On Bassian Modules

Abstract

A familiar description of torsion Bassian Abelian groups is extended to Bassian modules over non-primitive Dedekind prime rings.

1. Introduction

We only consider associative unital non-zero rings and unitary modules. The phrases <<a Noetherian ring A>> mean that the both modules $A_A$ and ${}_{A}A$ are Noetherian. The word <<A-module>> usually means <<right A-module>>. The notions that are not defined in the paper are standard; e.g., see [1,2,3,4,5,6].

A module M is said to be Bassian if M is not isomorphic to a submodule of any its proper homomorphic image; equivalently, $N=0$ for any submodule N of M such that there exists a monomorphism $M\to M/N$. In [7], the notion of a Bassian Abelian group was introduced. In [7], all Bassian Abelian groups, i.e., Bassian $\mathbb{Z}$-modules, are described; see (Main Theorem [7]).

The main result of this paper is Theorem 1, which describes all singular Bassian modules over non-primitive Dedekind prime rings (examples of such rings include all rings of algebraic integers and matrix rings over them). In addition, Section 3 contains some results (see Theorem 3) on Bassian modules $M\ne \mathrm{Sing}M$. A ring A is called a Dedekind prime ring if A is a Noetherian prime ring with classical ring of fractions Q and every non-zero ideal of the ring A is invertible in Q. Every Dedekind prime ring is a hereditary Noetherian prime ring.

Example 1. 

In Section 4 of [8], the following example of a hereditary Noetherian prime ring C, which is not a Dedekind prime ring, is constructed. Let F be a field of characteristic zero and $A=F[x,y]$ be the ring of formal polynomials in x and y over F subject to the relation $xy-yx=1$. The ring A is known to be a simple hereditary Noetherian domain, which is not a division ring. Consider the ring $B=F\left(y\right)\left[x\right]$, that is, polynomials in x with coefficients that are rational functions in y. Then, B is contained in the quotient ring of A. But,

$$xB\subseteq xB+(1/y)A\subseteq xB+(1/y^2)A\subseteq xB+(1/y^3)A\subseteq \dots \subseteq B$$

is a strictly ascending chain of A-modules between $xB$ and B. The subring $C=F+xB$ of B is, using Robson’s theory of idealizers [9], a hereditary Noetherian domain with a unique ideal $xB$. This ideal is obviously idempotent and does not contain an invertible ideal of C.

A ring A is said to be non-primitive if A is not right or left primitive, i.e., A has no faithful simple right or left modules. Other required definitions and notations are given in the Introduction and Section 2.

Remark 1. 

Since the ring $\mathbb{Z}$ is a non-primitive Dedekind prime ring, Theorem 1 generalizes the description of torsion Bassian Abelian groups from [7], where all Bassian Abelian groups are described.

Remark 2. 

We note that all Dedekind prime rings are hereditary Noetherian prime (HNP) rings. In [8], it is proved that any HNP ring A is (right and left) primitive or (right and left) bounded, and if A is a primitive bounded ring, then A is a simple Artinian ring. Therefore, non-primitive HNP rings coincide with non-Artinian bounded HNP rings.

Theorem 1. 

If A is a non-primitive Dedekind prime ring and M is a singular right A-module with primary components $M_i=M\left(P_i\right)$, $i\in I$, then the following conditions are equivalent.

(1) 

M is a Bassian module.

(2) 

Every primary component $M_i$ of the module M is a reduced finite-dimensional module.

(3) 

Every primary component $M_i$ of the module M is a Noetherian module.

The proof of Theorem 1 is divided into several assertions from Section 2 of this paper.

Remark 3. 

Assume that a non-zero module $M=M_0$ is not Bassian. It follows from the definition of a Bassian module that M has a non-zero submodule $N_1$ such that there exists an isomorphism $f_1:M\to M_1/N_1$, where $M_1$ is a submodule of $M=M_0$, and $M_1$ properly contains $N_1\ne 0$. Since the non-zero module $M_1/N_1$ is isomorphic to M, and it is not Bassian, there exists an isomorphism $f_2:M_1/N_1\to M_2/N_2$, where $M_2$ is a submodule of $M_1$, $M_2$ properly contains $N_2$, and $N_2$ properly contains $N_1$. We similarly continue this process and obtain a properly ascending chain of submodules $N_1\subset N_2\subset \dots$ of M and a descending chain of submodules $M\supseteq M_1\supseteq M_2\supseteq \dots$. Consequently, the module M is not Noetherian.

Remark 4. 

It follows from Remark 3 that all Noetherian modules are Bassian. We note that any infinite direct sum of pairwise non-isomorphic simple modules is a Bassian module that is not Noetherian. In particular, if M is the direct sum of the cyclic p-groups for every distinct prime integers p, then M is a Bassian $\mathbb{Z}$-module that is not Noetherian.

Remark 5. 

All direct summands of Bassian modules are Bassian modules.

Proof. 

Assume that M is a Bassian module, $M=X\oplus Y$, and the module X is not Bassian. There exists a monomorphism $f:X\to X/X_1$, such that $X_1$ is a non-zero submodule of X. Then, there exists a monomorphism $g=f\oplus 1_Y:M\to M/X_1=(X/X_1)\oplus B$, where $X_1\ne 0$. This is a contradiction. □

2. Singular Modules and Related Topics

A module M is said to be injective if for any monomorphism f from M into an arbitrary module X, the module $f\left(M\right)$ is a direct summand of X. A module is said to be reduced if it has no non-zero injective submodules. A submodule X of the module M is said to be essential if $X\cap Y\ne 0$ for every non-zero submodule Y of M. In this case, one says that M is an essential extension of the module X. The module M is said to be uniform if the intersection of any pair of its non-zero submodules is non-zero, i.e., any non-zero submodule of M is essential.

If N is a subset of the right (resp., left) A-module M, then $r\left(N\right)$ (resp., $\ell \left(N\right)$) denotes the right (resp., left) annihilator of the set N in A, i.e., $r\left(N\right)=\{a\in A|Na=0\}$ (resp., $\ell \left(N\right)=\{a\in A|aN=0\}$). We denote by $\mathrm{Sing}M$ the set of all elements m of the right (resp., left) A-module M such that $r\left(m\right)$ (resp., $\ell \left(m\right)$) is an essential right (resp., left) ideal of the ring A. It is well known that $\mathrm{Sing}M$ is a fully invariant submodule of M; it is called the singular submodule of M. If $\mathrm{Sing}M=M$ (resp., $\mathrm{Sing}M=0$), then the module M is said to be singular (resp., non-singular). An element a of the ring A is said to be regular if $r\left(a\right)=\ell \left(a\right)=0$. For a right A-module M, we denote by $T\left(M\right)$ the set of all elements $m\in M$ such that $r\left(m\right)$ contains a regular element of the ring A; this set is called the torsion part of the module M. A module M is called a torsion module (resp., a torsion-free module) if $T\left(M\right)=M$ (resp., $T\left(M\right)=0$). A module $M_A$ is said to be divisible if $M=Ma$ for every regular element $a\in A$. A submodule X of M is said to be closed in M if X has no proper essential extensions in M.

If the module M is an essential extension of a module ${\oplus }_{j\in J}{X}_j$, where all $X_j$ are non-zero uniform modules and $\tau$ is the cardinality of the set J, then $\tau$ is called the Goldie dimension or the uniform dimension of M; it is denoted by $\mathrm{Gdim}M$. It is well known that the cardinal number $\mathrm{Gdim}M$ is uniquely defined. The module M is said to be finite-dimensional if M has no submodules that are infinite direct sums of non-zero modules; in this case, its Goldie dimension is finite.

In the following Remark 6, assertion 1 is well known and is verified wia Zorn’s lemma.

Remark 6. 

Let M be a module, and let X be a submodule of M.

1. 

There exist two closed submodules $M_1$ and $M_2$ of M such that $M_1$ is an essential extension of X, $M_1\cap M_2=0$, and M is an essential extension of the module $M_1\oplus M_2$. In this case, $M_1$ and $M_2$ are called a closure and, respectively, an additive complement of X in M.

2. 

If $M_3$ is any closed essential submodule of M, and $X\subseteq M_3$, then $M_3/X$ is an essential submodule of $M/M_1$; (Proposition 1.4 [3]).

3. 

If $M_4$ is a closed submodule of M, and $X\subseteq M_4$, then $M_4/M_1$ is a closed submodule of $M/X$; (p. 20, Ex. 16 [3]).

4. 

If $M_1$ is a closed submodule of M, and $M_5$ is an essential submodule of M, then $M_1\cap M_5$ is a closed submodule of $M_5$; (p. 20, Ex. 17 [3]).

Remark 7. 

Let M be a module and let $M_1$ be a submodule of M.

1. 

If the kernel of a module homomorphism is essential, its image is singular; (Proposition 1.20 [3]). Consequently, if the quotient module $M/M_1$ is non-singular, then the submodule $M_1$ is closed in M.

2. 

If the module M is non-singular, then the module $M/M_1$ is singular if and only if M is an essential extension of the module $M_1$; (Proposition 1.21 [3]).

A ring A is said to be right (resp., left) bounded if any its essential right (resp., left) ideal contains a non-zero ideal of the ring A. A ring is said to be prime if the product of any two its non-zero ideals is not zero. A ring A is said to be semiprime if A has no non-zero nilpotent ideals. A right finite-dimensional ring with the maximum condition on right annihilators is called a right Goldie ring.

Proposition 1. 

Let A be a right Goldie semiprime ring.

1. 

The set of all essential right ideals of the ring A coincides with the set of all right ideals of A that contain regular elements. Consequently, the class of all singular (resp., non-singular) right A-modules coincides with the class of all torsion right A-modules (resp., torsion-free A-modules). All essential extensions of singular (resp., non-singular) right A-modules are singular (resp., non-singular) modules. For any module $M_A$, the module $M/\mathit{Sing}\left(M\right)$ is non-singular. The ring A has a semisimple artinian right classical ring of fractions Q, and non-zero injective non-singular uniform right A-modules coincide (up to an A-module isomorphism) with the minimal right ideals of the semisimple artinian ring Q; if A is prime, then the ring Q is simple.

2. 

Every non-torsion right A-module contains a non-zero non-singular submodule.

3. 

Every non-singular divisible right A-module is injective.

Proposition 1 is well known (see [3,4]), assertion 1 is proved in (5.9, 5.10, 6.14, 6.10(a) [4]), assertion 2 follows from 1, and assertion 3 is proved in (6.12 [4]).

Assume that a ring A has a classical right and left ring of fractions Q. It follows from Proposition 1(1) that Q is simple Artinian, i.e., it is isomorphic to a matrix ring over a division ring. In particular, every Noetherian prime ring has a simple Artinian classical right and left ring of fractions. An ideal B of the ring A is said to be invertible if there exists a sub-bimodule $B^{-1}$ of the bimodule ${}_{A}Q_A$ such that $B{B}^{-1}=B^{-1}B=A$. The maximal elements of the set of all proper invertible ideals of the ring A are called maximal invertible ideals. The set of all maximal invertible ideals of the ring A is denoted by $\mathcal{P}\left(A\right)$. If M is an A-module and $P\in \mathcal{P}\left(A\right)$, then the submodule $\{m\in M|m{P}^n=0,n=1,2,\dots \}$ is called the P-primary component of the module M; it is denoted by $M\left(P\right)$. If $M=M\left(P\right)$ for some $P\in \mathcal{P}\left(A\right)$, then the module M is said to be primary or P-primary.

Remark 8. 

Let A be a non-primitive HNP ring, M be a singular A-module and let ${\left\{M\left(P_i\right)\right\}}_{i\in I}$ be the set of all primary components of the module M. Many properties of the singular module M and its primary components are well known; for example, see [10,11,12,13,14,15]. These properties are similar to properties of primary components of torsion Abelian groups. For example, $M={\oplus }_{i\in I}M\left(P_i\right)$, all primary components $M\left(P_i\right)$ are fully invariant in M, $X={\oplus }_{i\in I}X\cap M\left(P_i\right)$ for any submodule X of the module M, where $X\cap M\left(P_i\right)$ are primary components of the module X, $M/X={\oplus }_{i\in I}M/(X\cap M\left(P_i\right))$, and so on.

A module M is said to be uniserial if the lattice of all its submodules is a chain, i.e., any two submodule of M are comparable under inclusion (equivalently, any two cyclic submodules are comparable under inclusion). Every uniserial module is uniform, and every uniform module is indecomposable.

Proposition 2 

(4.19, 4.20 [4]). A ring A is a right Noetherian ring if and only if every injective right A-module is a direct sum of uniform injective modules, and if and only if all direct sums of injective right A-modules are injective.

Proposition 3. 

Let A be a non-primitive HNP ring, M be a singular right A-module, and let $\left\{M\left(P_i\right)\right\}$ be the set of all primary components of the module M.

1. 

All finitely generated submodules of the module M are finite direct sums of cyclic uniserial modules of finite length.

2. 

Every non-injective submodule of the module M has a non-zero cyclic uniserial direct summand of finite length.

3. 

M is an injective module if and only if every primary component $M\left(P_i\right)$ is an injective module, and if and only if every primary component $M\left(P_i\right)$ is a direct sum of uniserial injective modules.

Proof. 

1, 2. The assertions are proved in [11,12].

3. The assertion follows from 1, 2, and Proposition 2. □

Proposition 4. 

Let A be a non-primitive HNP ring, M be an injective indecomposable non-zero singular module, and let $M\left(P\right)$ be the primary component of the module M, where P is a maximal invertible ideal of the ring A.

1. 

M is a uniserial non-cyclic primary module without maximal submodules. All proper submodules of the module M are cyclic modules of finite length. They form a countable chain $0=X_0\subset X_1\subset \dots \subset X_k\subset \dots$, where $X_k/X_{k-1}$ is a simple module for any k. There exists a positive integer n such that $X_j/X_{j-1}\cong X_k/X_{k-1}$ if and only if $j-k$ is a multiple of n. In addition, the module M has a surjective endomorphism with non-zero kernel. Consequently, M is not a Bassian module.

2. 

If $\overline{M}$ is any non-zero homomorphic image of the module M, then, for any cyclic submodule $\overline{X}$ of the module $\overline{M}$ of length $\ge k+n$, there exists an epimorphism $\overline{X}\to X_k$ with a non-zero kernel for an arbitrary cyclic submodule $X_k$ from 1.

3. 

If Y is an arbitrary cyclic uniserial P-primary module, then it is isomorphic to a subfactor of the module M and is annihilated by some power of the ideal P.

4. 

If ${\oplus }_{i\in I}{Y}_i$ is a P-primary module, where all $Y_i$ are cyclic uniserial modules whose composition lengths have a common upper bound, then the module is annihilated by some positive power of the ideal P.

5. 

Every finite-dimensional reduced singular A-module is a finite direct sum of cyclic uniserial modules of finite composition length.

6. 

Every A-module with non-zero annihilator is a direct sum of cyclic uniserial modules of finite composition lengths, which are bounded in totality.

Proposition 4 follows from Proposition 2 and the results of [11,12].

Proposition 5. 

Let A be a non-primitive Dedekind prime ring, M be an injective indecomposable non-zero singular module, and let $M\left(P\right)$ be the primary component of the module M, where P is a maximal ideal of the ring A.

1. 

M is a uniserial non-cyclic primary module, all non-zero quotient modules of M are isomorphic to M, and all proper submodules of M are cyclic modules of finite length and form a countable chain $0=X_0\subset X_1\subset \dots \subset X_k\subset \dots$, where for any $k\ge 1$, all modules $X_k/X_{k-1}$ are isomorphic simple modules.

2. 

If Y is an arbitrary cyclic uniserial P-primary module, then it is isomorphic to a subfactor of the module M and is annihilated by some power of the ideal P.

3. 

Every P-primary module $N_A$ of Goldie dimension τ is an essential extension of a direct sum ${\oplus }_{j\in J}{S}_j$ of isomorphic simple modules $S_j$, where $\left|J\right|=\tau$ and $\left({\oplus }_{j\in J}{S}_j\right)P=0$.

4. 

If ${\oplus }_{i\in I}{Y}_i$ is a P-primary module, where all $Y_i$ are cyclic uniserial modules whose composition length are limited in totality, then the module is annihilated by some positive power of the ideal P.

5. 

Let Q be an injective P-primary A-module of Goldie dimension τ, and let X be a P-primary A-module of Goldie dimension $\sigma \le \tau$. Then, there exists a monomorphism $f:X\to Q$.

Proof. 

Assertions 1–4 follow from Proposition 4 and the results of [11,12].

5. The module X is an essential extension of some module $Y={\oplus }_{j\in J}{Y}_j$, where all $Y_j$ are isomorphic simple modules and $\left|J\right|\sigma \le \tau$. Then, there exists a monomorphism $g:Y\to Q$. Since the module Q is injective, g can be extended to a homomorphism $f:X\to Q$, and $Y\cap \mathrm{Ker}f\subseteq \mathrm{Ker}g=0$. Since Y is an essential submodule of X, we have that f is a monomorphism. □

Remark 9 

(p. 45, Theorems 2.7, 1.3 [14]). Let A be a non-primitive HNP ring, and let M be a singular right A-module. Then, M satisfies the following conditions (I–III):

(I) 

Every finitely generated subfactor X of the module M is a finite direct sum of cyclic uniserial modules $X_i$ of finite composition length $d\left(X_i\right)$.

(II) 

If X is any subfactor of the module M containing cyclic uniserial submodules U and V, then, for any submodule W of U and every non-zero homomorphism $f:S\to S$ with the condition $d(U/W)<d(V/f(W\left)\right)$, the mapping f can be extended to a homomorphism $U\to V$.

(III) 

For any finitely generated submodule N of M, the ring $A/r\left(N\right)$ is right Artinian.

Further, M has a basis submodule B; this means that B is a pure submodule of M, B is a direct sum of cyclic uniserial modules of finite length, and $M/B$ is a direct sum of uniform modules $M_i/B$ of infinite length; it follows from Propositions 4(1) and 2 that all modules $M_i/B$ and $M/B$ are injective. In addition, any two basis submodules of the module M are isomorphic.

Remark 10 

(p. 139 [15] and Theorem 3.8 [15]). Let A be a non-primitive HNP ring, and let M be a singular right A-module. It follows from Remark 9 that M has a basis submodule, and its two basis submodules are isomorphic.

An element $x\in M$ is said to be uniform if $xA$ is a uniserial module. For any uniform element $x\in M$, we denote by $H_M\left(x\right)$ the upper bound of all lengths $d(T/xA)$, where T is an arbitrary cyclic uniserial submodule of M containing x. We denote by $H_k\left(M\right)$ the submodule of M generated by all uniform elements with $H\left(x\right)\ge k$. The minimum of the Goldie dimensions of all submodules $H_k\left(M\right)$ is called the final rank $\mathrm{fin}.\mathrm{rank}M$ of the module M. A basis submodule B of the module M is called a lower basis submodule if $\mathrm{Gdim}M=\mathrm{fin}.\mathrm{rank}M$. In general, M need not have a lower basis submodule. However, if the socle of the module M has only finitely many homogeneous components (for example, this is the case if the module M is primary), then the module M has a lower basis submodule (Theorem 3.8 [15]).

Theorem 2  

(Theorems 3.8, 3.10 [15]). Let A be a non-primitive HNP ring, and let M be a singular right A-module whose socle has only a finite number of homogeneous components. Then, the module M has a lower basis submodule B and $M=H_1\oplus H_2$, where $r\left(H_1\right)\ne 0$ and $\mathit{Gdim}{H}_2=\mathit{fin}.\mathit{rank}{H}_2=\mathit{fin}.\mathit{rank}M$.

Lemma 1. 

Let A be a non-primitive HNP ring, and let M be a singular right A-module with primary components $M_i=M\left(P_i\right)$, $i\in I$.

1. 

Every primary component $M_i$ of the module M is a reduced finite-dimensional module if and only if every primary component $M_i$ of the module M is a Noetherian module.

2. 

If the conditions in 1 hold, then the module M is Bassian.

3. 

If M is an injective non-zero module, then M is a direct sum of non-zero injective uniserial modules that are not Bassian.

4. 

If M is a Bassian module, then M is a reduced module.

5. 

If M is a Bassian module and $r\left(M\right)\ne 0$, then every primary component of the module M is a Noetherian Artinian module of finite length.

Proof. 

By Remarks 5 and 4, we may assume that $M=M_i$ is a primary module.

1. The assertion is directly verified with the use of induction and Proposition 4(1).

2. Let X be a submodule of M with primary components $X_i=X\cap M_i$, and suppose there exists an isomorphism $f:M\to Y/X$, where $X\subseteq Y$, and Y is a module with the primary components $Y_i=Y\cap M_i$. It follows from Remark 8 that $M/X={\oplus }_{i\in I}{M}_i/X_i$. By Remark 4, all Noetherian modules $M_i$ are Bassian modules. In addition, the isomorphism f induces isomorphisms $f_i:M_i\to Y_i/X_i$. Therefore, $X_i=0$ for all i, whence $X=0$, and M is a Bassian module.

3. By Proposition 2, M is a direct sum of injective singular indecomposable non-zero modules $M_i$. By Proposition 4(1), the modules $M_i$ are not Bassian.

4. By Remark 5, direct summands of Bassian modules are Bassian. Therefore, the assertion follows from 3.

5. By Proposition 4(6), $M={\oplus }_{j\in J}{M}_j$, where all $M_i$ are cyclic uniserial modules of finite composition length, which are bounded by some positive integer n. By 4, the module M is reduced. Now, it is sufficient to prove that $\left|J\right|<\infty$.

Assume that $\left|J\right|=\infty$. With the use of Proposition 4, we may verify that the Bassian module M has a direct summand $X={\oplus }_{i=1}^{\infty }{X}_i$, where all $X_i$ are non-zero isomorphic cyclic uniserial modules. Therefore, it is clear that the module X is not Bassian; this contradicts to Remark 5. □

The Proof of Theorem 1. 

By Remarks 4 and 5, we can assume that $M=M_i$ is a primary module.

(1) ⇒ (2). By Lemma 1(4), M is a reduced module. Assume that M has an infinite final rank $\mathrm{fin}.\mathrm{rank}M$ (see the text after Remark 10 about the final rank). By Theorem 2, $M=H_1\oplus H_2$, where $r\left(H_1\right)\ne 0$, and the module $H_2$ has an infinite final rank $\mathrm{fin}.\mathrm{rank}{H}_2$, which is equal to the Goldie dimension $\mathrm{Gdim}{H}_2=\tau$. By Lemma 1(5), the module $H_1$ is Noetherian. In particular, the module $H_1$ is finite-dimensional. By Theorem 2, the module $H_2$ has a lower basis submodule $B_2\ne 0$, and $\mathrm{Gdim}(H_2/B_2)=\tau$. By Proposition 5(3), the module $H_2$ is an essential extension of a direct sum ${\oplus }_{j\in J}{S}_j$ of isomorphic simple modules $S_j$, where $\left|J\right|=\tau$. Since $\mathrm{Gdim}(H_2/B_2)=\tau$, it follows from Proposition 5(3) that there exists a monomorphism f from the module ${\oplus }_{j\in J}{S}_j$ into M the module $H_2/B_2$. The module $H_2$ is an essential extension of the module ${\oplus }_{j\in J}{S}_j$, and the module $H_2/B_2$ is injective (see Remark 9). Therefore, the monomorphism f can be extended to a monomorphism $g:H_2\to H_2/B_2$ and $M=H_1\oplus H_2$. Then, there exists a monomorphism $h:M=H_1\oplus H_2\to H_1\oplus (H_2/B_2)=M/B_2$, which acts identically on $H_1$ and coincides with g on $H_2$. Since M is a Bassian module, and $B_2\ne 0$, we have a contradiction.

We proved that M has a finite final rank $\mathrm{fin}.\mathrm{rank}M=\mathrm{Gdim}{H}_2<\infty$ and $M=H_1\oplus H_2$, where $\mathrm{Gdim}{H}_i<\infty$, $i=1,2$. Therefore, the module M is finite-dimensional. By Proposition 4(5), the reduced finite-dimensional singular module M is a Noetherian module.

The equivalence (2) ⇔ (3) and the implication (2) ⇒ (1) follow from Lemma 1. □

Example 2. 

It is directly verified that $\mathbb{Q}$ is a Bassian $\mathbb{Z}$-module any its nontrivial (torsion) quotient module is not Bassian (e.g., see Theorem 1).

3. Supplement About Bassian Modules $\mathit{M}\ne \mathbf{Sing}\mathit{M}$

We denote by $Z_2\left(M\right)$ the Goldie radical of the module M, i.e., $Z_2\left(M\right)$ is the submodule of M such that $Z_2\left(M\right)$ contains $\mathrm{Sing}M$ and the module $Z_2\left(M\right)/\mathrm{Sing}M)=\mathrm{Sing}(M/\mathrm{Sing}M)$. A module M is called a Goldie-radical module if $M=Z_2\left(M\right)$.

Remark 11  

(p. 37, Exs. 20, 21, 19 [3]). Let A be a ring.

1. 

$\mathit{Sing}(M/Z_2\left(M\right))=0$ for any A-module M, $Z_2\left(A_A\right)$ is an ideal of the ring A, and the factor ring $A/Z_2\left(A_A\right)$ is right non-singular.

2. 

The class of all Goldie-radical A-modules M is closed with respect to submodules, quotient modules, direct sums, module extensions, and essential extensions.

3. 

The ring A is non-singular if and only if for any right A-module M, the Goldie radical of M coincides with the singular submodule of M.

Lemma 2. 

Let E be an injective right A-module not equal to its Goldie radical $E_1$, and let the module $E/E_1$ have finite Goldie dimension $n\ge 1$. Then, the module E does not contain a submodule ${\oplus }_{k=1}^{n+1}{H}_k$, where all $H_k$ are non-zero uniform non-singular modules.

Proof. 

Assume the contrary. Then, there exists a direct decomposition $E=E_1\oplus F\oplus G$, where the module F is non-singular (possibly, zero), $G={\oplus }_{k=1}^{n+1}{G}_k$, and $G_k$ is the injective hull of the module $H_k$, which is a non-zero non-singular uniform injective module, $k=1,\dots ,n+1$. Then, $\mathrm{Gdim}(F\oplus G)\ge \mathrm{Gdim}G=n+1$. Since $\mathrm{Gdim}(E/E_1)=\mathrm{Gdim}(F\oplus G)=n$, we have a contradiction. □

Remark 12.  

Let A be a right hereditary ring. Then, every homomorphic image of every injective right A-module is injective (e.g., see Theorem 5.4 [1]). In addition, the ring A is right non-singular, since its principal right ideal $aA$ is projective, and it is isomorphic to $A_A/r\left(a\right)$, whence $r\left(a\right)$ is a direct summand of the module $A_A$.

Proposition 6  

(Theorems 21.15, 21.6 [2], 4.19, 4.20 [4]). A ring A is right Noetherian if and only if every non-zero injective right A-module E is the direct sum ${\oplus }_{i\in I}{E}_i$ of non-zero uniform injective modules $E_i$ with local endomorphism rings, and if and only if all direct sums of injective right A-modules are injective. In addition, if $E={\oplus }_{i\in I}{E}_i={\oplus }_{j\in J}{F}_j$ are decompositions of the module E into direct sums of non-zero uniform injective modules, then there exists a bijection $f:I\to J$ such that $E_i\cong F_{f\left(i\right)}$.

Proposition 7. 

Let A be a right hereditary and right Noetherian ring, and let M be a right A-module such that $Z_2\left(M\right)$ is a Bassian module, and the module $M/Z_2\left(M\right)$ is finite-dimensional. Then, M is a Bassian module.

Proof. 

It follows from Remark 11(3) that, for any right A-module X, the Goldie radical of X is equal to its (largest) singular submodule of X. We set $T=Z_2\left(M\right)=\mathrm{Sing}M$. Without loss of generality, we may assume that $M\ne T$, i.e., $M/T$ is a non-zero non-singular module of finite positive Goldie dimension $n+1$, $n\ge 0$. Let X be a non-zero submodule of M. Assume that there exists a monomorphism $\alpha :M\to M/X$. In what follows, we deduce a contradiction.

Consider the two possible cases (1) $X\subseteq T$; (2)$X⊈T$.

Case (1): $X\subseteq T$. Since $T/X$ is a Goldie-radical module, and $(M/X)/(T/X)$ is isomorphic to the non-singular module $M/T$, we have $Z_2(M/X)=T/X$. Then, $\alpha \left(T\right)\subseteq T/X$, and there exists a monomorphism from the Bassian module T into its proper quotient module $T/X$. This is a contradiction.

Case (2): $X⊈T$. There exists a submodule $T_1$ of T such that $T_1\cap (T\cap X)=0$, and T is an essential extension of the module $T_1\oplus (T\cap X)$. Let $E\left(M\right)$ be the injective hull of the module M, and let $E(T+X)$, $E(T\cap X)$, $E\left(T\right)$, $E\left(T_1\right)$, $E\left(X\right)$ be the injective hulls of the modules $T+X$, $T\cap X$, T, $T_1$, X, respectively, contained in $E\left(M\right)$. Then, $E\left(T\right)$ equals the injective hull $E(T_1\oplus (T\cap X))=E\left(T_1\right)\oplus E(T\cap X)$ of the module $T_1\oplus (T\cap X)$. By Remark 12, the modules $E\left(M\right)/X$ and $E\left(X\right)/X$ are injective. By Remark 7(1), the module $E\left(X\right)/X$ is singular. Since $X⊈T$, modules $T+X$ and X are not essential extensions of the modules T and $T\cap X$, respectively. Therefore, there exists a non-zero non-singular submodule Y of X such that X is an essential extension of the module $(T\cap X)\oplus Y$. Then,

$$E\left(X\right)=E\left(\right(T\cap X)\oplus Y)=E(T\cap X)\oplus E\left(Y\right),$$

where $E\left(\right(T\cap X)\oplus Y)$, $E(T\cap X)$, $E\left(Y\right)$ are injective hulls of modules $(T\cap X)\oplus Y$, $T\cap X$, Y in the module $E\left(X\right)\subseteq E\left(M\right)$, and the module $E\left(Y\right)$ is non-singular. There exists a non-singular submodule Z of M such that $(T+X)\cap Z=0$, M is an essential extension of the module $(T+X)\oplus Z$, $E\left(M\right)=E(T+X)\oplus E\left(Z\right)$ is the injective hull of the module $(T+X)\oplus Z$, where $E\left(Z\right)$ is the injective hull of the module Z, and $E\left(Z\right)$ is a (possibly zero) non-singular finite-dimensional module of Goldie dimension $n<n+1=\mathrm{Gdim}E=\mathrm{Gdim}M$. Then,

$$E\left(M\right)=E\left(T\right)\oplus E\left(Y\right)\oplus E\left(Z\right)=E\left(T_1\right)\oplus E(T\cap X)\oplus E\left(Y\right)\oplus E\left(Z\right)=$$

$$=E\left(T_1\right)\oplus E\left(X\right)\oplus E\left(Z\right).$$

Since the module $E\left(M\right)/X$ is injective, the monomorphism $\alpha :M\to M/X$ can be extended to a homomorphism $\beta :E\left(M\right)\to E\left(M\right)/X$. In addition, $\beta$ is a monomorphism, since $M\cap \mathrm{Ker}\beta =\mathrm{Ker}\alpha =0$, and M is an essential submodule of $E\left(M\right)$. Therefore, the module $\beta \left(E\right(M\left)\right)$ is injective, and there exists a direct decomposition $E\left(M\right)/X=\beta \left(E\right(M\left)\right)\oplus W$. Since the ring A is right Noetherian, it follows from Proposition 6 that there exists a direct decomposition $\beta \left(E\left(M\right)\right)={\oplus }_{i\in I}{E}_i$ into a direct sum of non-zero uniform injective modules $E_i$. Since the module $M/T$ contains a direct sum of $n+1$ non-zero uniform non-singular modules, $E\left(M\right)$ has a direct summand ${\oplus }_{j=1}^{n+1}{W}_j$, where all $W_j$ are non-zero uniform non-singular modules. Therefore, the following holds:

$(\ast )$ the injective module $E\left(M\right)/X$ has a direct summand ${\oplus }_{j=1}^{n+1}\beta \left(W_j\right)$, where all $\beta \left(W_j\right)$ are non-zero uniform non-singular modules.

In addition, $E\left(M\right)/X=(E\left(T_1\right)\oplus E\left(X\right)\oplus E\left(Z\right))/X\cong E\left(T_1\right)\oplus (E\left(X\right)/X)\oplus E\left(Z\right)$, where the module $E\left(T_1\right)\oplus (E\left(X\right)/X)$ is singular, and $E\left(Z\right)$ is a non-singular module of finite Goldie dimension $n\ge 0$. This contradicts $(\ast )$. □

Theorem 3. 

Let A be a non-primitive HNP ring, and let M be a right A-module such that $M\ne \mathit{Sing}M$, every primary component of the singular module $\mathit{Sing}M$ is a Noetherian module, and the module $M/\mathit{Sing}M$ is finite-dimensional. Then, M is a Bassian module.

Theorem 3 follows from Theorem 1 and Proposition 7.

Lemma 3  

(Lemma 2.15 [16]). Let E be an injective right A-module that is not equal to the Goldie radical $E_1$ of E, and let the module $E/E_1$ be of finite Goldie dimension $n\ge 1$. Then, the module E contains no submodules of the form ${\oplus }_{k=1}^{n+1}{H}_k$ for $H_k$ non-zero uniform non-singular modules.

Lemma 4. 

Let A be a right Goldie prime ring.

1. 

Every non-zero ideal B of the ring A is an essential right ideal and contains a regular element.

2. 

There exists a positive integer n such that, for any non-zero elements $b_1,\dots b_n$ of the ring A, the module $b_{1}A\oplus \dots \oplus b_{n}A$ contains an isomorphic copy of the free cyclic module $A_A$.

3. 

If X is a non-torsion right A-module, and X contains a non-singular submodule Y of infinite Goldie dimension τ, then X contains a non-zero free submodule F of infinite rank τ; therefore, there exists an epimorphism from the module F onto any τ-generated right A-module.

Proof. 

1. By Proposition 1(1), it is sufficient to prove that B is an essential right ideal. Let C be a right ideal of the ring A with $B\cap C=0$. Then, $\left(AC\right)B\subseteq B\cap C=0$. Since the ring A is prime, $C\subseteq AC=0$, and B is an essential right ideal.

2. By Proposition 1(1), the right Goldie prime ring A has the classical right ring of fractions Q, which is isomorphic to the ring of all $n\times n$ matrices over a division ring for some positive integer n. We denote the module $b_{1}A\oplus \dots \oplus b_n$ by B. Then, the Q-module $b_{1}Q\oplus \dots \oplus b_{n}Q$ contains an isomorphic copy of the module $Q_Q$. Therefore, the module B contains an isomorphic copy of the module $A_A$.

3. By Lemma 3, X contains a submodule, which is isomorphic to a non-zero principal right ideal $bA$ of the ring A. By 2, there exists a positive integer n such that the direct sum of n isomorphic copies of the module $b{A}_A$ contains an isomorphic copy of the free cyclic module $A_A$. Then, X contains a non-zero free submodule of infinite rank $\left|J\right|$. □

Lemma 5. 

Let A be a right Goldie prime ring of infinite cardinality $\left|A\right|$, and let M be a non-zero non-singular right A-module of infinite Goldie dimension $\tau \ge \left|A\right|$. Then, the module M is not Bassian.

Proof. 

Let E be an injective hull of the module M. It is well known that $E={\oplus }_{j\in J}{E}_j$, where $\left|J\right|=\tau$, and for any j, the module $E_j$ is a uniform injective module such that there exists an isomorphism $E_j\cong S_A$, where S is some minimal right ideal of the simple Artinian right classical ring of fractions Q of the ring A. Since it is clear that the cardinality $\left|Q\right|$ of the right classical ring of fractions Q of the infinite ring A is equal to $\left|A\right|$, we have that $|E_j|=|A|$ and $\left|M\right|=\tau$. By Lemma 4(3), the module M contains a free submodule F of infinite rank $\tau$ and $\left|M\right|=\tau$. Therefore, there exist a submodule X of F and an epimorphism $f:M\to F/X\subseteq M_X$. If $X=0$, then $M=F$ is an infinite direct sum of isomorphic copies of the module $A_A$; in this case, it is clear that the module M is not Bassian. If $X\ne 0$, and M is isomorphic to the submodule $F/X$ of $M_X$, then the module M is not Bassian. □

Proposition 8. 

Let A be a right hereditary right and Noetherian prime countable ring, and let M be a non-zero non-singular right A-module. The following conditions are equivalent.

(1). 

M is a Bassian module.

(2). 

M is a finite-dimensional module.

Proof. 

Implication (2) ⇒ (1) follows from Theorem 3.

(1)⇒(2). Assume the contrary. Then, $\mathrm{Gdim}M\ge {\aleph }_0=\left|A\right|$. By Lemma 5, the module M is not Bassian. This is a contradiction. □

4. Open Questions

Open question 1. Let A be a primitive hereditary noetherian prime ring. Describe singular Bassian A-modules.

Open question 2. Let A be a right hereditary and right Noetherian prime countable ring. In Proposition 8, it is proved that a non-singular right A-module M is Bassian if and only if M is finite-dimensional. Is a similar statement true without the condition that the ring A is countable?