Remark 4.4

Let $R$ be a Dedekind domain and let $I$ be a nonzero ideal of $R$. By [14,
Lemma 6.12 and Theorem 6.14],
$I=\mathfrak{p}_{1}^{n_{1}}\mathfrak{p}_{2}^{n_{2}}\dots\mathfrak{p}_{k}^{n_{k}}=\mathfrak{p}_{1}^{n_{1}}\cap\mathfrak{p}_{2}^{n_{2}}\cap\cdots\cap\mathfrak{p}_{k}^{n_{k}}$
for some distinct maximal ideals $\mathfrak{p}_{1},\dots,\mathfrak{p}_{k}$ of
$R$ and some positive integers $n_{i}$ $(1\leq i\leq k)$. By the Chinese
Remainder Theorem, we have $R/I\cong
R/\mathfrak{p}_{1}^{n_{1}}\oplus\cdots\oplus R/\mathfrak{p}_{k}^{n_{k}}$. Now
using Theorem 4.3, we conclude that every cyclic $R$-module is weakly IN (see
also Theorem 2.11).

Combining Theorem 4.3 and Remark 4.4, we obtain the following corollary.

###### Corollary 4.5

Let $R$ be a Dedekind domain and let $M$ be a nonzero finitely generated
$R$-module. Then the following are equivalent:

(1) $M$ is weakly IN;

(2) $M$ is cyclic or $M\cong I$ where $I$ is a nonzero fractional ideal of
$R$.

##  5 CS Trivial Extensions

Let $R$ be a ring and let $M$ be an $R$-module. The abelian group $R\oplus M$
can be endowed with the following product: $(a,x)(b,y)=(ab,ay+bx)$. The result
is a ring called the trivial extension of $R$ by $M$ denoted by $R\varpropto
M$. $R$ becomes a subring of $R\varpropto M$ and $M$ an ideal such that
$M^{2}=0$. If $I$ is an ideal of $R$ and $N$ is a submodule of $M$ such that
$IM\subseteq N$, then $(I,N)=\\{(a,x)\in R\varpropto M\mid a\in I,\ x\in N\\}$
is an ideal of $A=R\varpropto M$ and we have $Ann_{A}(I,N)=(Ann_{R}(I)\cap
Ann_{R}(N),Ann_{M}(I)$). In this section our main result is a characterization
of CS trivial extensions. To prove it, we need the following three lemmas.

###### Lemma 5.1

Let $R$ be a ring and let $M$ be an $R$-module such that $A=R\propto M$ is a
CS ring. Then $M$ is a weakly IN $R$-module.

Proof. Let $N$ and $L$ be two submodules of $M$ such that $N\cap L=0$. Then
$(0,N)$ and $(0,L)$ are two ideals of $A$ satisfying $(0,N)\cap(0,L)=0$. By
Proposition 2.1, $Ann_{A}(0,N)+Ann_{A}(0,L)=A$. It follows that
$(Ann_{R}(N),M)+(Ann_{R}(L),M)=A$ and hence $Ann_{R}(N)+Ann_{R}(L)=R$.
Therefore $M$ is weakly IN. $\Box$

###### Lemma 5.2

Let $R$ be a ring and let $M$ be a faithful $R$-module. Then the following are
equivalent:

(1) $A=R\propto M$ is a CS ring;

(2) $M$ is a weakly IN $R$-module;

(3) $M$ is a strongly CS $R$-module.

Proof. (1) $\Rightarrow$ (2) See Lemma 5.1.

(2) $\Leftrightarrow$ (3) This follows from Corollary 2.10.

(3) $\Rightarrow$ (1) Let $I$ be an ideal of $A$ and let $V=\\{x\in M\ |\
(0,x)\in I\\}$. Then $V$ is a submodule of $M$ and $I\cap(0,M)=(0,V)$. Since
$M$ is strongly CS, there exists $e=e^{2}\in R$ such that
$V\subseteq^{ess}eM$. First let us show that $I\subseteq(e,0)A=(eR,eM)$.
Consider an element $(a,x)$ of $I$. Then, for any element $z$ of $M$,
$(a,x)(0,z)=(0,az)\in I\cap(0,M)=(0,V)$. Thus $az\in V$ and hence $aM\subseteq
V\subseteq eM$. Consequently, $(1-e)aM=0$. But $M$ is a faithful $R$-module,
so $(1-e)a=0$ and hence $a=ea\in eR$. Moreover,
$(a,x)(1-e,0)=(a(1-e),(1-e)x)=(0,(1-e)x)\in I\cap(0,M)=(0,V)$. This gives
$(1-e)x\in V\subseteq eM$. Thus $x-ex\in eM$ and so $x\in eM$. Therefore
$I\subseteq(eR,eM)$. Now let us show that $I\subseteq^{ess}(eR,eM)$. Let
$(0,0)\neq(ea,ex)\in(eR,eM)$, where $a\in R$ and $x\in M$. If $ea\neq 0$, then
$eaM\neq 0$ since $M$ is faithful. So there exists $y\in M$ such that $eay\neq
0$. But $0\neq eay\in eM$ and $V\subseteq^{ess}eM$, so there exists $t\in R$
such that $0\neq teay\in V$. Hence
$(0,0)\neq(0,ty)(ea,ex)=(0,teay)\in(0,V)=I\cap(0,M)$. Therefore
$(0,0)\neq(0,ty)(ea,ex)\in I$. Now suppose that $ea=0$. Then $ex\neq 0$. Since
$V\subseteq^{ess}eM$, there exists $u\in R$ such that $0\neq uex\in V$. So
$(0,0)\neq(u,0)(ea,ex)=(0,uex)\in(0,V)=I\cap(0,M)$. Hence
$0\neq(u,0)(ea,ex)\in I$. It follows that $I\subseteq^{ess}(e,0)A$. Note that
$(e,0)^{2}=(e,0)$. Consequently, $A$ is a CS ring. $\Box$

###### Lemma 5.3

Let $R$ be a ring and let $M$ be a nonzero $R$-module. Let $e^{2}=e\in R$.
Then the following hold true:

(1) $eR$ is strongly CS as $R$-module if and only if $eR$ is a CS ring.

(2) Assume that $Ann_{R}(M)=eR$. Then $(1-e)M$ is a faithful $(1-e)R$-module.
Moreover, the rings $A=R\propto M$ and $eR\times((1-e)R\propto(1-e)M)$ are
isomorphic.

Proof. (1) Note the $R$-submodules and the $eR$-submodules of $eR$ are the
same.

($\Rightarrow$) Let $I$ be an ideal of the ring $eR$. Since $eR$ is a strongly
CS $R$-module, there exists $f=f^{2}\in R$ such that $I$ is an essential
$R$-submodule of $feR$. Thus $I$ is an essential $eR$-submodule of $(fe)eR$.
Moreover, $fe=(fe)^{2}\in eR$. Therefore $eR$ is a CS ring.

($\Leftarrow$) Let $I$ be an ideal of $R$ contained in $eR$. Then $I=eI$ is an
ideal of $eR$. Since $eR$ is a CS ring, there exists an idempotent $f\in eR$
such that $I$ is an essential $eR$-submodule of $feR$. It is clear that $I$ is
also an essential $R$-submodule of $feR$.

(2) It is clear that $R=eR\oplus(1-e)R$ and $M=eM\oplus(1-e)M$. By [4, Theorem
4.4], $A=R\propto M\cong(eR\propto eM)\times((1-e)R\propto(1-e)M)$ (as rings).
But $eM=0$, so the ring $A$ is isomorphic to the ring
$eR\times((1-e)R\propto(1-e)M)$. The first assertion is obvious. $\Box$

###### Theorem 5.4

The following are equivalent for a ring $R$ and an $R$-module $M$:

(1) $A=R\propto M$ is a CS ring;

(2) $M$ satisfies the following two conditions:

(a) $Ann_{R}(M)$ is a direct summand of $R$ which is a CS ring, and

(b) $M$ is a weakly IN $R$-module;

(3) $M$ satisfies the following two conditions:

(a) $Ann_{R}(M)$ is a direct summand of $R$ which is a CS ring, and

(b) $M$ is a strongly CS $R$-module;

(4) $M\oplus Ann_{R}(M)$ is a weakly IN $R$-module;

(5) $M\oplus Ann_{R}(M)$ is a strongly CS $R$-module.

Proof. It is easily seen that $\\{(e,0)\in A\mid e^{2}=e\in R\\}$ is the set
of idempotents of $A$.

(1) $\Rightarrow$ (2) (a) Suppose that $Ann_{R}(M)\neq 0$. Clearly,
$(Ann_{R}(M),0)$ is an ideal of $A$. Since $A$ is a CS ring, there exists an
idempotent $e$ of $R$ such that $(Ann_{R}(M),0)\subseteq^{ess}(e,0)A=(eR,eM)$.
Hence, $Ann_{R}(M)\subseteq eR$. We claim that $eM=0$. Suppose, on the
contrary, that $eM\neq 0$. Then $(0,eM)$ is a nonzero ideal of $A$ which is
contained in $(eR,eM)$. Thus $(0,eM)\cap(Ann_{R}(M),0)\neq 0$, a
contradiction. Therefore $eR\subseteq Ann_{R}(M)$. It follows that
$Ann_{R}(M)=eR$ is a direct summand of $R$. Moreover, $eR$ is a CS ring by
Lemmas 3.2(2) and 5.3(2).

(b) follows from Lemma 5.1.

(2) $\Rightarrow$ (3) Use Corollary 2.10.

(3) $\Rightarrow$ (5) By hypothesis, there exists $e^{2}=e\in R$ such that
$Ann_{R}(M)=eR$. Thus $Ann_{R}(M)=eAnn_{R}(M)$. Moreover, since
$R=eR\oplus(1-e)R$, we have $M=(1-e)M$. Also, $M$ and $Ann_{R}(M)$ are
strongly CS $R$-modules by Lemma 5.3(1). So, by Theorem 3.3, $M\oplus
Ann_{R}(M)$ is a strongly CS $R$-module.

(5) $\Rightarrow$ (4) This follows from Theorem 2.8.

(4) $\Rightarrow$ (1) By Theorem 3.1, $Ann_{R}(M)+Ann_{R}(Ann_{R}(M))=R$. Then
there exist $a\in Ann_{R}(M)$ and $b\in Ann_{R}(Ann_{R}(M))$ such that
$a+b=1$. Let $c\in Ann_{R}(M)\cap Ann_{R}(Ann_{R}(M))$. Thus $c=ca+cb=0$ and
hence $Ann_{R}(M)\cap Ann_{R}(Ann_{R}(M))=0$. It follows that $Ann_{R}(M)$ is
a direct summand of $R$. So $Ann_{R}(M)=eR$ for some idempotent $e$ of $R$.
Note that $eR$ is a weakly IN $R$-module by Proposition 2.4. Moreover, the
$R$-submodules and the $eR$-submodules of $eR$ are exactly the same. Also, we
have $Ann_{eR}(L)=eAnn_{R}(L)$ for any $R$-submodule $L$ of $eR$. We thus
deduce that $eR$ is a CS ring by Proposition 2.1. In addition, using
Proposition 2.4 again, we see that $(1-e)M$ is a weakly IN $(1-e)R$-module.
Note that $(1-e)M$ is a faithful $(1-e)R$-module (see Lemma 5.3(2)). Then
$(1-e)R\propto(1-e)M$ is a CS ring by Lemma 5.2. Since the rings $R\propto M$
and $eR\times((1-e)R\propto(1-e)M)$ are isomorphic (see Lemma 5.3(2)), it
follows that $R\propto M$ is a CS ring by Lemma 3.2(2). $\Box$

###### Remark 5.5

Let $R$ be a ring and let $M$ be an $R$-module such that $R\propto M$ is a CS
ring. By Theorem 5.4, $M$ is a strongly CS $R$-module. However, the ring $R$
need not be a CS ring, in general, as it is shown in the following example.
Let $R$ be a local ring which not uniform (for example, we can take
$R=K\propto(K\oplus K)$, where $K$ is a field) and let ${\mathfrak{m}}$ be the
maximal ideal of $R$. Consider the $R$-module $M=E(R/{\mathfrak{m}})$ and the
ring $A=R\propto M$. By [14, Corollary 2 of Proposition 2.26], $M$ is a
faithful $R$-module. Moreover, since $M$ is a uniform $R$-module, $M$ is
strongly CS. Thus $A$ is a CS ring by Lemma 5.2 (indeed, $A$ is uniform).
Since $R$ is local, $R$ is indecomposable and so $R$ cannot be a CS ring since
it is not uniform (see Remark 2.3(3)). However, when the $R$-module $M$ is
flat, then the condition $R\propto M$ is CS forces $R$ to become CS as it is
shown in the following result.

###### Corollary 5.6

Let $R$ be a ring and let $M$ be a flat $R$-module. Then the following are
equivalent:

(1) $A=R\propto M$ is a CS ring;

(2) $R$ and $M$ are weakly IN $R$-modules and $Ann_{R}(M)$ is a direct summand
of $R$;

(3) $R$ and $M$ are strongly CS $R$-modules and $Ann_{R}(M)$ is a direct
summand of $R$.

Proof. (1) $\Rightarrow$ (3) By Theorem 5.4, $M$ is a strongly CS $R$-module
and $Ann_{R}(M)$ is a direct summand of $R$. To show that $R$ is CS, take two
ideals $I$ and $J$ of $R$ such that $I\cap J=0$. Since $M$ is a flat
$R$-module, we have $IM\cap JM=(I\cap J)M$ by [16, Proposition 8.5]. Thus
$IM\cap JM=0$. Consider the ideals $I^{\prime}=(I,IM)$ and $J^{\prime}=(J,JM)$
of $A$. Since $I^{\prime}\cap J^{\prime}=0$ and $A$ is CS, we have
$Ann_{A}(I^{\prime})+Ann_{A}(J^{\prime})=A$ (see Proposition 2.1). It follows
that

$(Ann_{R}(I),Ann_{M}(I))+(Ann_{R}(J),Ann_{M}(J))=A.$

So $Ann_{R}(I)+Ann_{R}(J)=R$. Using again Proposition 2.1, we conclude that
$R$ is a CS ring. Hence $R$ is strongly CS.

(3) $\Rightarrow$ (2) This follows from Theorem 2.8.

(2) $\Rightarrow$ (1) Since $R$ is weakly IN, $R$ is a strongly CS ring by
Remark 2.3(1). As $Ann_{R}(M)$ is a direct summand of $R$, it follows that
$Ann_{R}(M)$ is a CS ring (Proposition 2.4 and Lemma 5.3(1)). Now use Theorem
5.4. $\Box$

###### Corollary 5.7

Let $H$ be a faithful ideal of $R$ and let $T(R)$ be the total ring of
fractions of $R$. Then the following are equivalent:

(1) $R\propto H$ is a CS ring;

(2) $R\propto T(R)$ is a CS ring;

(3) $R$ is a CS ring.

Proof. Note that $T(R)$ is a flat $R$-module (see [5, Corollary 3.6]).

(1) $\Rightarrow$ (3) Let $I$ and $J$ be two ideals of $R$ such that $I\cap
J=0$. So $IH\cap JH=0$. But $IH$ and $JH$ are two $R$-submodules of the
faithful $R$-module $H$. Moreover, $H$ is weakly IN $R$-module by Lemma 5.2.
Therefore $Ann_{R}(IH)+Ann_{R}(JH)=R$. But $Ann_{R}(IH)=Ann_{R}(I)$ and
$Ann_{R}(JH)=Ann_{R}(J)$ as $H$ is faithful. Then $Ann_{R}(I)+Ann_{R}(J)=R$.
By Proposition 2.1, we see that $R$ is a CS ring.

(3) $\Rightarrow$ (2) It is clear that $Ann_{R}(T(R))=0$. To prove that
$R\propto T(R)$ is a CS ring, we only need to show that $T(R)$ is a strongly
CS $R$-module (see Corollary 5.6). Let $S$ denote the set of all non-zero-
divisors in $R$. Then $T(R)=S^{-1}R$. Let $I^{\prime}$ be an ideal of $T(R)$.
Then $I^{\prime}=S^{-1}I$ for some ideal $I$ of $R$. But $R$ is a CS ring, so
there exists $e^{2}=e\in R$ such that $I\subseteq^{ess}eR$. It is easily seen
that $I^{\prime}=S^{-1}I\subseteq^{ess}eT(R)$. Consequently, $T(R)$ is a
strongly CS $R$-module.

(2) $\Rightarrow$ (1) By Corollary 5.6, $R$ is a weakly IN $R$-module. Since
$H$ is an $R$-submodule of $R$, $H$ is also a weakly IN $R$-module. According
to Lemma 5.2, it follows that $R\propto H$ is a CS ring. $\Box$

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