Harmanci Injectivity of Modules

TL;DR

This paper explores Harmanci injective modules, a new class related to injective, flat, and cotorsion modules, providing characterizations, properties, and conditions under which the injective envelope of a ring is flat.

Contribution

It introduces and studies Harmanci injective modules, establishing their properties, their role in cotorsion theory, and conditions for the flatness of injective envelopes of rings.

Findings

Harmanci injective modules form an enveloping class.

They establish a perfect cotorsion theory with Matlis injective modules.

Conditions identified for the injective envelope of a ring to be flat.

Abstract

In this paper, we are interested in a class of modules partaking in the hierarchy of injective and cotorsion modules, so-called Harmanci injective modules, which turn out by the motivation of relations among the concepts of injectivity, flatness and cotorsionness. We give some characterizations and properties of this class of modules. It is shown that the class of all Harmanci injective modules is enveloping, and forms a perfect cotorsion theory with the class of modules whose character modules are Matlis injective. One of the main objectives we pursue is to know when the injective envelope of a ring as a module over itself is a flat module.

Equations

$$\cdots \to \text{mbox}{Tor}_1^R(A,E{(}_{R}R))\to \text{mbox}{Tor}_1^R(B,E{(}_{R}R))\to \text{mbox}{Tor}_1^R(C,E{(}_{R}R))\to A{\otimes }_{R}E{(}_{R}R)\to \cdots .$$

Full text

2010 MSC: 16D10, 16D40, 16D50, 16E30

Keywords: Injective module, Matlis injective module, Harmanci injective module, cotorsion module, flat module, character module, envelope

1. Introduction

Throughout this paper $R$ denotes an associative ring with identity and modules are unitary $R$-modules. The notion of cotorsion abelian groups introduced by Harrison in [8], that is, an abelian group $G$ is called cotorsion if Ext${}^{1}_{\mathbb{Z}}(\mathbb{Q},G)=0$. This notion extended to modules by Enochs in [6], namely, a right $R$-module $M$ is said to be (Enochs)* cotorsion* if Ext${}^{1}_{R}(F,M)=0$ for every flat right $R$-modules $F$. Let $\mathcal{A}$ be a class of right $R$-modules. Then $\mathcal{A}^{\perp}=\{M_{R}\mid\mbox{Ext}^{1}_{R}(A,M)=0,A\in\mathcal{A}\}$ is called the right orthogonal class of $\mathcal{A}$ and ${}^{\perp}\mathcal{A}=\{M_{R}\mid\mbox{Ext}^{1}_{R}(M,A)=0,A\in\mathcal{A}\}$ is called the left orthogonal class of $\mathcal{A}$ (see [15, p.29]). For any classes of right $R$-modules $\mathcal{A}$ and $\mathcal{B}$, if $\mathcal{A}=\hskip 0.28436pt^{\perp}\mathcal{B}$ and $\mathcal{B}=\mathcal{A}^{\perp}$, then the pair $(\mathcal{A},\mathcal{B})$ is called a cotorsion theory. It is well known that $(\mathcal{F},\mathcal{EC})$ is a cotorsion theory where $\mathcal{F}$ and $\mathcal{EC}$ are the classes of all flat modules and all cotorsion modules, respectively.

The connection between a flat module and its character module was observed by Lambek. He proved in [10] that a left $R$-module $M$ is flat if and only if its character module Hom${}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$ is an injective right $R$-module. Thus the class of all cotorsion modules is the right orthogonal class of modules whose character modules are injective. Injective modules, cotorsion modules and various generalizations of these modules have been investigated in the literature by many authors. In [4], a right $R$-module $M$ is called Whitehead if Ext${}^{1}_{R}(M,R)=0$. As a dual notion of Whitehead modules, Yan defined Matlis injective modules in [16], namely, a right $R$-module $M$ is said to be Matlis injective if Ext${}^{1}_{R}(E(R_{R}),M)=0$ where $E(R_{R})$ denotes the injective envelope of the ring $R$ as a right $R$-module. Motivated by the studies on the modules which belong to the right orthogonal class of flat modules, i.e., cotorsion modules, the goal of this paper is to provide an initial contribution to the study of the right orthogonal class of modules whose character modules are Matlis injective. We study the behavior of modules that belong to this right orthogonal class, so-called Harmanci injective modules. We observe that the pair consisting of all modules whose character modules are Matlis injective and all Harmanci injective modules is a cotorsion theory. We are interested in the hierarchy of injective modules and cotorsion modules, in this direction, we show that the class of Harmanci injective modules lies strictly between the classes of injective modules and cotorsion modules. For a commutative Noetherian ring $R$, $E(R)$ being flat is characterized in [15, Theorem 5.1.3]. A natural question arises: When is $E(_{R}R)$ a flat left $R$-module for any ring $R$? One of our main concerns is this question. We give an answer, that is, the notions of injectivity and Harmanci injectivity coincide if and only if $E(_{R}R)$ is flat. It is also known that Hom${}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$ is always pure injective and so cotorsion for any module $M$ (see [15, p.39]). As an application, we deal with character modules and also approximations of modules in terms of Harmanci injectivity.

Let $\mathcal{C}$ be a class of right $R$-modules and $M$ a right $R$-module. Following [5], a homomorphism $f\colon C\rightarrow M$ with $C\in\mathcal{C}$ is said to be a $\mathcal{C}$-precover of $M$ if for any homomorphism $g\colon C^{\prime}\rightarrow M$ with $C^{\prime}\in\mathcal{C}$, there exists a homomorphism $h\colon C^{\prime}\rightarrow C$ such that $fh=g$. The $\mathcal{C}$-precover $f$ is called a $\mathcal{C}$-cover of $M$ if any endomorphism $\alpha\colon C\rightarrow C$ with $f\alpha=f$ is an isomorphism. The concepts of a $\mathcal{C}$-preenvelope and $\mathcal{C}$-envelope are defined dually. Bican, Bashir and Enochs proved the existence of a flat cover and a cotorsion envelope for any module in [1]. It is also well known that every module has an injective envelope. Thus the following question seems natural to wonder: what can we say about the existence of Harmanci injective envelopes and covers with respect to the class of modules whose character modules are Matlis injective and their unique mapping properties? In [3], a $\mathcal{C}$-envelope $f\colon M\rightarrow C$ with $C\in\mathcal{C}$ of a module $M$ said to have the unique mapping property if for any homomorphism $g\colon M\rightarrow C^{\prime}$ with $C^{\prime}\in\mathcal{C}$, there exists a unique $h\colon C\rightarrow C^{\prime}$ such that $hf=g$.

Briefly, we devote the first part of this paper to study another class of modules, so-called Harmanci injective modules, which turn out in the light of relations among the concepts of injectivity, flatness and cotorsionness by addressing several aforementioned questions. We devote the second part of this paper to investigate the existence of Harmanci injective envelopes, its unique mapping property and a cotorsion theory arising from Harmanci injectivity.

In what follows, $\mathbb{Q}$, $\mathbb{Z}$, $\mathbb{Z}/n\mathbb{Z}$ and $E(M)$ denote the ring of rational numbers, the ring of integers, the $\mathbb{Z}$-module of integers modulo $n$ for a positive integer $n$ and the injective envelope of a module $M$, respectively. Also, id$(M)$, pd$(M)$ and fd$(M)$ stand for the injective dimension, projective dimension and flat dimension, respectively.

2. Harmanci Injective Modules

In this section, we consider the right orthogonal class of modules whose character modules are Matlis injective, and call any module in this class as Harmanci injective. Let us start the following example for a ring $R$ whose its injective envelope $E(_{R}R)$ is not a flat left $R$-module.

Example 2.1**.**

Let $D$ be a division ring, $n$ be a positive integer, $U_{n}(D)$ be the ring of $n\times n$ upper triangular matrices over $D$ and $M_{n}(D)$ be the left $U_{n}(D)$-module of $n\times n$ full matrices over $D$. Then the injective envelope of $U_{n}(D)$, considered as a left $U_{n}(D)$-module, is $M_{n}(D)$ which is not flat.**

Proof.

The left $U_{n}(D)$-module $M_{n}(D)$ being an injective envelope of $U_{n}(D)$ is proved in [2]. Next we claim that $M_{n}(D)$ is not a flat left $U_{n}(D)$-module. We prove it by contradiction. Assume that $M_{n}(D)$ is a flat left $U_{n}(D)$-module. Note that $M_{n}(D)$ is finitely generated over the Noetherian ring $U_{n}(D)$, so is finitely presented by [11, Corollary 3.19] and, then, it is also projective by [11, Theorem 3.56]. Hence it is a direct summand of a direct sum of copies of $U_{n}(D)$, say $\bigoplus\limits_{\mathcal{I}}U_{n}(D)=M_{n}(D)\oplus K$ for some submodule $K$ of $\bigoplus\limits_{\mathcal{I}}U_{n}(D)$ where $\mathcal{I}$ is an index set. Also, $\bigoplus\limits_{\mathcal{I}}U_{n}(D)=U_{n}(D)\oplus L$ for some submodule $L$ of $\bigoplus\limits_{\mathcal{I}}U_{n}(D)$. Since $U_{n}(D)$ is a submodule of $M_{n}(D)$, the modularity condition entails that $U_{n}(D)$ is a direct summand of $M_{n}(D)$. So there exists the natural epimorphism $\pi\colon M_{n}(D)\rightarrow U_{n}(D)$. Then $M_{n}(D)/$Ker$\pi\cong U_{n}(D)$. Since $M_{n}(D)$ is an injective left $U_{n}(D)$-module and $U_{n}(D)$ is left hereditary by [12, Example 2.8.13], $M_{n}(D)/$Ker$\pi$ is also injective, so $U_{n}(D)$ is left self-injective, but this is a contradiction, because the homomorphism g\colon U_{n}(D)\left[\begin{array}[]{cc}0&1\\ 0&0\\ \end{array}\right]\rightarrow U_{n}(D) defined by g\left[\begin{array}[]{cc}0&x\\ 0&0\\ \end{array}\right]=\left[\begin{array}[]{cc}x&0\\ 0&0\\ \end{array}\right] can not be extended to $U_{n}(D)$. Therefore $M_{n}(D)$ is not a flat left $U_{n}(D)$-module. ∎

By [11, Proposition 8.18], a left $R$-module $M$ is flat if and only if Tor${}^{R}_{1}(N,M)=0$ for every right $R$-module $N$. By means of Example 2.1, the left $R$-module $E(_{R}R)$ need not be flat in general. In the following, we characterize right $R$-modules which satisfy Tor${}^{R}_{1}(N,E(_{R}R))=0$.

Lemma 2.2**.**

Let $N$ be a right $R$-module. Then Tor${}^{R}_{1}(N,E(_{R}R))=0$ if and only if Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ is Matlis injective.

Proof.

According to [7, Theorem 3.2.1],

Hom${}_{\mathbb{Z}}($Tor${}^{R}_{1}(N,E(_{R}R)),\mathbb{Q}/\mathbb{Z})\cong$ Ext${}^{1}_{R}(E(_{R}R),$ Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z}))$.

This implies Ext${}^{1}_{R}(E(_{R}R),$ Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z}))=0$ if and only if Tor${}^{R}_{1}(N,E(_{R}R))=0$. ∎

We now give our main definition, namely, Harmanci injective modules.

Definition 2.3**.**

A right $R$-module $M$ is said to be Harmanci injective if Tor{}^{R}_{1}\big{(}N,E(_{R}R)\big{)}=0 implies Ext${}^{1}_{R}(N,M)=0$ for every right $R$-module $N$.**

So clearly we have the next result.

Proposition 2.4**.**

Let $R$ be a ring. Then Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z})$ is a Harmanci injective right $R$-module.

Obviously, every injective module is Harmanci injective. We now observe when the converse of this statement holds. For a commutative Noetherian ring $R$, $E(R)$ being flat is characterized in [15, Theorem 5.1.3]. Via the next theorem, we characterize $E(_{R}R)$ being flat for an arbitrary ring $R$. On the other hand, this result and Example 2.1 make sure that there exists a Harmanci injective right $U_{n}(D)$-module which is not injective.

Theorem 2.5**.**

Let $R$ be a ring. Then every Harmanci injective right $R$-module is injective if and only if $E(_{R}R)$ is flat.

Proof.

Assume that Harmanci injectivity implies injectivity. By Proposition 2.4, Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z})$ is Harmanci injective, and so is injective. This yields that $E(_{R}R)$ is flat. Suppose now that $E(_{R}R)$ is flat and $M$ is a Harmanci injective right $R$-module. Let $N$ be any right $R$-module. The module $E(_{R}R)$ being flat implies Tor${}^{R}_{1}(N,E(_{R}R))=0$. Since $M$ is Harmanci injective, Ext${}^{1}_{R}(N,M)=0$. Therefore $M$ is injective. ∎

We immediately get the next consequences from Theorem 2.5.

Corollary 2.6**.**

The following hold.

1. (1)

Injectivity and Harmanci injectivity coincide for the modules over a commutative domain. 2. (2)

If $R$ is a left self-injective ring or a von Neumann regular ring, then a right $R$-module $M$ is injective if and only if $M$ is Harmanci injective.

Proof.

(1) Let $R$ be a commutative domain. Then $E(_{R}R)$ is the field of fractions of $R$ and it is a flat $R$-module by [11, Corollary 5.35(i)]. Hence Theorem 2.5 completes the proof.

(2) If $R$ is a left self-injective ring, then the assertion is obtained immediately from Theorem 2.5. If $R$ is a von Neumann regular ring, then the proof is clear from the fact that every module over $R$ is flat and Theorem 2.5. ∎

The converse statement of (2) in Corollary 2.6 need not be true in general as shown below.

Example 2.7**.**

By Corollary 2.6(1), every Harmanci injective $\mathbb{Z}$-module is injective but $\mathbb{Z}$ is neither self-injective nor von Neumann regular.**

In the next result, we are interested in the flat dimension of $E(_{R}R)$ for any ring $R$ in terms of Harmanci injectivity.

Theorem 2.8**.**

Let $R$ be a ring. Then the flat dimension of $E(_{R}R)$ is exactly $1$ if and only if the injective dimension of every Harmanci injective right $R$-module is exactly $1$.

Proof.

Firstly, we assert that fd$(E(_{R}R))$ is at most $1$ if and only if id$(M)$ is at most $1$ for every Harmanci injective right $R$-module $M$. In an attempt to prove the necessity of this assertion, let $M$ and $N$ be right $R$-modules with $M$ Harmanci injective. There exists an exact sequence $0\rightarrow K\rightarrow F\rightarrow N\rightarrow 0$ where $F$ is a free right $R$-module. Applying the functor $-\otimes_{R}E(_{R}R)$ to the sequence, we get the exactness of

$\cdots\rightarrow$ Tor${}^{R}_{2}(N,E(_{R}R))\rightarrow$ Tor${}^{R}_{1}(K,E(_{R}R))\rightarrow$ Tor${}^{R}_{1}(F,E(_{R}R))\rightarrow\cdots$.

Being fd$(E(_{R}R))\leq 1$ and flatness of $F$ imply that Tor${}^{R}_{1}(K,E(_{R}R))=0$. Since $M$ is Harmanci injective, Ext${}^{1}_{R}(K,M)=0$. On the other hand, we also have the exact sequence $0=$ Ext${}^{1}_{R}(K,M)\rightarrow$ Ext${}^{2}_{R}(N,M)\rightarrow$ Ext${}^{2}_{R}(F,M)\rightarrow\cdots$. As $F$ is projective, Ext${}^{2}_{R}(F,M)=0$. Thus Ext${}^{2}_{R}(N,M)=0$. This implies id$(M)\leq 1$. For the sufficiency, let $N$ be a right $R$-module. By Proposition 2.4 and hypothesis, id$($Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z}))\leq 1$. This yields Ext${}^{2}_{R}(N,$ Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z}))=0$. So by the isomorphism Ext${}^{2}_{R}(N,$ Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z}))\cong$ Hom${}_{\mathbb{Z}}($Tor${}^{R}_{2}(N,E(_{R}R)),\mathbb{Q}/\mathbb{Z})$, we have Tor${}^{R}_{2}(N,E(_{R}R))=0$. Therefore fd$(E(_{R}R))\leq 1$. Now we complete the proof in the light of this authenticated assertion and Theorem 2.5. ∎

The next result shows that the class of Harmanci injective modules lies between those of injective modules and cotorsion modules.

Proposition 2.9**.**

Every Harmanci injective right $R$-module is cotorsion.

Proof.

Let $M$ and $F$ be right $R$-modules with $M$ Harmanci injective and $F$ flat. Then Tor${}^{R}_{1}(F,E(_{R}R))=0$. Hence Harmanci injectivity of $M$ implies Ext${}^{1}_{R}(F,M)=0$. Thus $M$ is cotorsion. ∎

The next examples show that the converse of Proposition 2.9 need not be hold in general.

Example 2.10**.**

(1) The $\mathbb{Z}$-module $M=$ Hom${}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Q}/\mathbb{Z})$ is pure-injective as it is the character module of $\mathbb{Z}/2\mathbb{Z}$ and so it is cotorsion. On the other hand, $M$ is not injective because $\mathbb{Z}/2\mathbb{Z}$ is not flat. Then Corollary 2.6(1) implies that $M$ is not Harmanci injective.

(2) Let $R$ be a quasi-Frobenius (shortly, QF) ring which is not right pure-semisimple. Then $R$ is right perfect. Since every flat right $R$-module is projective, every right $R$-module is cotorsion. On the other hand, there is a right $R$-module $M$ which is not pure-injective as $R$ is not right pure-semisimple. Hence $M$ is not injective. The ring $R$ being left self-injective implies $M$ is not Harmanci injective by Corollary 2.6(2).**

Recall that a right $R$-module $M$ is said to be divisible if Ext${}^{1}_{R}(R/aR,M)=0$ for all $a\in R$. By Corollary 2.6(1) and [11, Corollary 3.35(i)], if $R$ is a principal ideal domain, then an $R$-module $M$ is Harmanci injective if and only if $M$ is injective if and only if $M$ is divisible. In the next result, we investigate when Harmanci injectivity implies being divisible.

Proposition 2.11**.**

Let $R$ be a ring. If every principal right ideal of $R$ is pure, then every Harmanci injective right $R$-module is divisible.

Proof.

Let $M$ be a Harmanci injective right $R$-module and $a\in R$. Consider the short exact sequence $0\rightarrow aR\rightarrow R\rightarrow R/aR\rightarrow 0$. Then we have the exactness of $0=$ Tor${}_{1}^{R}(R,E(_{R}R))\rightarrow$ Tor${}_{1}^{R}(R/aR,E(_{R}R))\rightarrow aR\otimes E(_{R}R)\rightarrow R\otimes E(_{R}R)\rightarrow R/aR\otimes E(_{R}R)\rightarrow 0$. Since $aR$ is pure in $R$, the homomorphism $aR\otimes E(_{R}R)\rightarrow R\otimes E(_{R}R)$ is monic. It follows that Tor${}_{1}^{R}(R/aR,E(_{R}R))=0$. The module $M$ being Harmanci injective implies Ext${}^{1}_{R}(R/aR,M)=0$. Therefore $M$ is divisible. ∎

For any ring $R$, owing to Proposition 2.4, the character module of $E(_{R}R)$ is always Harmanci injective. In the next result, we investigate Harmanci injectivity of a character module of any module.

Theorem 2.12**.**

Let $S$ be a ring and $N$ a left $S$-module. Then the following are equivalent.

1. (1)

Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$* is a Harmanci injective right $S$-module.* 2. (2)

For every ring $R$ with $N$ a left $S$-right $R$-bimodule and every injective right $R$-module $M$, Hom${}_{R}(N,M)$* is Harmanci injective as a right $S$-module.*

Proof.

(1) $\Rightarrow$ (2) Let $N$ be a left $S$-right $R$-bimodule, $M$ an injective right $R$-module and $K$ a right $S$-module with Tor{}^{S}_{1}\big{(}K,E(_{S}S)\big{)}=0. We claim that Ext${}^{1}_{S}(K,$ Hom${}_{R}(N,M))=0$. Harmanci injectivity of the right $S$-module Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ yields Ext${}^{1}_{S}(K,$ Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z}))=0$. It follows Hom${}_{\mathbb{Z}}($Tor${}_{1}^{S}(K,N),\mathbb{Q}/\mathbb{Z})=0$, and so Tor${}_{1}^{S}(K,N)=0$. Hence Hom${}_{R}($Tor${}_{1}^{S}(K,N),M)=0$. Thus injectivity of $M$ implies Ext${}^{1}_{S}(K,$ Hom${}_{R}(N,M))=0$. Therefore the right $S$-module Hom${}_{R}(N,M)$ is Harmanci injective.

(2) $\Rightarrow$ (1) Obvious by taking $R=\mathbb{Z}$ and $M=\mathbb{Q}/\mathbb{Z}$. ∎

Owing to Theorem 2.12 and Proposition 2.4, we acquire the next result.

Corollary 2.13**.**

Let $S$ be a ring. Then for every ring $R$ with $E(_{S}S)$ a left $S$-right $R$-bimodule and every injective right $R$-module $M$, Hom${}_{R}(E(_{S}S),M)$ is a Harmanci injective right $S$-module.

In the sequel, let $\mathcal{HI}$ denote the class of all Harmanci injective right $R$-modules and $\mathcal{CMI}$ stand for the class of all right $R$-modules whose character modules are Matlis injective. In the following, we mention some properties of the class $\mathcal{HI}$.

Lemma 2.14**.**

Harmanci injective modules satisfy the following properties.

1. (1)

Any direct product of Harmanci injective modules is Harmanci injective. 2. (2)

A finite direct sum of Harmanci injective modules is Harmanci injective. 3. (3)

Direct summands of Harmanci injective modules are Harmanci injective. 4. (4)

The class of Harmanci injective modules is closed under extensions.

Proof.

(1) Let $\{M_{i}\}_{i\in I}$ be a collection of Harmanci injective right $R$-modules, $M=\prod\limits_{i\in I}M_{i}$ and $N$ be a right $R$-module with Tor${}^{R}_{1}(N,E(_{R}R))=0$. Then Ext${}^{1}_{R}(N,M_{i})=0$ for each $i\in I$. Since Ext${}^{1}_{R}(N,M)\cong\prod\limits_{i\in I}$ Ext${}^{1}_{R}(N,M_{i})$, we have Ext${}^{1}_{R}(N,M)=0$.

(2) It is clear by (1).

(3) Let $M=M_{1}\oplus M_{2}$ be a Harmanci injective right $R$-module. Let Tor${}^{R}_{1}(N,E(_{R}R))=0$ for some right $R$-module $N$. By hypothesis, Ext${}^{1}_{R}(N,M)=0$. Since Ext${}^{1}_{R}(N,M)\cong$ Ext${}^{1}_{R}(N,M_{1})\oplus$ Ext${}^{1}_{R}(N,M_{2})$, Ext${}^{1}_{R}(N,M_{1})=$ Ext${}^{1}_{R}(N,M_{2})=0$.

(4) Let $0\rightarrow M\rightarrow N\rightarrow K\rightarrow 0$ be an exact sequence of right $R$-modules with $M$ and $K$ Harmanci injective. Let $L$ be a right $R$-module with Tor${}^{R}_{1}(L,E(_{R}R))=0$. Then we obtain $\cdots\rightarrow$ Ext${}_{R}^{1}(L,M)\rightarrow$ Ext${}_{R}^{1}(L,N)\rightarrow$ Ext${}_{R}^{1}(L,K)\rightarrow\cdots$. By hypothesis, Ext${}_{R}^{1}(L,M)=$ Ext${}_{R}^{1}(L,K)=0$. This yields Ext${}_{R}^{1}(L,N)=0$. ∎

Proposition 2.15**.**

Let $M$ be a Harmanci injective right $R$-module and $K$ a submodule of $M$ with injective dimension at most $1$. Then $M/K$ is Harmanci injective.

Proof.

Consider the exact sequence $0\rightarrow K\rightarrow M\rightarrow M/K\rightarrow 0$. Then we have the long exact sequence $\cdots\rightarrow$ Ext${}^{1}_{R}(N,M)\rightarrow$ Ext${}^{1}_{R}(N,M/K)\rightarrow$ Ext${}^{2}_{R}(N,K)\rightarrow\cdots$

by means of the functor Hom${}_{R}(N,-)$ where $N$ is a right $R$-module such that Tor${}^{R}_{1}(N,E(_{R}R))=0$. Since $M$ is Harmanci injective, Ext${}^{1}_{R}(N,M)=0$, also Ext${}^{2}_{R}(N,K)=0$ by hypothesis. Therefore Ext${}^{1}_{R}(N,M/K)=0$. ∎

We now characterize quotients of Harmanci injective modules being Harmanci injective. Recall that a module is said to be h-divisible if it is an epic image of an injective module. Let $M$ be a right $R$-module and $K$ a submodule of $M$. We call $K$ an $E(_{R}R)$-pure submodule of $M$ if the sequence $0\rightarrow K\otimes_{R}E(_{R}R)\rightarrow M\otimes_{R}E(_{R}R)$ is exact. Obviously, every pure submodule of a right $R$-module is $E(_{R}R)$-pure.

Proposition 2.16**.**

The following are equivalent.

1. (1)

The class $\mathcal{HI}$ is closed under homomorphic images. 2. (2)

Every h-divisible right $R$-module is Harmanci injective. 3. (3)

Every module which belongs to $\mathcal{CMI}$ has projective dimension at most $1$. 4. (4)

Every $E(_{R}R)$-pure submodule of projective modules is projective.

In this case, the projective dimension of a flat right $R$-module is at most $1$, equivalently, pure submodules of projective modules are also projective.

Proof.

(1) $\Rightarrow$ (2) Obvious.

(2) $\Rightarrow$ (3) Let $N\in\mathcal{CMI}$ and $M$ any right $R$-module. We claim that Ext${}_{R}^{2}(N,M)=0$. Applying the functor Hom${}_{R}(N,-)$ to the exact sequence $0\rightarrow M\rightarrow E(M)\rightarrow E(M)/M\rightarrow 0$, we have the exactness of

$\cdots\rightarrow$ Ext${}_{R}^{1}(N,E(M)/M)\rightarrow$ Ext${}_{R}^{2}(N,M)\rightarrow$ Ext${}_{R}^{2}(N,E(M))\rightarrow\cdots$.

By (2), Ext${}_{R}^{1}(N,E(M)/M)=0$, also the injectivity of $E(M)$ implies Ext${}_{R}^{2}(N,E(M))=0$. Thus Ext${}_{R}^{2}(N,M)=0$, as desired.

(3) $\Rightarrow$ (1) Let $M$ be a Harmanci injective right $R$-module, $K$ a submodule of $M$ and $N$ a right $R$-module with Tor${}^{R}_{1}(N,E(_{R}R))=0$. Consider the exact sequence $0\rightarrow K\rightarrow M\rightarrow M/K\rightarrow 0$. The functor Hom${}_{R}(N,-)$ yields the exact sequence

$\cdots\rightarrow$ Ext${}_{R}^{1}(N,M)\rightarrow$ Ext${}_{R}^{1}(N,M/K)\rightarrow$ Ext${}_{R}^{2}(N,K)\rightarrow\cdots$.

Harmanci injectivity of $M$ and (3) imply Ext${}_{R}^{1}(N,M)=$ Ext${}_{R}^{2}(N,K)=0$. Therefore Ext${}_{R}^{1}(N,M/K)=0$.

(3) $\Rightarrow$ (4) Let $P$ be a projective right $R$-module and $K$ be an $E(_{R}R)$-pure submodule of $P$. Applying the functor $-\otimes_{R}E(_{R}R)$ to the short exact sequence $0\rightarrow K\rightarrow P\rightarrow P/K\rightarrow 0$, we have $\cdots\rightarrow$ Tor${}^{R}_{1}(P,E(_{R}R))\rightarrow$ Tor${}^{R}_{1}(P/K,E(_{R}R))\rightarrow K\otimes_{R}E(_{R}R)\rightarrow P\otimes_{R}E(_{R}R)\rightarrow\cdots$. The module $P$ being projective and $K$ being an $E(_{R}R)$-pure submodule of $P$ imply Tor${}^{R}_{1}(P/K,E(_{R}R))=0$. Then pd$(P/K)\leq 1$ by (3). If pd$(K)>0=$ pd$(P)$, then pd$(P/K)\geq 2$ by [11, p.466, Ex.8.5(ii)]. This contradiction yields pd$(K)=0$, i.e, $K$ is projective.

(4) $\Rightarrow$ (3) Let $M\in\mathcal{CMI}$. There exists a short exact sequence $0\rightarrow K\rightarrow F\rightarrow M\rightarrow 0$ where $F$ is free. Since Tor${}^{R}_{1}(M,E(_{R}R))=0$, $K$ is an $E(_{R}R)$-pure submodule of $F$. By (4), $K$ is projective. In the light of [11, p.466, Ex.8.5(iii)], pd$(M)\leq 1$. ∎

Recall that a class $\mathcal{C}$ is called coresolving provided that $\mathcal{C}$ is closed under extensions, every injective module is in $\mathcal{C}$ and $C\in\mathcal{C}$ whenever $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence such that $A,B\in\mathcal{C}$. In the light of the fact that injectivity implies Harmanci injectivity and Lemma 2.14(4), we now address the following question: When is the class $\mathcal{HI}$ coresolving?

Proposition 2.17**.**

The following are equivalent.

1. (1)

For every exact sequence $0\rightarrow M\rightarrow N\rightarrow K\rightarrow 0$ of right $R$-modules, if $M$ and $N$ are Harmanci injective, then $K$ is Harmanci injective. 2. (2)

If $M$ is a Harmanci injective right $R$-module, then $E(M)/M$ is Harmanci injective. 3. (3)

If $M$ is a Harmanci injective right $R$-module, then for any right $R$-module $N$, being Tor${}^{R}_{1}(N,E(_{R}R))=0$* implies Ext*${}^{n}_{R}(N,M)=0$* where $n\geq 2$.*

In this case, $\mathcal{HI}$ is a coresolving class.

Proof.

(1) $\Rightarrow$ (2) Clear.

(2) $\Rightarrow$ (3) Let $M$ and $N$ be right $R$-modules with $M$ Harmanci injective and Tor{}^{R}_{1}\big{(}N,E(_{R}R)\big{)}=0. If we apply the functor Hom${}_{R}(N,-)$ to the exact sequence $0\rightarrow M\rightarrow E(M)\rightarrow E(M)/M\rightarrow 0$, then we obtain

$\cdots\rightarrow$ Ext${}_{R}^{1}(N,E(M)/M)\rightarrow$ Ext${}_{R}^{2}(N,M)\rightarrow$ Ext${}_{R}^{2}(N,E(M))\rightarrow\cdots$.

By (2), Ext${}_{R}^{1}(N,E(M)/M)=0$ and by the injectivity of $E(M)$, Ext${}_{R}^{2}(N,E(M))=0$. Hence Ext${}_{R}^{2}(N,M)=0$. Since $E(M)/M$ is Harmanci injective, by the similar discussion above, we have Ext${}_{R}^{2}(N,E(M)/M)=0$. This yields Ext${}_{R}^{3}(N,M)=0$. Continuing in this way, by induction on $n\geq 2$, Ext${}^{n}_{R}(N,M)=0$ is obtained.

(3) $\Rightarrow$ (1) Consider an exact sequence $0\rightarrow M\rightarrow N\rightarrow K\rightarrow 0$ of right $R$-modules with $M$ and $N$ Harmanci injective. Let $L$ be a right $R$-module with Tor${}^{R}_{1}(L,E(_{R}R))=0$. Then we have the exact sequence $\cdots\rightarrow$ Ext${}_{R}^{1}(L,N)\rightarrow$ Ext${}_{R}^{1}(L,K)\rightarrow$ Ext${}_{R}^{2}(L,M)\rightarrow\cdots$. By hypothesis, Ext${}_{R}^{1}(L,N)=$ Ext${}_{R}^{2}(L,M)=0$. It follows that Ext${}_{R}^{1}(L,K)=0$, establishing the result. ∎

We now investigate when the character module of a Harmanci injective module is Matlis injective. We need the next lemma for this investigation. By this means, some properties of the class $\mathcal{CMI}$ are acquired.

Lemma 2.18**.**

The following hold.

1. (1)

$\mathcal{CMI}$* is closed under extensions.* 2. (2)

$\mathcal{CMI}$* is closed under direct summands.* 3. (3)

$\mathcal{CMI}$* is closed under direct sums.* 4. (4)

$\mathcal{CMI}$* is closed under pure quotients.* 5. (5)

$\mathcal{CMI}$* is a covering class.* 6. (6)

The kernel of every $\mathcal{CMI}$-cover is Harmanci injective. 7. (7)

Every right $R$-module has a $\mathcal{CMI}$-cover with the unique mapping property if and only if for every exact sequence $A\rightarrow B\rightarrow C\rightarrow 0$ of right $R$-modules, being $A,B\in\mathcal{CMI}$ implies $C\in\mathcal{CMI}$.

Proof.

Note that by Lemma 2.2, $\mathcal{CMI}=\{N_{R}\mid$ Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ is Matlis injective$\}=\{N_{R}\mid$ Tor${}^{R}_{1}(N,E(_{R}R))=0\}$.

(1) Let $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be an exact sequence of right $R$-modules with $A,C\in\mathcal{CMI}$. We claim that $B\in\mathcal{CMI}$. If we apply the functor $-\otimes_{R}E(_{R}R)$ to the sequence, we get the long exact sequence

[TABLE]

Since $A,C\in\mathcal{CMI}$, Tor${}^{R}_{1}(A,E(_{R}R))=$ Tor${}^{R}_{1}(C,E(_{R}R))=0$. This implies that Tor${}^{R}_{1}(B,E(_{R}R))=0$. Hence $B\in\mathcal{CMI}$, as desired.

(2) Let $N\in\mathcal{CMI}$ and assume that $N$ has a decomposition $N=K\oplus L$. Then $0=$ Tor${}^{R}_{1}(N,E(_{R}R))\cong$ Tor${}^{R}_{1}(K,E(_{R}R))\oplus$ Tor${}^{R}_{1}(L,E(_{R}R))$. Hence Tor${}^{R}_{1}(K,E(_{R}R))=$ Tor${}^{R}_{1}(L,E(_{R}R))=0$. Hence $K,L\in\mathcal{CMI}$.

(3) Let $\{M_{i}\}_{i\in I}$ be a family of modules for an index set $I$ with $M_{i}\in\mathcal{CMI}$ for each $i\in I$. It is known that Tor${}^{R}_{1}(\bigoplus\limits_{i\in I}M_{i},E(_{R}R))\cong\bigoplus\limits_{i\in I}$ Tor${}^{R}_{1}(M_{i},E(_{R}R))$. For each $i\in I$, being Tor${}^{R}_{1}(M_{i},E(_{R}R))=0$ implies Tor${}^{R}_{1}(\bigoplus\limits_{i\in I}M_{i},E(_{R}R))=0$. Therefore $\bigoplus\limits_{i\in I}M_{i}\in\mathcal{CMI}$.

(4) Let $M\in\mathcal{CMI}$ and $N$ a pure submodule of $M$. We show that $M/N$ belongs to $\mathcal{CMI}$. If we apply the functor $-\otimes_{R}E(_{R}R)$ to the exact sequence $0\rightarrow N\rightarrow M\rightarrow M/N\rightarrow 0$, the long exact sequence $\cdots\rightarrow$ Tor${}^{R}_{1}(M,E(_{R}R))\rightarrow$ Tor${}^{R}_{1}(M/N,E(_{R}R))\rightarrow N\otimes_{R}E(_{R}R)\rightarrow M\otimes_{R}E(_{R}R)\rightarrow M/N\otimes_{R}E(_{R}R)\rightarrow 0$ is obtained. Since $M\in\mathcal{CMI}$ and $N\otimes_{R}E(_{R}R)\rightarrow M\otimes_{R}E(_{R}R)$ is monic, Tor${}^{R}_{1}(M/N,E(_{R}R))=0$.

(5) Clear by (3), (4) and [9, Theorem 2.5].

(6) It follows from (1) and Wakamatsu’s Lemma [15, Lemma 2.1.1].

(7) Assume that every right $R$-module has a $\mathcal{CMI}$-cover with the unique mapping property and consider an exact sequence $A\overset{f}{\rightarrow}B\overset{g}{\rightarrow}C\rightarrow 0$ where $A,B\in\mathcal{CMI}$. Let $h\colon D\rightarrow C$ be a $\mathcal{CMI}$-cover of $C$. Then there exists a homomorphism $\alpha\colon B\rightarrow D$ such that $h\alpha=g$ as shown in the following diagram.

\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\scriptstyle{\alpha}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\textstyle{0}

Hence $h\alpha f=gf=0$ because of the exactness of the sequence. The unique mapping property and being $h(\alpha f)=h0$ imply $\alpha f=0$. Thus Ker$g=$ Im$f\subseteq$ Ker$\alpha$. So Factor Theorem yields a homomorphism $\beta\colon C\rightarrow D$ such that $\beta g=\alpha$. Then $h\beta g=h\alpha=g=1_{C}g$, and so $h\beta=1_{C}$ because $g$ is an epimorphism. It follows that $D=$ Ker$h~{}\oplus$ Im$\beta$ and $\beta$ is a monomorphism. Thus $C\cong$ Im$\beta$. Being $D\in\mathcal{CMI}$ implies $C\in\mathcal{CMI}$ by (2). Conversely, let $M$ be a right $R$-module. By (5), there is a $\mathcal{CMI}$-cover $f\colon F\rightarrow M$ of $M$. Suppose that for any $G\in\mathcal{CMI}$ and $g\colon G\rightarrow M$, there exist $h_{1},h_{2}\colon G\rightarrow F$ such that $fh_{1}=fh_{2}=g$. Then $f(h_{1}-h_{2})=0$, so Im$(h_{1}-h_{2})\subseteq$ Ker$f$. Hence there is a homomorphism $\alpha\colon F/$Im$(h_{1}-h_{2})\rightarrow M$ with $\alpha\pi=f$ by the Factor Theorem where $\pi\colon F\rightarrow F/$Im$(h_{1}-h_{2})$ is the natural projection. On the other hand, the exactness of $G\rightarrow F\rightarrow F/$Im$(h_{1}-h_{2})\rightarrow 0$ implies $F/$Im$(h_{1}-h_{2})\in\mathcal{CMI}$ by hypothesis. It follows that there exists $\beta\colon F/$Im$(h_{1}-h_{2})\rightarrow F$ with $f\beta=\alpha$. Thus $f(\beta\pi)=\alpha\pi=f$. Since $f$ is a $\mathcal{CMI}$-cover, $\beta\pi$ is an isomorphism. This yields $\pi$ is a monomorphism, and so Ker$\pi=$ Im$(h_{1}-h_{2})=0$. This implies that $h_{1}=h_{2}$. This completes the proof. ∎

Theorem 2.19**.**

Consider the following conditions.

1. (1)

$\mathcal{CMI}$* is closed under homomorphic images.* 2. (2)

Every Harmanci injective right $R$-module belongs to $\mathcal{CMI}$.

Then (1) $\Rightarrow$ (2). Moreover, if every $\mathcal{CMI}$-cover satisfies the unique mapping property, then (2) $\Rightarrow$ (1).

Proof.

(1) $\Rightarrow$ (2) Let $M$ be a Harmanci injective right $R$-module. Then $M$ is a homomorphic image of a flat module $F$. Since the character module of $F$ is injective, so is Matlis injective, $F\in\mathcal{CMI}$. Hence $M\in\mathcal{CMI}$ by (1).

Now assume that every $\mathcal{CMI}$-cover satisfies the unique mapping property.

(2) $\Rightarrow$ (1) Let $M$ be a homomorphic image of a module in $\mathcal{CMI}$. We claim that $M\in\mathcal{CMI}$. By Lemma 2.18(5) and assumption, $M$ has a $\mathcal{CMI}$-cover with the unique mapping property, say $f\colon F\rightarrow M$. Then we have an exact sequence $0\rightarrow\mbox{Ker}f\rightarrow F\overset{f}{\rightarrow}M\rightarrow 0$ with $F\in\mathcal{CMI}$. By Lemma 2.18(6), Ker$f$ is Harmanci injective. Hence (2) implies Ker$f\in\mathcal{CMI}$. By taking into account of Lemma 2.18(7), we have $M\in\mathcal{CMI}$. ∎

We close this section by observing some characterizations of Harmanci injectivity.

Proposition 2.20**.**

The following are equivalent for a right $R$-module $M$.

1. (1)

$M$* is Harmanci injective.* 2. (2)

$M$* is injective relative to every exact sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ of right $R$-modules with $C\in\mathcal{CMI}$.* 3. (3)

Every right $R$-module $N\in\mathcal{CMI}$ is projective relative to any exact sequence $0\rightarrow M\rightarrow B\rightarrow C\rightarrow 0$ of right $R$-modules.

Proof.

(1) $\Rightarrow$ (2) Assume that $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is an exact sequence where $C\in\mathcal{CMI}$. Being Ext${}^{1}_{R}(C,M)=0$ gives rise to the exactness of Hom${}_{R}(B,M)\rightarrow$ Hom${}_{R}(A,M)\rightarrow 0$. Therefore $M$ is injective relative to the aforementioned sequence.

(2) $\Rightarrow$ (1) Let $N\in\mathcal{CMI}$ be a right $R$-module and $0\rightarrow M\overset{f}{\rightarrow}X\rightarrow N\rightarrow 0$ an exact sequence for a right $R$-module $X$. (2) yields the exactness of Hom${}_{R}(X,M)\rightarrow$ Hom${}_{R}(M,M)\rightarrow 0$. Hence there exists a homomorphism $g\colon X\rightarrow M$ such that $gf=1_{M}$. It follows that the sequence $0\rightarrow M\rightarrow X\rightarrow N\rightarrow 0$ is split. Therefore Ext${}^{1}_{R}(N,M)=0$.

(1) $\Leftrightarrow$ (3) It is a dual of the proof of (1) $\Leftrightarrow$ (2). ∎

Theorem 2.21**.**

Let $R$ be a commutative ring. Then the following are equivalent for an $R$-module $M$.

1. (1)

$M$* is Harmanci injective.* 2. (2)

Hom${}_{R}(N,M)$* is Harmanci injective for every flat $R$-module $N$.* 3. (3)

Hom${}_{R}(N,M)$* is Harmanci injective for every projective $R$-module $N$.* 4. (4)

Hom${}_{R}(N,M)$* is Harmanci injective for every free $R$-module $N$.*

Proof.

(1) $\Rightarrow$ (2) Let $N$ and $K$ be $R$-modules with $N$ flat and Tor${}^{R}_{1}(K,E(_{R}R))=0$. There exists an exact sequence $0\rightarrow L\rightarrow F\rightarrow K\rightarrow 0$ where $F$ is free. Since $N$ is flat, the sequence $0\rightarrow N\otimes_{R}L\rightarrow N\otimes_{R}F\rightarrow N\otimes_{R}K\rightarrow 0$ is exact. It follows that

$\cdots\rightarrow$ Hom${}_{R}(N\otimes_{R}F,M)\rightarrow$ Hom${}_{R}(N\otimes_{R}L,M)\rightarrow$ Ext${}^{1}_{R}(N\otimes_{R}K,M)\rightarrow\cdots$

is also exact. Since $N$ is flat and Tor${}^{R}_{1}(K,E(_{R}R))=0$, Tor${}^{R}_{1}(N\otimes_{R}K,E(_{R}R))\cong N\otimes_{R}$ Tor${}^{R}_{1}(K,E(_{R}R))=0$ by [11, p. 667]. This implies that Ext${}^{1}_{R}(N\otimes_{R}K,M)=0$ by (1). Hence via the Adjoint Isomorphism, we have the exactness of the sequence

Hom${}_{R}(F,$ Hom${}_{R}(N,M))\rightarrow$ Hom${}_{R}(L,$ Hom${}_{R}(N,M))\rightarrow 0$.

Then again, according to the short exact sequence $0\rightarrow L\rightarrow F\rightarrow K\rightarrow 0$ and the functor Hom${}_{R}(-,$ Hom${}_{R}(N,M))$, we obtain

$\cdots\rightarrow$ Hom${}_{R}(F,$ Hom${}_{R}(N,M))\rightarrow$ Hom${}_{R}(L,$ Hom${}_{R}(N,M))\rightarrow$ Ext${}^{1}_{R}(K,$ Hom${}_{R}(N,M))\rightarrow$ Ext${}^{1}_{R}(F,$ Hom${}_{R}(N,M))=0$.

Therefore Ext${}^{1}_{R}(K,$ Hom${}_{R}(N,M))=0$, as desired.

(2) $\Rightarrow$ (3) $\Rightarrow$ (4) Obvious.

(4) $\Rightarrow$ (1) Clear from the fact that $M\cong$ Hom${}_{R}(R,M)$. ∎

We illustrate some consequences of Theorem 2.21 as follows.

Corollary 2.22**.**

Let $R$ be a commutative ring and $M$ a flat $R$-module. If $M$ is fully invariant in a Harmanci injective $R$-module, then End$(M)$ is also a Harmanci injective $R$-module.

Corollary 2.23**.**

The following are equivalent for a ring $R$.

1. (1)

$R$* is Harmanci injective as a right $R$-module.* 2. (2)

Hom${}_{R}(N,R)$* is a Harmanci injective right $R$-module for every $R$-$R$ bimodule $N$ with $N$ flat as a left $R$-module.* 3. (3)

Hom${}_{R}(N,R)$* is a Harmanci injective right $R$-module for every $R$-$R$ bimodule $N$ with $N$ projective as a left $R$-module.* 4. (4)

Hom${}_{R}(N,R)$* is a Harmanci injective right $R$-module for every $R$-$R$ bimodule $N$ with $N$ free as a left $R$-module.* 5. (5)

Every direct product of the left $R$-module $R$ is Harmanci injective as a right $R$-module.

3. The pair $(\mathcal{CMI},\mathcal{HI})$

In this section, we are particularly interested in some properties of the pair $(\mathcal{CMI},\mathcal{HI})$ and approximations of modules in terms of Harmanci injectivity.

Theorem 3.1**.**

The pair $(\mathcal{CMI},\mathcal{HI})$ is a cotorsion theory.

Proof.

Obviously, $\mathcal{HI}=\mathcal{CMI}^{\bot}$. We need to show that $\mathcal{CMI}=$ ${}^{\bot}\mathcal{HI}$. By the definition, it is clear that $\mathcal{CMI}\subseteq$ ${}^{\bot}\mathcal{HI}$. For the reverse inclusion, let $N\in$ ${}^{\bot}\mathcal{HI}$. By Proposition 2.4, Ext${}^{1}_{R}(N,$ Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z}))=0$. So being Ext${}^{1}_{R}(N,$ Hom${}_{\mathbb{Z}}(E(_{R}R),\mathbb{Q}/\mathbb{Z}))\cong$ Hom${}_{\mathbb{Z}}($Tor${}^{R}_{1}(N,E(_{R}R)),\mathbb{Q}/\mathbb{Z})$ implies Tor${}^{R}_{1}(N,E(_{R}R))=0$, that is Hom${}_{\mathbb{Z}}(N,\mathbb{Q}/\mathbb{Z})$ is Matlis injective by Lemma 2.2. Thus $N\in\mathcal{CMI}$. Therefore ${}^{\bot}\mathcal{HI}\subseteq\mathcal{CMI}$. This completes the proof. ∎

Remark 3.2**.**

In the light of Theorem 3.1, we have the following.

1. (1)

The following are equivalent.

1. (i)

Every cotorsion right $R$-module is Harmanci injective. 2. (ii)

Every module which belongs to $\mathcal{CMI}$ is flat. 2. (2)

The following are equivalent.

1. (i)

Every right $R$-module is Harmanci injective. 2. (ii)

Every module which belongs to $\mathcal{CMI}$ is projective. 3. (3)

If every right $R$-module is Harmanci injective, then $R$ is right perfect.

Let $(\mathcal{F},\mathcal{T})$ be a cotorsion pair. It is called perfect if $\mathcal{F}$ is a covering class and $\mathcal{T}$ is an enveloping class. Also, $(\mathcal{F},\mathcal{T})$ is complete if each module has a special $\mathcal{F}$-precover, equivalently, each module has a special $\mathcal{T}$-preenvelope (see [14, Lemma 1.17]). Let $\mathcal{C}\subseteq$ Mod-$R$ and $M\in$ Mod-$R$ and $f\colon M\rightarrow C$ be a $\mathcal{C}$-preenvelope of $M$. It is called special if $f$ is injective and Ext${}^{1}_{R}(\mbox{Coker}f,T)=0$ for all $T\in\mathcal{C}$. Let $f\colon C\rightarrow M$ be a $\mathcal{C}$-precover. It is called special if $f$ is surjective and Ext${}^{1}_{R}(F,\mbox{Ker}f)=0$ for all $F\in\mathcal{C}$. Given the above concepts, we now address the following question: What can be said about such properties for the cotorsion pair $(\mathcal{CMI},\mathcal{HI})$?

Theorem 3.3**.**

The following hold.

1. (1)

The cotorsion theory $(\mathcal{CMI},\mathcal{HI})$ is complete. 2. (2)

$\mathcal{CMI}$* is a special covering class.* 3. (3)

$\mathcal{HI}$* is a special enveloping class.* 4. (4)

The cotorsion theory $(\mathcal{CMI},\mathcal{HI})$ is perfect.

Proof.

(1) Follows from [13, Lemma 1.9(2) and Lemma 1.13] and Lemma 2.18(1) and (5).

(2) By Lemma 2.18(5) each module has a $\mathcal{CMI}$-cover. Since the class of all projective modules is contained in $\mathcal{CMI}$, every $\mathcal{CMI}$-cover is an epimorphism. On the other hand, for a $\mathcal{CMI}$-cover $f$, Lemma 2.18(6) implies that Ext${}^{1}_{R}(F,\mbox{Ker}f)=0$ for all $F\in\mathcal{CMI}$. Thus every $\mathcal{CMI}$-cover is special.

(3) Since Tor commutes with direct limits (see [11, Proposition 7.8]), the class $\mathcal{CMI}$ is closed under direct limits. Hence (1) and [13, Corollary 1.19] imply that $\mathcal{HI}$ is an enveloping class. As the class of all injective modules is contained in $\mathcal{HI}$, every $\mathcal{HI}$-envelope is a monomorphism. For a $\mathcal{HI}$-envelope $f$, [15, Lemma 2.1.2] and Lemma 2.14(4) yield that Ext${}^{1}_{R}(\mbox{Coker}f,X)=0$ for all $X\in\mathcal{HI}$. Hence every $\mathcal{HI}$-envelope is special.

(4) Since $\mathcal{CMI}$ is a covering class and $\mathcal{HI}$ is an enveloping class, $(\mathcal{CMI},\mathcal{HI})$ is perfect. ∎

Let $M$ be a module and $HIE(M)$ denote the Harmanci injective envelope of $M$. By Theorem 3.3(3), $HIE(M)/M$ belongs to $\mathcal{CMI}$. We now investigate the case for injective envelopes for Harmanci injective modules.

Proposition 3.4**.**

Let $M$ be a Harmanci injective module and $N$ is an essential extension of $M$. Then $N/M$ belongs to $\mathcal{CMI}$ if and only if $M=N$.

Proof.

Let $N/M\in\mathcal{CMI}$ and consider the exact sequence $0\rightarrow M\rightarrow N\rightarrow N/M\rightarrow 0$. If we apply the functor Hom${}_{R}(-,M)$ to the sequence, then we obtain the exactness of Hom${}_{R}(N,M)\rightarrow$ Hom${}_{R}(M,M)\rightarrow$ Ext${}^{1}_{R}(N/M,M)=0$ by the Harmanci injectivity of $M$. It follows that $M$ is a direct summand of $N$. So the essentiality of $M$ in $N$ yields $M=N$. The converse is evident. ∎

Theorem 3.5**.**

A module $M$ is injective if and only if $M$ is Harmanci injective and $E(M)/M\in\mathcal{CMI}$.

Proof.

The necessity is obvious. For the sufficiency, we have $M=E(M)$ by Proposition 3.4. This completes the proof. ∎

We now apply Theorem 3.5 to rings.

Corollary 3.6**.**

The following are equivalent for a ring $R$.

1. (1)

$R$* is right self-injective.* 2. (2)

$R$* is Harmanci injective as a right $R$-module, $E(R_{R})$ is flat and $R$ is a pure submodule of $E(R_{R})$.* 3. (3)

$R$* is a Harmanci injective right $R$-module and $E(R_{R})/R\in\mathcal{CMI}$.*

Proof.

(1) $\Rightarrow$ (2) Obvious. (3) $\Rightarrow$ (1) By Theorem 3.5.

(2) $\Rightarrow$ (3) Applying the functor $-\otimes_{R}E(_{R}R)$ to the exact sequence $0\rightarrow R\rightarrow E(R_{R})\rightarrow E(R_{R})/R\rightarrow 0$, we acquire the long exact sequence

$\cdots\rightarrow$ Tor${}^{R}_{1}(E(R_{R}),E(_{R}R))\rightarrow$ Tor${}^{R}_{1}(E(R_{R})/R,E(_{R}R))\rightarrow R\otimes_{R}E(_{R}R)\rightarrow E(R_{R})\otimes_{R}E(_{R}R)\rightarrow E(R_{R})/R\otimes_{R}E(_{R}R)\rightarrow 0$. According to flatness of $E(R_{R})$, we have Tor${}^{R}_{1}(E(R_{R}),E(_{R}R))=0$. Also by (2), the purity of $R$ in $E(R_{R})$ yields that Tor${}^{R}_{1}(E(R_{R})/R,E(_{R}R))=0$ as asserted. ∎

Proposition 3.7**.**

$\mathcal{CMI}$-covers of Harmanci injective modules are Harmanci injective.

Proof.

Let $M$ be a Harmanci injective module and $f\colon C\rightarrow M$ a $\mathcal{CMI}$-cover of $M$. According to Theorem 3.3(2), we have the exact sequence $0\rightarrow$ Ker$f\rightarrow C\rightarrow M\rightarrow 0$. Lemma 2.18(6) implies that Ker$f$ is Harmanci injective. Hence Lemma 2.14(4) completes the proof. ∎

Proposition 3.8**.**

Let $M$ be a right $R$-module and consider the following conditions.

1. (1)

$M$* is Harmanci injective.* 2. (2)

For every exact sequence $0\rightarrow M\rightarrow B\overset{f}{\rightarrow}C\rightarrow 0$ of right $R$-modules with $B\in\mathcal{CMI}$, $f$ is a $\mathcal{CMI}$-precover of $C$. 3. (3)

There exists a $\mathcal{CMI}$-precover $f\colon B\rightarrow C$ with $B$ Harmanci injective and Ker$f=M$.

Then (3) $\Rightarrow$ (1) $\Rightarrow$ (2). Furthermore if $\mathcal{CMI}$ is closed under homomorphic images, then all of them are equivalent.

Proof.

(3) $\Rightarrow$ (1) Let $N\in\mathcal{CMI}$ and consider the exact sequence $0\rightarrow M\rightarrow B\overset{f}{\rightarrow}C\rightarrow 0$ where $f$ is a $\mathcal{CMI}$-precover and $B$ is Harmanci injective by (3). Applying the functor Hom${}_{R}(N,-)$, we behold the exactness of Hom${}_{R}(N,B)\rightarrow$ Hom${}_{R}(N,C)\rightarrow$ Ext${}^{1}_{R}(N,M)\rightarrow$ Ext${}^{1}_{R}(N,B)\rightarrow\cdots$. The homomorphism $f$ being a $\mathcal{CMI}$-precover and Harmanci injectivity of $B$ imply Ext${}^{1}_{R}(N,M)=0$.

(1) $\Rightarrow$ (2) Obvious by Proposition 2.20.

Now assume that the class $\mathcal{CMI}$ is closed under homomorphic images.

(2) $\Rightarrow$ (3) The natural projection $HIE(M)\rightarrow HIE(M)/M$ is the required $\mathcal{CMI}$-precover due to Theorem 2.19. ∎

We end this paper by investigating the unique mapping property of $\mathcal{HI}$-envelopes.

Theorem 3.9**.**

The following are equivalent for a ring $R$.

1. (1)

Every right $R$-module has an $\mathcal{HI}$-envelope with the unique mapping property. 2. (2)

For every exact sequence $0\rightarrow A\rightarrow B\rightarrow C$ of right $R$-modules, being $B,C\in\mathcal{HI}$ implies $A\in\mathcal{HI}$.

Proof.

(1) $\Rightarrow$ (2) Let $0\rightarrow A\overset{f}{\rightarrow}B\overset{g}{\rightarrow}C$ be an exact sequence of right $R$-modules with $B,C\in\mathcal{HI}$ and $h\colon A\rightarrow D$ an $\mathcal{HI}$-envelope of $A$. Since $B$ is Harmanci injective, there exists a unique homomorphism $\alpha\colon D\rightarrow B$ such that $\alpha h=f$. Consider the following diagram:

\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\scriptstyle{\beta}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{h}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\textstyle{C}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}

The equality $gf=0$ implies $g\alpha h=0=0h$, and so $g\alpha=0$ due to the unique mapping property of $h$. Then Im$\alpha\subseteq$ Ker$g=$ Im$f$. By Factor Theorem, there exists a homomorphism $\beta\colon D\rightarrow A$ with $f\beta=\alpha$. Hence $f\beta h=\alpha h=f=f1_{A}$. Since $f$ is monic, $\beta h=1_{A}$, this yields $D=$ Ker$\beta\oplus$ Im$h$. By Theorem 3.3(3), it is known that $h$ is monic, so $A\cong$ Im$h$. Thus this isomorphism and Lemma 2.14(3) imply that $A$ is Harmanci injective.

(2) $\Rightarrow$ (1) Let $M$ be a right $R$-module and $0\rightarrow M\overset{f}{\rightarrow}F$ an $\mathcal{HI}$-envelope of $M$. Assume that for any $G\in\mathcal{HI}$ and any homomorphism $g\colon M\rightarrow G$, there exist $h_{1},h_{2}\colon F\rightarrow G$ such that $h_{1}f=h_{2}f=g$. Since $(h_{1}-h_{2})f=0$, Im$f\subseteq$ Ker$(h_{1}-h_{2})$. Consider the exact sequence $0\rightarrow$ Ker$(h_{1}-h_{2})\rightarrow F\overset{h_{1}-h_{2}}{\rightarrow}G$. By (2), Ker$(h_{1}-h_{2})$ is Harmanci injective. Then there exists a homomorphism $\beta\colon F\rightarrow$ Ker$(h_{1}-h_{2})\subseteq F$ satisfying $\beta f=if=f$ where $i\colon$ Im$f\rightarrow$ Ker$(h_{1}-h_{2})$ is inclusion. Since $f\colon M\rightarrow F$ is an $\mathcal{HI}$-envelope, $\beta$ is an isomorphism. Hence $\beta(F)=$ Ker$(h_{1}-h_{2})$. This yields $(h_{1}-h_{2})\beta=0$. As $\beta$ has an inverse, $h_{1}-h_{2}=0$, and so $h_{1}=h_{2}$. Therefore $f$ has the unique mapping property. ∎

By Theorem 3.9, we immediately get the next result.

Corollary 3.10**.**

Let $M$ be a Harmanci injective module and $N$ a submodule of $M$ with $M/N$ Harmanci injective. If the $\mathcal{HI}$-envelope of $N$ satisfies the unique mapping property, then $N$ is also Harmanci injective.