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##  eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2021; 61(2): 213-222

Published online June 30, 2021

### Baer–Kaplansky Theorem for Modules over Non-commutative Algebras

Gabriella D'Este, Derya Keskİn Tütüncü*

Department of Mathematics, Milano University, Milano, Italy
e-mail: gabriella.deste@unimi.it

Department of Mathematics, Hacettepe University, 06800 Beytepe, Ankara, Turkey
e-mail : keskin@hacettepe.edu.tr

Received: December 3, 2019; Revised: November 30, 2020; Accepted: December 14, 2020

### Abstract

In this paper we investigate the Baer-Kaplansky theorem for module classes on algebras of finite representation types over a field. To do this we construct finite dimensional quiver algebras over any field.

Keywords: Baer-Kaplansky theorem, quivers and representations

### 1. Introduction

We consider associative rings R with identity; all modules considered are unitary left R-modules. Throughout this paper K will be any field.

For a vertex x of a quiver Q, S(x) denotes the simple representation corresponding to the vertex x. Moreover, P(x) (resp. I(x)) denotes the indecomposable projective (resp. injective) representation corresponding to the vertex x. For short, S(x) is replaced by x. With this convention a chain of length $n≥2$ of the form

$12...n$

describes a uniserial module of length n with composition factors $n,⋯,2,1$. Moreover, a picture of the form $112$ $(resp.122)$describes an indecomposable module M of length three such that the socle of M is isomorphic to $1⊕2$ (resp. 2), while the factor module $M/socM$ is isomorphic to 1 (resp. $1⊕2$). For more background on quivers we refer to  and .

The aim of this paper is to construct classes of modules which satisfy or do not satisfy the Baer-Kaplansky theorem defined on K-algebras, where K is any field. When we look at the literature, we see that the Baer-Kaplansky theorem states that any two torsion abelian groups having isomorphic endomorphism rings are isomorphic [4, Theorem 108.1]. Finding other classes of abelian groups, and more generally, of modules, for which a Baer-Kaplansky-type theorem is still true remains an interesting problem. In , Ivanov and Vámos called such classes Baer-Kaplanksy classes. For example, the class of finitely generated abelian groups is Baer-Kaplansky (e.g., see [7, Example 1.3]). Over commutative rings, there are several Baer-Kaplansky classes of modules, but there are relatively few known over non-commutative rings. In particular, we know from Morita's paper ([8, Lemma 7.4]) that the class of all modules over a primary artinian uniserial ring is Baer-Kaplansky. Moreover, we know from Ivanov's paper ([5, Theorem 9]) that the class of all modules over a non-singular artinian serial ring is Baer-Kaplansky.

These are the motivating and leading ideas in our investigation of Baer-Kaplansky classes over non-commutative algebras.

This paper is organized as follows. In Section 1 we recall some definitions and conventions. In Section 2 we collect all the results. We begin with some negative results. As we shall see, rather few classes of modules over non-commutative algebras fail to be Baer-Kaplansky. In Example 2.1, we construct a class of simple injective left R-modules which is not Baer-Kaplansky over a hereditary K-algebra R of finite representation type. In Example 2.3, we construct a class of simple left R-modules which is not Baer-Kaplansky over a non-hereditary K-algebra R of finite representation type. Also we obtain some positive results by dealing with classes of modules with a rigid structure, containing two indecomposable modules and closed under finite direct sums. Indeed the endomorphism rings of the two indecomposable modules always have dimension 1 and 2, and the vector spaces of the morphisms between two indecomposable non-isomorphic modules have dimension ≤ 2. As for the invariants, some of these classes admit the number of indecomposable direct summands and the dimension of the endomorphism ring as a complete set of invariants (Example 2.8 and Example 2.10). However this property is not always true for a Baer-Kaplansky class of finitely generated projective (resp. injective) modules over a finite dimensional algebra (Example 2.6.)

### Example 2.1.

There is a hereditary K-algebra R of finite representation type and a class of R-modules such that Baer-Kaplansky theorem fails.

Construction: Let R be the K-algebra given by the quiver 1 &#_10230; 3 &#_10229; 2. Then 1, 2, 3, $13,23$ and $123$ are the indecomposable left R-modules. Let M be the R-module I(3)=$123$. Note that $P(1)=13,P(2)=23$ and P(3)=3. The lattice of submodules of M is and we have $I(3)/P(1)≇I(3)/P(2)$. Also $EndR(I(3)/P(1))≅EndR(I(3)/P(2))≅K$ because $I(3)/P(1)$ and $I(3)/P(2)$ are one dimensional vector spaces. Therefore the class of simple injective left R-modules ${I(3)/P(1),I(3)/P(2)}={S(2),S(1)}$ is not Baer-Kaplansky.

### Remark 2.2.

Let R be a K-algebra of finite representation type such that K is the endomorphism ring of any indecomposable left R-module. Then Baer-Kaplansky theorem fails for any class with more than one indecomposable module. Moreover Baer-Kaplansky theorem holds for any class of the form ${Mn∣n≥1}$, where M is an indecomposable left R-module.

### Example 2.3.

There is a non-hereditary K-algebra R of finite representation type and a class of R-modules such that Baer-Kaplansky theorem fails.

Construction: Let R be the K-algebra given by the quiver with relations $ba=b2=0$. Then the indecomposable left R-modules are 1, 2, $12$, $22$ and $122$. Let M=I(2)=

$122$. Note that $P(1)=12$, $P(2)=22$ and S(2)=2. Also in this case the lattice of submodules of M is of the form with $I(2)/P(1)≇I(2)/P(2)$. Since $I(2)/P(1)$ and $I(2)/P(2)$ are one dimensional vector spaces, $EndR(I(2)/P(1))≅EndR(I(2)/P(2))≅K$. Then the class of simple left R-modules ${I(2)/P(1),I(2)/P(2)}={S(2),S(1)}$ is not Baer-Kaplansky. Note that S(1) is injective, while S(2) has infinite injective dimension. Here S(2) has a minimal injective resolution of the form

$0→S(2)=2→122→1⊕122→1⊕122→⋯.$

Moreover both simple modules have infinite projective dimension. The minimal projective resolutions of S(1) and S(2) are of the form

$⋯→22→22→12→1=S(1)→0$

and

$⋯→22→22→22→2=S(2)→0.$

### Proposition 2.4.

Let R be a K-algebra admitting three non-isomorphic modules M with the following properties:

• (1) M has exactly three non-zero proper submodules N1, N2 and N1∩ N2.

• (2) M/N1 is not isomorphic to M/N2.

• (3) $EndR(M/Ni)≅K$ for i=1, 2.

Then R is not commutative and there is a class (of non-uniserial modules) which is not Baer-Kaplansky.

Proof. The existence of a module satisfying (1), (2) and (3) implies that R is not commutative [3, Remark 2.2]. On the other hand the endomorphism ring of a module satisfying (1), (2) and (3) is either isomorphic to K or isomorphic to K[x]/(x2) [2, Theorem 3.8]. Since there exist three non-isomorphic modules M1, M2 and M3 satisfying (1), (2) and (3), without loss of generality we may assume that {M1, M2} is not a Baer-Kaplansky class.

We will use the next lemma to construct Baer-Kaplansky classes with infinitely many modules and exactly two indecomposable modules. In the sequel given a module M we will denote by add M the class of all finite direct sums of direct summands of M.

### Lemma 2.5.

Let R be a K-algebra of finite dimension and let U and V be two finite dimensional left R-modules with the following properties: $EndR(U)≅K$, $EndR(V)≅K[x]/(x2)$, $HomR(U,V)=0$, $HomR(V,U)≅K$. Then the class $add(U⊕V)$is a Baer-Kaplansky class. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are not a complete set of invariants for the modules in $add(U⊕V)$.

Proof. Since U and V are indecomposable modules, it follows that $add(U⊕V)$ consists of finite direct sums of copies of U and V. Let X be a non-zero left R-module of the form $Um⊕Vn$ with $m,n∈ℕ$. Let $A=EndR(Um)$, $B=EndR(Vn)$ and let $H=HomR(Vn,Um)$. Then A is isomorphic to the full matrix algebra Mm(K) and B is isomorphic to the full matrix algebra $Mn(K[x]/(x2))$. Finally H is an A-B-bimodule of dimension mn. Let $T=EndR(X)$. Then our hypotheses on U, V and X imply that

(1) $T=EndR(X)$ is isomorphic to the matrix algebra $AH0B$.

This means that the identity of T is the sum of primitive idempotents $e1,⋯,em$, $ϵ1,⋯,ϵn$ such that

(2) The regular module TT is the direct sum of the simple isomorphic modules $Te1,⋯,Tem$ (of dimension m) and of the indecomposable non-simple isomorphic modules $Tϵ1,⋯,Tϵn$ (of dimension m+2n).

Let Y be a module in add$(U⊕V)$ such that End$R(X)≅EndR(Y)$. Then there exist $p,q∈ℕ$ such that $Y≅Up⊕Vq$, and so

(3) EndR(Y) is the direct sum of p simple isomorphic modules (of dimension p) and q indecomposable non-simple isomorphic modules (of dimension p+2q).

Since T is a finite dimensional algebra, we know from [9, p. 66] that the category modT of finitely generated T-modules is a Krull-Schmidt category, that is a category where the Krull-Remark-Schmidt theorem [10, p. 3] holds. Consequently we deduce from (2) and (3) that m=p and n=q, and so $X≅Y$.

We finally note that U2 and $U⊕V$ are non-isomorphic modules with endomorphism ring of dimension 4. More generally, if $m,n,s∈ℕ$ and s ≤ m, then $Um⊕Vn$ and $Um−s⊕Vn+s$ have endomorphism ring of the same dimension $m2+mn+2n2$ if and only if $ms−3ns−2s2=0$. The lemma is proved.

### Example 2.6.

There is a non-commutative K-algebra R of finite representation type such that the class of finitely generated projective (resp. injective) modules is a Baer-Kaplansky class with the property described in Lemma 2.5.

Construction: Let R be the algebra considered in Example 2.3, given by the quiver with relations $ba=b2=0$. Then the classes of finitely generated projective and injective modules are add$(12⊕22)$ and add$(1⊕122)$ respectively. Moreover we clearly have End$R(12)≅K≅EndR(1)$, End$R(22)≅K[x]/(x2)≅EndR(122)$, Hom$R(12,22)=0=HomR(1,122)$ and Hom$R(22,12)≅K≅HomR(122,1)$. Hence the conclusion that add$(12⊕22)$ and add$(1⊕122)$ are Baer-Kaplansky classes with the desired property follows from Lemma 2.5.

We will use the next lemma to obtain Baer-Kaplansky classes $C$ with the property that Hom$R(L,M)≠0$ if L and M are two non-zero modules in $C$.

### Lemma 2.7.

Let R be a K-algebra of finite dimension and let U and V be two finite dimensional left R-modules with the following properties: $EndR(U)≅K,EndR(V)≅K[x]/(x2),HomR(U,V)≅K≅HomR(V,U)$. Then the class $add(U⊕V)$ is a Baer-Kaplansky class. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are a complete set of invariants for the modules in $add(U⊕V)$.

Proof. We first note that for any $m,n∈ℕ$ we have dim$EndR(Um⊕Vn)=m2+2mn+2n2$. Assume that s is a natural number ≤ m such that dim$EndR(Um−s⊕Vn+s)=dim(Um⊕Vn)$. Then we have $2ns+s2=0$. Consequently s=0. Hence add$(U⊕V)$ is a Baer-Kaplansky class with the desired property.

### Example 2.8.

There is a non-hereditary K-algebra R of finite representation type, such that any indecomposable module is uniserial, with the following properties:

• (1) The class of simple modules is not Baer-Kaplansky.

• (2) Let $P$ and $I$ be the classes of finitely generated projective modules and finitely generated injective modules, respectively and let $C$ be the class of finitely generated modules of projective and injective dimension at most one. Then $P$, $I$ and $C$ are Baer-Kaplansky classes. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are a complete set of invariants for the modules in $P,I$ and $C$.

Construction: Let R be the K-algebra given by the quiver $1⇄ba2$ with relation ab=0. Then $1,2,12,21$ and $121$ are the indecomposable modules. Since 1 and 2 are non-isomorphic one dimensional modules, (1) clearly holds. On the other hand we have $P=add(21⊕121)$, $I=add(12⊕121)$ and $C=add(1⊕121)$. Consequently (2) follows from Lemma 2.7.

We will use the next lemma to obtain Baer-Kaplansky classes $C$ with more complicated Hom spaces between indecomposable non-isomorphic modules in $C$.

### Lemma 2.9.

Let R be a K-algebra of finite dimension and let U and V be two finite dimensional left R-modules with the following properties: $EndR(U)≅K,EndR(V)≅K[x]/(x2),HomR(U,V)=0$ and $dimHomR(V,U)=2$. Then the class $add(U⊕V)$ is a Baer-Kaplansky class with the property described in Lemma 2.7.

Proof. Our hypotheses imply that $dimEndR(Um⊕Vn)=m2+2mn+2n2$ for any $m,n∈ℕ$. Hence the conclusion follows from the proof of Lemma 2.7.

### Example 2.10.

There is a K-algebra R of finite dimension such that the classes $P$ and $I$ of finitely generated projective and finitely generated injective modules are Baer-Kaplansky, but the class $C$ of finitely generated modules of projective and injective dimension at most one is not Baer-Kaplansky. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are a complete set of invariants for the modules in $P$ and $I$.

Construction: Let R be the K-algebra given by the quiver with relation b2=0. Then we have $P=add(122⊕22)$ and $I=add(1⊕1122)$. This observation and Lemma 2.9 imply that $P$ and $I$ are Baer-Kaplansky classes with the desired property. We also note that there exist exact sequences of the form

$0→22→1122→1⊕1→0,$ $0→122→1122→1→0$

and $0→22→122⊕22→122→0$.

Hence $22$ and $122$ are in $C$. Since End$R(22)≅K[x]/(x2)≅EndR(122)$, we conclude that $C$ is not Baer-Kaplansky.

### Proposition 2.11.

Let A and B be finite dimensional K-algebras such that there is an epimorphism from A to B. For any algebra R let $PR$ and $IR$ denote the classes of finitely generated projective and finitely generated injective left R-modules, respectively. Among others the following cases are possible:

• (1) $PA=IA$ is not a Baer-Kaplansky class, while $PB$ and $IB$ are Baer-Kaplansky classes. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are a complete set of invariants for the modules in $PB$ and $IB$.

• (2) $PA,IA,PB$ and $IB$ are Baer-Kaplansky classes. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are (resp. are not) a complete set of invariants for the modules in $PA$ and $IA$ (resp. $PB$ and $IB$).

Proof. (1) Let A be the K-algebra given by the quiver $1⇄ba2$ with relations aba=bab=0. Then we have $PA=IA=add(121⊕212)$. Since End$A(121)≅K[x]/(x2)≅EndA(212)$, it follows that $PA=IA$ is not Baer-Kaplansky. Let B denote the algebra considered in Example 2.8, given by the quiver $1⇄ba2$ with relation ab=0. Then there is an epimorphism $A→B$ and Lemma 2.7 implies that (1) holds.

(2) Let A be the algebra considered in Example 2.10, given by the quiver with relation b2=0. Next let B be the algebra considered in Example 2.6, given by the quiver with relations ba=b2=0. Also in this case there is an epimorphism $A→B$. Hence (2) holds.

We will use the next lemma to investigate classes of modules of finite projective or injective dimension.

### Lemma 2.12.

Let R be the K-algebra of finite dimension and let U and V be two finite dimensional left R-modules such that $EndR(U)≅K,EndR(V)≅K[x]/(x2)$, $HomR(U,V)≅K$ and $HomR(V,U)=0$. Then the class $add(U⊕V)$ is a Baer-Kaplansky class. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are not a complete set of invariants for the modules in $add(U⊕V)$.

Proof. The proof is similar to the proof of Lemma 2.5. More precisely, let X be a non-zero left R-module of the form $Um⊕Vn$ with $m,n∈ℕ$. Let $A=EndR(Um)$, $B=EndR(Vn)$, $H=HomR(Um,Vn)$, $T=EndR(X)$. Then A is isomorphic to the full matrix algebra Mm(K), B is isomorphic to the full matrix algebra $Mn(K[x]/(x2))$. H is a B-A-bimodule of dimension mn and T is isomorphic to the matrix algebra $A0HB$. Consequently, the following facts hold:

• (1) $dimT=m2+2n2+mn$.

• (2) The identity of T is the sum of m+n primitive idempotents.

• (3) The category of finitely generated left T-modules is a Krull-Schmidt category category [9, page 66].

• (4) The regular module TT is the direct sum of m indecomposable non-simple isomorphic left T-modules (of dimension m+n) and of n simple isomorphic left T-modules (of dimension 2n).

From now on we continue as in the last part of the proof of Lemma 2.5..

### Example 2.13.

There exist finite dimensional K-algebras A and B with the following properties:

• (1) The classes of finitely generated projective (resp. injective) left modules over A and B are Baer-Kaplansky classes.

• (2) Any finitely generated left A-module of finite projective dimension is projective.

• (3) Any finitely generated left B-module of finite injective dimension is injective.

• (4) The class of finitely generated left A-modules of finite injective dimension is not a Baer-Kaplansky class.

• (5) The class of finitely generated left B-modules of finite projective dimension is not a Baer-Kaplansky class.

Construction: Let A be the algebra of Example 2.6, given by the quiver with relations ba=b2=0. Next let B be the algebra, isomorphic to $Aop$, given by the quiver with relations a2=ba=0. Then $1,2,11,12,112$ are the indecomposable left B-modules, while

$add(2⊕112) and add(12⊕11)$

are the classes of finitely generated projective and finitely generated injective left B-modules, respectively. Hence (1) immediately follows from Example 2.6 (or Lemma 2.5) and Lemma 2.12. Since the left A-modules $1,2,122$ have infinite projective dimension, we conclude that (2) holds. Dually, the left B-modules $1,2,112$ have infinite injective dimension. Hence also (3) holds. Moreover the projective left A-module $22$ has injective dimension one, and we clearly have End$A(22)≅K[x]/(x2)≅EndA(122)$. Consequently (4) holds. Finally the injective left B-module $11$ has projective dimension one, and we obviously have End$B(11)≅K[x]/(x2)≅EndB(112)$. Hence also (5) holds.

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