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

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 [1] and [11].

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 [6], 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 ⟶ 3 ⟵ 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

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

and

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 M_m(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 EndR(X)≅EndR(Y). Then there exist p,q∈ℕ such that Y≅Up⊕Vq, and so

(3) End_R(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 EndR(12)≅K≅EndR(1), EndR(22)≅K[x]/(x2)≅EndR(122), HomR(12,22)=0=HomR(1,122) and HomR(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 HomR(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 dimEndR(Um⊕Vn)=m2+2mn+2n2. Assume that s is a natural number ≤ m such that dimEndR(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

and    0→22→122⊕22→122→0.

Hence 22 and 122 are in C. Since EndR(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 EndA(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 M_m(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

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 EndA(22)≅K[x]/(x2)≅EndA(122). Consequently (4) holds. Finally the injective left B-module 11 has projective dimension one, and we obviously have EndB(11)≅K[x]/(x2)≅EndB(112). Hence also (5) holds.