EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 4, 2023, 2145-2155
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Derivations and representations of commutative
algebras verifying a polynomial identity of degree five

Hamed Ouédraogo1, Abdoulaye Dembega1, André Conseibo1,∗

1 Départment de Mathematiques/Université Norbert Zongo, Koudougou, Burkina Faso

Abstract. In this paper we study a class of commutative non associative algebras satisfying a
polynomial identity of degree five. We show that under the assumption of the existence of a non-zero
idempotent, any commutative algebra verifying such an identity admits a Peirce decomposition.
Using this decomposition we proceeded to the study of the derivations and representations of
algebras of this class.

2020 Mathematics Subject Classifications: 17A30, 17A36

Key Words and Phrases: Generalized Almost-Jordan algebra, Identity of degree five, Peirce
decomposition, Idempotent, Derivation, Representation

1. Introduction

Albert[1] was one of the authors who made a major contribution to the study of Jordan
algebras in the algebra theory of population genetics. Following him, several authors
[4] studied algebras verifying more general polynomial identities, such as almost Jordan
algebras. The irreducible identities of degree five, not a consequence of commutativity,
have all been determined by Osborn in [8], in characteristics other than 2, 3, and 5. In[7],
he proceeds to study his first identity

2((x2x)x)x− 3(x2x2)x+ (x2x)x2 = 0. (1)

More recently, in [2], A. Dembega study the second identity under certain conditions

β1[yx
4 − 4(yx3)x+ 6((yx2)x)x− 3(((yx)x)x)x]+

β2[−y(x2.x2) + 5(yx3)x− 9((yx2)x)x+ 4(((yx)x)x)x+ ((yx)x2)x+ (2)

(yx2)x2 − (yx)x3] + β3[((yx
2)x)x− (((yx)x)x)x− (yx2)x2 + ((yx)x)x2] = 0

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4924

Email addresses: ouedraogohamed557@gmail.com (H. Ouédraogo), doulaydem@yahoo.fr (A. Dembega),
andreconsebo@yahoo.fr (A. Conseibo)

https://www.ejpam.com 2145 © 2023 EJPAM All rights reserved.



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2146

where β1, β2 and β3 ∈ K, where K is a commutative field.
The aim of the this paper is to determine the representations and the derivations of
commutative algebras A defined by

yx4 − 4(yx3)x+ 6((yx2)x)x− 3(((yx)x)x)x = 0, (3)

corresponding to β1 = 1, β2 = β3 = 0 in the identity (2). Such algebras can be illustrated
by the following examples given in [2]. In the rest of the paper, K refers to the field C of
complex numbers.

Example 1.1. Let A be the commutative five-dimensional algebra whose the multipli-
cation table in the basis {e1, e2, e3, e4, e5} is e21 = e2, e1e2 = e3, e22 = e4, e1e3 = e5,
the other products not mentioned being zero. This algebra verifies (3). Indeed, A2 =<
e2, e3, e4, e5 >, A3 =< e3, e4, e5 >, A4 =< e5 > and A5 = 0.

Example 1.2. Let A be the commutative 5-dimensional algebra whose the multiplication
table in the basis {e, x0, y1, zλ1 , wλ2} is e2 = e, ey1 = y1, ezλ1 = λ1zλ1 , ewλ2 = λ2wλ2 ,

x0zλ1 = zλ1 , y1wλ2 = wλ2 with λ1 = 3+i
√
3

6 and λ2 = 3−i
√
3

6 , all other products not
mentioned being zero.

2. Peirce decomposition

In the rest of the paper we study commutative algebras verifying the identity (3) and
admitting a non-zero idempotent e. By setting x = e in (3), we obtain

[Le(Le − idA)(3L
2
e − 3Le + I)](y) = 0, (4)

where y ∈ A, idA : A → A, x 7→ x and Le : A → A, y 7→ ey is the multiplication by e.
Consider p(t) = −3t4 + 6t3 − 4t2 + t which factorises to p(t) = −t(t − 1)(3t2 − 3t + 1).
Since K is algebraically closed, then the polynomial 3t2 − 3t + 1 has two roots λ1 and
λ2, i.e p(t) = −t(t − 1)(t − λ1)(t − λ2), in which case A = A0 ⊕ A1 ⊕ Aλ1 ⊕ Aλ2 , where

Aµ = {x ∈ A|ex = µx}, λ1 =
3+i

√
3

6 and λ2 =
3−i

√
3

6 .

The products of the components of the Peirce decomposition are given by the following
theorem.

Theorem 2.1. ([2],Théorème 5.1.1) Let A be an algebra satisfying the identity (3). Sup-
pose that the polynomial 3t2 − 3t + 1 = 3(t − λ1)(t − λ2) in K[t]. Then the Peirce de-

composition of A is A = A0 ⊕ A1 ⊕ Aλ1 ⊕ Aλ2 , where λ1 = 3+i
√
3

6 and λ2 = 3−i
√
3

6 . We
have:

i) AiAλj
⊆ Aλj

, with i ∈ {0, 1} and j ∈ {1; 2};

ii) Aλi
Aλj

= {0}, with i, j ∈ {1; 2};



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2147

iii) AiAi ⊆ Ai with i ∈ {0; 1};

iv) A0A1 = {0}.

The following theorem gives us the identities used in the rest of the document.

Theorem 2.2. ([2], Théorème 5.1.2) Let A be an algebra satisfying the identity (3),
whose Peirce decomposition of A is written A = A0 ⊕A1 ⊕Aλ1 ⊕Aλ2 , then

i) x0(y0xλk
) + y0(x0xλk

) = 2λkxλk
(x0y0);

ii) x1(y1xλk
) + y1(x1xλk

) = (1− 3(λk)
3)xλk

(x1y1);

iii) x1(xλk
x0) = x0(x1xλk

),

where λk is the conjugate of λk, and xi, yi ∈ Ai, xλk
∈ Aλk

for i ∈ {0; 1} and k ∈ {1; 2}.

3. Derivations

Let A be aK-algebra satisfying (3). We say that an endomorphism d of A is a derivation
of A if for all x, y ∈ A, d(xy) = d(x)y+xd(y). The set DerK(A) of all derivations of A with
the Lie bracket [, ], is a Lie subalgebra of EndK(A)−, where EndK(A) is the associative
algebra of endomorphisms of A. The Lie bracket is defined by: [d, d′] = dd′− d′d ([5], [6]).
The following theorems characterise the derivations of an algebra satisfying (3).

Theorem 3.1. Let A = A0 ⊕ A1 ⊕ Aλ1 ⊕ Aλ2 be the Peirce decomposition of an alge-
bra satisfying (3). If d is a derivation of A, then d satisfies the following properties, for
i ∈ {0, 1, λ1, λ2}; α ∈ {λ1, λ2} ; β ∈ {0, 1}:

i) d(e) = 0;

ii) d(xi) ∈ Ai;

iii) fi = d/Ai
;

iv) fα(xβyα) = xβfα(yα) + yαfβ(xβ).

Proof. i) Since e is an idempotent of A then 2ed(e) = d(e). By setting d(e) = x0+x1+
xλ1 +xλ2 , we have −x0+x1+(2λ1−1)xλ1 +(2λ2−1)xλ2 = 0 i.e x0 = x1 = xλ1 = xλ2 = 0;
hence d(e) = 0.
ii) and iii) Let xi ∈ Ai for i ∈ {0, 1, λ1, λ2}; then exi = ixi i.e ed(xi) = id(xi) because
d(e) = 0. So we have d(xi) ∈ Ai, and fi = d/Ai

.

iv) We have xβyα ∈ Aα for α ∈ {λ1, λ2} and β ∈ {0, 1} according to the products of
the Peirce subspaces of A, since d(xβyα) = xβd(yα)+ yαd(xβ) then, applying iii), we have
fα(xβyα) = xβfβ(yα) + yαfα(xβ).



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2148

Theorem 3.2. Let A = A0 ⊕ A1 ⊕ Aλ1 ⊕ Aλ2 be the Peirce decomposition of an algebra
satisfying (3). Let d and d′ be derivations of A, then the derivation [d, d′] satisfies for
i ∈ {0; 1;λ1;λ2}; α ∈ {λ1;λ2} and β ∈ {0; 1} the following conditions:

i) [d, d′](e) = 0;

ii) [d, d′](xi) ∈ Ai;

iii) f[d,d′](xi) = [d, d′]/Ai
= fd(fd′(xi))− fd′(fd(xi)).

Proof. We have d(e) = 0 by the theorem(3.1), so [d, d′](e) = d(d′(e)) − d′(d(e)) = 0.
Hence i).

Also [d, d′](xi) = fd(fd′(xi))− fd′(fd(xi)), hence ii) and iii).

4. Representations

We are interested here in the representations of algebras defined by the identity (3)
([4]). Following Eilenberg [3], if A is a commutative K-algebra belonging to a class C
and M a vector space on K, a linear application ρ: A −→ End(M) is a representation
of A in the class C if the extension S = A ⊕ M of M with multiplication given by
(a+m)(b+ n) = ab+ ρ(a)(n) + ρ(b)(m) for all a, b ∈ A,m, n ∈ M ; belong to the class C.
In the following section, we give some results on the representations of algebras verifying
the identity (3).

4.1. General results

Lemma 4.1. Let A be an algebra verifying the identity (3) and ρ: A −→ End(M) a
linear application. The application ρ is a representation of A if and only if for all a, b ∈ A
the following identities are verified:

ρa4 − 4ρaρa3 + 6ρ2aρa2 − 3ρ4a = 0; (5)

ρbρaρa2+2ρbρ
3
a+ρbρa3−4ρaρbρa2−8ρaρbρ

2
a−4ρa3b+12ρ2aρbρa+6ρaρa2b+6ρa(a2b)−3ρ3aρb

− 3ρ2aρab − 3ρaρa(ab) − 3ρa(a(ab)) = 0 (6)

where ρa := ρ(a) ∈ End(M) and for any a ∈ A.

Proof. The application ρ is a representation of A if and only if for all x = a+m, and
y = b+ n ∈ A⊕M , the equality (3) is satisfied.

Since:
yx4 = a4b+ ρbρaρa2(m) + 2ρbρ

3
a(m) + ρa4(n) + ρbρa3(m);

x(yx3) = a(a3b) + ρaρbρa2(m) + 2ρaρbρ
2
a(m) + ρaρa3(n) + ρa3b(m);

x(x(yx2)) = a(a(a2b)) + 2ρ2aρbρa(m) + ρ2aρa2(n) + ρaρa2b(m) + ρa(a2b)(m);



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2149

x(x(x(yx))) = a(a(a(ab))) + ρ4a(n) + ρ3aρb(m) + ρ2aρab(m) + ρaρa(ab)(m) + ρa(a(ab))(m);
Equality (3) becomes:

[a4b−4a(a3b)+6a(a(a2b))−3a(a(a(ab)))]+[ρbρaρa2+2ρbρ
3
a+ρbρa3−4ρaρbρa2−8ρaρbρ

2
a−4ρa3b

+12ρ2aρbρa+6ρaρa2b+6ρa(a2b)−3ρ3aρb−3ρ2aρab−3ρaρa(ab)−3ρa(a(ab))](m)+ [ρa4 −4ρaρa3

+ 6ρ2aρa2 − 3ρ4a](n) = 0. (7)

Since a and b are elements of the algebra A, then:

a4b− 4a(a3b) + 6a(a(a2b))− 3a(a(a(ab))) = 0.

We then obtain,

[ρbρaρa2+2ρbρ
3
a+ρbρa3−4ρaρbρa2−8ρaρbρ

2
a−4ρa3b+12ρ2aρbρa+6ρaρa2b+6ρa(a2b)−3ρ3aρb

− 3ρ2aρab − 3ρaρa(ab) − 3ρa(a(ab))](m) + [ρa4 − 4ρaρa3 + 6ρ2aρa2 − 3ρ4a](n) = 0. (8)

Then finally:
ρa4 − 4ρaρa3 + 6ρ2aρa2 − 3ρ4a = 0;

ρbρaρa2+2ρbρ
3
a+ρbρa3−4ρaρbρa2−8ρaρbρ

2
a−4ρa3b+12ρ2aρbρa+6ρaρa2b+6ρa(a2b)−3ρ3aρb

− 3ρ2aρab − 3ρaρa(ab) − 3ρa(a(ab)) = 0.

Proposition 4.2. Let A be an algebra verifying the identity (3). Suppose that A has an
idempotent e ̸= 0. Let ρ : A −→ End(M) be a representation of A. Then:

M = M0 ⊕M1 ⊕Mλ1 ⊕Mλ2 ,

with Mi = {m ∈ M,ρe(m) = im} for i ∈ {0, 1, λ1, λ2} where λ1 = 3+i
√
3

6 and λ2 = 3−i
√
3

6
are roots of the polynomial 3t2 − 3t+ 1.

Proof. Putting a = e into the identity (5), we obtain −ρe(ρe − I)(3ρ2e − 3ρe + I) = 0.
The kernel lemma gives us: M = M0⊕M1⊕Mλ1 ⊕Mλ2 , with Mi = {m ∈ M |ρe(m) =

im} for i ∈ {0, 1, λ1, λ2}.

Let’s study the action of A on M . In this section, we’ll focus on the products of Ai.Mj ,
for i; j ∈ {0, 1, λ1, λ2}.

Theorem 4.3. Let A be an algebra verifying the identity (3) admitting an idempotent
e ̸= 0. Let ρ : A −→ End(M) be a representation of A. Then the action of A on M
satisfies the following relations:

i) Ai.Mλj
⊆ Mλj

, where i ∈ {0; 1} and j ∈ {1; 2};

ii) Aλj
.Mi ⊆ Mλj

, where i ∈ {0, 1} and j ∈ {1, 2};



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2150

iii) Aλi
.Mλj

= {0}, with i, j ∈ {1, 2};

iv) Ai.Mi ⊆ Mi with i ∈ {0, 1};

v) Ai.Mj = {0} with i, j ∈ {0, 1}, i ̸= j.

Let A = A0 ⊕ A1 ⊕ Aλ1 ⊕ Aλ2 be the Peirce decomposition of an algebra satisfying
(3) relative to an idempotent e and that of the modulus M = M0 ⊕ M1 ⊕ Mλ1 ⊕ Mλ2

then the Peirce decomposition of the extension S is S = S0 ⊕ S1 ⊕ Sλ1 ⊕ Sλ2 where
Sk = {a + m ∈ S : e(a + m) = k(a + m)} and k = {0, 1, λ1, λ2} such that the following
result is verified.

Proposition 4.4. The subspaces Si of the extension S satisfy the relations:

i) Si.Sλj
⊆ Sλj

, and Sλj
.Si ⊆ Sλj

with i ∈ {0, 1} and j ∈ {1, 2};

ii) Sλi
.Sλj

= {0}, with i, j ∈ {1, 2};

iii) Si.Si ⊆ Si with i ∈ {0, 1};

iv) Si.Sj = {0} with i, j ∈ {0, 1}, i ̸= j.

Proof. This follows from the fact that the algebra S must belong to the same class as A,
i.e it must verify the identity (3). Consequently, its subspaces verify the same properties
as those of A.

4.2. Irreducible representations

As in the case of groups, rings or vector spaces, a submodule is a non-empty part of a
module, stable for the laws of modules, hence the following definition:

Definition 4.5. Let A be a K-algebra and ρ : A −→ End(M) a representation of A.

i) Let N be a subspace of M , N is a submodule of M if and only if A.N ⊆ N .

ii) M is an irreducible module or ρ is an irreducible representation of A, if M ̸= 0 and
the only submodules of M are its trivial submodules.

iii) An A-module M is said to be simple or irreducible if M is not the null module and
there are no submodules outside {0} and M .

Lemma 4.6. Let A be an algebra satisfying the identity (3), then Mλ1 and Mλ2 are
submodules of M .

Proof. Using Theorem (4.3), we observe that for all a ∈ A and m ∈ Mλ, we have
a.m ∈ Mλ for any λ ∈ {λ1, λ2}.



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2151

Proposition 4.7. Let A be an algebra satisfying the identity (3) admitting an idempotent
e ̸= 0 and ρ : A −→ End(M) an irreducible representation of A. Then one of the
conditions below holds:

i) M = M0 or M = M1;

ii) M = Mλ1;

iii) M = Mλ2.

Proof. Using the Lemma 4.6, we have Mλ1 and Mλ2 are submodules of M . Since M
is irreducible, then: Mλ1 = M or Mλ1 = {0}. In other words, Mλ2 = M or Mλ2 = {0}.

Theorem 4.8. Let A be an algebra verifying the identity (3). Suppose that A has an
idempotent e ̸= 0 and M an irreducible module. If M = Mi, with i ∈ {0; 1}, then-
∀a, b ∈ A;m ∈ M

1) (a, b,m) = (ai, bi,mi);

2) (a,m, b) = (ai,mi, bi).

Proof. Suppose M = Mi, i ∈ {0; 1} then Mλ1 = Mλ2 = {0}. We’ll do the proof for
M = M0; for the case where M = M1, the proof is done simply. Let a, b ∈ A be such that
a = a0+a1+aλ1 +aλ2 and b = b0+b1+bλ1 +bλ2 , where al, bl ∈ Al, l ∈ {0; 1;λ1;λ2}. Take
m ∈ M = M0, so m = m0, where m0 ∈ M0, i.e ρe(m0) = 0. Let’s calculate (a, b,m0),
(a,m0, b) ∀a, b ∈ A and m ∈ M . We have:

(a, b,m) = (a0 + a1 + aλ1 + aλ2 , b0 + b1 + bλ1 + bλ2 ,m0)

= (a0, b0,m0) + (a0, b1,m0) + (a0, bλ1 ,m0) + (a0, bλ2 ,m0) + (a1, b0,m0) + (a1, b1,m0)

+ (a1, bλ1 ,m0) + (a1, bλ2 ,m0) + (aλ1 , b0,m0) + (aλ1 , b1,m0) + (aλ1 , bλ1 ,m0)

+ (aλ1 , bλ2 ,m0) + (aλ2 , b0,m0) + (aλ2 , b1,m0) + (aλ2 , bλ1 ,m0) + (aλ2 , bλ2 ,m0)

= (a0b0)m0 − a0(b0m0) + (a0b1)m0 − a0(b1m0) + (a0bλ1)m0 − a0(bλ1m0) + (a0bλ2)m0

− a0(bλ2m0) + (a1b0)m0 − a1(b0m0) + (a1b1)m0 − a1(b1m0) + (a1bλ1)m0 − a1(bλ1m0)

+ (a1bλ2)m0 − a1(bλ2m0) + (aλ1b0)m0 − aλ1(b0m0) + (aλ1b1)m0 − aλ1(b1m0)

+ (aλ1bλ1)m0 − aλ1(bλ1m0) + (aλ1bλ2)m0 − aλ1(bλ2m0) + (aλ1b0)m0 − aλ2(b0m0)

+ (aλ2b1)m0 − aλ2(b1m0) + (aλ2bλ1)m0 − aλ2(bλ1m0) + (aλ2bλ2)m0 − aλ2(bλ2m0)

= (a0, b0,m0)

This result is found by using the multiplication table of A (see theorem (2.1)) and the
action of A on M (see theorem (4.3)), for all m ∈ M , we have: (a, b,m) = (a0, b0,m0).
Similarly, if m ∈ M = M0, then we have:

(a,m, b) = (a0 + a1 + aλ1 + aλ2 ,m0, b0 + b1 + bλ1 + bλ2)

= (a0,m0, b0) + (a0,m0, b1) + (a0,m0, bλ1) + (a0,m0, bλ2) + (a1,m0, b0) + (a1,m0, b1)



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2152

+ (a1,m0, bλ1) + (a1,m0, bλ2) + (aλ1 ,m0, b0) + (aλ1 ,m0, b1) + (aλ1 ,m0, bλ1)

+ (aλ1 ,m0, bλ2) + (aλ2 ,m0, b0) + (aλ2 ,m0, b1) + (aλ2 ,m0, bλ1) + (aλ2 ,m0, bλ2)

= (a0m0)b0 − a0(m0b0) + (a0m0)b1 − a0(m0b1) + (a0m0)bλ1 − a0(m0bλ1) + (a0m0)bλ2

− a0(m0bλ2) + (a1m0)b0 − a1(m0b0) + (a1m0)b1 − a1(m0b1) + (a1m0)bλ1 − a1(m0bλ1)

+ (a1m0)bλ2 − a1(m0bλ2) + (aλ1m0)b0 − aλ1(m0b0) + (aλ1m0)b1 − aλ1(m0b1)

+ (aλ1m0)bλ1 − aλ1(m0bλ1) + (aλ1m0)bλ2 − aλ1(m0bλ2) + (aλ2m0)b0 − aλ2(m0b0)

+ (aλ2m0)b1 − aλ2(m0b1) + (aλ2m0)bλ1 − aλ2(m0bλ1) + (aλ2m0)bλ2 − aλ2(m0bλ2)

= (a0,m0, b0)

Theorem 4.9. Let A be an algebra verifying the identity (3). Suppose that A has an
idempotent e ̸= 0 and M an irreducible module. If M = Mλ, with λ ∈ {λ1, λ2} then

(i)(a, b,m) = 0 , ∀a, b ∈ A;m ∈ M (ii)ρbρc =
3λ2 − 1

9λ3
ρcb , ∀a, b ∈ A;m ∈ M (9)

Proof. Suppose M = Mλ, λ ∈ {λ1, λ2} then M0 = M1 = {0}. Let a, b ∈ A be such
that a = a0+a1+aλ1 +aλ2 and b = b0+ b1+ bλ1 + bλ2 , where ai, bi ∈ Ai, i ∈ {0; 1;λ1;λ2}.
Take m ∈ M = Mλ, so m = mλ, where mλ ∈ Mλ, i.e ρe(mλ) = λmλ. Let’s calculate
(a, b,mλ), (a,mλ, b) ∀a, b ∈ A and m ∈ M . We have:

(a, b,m) = (a0 + a1 + aλ1 + aλ2 , b0 + b1 + bλ1 + bλ2 ,mλ)

= (a0, b0,mλ) + (a0, b1,mλ) + (a0, bλ1 ,mλ) + (a0, bλ2 ,mλ) + (a1, b0,mλ) + (a1, b1,mλ)

+ (a1, bλ1 ,mλ) + (a1, bλ2 ,mλ) + (aλ1 , b0,mλ) + (aλ1 , b1,mλ) + (aλ1 , bλ1 ,mλ)

+ (aλ1 , bλ2 ,mλ) + (aλ2 , b0,mλ) + (aλ2 , b1,mλ) + (aλ2 , bλ1 ,mλ) + (aλ2 , bλ2 ,mλ)

= (a0b0)mλ − a0(b0mλ) + (a0b1)mλ − a0(b1mλ) + (a0bλ1)mλ − a0(bλ1mλ) + (a0bλ2)mλ

− a0(bλ2mλ) + (a1b0)mλ − a1(b0mλ) + (a1b1)mλ − a1(b1mλ) + (a1bλ1)mλ − a1(bλ1mλ)

+ (a1bλ2)mλ − a1(bλ2mλ) + (aλ1b0)mλ − aλ1(b0mλ) + (aλ1b1)mλ − aλ1(b1mλ)

+ (aλ1bλ1)mλ − aλ1(bλ1mλ) + (aλ1bλ2)mλ − aλ1(bλ2mλ) + (aλ2b0)mλ − aλ2(b0mλ)

+ (aλ2b1)mλ − aλ2(b1mλ) + (aλ2bλ1)mλ − aλ2(bλ1mλ) + (aλ2bλ2)mλ − aλ2(bλ2mλ)

= (a0b0)mλ − a0(b0mλ)− a0(b1mλ)− a1(b0mλ) + (a1b1)mλ − a1(b1mλ)

= (a0, b0,mλ) + (a1, b1,mλ)− a0(b1mλ)− a1(b0mλ)

Similarly, if ∀m ∈ M = Mλ, we have:

(a,m, b) = (a0 + a1 + aλ1 + aλ2 ,mλ, b0 + b1 + bλ1 + bλ2)

= (a0,mλ, b0) + (a0,mλ, b1) + (a0,mλ, bλ1) + (a0,mλ, bλ2) + (a1,mλ, b0) + (a1,mλ, b1)



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2153

+ (a1,mλ, bλ1) + (a1,mλ, bλ2) + (aλ1 ,mλ, b0) + (aλ1 ,mλ, b1) + (aλ1 ,mλ, bλ1)

+ (aλ1 ,mλ, bλ2) + (aλ2 ,mλ, b0) + (aλ2 ,mλ, b1) + (aλ2 ,mλ, bλ1) + (aλ2 ,mλ, bλ2)

= (a0mλ)b0 − a0(mλb0) + (a0mλ)b1 − a0(mλb1) + (a0mλ)bλ1 − a0(mλbλ1) + (a0mλ)bλ2

− a0(mλbλ2) + (a1mλ)b0 − a1(mλb0) + (a1mλ)b1 − a1(mλb1) + (a1mλ)bλ1 − a1(mλbλ1)

+ (a1mλ)bλ2 − a1(mλbλ2) + (aλ1mλ)b0 − aλ1(mλb0) + (aλ1mλ)b1 − aλ1(mλb1)

+ (aλ1mλ)bλ1 − aλ1(mλbλ1) + (aλ1mλ)bλ2 − aλ1(mλbλ2) + (aλ2mλ)b0 − aλ2(mλb0)

+ (aλ2mλ)b1 − aλ2(mλb1) + (aλ2mλ)bλ1 − aλ2(mλbλ1) + (aλ2mλ)bλ2 − aλ2(mλbλ2)

= (a0mλ)b0 − a0(mλb0) + (a0mλ)b1 − a0(mλb1) + (a1mλ)b0 − a1(mλb0) + (a1mλ)b1

− a1(mλb1)

= (a0,mλ, b0) + (a0,mλ, b1) + (a1,mλ, b0) + (a1,mλ, b1)

Partial linearisation of the terms of the identity (6) gives us:

ρbρaρa2 = ρbρcρa2 + 2ρbρaρac;

2ρbρ
3
a = 2(ρbρcρ

2
a + ρbρaρcρa + ρbρ

2
aρc);

ρbρa3 = ρbρca2 + 2ρbρa(ac);

4ρaρbρa2 = 4(ρcρbρa2 + 2ρaρbρac);

8ρaρbρ
2
a = 8(ρcρbρ

2
a + ρaρbρcρa + ρaρbρaρc);

4ρa3b = 4(ρ(ca2)b + 2ρ(a(ac))b);

12ρ2aρbρa = 12(ρaρcρbρa + ρcρaρbρa + ρ2aρbρc);

6ρaρa2b = 6(ρcρa2b + 2ρaρ(ca)b);

6ρa(a2b) = 6(ρc(a2b) + 2ρa((ac)b));

3ρ3aρb = 3(ρcρ
2
aρb + ρaρcρaρb + ρ2aρcρb);

3ρ2aρab = 3(ρcρaρab + ρaρcρab + ρ2aρcb);

3ρaρa(ab) = 3(ρcρa(ab) + ρaρc(ab) + ρaρa(cb));

3ρa(a(ab)) = 3(ρc(a(ab)) + ρa(c(ab)) + ρa(a(cb))).

Putting it all together gives:

[ρbρcρa2 + 2ρbρaρac + 2ρbρcρ
2
a + 2ρbρaρcρa + 2ρbρ

2
aρc + ρbρca2 + 2ρbρa(ac) − 4ρcρbρa2

−8ρaρbρac−8ρcρbρ
2
a−8ρaρbρcρa−8ρaρbρaρc−4ρ(ca2)b−8ρ(a(ac))b+12ρaρcρbρa+12ρcρaρbρa+12ρ2aρbρc

+6ρcρa2b+12ρaρ(ca)b+6ρc(a2b)+12ρa((ac)b)−3ρcρ
2
aρb−3ρaρcρaρb−3ρ2aρcρb−3ρcρaρab−3ρaρcρab−3ρ2aρcb

− 3ρcρa(ab) − 3ρaρc(ab) − 3ρaρa(cb) − 3ρc(a(ab)) − 3ρa(c(ab)) − 3ρa(a(cb))](m) = 0. (10)

If we put a = e in (10) then

[ρbρcρe2 + 2ρbρeρec + 2ρbρcρ
2
e + 2ρbρeρcρe + 2ρbρ

2
eρc + ρbρce2 + 2ρbρe(ec) − 4ρcρbρe2



H. Ouédraogo, A. Dembega, A. Conseibo / Eur. J. Pure Appl. Math, 16 (4) (2023), 2145-2155 2154

−8ρeρbρec−8ρcρbρ
2
e−8ρeρbρcρe−8ρeρbρeρc−4ρ(ce2)b−8ρ(e(ec))b+12ρeρcρbρe+12ρcρeρbρe+12ρ2eρbρc

+6ρcρe2b+12ρeρ(ce)b+6ρc(e2b)+12ρe((ec)b)−3ρcρ
2
eρb−3ρeρcρeρb−3ρ2eρcρb−3ρcρeρeb−3ρeρcρeb−3ρ2eρcb

− 3ρcρe(eb) − 3ρeρc(eb) − 3ρeρe(cb) − 3ρc(e(eb)) − 3ρe(c(eb)) − 3ρe(e(cb))](m) = 0. (11)

We will demonstrate (i) and (ii) by taking a, b ∈ A1 and m ∈ Mλ, with λ ∈ {λ1, λ2}. If
M = Mλi

, i = 1, 2, we have ρe = λiidM . By replacing b, c ∈ A1 and m ∈ M in (11) we
obtain:

[λρbρc+2λρbρc+2λ2ρbρc+2λ2ρbρc+2λ2ρbρc+ρbρc+2ρbρc−4λρcρb−8λρbρc−8λ2ρcρb−8λ2ρbρc

− 8λ2ρbρc − 4ρcb − 8ρcb + 12λ2ρcρb + 12λ2ρcρb + 12λ2ρbρc + 6ρcρb + 12λρcb + 6ρcb + 12ρcb

−3λ2ρcρb−3λ2ρcρb−3λ2ρcρb−3λρcρb−3λρcρb−3λ2ρcb−3ρcρb−3λρcb−3λρcb−3ρcb−3ρcb−3ρcb](m) = 0.

After reduction we have:

(2λ2 − 5λ+ 3)ρbρc + (7λ2 − 10λ+ 3)ρcρb + (−3λ2 + 6λ− 3)ρcb = 0. (12)

Since 3λ2 = 3λ− 1, then

1

3
(−9λ+ 7)ρbρc +

1

3
(−9λ+ 2)ρcρb +

1

3
(9λ− 6)ρcb = 0.

(−9λ+ 7)ρbρc + (−9λ+ 2)ρcρb + (9λ− 6)ρcb = 0 (13)

By interchanging b and c in (13) we obtain:

(−9λ+ 7)ρcρb + (−9λ+ 2)ρbρc + (9λ− 6)ρbc = 0 (14)

The difference between (13) and (14) gives:

5ρbρc − 5ρcρb = 0 ⇔ ρbρc = ρcρb (15)

So (14) becomes

−3(2λ− 1)ρcρb + (3λ− 2)ρbc = 0 ⇔ −9λ3ρbρc + (3λ2 − 1)ρcb = 0.

hence

ρbρc =
3λ2 − 1

9λ3
ρcb

Remark 4.10. Taking a, b ∈ A0 and m ∈ Mλ, then replacing them in (11), we have:

(9λ− 2)ρbρc + (9λ− 7)ρcρb − 9λ2ρcb = 0 (16)

Here again, by interchanging b and c and then subtracting, we obtain ρbρc = ρcρb ⇒
(a, b,m) = 0. On the other hand, here we have: ρbρc =

1
3λρcb.



REFERENCES 2155

Acknowledgements

The authors would like to thank the referees whose suggestions helped to improve this
paper.

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