The Rational Hull of Modules

Abstract

In this paper, we provide several new characterizations of the maximal right ring of quotients of a ring by using the relatively dense property. As a ring is embedded in its maximal right ring of quotients, we show that the endomorphism ring of a module is embedded into that of the rational hull of the module. In particular, we obtain new characterizations of rationally complete modules. The equivalent condition for the rational hull of the direct sum of modules to be the direct sum of the rational hulls of those modules under certain assumption is presented. For a right H-module M where H is a right ring of quotients of a ring R, we provide a sufficient condition under which $\text {End}_R(M)=\text {End}_H(M)$. Also, we give a condition for the maximal right ring of quotients of the endomorphism ring of a module to be the endomorphism ring of the rational hull of the module.

1 Introduction

The theory of rings of quotients has its origin in the work of Ø. Ore [11] and K. Asano [2] on the construction of the total ring of fractions, in the 1930’s and 40’s. But the subject did not really develop until the end of the 1950’s, when a number of important papers appeared (by R.E. Johnson [6], Y. Utumi [15], A.W. Goldie [5], J. Lambek [8] et al). In particular, Johnson(1951), Utumi(1956), and Findlay & Lambek(1958) have studied the maximal right ring of quotients of a ring which is an extended ring of the base ring. For example, the maximal right ring of quotients of $\mathbb {Z}$ is $\mathbb {Q}$, which is also the injective hull of $\mathbb {Z}$. Here, $\mathbb {Z}$ stands for the ring of integers and $\mathbb {Q}$ is the field of rational numbers. For a commutative ring R, its classical right ring of quotients coincides with its total quotient ring as the maximal right ring of quotients of R coincides with the complete ring of quotients of R.

As we know, the study of the rational hull of a module is the same as that of the maximal right ring of quotients in a different way. Also, like every module has the injective hull, it is known that every module has the rational hull in [4, Theorem 2.6]. Now, we recall the definition of the rational hull of a module and present its well-known results, briefly. Let M be a right R-module and $T=\text {End}_R(E(M))$. Put Then ${\widetilde{E}}(M)$ is the unique maximal rational extension of M. We call it the rational hull of M. Also, it is known that $\textbf{r}_{E(M)}(J(T)) \le \textbf{r}_{E(M)}\left( \textbf{l}_T(M)\right) ={\widetilde{E}}(M)$ because $\textbf{l}_T(M)\subseteq J(T)$ where $J(T) = \{ \alpha \in T\, |\, \text {Ker}\alpha \le ^\text {ess} E(M)\}$ is the Jacobson radical of the ring T. Note that the maximal right ring of quotients of R is $Q(R)=\textbf{r}_{E(R)}(\textbf{l}_H(R))$ where $H=\text {End}_R(E(R))$ (see [8, Proposition 2]).

After the necessary background history, notations, and results in this section and the next section, we provide several characterizations of the rational hull of a module in Sect. 3 (see Theorem 3.3 and Corollary 3.10). In addition, characterizations of rationally complete modules are presented. As a corollary, we obtain several new characterizations of the maximal right ring of quotients of a ring. In particular, we show that the endomorphism ring of a module is embedded into that of the rational hull of the module as the inherited property of its maximal right ring of quotients (see Theorem 3.15). Our focus, in Sect. 4, is on the question of when is the rational hull of the direct sum of modules the direct sum of the rational hulls of those modules. For $M=\bigoplus _{k\in \Lambda } M_k$, we prove that $\widetilde{E}(M)=\bigoplus _{k\in \Lambda } \widetilde{E}(M_k)$ if and only if $M_i$ is $M_j$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$ when either R is right noetherian or $|\Lambda |$ is finite (see Theorem 4.6). In the last section, we obtain a condition under which $\text {End}_R(M)=\text {End}_H(M)$ where H is a right ring of quotients of a ring R (Theorem 5.1). This condition is called the relatively dense property for a module. Also, we provide a sufficient condition for the maximal right ring of quotients of the endomorphism ring of a module to be the endomorphism ring of the rational hull of the module (see Theorem 5.5).

Throughout this paper, R is a ring with unity and M is a unital right R-module. For a right R-module M, $S=\text {End}_R(M)$ denotes the endomorphism ring of M; thus M can be viewed as a left S- right R-bimodule. For $\varphi \in S$, $\text {Ker} \varphi$ and $\text {Im}\varphi$ stand for the kernel and the image of $\varphi$, respectively. The notations $N\le M$, $N\le ^\text {ess} M$, $N\le ^\text {den} M$ or $N\le ^\oplus M$ mean that N is a submodule, an essential submodule, a dense submodule, or a direct summand of M, respectively. By E(M), ${\widehat{M}}$, and ${\widetilde{E}}(M)$ we denote the injective hull, the quasi-injective hull, and the rational hull of M, respectively, and $T=\text {End}_{R}(E(M))$. Q(R) denotes the maximal right ring of quotients of R. The direct sum of $\Lambda$ copies of M is denoted by $M^{(\Lambda )}$ where $\Lambda$ is an arbitrary index set. $\textsf{CFM}_{\mathbb {N}}(F)$ denotes the $\mathbb {N}\times \mathbb {N}$ column finite matrix ring over a field F. By $\mathbb {Q}$, $\mathbb {Z}$, and $\mathbb {N}$ we denote the set of rational, integer, and natural numbers, respectively. $\mathbb {Z}_n$ denotes the $\mathbb {Z}$-module $\mathbb {Z}/n\mathbb {Z}$. For $x\in M$, $x^{-1}K=\{r\in R\,|\,xr\in K\}\le R_R$ with a right R-submodule K of M. We also denote $\textbf{r}_M(I)=\{m\in M\,|\, Im=0\}$ for $I\le S$ and $\textbf{l}_S(N)=\{\varphi \in S\,|\, \varphi N=0\}$ for $N\le M$ where $\varphi N$ is the image of N under $\varphi$.

2 Some Well-Known Results

We give some properties of dense submodules. Recall that a submodule N of M is said to be dense in M if for any $x, 0\ne y\in M,$ there exists $r\in R$ such that $xr\in N$ and $0\ne yr$.

Proposition 2.1

([3, Proposition 1.3.6]) Let $N\le M$ be right R-modules. Then the following conditions are equivalent:

• (a) N is dense in M;

• (b) $Hom _R(M/N, E(M))=0$;

• (c) for any submodule P such that $N\le P\le M$, $Hom _R(P/N, M)=0$.

Proposition 2.2

([7, Proposition 8.7]) Let L, N be submodules of a right R-module M:

• (i) If $L\le ^den M$ and $N\le ^den M$ then $L\cap N\le ^den M$.

• (ii) Let $L\le V \le M$. Then $L\le ^den M$ if and only if $L\le ^den V$ and $V\le ^den M$.

Proposition 2.3

([3, Proposition 1.3.7]) Let M be a right R-module and $M\le V\le {E}(M)$. Then $M\le ^den V$ if and only if $V\le \widetilde{E}(M)$.

We remind the reader of some important characterizations of the rational hull of a module.

Proposition 2.4

The following statements hold true for a right R-module M and $T=End _R(E(M))$:

• (i) ([9, Exercises 5]) ${\widetilde{E}}(M)=\{x \in E(M)\,|\,~ \vartheta |_M=1_M ~with ~\vartheta \in T\ ~ \Rightarrow \ \vartheta (x)=x\}$.

• (ii) ([7, Proposition 8.16]) ${\widetilde{E}}(M)=\{x\in E(M)\,|\,~ \forall y\in E(M)\! {\setminus }\!\{0\},~ y\cdot x^{-1}M\ne 0\}$.

3 The Rational Hull of a Module

As the injective hull of a module M is the minimal injective module including M, the next result shows that the rational hull of a module M is the minimal rationally complete module including M. Recall that a right R-module M is said to be rationally complete if it has no proper rational (or dense) extensions, or equivalently $\widetilde{E}(M)=M$. Thus, the rational hull $\widetilde{E}(M)$ of a module M is rationally complete.

Theorem 3.1

The following conditions are equivalent for right R-modules M and F:

• (a) F is maximal dense over M;

• (b) F is rationally complete, and is dense over M;

• (c) F is minimal rationally complete, and is essential over M.

Note that a right R-module F is exactly the rational hull of a module M if F satisfies any one of the above equivalent conditions.

Proof

(a)$\Rightarrow$(b) From Proposition 2.3, it is easy to see that F has no proper dense extension. So, F is a rationally complete module.

(b)$\Rightarrow$(c) Let $F'$ be rationally complete such that $M\le F'\le F$. Since $M\le ^\text {den}F$, from Proposition 2.2(ii) $M\le ^\text {den} F'\le ^\text {den}F$. Thus, from Proposition 2.3$F\le ^\text {den}\widetilde{E}(F')=F'$ because $F'$ is rationally complete. Therefore $F=F'$.

(c)$\Rightarrow$(a) Let F be minimal rationally complete over M. Since F is essential over M, $M\le F\le E(M)$. Since $M\le ^\text {den}\widetilde{E}(M)$, $\text {Hom}_R(\widetilde{E}(M)/M, E(M))=0$. Also, since $E(F)=E(M)$, $\text {Hom}_R(\widetilde{E}(M)/M, E(F))=0$. From [7, Theorem 8.24], an inclusion map $\iota : M\rightarrow F$ extends to $\rho : \widetilde{E}(M)\rightarrow F$ as F is rationally complete (see also Proposition 3.13). Note that $\rho$ is a monomorphism. Since $\widetilde{E}(M)$ is rationally complete and F is minimal, $\widetilde{E}(M)=F$. $\square$

The next example shows that the condition “essential over M” in Theorem 3.1(c) is not superfluous.

Example 3.2

Let $M = \mathbb {Z}$ and $F=\mathbb {Z}_{(p)}\oplus \mathbb {Z}_p$ be right $\mathbb {Z}$-modules where $\mathbb {Z}_{(p)}$ is the localization of $\mathbb {Z}$ at the prime ideal (p) where p is prime. It is easy to see that M is not essential in F, so F is not a rational hull of M. In fact, F is minimal rationally complete over M. From [7, Example 8.21], F is rationally complete because F is the rational hull of $L=\mathbb {Z} \oplus \mathbb {Z}_p$. It is enough to show that F is minimal over M. Let K be a rationally complete module such that $M \le K \le F$. Hence $1=\text {u.dim}(M) \le \text {u.dim}(K) \le \text {u.dim}(F)=2$. Assume that $\text {u.dim}(K)=1.$ Then $M \le ^\text {ess} K$, and hence K is nonsingular since M is nonsingular. Thus $M \le ^\text {den} K$, which implies that $K \cong \mathbb {Q}$ since K is rationally complete and $\widetilde{E}(M)=\mathbb {Q}$. It follows that $\mathbb {Q}$ can be embedded into $F=\mathbb {Z}_{(p)} \oplus \mathbb {Z}_p$, a contradiction. Therefore, $\text {u.dim}(K)=2.$ Then $K \le ^\text {ess} F$, and hence $K \cap \mathbb {Z}_p \ne 0$. Thus $\mathbb {Z}_p \le K$, which implies that $L=\mathbb {Z} \oplus \mathbb {Z}_p \le K$. Note that $L \le ^\text {den} F$ since $F=\widetilde{E}(L)$. Hence $K \le ^\text {den} F$, so that $K=F$ due to the fact that K is rationally complete.

We provide another characterization for the rational hull of a module using the relatively dense property. A right ideal I of a ring R is called relatively dense to a right R-module M (or M-dense) in R if for any $r\in R$ and $0\ne m\in M$, $m\cdot r^{-1}I\ne 0$. It is denoted by $I\le ^\text {den}_M R$.

Theorem 3.3

For a right R-module M, $\widetilde{E}(M)=\{x \in E(M)\,|\,x^{-1}M \le ^{den }_M R\}$.

Proof

Let $x \in \widetilde{E}(M)$ be arbitrary. Consider a right ideal $x^{-1}M \le R$. Let $0 \ne m \in M$ and $r \in R$. Since $M \le ^{\text {den}} \widetilde{E}(M)$, there exists $s \in R$ such that $ms \ne 0$ and $(xr)s=x(rs) \in M$, that is, $rs\in x^{-1}M$. Hence $x^{-1}M \le ^{\text {den}}_M R$.

For the reverse inclusion, let $x \in E(M)$ such that $x^{-1}M \le ^{\text {den}}_M R$. For an arbitrary nonzero element $y \in E(M)$, it suffices to show that $y\cdot x^{-1}M \ne 0$. As $M \le ^{\text {ess}} E(M)$, $0\ne yr\in M$ for some $r \in R$. Since $x^{-1}M \le ^{\text {den}}_M R$, there exists $s \in R$ such that $yrs \ne 0$ and $rs \in x^{-1}M$. Hence $0 \ne yrs \in y\cdot x^{-1}M$. Therefore $x \in \widetilde{E}(M)$. $\square$

The next definition was shown in [4, pp 79] as $N\le M (K)$, so we call a submodule N relatively dense to a module K in a module M. (For details, see [17].)

Definition 3.4

A submodule N of a right R-module M is said to be relatively dense to a right R-module K (or K-dense) in M if for any $m\in M$ and $0\ne x\in K$, $x\cdot m^{-1}N\ne 0$, denoted by $N \le ^\text {den}_K M$.

Note that N is M-dense in M if and only if N is dense in M.

We provide some characterizations of the relative density property. One can compare the following characterizations to Proposition 2.1. The equivalence (a)$\Leftrightarrow$(c) in the following proposition is provided by [4, pp79].

Proposition 3.5

The following are equivalent for right R-modules M, K and $N\le M$:

• (a) N is K-dense in M;

• (b) $Hom _R(M/N, E(K))=0$;

• (c) for any submodule P such that $N\le P\le M$, $Hom _R(P/N, K)=0$.

Proof

(a)$\Rightarrow$(b) Assume that there exists $0\ne \alpha \in \text {Hom}_R(M, E(K))$ with $\alpha N=0$. Since $\alpha M\cap K\ne 0$ because $K\le ^\text {ess} E(K)$, there exist $x\in M$ and $0\ne y\in K$ such that $\alpha (x)=y$. Since N is K-dense in M, there exists $r\in R$ such that $xr\in N$ and $0\ne yr$. However, $0=\alpha (xr)=\alpha (x)r=yr\ne 0$, a contradiction. Hence $\text {Hom}_R(M/N, E(K))=0$.

(b)$\Rightarrow$(c) Assume that for any submodule P such that $N\le P\le M$, there exists $0\ne \eta \in \text {Hom}_R(P/N, K)$. Then by the injectivity of E(K), we can extend $\eta$ to a nonzero homomorphism from M/N to E(K), a contradiction. Hence $\text {Hom}_R(P/N, K)=0$.

(c)$\Rightarrow$(a) Assume that $y\cdot x^{-1}N=0$ for some $x\in M$ and $0\ne y\in K$. We define $\gamma : N+xR \rightarrow K$ given by $\gamma (n+xr)=yr$ for $n\in N$ and $r\in R$. It is easy to see that $\gamma$ is a well-defined R-homomorphism vanishing on N. Since $N\le N+xR\le M$, by hypothesis $0=\gamma (x)=y\ne 0$, a contradiction. Thus N is K-dense in M. $\square$

We obtain another characterization of the relative density property related to homomorphisms.

Proposition 3.6

Let M, K be right R-modules. Then a submodule N is K-dense in M if and only if $\textbf{l}_{H}(N)=0$ where $H=Hom _R(M, E(K))$.

Proof

Suppose N is K-dense in M. Assume that $0\ne \varphi \in H$ such that $\varphi N=0$. Then there exists $m\in M\setminus N$ such that $\varphi (m)\ne 0$. Since $\varphi (m)\in E(K)$, $0\ne \varphi (m)r\in K$ for some $r\in R$. Hence there exists $s\in R$ such that $mrs\in N$ and $\varphi (m)rs\ne 0$ because $N\le ^\text {den}_K M$. That yields a contradiction that $0\ne \varphi (m)rs=\varphi (mrs)\in \varphi N=0$. Therefore $\textbf{l}_{H}(N)=0$. Conversely, assume that $x \cdot m^{-1}N = 0$ for some $0\ne x\in K$ and $m\in M$. We define $\gamma : N+mR\rightarrow E(K)$ by $\gamma (n+mt)=xt$ for $n\in N$ and $t\in R$. Clearly, $\gamma$ is a nonzero R-homomorphism vanishing on N. Also, there exists ${\overline{\gamma }}: M\rightarrow E(K)$ such that ${\overline{\gamma }}|_{N+mR}=\gamma$. Since $0={\overline{\gamma }}N$, $0\ne {\overline{\gamma }}\in \textbf{l}_{H}(N)$, a contradiction. Therefore $x \cdot m^{-1}N \ne 0$. $\square$

If $M=R$, the following result is directly provided.

Corollary 3.7

([14, Proposition 1.1]) Let K be a right R-module and I be a right ideal of a ring R. Then I is K-dense in R if and only if $\textbf{l}_{E(K)}(I)=0$.

Proposition 3.8

Let K be a right R-module and I be an ideal of a ring R. Then $\textbf{l}_{K}(I)=0$ if and only if $\textbf{l}_{E(K)}(I)=0$.

Proof

Since one direction is trivial, we need to show the other direction. Suppose $\textbf{l}_{K}(I)=0$. Assume that $\textbf{l}_{E(K)}(I)\ne 0$. Then there exists $0\ne x\in E(K)$ such that $xI=0$. Also, $0\ne xr\in K$ for some $r\in R$ because $K\le ^\text {ess}E(K)$. Since $xrI\subseteq xI=0$, $0\ne xr\in \textbf{l}_{K}(I)$, a contradiction. Therefore $\textbf{l}_{E(K)}(I)=0$. $\square$

Corollary 3.9

([3, Proposition 1.3.11(iv)]) Let I be an ideal of a ring R. Then $I\le ^den R_R$ if and only if $\textbf{l}_{R}(I)=0$.

Proof

The proof also follows from Corollary 3.7 and Proposition 3.8. $\square$

Using Theorem 3.3 and Corollary 3.7, we obtain another characterization for the rational hull of a module. Also, using the characterization of the relatively dense property, a new characterization for the rational hull of a module is provided.

Corollary 3.10

Let M be a right R-module. Then the following statements hold true:

• (i) ( [14, Proposition 1.4(b)]) $\widetilde{E}(M)=\{x \in E(M)\,|\,\textbf{l}_{E(M)}(x^{-1}M)=0\}$.

• (ii) $\widetilde{E}(M) =\{x\in E(M)\,|\,Hom _R(R/x^{-1}M, E(M))=0\}$.

Proof

It directly follows from Theorem 3.3, Proposition 3.5, and Corollary 3.7. $\square$

Several new characterizations for the maximal right ring of quotients of a ring are provided as the following.

Theorem 3.11

Let R be a ring. Then the following statements hold true:

• (i) A right ideal I is dense in R if and only if $\textbf{l}_{E(R)}(I)=0$.

• (ii) $Q(R)=\{x \in E(R)\,|\,x^{-1}R \le ^{den } R\}$.

• (iii) $Q(R)=\{x \in E(R)\,|\,\textbf{l}_{E(R)}(x^{-1}R)=0\}$.

• (iv) $Q(R)=\{x\in E(R)\,|\,Hom _R(R/x^{-1}R, E(R))=0\}$.

We give characterizations for a rationally complete module.

Theorem 3.12

The following conditions are equivalent for a right R-module M:

• (a) M is a rationally complete module;

• (b) $\{\overline{x}\in E(M)/M\,|\, \textbf{l}_{E(M)}(\textbf{r}_R(\overline{x}))=0\}=\overline{0}$;

• (c) For any $I\le ^den _MR$, $\varphi \in Hom _R(I, M)$ can be uniquely extended to ${\widetilde{\varphi }}\in Hom _R(R, M)$.

Proof

Take $A:=\{\overline{x}\in E(M)/M\,|\, \textbf{l}_{E(M)}(\textbf{r}_R(\overline{x}))=0\}$.

(a)$\Rightarrow$(b) Assume that $x\in E(M)\setminus M$ such that $\overline{x}\in A$. From Corollary 3.7, $\textbf{r}_R(\overline{x})\le ^\text {den}_M R$. Since $\textbf{r}_R(\overline{x})=x^{-1}M$, $x^{-1}M\le ^\text {den}_M R$. Hence from Theorem 3.3$x\in {\widetilde{E}}(M)$. As M is rationally complete, $M={\widetilde{E}}(M)$. Thus $x\in M$, a contradiction. Therefore $A=\overline{0}$.

(b)$\Rightarrow$(c) Assume to the contrary of the condition (c). For $I\le ^\text {den}_MR$, since $M \subseteq E(M)$, there exists $\varphi \in \text {Hom}_R(I, M)$ such that ${\widetilde{\varphi }}\in \text {Hom}_R(R, E(M))$, ${\widetilde{\varphi }}|_I=\varphi$, and ${\widetilde{\varphi }}(1)\notin M$. Since $\overline{0}\ne {\widetilde{\varphi }}(1)+M\in E(M)/M$ and $I\subseteq \textbf{r}_R({\widetilde{\varphi }}(1)+M)\le ^\textrm{den}_\textrm{M} R$, $\textbf{l}_{E(M)}(\textbf{r}_R({\widetilde{\varphi }}(1)+M))=0$ from Corollary 3.7, a contradiction that $A=\overline{0}$. Therefore $\varphi \in \text {Hom}_R(I, M)$ is extended to ${\widetilde{\varphi }}\in \text {Hom}_R(R, M)$. For the uniqueness, the proof is similar to that of Proposition 3.13.

(c)$\Rightarrow$(a) Assume that M is not rationally complete. Then there exists $x\in {\widetilde{E}}(M)\setminus M$ such that $x^{-1}M\le ^\text {den}_M R$ from Theorem 3.3. Define $\varphi : x^{-1}M \rightarrow M$ given by $\varphi (r)=xr$. By hypothesis, ${\widetilde{\varphi }} (1)=x1=x\in M$, a contradiction. Therefore M is rationally complete. $\square$

Next, as a ring is embedding into its maximal right ring of quotients, we provide the relationship between the endomorphism rings of a module and its rational hull.

Proposition 3.13

Let M and K be right R-modules. For any $N\le ^den _KM$, $\varphi \in Hom _R(N, K)$ is uniquely extended to ${\widetilde{\varphi }}\in Hom _R(M, \widetilde{E}(K))$ and ${\widetilde{\varphi }}|_N=\varphi$. In addition, $\varphi N\le ^den {\widetilde{\varphi }} M$.

Proof

(Existence) Let $\varphi \in \text {Hom}_R(N, K)$ be arbitrary. Then there exists ${\widetilde{\varphi }}\in \text {Hom}_R(M, E(K))$ such that ${\widetilde{\varphi }}|_N=\varphi$. Since ${\widetilde{\varphi }}$ induces a surjection from M/N to $({\widetilde{\varphi }}M+\widetilde{E}(K))/\widetilde{E}(K)$ and $\text {Hom}_R(M/N, E(K))=0$ (see Proposition 3.5), $\text {Hom}_R\left( \frac{{\widetilde{\varphi }}M+\widetilde{E}(K)}{\widetilde{E}(K)}, E(K)\right) =0$. Hence $\widetilde{E}(K)\le ^\text {den}{\widetilde{\varphi }}M+\widetilde{E}(K)$ by Proposition 2.1. As $\widetilde{E}(K)$ is rationally complete, ${\widetilde{\varphi }}M\subseteq \widetilde{E}(K)$.

(Uniqueness) Suppose ${\widetilde{\varphi }}$ and ${\widetilde{\psi }}$ are in $\text {Hom}_R(M, \widetilde{E}(K))$ such that ${\widetilde{\varphi }}|_N={\widetilde{\psi }}|_N$. It is enough to show that ${\widetilde{\varphi }}={\widetilde{\psi }}$. Assume that ${\widetilde{\varphi }}(x)\ne {\widetilde{\psi }}(x)$ for some $x\in M$. Take $0\ne y:=({\widetilde{\varphi }}-{\widetilde{\psi }})(x)\in \widetilde{E}(K)$. Thus, there exists $r\in R$ such that $0\ne yr\in K$. Since $N\le ^\text {den}_KM$, there exists $s\in R$ such that $xrs\in N$ and $yrs\ne 0$. This yields a contradiction that $0\ne yrs=({\widetilde{\varphi }}-{\widetilde{\psi }})(xrs)=({\widetilde{\varphi }}|_N-{\widetilde{\psi }}|_N)(xrs)=0$. Therefore ${\widetilde{\varphi }}={\widetilde{\psi }}$.

In addition, let $x_1\in {\widetilde{\varphi }}M$ and $0\ne x_2\in {\widetilde{\varphi }}M$. Then ${\widetilde{\varphi }}(m_1)=x_1, {\widetilde{\varphi }}(m_2)=x_2$ for some $m_1, m_2\in M$. As ${\widetilde{\varphi }}M\subseteq \widetilde{E}(K)$, $0\ne x_2r\in K$ for some $r\in R$. Since $N\le ^\text {den}_KM$ and $m_1r\in M$, there exists $s\in R$ such that $m_1rs\in N$ and $0\ne x_2rs$. Thus $x_1rs={\widetilde{\varphi }}(m_1rs)\in \varphi N$ and $0\ne x_2rs$. Therefore $\varphi N\le ^\text {den}{\widetilde{\varphi }}M$. $\square$

Noting that the dense property implies the essential property, however the relatively dense property does not imply the essential property in general. See $\mathbb {Z}_p\le ^\text {den}_\mathbb {Z} \mathbb {Z}_p\oplus \mathbb {Z}_p$ but $\mathbb {Z}_p\nleq ^\text {ess} \mathbb {Z}_p\oplus \mathbb {Z}_p$ as a $\mathbb {Z}$-module. However, Proposition 3.13 shows that $\varphi N\le ^\text {den}{\widetilde{\varphi }} M$ when $N\le ^\text {den}_KM$ for any $\varphi \in \text {Hom}_R(N, K)$. As a corollary, we have a generalized result of Theorem 3.12((a)$\Rightarrow$(c)).

Corollary 3.14

Let M be a right R-module. If K is rationally complete, then for any $N\le ^den _KM$, $\varphi \in Hom _R(N, K)$ is uniquely extended to ${\widetilde{\varphi }}\in Hom _R(M, K)$ and ${\widetilde{\varphi }}|_N=\varphi$.

Theorem 3.15

Let M be a right R-module. Then $End _R(M)$ is considered as a subring of $End _R(\widetilde{E}(M))$.

Proof

Since $M\le ^\text {den} \widetilde{E}(M)$, from Proposition 3.13$\varphi \in \text {End}_R(M)$ can be uniquely extended to ${\widetilde{\varphi }}\in \text {End}_R(\widetilde{E}(M))$ because $\text {End}_R(M)\subseteq \text {Hom}_R(M, \widetilde{E}(M))$. Thus we have a one-to-one correspondence between $\text {End}_R(M)$ and $\{{\widetilde{\varphi }}\in \text {End}_R(\widetilde{E}(M))\,|\,{\widetilde{\varphi }}|_M={\varphi }\in \text {End}_R(M)\}$ given by $\Omega (\varphi )={\widetilde{\varphi }}$. We need to check that $\Omega$ is a ring homomorphism.

(i) Since $\Omega (\varphi +\psi )|_M=(\widetilde{\varphi +\psi })|_M=\varphi +\psi =\Omega (\varphi )|_M+\Omega (\psi )|_M=(\Omega (\varphi )+\Omega (\psi ))|_M$, from the uniqueness of Proposition 3.13 we have $\Omega (\varphi +\psi )= \Omega (\varphi )+\Omega (\psi )$.

(ii) Since $\Omega {(\varphi \circ \psi )}|_M=(\widetilde{\varphi \circ \psi })|_M=\varphi \circ \psi =\Omega (\varphi )|_M\circ \Omega (\psi )|_M=(\Omega (\varphi )\circ \Omega (\psi ))|_M$ because $\Omega (\varphi )|_M\le M$, from the uniqueness of Proposition 3.13 we have $\Omega (\varphi \circ \psi )= \Omega (\varphi )\circ \Omega (\psi )$.

Thus $\text {End}_R(M)$ is isomorphic to a subring of $\text {End}_R(\widetilde{E}(M))$. Therefore we consider $\text {End}_R(M)$ as a subring of $\text {End}_R(\widetilde{E}(M))$. $\square$

We conclude this section with results for the rational hulls of quasi-continuous modules and quasi-injective modules. Recall that a module M is called quasi-continuous if every submodule of M is essential in a direct summand of M, and for any direct summands $M_1$ and $M_2$ of M such that $M_1\cap M_2=0$, $M_1\oplus M_2$ is a direct summand of M. Also, a ring R is called right  quasi-continuous if $R_R$ is quasi-continuous.

Theorem 3.16

The following statements hold true for a module M:

• (i) If M is a quasi-continuous module then $\widetilde{E}(M)$ is a quasi-continuous module.

• (ii) If M is a quasi-injective module then $\widetilde{E}(M)$ is a quasi-injective module.

Proof

(i) Let $T=\text {End}_R(E(\widetilde{E}(M)))=\text {End}_R(E(M))$. From [10, Theorem 2.8], we need to show that $f\widetilde{E}(M)\le \widetilde{E}(M)$ for all idempotents $f^2=f\in T$. Assume that $f\widetilde{E}(M)\nleq \widetilde{E}(M)$ for some idempotent $f^2=f\in T$. Then there exists $x\in \widetilde{E}(M)$ such that $f(x)\notin \widetilde{E}(M)$. Thus, there exists $g\in T$ such that $g M=0$ and $g f(x)\ne 0$. Since $g f(x)\in E(M)$, there exists $r\in R$ such that $0\ne gf(xr)\in \widetilde{E}(M)$. Thus, as $M\le ^\text {den}\widetilde{E}(M)$ and $xr\in \widetilde{E}(M)$, there exists $s\in R$ such that $0\ne g f(xrs)$ and $xrs\in M$. Note that $fM\le M$ for all idempotents $f^2=f\in T$ because M is quasi-continuous. However, $0\ne gf(xrs)\in g fM\le gM=0$, a contradiction. Therefore $\widetilde{E}(M)$ is a quasi-continuous module.

(ii) The proof is similar to that of part (i) by using [10, Corollary 1.14]. $\square$

Remark 3.17

([1, Theorem 5.3]) The rational hull of every extending module is an extending module.

Note that if M is an injective module then $M=\widetilde{E}(M)$ (see [7, Examples 8.18(1)]). The next examples exhibit that the converses of Theorem 3.16 and Remark 3.17 do not hold true, in general.

Example 3.18

(i) Consider $\mathbb {Z}$ as a $\mathbb {Z}$-module. Then $\widetilde{E}(\mathbb {Z})=\mathbb {Q}$ is (quasi-)injective, while $\mathbb {Z}$ is not quasi-injective.

(ii)( [10, Example 2.9]) Consider a ring $R=\left( {\begin{matrix} F&{} F \\ 0 &{} F\end{matrix}}\right)$ where F is a field. Then $\widetilde{E}(R_R)=\left( {\begin{matrix} F&{} F \\ F &{} F\end{matrix}}\right)$ is injective (hence, quasi-continuous), while $R_R$ is not right quasi-continuous.

(iii) Consider $\mathbb {Z}^{(\mathbb {N})}$ as a $\mathbb {Z}$-module. Then $\widetilde{E}(\mathbb {Z}^{(\mathbb {N})})=\mathbb {Q}^{(\mathbb {N})}$ is injective (hence, extending), while $\mathbb {Z}^{(\mathbb {N})}$ is not extending.

Corollary 3.19

The maximal right ring of quotients of a right quasi-continuous ring is also a right quasi-continuous ring.

Remark 3.20

([7, Exercises 13.8]) The maximal right ring of quotients of a simple (resp., prime, semiprime) ring is also a simple (resp., prime, semiprime) ring.

Open Question 1

Is the rational hull of a continuous module always a continuous module?

4 Direct Sum of Rational Hulls of Modules

As we know, the injective hull of the direct sum of two modules is the direct sum of the injective hulls of each module without any condition. However, the rational hull case is different from the injective hull case. In this section, we discuss the condition for the rational hull of the direct sum of two modules to be the direct sum of the rational hulls of those modules. The next example shows that the rational hull of the direct sum of two modules is not the direct sum of the rational hulls of each module, in general.

Example 4.1

Consider $M=\mathbb {Z}\oplus \mathbb {Z}_p$ as a $\mathbb {Z}$-module where p is prime. Then $\widetilde{E}(\mathbb {Z})=\mathbb {Q}$ and $\widetilde{E}(\mathbb {Z}_p)=\mathbb {Z}_p$. However, by [7, Example 8.21] $\widetilde{E}(M)=\mathbb {Z}_{(p)}\oplus \mathbb {Z}_p\ne \mathbb {Q}\oplus \mathbb {Z}_p$ where $\mathbb {Z}_{(p)}=\{\frac{m}{n}\in \mathbb {Q}\,|\,{m, n}\in \mathbb {Z}, (n, p)=1\}$. Hence M is not a dense submodule of $\mathbb {Q}\oplus \mathbb {Z}_p$. For $(\frac{1}{p}, \overline{0})$ and $0\ne (0, \overline{1})\in \mathbb {Q}\oplus \mathbb {Z}_p$, there is no $n\in \mathbb {Z}$ such that $n(\frac{1}{p}, \overline{0})\in \mathbb {Z}\oplus \mathbb {Z}_p$ and $n(0, \overline{1})\ne 0$.

Proposition 4.2

Let $M=\bigoplus _{k\in \Lambda } M_k$ where $M_k$ be a right R-module and $\Lambda$ is any index set. If either R is right noetherian or $|\Lambda |$ is finite, then $\widetilde{E}(M)\le \bigoplus _{k\in \Lambda }\widetilde{E}(M_k)$.

Proof

Suppose $0\ne m\in \widetilde{E}(M)$. Since $\widetilde{E}(M)\subseteq E(M)=\oplus _{k\in \Lambda }E(M_k)$ because R is right noetherian or $|\Lambda |$ is finite, there exists $\ell \in \mathbb {N}$ such that $m\in \oplus _{i=1}^\ell E(M_i)$. Thus, $m=(m_1,\dots , m_\ell )$ where $m_i\in E(M_i)$. Since $(0, \dots , 0, y_i, 0, \dots , 0)\cdot m^{-1}M\ne 0$ for all $0\ne y_i\in E(M_i)$ and $m^{-1}M=m_1^{-1}M_1\cap \cdots \cap m_\ell ^{-1}M_\ell$, $y_i\cdot m_i^{-1}M_i\ne 0$ for all $0\ne y_i\in E(M_i)$. Thus, $m_i\in \widetilde{E}(M_i)$ for all $1\le i\le \ell$ from Proposition 2.4. So, $m=(m_1,\dots , m_\ell )\in \oplus _{i=1}^\ell \widetilde{E}(M_i)\subseteq \oplus _{k\in \Lambda }\widetilde{E}(M_k)$. Therefore $\widetilde{E}(M)\le \oplus _{k\in \Lambda }\widetilde{E}(M_k)$. $\square$

Remark 4.3

Example 4.1 illustrates Proposition 4.2 because $R=\mathbb {Z}$ is a noetherian ring, that is, $\widetilde{E}(\mathbb {Z}\oplus \mathbb {Z}_p)=\mathbb {Z}_{(p)}\oplus \mathbb {Z}_p\lneq \mathbb {Q}\oplus \mathbb {Z}_p=\widetilde{E}(\mathbb {Z})\oplus \widetilde{E}(\mathbb {Z}_p)$. However, Example 4.7 shows that the condition “either R is right noetherian or $|\Lambda |$ is finite” is not superfluous because $\widetilde{E}(\oplus _{k\in \Lambda }\mathbb {Z}_2)=\prod _{k\in \Lambda }\mathbb {Z}_2\gneq \oplus _{k\in \Lambda }\mathbb {Z}_2=\oplus _{k\in \Lambda }\widetilde{E}(\mathbb {Z}_2)$ with a non-noetherian ring $R=\langle \oplus _{k\in \Lambda }\mathbb {Z}_2, 1\rangle$.

To get the reverse inclusion of Proposition 4.2, first we provide the properties of the relatively dense property.

Lemma 4.4

Let $N\le M$ and $K_i$ be right R-modules for all $i\in \Lambda$. Then the following conditions are equivalent:

• (a) N is $K_i$-dense in M for all $i\in \Lambda$;

• (b) N is $\bigoplus _{i\in \Lambda }K_i$-dense in M;

• (c) N is $\bigoplus _{i\in \Lambda }\widetilde{E}(K_i)$-dense in M.

Proof

(a)$\Rightarrow$(b) Let P be any submodule such that $N\le P\le M$. Since N is $K_i$-dense in M, $\text {Hom}_R(P/N, K_i)=0$ for all $i\in \Lambda$ from Proposition 3.5. Consider the canonical embedding $\oplus _{i\in \Lambda }K_i\rightarrow \prod _{i\in \Lambda }K_i$. Then we have $0\rightarrow \text {Hom}_R(P/N, \oplus _{i\in \Lambda }K_i)\rightarrow \text {Hom}_R(P/N,\prod _{i\in \Lambda }K_i)\cong \prod _{i\in \Lambda }\text {Hom}_R(P/N, K_i)=0$. Thus $\text {Hom}_R(P/N, \oplus _{i\in \Lambda }K_i)=0$. Therefore N is $\oplus _{i\in \Lambda }K_i$-dense in M from Proposition 3.5.

(b)$\Rightarrow$(a) Since $\text {Hom}_R(P/N, \oplus _{i\in \Lambda }K_i)=0$, $\text {Hom}_R(P/N, K_i)=0$ for each $i\in \Lambda$. Hence N is $K_i$-dense in M for all $i\in \Lambda$.

(a)$\Leftrightarrow$(c) Since $E(\widetilde{E}(K_i))=E(K_i)$, from Proposition 3.5 it is easy to see that N is $K_i$-dense in M if and only if N is $\widetilde{E}(K_i)$-dense in M, for all $i\in \Lambda$. The proof is similar to that of the equivalence (a)$\Leftrightarrow$(b). $\square$

Using Lemma 4.4, we obtain a characterization for $\oplus _{k\in \Lambda } N_k$ to be a dense submodule of $\oplus _{k\in \Lambda } M_k$ where $N_i$ is a submodule of $M_i$ for each $i\in \Lambda$.

Proposition 4.5

Let $N_i\le M_i$ be right R-modules for all $i\in \Lambda$ where $\Lambda$ is any index set. Let $N=\bigoplus _{k\in \Lambda } N_k$ and $M=\bigoplus _{k\in \Lambda } M_k$. Then $N\le ^den M$ if and only if $N_i$ is $M_j$-dense in $M_i$ for all $i, j\in \Lambda$.

Proof

Suppose $N\le ^\text {den}M$. Then N is M-dense in M by the definition. From Lemma 4.4N is $M_j$-dense in M for all $j\in \Lambda$. Let $x_i\in M_i$ and $0\ne y_j\in M_j$ be arbitrary for each $i, j\in \Lambda$. Consider the canonical embedding $\iota : M_i \rightarrow M$. Since $\iota (x_i)=(0,\dots ,0,x_i,0,\dots ) \in M$ and $0\ne y_j\in M_j$, there exists $r\in R$ such that $\iota (x_i)r=\iota (x_i r)\in N$ and $y_jr\ne 0$. Since $x_ir\in N_i$ and $y_jr\ne 0$, $N_i$ is $M_j$-dense in $M_i$ for all $i, j\in \Lambda$.

Conversely, suppose $N_i$ is $M_j$-dense in $M_i$ for all $i, j \in \Lambda$. From Lemma 4.4, $N_i$ is $\oplus _{k\in \Lambda } M_k$-dense in $M_i$ for all $i \in \Lambda$. Let $x\in M$ and $0\ne y\in M$ be arbitrary. Then there exists $\ell \in \mathbb {N}$ such that $x=(x_1, \dots , x_\ell )\in \oplus _{k=1}^\ell M_k\le M$. Since $N_1$ is M-dense in $M_1$, there exists $r_1\in R$ such that $x_1r_1\in N_1$ and $0\ne yr_1\in M$. Also, since $N_2$ is M-dense in $M_2$, there exists $r_2\in R$ such that $x_2r_1r_2\in N_2$ and $0\ne yr_1r_2\in M$. By the similar processing, we have $r=r_1r_2\cdots r_\ell \in R$ such that $xr\in \oplus _{k=1}^\ell N_k\le N$ and $yr\ne 0$. Therefore $N\le ^\text {den}M$. $\square$

From Propositions 4.2 and 4.5, we have a characterization for the rational hull of the direct sum of modules to be the direct sum of the rational hulls of each module.

Theorem 4.6

Let $M=\bigoplus _{k\in \Lambda } M_k$ where $M_k$ is a right R-module and $\Lambda$ is any index set. If either R is right noetherian or $|\Lambda |$ is finite, then $\widetilde{E}(M)=\bigoplus _{k\in \Lambda } \widetilde{E}(M_k)$ if and only if $M_i$ is $M_j$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$.

Proof

Suppose $\widetilde{E}(M)=\oplus _{k\in \Lambda } \widetilde{E}(M_k)$. Since $M\le ^\text {den}\oplus _{k\in \Lambda } \widetilde{E}(M_k)$, from Proposition 4.5$M_i$ is $\widetilde{E}(M_j)$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$. Thus, $M_i$ is $M_j$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$ from Lemma 4.4.

Conversely, suppose $M_i$ is $M_j$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$. Then $M_i$ is $\widetilde{E}(M_j)$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$ from Lemma 4.4. Thus, from Proposition 4.5$M\le ^{\text {den}}\oplus _{k\in \Lambda } \widetilde{E}(M_k)$. Hence $\oplus _{k\in \Lambda } \widetilde{E}(M_k)\le \widetilde{E}(M)$ from Proposition 2.3. Also, from Proposition 4.2$\widetilde{E}(M)\le \oplus _{k\in \Lambda } \widetilde{E}(M_k)$. Therefore $\widetilde{E}(M)=\oplus _{k\in \Lambda } \widetilde{E}(M_k)$. $\square$

The next examples show that the condition “R is right noetherian or $|\Lambda |$ is finite” in Theorem 4.6 is not superfluous.

Example 4.7

(i) Let $R=\langle \oplus _{k\in \Lambda }\mathbb {Z}_2, 1\rangle$ and $M=\oplus _{k\in \Lambda }M_k$ where $M_k=\mathbb {Z}_2$. Note that R is not noetherian. Since $\mathbb {Z}_2$ is an injective R-module, $\widetilde{E}(\mathbb {Z}_2)=\mathbb {Z}_2$. Thus $M_i$ is $M_j$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$. However, $\widetilde{E}(\oplus _{k\in \Lambda }\mathbb {Z}_2)=\prod _{k\in \Lambda }\mathbb {Z}_2\gneq \oplus _{k\in \Lambda }\mathbb {Z}_2=\oplus _{k\in \Lambda }\widetilde{E}(\mathbb {Z}_2)$.

(ii) Let $R=\{(a_k)\in \prod _{k\in \Lambda }\mathbb {Z}\,|\, a_k~ \text {is eventually constant}\}$ and $M=\oplus _{k\in \Lambda }\mathbb {Z}$. Note that R is not noetherian. Then $\widetilde{E}(\mathbb {Z})=\mathbb {Q}$ and $\mathbb {Z}$ is $\mathbb {Z}$-dense in $\widetilde{E}(\mathbb {Z})$. However, $\widetilde{E}(\oplus _{k\in \Lambda }\mathbb {Z})=\prod _{k\in \Lambda }\mathbb {Q}\gneq \oplus _{k\in \Lambda }\mathbb {Q}=\oplus _{k\in \Lambda }\widetilde{E}(\mathbb {Z})$.

The next example illustrates Theorem 4.6.

Example 4.8

Consider $M=\mathbb {Z}\oplus \mathbb {Z}_p$ as a $\mathbb {Z}$-module where p is prime. Then $\mathbb {Z}_p$ is $\mathbb {Z}$-dense in $\widetilde{E}(\mathbb {Z}_p)=\mathbb {Z}_p$, but $\mathbb {Z}$ is not $\mathbb {Z}_p$-dense in $\mathbb {Q}$ because for $\frac{1}{p}\in \mathbb {Q}, \overline{1}\in \mathbb {Z}_p$, there is no element $t\in \mathbb {Z}$ such that $t\frac{1}{p}\in \mathbb {Z}$ and $t\overline{1}\ne 0$. Thus, from Theorem 4.6$\widetilde{E}(M)=\mathbb {Z}_{(p)}\oplus \mathbb {Z}_p\lneq \mathbb {Q}\oplus \mathbb {Z}_p=\widetilde{E}(\mathbb {Z})\oplus \widetilde{E}(\mathbb {Z}_p)$. (See Example 4.1 for details.)

Corollary 4.9

Let M be a right R-module. If either R is right noetherian or $\Lambda$ is a finite index set, then $\widetilde{E}(M^{(\Lambda )})=(\widetilde{E}(M))^{(\Lambda )}$.

Corollary 4.10

Let $\{M_k\}_{k\in \Lambda }$ be a class of rationally complete right R-module for any index set $\Lambda$. If either R is right noetherian or $|\Lambda |$ is finite, then $M=\bigoplus _{k\in \Lambda } M_k$ is rationally complete.

Proof

Since $\widetilde{E}(M_i)=M_i$, $M_i$ is $M_j$-dense in $\widetilde{E}(M_i)$ for all $i, j \in \Lambda$. From Theorem 4.6, $\widetilde{E}(M)=\oplus _{k\in \Lambda } \widetilde{E}(M_k)=\oplus _{k\in \Lambda } M_k=M$. Therefore $\oplus _{k\in \Lambda } M_k$ is rationally complete. $\square$

Proposition 4.11

([14, Proposition 1.9]) Let $\{S_i\}_{i\in \Lambda }$ be a set of nonisomorphic simple modules, representing all singular simple modules. Then every module containing the module $P=\bigoplus _{i\in \Lambda } S_i$ is rationally complete.

5 The Endomorphism Ring of a Module Over a Right Ring of Quotients of a Ring

In this section, we obtain some condition under which $\text {End}_R(M)=\text {End}_H(M)$ where H is a right ring of quotients of a ring R. Recall that an extension ring H of a ring R is called a right ring of quotients of R if for any two elements $x\ne 0$ and y of H, there exists an element $r\in R$ such that $xr\ne 0$ and $yr\in R$.

Theorem 5.1

Let M be a right H-module where H is a right ring of quotients of a ring R. If R is $M_R$-dense in $H_R$ then $End _R(M)=End _H(M)$.

Proof

Since $\text {End}_H(M) \subseteq \text {End}_R(M)$, it suffices to show that $\text {End}_R(M) \subseteq \text {End}_H(M)$. Let $\varphi \in \text {End}_R(M)$ be arbitrary. Assume that $\varphi \notin \text {End}_H(M)$. Then there exist $m \in M, t \in H$ such that $\varphi (mt)-\varphi (m)t \ne 0.$ Since R is $M_R$-dense in $H_R$, there exists $r \in R$ such that $(\varphi (mt)-\varphi (m)t)r \ne 0$ and $tr \in R$. Hence $0\ne (\varphi (mt)-\varphi (m)t)r=\varphi (mt)r-\varphi (m)(tr)=\varphi (mtr)-\varphi (mtr)=0$, a contradiction. Therefore $\text {End}_R(M)=\text {End}_H(M)$. $\square$

Remark 5.2

(i) A ring R is always E(R)-dense in $H_R$ where H is a right ring of quotients of R. Let $x\in H_R$ and $0\ne y\in E(R)$. Since $H\le ^\text {ess} E(R)_R$, there exists $s\in R$ such that $0\ne ys\in H$. Also, $xs\in H$. Since $R\le ^\text {den} H_R$, there exists $t\in R$ such that $xst\in R$ and $0\ne yst$. Therefore R is E(R)-dense in $H_R$.

(ii) If M is a nonsingular R-module, then R is $M_R$-dense in $H_R$. Let $0\ne m\in M$ and $t\in H$ be arbitrary. Take $t^{-1}R=\{r\in R\,|\,tr\in R\}$ a right ideal of R. Note that $t^{-1}R\le ^{\text {ess}}R_R$. Since $t^{-1}R \nleq \textbf{r}_R(m)$, there exists $r\in t^{-1}R$ and $r\notin \textbf{r}_R(m)$. Thus, $tr\in R$ and $mr\ne 0$. Therefore R is $M_R$-dense in $H_R$.

(iii) If M is a submodule of a projective right H-module, then R is $M_R$-dense in $H_R$. Let P be a projective right H-module including M, that is, $M \le P$ where $P \le ^{\oplus } H^{(\Lambda )}$ with some index set $\Lambda$. Then there is a right R-module $K \le E(P)$ such that $E(P)=E(M) \oplus K.$ Since $R \le ^\text {den} H_R$, we get that R is $H^{(\Lambda )}$-dense in $H_R$ from Lemma 4.4. Hence R is P-dense in $H_R$. Thus $\text {Hom}_R(H/R,E(P))=0$ from Proposition 3.5. Since $\text {Hom}_R(H/R,E(P)) \cong \text {Hom}_R(H/R,E(M)) \oplus \text {Hom}_R(H/R,K)$, we obtain $\text {Hom}_R(H/R,E(M))=0$. It follows that R is $M_R$-dense in $H_R.$

Corollary 5.3

Let M be a projective right H-module where H is a right ring of quotients of R. Then $End _R(M)=End _H(M)$.

The next example illustrates Corollary 5.3.

Example 5.4

Let $H=\prod _{n=1}^\infty \mathbb {Z}_2$ and $R=\{(a_n) \in H~|~a_n$ is eventually constant$\}$. Then H is a maximal right ring of quotients of R. Hence from Theorem 5.1, End$_R(H^{(\Lambda )})$= End$_H(H^{(\Lambda )})=\textsf{CFM}_\Lambda (H)$.

Theorem 5.5

Let M be a finitely generated free R-module with $S=End _R(M)$. If either R is right noetherian or $\Lambda$ is any finite index set, then $End _R\left( \widetilde{E}(M^{(\Lambda )})\right) =\textsf{CFM}_{\Lambda }\left( Q(S)\right)$.

Proof

Let $M=R^{(n)}$ for some $n\in \mathbb {N}$. From Corollary 4.9, $\widetilde{E}(R^{(n)})=\widetilde{E}(R)^{(n)}=Q(R)^{(n)}$ as $\widetilde{E}(R)=Q(R)$. Hence $\text {End}_R\left( \widetilde{E}(M^{(\Lambda )})\right) =\text {End}_R\left( \widetilde{E}(M)^{(\Lambda )}\right) =\text {End}_R\left( (Q(R)^{(n)})^{(\Lambda )}\right) =\text {End}_{Q(R)}\left( (Q(R)^{(n)})^{(\Lambda )}\right) =\textsf{CFM}_{\Lambda }\left( \text {End}_{Q(R)}(Q(R)^{(n)})\right) =\textsf{CFM}_{\Lambda }\left( \textsf{Mat}_n(Q(R))\right)$ from Theorem 5.1. Therefore $\text {End}_R\left( \widetilde{E}(M^{(\Lambda )})\right) =\textsf{CFM}_{\Lambda }\left( Q(\text {End}_R(M))\right)$ because $\textsf{Mat}_n(Q(R))=Q(\textsf{Mat}_n(R))$ by [15, 2.3] and $\text {End}_R(M)=\textsf{Mat}_n(R)$. $\square$

The next result is generalized from [15, 2.3].

Corollary 5.6

Let M be a finitely generated free R-module. Then $Q(End _R(M))=End _R(\widetilde{E}(M))$.

The following example shows that the above result can not be extended to flat modules. This example also shows that a ring R is not M-dense in Q as a right R-module where Q is a right ring of quotients of R.

Example 5.7

Let $H=\prod _{n=1}^\infty \mathbb {Z}_2$, $R=\{(a_n) \in H~|~a_n$ is eventually constant$\}$, and $I=\{ (a_n) \in H~|~a_n=0~\text {eventually}\}$. Note that $H=Q(R)$. Let $M=H/I$, which is a flat H-module but not projective. We claim that End$_H(M) \subsetneq$ End$_R(M)$. Indeed, define $f: M\rightarrow M$ via

$$\begin{aligned} f[(a_1, a_2, \dots , a_n, a_{n+1}, \dots )+I] = (a_1, 0, a_2, 0, \dots , a_n, 0, a_{n+1}, 0, \dots )+I, \end{aligned}$$

for any $\overline{a} = a+I=(a_1, a_2, \dots , a_n, a_{n+1}, \dots )+I \in M$. It is easy to see that $f(\overline{a}+\overline{b})$ = $f(\overline{a})$ + $f(\overline{b})$ for any $\overline{a}$, $\overline{b} \in M$. Meanwhile, for any $r = (r_1, r_2, \dots , r_n, r_{n+1}, \dots )\in R$, we have

$$\begin{aligned} (a+I)r = ar + I = \left\{ \begin{array}{ll} (0, \dots , 0, a_n, a_{n+1}, \dots )+I, &{} \text{ if } r_n = r_{n+1} = \dots = 1; \\ (0, \dots , 0, 0, 0, \dots )+I, &{} \text{ if } r_n = r_{n+1} = \dots = 0. \end{array} \right. \end{aligned}$$

Note that $a+I=(0, 0, \dots , 0, a_n, a_{n+1}, \dots )+I$ for some $n\in \mathbb {N}$. One can easily see that $f[(a+I)r]$ = $[f(a+I)]r$ for all $a\in H$, $r\in R$. This shows $f\in$ End$_R(M)$. However, for $q = (0, 1, 0, 1, \dots ) = q^2 \in H$, we have $[f(q+I)]q = 0+I$ while $f[(q+I)q] = f(q+I) \ne 0+I$. This means $f\not \in$ End$_H(M)$. Thus, End$_H(M) \subsetneq$ End$_R(M)$. Note that R is not $M_R$-dense in H. Let $h\in H\setminus R$ and $m=1+I\in M$. Since $(1+I)r=0+I$ for all $r\in I, \mathrm{it~has~to~be}~r\in R\setminus I$ to get $mr\ne 0+I$. However, $hr \notin R$.

Recall that a module M is said to be polyform if every essential submodule of M is a dense submodule.

Lemma 5.8

A module M is polyform if and only if $\widetilde{E}(M)$ is a polyform quasi-injective module.

Proof

Let X be essential in $\widetilde{E}(M)$. Then $X\cap M\le ^\text {ess} M$. Hence $X\cap M$ is a dense submodule of M because M is polyform. Since $X\cap M\le ^\text {den} M\le ^\text {den}\widetilde{E}(M)$, $X\cap M\le ^\text {den}\widetilde{E}(M)$. Thus X is a dense submodule of $\widetilde{E}(M)$ from Proposition 2.2(ii). Therefore $\widetilde{E}(M)$ is a polyform module. In addition, $\widehat{M}$ is also a polyform module from [16, 11.1]. Since $M\le ^\text {ess}\widehat{M}$, $M\le ^\text {den}\widehat{M}$. Thus $\widetilde{E}(M)=\widetilde{E}(\widehat{M})$. Since the rational hull of a quasi-injective module is also quasi-injective from Theorem 3.16(ii), $\widetilde{E}(M)$ is a quasi-injective module. Therefore $\widetilde{E}(M)$ is a polyform quasi-injective module.

Conversely, let N be any essential submodule of M. Then N is also essential in $\widetilde{E}(M)$. Hence N is a dense submodule of $\widetilde{E}(M)$ as $\widetilde{E}(M)$ is polyform. So N is a dense submodule of M. Therefore M is polyform. $\square$

We show from Theorem 3.15 that there is a monomorphism from the ring $\text {End}_R(M)$ into the ring $\text {End}_R(\widetilde{E}(M))$. Next, we obtain a condition when $\text {End}_R(M)$ and $\text {End}_R(\widetilde{E}(M))$ are isomorphic. It is a generalization of [7, Exercises 7.32].

Proposition 5.9

If M is a quasi-injective module then there is an isomorphism $\Omega : End _R(M) \rightarrow End _R(\widetilde{E}(M))$. In particular, if M is a polyform module, then the converse holds true.

Proof

In the proof of Theorem 3.15, we only need to show that $\Omega : \text {End}_R(M) \rightarrow \text {End}_R(\widetilde{E}(M))$ given by $\Omega (\varphi )={\widetilde{\varphi }}$, is surjective. Let $\psi \in \text {End}_R(\widetilde{E}(M))$ be arbitrary. Then there exists ${\widehat{\psi }}\in \text {End}_R(E(M))$ such that ${\widehat{\psi }}|_{\widetilde{E}(M)}=\psi$. Since ${\widehat{\psi }}M\le M$ as M is quasi-injective, ${\widehat{\psi }}|_M=\psi |_M\in \text {End}_R(M)$. Thus, $\Omega (\psi |_M)=\psi$, which shows that $\Omega$ is surjective.

In addition, suppose that M is a polyform module. Then from Lemma 5.8, $\widetilde{E}(M)$ is quasi-injective. Thus, for any $\vartheta \in \text {End}_R(E(M))$, $\vartheta \widetilde{E}(M)\le \widetilde{E}(M)$. Since $\vartheta |_{\widetilde{E}(M)}\in \text {End}_R(\widetilde{E}(M))$ and $\text {End}_R(M)\cong \text {End}_R(\widetilde{E}(M))$ via ${\Omega }$, there exists $\varphi \in \text {End}_R(M)$ such that $\Omega (\varphi )=\vartheta |_{\widetilde{E}(M)}$. Also by Theorem 3.15, $\vartheta |_M=\varphi$. Thus, $\vartheta M=\varphi M\le M$. Therefore M is a quasi-injective module. $\square$

Corollary 5.10

If M is a quasi-injective module, then $Q(End _R(M))\cong End _R(\widetilde{E}(M))$.

Proof

Since M is a quasi-injective module $\text {End}_R(M)$ is a right self-injective ring. So, $Q(\text {End}_R(M))=\text {End}_R(M)$. Thus, $Q(\text {End}_R(M))\cong \text {End}_R(\widetilde{E}(M))$ by Proposition 5.9. $\square$

Remark that if M is a quasi-injective module then $\widetilde{E}(M)$ is a quasi-injective module from Theorem 3.16(ii) and $\text {End}_R(M)\cong \text {End}_R(\widetilde{E}(M))$ from Proposition 5.9. However, the next example shows that there exists a quasi-injective module M such that $M\ne \widetilde{E}(M)$.

Example 5.11

Let $R=\left( {\begin{matrix}F&{}F\\ 0&{}F\end{matrix}}\right)$ and $M=\left( {\begin{matrix}0&{}0\\ 0&{}F\end{matrix}}\right)$ where F is a field. Then M is a quasi-injective R-module. However, $\widetilde{E}(M)=E(M)=\left( {\begin{matrix}0&{}0\\ F&{}F\end{matrix}}\right)$ because M is nonsingular. Thus M is a quasi-injective R-module such that $M\lneq \widetilde{E}(M)$ and $\text {End}_R(M)\cong \left( {\begin{matrix}0&{}0\\ 0&{}F\end{matrix}}\right) \cong \text {End}_R(\widetilde{E}(M))$.

Because $\widetilde{E}(M)=E(M)$ for a right nonsingular module M, we have the following well-known results as a consequence of Proposition 5.9.

Corollary 5.12

([7, Exercises 7.32]) For any nonsingular module M, the following statements hold true:

• (i) there is a canonical embedding $\Omega : End _R(M) \rightarrow End _R({E}(M))$.

• (ii) M is a quasi-injective R-module if and only if $\Omega$ is an isomorphism.

Corollary 5.13

Let M be a right H-module where H is a right ring of quotients of a ring R. Then the following statements hold true:

• (i) If M is a nonsingular R-module then $End _R(M)=End _H(M)$.

• (ii) If M is a submodule of a projective H-module, then $End _R(M)=End _H(M)$.

• (iii) If M is a nonsingular quasi-injective R-module then $End _R(M)\cong End _R(E(M))$ and $End _H(M)\cong End _H(E(M))$.

• (iv) If M is a quasi-injective R-module and is a submodule of a projective H-module then $End _R(M)\cong End _R(\widetilde{E}(M))$ and $End _H(M)\cong End _H(\widetilde{E}(M))$.