IRREDUCIBLE UNITAL RIGHT ALTERNATIVE BIMODULES

Lucia Ikemoto Murakami^{1} Instituto de Matemática e Estatística, Universidade de São Paulo, Caixa Postal 66281, São Paulo - SP, 05315–970, Brazil e-mail: ikemoto@ime.usp.br and Ivan Shestakov^{2} Instituto de Matemática e Estatística, Universidade de São Paulo, Caixa Postal 66281, São Paulo - SP, 05315–970, Brazil and Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia e-mail:shestak@ime.usp.br

Abstract. Right alternative bimodules over the matrix algebra M_{2}(F) are investigated. It is shown that, contrary to the Jordan and alternative cases, there are infinitely many irreducible right alternative bimodules over this algebra. The irreducible unital bimodules up to dimension 6 are classified. An example of an indecomposable bimodule that is not irreducible is also constructed.

1. Introduction

The theory of representations forms an essential part of the structure theory in various classes of algebras. It suffices to mention here the Lie algebras, where there exists an enormous amount of papers dedicated to this subject.

The concepts of bimodule and birepresentation for a general class of algebras were introduced by S. Eilenberg [2]. Besides Lie algebras, the theory of birepresentations has been developed in other classical varieties of non-associative algebras. For instance, for alternative algebras. It was developed by R.Schafer [10], for Jordan algebras, by N.Jacobson [4], [5], for Malcev algebras and superalgebras, by R.Carlsson [1], E.Kuzmin [8], and A.Elduque with I.Shestakov [3].

2000 Mathematics Subject Classification. 17D15, 17C50.

Key words and phrases. right alternative bimodule, irreducible bimodule, Jordan bimodule, right alternative right module.

1Supported by Fapesp grant 99/09135-7

2Supported by CNPq grant 300528/99-0

In the case of right alternative algebras, only the theory of right representations has been developed, by A.Slin’ko and I.Shestakov. In our paper, we start to investigate irreducible right alternative bimodules. More exactly, we study the unital irreducible right alternative bimodules over the matrix algebra M_{2}(F). This problem was formulated in [7, Problem 2.132]. We classify all the irreducible unital bimodules up to dimension 6. It occurs that, contrary to the alternative or Jordan cases, where every simple finite dimensional algebra has a finite number of non-isomorphic irreducible bimodules, there are infinitely many such right alternative bimodules over M_{2}(F). The infinite number of non-isomorphic irreducible bimodules appears already in dimension 4, and we also show that there are irreducible bimodules of arbitrary big dimension. On the other hand, in dimensions 2 and 6 there are unique irreducible non-associative bimodules.

Another difference between the classical varieties and right alternative algebras is that not every unital right alternative bimodule over M_{2}(F) is completely reducible; we construct an indecomposable bimodule of dimension 6 that is not irreducible.

2. Basic concepts

We shall be concerned with algebras over a field F of characteristic not 2. Let A be an algebra over F. A vector space M over F is called A-bimodule if there are bilinear mappings A × M → M and M × A → M sending (a, m) to a.m and (m, a) to m.a, respectively. If the algebra A belongs to a class M of algebras, we say that M is a M-bimodule for A or an A-bimodule in the class M if the algebra E = A ∔ M with the multiplication given by (a + m)(b + n) = ab + (m.b + a.n), for all a, b in A and m, n in M, also belongs to M. The algebra E = A ∔ M is called the split null extension determined by M. For instance, in the associative case, we have E = A ∔ M is associative if and only if (a_{1}, a_{2}, m) = (a_{1}, m, a_{2}) = (m, a_{1}, a_{2}) = 0, for all a_{1}, a_{2} ∈ A and m ∈ M, where (x, y, z) = (xy)z − x(yz) is the associator of x, y and z (note that xy represents multiplication in A or the action of A on the bimodule M). Therefore this definition of associative bimodule coincides with the usual one. We are interested in right alternative bimodules. Recall that an algebra A over F is right alternative if (x, y, y) = 0, for all x, y ∈ A. Then M is a right alternative bimodule if

(m, a_{1}, a_{2}) + (m, a_{2}, a_{1}) = 0,  (1)
(a_{1}, m, a_{2}) + (a_{1}, a_{2}, m) = 0,  (2)

for all m ∈ M and a_{1}, a_{2} ∈ A.

Given an algebra A and a bimodule M for A, we have the applications ρ, λ: A → End M such that ρ(a)(m) = m.a and λ(a)(m) = a.m, for all a ∈ A and m ∈ M. If we denote by (λ, ρ)(A) the subalgebra of End M generated by {λ(a), ρ(a): a ∈ A} then M will be an associative left module for (ρ, λ)(A). In this sense, for every algebra A in a variety M, there is a universal algebra, more specifically, there is an associative algebra, denoted by U_{M}(A), such that each M-bimodule for A is an associative left module for U_{M}(A) and reciprocally, each associative left module for U_{M}(A) is a Mbimodule for A. The algebra U_{M}(A) is called M-multiplicative envelope for A. Let Ā be an algebra isomorphic to A and let T (A⊕Ā) be the tensor algebra of A⊕Ā. Then U_{M}(A) = T (A ⊕ Ā)/I for a suitable ideal I of T (A ⊕ Ā). We give some examples. If M is the variety of associative algebras and A ∈ M then I is the ideal of T (A ⊕ Ā) generated by \{\overline{ab}−b̄⊗ā, ab−a⊗b, ā⊗b−b⊗ā : a, b ∈ A\}. For Jordan algebras, we take the elements a−ā, a^{2}⊗a−a⊗a^{2}, 2a⊗(b⊗a−ba)+a^{2}b−b⊗a^{2} and for right alternative algebras, I is the ideal generated by the set {ā⊗ā−ā^{2}, ā⊗b−b⊗ā+a⊗b−ab : a, b ∈ A}. We can also consider the unital M-multiplicative envelope U_{M}^{1}(A) for an algebra A with unity that establishes a correspondence between unital M-bimodules for A and unital associative left modules for U_{M}^{1}(A).

The description of the M-multiplicative envelope for algebras in an arbitrary variety M may be very difficult. When A is a finite dimensional alternative or Jordan algebra then Schafer [10] and Jacobson [4] have proved that its multiplicative envelope will also have finite dimension. On the other hand, the unital multiplicative envelope for the right alternative algebra M_{2}(F) will already be of infinite dimension, since we will show that there are infinitely many non isomorphic irreducible unital right alternative bimodules for M_{2}(F).

Let A be a right alternative algebra. Then the algebra A^{+} with multiplication x ∘ y = \frac{1}{2}(xy + yx) is a Jordan algebra. Let M be a right alternative bimodule for A. We have two structures related to this bimodule.

If we consider only the right action of A in M, we will have a right alternative right module for A, that is, (m, a, a) = 0, for all a ∈ A and m ∈ M. This module is denoted by M^{r}. The structure of right alternative right modules has been described in [11]. It is proved, in particular, that the matrix algebra M_{n}(D), for a division algebra D, has two non isomorphic irreducible right alternative right modules, I_{n}^{r} and I_{n}^{l}, where I_{n}^{r} is isomorphic to the irreducible associative right module for M_{n}(D) and I_{n}^{l} is isomorphic to the irreducible associative left module for M_{n}(D). Besides, it is also shown that the right M_{n}(D)-modules are completely reducible. In the case n = 2, the irreducible right alternative right modules are I_{2}^{r} generated by {m′_{1}, m′_{2}}, with action

m_{i}e_{jk} = δ_{ij}m_{k}, for i, j, k = 1, 2,  (3)

and I_{2}^{l}, generated by {m_{1}, m_{2}}, with action

m_{i}e_{jk} = δ_{ik}m_{j}, for i, j, k = 1, 2.  (4)

Observe that these bimodules are alternative and isomorphic to sub-bimodules of Reg M_{2}(F), the regular bimodule, and Cay M_{2}(F), the Cayley bimodule (see [6]).

We can also consider another action on M. For each m ∈ M and a ∈ A^{+}, we define m ∘ a = \frac{1}{2}(m.a + a.m). In this way, we have a Jordan module for A^{+}. This module is denoted by M^{J}. Note that if A is a unital bimodule then M^{r} and M^{J} are unital too. The structure of irreducible Jordan modules for M_{2}(F) is also known, since M_{2}(F)^{+} is a Jordan algebra of a nondegenerate symmetric bilinear form.

We recall that, given a vector space V over F and a symmetric bilinear form f: V × V → F, the Jordan algebra J(V, f) of the form f is defined as the vector space direct sum J(V, f) = F.1 ⊕ V, where 1 is the unit of J(V, f) and, for u, v ∈ V,

u.v = f(u, v)1.

It is known that the unital multiplicative envelope of J = J(V, f) is the Meson algebra D(V, f), which is the quotient of the tensor algebra T (V) by the ideal of T (V) generated by {x ⊗ y ⊗ x − f(x, y)x : x, y ∈ V }. The structure of these algebras has been described by Jacobson [6] for a nondegenerate symmetric form f on a finite dimensional vector space V.

In our case, M_{2}(F)^{+} is the Jordan algebra of the nondegenerate symmetric form f over the vector space V generated by {e_{11} −e_{22}, e_{12}, e_{21}}, where e_{ij} are the elementary matrices, and f: V × V → F satisfies f(v_{1}, v_{2}).1 = v_{1} ∘ v_{2} = \frac{1}{2}(v_{1}v_{2} + v_{2}v_{1}) ∈ F.1, for all v_{1}, v_{2} ∈ V. By [6, VII.2,Theorem 5], the unital multiplicative envelope for J is isomorphic to F ⊕ M_{4}(F) ⊕ M_{3}(F) ⊕ M_{3}(F), and there are 4 non isomorphic irreducible unital Jordan bimodules for M_{2}(F)^{+}:

1. F n, with the action given by

n ∘ e_{11} = n ∘ e_{22} = \frac{1}{2}n; n ∘ e_{12} = n ∘ e_{21} = 0;  (5)

2. Reg M_{2}(F)^{+};

3. N = F n_{1} + F n_{2} + F n_{3}, with the action

\begin{matrix}n_{1} ∘ e_{11} = n_{1} ∘ e_{22} = n_{2} ∘ e_{12} = −n_{2} ∘ e_{21} = −n_{3} ∘ e_{12} = −n_{3} ∘ e_{21} = \frac{1}{2}n_{1}; \\ −n_{1} ∘ e_{12} = n_{2} ∘ e_{11} = n_{3} ∘ e_{11} = \frac{1}{2}(n_{2} + n_{3}); \\ n_{1} ∘ e_{21} = n_{2} ∘ e_{22} = −n_{3} ∘ e_{22} = \frac{1}{2}(n_{2} − n_{3}); \end{matrix} (6)

4. N′ = F n_{1} + F n_{2} + F n_{3}, with the action

\begin{matrix}n_{1} ∘ e_{11} = n_{1} ∘ e_{22} = −n_{2} ∘ e_{12} = n_{2} ∘ e_{21} = n_{3} ∘ e_{12} = n_{3} ∘ e_{21} = \frac{1}{2}n_{1}; \\ n_{1} ∘ e_{12} = n_{2} ∘ e_{22} = n_{3} ∘ e_{22} = \frac{1}{2}(n_{2} + n_{3}); \\ n_{1} ∘ e_{21} = −n_{2} ∘ e_{11} = n_{3} ∘ e_{11} = \frac{1}{2}(n_{3} − n_{2}). \end{matrix}  (7)

3. Some properties of irreducible unital right alternative bimodules

Using the known structures of unital Jordan bimodules for M_{2}(F)^{+} and unital right alternative right modules for M_{2}(F), we investigate some properties of irreducible unital right alternative bimodules. Let M be an irreducible unital right alternative bimodule over M_{2}(F). The underlying right alternative right module M^{r} has decomposition

M^{r} = ( ⨁_{i=1}^{k_{1}} M_{i}) ⊕ ( ⨁_{j=1}^{k_{2}} M′_{j}),

where M_{i} ≅ I_{2}^{r} and M′_{j} ≅ I_{2}^{l}, for all i and j. We denote M_{r}^{r} = ⨁_{i=1}^{k_{1}} M_{i} and M_{l}^{r} = ⨁_{j=1}^{k_{2}} M′_{j}.

If we consider the Jordan module M^{J} then

M^{J} ≅ ( ⨁_{i=1}^{s_{1}} F_{i}) ⊕ ( ⨁_{j=1}^{s_{2}} N_{j}) ⊕ ( ⨁_{k=1}^{s_{3}} N′_{k}) ⊕ ( ⨁_{l=1}^{s_{4}} A_{l}),

where F_{i} ≅ F, N_{j} ≅ N, N′_{k} ≅ N′ and A_{l} ≅ M_{2}(F)^{+}. We describe how these structures are related.

Proposition 1. If M^{J} has a component isomorphic to F then this component is contained in M_{l}^{r}.

Proof: If M^{J} has a component isomorphic to F then there is n ∈ M satisfying relations (5), that is,

e_{11}n = n − ne_{11}, e_{12}n = −ne_{12}, e_{21}n = −ne_{21}, e_{22}n = n − ne_{22}.

We decompose n = m_{1} + m_{2}, with m_{1} ∈ M_{r}^{r} and m_{2} ∈ M_{l}^{r}. Then, for any i, j, k, l ∈ {1, 2}, we have

(m_{1}e_{ij})e_{kl} = m_{1}(e_{ij}e_{kl}), (m_{2}e_{ij})e_{kl} = m_{2}(e_{kl}e_{ij}).

Using these relations and identity (2), we have −e_{11}(ne_{12}) = e_{11}(e_{12}n) = (e_{11}n)e_{12} − e_{11}(ne_{12})+e_{12}n = (n−ne_{11})e_{12} −e_{11}(ne_{12})−ne_{12} = −((m_{1} +m_{2})e_{11})e_{12} −e_{11}(ne_{12}) = −m_{1}e_{12} − e_{11}(ne_{12}), so m_{1}e_{12} = 0. In a similar way, considering −e_{22}(ne_{21}), we conclude that m_{1}e_{21} = 0. The component M_{r}^{r} has decomposition M_{r}^{r} = ⨁_{i=1}^{k} M_{i}, M_{i} ≅ I_{2}^{r}, and each M_{i} has a basis {m_{i1}, m_{i2}} satisfying (3). Writing m_{1} = ∑_{i=1}^{k} α_{i}m_{i1} + β_{i}m_{i2}, we have 0 = m_{1}e_{12} = ∑ α_{i}m_{i2} and 0 = m_{1}e_{21} = ∑ β_{i}m_{i1}, so α_{1} = … = α_{k} = 0 = β_{1} = ⋯ = β_{k}, that is, m_{1} = 0. Consequently, the component F.n is contained in M_{l}^{r}. □

Proposition 2. If M^{J} has a component isomorphic to N then this component is contained in M_{l}^{r}.

Proof: Let n_{1}, n_{2}, n_{3} ∈ M satisfy (6). Then

\begin{matrix}n_{1} = n_{2}e_{12} + e_{12}n_{2} = −n_{2}e_{21} − e_{21}n_{2}, \\ n_{3} = −n_{2} + n_{2}e_{11} + e_{11}n_{2} = n_{2} − n_{2}e_{22} − e_{22}n_{2}, \end{matrix}  (8)

and the relation (n_{2}e_{ij})e_{ij} = n_{2}e_{ij}^{2} gives us the following identities:

\begin{matrix}n_{1}e_{12} = (e_{12}n_{2})e_{12}, n_{1}e_{21} = −(e_{21}n_{2})e_{21}, \\ n_{3}e_{11} = (e_{11}n_{2})e_{11}, n_{3}e_{22} = −(e_{22}n_{2})e_{22}. \end{matrix}  (9)

Using (2), (8), and (9), we have

e_{12}n_{1} = e_{21}(n_{2}e_{12} + e_{12}n_{2}) = (e_{12}n_{2})e_{12} = n_{1}e_{12}.

In the same way,

e_{21}n_{1} = n_{1}e_{21}, e_{11}n_{3} = n_{3}e_{11}, e_{22}n_{3} = n_{3}e_{22}.

Now, comparing this with relations (6), we obtain n_{3}e_{11} = \frac{1}{2}(n_{3}+n_{2}) = e_{11}n_{3}, n_{1}e_{12} = \frac{1}{2}(−n_{3} − n_{2}) = e_{12}n_{1}, n_{1}e_{21} = \frac{1}{2}(−n_{3} + n_{2}) = e_{21}n_{1}, and n_{3}e_{22} = \frac{1}{2}(n_{3} − n_{2}) = e_{22}n_{3}, which provides us with four relations involving n_{2}:

n_{2} = 2n_{3}e_{11} − n_{3} = −2n_{1}e_{12} − n_{3} = 2n_{1}e_{21} + n_{3} = −2n_{3}e_{22} + n_{3}.  (10)

The first one gives us

n_{2}e_{11} = 2(n_{3}e_{11})e_{11} − n_{3}e_{11} = n_{3}e_{11} = \frac{1}{2}(n_{2} + n_{3}) = e_{11}n_{2}.  (11)

Analogously,

n_{2}e_{12} = −n_{3}e_{12}, n_{2}e_{21} = n_{3}e_{21}, n_{2}e_{22} = e_{22}n_{2} = −n_{3}e_{22}.  (12)

As in the previous proposition, we decompose M_{l}^{r} = ⨁_{i=1}^{k_{1}}⟨m_{i1}, m_{i2}⟩ and M_{r}^{r} = ⨁_{i=1}^{k_{2}}⟨m′_{i1}, m′_{i2}⟩. Let M_{1} = ⟨m_{i1}, i = 1, …, k_{1}⟩, M_{2} = ⟨m_{i2}, i = 1, …, k_{1}⟩, M′_{1} = ⟨m′_{i1}, i = 1, …, k_{2}⟩, and M′_{2} = ⟨m′_{i2}, i = 1, …, k_{2}⟩. Then we can write n_{2} = m_{1} + m_{2} + m′_{1} + m′_{2}, m_{i} ∈ M_{i}, and m′_{i} ∈ M′_{i}, i = 1, 2. Using the relations (10) – (12), with (3) and (4) we have

m_{1} + m′_{1} = n_{2}e_{11} = −2(n_{1}e_{12})e_{11} − n_{3}e_{11} = −2(n_{1}e_{12})e_{11} − n_{2}e_{11}.

So n_{2}e_{11} = −(n_{1}e_{12})e_{11} ∈ (M_{1} + M′_{2})e_{11} = M_{1}e_{11} = M_{1}, which implies m′_{1} = 0. In the same way, m_{2} + m′_{2} = n_{2}e_{22} = 2(n_{1}e_{21})e_{22} + n_{3}e_{22} = 2(n_{1}e_{21})e_{22} − n_{2}e_{22}; that is, n_{2}e_{22} = (n_{1}e_{21})e_{22} ∈ (M_{2} + M′_{1})e_{22} = M_{2}, which implies m′_{2} = 0. Then n_{2} = m_{1} + m_{2} ∈ M_{1} + M_{2} = M_{l}^{r}. Also we have m_{1} = n_{2}e_{11} = \frac{1}{2}(n_{2} + n_{3}), so

n_{3} = 2m_{1} − n_{2} = 2m_{1} − m_{1} − m_{2} = m_{1} − m_{2} ∈ M_{1} + M_{2} = M_{l}^{r}.

It remains to show that n_{1} ∈ M_{l}^{r}. Using (8), (2), (11), and the fact that n_{2} ∈ M_{l}^{r}, we have e_{11}n_{1} = −e_{11}(n_{2}e_{21}) − e_{11}(e_{21}n_{2}) = −(e_{11}n_{2})e_{21} = −(n_{2}e_{11})e_{21} = −n_{2}e_{21} and e_{11}n_{1} = e_{11}(n_{2}e_{12}) + e_{11}(e_{12}n_{2}) = (e_{11}n_{2})e_{12} + e_{12}n_{2} = (n_{2}e_{11})e_{12} + e_{12}n_{2} = e_{12}n_{2}. Then

e_{12}n_{2} = e_{11}n_{1} = −n_{2}e_{21}.  (13)

In the same way, we can conclude that

−e_{21}n_{2} = e_{22}n_{1} = n_{2}e_{12}.  (14)

Finally, n_{1} = n_{2}e_{12} +e_{12}n_{2} = n_{2}e_{12} −n_{2}e_{21} ∈ M_{1} +M_{2} = M_{l}^{r}, since M_{1} +M_{2} is a right submodule of M. Therefore the component isomorphic to N is contained in M_{l}^{r}. □

Remark 1. The above proof gives us some additional information about the structure of M when M^{J} has a component isomorphic to N. We have proved that n_{2} = m_{1}+m_{2} and n_{3} = m_{1} − m_{2}. As {n_{2}, n_{3}} is a linearly independent set, we must have m_{1} ≠ 0 and m_{2} ≠ 0. Let m_{3} = n_{2}e_{12} = m_{2}e_{12} ∈ M_{1} and m_{4} = n_{2}e_{21} = m_{1}e_{21} ∈ M_{2}. Then n_{1} = n_{2}e_{12} + e_{21}n_{2} = n_{2}e_{12} − n_{2}e_{21} = m_{3} − m_{4}. We note that {m_{1}, m_{2}, m_{3}, m_{4}} is a linearly independent set. In fact, if m_{4} = αm_{2} then αm_{3} = αn_{2}e_{12} = αm_{2}e_{12} = m_{4}e_{12} = (n_{2}e_{21})e_{12} = n_{2}e_{11} = m_{1}; that is, {n_{1}, n_{2}, n_{3}} would have dimension 2. In the same way, we prove that m_{3} is not a multiple of m_{1}. So {m_{1}, m_{2}, m_{3}, m_{4}} is a linearly independent set. The relations (11) – (14) and the structure of right alternative right module give us the following table the actions of M_{2}(F) on these elements:

\begin{matrix}m_{1}e_{11} = m_{1}, & m_{2}e_{11} = 0, & m_{3}e_{11} = m_{3}, & m_{4}e_{11} = 0, \\ m_{1}e_{12} = 0, & m_{2}e_{12} = m_{3}, & m_{3}e_{12} = 0, & m_{4}e_{12} = m_{1}, \\ m_{1}e_{21} = m_{4}, & m_{2}e_{21} = 0, & m_{3}e_{21} = m_{2}, & m_{4}e_{21} = 0, \\ m_{1}e_{22} = 0, & m_{2}e_{22} = m_{2}, & m_{3}e_{22} = 0, & m_{4}e_{22} = m_{4}, \end{matrix}  (15)

and

\begin{matrix}e_{11}m_{1} = m_{1}, & e_{11}m_{2} = 0, & e_{11}m_{3} − e_{11}m_{4} = −m_{4}, \\ e_{12}m_{1} = 0, & e_{12}m_{2} = −m_{4}, & e_{12}m_{3} − e_{12}m_{4} = −m_{1}, \\ e_{21}m_{1} = −m_{3}, & e_{21}m_{2} = 0 & e_{21}m_{3} − e_{21}m_{4} = m_{2}, \\ e_{22}m_{1} = 0, & e_{22}m_{2} = m_{2} & e_{22}m_{3} − e_{22}m_{4} = m_{3}. \end{matrix}  (16)

Proposition 3. The Jordan module M^{J} does not have any irreducible component isomorphic to N′.

Proof: Let n_{1}, n_{2}, n_{3} ∈ M satisfy (7), then

\begin{matrix}n_{1} = −n_{2}e_{12} − e_{12}n_{2} = n_{2}e_{21} + e_{21}n_{2}, \\ n_{3} = n_{2} − n_{2}e_{11} − e_{11}n_{2} = n_{2}e_{22} + e_{22}n_{2} − n_{2}. \end{matrix}  (17)

Using (n_{2}e_{ij})e_{ij} = n_{2}e_{ij}^{2} and (2), we conclude that n_{3}e_{11} = e_{11}n_{3} = \frac{1}{2}(n_{3} − n_{2}), n_{3}e_{22} = e_{22}n_{3} = \frac{1}{2}(n_{3} + n_{2}),

n_{1}e_{12} = e_{12}n_{1} = \frac{1}{2}(n_{3} + n_{2}) and n_{1}e_{21} = e_{21}n_{1} = \frac{1}{2}(n_{3} − n_{2}).

Again we have some relations for n_{2}:

n_{2} = −2n_{3}e_{11} + n_{3} = 2n_{1}e_{12} − n_{3} = −2n_{1}e_{21} + n_{3} = 2n_{3}e_{22} − n_{3},  (18)

which implies

n_{2}e_{11} = −n_{3}e_{11}, n_{2}e_{12} = −n_{3}e_{12}, n_{2}e_{21} = n_{3}e_{21}, n_{2}e_{22} = n_{3}e_{22}.

As n_{2} ∘ e_{11} = −n_{3} ∘ e_{11}, n_{2} ∘ e_{22} = n_{3} ∘ e_{22}, n_{3}e_{11} = e_{11}n_{3} and n_{3}e_{22} = e_{22}n_{3}, we also have

n_{2}e_{11} = e_{11}n_{2} and n_{2}e_{22} = e_{22}n_{2}.

Decomposing n_{2} = m_{1} +m_{2} +m′_{1} +m′_{2}, m_{i} ∈ M_{i}, m′_{i} ∈ M′_{i}, i = 1, 2, as in the previous proposition, and using the above relations, we have

m_{1} + m′_{1} = n_{2}e_{11} = −2(n_{1}e_{21})e_{11} + n_{3}e_{11} = −2(n_{1}e_{21})e_{11} − n_{2}e_{11}.

Thus n_{2}e_{11} = −(n_{1}e_{21})e_{11} ∈ (M_{2} + M′_{1})e_{11} = M′_{1}, and then m_{1} = 0. In the same way, m_{2} + m′_{2} = n_{2}e_{22} = 2(n_{1}e_{12})e_{22} − n_{3}e_{22} = 2(n_{1}e_{12})e_{22} − n_{2}e_{22} implies n_{2}e_{22} = (n_{1}e_{12})e_{22} ∈ (M_{1} + M′_{2})e_{22} = M′_{2}, so m_{2} = 0. Therefore n_{2} = m′_{1} + m′_{2} ∈ M_{r}^{r}, then

e_{11}n_{1} = e_{11}(n_{2}e_{21}) + e_{11}(e_{21}n_{2}) = (e_{11}n_{2})e_{21} = (n_{2}e_{11})e_{21} = 0.

By (7) we conclude that n_{1}e_{11} = n_{1}. Since M is unital, we must have n_{1}e_{22} = 0, and so e_{22}n_{1} = n_{1}. However, as n_{2} ∈ M_{r}^{r}, e_{22}n_{1} = −e_{22}(n_{2}e_{12}) − e_{22}(e_{12}n_{2}) = −(e_{22}n_{2})e_{12} = −(n_{2}e_{22})e_{12} = 0, which is a contradiction. Consequently, M^{J} does not have any component isomorphic to N′. □

Proposition 4. If M is a unital right alternative bimodule for M_{2}(F) such that its correspondent Jordan module M^{J} has a component isomorphic to Reg M_{2}(F)^{+} then M has a sub-bimodule isomorphic to Reg M_{2}(F).

Proof: Suppose M^{J} has a component isomorphic to M_{2}(F)^{+}, and let n_{1}, n_{2}, n_{3}, n_{4} ∈ M be the elements corresponding to e_{11}, e_{12}, e_{21} and e_{22}, respectively. We intend to prove that this correspondence is also an isomorphism between M and Reg M_{2}(F). These elements satisfy

\begin{matrix}n_{1} ∘ e_{11} = n_{1} & n_{2} ∘ e_{11} = \frac{1}{2}n_{2} & n_{3} ∘ e_{11} = \frac{1}{2}n_{3} & n_{4} ∘ e_{11} = 0 \\ n_{1} ∘ e_{12} = \frac{1}{2}n_{2} & n_{2} ∘ e_{12} = 0 & n_{3} ∘ e_{12} = \frac{1}{2}(n_{1} + n_{4}) & n_{4} ∘ e_{12} = \frac{1}{2}n_{2} \\ n_{1} ∘ e_{21} = \frac{1}{2}n_{3} & n_{2} ∘ e_{21} = \frac{1}{2}(n_{1} + n_{4}) & n_{3} ∘ e_{21} = 0 & n_{4} ∘ e_{21} = \frac{1}{2}n_{3} \\ n_{1} ∘ e_{22} = 0 & n_{2} ∘ e_{22} = \frac{1}{2}n_{2} & n_{3} ∘ e_{22} = \frac{1}{2}n_{3} & n_{4} ∘ e_{22} = n_{4} \end{matrix} (19)

By the first relation, we have e_{11}n_{1} = 2n_{1} − n_{1}e_{11} and then (e_{11}n_{1})e_{11} = n_{1}e_{11}, which implies 2n_{1} = e_{11}n_{1} + n_{1}e_{11} = (e_{11}e_{11})n_{1} + (e_{11}n_{1})e_{11} = e_{11}(n_{1}e_{11}) + e_{11}(e_{11}n_{1}) = 2e_{11}n_{1}; that is,

e_{11}n_{1} = n_{1} and n_{1}e_{11} = n_{1}.

Similarly, one can show that

e_{11}n_{2} = n_{2}, n_{2}e_{11} = 0, e_{11}n_{3} = 0, n_{3}e_{11} = n_{3}, e_{11}n_{4} = 0, n_{4}e_{11} = 0, and also e_{22}n_{1} = n_{1}e_{22} = e_{22}n_{2} = 0, n_{2}e_{22} = n_{2}, e_{22}n_{3} = n_{3}, n_{3}e_{22} = 0, e_{22}n_{4} = n_{4}e_{22} = n_{4}. With the relation (n_{1}e_{ij})e_{ij} = n_{1}e_{ij}^{2} and (19), we prove that

n_{2}e_{12} = (e_{12}n_{1})e_{12} and n_{3}e_{21} = (e_{21}n_{1})e_{21}.

Then e_{12}n_{2} = e_{12}(n_{1}e_{12}) + e_{12}(e_{12}n_{1}) = (e_{12}n_{1})e_{12} = n_{2}e_{12} and e_{21}n_{3} = e_{21}(n_{1}e_{21}) + e_{21}(e_{21}n_{1}) = (e_{21}n_{1})e_{21} = n_{3}e_{21}. So

e_{12}n_{2} = n_{2}e_{12} = 0 = e_{21}n_{3} = n_{3}e_{21}.

Now 0 = (n_{2}e_{21})e_{21} = (n_{1} + n_{4} − e_{21}n_{2})e_{21}; that is, (n_{1} + n_{4})e_{21} = (e_{21}n_{2})e_{21} = e_{21}(n_{2}e_{21})+e_{21}(e_{21}n_{2}) = e_{21}(n_{1}+n_{4}). Since e_{21}n_{1}+n_{1}e_{21}+e_{21}n_{4}+n_{4}e_{21} = 2n_{3}, we have e_{21}(n_{1} +n_{4}) = n_{3} = (n_{1} +n_{4})e_{21}. Analogously, e_{12}(n_{1} +n_{4}) = n_{2} = (n_{1} +n_{4})e_{12}. Thus 0 = n_{2}e_{11} = (e_{12}n_{1})e_{11}+(e_{12}n_{4})e_{11} = e_{12}(n_{1}e_{11})+e_{12}(e_{11}n_{1})+e_{12}(n_{4}e_{11})+e_{12}(e_{11}n_{4}) = 2e_{12}n_{1}, which implies

e_{12}n_{1} = 0 and n_{1}e_{12} = n_{2}.

In the same way, we prove that

e_{12}n_{4} = n_{2}, n_{4}e_{12} = 0, e_{21}n_{1} = n_{3}, n_{1}e_{21} = 0, e_{21}n_{4} = 0, n_{4}e_{21} = n_{3}. Finally, n_{2}e_{21} = (n_{1}e_{12})e_{21} = n_{1}(e_{12}e_{21}) − (n_{1}e_{21})e_{12} + n_{1}(e_{21}e_{12}) = n_{1}e_{11} + n_{1}e_{22} = n_{1} and n_{3}e_{12} = n_{4}. Then, by (19), we have e_{21}n_{2} = n_{4} and n_{12}n_{3} = n_{1}. Hence we have proved that n_{1}, n_{2}, n_{3}, n_{4} generates a sub-bimodule isomorphic to M_{2}(F). □

Corollary 1. Let M be a irreducible unital right alternative bimodule for M_{2}(F) such that its correspondent Jordan module M^{J} has a component isomorphic to Reg M_{2}(F)^{+}. Then M = RegM_{2}(F) and, consequently, M is associative.

Corollary 2. If M is an irreducible unital right alternative bimodule for M_{2}(F) and M is not associative then M^{J} ≅ ( ⨁_{i=1}^{s_{1}} F_{i}) ⊕ ( ⨁_{j=1}^{s_{2}} N_{j}) and M_{r}^{r} = 0.

4. Irreducible unital right alternative bimodules of low dimension

In this section we describe all non isomorphic non associative irreducible unital right alternative bimodules of dimension less than or equal to 6. We will see that, already in dimension 4, there are infinitely many non isomorphic irreducible bimodules. This, in particular, shows that the unital multiplication envelope for the right alternative algebra M_{2}(F) has infinite dimension.

We note that unital right alternative bimodules have even dimension because the irreducible right alternative right modules have dimension 2. Moreover, we have seen in the previous section that every irreducible unital right alternative bimodule M for M_{2}(F), which is not the regular bimodule, satisfies M_{r}^{r} = 0. Then M^{r} = M_{l}^{r} = ⨁_{i=1}^{k_{1}}⟨m_{i1}, m_{i2}⟩. As before we denote M_{1} = ⟨m_{i1}, i = 1, …, k_{1}⟩ and M_{2} = ⟨m_{i2}, i = 1, …, k_{1}⟩.

4.1. Bimodules of dimension 2. By the previous considerations about the structure of unital right alternative bimodules we have only one possibility for dimension two: M^{r} ≅ I_{2}^{l} and M^{J} ≅ F ⊕ F. Comparing these two structures we conclude that M = F m_{1} + F m_{2} with action

m_{i}e_{jk} = δ_{ik}m_{j}, for i, j, k = 1, 2;

e_{ij}m = −me_{ij} and e_{jj}m = me_{ii}, for i ≠ j. (20)

A straightforward calculation shows that this action, in fact, determines a right alternative bimodule, which is alternative, isomorphic to the Cayley bimodule over M_{2}(F).

4.2. Bimodules of dimension 4. To study irreducible unital bimodules of dimension 4, we need the following result:

Proposition 5. If M is a unital right alternative bimodule such that M^{J} ≅ ⨁_{i=1}^{r} F with r ≥ 3 then M is reducible.

Proof: By Proposition 1 we have that M^{r} ≅ ⨁ I_{2}^{l}. As M^{J} ≅ ⨁_{i=1}^{r} F, every element n ∈ M satisfy (5). Therefore M must be a sum of bimodules of dimension 2, which has been described in Section 4.1. So, M is reducible. □

Let M be an irreducible unital right alternative bimodule of dimension 4 and suppose M is not associative. Then, by the previous propositions, we have M^{r} ≅ I_{2}^{l} ⊕ I_{2}^{l} and M^{J} ≅ F n ⊕ N where n satisfies (5). We have already seen that there are linearly independent elements m_{1}, m_{2}, m_{3}, m_{4} ∈ M satisfying relations (15) and (16) such that {m_{1} + m_{2}, m_{1} − m_{2}, m_{3} − m_{4}} generates N. As M has dimension 4, we have that n = α_{1}m_{1} + α_{2}m_{2} + α_{3}m_{3} + α_{4}m_{4} with α_{3} + α_{4} ≠ 0. We can assume α_{3}+α_{4} = 1. By (15) and (16), it remains to determine only the left action of e_{ij} on m_{3} and m_{4}. From ne_{11} +e_{11}n = n we obtain α_{1}m_{1} +α_{3}m_{3} +α_{1}m_{1} +α_{3}e_{11}m_{3} +α_{4}e_{11}m_{4} = α_{1}m_{1} + α_{2}m_{2} + α_{3}m_{3} + α_{4}m_{4}; that is, α_{3}e_{11}m_{3} + α_{4}e_{11}m_{4} = −α_{1}m_{1} + α_{2}m_{2} + α_{4}m_{4}. By (16), e_{11}m_{3} − e_{11}m_{4} = −m_{4}, therefore (α_{3} + α_{4})e_{11}m_{3} = −α_{1}m_{1} + α_{2}m_{2} and (α_{3} + α_{4})e_{11}m_{4} = −α_{1}m_{1} + α_{2}m_{2} + m_{4}. Similarly, using also that α_{3} + α_{4} = 1,

e_{11}m_{3} = −α_{1}m_{1} + α_{2}m_{2}  e_{11}m_{4} = −α_{1}m_{1} + α_{2}m_{2} + m_{4}  (21)
e_{12}m_{3} = −2α_{4}m_{1} − α_{2}m_{3} + α_{2}m_{4}  e_{12}m_{4} = (1 − 2α_{4})m_{1} − α_{2}m_{3} + α_{2}m_{4} (22)
e_{21}m_{3} = (2α_{4} − 1)m_{2} + α_{1}m_{3} − α_{1}m_{4} e_{21}m_{4} = 2(α_{4} − 1)m_{2} + α_{1}m_{3} − α_{1}m_{4} (23)
e_{22}m_{3} = α_{1}m_{1} − α_{2}m_{2} + m_{3}  e_{22}m_{4} = α_{1}m_{1} − α_{2}m_{2}  (24)

This action, in fact, determine a right alternative bimodule. It is possible to check the identities (1) and (2) by a straightforward calculation or using a computer program, like MAPLE. Hence we have a family of irreducible unital bimodules which depends on three parameters α_{1}, α_{2}, α_{4}. We observe that these bimodules are irreducible since the two Jordan components have odd dimension and the right alternative right submodules have even dimension.

Definition 1. We denote by M(α, β, γ) the four dimensional right alternative bimodule F n ⊕ N, where n = αm_{1} + βm_{2} + (1 − γ)m_{3} + γm_{4} and m_{i} satisfy relations (15), (16), and (21) – (24), with α_{1} = α, α_{2} = β, and α_{4} = γ.

Theorem 1. A unital right alternative bimodule M of dimension 4 is irreducible if and only if it is isomorphic to a bimodule M(α, β, γ), for some α, β, γ ∈ F. Moreover M(α, β, γ) and M(α′, β′, γ′) are isomorphic if and only if α = α′, β = β′ and γ = γ′.

Proof: By the previous considerations we have already proved the first part of the theorem. It remains to show the second one. Let M = M(α, β, γ) generated by {m_{1}, m_{2}, m_{3}, m_{4}} and M = M(α′, β′, γ′) generated by {m′_{1}, m′_{2}, m′_{3}, m′_{4}} be isomorphic bimodules, where m_{i} and m′_{j} satisfy (15), (16), and (21) – (24) and let φ: M′ → M be an isomorphism. Writing φ(m′_{i}) = n_{i} = a_{i1}m_{1}+a_{i2}m_{2}+a_{i3}m_{3}+a_{i4}m_{4}, with a_{ij} ∈ F we have n_{1} = φ(m′_{1}) = φ(m′_{1}e_{11}) = φ(m′_{1})e_{11} = n_{1}e_{11} = a_{11}m_{1} + a_{13}m_{3}, then a_{12} = a_{14} = 0. In the same way, we prove that n_{i} are of the form

n_{1} = a_{1}m_{1} + a_{3}m_{3}  n_{2} = a_{2}m_{2} + a_{4}m_{4}
n_{3} = b_{1}m_{1} + b_{3}m_{3}  n_{4} = b_{2}m_{2} + b_{4}m_{4}

where a_{i}, b_{j} ∈ F. Moreover, b_{2}m_{2} + b_{4}m_{4} = n_{4} = φ(m′_{4}) = φ(m′_{1}e_{21}) = φ(m′_{1})e_{21} = (a_{1}m_{1} +a_{3}m_{3})e_{21} = a_{1}m_{4} +a_{3}m_{2}. Analogously we prove b_{1}m_{1} +b_{3}m_{3} = a_{2}m_{3} +a_{4}m_{1}. Then b_{4} = a_{1}, b_{2} = a_{3}, b_{1} = a_{4}, and b_{3} = a_{2}. Besides e_{11}m′_{2} = 0 implies 0 = e_{11}φ(m′_{2}) = e_{11}(a_{2}m_{2} + a_{4}m_{4}) = a_{4}e_{11}m_{4} = a_{4}(αm_{1} + βm_{2} + m_{4}). Thus a_{4} = 0. Using e_{22}m′_{1} = 0 we conclude that a_{3} = 0, so

n_{1} = a_{1}m_{1}, n_{2} = a_{2}m_{2}, n_{3} = a_{2}m_{3}, n_{4} = a_{1}m_{4}.

Finally e_{21}m′_{1} = −m′_{3} implies −a_{2}m_{3} = −n_{3} = φ(−m′_{3}) = φ(e_{21}m′_{1}) = e_{21}(a_{1}m_{1}) = −a_{1}m_{3}, that is, a_{1} = a_{2}. Therefore m′_{i} ↦ m_{i} is an isomorphism and consequently α = α′, β = β′, and γ = γ′. □

Remark 2. With the same proof as in 4.2 we can show that if M is a unital bimodule such that M^{J} has a component isomorphic to N, generated by {m_{1} + m_{2}, m_{1} − m_{2}, m_{3} − m_{4}} and also a component isomorphic to F, generated by n ∈ M such that n is a linear combination of m_{1}, m_{2}, m_{3}, and m_{4} then M has a sub-bimodule of dimension 4.

Remark 3. Some examples of right alternative bimodules over M_{2}(F) of dimension 4 appear in [12]. In this paper, the author has constructed a family of irreducible bimodules depending on parameters κ, τ and ξ. These ones are particular cases of the bimodules obtained above, just taking α = κτ, β = 0, and 2γ = 1 + ξ.

4.3. Bimodules of dimension 6. To classify all irreducible unital bimodules of dimension 6 we give a general result about reducibility of unital right alternative bimodules analysing the decomposition of the Jordan module associated to them.

Proposition 6. If M is a unital right alternative bimodule such that M^{J} ≅ N ⊕ ( ⨁_{i=1}^{k} F ) with k ≥ 3 then M is reducible.

Proof: We intend to prove that M has a sub-bimodule of dimension 4. We observe that any element in ⨁_{i=1}^{k} F n_{i} ≅ ⨁_{i=1}^{k} F generates a Jordan module isomorphic to F. Then, by Remark 1, it is enough to show that there is n ∈ ⨁_{i=1}^{k} F n_{i} which is a linear combination of m_{1}, m_{2}, m_{3} and m_{4}, where {m_{1} + m_{2}, m_{1} − m_{2}, m_{3} − m_{4}} generates the component of M^{J} isomorphic to N and satisfy the relations given in (15) and (16). As {m_{1}, m_{2}, m_{3}, m_{4}} and {n_{1}, …, n_{k}} generates subspaces V and W of dimensions 4 and k, respectively, and M is a bimodule of dimension k +3, we must have V ∩ W ≠ 0, and so, the proposition is proved. □

By Propositions 4, 5 and 6, the only possibility for a unital bimodule M of dimension 6 be irreducible is when M^{J} ≅ N ⊕ N = ⟨n_{1}, n_{2}, n_{3}⟩ + ⟨n′_{1}, n′_{2}, n′_{3}⟩. As in the previous cases, we have m_{1}, m_{3}, m′_{1}, m′_{3} ∈ M_{1} and m_{2}, m_{4}, m′_{2}, m′_{4} ∈ M_{2} such that

\begin{matrix}n_{2} = m_{1} + m_{2}, n_{3} = m_{1} − m_{2}, n_{1} = m_{3} − m_{4}, \\ n′_{2} = m′_{1} + m′_{2}, n′_{3} = m′_{1} − m′_{2}, n′_{1} = m′_{3} − m′_{4} \end{matrix}  (25)

and satisfy (15) and (16). First we observe that {m_{2}, m_{3}, m′_{2}, m′_{3}} is a linearly independent set. In fact, 2m_{2} = n_{2} − n_{3} ∈ N and 2m′_{2} = n′_{2} − n′_{3} ∈ N′, therefore m_{2} and m′_{2} are linearly independent. Consequently m_{3} and m′_{3} are also linearly independent since m_{3}e_{21} = m_{2} and m′_{3}e_{21} = m′_{2}. As m_{2}, m′_{2} ∈ M_{2} and m_{3}, m′_{3} ∈ M_{1}, {m_{2}, m_{3}, m′_{2}, m′_{3}} is linearly independent. Let P be the subspace generated by these elements. Since P is closed by right action of M_{2}(F), it is a proper right alternative right bimodule of M. If m_{1}, m′_{1} ∈ P then we would have m_{4}, m′_{4} ∈ P, which would imply P = M, a contradiction. So we can suppose that m_{1} ∉ P and then

M = ⟨m_{2}, m_{3}⟩ ⊕ ⟨m_{1}, m_{4}⟩ ⊕ ⟨m′_{2}, m′_{3}⟩.

Now we write m′_{1} = αm_{1} + βm_{3} + γm′_{3}, with α, β, γ ∈ F and so m′_{4} = m′_{1}e_{21} = αm_{4} + βm_{2} + γm′_{2}. If α ≠ 0 then we substitute each m′_{i} by m̃′_{i} = m′_{i} − αm_{i}. These new elements satisfy relations (15) and (16),

M = ⟨m_{2}, m_{3}⟩ ⊕ ⟨m_{1}, m_{4}⟩ ⊕ ⟨m̃′_{2}, m̃′_{3}⟩,

m̃′_{1} = β′m_{3} + γm̃′_{3} and m̃′_{4} = β′m_{2} + γm̃′_{2}, where β′ = αγ + β. As {m̃′_{1}, m̃′_{2}, m̃′_{3}, m̃′_{4}} is a linearly independent set we have β′ ≠ 0. Now we substitute each m_{i} by m̃_{i} = β′m_{i} + γm̃′_{i}. As before, these elements satisfy relations (15) and (16). Moreover we have

M = ⟨m̃_{2}, m̃_{3}⟩ ⊕ ⟨m̃_{1}, m̃_{4}⟩ ⊕ ⟨m̃′_{2}, m̃′_{3}⟩,

m̃′_{1} = m̃_{3} and m̃′_{4} = m̃_{2}. Therefore we can assume that the elements m_{i}, m′_{j} in (25) satisfy

M = ⟨m_{1}, m_{2}, m_{3}, m_{4}⟩ ⊕ ⟨m′_{2}, m′_{3}⟩ with m′_{1} = m_{3} and m′_{4} = m_{2}.

Then there is at most one irreducible unital right alternative bimodule of dimension 6. If we rename m′_{3} by m_{5} and m′_{2} by m_{6}, we have M = ⟨m_{1}, …, m_{6}⟩ where m_{i} satisfy (15), (16) and

\begin{matrix}m_{5}e_{11} = m_{5}, & m_{6}e_{11} = 0, & e_{11}m_{3} = m_{3}, \\ m_{5}e_{12} = 0, & m_{6}e_{12} = m_{5}, & e_{12}m_{3} = 0, \\ m_{5}e_{21} = m_{6}, & m_{6}e_{21} = 0, & e_{21}m_{3} = −m_{5}, \\ m_{5}e_{22} = 0, & m_{6}e_{22} = m_{6}, & e_{22}m_{3} = 0, \end{matrix}  (26)

\begin{matrix}e_{11}m_{5} = −m_{2}, & e_{11}m_{6} = 0, \\ e_{12}m_{5} = −m_{3} − m_{4}, & e_{12}m_{6} = −m_{2}, \\ e_{21}m_{5} = m_{6} & e_{21}m_{6} = 0, \\ e_{22}m_{5} = m_{2} + m_{5}, & e_{22}m_{6} = m_{6}, \end{matrix}  (27)

These relations in fact determine a unital right alternative bimodule. We can check this by a straightforward calculation or using a MAPLE program. Thus we have just proved

Theorem 2. Any irreducible unital right alternative bimodule of dimension 6 is isomorphic to the bimodule M = ⟨m_{1}, …, m_{6}⟩, where m_{i} satisfy relations (15), (16),

(26) and (27).

Remark 4. Let M a unital right alternative bimodule such that M^{J} has a subbimodule isomorphic to N ⊕ N. Let {m_{1}, …, m_{4}} and {m′_{1}, …, m′_{4}} be the elements corresponding to these components, which satisfy (15) and (16). If the set {m_{1}, …, m_{4}, m′_{1}, …, m′_{4}} is linearly dependent, then M has a sub-bimodule of dimension 6. In fact, the proof is analogous to the previous theorem.

5. Other examples of unital bimodules

In this section we construct a family of irreducible unital bimodules of dimension 4h, for each h ≥ 2 and also an example of a bimodule which is indecomposable but not irreducible.

5.1. Irreducible bimodules of dimension 4h. Let M = ⨁_{i=1}^{h}⟨m_{1}^{(i)}, m_{2}^{(i)}, m_{3}^{(i)}, m_{4}^{(i)}⟩, h ≥ 2 and define, for i = 1, …, h:

m_{1}^{(i)}e_{11} = m_{1}^{(i)},  m_{2}^{(i)}e_{11} = 0,  m_{3}^{(i)}e_{11} = m_{3}^{(i)},
m_{1}^{(i)}e_{12} = 0,  m_{2}^{(i)}e_{12} = m_{3}^{(i)},  m_{3}^{(i)}e_{12} = 0,
m_{1}^{(i)}e_{21} = m_{4}^{(i)},  m_{2}^{(i)}e_{21} = 0,  m_{3}^{(i)}e_{21} = m_{2}^{(i)},
m_{1}^{(i)}e_{22} = 0,  m_{2}^{(i)}e_{22} = m_{2}^{(i)},  m_{3}^{(i)}e_{22} = 0,
m_{4}^{(i)}e_{11} = 0,  e_{11}m_{1}^{(i)} = m_{1}^{(i)},  e_{11}m_{2}^{(i)} = 0,
m_{4}^{(i)}e_{12} = m_{1}^{(i)},  e_{12}m_{1}^{(i)} = 0,  e_{12}m_{2}^{(i)} = −m_{4}^{(i)},
m_{4}^{(i)}e_{21} = 0,  e_{21}m_{1}^{(i)} = −m_{3}^{(i)},  e_{21}m_{2}^{(i)} = 0,
m_{4}^{(i)}e_{22} = m_{4}^{(i)},  e_{22}m_{1}^{(i)} = 0,  e_{22}m_{2}^{(i)} = m_{2}^{(i)},

for i = 1, …, h − 1:

e_{11}m_{3}^{(i)} = −m_{1}^{(i+1)},  e_{11}m_{4}^{(i)} = m_{4}^{(i)} − m_{1}^{(i+1)},
e_{12}m_{3}^{(i)} = 0,  e_{12}m_{4}^{(i)} = m_{1}^{(i)},
e_{21}m_{3}^{(i)} = −m_{2}^{(i)} + m_{3}^{(i+1)} − m_{4}^{(i+1)},  e_{21}m_{4}^{(i)} = −2m_{2}^{(i)} + m_{3}^{(i+1)} − m_{4}^{(i+1)},
e_{22}m_{3}^{(i)} = m_{3}^{(i)} + m_{1}^{(i+1)},  e_{22}m_{4}^{(i)} = m_{1}^{(i+1)},

and

e_{11}m_{3}^{(h)} = m_{2}^{(1)},  e_{11}m_{4}^{(h)} = m_{2}^{(1)} + m_{4}^{(h)},
e_{12}m_{3}^{(h)} = −m_{3}^{(1)} + m_{4}^{(1)},  e_{12}m_{4}^{(h)} = −m_{3}^{(1)} + m_{4}^{(1)} + m_{1}^{(h)},
e_{21}m_{3}^{(h)} = −m_{2}^{(h)},  e_{21}m_{4}^{(h)} = −2m_{2}^{(h)},
e_{22}m_{3}^{(h)} = −m_{2}^{(1)} + m_{3}^{(h)},  e_{22}m_{4}^{(h)} = −m_{2}^{(1)}.

This bimodule is right alternative. In fact, for h = 2 and h = 3, one can check directly or using a MAPLE program. For h ≥ 4, we observe that the action of M_{2}(F) on m_{j}^{(i)}, M_{2}(F)m_{j}^{(i)} and m_{j}^{(i)}M_{2}(F), for i = 1, …, h − 2, are analogous to the action of M_{2}(F) on the elements m_{j}^{(1)}, M_{2}(F)m_{j}^{(1)} and m_{j}^{(1)}M_{2}(F) in the right alternative bimodule obtained for h = 3, for each j = 1, …, 4. Therefore identities (1) and (2) are valid for m_{j}^{(i)}, i = 1, …, h − 2, j = 1, …, 4. The same conclusion is true for m_{j}^{(h−1)} and m_{j}^{(h)} because they correspond, respectively, to the elements m_{j}^{(2)} and m_{j}^{(3)} in the right alternative bimodule for h = 3. Thus M is right alternative. Now let m be a nonzero element of M and let H be the sub-bimodule of M generated by m. To prove that M is irreducible it is sufficient to show that m_{1}^{(1)} ∈ H because m_{1}^{(1)} generates M. This is done in two steps. We prove:

1. If n = ∑_{j=1}^{4} ∑_{i=1}^{h} α_{j}^{(i)}m_{j}^{(i)} ∈ H and α_{j}^{(h)} ≠ 0 then m_{1}^{(1)} ∈ H,

2. If H ≠ 0 then there is n = ∑_{j=1}^{4} ∑_{i=1}^{h} α_{j}^{(i)}m_{j}^{(i)} ∈ H with α_{j}^{(h)} ≠ 0.

The first part is proved analysing the four possibilities: if α_{3}^{(h)} ≠ 0, we have

e_{12}(e_{12}(ne_{11})) = e_{12}(e_{12}(∑_{i=1}^{h} α_{1}^{(i)}m_{1}^{(i)} +α_{3}^{(i)}m_{3}^{(i)})) = e_{12}(α_{3}^{(h)}(−m_{3}^{(1)} +m_{4}^{(1)})) = α_{3}^{(h)}m_{1}^{(1)}.

Then m_{1}^{(1)} = (α_{3}^{(h)})^{−1}e_{12}(e_{12}(ne_{11})). If α_{2}^{(h)} ≠ 0 then n′ = ne_{12} = ∑ β_{j}^{(i)}m_{j}^{(i)} has β_{3}^{(h)} = α_{2}^{(h)} and we can use the previous case. If α_{1}^{(h)} ≠ 0 we consider n′ = e_{21}(ne_{21}) = e_{21}(∑_{i=1}^{h} α_{1}^{(i)}m_{4}^{(i)} +α_{3}^{(i)}m_{2}^{(i)}) = (∑_{i=1}^{h−1} α_{1}^{(i)}(−2m_{2}^{(i)} +m_{3}^{(i+1)} −m_{4}^{(i+1)}))−2α_{1}^{(h)}m_{2}^{(h)}. The coefficient of m_{2}^{(h)} in n′ is −2α_{1}^{(h)}. By the second case, we have m_{1}^{(1)} ∈ H. If α_{4}^{(h)} ≠ 0, we take n′ = e_{21}(ne_{22}) = e_{21}(∑_{i=1}^{h} α_{2}^{(i)}m_{2}^{(i)} + α_{4}^{(i)}m_{4}^{(i)}) = (∑_{i=1}^{h−1} α_{4}^{(i)}(−2m_{2}^{(i)} + m_{3}^{(i+1)} − m_{4}^{(i+1)})) − 2α_{4}^{(h)}m_{2}^{(h)}. As above, the coefficient of m_{2}^{(h)} in n′ is −2α_{4}^{(h)}. Again, by the second case, m_{1}^{(1)} ∈ H.

For the second part we use the following operators:

(i) g_{1} : M → M, defined by g_{1}(n) = e_{22}(ne_{21}); observe that g_{1}(n) has α_{1}^{(i−1)} as coefficient of m_{1}^{(i)}, for all i = 2, …, h,

(ii) g_{2}(n) = e_{12}(e_{21}(ne_{12})), which has −α_{2}^{(i−1)} as coefficient of m_{2}^{(i)}, for i = 2, …, h, (iii) g_{3}(n) = e_{12}(e_{21}(ne_{11})), which has −α_{3}^{(i−1)} as coefficient of m_{3}^{(i)}, for i = 2, …, h, (iv) g_{4}(n) = e_{11}(ne_{22}), which has −α_{4}^{(i−1)} as coefficient of m_{4}^{(i)}, i = 2, …, h.

So if n = ∑_{j=1}^{4} ∑_{i=1}^{h} α_{j}^{(i)}m_{j}^{(i)} ∈ H is a nonzero element and α_{j}^{(i)} ≠ 0 then g_{j}^{h−i}(n) satisfies the condition 2. Therefore M is irreducible.

5.2. An indecomposable bimodule which is not irreducible. Let m_{1}, m_{2}, m_{3} and m_{4} satisfy (15), (16) and (21)-(24), for α_{1} = α_{2} = 0 and α_{4} = 1. We construct a bimodule M of dimension 6 with these elements in the following way: M = ⟨m_{1}, m_{2}, m_{3}, m_{4}⟩ ⊕ F v_{1} ⊕ F v_{2}, where v_{1} = m_{2} + m_{5}, v_{2} = m_{6} satisfy (5), m_{5} ∈ M_{1}, and m_{6} = m_{5}e_{21}. Then we must have

\begin{matrix}e_{11}m_{3} = 0, & e_{11}m_{4} = m_{4}, & e_{11}m_{5} = m_{2}, & e_{11}m_{6} = m_{6}, \\ e_{12}m_{3} = −2m_{1}, & e_{12}m_{4} = −m_{1}, & e_{12}m_{5} = m_{4} − m_{3}, & e_{12}m_{6} = −m_{5}, \\ e_{21}m_{3} = m_{2}, & e_{21}m_{4} = 0, & e_{21}m_{5} = −m_{6}, & e_{21}m_{6} = 0, \\ e_{22}m_{3} = m_{3}, & e_{22}m_{4} = 0, & e_{22}m_{5} = m_{5} − m_{2}, & e_{22}m_{6} = 0, \end{matrix}

These relations, (15) and (16), in fact, determine a unital right alternative bimodule. Suppose that M has a sub-bimodule of dimension 2. Then there would exist n_{1}, n_{2} ∈ M that satisfy (20). If we write n_{1} = λ_{1}m_{1} + λ_{3}m_{3} + λ_{5}m_{5}, we have 0 = −n_{1}e_{12} = e_{12}n_{1} = −2λ_{3}m_{1} + λ_{5}(m_{4} − m_{3}). Therefore, λ_{3} = λ_{5} = 0, and so, n_{1} = λ_{1}m_{1}, which is a contradiction, since m_{1} generates a sub-bimodule of dimension 4. Thus we have an example of a unital right alternative bimodule of dimension 6 which is indecomposable but not irreducible.

Observe that every bimodule of dimension less that 6 is completely reducible. Therefore, the above example is of minimal dimension with that property.

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