FEW REMARKS ON ESSENTIAL SUBMODULES

Let $M=\bigoplus_{\alpha\in\Lambda}N_{\alpha}$, where each $N_{\alpha}$ is a simple submodule of $M$. If a submodule $E$ of $M$ intersects every other non-trivial submodule of $M$ non-trivially, then $N_{\alpha}\subseteq E$ for all $\alpha\in\Lambda$. But this implies $\bigoplus_{\alpha\in\Lambda}N_{\alpha}\subseteq E$, that is, $M\subseteq E$.

Let $\{v_{\alpha}\}_{\alpha\in\Lambda}$ be a basis for the $\mathbbm{k}$-vector space $V$. Then we have $V=\bigoplus_{\alpha\in\Lambda}V_{\alpha}$, where $V_{\alpha}$ is the subspace generated by $v_{\alpha}$. As each $V_{\alpha}$ is a simple $\mathbbm{k}$-subspace of $V$, we can use the first part of this proposition to conclude that $V$ does not contain any proper essential subspace. ∎

Equivalency of (a), (c) and (d) are found in [TL (1999)]. Here, we show the equivalence of (a) and (b), that is, $\mathrm{(a)}\Leftrightarrow\mathrm{(b)}$.

$\mathrm{(a)}\Rightarrow\mathrm{(b)}:$ Suppose $M$ has a proper essential submodule. Then from the definition, it trivially follows that the proper essential submodule can not be the direct summand.

$\mathrm{(b)}\Rightarrow\mathrm{(a)}:$ Conversely let $N$ be a proper submodule of $M$. Then $N$ is not proper essential. There exists a submodule $T$ such that $N\cap T=\{0\}$. Choose a maximal such $T$, then $N\bigoplus T$ is either proper essential in $M$ or $N\bigoplus T=M$. But as per hypothesis, $N\bigoplus T=M$, so the result follows. ∎

Let $M=\bigoplus_{\omega\in\Lambda}M_{\omega}$ and $\omega_{0}\in\Lambda$ be such that $M_{\omega_{0}}$ has a proper essential submodule $E_{\omega_{0}}$. Let $X=\bigoplus_{\omega\in\Lambda}X_{\omega}$, where $X_{\omega_{0}}=E_{\omega_{0}}$ and $X_{\omega}=M_{\omega}$ if $\omega\neq\omega_{0}$. We claim that $X$ is a proper essential submodule of $M$. Let $N$ be any nontrivial submodule of $M$ and let $(n_{\omega})_{\omega\in\Lambda}$ be a nonzero element of $N$. Clearly, $(n_{\omega})_{\omega\in\Lambda}\in X$ if $n_{\omega_{0}}=0_{M_{\omega_{0}}}$. If $n_{\omega_{0}}\neq 0_{M_{\omega_{0}}}$, consider the cyclic module $\langle n_{\omega_{0}}\rangle$. As $E_{\omega_{0}}\cap\langle n_{\omega_{0}}\rangle\neq\{0_{M_{\omega_{0}}}\}$, we conclude that $X\cap N\neq\{0_{M}\}$.

For the converse, assume that $M$ has a proper essential submodule $E$ but $M_{\omega}$ does not have a proper essential submodule for all $\omega\in\Lambda$. Observe that, for all $\omega\in\Lambda$, we have either $E\cap M_{\omega}$ is proper essential in $M_{\omega}$ or $E\cap M_{\omega}=M_{\omega}$. As $M_{\omega}$ does not have any proper essential submodule, it must be that $E\cap M_{\omega}=M_{\omega}$ for all $\omega\in\Lambda$, that is, $M\subseteq E$, a contradiction. ∎

Since $R(A)$ is free on the subset $A$, we have $R(A)\simeq\bigoplus_{A}R$. The proof is now completed using theorem (2.1). ∎

Let $\mathcal{M}$ be a maximal ideal of $R$ that contains $\mathrm{Ann}_{R}(m)$ and $I$ be any ideal of $R$ containing $\mathrm{Ann}_{R}(m)$. If for some $\alpha\in\mathcal{M}\setminus\mathrm{Ann}_{R}(m)$ and for some $\beta\in I\setminus\mathrm{Ann}_{R}(m)$ we have $\alpha\beta\notin\mathrm{Ann}_{R}(m)$, then we see that $\mathcal{M}/\mathrm{Ann}_{R}(m)$ is proper essential in $R/\mathrm{Ann}_{R}(m)$. As $R/\mathrm{Ann}_{R}(m)\simeq\langle m\rangle$, we conclude that $\langle m\rangle$ has a proper essential submodule.

Conversely, suppose $\langle m\rangle$ has a proper essential submodule $E$. Since we have $\langle m\rangle\simeq R/\mathrm{Ann}_{R}(m)$, there exists some ideal $J$ of $R$ containing $\mathrm{Ann}_{R}(m)$ such that $J/\mathrm{Ann}_{R}(m)\simeq E$. Put $J$ in some maximal ideal $\mathcal{M}$. As $J/\mathrm{Ann}_{R}(m)$ is proper essential in $R/\mathrm{Ann}_{R}(m)$, for any ideal $I$ of $R$ containing $\mathrm{Ann}_{R}(m)$, we have $J/\mathrm{Ann}_{R}(m)\cap I/\mathrm{Ann}_{R}(m)\neq\{0_{\langle m\rangle}\}$. As $J$ is contained in $\mathcal{M}$, we get that $\mathcal{M}/\mathrm{Ann}_{R}(m)\cap I/\mathrm{Ann}_{R}(m)\neq\{0_{\langle m\rangle}\}$.

Let $\mathrm{Ann}_{R}(m)=\langle a\rangle$ and let $a=up_{1}p_{2}\cdots p_{n}$ where $p_{i}$’s are distinct primes and $u$ is a unit. The only maximal ideals that contain $\langle a\rangle$ are the ideals $\langle p_{i}\rangle$ where $1\leq i\leq n$. If $\alpha\in\langle p_{i}\rangle\setminus\langle a\rangle$ and $\beta\in\langle q_{i}\rangle\setminus\langle a\rangle$ where $q_{i}=p_{1}p_{2}\cdots p_{i-1}p_{i+1}\cdots p_{n}$, we see that $\alpha\beta\in\langle a\rangle$. Now, the first part of this lemma shows that $\langle m\rangle$ does not have a proper essential submodule.

Assume now that $a=up_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{n}^{\alpha_{n}}$, where $p_{i}$’s are distinct primes, $u$ is a unit and $\alpha_{i}>1$ for some $i$. It is now evident that the maximal ideal $\langle p_{i}\rangle$ of $R$ satisfies the property of the first of this lemma. ∎

The fundamental theorem of finitely-generated modules over principal ideal domain yields the invariant-factor decomposition $M=R^{n}\bigoplus R/\langle a_{{1}}\rangle\bigoplus\cdots\bigoplus R/\langle a_{{m}}\rangle$. The proof is now completed using theorems (2.1), (2.2), and lemma (2.3) ∎

At first, assume that $R$-module $M$ has a proper essential submodule $E$. Now, let $N$ be any submodule of $R/\mathrm{Ann}_{R}(M)$-module $M$. Consider the cyclic $R/\mathrm{Ann}_{R}(M)$-module $\langle n\rangle$, where $n\in N$ be any element. As, $\langle n\rangle\cap E\neq\{0_{M}\}$, there exists $r\in R$ such that $rn\neq 0$ and $rn\in\langle n\rangle\cap E$. Then $(r+\mathrm{Ann}_{R}(M))n\in\langle n\rangle\cap E$ (as $R/\mathrm{Ann}_{R}(M)$-module). Hence, this shows that $R/\mathrm{Ann}_{R}(M)$-module $M$ has a proper essential submodule $E$.

Conversely, assume that $R/\mathrm{Ann}_{R}(M)$-module $M$ has a proper essential submodule $E$. Let $N$ be any submodule of $M$. Then, $\langle n\rangle\cap E\neq\{0_{M}\}$ for $n\in N$ with $n\neq 0_{M}$. This implies, for some $(r+\mathrm{Ann}_{R}(M))\in R/\mathrm{Ann}_{R}(M)$ with $r+\mathrm{Ann}_{R}(M)\neq\mathrm{Ann}_{R}(M)$, $(r+\mathrm{Ann}_{R}(M))n\in E$, that is, $rn\in E$. Thus, $E$ is proper essential in $M$.

Now, applying proposition (1.2) and the first part of this lemma, we get that $\mathrm{Ann}_{R}(M)$ can not be a maximal ideal. ∎

Let $\mathrm{Ann}_{R}(M)=\langle a\rangle$ and $a=up_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{n}^{\alpha_{n}}$, where $p_{i}$’s are distinct primes and $u$ is a unit. For $1\leq i\leq n$, if we set $q_{i}=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{i-1}^{\alpha_{i-1}}p_{i+1}^{\alpha_{i+1}}\cdots p_{n}^{\alpha_{n}}$, then the homomorphism $\varphi_{i}:M\rightarrow q_{i}M$ shows that $M/\mathrm{Ker}(\varphi_{i})\simeq q_{i}M$. Since we have $\mathrm{Ker}(\varphi_{i})=\langle p_{i}^{\alpha_{i}}\rangle M$ and $q_{i}M=\mathrm{Ann}_{M}(p_{i}^{\alpha_{i}})$, we conclude that $M/\langle p_{i}^{\alpha_{i}}\rangle M\simeq\mathrm{Ann}_{M}(p_{i}^{\alpha_{i}})$. As the ideals $\langle p_{1}^{\alpha_{1}}\rangle$, $\langle p_{2}^{\alpha_{2}}\rangle$, …, $\langle p_{n}^{\alpha_{n}}\rangle$ are pairwise comaximal, the chinese remainder theorem yields $M\simeq\bigoplus_{i=1}^{n}\mathrm{Ann}_{M}(p_{i}^{\alpha_{i}})$. The result now follows from theorem (2.1). ∎

Let $m\in M$ be such that $rm=0$ for some $r\in R$. Let $r=up_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{n}^{\alpha_{n}}$ be the prime factorisation of $r$ into primes, where $u$ is some unit. Consider the cyclic submodule $\langle m\rangle$. As the submodule $\langle m\rangle$ is annihilated by the ideal $\langle r\rangle$, we can apply lemma (2.6) to conclude that $\langle m\rangle=\bigoplus_{i=1}^{n}\mathrm{Ann}_{\langle m\rangle}(p_{i}^{\alpha_{i}})$. Thus, $M=\sum_{{p{\text{-}\mathrm{prime}}}}\mathrm{Ann}^{\ast}_{M}(p)$. Now choose finite number of primes $q_{1},q_{2},\ldots,q_{l}$ in $R$ and consider the submodules $\mathrm{Ann}^{\ast}_{M}(q_{1}),\mathrm{Ann}^{\ast}_{M}(q_{2}),\ldots,\mathrm{Ann}^{\ast}_{M}(q_{l})$. If $m\in\mathrm{Ann}^{\ast}_{M}(q_{j})\bigcap\sum_{i\neq j}\mathrm{Ann}^{\ast}_{M}(q_{i})$ for some fixed $j$ with $1\leq j\leq l$, then we have $q_{j}^{s_{j}}m=0$ for some $s_{j}\in\mathbb{N}$ and $m=m_{1}+\cdots+m_{j-1}+m_{j+1}+\cdots+m_{l}$. Also, for all $1\leq i\leq l$ with $i\neq j$, we have $q_{i}^{s_{i}}m_{i}=0$ for some $s_{i}\in\mathbb{N}$. As $R$ is a principal ideal domain, there exists $x,y\in R$ such that $q_{j}^{s_{j}}x+\xi y=1$, where $\xi=q_{1}^{s_{1}}q_{2}^{s_{2}}\cdots q_{j-1}^{s_{j-1}}q_{j+1}^{s_{j+1}}\cdots q_{l}^{s_{l}}$. But this implies $m=q_{j}^{s_{j}}xm+\xi y(m_{1}+\cdots+m_{j-1}+m_{j+1}+\cdots+m_{l})$, that is, $m=0$. Hence, we have that $M=\bigoplus_{{p{\text{-}\mathrm{prime}}}}\mathrm{Ann}^{\ast}_{M}(p)$. Now, if $M$ has a proper essential submodule $E$, then for some prime $p$, $E\cap\mathrm{Ann}^{\ast}_{M}(p)$ is proper essential in $\mathrm{Ann}^{\ast}_{M}(p)$ (see theorem (2.1)). If $\mathrm{Ann}_{M}(p^{i})=\mathrm{Ann}_{M}(p^{j})$ for all positive integers $i,j$, then observe that $\mathrm{Ann}_{R}(\mathrm{Ann}_{M}(p))=\langle p\rangle$. Thus, using lemma (2.5), we conclude $R/\langle p\rangle$-vector space $\mathrm{Ann}_{M}(p)$ has a proper essential subspace $E\cap\mathrm{Ann}_{M}(p)$. Again, applying the second part of lemma (2.5), we get a contradiction.

Conversely, assume that $\mathrm{Ann}_{M}(p^{i})\subsetneq\mathrm{Ann}_{M}(p^{j})$ for some prime $p$ and for some positive integers $i,j$ with $i<j$. We claim that $\mathrm{Ann}_{M}(p^{i})$ is proper essential in $\mathrm{Ann}_{M}(p)$. Let $N$ be any submodule of $\mathrm{Ann}_{M}(p)$ and let $n\in N$. If $n\notin\mathrm{Ann}_{M}(p^{i})$, then suppose $l$ be the smallest positive integer such that $n\in\mathrm{Ann}_{M}(p^{l})$. Observe now that $p^{l-i}n\in\mathrm{Ann}_{M}(p^{i})$ and $p^{l-i}n\neq 0$. This implies $p^{l-i}n\in N\cap\mathrm{Ann}_{M}(p^{i})$. Now apply theorem (2.1) to conclude that $M$ has a proper essential submodule. ∎

Let $\frac{r}{d}\in D^{-1}R$ be such that $\frac{r}{d}\neq 0_{M}$. Now, consider the collection $\mathcal{C}=\{N\leq M\mid R\subseteq N,\frac{r}{d}\notin N\}$ of submodules of $M$. Clearly $\mathcal{C}$ is a partially ordered set under inclusion. By Zorn’s Lemma, it has a maximal element $\mathbbm{E}$. We claim that $\mathbbm{E}$ is a proper essential submodule of $M$. If not, there exists a non-trivial submodule $K$ that intersects $\mathbbm{E}$ trivially, that is, $\mathbbm{E}\cap K=\{0_{M}\}$. Observe that $K$ must not contain any nonzero element in $D^{-1}R$, because, if $\frac{r^{\prime}}{d^{\prime}}\in D^{-1}R$ is such that $\frac{r^{\prime}}{d^{\prime}}\neq 0_{M}$ and $\frac{r^{\prime}}{d^{\prime}}\in K$, then $r^{\prime}\in K$, contrary to our assumption that $\mathbbm{E}\cap K=\{0_{M}\}$. Let $\alpha\in K$ be such that $\alpha\neq 0_{M}$. Clearly, we have $\mathbbm{E}\cap\langle\alpha\rangle=\{0_{M}\}$. Consider the submodule $\mathbbm{E}\oplus\langle\alpha\rangle$. By maximality of $\mathbb{E}$, it must be that $\frac{r}{d}\in\mathbbm{E}\oplus\langle\alpha\rangle$. This implies $r/d+\alpha\in\mathbbm{E}\oplus\langle\alpha\rangle$. If $\frac{r}{d}+\alpha\in\mathbbm{E}$, then $r+d\alpha\in\mathbbm{E}$. But this gives, $d\alpha\in\mathbbm{E}$, a contradiction. Also, if $\frac{r}{d}+\alpha\in\langle\alpha\rangle$, then $r\in\langle\alpha\rangle$, a contradiction. Thus, we can write $\frac{r}{d}+\alpha=e+r^{\ast}\alpha$, where $e\in\mathbbm{E}\setminus\{0_{M}\}$ and $r^{\ast}\in R\setminus\{0_{R}\}$. But this implies $r+d\alpha=de+dr^{\ast}\alpha$. Hence, we get that $r-de=d(r^{\ast}-1)\alpha$. Observe that $r-de\in\mathbbm{E}$ whereas $d(r^{\ast}-1)\alpha\in\langle\alpha\rangle$. Hence, it must be that $r-de=0_{M}$ and $d(r^{\ast}-1)=0_{R}$. As $d\neq 0_{R}$, it must be that $r^{\ast}=1$. Plugging the value of $r^{\ast}$ in the equation $\frac{r}{d}+\alpha=e+r^{\ast}\alpha$ yields $\frac{r}{d}=e$. But this is absurd because $\mathbbm{E}$ being a member of $\mathcal{C}$ cannot contain $\frac{r}{d}$. ∎

If $N=\{0_{M}\}$, there is nothing to show. Assume $N\neq\{0_{M}\}$. Let $L$ be a nontrivial proper submodule of $M$ and $u\in(L+N)/N\cap E/N$. Clearly, $u=l+N=e+N$ for some $l\in L\setminus N$ and $e\in E\setminus N$. But this shows that $l\in E$. Hence, we get that $E$ is proper essential in $M$.

If $N=\{0_{M}\}$, there is nothing to show. Therefore, assume $N\neq\{0_{M}\}$. Let $L/N$ be a non-trivial proper submodule of $M/N$. Let $l\in L\setminus N$ be such $rl\in E$, for some nonzero $r\in R$ and $rl\neq 0_{M}$. As $rl\notin N$, we conclude that $rl+N\in(E+N)/N\cap L/N$, that is, $(E+N)/N$ is proper essential in $M/N$. ∎

Consider the quotient module $M/\mathrm{Tor}(M)$. As $M/\mathrm{Tor}(M)$ is torsion-free, so applying theorem (2.8) we get that $M/\mathrm{Tor}(M)$ has a proper essential submodule. Again, applying lemma (2.9), we see that $M$ has a proper essential submodule. ∎

Let $M$ and $N$ are the free $R$ modules on the sets of $n$ and $m$ elements respectively. Clearly, we have $M\simeq R^{n}$ and $N\simeq R^{m}$. As $\mathrm{Hom}_{R}(R^{n},R^{m})\simeq\mathrm{Hom}_{R}(M,N)$ and $\mathrm{Hom}_{R}(M,N)\simeq\mathcal{M}_{m\times n}(R)$, we get that $\mathrm{Hom}_{R}(R^{n},R^{m})\simeq\mathcal{M}_{m\times n}(R)$. Since we have $\mathrm{Hom}_{R}(R^{n},R^{m})\simeq\bigoplus_{i=1}^{n}\bigoplus_{j=1}^{m}\mathrm{Hom}_{R}(R,R)$, we obtain $\mathcal{M}_{m\times n}(R)\simeq R^{mn}$. Now, we can apply theorem (2.2) to obtain the desired result. ∎

Let $M$ be an Artinian module. If $M$ is simple there is nothing to show. Therefore assume $M$ is not simple. Let $N$ be any submodule of $M$. If we consider any strictly descending chain $C_{N}:N_{0}=N\supsetneq N_{1}\supseteq N_{2}\supsetneq\cdots\supsetneq N_{s}$ of submodules of $N$ that terminates with the nontrivial proper submodule $N_{s}$, then $N_{s}$ is a simple module contained in $N$. Because, otherwise one can enlarge the chain $C_{N}$.

Now assume $M$ is a locally finite module. Again, if $M$ is simple then $M$ must be finite and there is nothing to show. So, assume $M$ is not simple. Let $N$ be any submodule of $M$ and let $n\in N$. Let $\mathcal{G}$ be the collection of all submodules of $\langle n\rangle$. As $\langle n\rangle$ is finite, we can choose $S\in\mathcal{G}$ such that $|S|$ is least. Clearly, $S$ can not contain any nontrivial proper submodule $L$, because then $L$ being a member of $\mathcal{G}$ contradicts the minimality of $|S|$. This shows that $S$ is a simple submodule contained in $N$.

Finally, assume that the length of $M$ is finite. If $l(M)=1$, then $M$ is simple and there is nothing to show. So, assume $l(M)>1$. Let $N$ be any submodule of $M$. Let $\mathcal{H}$ be the collection of all nontrivial proper submodule of $N$. Choose $K\in\mathcal{H}$ such that $l(K)$ is minimal. If $K$ would not be simple and $P$ is a nontrivial proper submodule of $K$, then $l(P)<l(K)$ contradicts the minimality of $l(K)$. Thus, $K$ is a simple submodule contained in $N$. ∎

Let $\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}$, and $\mathcal{C}_{4}$ denote the class of all $\mathrm{SM}$ modules, modules of finite lengths, locally finite modules, and Artinian modules respectively. Consider the free $\mathbb{Z}_{p^{n}}$-module $M=\bigoplus_{\mathbb{N}}\mathbb{Z}_{p^{n}}$ on the set $\mathbb{N}$, where $p$ is some fixed prime and $n$ is a natural number such that $n>1$. Let $N$ be any nontrivial proper submodule of $M$ and let $n\in N$. Observe that $|\langle n\rangle|\leq p^{n}$. If we choose a submodule $L$ of $\langle n\rangle$ such that $|L|=p$, then clearly $L$ is simple and contained in $N$. This shows that $M$ is an $\mathrm{SM}$ module.

As $l(\mathbb{Z}_{p^{n}})=n$, therefore the divergence of the series $\sum_{i=1}^{\infty}n$ shows that the length of the free $\mathbb{Z}_{p^{n}}$-module $\bigoplus_{\mathbb{N}}\mathbb{Z}_{p^{n}}$ is not finite. Hence, we have that $\mathcal{C}_{2}\subsetneq\mathcal{C}_{1}$.

For each natural number $i$, we define the submodules $L_{i}=\bigoplus_{\mathbb{N}}F_{j}$, where $F_{j}=\mathbb{Z}_{p^{n}}$ if $j>i$, otherwise, $F_{j}=\{0_{\mathbb{Z}_{p^{n}}}\}$. Now, observe that the strictly descending chain of submodules $L_{0}=M\supsetneq L_{1}\supsetneq L_{2}\supsetneq\cdots L_{s}\supsetneq\cdots$ do not terminate. Thus, $\mathcal{C}_{4}\subsetneq\mathcal{C}_{1}$.

Now, consider an infinite $R$-module $W$ of finite length $n$ (for example finite dimensional vector space). If $W_{0}=W\supsetneq W_{1}\supsetneq W_{2}\supseteq\cdots\supsetneq W_{n}=\{0_{W}\}$, then observe that $W=R(A)$, where $A=\{w_{i}\mid w_{i}\in W_{i}\setminus W_{i+1},0\leq i\leq n-2\}$ and $R(A)$ is the module generated by the subset $A$. As $W\notin\mathcal{C}_{3}$, therefore, we get that $\mathcal{C}_{3}\subsetneq\mathcal{C}_{1}$. ∎

Assume first that $\mathrm{Soc}(M)$ is proper essential in $M$. Then, clearly we have $\mathrm{Soc}(M)\subsetneq M$. Hence, applying theorem (1.4), we conclude that $M$ is not semisimple.

Conversely, assume that $M$ is not semisimple. Again, applying theorem (1.4), we see that $M$ contains a proper essential submodule. Now, let $N$ be any submodule of $M$. As $N$ contains a simple submodule, we get that $\mathrm{Soc}(M)\cap N\neq\{0_{M}\}$. Further, as $\mathrm{Soc}(M)$ is contained in every proper essential submodule of $M$, we conclude that $\mathrm{Soc}(M)$ is proper essential in $M$. ∎

$\mathrm{(a)}:$ $\mathrm{Cl}_{M}(\{0_{M}\})$ is the set of all those elements $m$ in $M$ such that $rm=0$ for some nonzero $r\in R$. As $M$ is torsion-free, we see that $\mathrm{Cl}_{M}(\{0_{M}\})=\{0_{M}\}$. If $m_{1},m_{2}$ are in $\mathrm{Cl}_{M}(N)$, then $r_{1}m_{1}\in N$ and $r_{2}m_{2}\in N$ for some nonzero elements $r_{1},r_{2}$ in $R$. Since $r_{1}r_{2}(m_{1}-m_{2})\in N$ and $r_{1}r_{2}\neq 0$, we conclude that $(m_{1}-m_{2})\in\mathrm{Cl}_{M}(N)$. Now, let $m\in\mathrm{Cl}_{M}(N)$ and $r\in R$ be such that $r\neq 0$ and $rm\in N$. For any $s\in R$, we have $r(sm)\in N$. Thus, $sm\in\mathrm{Cl}_{M}(N)$. Hence, $\mathrm{Cl}_{M}(N)$ is a submodule of $M$.

$\mathrm{(b)}:$ For any element $n\in N$, we have $1n\in N$, so $N\subseteq\mathrm{Cl}_{M}(N)$. It is evident that $\mathrm{Cl}_{M}(N)\subseteq\mathrm{Cl}_{M}(\mathrm{Cl}_{M}(N))$. Now, suppose $m\in\mathrm{Cl}_{M}(\mathrm{Cl}_{M}(N))$. This implies, for some nonzero $r\in R$, $rm$ lie in $\mathrm{Cl}_{M}(N)$. Again, for some nonzero $s\in R$, $(sr)m\in N$. As, $sr\neq 0$, we conclude that $m\in\mathrm{Cl}_{M}(N)$. Hence, $\mathrm{Cl}_{M}(\mathrm{Cl}_{M}(N))\subseteq\mathrm{Cl}_{M}(N)$.

$\mathrm{(c)}:$ Let $m\in\mathrm{Cl}_{M}(N_{1})$. Then, $rm\in N_{1}$ for some nonzero $r\in R$. As $rm\in N_{2}$ as well, we conclude that $m\in\mathrm{Cl}_{M}(N_{2})$, that is, $\mathrm{Cl}_{M}(N_{1})\subseteq\mathrm{Cl}_{M}(N_{2})$.

$\mathrm{(d)}:$ Let $s\in f(\mathrm{Cl}_{M}(T))$. Then, there exists $m\in\mathrm{Cl}_{M}(T)$ such that $f(m)=s$. Also, there exists $r\in R$ such that $r\neq 0$ and $rm\in T$. This shows that $rf(m)\in f(T)$, that is, $s\in\mathrm{Cl}_{S}(f(T))$. Hence, we get that $f(\mathrm{Cl}_{M}(T))\leq\mathrm{Cl}_{S}(f(T))$. ∎

It is enough to show that $\mathrm{Cl}_{D^{-1}M}(L\bigoplus S)=\mathrm{Cl}_{D^{-1}M}(L)\bigoplus\mathrm{Cl}_{D^{-1}M}(S)$ for any two submodules $L,S$ of $M$. Observe that if $\frac{m}{d}\in\mathrm{Cl}_{D^{-1}M}(L)\cap\mathrm{Cl}_{D^{-1}M}(S)$, then $\frac{r_{1}m}{d}\in L$ and $\frac{r_{2}m}{d}\in S$ for some nonzero $r_{1},r_{2}\in R$. As $(r_{1}r_{2})\frac{m}{d}\in L\cap S$, we see that $\frac{m}{d}=0_{M}$. Now, suppose $(\frac{m_{1}}{d_{1}}+\frac{m_{2}}{d_{2}})\in\mathrm{Cl}_{D^{-1}M}(L)\bigoplus\mathrm{Cl}_{D^{-1}M}(S)$. We have then $\frac{r_{1}m_{1}}{d_{1}}\in L$ and $\frac{r_{2}m_{2}}{d_{2}}\in S$ for some $r_{1},r_{2}\in R\setminus\{0\}$. As $(r_{1}r_{2})(\frac{m_{1}}{d_{1}}+\frac{m_{2}}{d_{2}})\in L\bigoplus S$, we see that $(\frac{m_{1}}{d_{1}}+\frac{m_{2}}{d_{2}})\in\mathrm{Cl}_{D^{-1}M}(L\bigoplus S)$. Conversely, let $\frac{m}{d}\in\mathrm{Cl}_{D^{-1}M}(L\bigoplus S)$. Therefore, for some nonzero $r\in R$, we have $\frac{rm}{d}\in L\bigoplus S$. Let $\frac{rm}{d}=l+s$. Then, $r(\frac{m}{d}-\frac{l}{r})=s$, that is, $(\frac{m}{d}-\frac{l}{r})\in\mathrm{Cl}_{D^{-1}M}(S)$. Also, observe that $\frac{l}{r}\in\mathrm{Cl}_{D^{-1}M}(L)$. Hence, we get that $\frac{m}{d}=\frac{l}{r}+(\frac{m}{d}-\frac{l}{r})$, that is, $\frac{m}{d}\in\mathrm{Cl}_{D^{-1}M}(L)\bigoplus\mathrm{Cl}_{D^{-1}M}(S)$. ∎