Example 2.1.
There is a hereditary Kalgebra R of finite representation type and a class of Rmodules such that BaerKaplansky theorem fails.
Construction: Let R be the Kalgebra given by the quiver 1 &#_10230; 3 &#_10229; 2. Then 1, 2, 3, $\begin{array}{c}1\\ 3\end{array},\begin{array}{c}2\\ 3\end{array}$ and $\begin{array}{ccc}1& & 2\\ & 3& \end{array}$ are the indecomposable left Rmodules. Let M be the Rmodule I(3)=$\begin{array}{ccc}1& & 2\\ & 3& \end{array}$. Note that $P(1)=\begin{array}{c}1\\ 3\end{array},P(2)=\begin{array}{c}2\\ 3\end{array}$ and P(3)=3. The lattice of submodules of M is
and we have $I(3)/P(1)\ncong I(3)/P(2)$. Also ${\text{End}}_{R}(I(3)/P(1))\cong {\text{End}}_{R}(I(3)/P(2))\cong K$ because $I(3)/P(1)$ and $I(3)/P(2)$ are one dimensional vector spaces. Therefore the class of simple injective left Rmodules $\{I(3)/P(1),I(3)/P(2)\}=\{S(2),S(1)\}$ is not BaerKaplansky.
Remark 2.2.
Let R be a Kalgebra of finite representation type such that K is the endomorphism ring of any indecomposable left Rmodule. Then BaerKaplansky theorem fails for any class with more than one indecomposable module. Moreover BaerKaplansky theorem holds for any class of the form $\{{M}^{n}\mid n\ge 1\}$, where M is an indecomposable left Rmodule.
Example 2.3.
There is a nonhereditary Kalgebra R of finite representation type and a class of Rmodules such that BaerKaplansky theorem fails.
Construction: Let R be the Kalgebra given by the quiver with relations $ba={b}^{2}=0$. Then the indecomposable left Rmodules are 1, 2, $\begin{array}{c}1\\ 2\end{array}$, $\begin{array}{c}2\\ 2\end{array}$ and $\begin{array}{ccc}1& & 2\\ & 2& \end{array}$. Let M=I(2)=
$\begin{array}{ccc}1& & 2\\ & 2& \end{array}$. Note that $P(1)=\begin{array}{c}1\\ 2\end{array}$, $P(2)=\begin{array}{c}2\\ 2\end{array}$ and S(2)=2. Also in this case the lattice of submodules of M is of the form
with $I(2)/P(1)\ncong I(2)/P(2)$. Since $I(2)/P(1)$ and $I(2)/P(2)$ are one dimensional vector spaces, ${\text{End}}_{R}(I(2)/P(1))\cong {\text{End}}_{R}(I(2)/P(2))\cong K$. Then the class of simple left Rmodules $\{I(2)/P(1),I(2)/P(2)\}=\{S(2),S(1)\}$ is not BaerKaplansky. Note that S(1) is injective, while S(2) has infinite injective dimension. Here S(2) has a minimal injective resolution of the form
$$0\to S(2)=2\to \begin{array}{ccc}1& & 2\\ & 2& \end{array}\to 1\oplus \begin{array}{ccc}1& & 2\\ & 2& \end{array}\to 1\oplus \begin{array}{ccc}1& & 2\\ & 2& \end{array}\to \cdots .$$
Moreover both simple modules have infinite projective dimension. The minimal projective resolutions of S(1) and S(2) are of the form
$$\cdots \to \begin{array}{c}2\\ 2\end{array}\to \begin{array}{c}2\\ 2\end{array}\to \begin{array}{c}1\\ 2\end{array}\to 1=S(1)\to 0$$
and
$$\cdots \to \begin{array}{c}2\\ 2\end{array}\to \begin{array}{c}2\\ 2\end{array}\to \begin{array}{c}2\\ 2\end{array}\to 2=S(2)\to 0.$$
Proposition 2.4.
Let R be a Kalgebra admitting three nonisomorphic modules M with the following properties:

(1) M has exactly three nonzero proper submodules N_{1}, N_{2} and N_{1}∩ N_{2}.

(2) M/N_{1} is not isomorphic to M/N_{2}.

(3) ${\text{End}}_{R}(M/{N}_{i})\cong K$ for i=1, 2.
Then R is not commutative and there is a class (of nonuniserial modules) which is not BaerKaplansky.
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]/(x^{2}) [2, Theorem 3.8]. Since there exist three nonisomorphic modules M_{1}, M_{2} and M_{3} satisfying (1), (2) and (3), without loss of generality we may assume that {M_{1}, M_{2}} is not a BaerKaplansky class.
We will use the next lemma to construct BaerKaplansky 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 Kalgebra of finite dimension and let U and V be two finite dimensional left Rmodules with the following properties: ${\text{End}}_{R}(U)\cong K$, ${\text{End}}_{R}(V)\cong K[x]/({x}^{2})$, ${\text{Hom}}_{R}(U,V)=0$, ${\text{Hom}}_{R}(V,U)\cong K$. Then the class $\text{add}(U\oplus V)$is a BaerKaplansky 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 $\text{add}(U\oplus V)$.
Proof. Since U and V are indecomposable modules, it follows that $\text{add}(U\oplus V)$ consists of finite direct sums of copies of U and V. Let X be a nonzero left Rmodule of the form ${U}^{m}\oplus {V}^{n}$ with $m,n\in \mathbb{N}$. Let $A={\text{End}}_{R}({U}^{m})$, $B={\text{End}}_{R}({V}^{n})$ and let $H={\text{Hom}}_{R}({V}^{n},{U}^{m})$. Then A is isomorphic to the full matrix algebra M_{m}(K) and B is isomorphic to the full matrix algebra ${M}_{n}(K[x]/({x}^{2}))$. Finally H is an ABbimodule of dimension mn. Let $T={\text{End}}_{R}(X)$. Then our hypotheses on U, V and X imply that
(1) $T={\text{End}}_{R}(X)$ is isomorphic to the matrix algebra $\left[\begin{array}{cc}A& H\\ 0& B\end{array}\right]$.
This means that the identity of T is the sum of primitive idempotents ${e}_{1},\cdots ,{e}_{m}$, ${\u03f5}_{1},\cdots ,{\u03f5}_{n}$ such that
(2) The regular module _{T}T is the direct sum of the simple isomorphic modules $T{e}_{1},\cdots ,T{e}_{m}$ (of dimension m) and of the indecomposable nonsimple isomorphic modules $T{\u03f5}_{1},\cdots ,T{\u03f5}_{n}$ (of dimension m+2n).
Let Y be a module in add$(U\oplus V)$ such that End${}_{R}(X)\cong {\text{End}}_{R}(Y)$. Then there exist $p,q\in \mathbb{N}$ such that $Y\cong {U}^{p}\oplus {V}^{q}$, and so
(3) End_{R}(Y) is the direct sum of p simple isomorphic modules (of dimension p) and q indecomposable nonsimple 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 Tmodules is a KrullSchmidt category, that is a category where the KrullRemarkSchmidt theorem [10, p. 3] holds. Consequently we deduce from (2) and (3) that m=p and n=q, and so $X\cong Y$.
We finally note that U^{2} and $U\oplus V$ are nonisomorphic modules with endomorphism ring of dimension 4. More generally, if $m,n,s\in \mathbb{N}$ and s ≤ m, then ${U}^{m}\oplus {V}^{n}$ and ${U}^{ms}\oplus {V}^{n+s}$ have endomorphism ring of the same dimension ${m}^{2}+mn+2{n}^{2}$ if and only if $ms3ns2{s}^{2}=0$. The lemma is proved.
Example 2.6.
There is a noncommutative Kalgebra R of finite representation type such that the class of finitely generated projective (resp. injective) modules is a BaerKaplansky 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 $ba={b}^{2}=0$. Then the classes of finitely generated projective and injective modules are add$(\begin{array}{c}1\\ 2\end{array}\oplus \begin{array}{c}2\\ 2\end{array})$ and add$(1\oplus \begin{array}{ccc}1& & 2\\ & 2& \end{array})$ respectively. Moreover we clearly have End${}_{R}(\begin{array}{c}1\\ 2\end{array})\cong K\cong {\text{End}}_{R}(1)$, End${}_{R}(\begin{array}{c}2\\ 2\end{array})\cong K[x]/({x}^{2})\cong {\text{End}}_{R}(\begin{array}{ccc}1& & 2\\ & 2& \end{array})$, Hom${}_{R}(\begin{array}{c}1\\ 2\end{array},\begin{array}{c}2\\ 2\end{array})=0={\text{Hom}}_{R}(1,\begin{array}{ccc}1& & 2\\ & 2& \end{array})$ and Hom${}_{R}(\begin{array}{c}2\\ 2\end{array},\begin{array}{c}1\\ 2\end{array})\cong K\cong {\text{Hom}}_{R}(\begin{array}{ccc}1& & 2\\ & 2& \end{array},1)$. Hence the conclusion that add$(\begin{array}{c}1\\ 2\end{array}\oplus \begin{array}{c}2\\ 2\end{array})$ and add$(1\oplus \begin{array}{ccc}1& & 2\\ & 2& \end{array})$ are BaerKaplansky classes with the desired property follows from Lemma 2.5.
We will use the next lemma to obtain BaerKaplansky classes $\mathcal{C}$ with the property that Hom${}_{R}(L,M)\ne 0$ if L and M are two nonzero modules in $\mathcal{C}$.
Lemma 2.7.
Let R be a Kalgebra of finite dimension and let U and V be two finite dimensional left Rmodules with the following properties: ${\text{End}}_{R}(U)\cong K,{\text{End}}_{R}(V)\cong K[x]/({x}^{2}),{\text{Hom}}_{R}(U,V)\cong K\cong {\text{Hom}}_{R}(V,U)$. Then the class $\text{add}(U\oplus V)$ is a BaerKaplansky 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 $\text{add}(U\oplus V)$.
Proof. We first note that for any $m,n\in \mathbb{N}$ we have dim${\text{End}}_{R}({U}^{m}\oplus {V}^{n})={m}^{2}+2mn+2{n}^{2}$. Assume that s is a natural number ≤ m such that dim${\text{End}}_{R}({U}^{ms}\oplus {V}^{n+s})=\text{dim}({U}^{m}\oplus {V}^{n})$. Then we have $2ns+{s}^{2}=0$. Consequently s=0. Hence add$(U\oplus V)$ is a BaerKaplansky class with the desired property.
Example 2.8.
There is a nonhereditary Kalgebra R of finite representation type, such that any indecomposable module is uniserial, with the following properties:

(1) The class of simple modules is not BaerKaplansky.

(2) Let $\mathcal{P}$ and $\mathcal{I}$ be the classes of finitely generated projective modules and finitely generated injective modules, respectively and let $\mathcal{C}$ be the class of finitely generated modules of projective and injective dimension at most one. Then $\mathcal{P}$, $\mathcal{I}$ and $\mathcal{C}$ are BaerKaplansky 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 $\mathcal{P},\mathcal{I}$ and $\mathcal{C}$.
Construction: Let R be the Kalgebra given by the quiver $\text{1}\underset{b}{\overset{a}{\rightleftarrows}}2$ with relation ab=0. Then $1,2,\begin{array}{c}1\\ 2\end{array},\begin{array}{c}2\\ 1\end{array}$ and $\begin{array}{c}1\\ 2\\ 1\end{array}$ are the indecomposable modules. Since 1 and 2 are nonisomorphic one dimensional modules, (1) clearly holds. On the other hand we have $\mathcal{P}=\text{add}(\begin{array}{c}2\\ 1\end{array}\oplus \begin{array}{c}1\\ 2\\ 1\end{array})$, $\mathcal{I}=\text{add}(\begin{array}{c}1\\ 2\end{array}\oplus \begin{array}{c}1\\ 2\\ 1\end{array})$ and $\mathcal{C}=\text{add}(1\oplus \begin{array}{c}1\\ 2\\ 1\end{array})$. Consequently (2) follows from Lemma 2.7.
We will use the next lemma to obtain BaerKaplansky classes $\mathcal{C}$ with more complicated Hom spaces between indecomposable nonisomorphic modules in $\mathcal{C}$.
Lemma 2.9.
Let R be a Kalgebra of finite dimension and let U and V be two finite dimensional left Rmodules with the following properties: ${\text{End}}_{R}(U)\cong K,{\text{End}}_{R}(V)\cong K[x]/({x}^{2}),{\text{Hom}}_{R}(U,V)=0$ and ${\text{dimHom}}_{R}(V,U)=2$. Then the class $\text{add}(U\oplus V)$ is a BaerKaplansky class with the property described in Lemma 2.7.
Proof. Our hypotheses imply that ${\text{dimEnd}}_{R}({U}^{m}\oplus {V}^{n})={m}^{2}+2mn+2{n}^{2}$ for any $m,n\in \mathbb{N}$. Hence the conclusion follows from the proof of Lemma 2.7.
Example 2.10.
There is a Kalgebra R of finite dimension such that the classes $\mathcal{P}$ and $\mathcal{I}$ of finitely generated projective and finitely generated injective modules are BaerKaplansky, but the class $\mathcal{C}$ of finitely generated modules of projective and injective dimension at most one is not BaerKaplansky. Moreover the number of indecomposable direct summands and the dimension of the endomorphism ring are a complete set of invariants for the modules in $\mathcal{P}$ and $\mathcal{I}$.
Construction: Let R be the Kalgebra given by the quiver
with relation b^{2}=0. Then we have $\mathcal{P}=\text{add}(\begin{array}{c}1\\ 2\\ 2\end{array}\oplus \begin{array}{c}2\\ 2\end{array})$ and $\mathcal{I}=\text{add}(1\oplus \begin{array}{ccc}& & 1\\ 1& & 2\\ & 2& \end{array})$. This observation and Lemma 2.9 imply that $\mathcal{P}$ and $\mathcal{I}$ are BaerKaplansky classes with the desired property. We also note that there exist exact sequences of the form
$$0\to \begin{array}{c}2\\ 2\end{array}\to \begin{array}{ccc}& & 1\\ 1& & 2\\ & 2& \end{array}\to 1\oplus 1\to 0,$$ $$0\to \begin{array}{ccc}1& & 2\\ & 2& \end{array}\to \begin{array}{ccc}& & 1\\ 1& & 2\\ & 2& \end{array}\to 1\to 0$$
and $\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}0\to \begin{array}{c}2\\ 2\end{array}\to \begin{array}{c}1\\ 2\\ 2\end{array}\oplus \begin{array}{c}2\\ 2\end{array}\to \begin{array}{ccc}1& & 2\\ & 2& \end{array}\to 0$.
Hence $\begin{array}{c}2\\ 2\end{array}$ and $\begin{array}{ccc}1& & 2\\ & 2& \end{array}$ are in $\mathcal{C}$. Since End${}_{R}(\begin{array}{c}2\\ 2\end{array})\cong K[x]/({x}^{2})\cong {\text{End}}_{R}(\begin{array}{ccc}1& & 2\\ & 2& \end{array})$, we conclude that $\mathcal{C}$ is not BaerKaplansky.
Proposition 2.11.
Let A and B be finite dimensional Kalgebras such that there is an epimorphism from A to B. For any algebra R let ${\mathcal{P}}_{R}$ and ${\mathcal{I}}_{R}$ denote the classes of finitely generated projective and finitely generated injective left Rmodules, respectively. Among others the following cases are possible:

(1) ${\mathcal{P}}_{A}={\mathcal{I}}_{A}$ is not a BaerKaplansky class, while ${\mathcal{P}}_{B}$ and ${\mathcal{I}}_{B}$ are BaerKaplansky 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 ${\mathcal{P}}_{B}$ and ${\mathcal{I}}_{B}$.

(2) ${\mathcal{P}}_{A},{\mathcal{I}}_{A},{\mathcal{P}}_{B}$ and ${\mathcal{I}}_{B}$ are BaerKaplansky 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 ${\mathcal{P}}_{A}$ and ${\mathcal{I}}_{A}$ (resp. ${\mathcal{P}}_{B}$ and ${\mathcal{I}}_{B}$).
Proof. (1) Let A be the Kalgebra given by the quiver $1\underset{b}{\overset{a}{\rightleftarrows}}2$ with relations aba=bab=0. Then we have ${\mathcal{P}}_{A}={\mathcal{I}}_{A}=\text{add}(\begin{array}{c}1\\ 2\\ 1\end{array}\oplus \begin{array}{c}2\\ 1\\ 2\end{array})$. Since End${}_{A}(\begin{array}{c}1\\ 2\\ 1\end{array})\cong K[x]/({x}^{2})\cong {\text{End}}_{A}(\begin{array}{c}2\\ 1\\ 2\end{array})$, it follows that ${\mathcal{P}}_{A}={\mathcal{I}}_{A}$ is not BaerKaplansky. Let B denote the algebra considered in Example 2.8, given by the quiver $1\underset{b}{\overset{a}{\rightleftarrows}}2$ with relation ab=0. Then there is an epimorphism $A\to B$ 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 b^{2}=0. Next let B be the algebra considered in Example 2.6, given by the quiver with relations ba=b^{2}=0. Also in this case there is an epimorphism $A\to B$. 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 Kalgebra of finite dimension and let U and V be two finite dimensional left Rmodules such that ${\text{End}}_{R}(U)\cong K,{\text{End}}_{R}(V)\cong K[x]/({x}^{2})$, ${\text{Hom}}_{R}(U,V)\cong K$ and ${\text{Hom}}_{R}(V,U)=0$. Then the class $\text{add}(U\oplus V)$ is a BaerKaplansky 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 $\text{add}(U\oplus V)$.
Proof. The proof is similar to the proof of Lemma 2.5. More precisely, let X be a nonzero left Rmodule of the form ${U}^{m}\oplus {V}^{n}$ with $m,n\in \mathbb{N}$. Let $A={\text{End}}_{R}({U}^{m})$, $B={\text{End}}_{R}({V}^{n})$, $H={\text{Hom}}_{R}({U}^{m},{V}^{n})$, $T={\text{End}}_{R}(X)$. Then A is isomorphic to the full matrix algebra M_{m}(K), B is isomorphic to the full matrix algebra ${M}_{n}(K[x]/({x}^{2}))$. H is a BAbimodule of dimension mn and T is isomorphic to the matrix algebra $\left[\begin{array}{cc}A& 0\\ H& B\end{array}\right]$. Consequently, the following facts hold:

(1) $\text{dim}T={m}^{2}+2{n}^{2}+mn$.

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

(3) The category of finitely generated left Tmodules is a KrullSchmidt category category [9, page 66].

(4) The regular module _{T}T is the direct sum of m indecomposable nonsimple isomorphic left Tmodules (of dimension m+n) and of n simple isomorphic left Tmodules (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 Kalgebras A and B with the following properties:

(1) The classes of finitely generated projective (resp. injective) left modules over A and B are BaerKaplansky classes.

(2) Any finitely generated left Amodule of finite projective dimension is projective.

(3) Any finitely generated left Bmodule of finite injective dimension is injective.

(4) The class of finitely generated left Amodules of finite injective dimension is not a BaerKaplansky class.

(5) The class of finitely generated left Bmodules of finite projective dimension is not a BaerKaplansky class.
Construction: Let A be the algebra of Example 2.6, given by the quiver with relations ba=b^{2}=0. Next let B be the algebra, isomorphic to ${A}^{\text{op}}$, given by the quiver with relations a^{2}=ba=0. Then $1,2,\begin{array}{c}1\\ 1\end{array},\begin{array}{c}1\\ 2\end{array},\begin{array}{ccc}& 1& \\ 1& & 2\end{array}$ are the indecomposable left Bmodules, while
$$add(2\oplus \begin{array}{ccc}& 1& \\ 1& & 2\end{array})\text{\hspace{1em}}and\text{\hspace{1em}}add(\begin{array}{c}1\\ 2\end{array}\oplus \begin{array}{c}1\\ 1\end{array})$$
are the classes of finitely generated projective and finitely generated injective left Bmodules, respectively. Hence (1) immediately follows from Example 2.6 (or Lemma 2.5) and Lemma 2.12. Since the left Amodules $1,2,\begin{array}{ccc}1& & 2\\ & 2& \end{array}$ have infinite projective dimension, we conclude that (2) holds. Dually, the left Bmodules $1,2,\begin{array}{ccc}& 1& \\ 1& & 2\end{array}$ have infinite injective dimension. Hence also (3) holds. Moreover the projective left Amodule $\begin{array}{c}2\\ 2\end{array}$ has injective dimension one, and we clearly have End${}_{A}(\begin{array}{c}2\\ 2\end{array})\cong K[x]/({x}^{2})\cong {\text{End}}_{A}(\begin{array}{ccc}1& & 2\\ & 2& \end{array})$. Consequently (4) holds. Finally the injective left Bmodule $\begin{array}{c}1\\ 1\end{array}$ has projective dimension one, and we obviously have End${}_{B}(\begin{array}{c}1\\ 1\end{array})\cong K[x]/({x}^{2})\cong {\text{End}}_{B}(\begin{array}{ccc}& 1& \\ 1& & 2\end{array})$. Hence also (5) holds.