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, and are the indecomposable left R-modules. Let M be the R-module I(3)=. Note that and P(3)=3. The lattice of submodules of M is

and we have . Also because and are one dimensional vector spaces. Therefore the class of simple injective left R-modules 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 , 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 . Then the indecomposable left R-modules are 1, 2, , and . Let M=I(2)=
. Note that , and S(2)=2. Also in this case the lattice of submodules of M is of the form

with . Since and are one dimensional vector spaces, . Then the class of simple left R-modules 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:
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(1) M has exactly three non-zero proper submodules N1, N2 and N1∩ N2.
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(2) M/N1 is not isomorphic to M/N2.
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(3) 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: , , , . Then the class 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 .
Proof. Since U and V are indecomposable modules, it follows that consists of finite direct sums of copies of U and V. Let X be a non-zero left R-module of the form with . Let , and let . Then A is isomorphic to the full matrix algebra Mm(K) and B is isomorphic to the full matrix algebra . Finally H is an A-B-bimodule of dimension mn. Let . Then our hypotheses on U, V and X imply that
(1) is isomorphic to the matrix algebra .
This means that the identity of T is the sum of primitive idempotents , such that
(2) The regular module TT is the direct sum of the simple isomorphic modules (of dimension m) and of the indecomposable non-simple isomorphic modules (of dimension m+2n).
Let Y be a module in add such that End. Then there exist such that , 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 .
We finally note that U2 and are non-isomorphic modules with endomorphism ring of dimension 4. More generally, if and s ≤ m, then and have endomorphism ring of the same dimension if and only if . 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 . Then the classes of finitely generated projective and injective modules are add and add respectively. Moreover we clearly have End, End, Hom and Hom. Hence the conclusion that add and add are Baer-Kaplansky classes with the desired property follows from Lemma 2.5.
We will use the next lemma to obtain Baer-Kaplansky classes with the property that Hom if L and M are two non-zero modules in .
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: . Then the class 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 .
Proof. We first note that for any we have dim. Assume that s is a natural number ≤ m such that dim. Then we have . Consequently s=0. Hence add 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:
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(1) The class of simple modules is not Baer-Kaplansky.
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(2) Let and be the classes of finitely generated projective modules and finitely generated injective modules, respectively and let be the class of finitely generated modules of projective and injective dimension at most one. Then , and 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 and .
Construction: Let R be the K-algebra given by the quiver with relation ab=0. Then and are the indecomposable modules. Since 1 and 2 are non-isomorphic one dimensional modules, (1) clearly holds. On the other hand we have , and . Consequently (2) follows from Lemma 2.7.
We will use the next lemma to obtain Baer-Kaplansky classes with more complicated Hom spaces between indecomposable non-isomorphic modules in .
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: and . Then the class is a Baer-Kaplansky class with the property described in Lemma 2.7.
Proof. Our hypotheses imply that for any . 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 and of finitely generated projective and finitely generated injective modules are Baer-Kaplansky, but the class 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 and .
Construction: Let R be the K-algebra given by the quiver 
with relation b2=0. Then we have and . This observation and Lemma 2.9 imply that and are Baer-Kaplansky classes with the desired property. We also note that there exist exact sequences of the form
and .
Hence and are in . Since End, we conclude that 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 and denote the classes of finitely generated projective and finitely generated injective left R-modules, respectively. Among others the following cases are possible:
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(1) is not a Baer-Kaplansky class, while and 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 and .
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(2) and 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 and (resp. and ).
Proof. (1) Let A be the K-algebra given by the quiver with relations aba=bab=0. Then we have . Since End, it follows that is not Baer-Kaplansky. Let B denote the algebra considered in Example 2.8, given by the quiver with relation ab=0. Then there is an epimorphism 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 . 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 , and . Then the class 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 .
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 with . Let , , , . Then A is isomorphic to the full matrix algebra Mm(K), B is isomorphic to the full matrix algebra . H is a B-A-bimodule of dimension mn and T is isomorphic to the matrix algebra . Consequently, the following facts hold:
-
(1) .
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(2) The identity of T is the sum of m+n primitive idempotents.
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(3) The category of finitely generated left T-modules is a Krull-Schmidt category category [9, page 66].
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(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:
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(1) The classes of finitely generated projective (resp. injective) left modules over A and B are Baer-Kaplansky classes.
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(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.
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(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 , given by the quiver
with relations a2=ba=0. Then 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 have infinite projective dimension, we conclude that (2) holds. Dually, the left B-modules have infinite injective dimension. Hence also (3) holds. Moreover the projective left A-module has injective dimension one, and we clearly have End. Consequently (4) holds. Finally the injective left B-module has projective dimension one, and we obviously have End. Hence also (5) holds.