EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 3, 2023, 1480-1490
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Algebras satisfying a polynomial identity of degree six
that are principal train

Daouda Kabre1, André Conseibo1,∗

1 Départment de Mathématiques, Université Norbert ZONGO, Koudougou, Burkina Faso

Abstract. In this paper we study the class of algebras satisfying a polynomial identity of degree
six that are principal train algebras of rank 3 or 4, for which we give the explicit form of the train
equation. If the rank of A is n ≥ 5 in general, we provide the form of the train equation in some
cases.
2020 Mathematics Subject Classifications: 17D92, 17A05
Key Words and Phrases: Peirce decomposition, principal train algebra, polynomial identity,
idempotent

1. Introduction

In 1923, Serge Bernstein gave a mathematical proof of the principle of stationarity of
Hardy-Weinberg ([3]). From 1939 onwards, Etherington introduced the notion of weighted
algebra and principal train algebra for an algebraic model of genetics. However, it was
not until 1975 ([5])that Philip Holgate defined algebraically the so-called Bernstein ([6]).
Following him, several authors studied various classes of algebras satisfying polynomial
identities, in order to model the process of genetic transmission. (see, for instances, [9],[1],
[2]). The aim of this paper is to study the algebras verifying the polynomial identity
2x2x4 = ω(x)2x4 + ω(x)4x2 that are principal train algebras. In ([8], the authors prove
that such an algebra,assuming the existence of nonzero idempotent, admits a Peirce de-
composition. The use of the Peirce decomposition will allow us to finally establish links
between this class of algebras and principal train algebras.

2. Preliminaries

Let K be a commutative field and A a commutative K-algebra, not necessarily as-
sociative. For any element x of A we define the principal powers of x by x1 = x and
xk+1 = xxk for any integer k ≥ 1. An idempotent is any element e of A such that e2 = e.
In this paper the idempotents considered are all non-zero.

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i3.4770

Email addresses: daoudakabre@yahoo.fr (D. Kabre), andreconsebo@yahoo.fr (A. Conseibo)

https://www.ejpam.com 1480 © 2023 EJPAM All rights reserved.



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1481

Definition 1. We will say that the algebra A is a baric if there exists a non-zero morphism
of algebras ω : A → K. The morphism ω is then called the weight function of the algebra
A. The weight of an element x of A is the scalar ω(x).

Definition 2. A baric K-algebra (A,ω) is a principal train algebra of rank n ≥ 2 if there
are scalars γ1, . . . , γn−1 ∈ K such that xn + γ1ω(x)x

n−1 + · · ·+ γn−1ω(x)
n−1x = 0, where

the integer n ≥ 2 is the smallest having this property.

Definition 3. A baric K-algebra (A,ω) is a Bernstein algebra if (x2)2 = ω(x)2x2 for any
x in A.

In the rest of the document, K denotes an algebraically closed infinite commutative
field with characteristic different from 2.

In [11], it is shown that if A denotes a Bernstein algebra, then for any x in A, 2xixj =
ω(x)ixj + ω(x)jxi, ∀i, j ≥ 2; in particular, for i = 2 and j = 4, 2x2x4 = ω(x)2x4 +
ω(x)4x2,∀x ∈ A. In this paper, our attention will be focused on the structure of algebras
satisfying the latter polynomial identity and that are principal train. Let us consider the
identity

2x2x4 = ω(x)2x4 + ω(x)4x2 (1)

In the rest of the paper, K = C, i.e the field of complex numbers. In [8], the authors
obtained the following two theorems. .

Theorem 1. [8] Let (A,ω) be a K-algebra verifying (1) and e be a non-zero idempotent
of A. Then A admits a Peirce decomposition relative to e: A = Ke⊕A0 ⊕A 1

2
⊕Aλ ⊕Aλ̄

where Aα = {x ∈ Kerω, ex = αx}, with α ∈ {0; 12 ;λ = −1−i
√
23

4 ; λ̄ = −1+i
√
23

4 }.

Theorem 2. [8] Let A = Ke ⊕ A0 ⊕ A 1
2
⊕ Aλ ⊕ Aλ̄ be the Peirce decomposition of an

algebra verifying (1), then:

i) A0A0 ⊂ A 1
2
;

ii) A 1
2
A 1

2
⊂ A0 ⊕Aλ ⊕Aλ̄;

iii) AλAλ̄ = 0;

iV) AλAλ = 0;

V̇ ) Aλ̄Aλ̄ = 0:

Vi) A0A 1
2
⊂ A 1

2
⊕Aλ ⊕Aλ̄;

Vii) AλA 1
2
⊂ A 1

2
⊕A0 ⊕Aλ̄;

Viii) Aλ̄A 1
2
⊂ A 1

2
⊕A0 ⊕Aλ;

iX) A0Aλ ⊂ A 1
2
;



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1482

X) A0Aλ̄ ⊂ A 1
2
.

Let (A,ω) be a baric commutative K-algebra not necessarily associative verifying the
identity (1). The partial linearisation of this identity gives the following result:

Proposition 1. Let (A,ω) be a K-algebra verifying (1). For all x, y, z, t in A we have:
4x2[z(t(xy)) + z(x(ty)) + t(z(xy)) + x(z(ty)) + t(x(yz)) + x(t(yz)) + z(y(tx)) + t(y(xz)) +
x(y(tz))+y(z(tx))+y(t(xz))+y(x(tz))]+4xt[2z(x(xy))+2x(z(xy))+2x(x(yz))+2x(y(xz))+
2y(x(xz)) + z(x2y) + y(x2z)] + 4ty[zx3 + x(zx2) + 2x(x(xz))] + 4xy[z(tx2) + 2z(x(xt)) +
t(zx2) + 2x(z(xt)) + 2t(x(xz)) + 2x(t(xz)) + 2x(x(tz))] + 4tz[2x(x(xy)) + x(yx2) + yx3] +
4yz[tx3+x(tx2)+2x(x(xt))]+4xz[2t(x(xy))+2x(t(xy))+2x(x(ty))+t(yx2)+2x(y(xt))+
y(tx2) + 2y(x(xt))] = 2ω(tz)[2x(x(xy) + x(x2y) + x3y] + 2ω(xz)[2t(x(xy) + 2x(t(xy) +
2x(x(ty)+t(x2y)+2x(y(xt)+y(tx2)+2y(x(tx)]+2ω(xt)[2z(x(xy)+2x(z(xy)+2x(x(zy)+
z(x2y)+2x(y(xz)+y(zx2)+2y(x(zx)]+2ω(x2)[z(t(xy))+z(x(ty))+ t(z(xy))+x(z(ty))+
t(x(yz)) + x(t(yz)) + z(y(xt)) + t(y(xz)) + x(y(tz)) + y(z(xt)) + y(t(xz)) + y(x(tz))] +
24[ω(xyzt)x2+ω(yzx2)xt+ω(ytx2)xz+ω(ztx2)xy]+8[ω(x3y)zt+ω(x3z)yt+ω(x3t)zy]+
2ω(yz)[tx3 + x(tx2) + 2x(x(xt))] + 2ω(ty)[zx3 + x(zx2) + 2x(x(xz))] + 2ω(xy)[z(tx2) +
t(zx2) + 2z(x(xt)) + 2x(z(xt)) + 2t(x(xz)) + 2x(t(xz)) + 2x(x(zt))]

The previous proposition allows us to establish the following lemma.

Lemma 1. For all x0, y0, z0 ∈ A0; x 1
2
, y 1

2
, z 1

2
∈ A 1

2
; xλ, yλ, zλ ∈ Aλ; xλ̄, yλ̄, zλ̄ ∈ Aλ̄ the

following identities are verified:

1) [z0(x0y0) + x0(y0z0) + y0(x0z0)]λ = [z0(x0y0) + x0(y0z0) + y0(x0z0)]λ̄ = 0;

2) [4(x 1
2
(y 1

2
z 1

2
)0+y 1

2
(x 1

2
z 1

2
)0+z 1

2
(y 1

2
x 1

2
)0)+(λ+1)(x 1

2
(y 1

2
z 1

2
)λ+y 1

2
(x 1

2
z 1

2
)λ+z 1

2
(y 1

2
x 1

2
)λ)+

(λ̄+ 1)(x 1
2
(y 1

2
z 1

2
)λ̄ + y 1

2
(x 1

2
z 1

2
)λ̄ + z 1

2
(y 1

2
x 1

2
)λ̄)] 1

2
= 0;

3) (x 1
2
(y0z0))0 = 0;

4) (λ+ 1)(x 1
2
(y0z0))λ = −4(y0(x 1

2
z0) 1

2
+ z0(y0x 1

2
) 1
2
)λ;

5) (λ̄+ 1)(x 1
2
(y0z0))λ̄ = −4(y0(x 1

2
z0) 1

2
+ z0(y0x 1

2
) 1
2
)λ̄;

6) [xλ̄(y0z0)]0 = [xλ(y0z0)]0 = 0;

7) [yλ(x0zλ) 1
2
]0 = [yλ̄(x0zλ̄) 1

2
]0 = 0;

8) [xλ(y0z0)] 1
2
= [xλ̄(y0z0)] 1

2
= 0;

9) [yλ(x0zλ) 1
2
] 1
2
= [yλ̄(x0zλ̄) 1

2
] 1
2
= 0;

10) [xλ(y0z0) 1
2
]λ̄ = [xλ̄(y0z0) 1

2
]λ = 0;

11) [yλ(x0zλ) 1
2
]λ̄ = [yλ̄(x0zλ̄) 1

2
]λ̄ = 0;



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1483

12) [y0(xλ̄z0) 1
2
+ z0(xλ̄y0) 1

2
]λ̄ = 0;

13) [y0(xλz0) 1
2
+ z0(xλy0) 1

2
]λ = 0;

14) (−2λ− 1)((y 1
2
(x0z 1

2
)λ)0+(z 1

2
(x0y 1

2
)λ)0)+ (−2λ̄− 1)((y 1

2
(x0z 1

2
)λ̄)0+(z 1

2
(x0y 1

2
)λ̄)0)+

6(y 1
2
(x0z 1

2
) 1
2
)0 + (z 1

2
(x0y 1

2
) 1
2
)0 = 0;

15) (xλ(y 1
2
z 1

2
)0) 1

2
= (xλ̄(y 1

2
z 1

2
)0) 1

2
;

16) (−6λ−13)((y 1
2
(xλz 1

2
)0)λ+(z 1

2
(xλy 1

2
)0)λ)+(−6λ−2λ̄−7)((y 1

2
(xλz 1

2
)λ̄)λ+(z 1

2
(xλy 1

2
)λ̄)λ)+

(−2λ− 12)((y 1
2
(xλz 1

2
) 1
2
)λ + (z 1

2
(xλy 1

2
) 1
2
)λ) = 0;

17) (−6λ̄−13)((y 1
2
(xλ̄z 1

2
)0)λ̄+(z 1

2
(xλ̄y 1

2
)0)λ̄)+(−6λ̄−2λ−7)((y 1

2
(xλ̄z 1

2
)λ)λ̄+(z 1

2
(xλ̄y 1

2
)λ)λ̄)+

(−2λ̄− 12)((y 1
2
(xλ̄z 1

2
) 1
2
)λ̄ + (z 1

2
(xλ̄y 1

2
) 1
2
)λ̄) = 0;

18) (yλ(x 1
2
zλ) 1

2
)0+(zλ(x 1

2
yλ) 1

2
)0 = (yλ̄(x 1

2
zλ̄) 1

2
)0+(zλ̄(x 1

2
yλ̄) 1

2
)0 = (yλ(x 1

2
zλ) 1

2
)λ̄+(zλ(x 1

2
yλ) 1

2
)λ̄ =

(yλ̄(x 1
2
zλ̄) 1

2
)λ + (zλ̄(x 1

2
yλ̄) 1

2
)λ = 0;

19) (−14λ− 6)((yλ(x 1
2
zλ) 1

2
) 1
2
+ (zλ(x 1

2
yλ) 1

2
) 1
2
)− 16λ((yλ(x 1

2
zλ)0) 1

2
+ (zλ(x 1

2
yλ)0) 1

2
= 0;

20) (−14λ̄− 6)((yλ̄(x 1
2
zλ̄) 1

2
) 1
2
+ (zλ̄(x 1

2
yλ̄) 1

2
) 1
2
)− 16λ̄((yλ̄(x 1

2
zλ̄)0) 1

2
+ (zλ̄(x 1

2
yλ̄)0) 1

2
) = 0;

21) (xλ(y0z 1
2
) 1
2
)0 + (z 1

2
(xλy0) 1

2
)0 = (xλ̄(y0z 1

2
) 1
2
)0 + (z 1

2
(xλ̄y0) 1

2
)0 = 0;

22) (z 1
2
(xλy0) 1

2
)λ + (y0(xλz 1

2
) 1
2
)λ = (z 1

2
(xλ̄y0) 1

2
)λ̄ + (y0(xλ̄z 1

2
) 1
2
)λ̄ = 0;

23) (z 1
2
(xλy0) 1

2
)λ̄ + (y0(xλz 1

2
) 1
2
)λ̄ + (xλ(z 1

2
y0) 1

2
)λ̄ = 0;

24) (z 1
2
(xλ̄y0) 1

2
)λ + (y0(xλ̄z 1

2
) 1
2
)λ + (xλ̄(z 1

2
y0) 1

2
)λ = 0;

25) (−18λ− 2λ̄+ 5)(zλ(xλ̄y0) 1
2
)0 + (−18λ̄− 2λ+ 5)(xλ̄(zλy0) 1

2
)0 = 0;

26) (−3λ+ 1)(zλ(xλ̄y0) 1
2
) 1
2
+ (−3λ̄+ 1)(xλ̄(zλy0) 1

2
) 1
2
= 0;

27) (zλ(xλ̄y0) 1
2
)λ̄ = (xλ̄(zλy0) 1

2
)λ = 0;

28) (−12λ− 2λ̄+ 4)(zλ(xλ̄y 1
2
) 1
2
)0 + (−12λ̄− 2λ+ 4)(xλ̄(zλy 1

2
) 1
2
)0 = 0;

29) (−12λ+6)(zλ(xλ̄y 1
2
) 1
2
) 1
2
+(−12λ̄+6)(xλ̄(zλy 1

2
) 1
2
) 1
2
−16λ(zλ(xλ̄y 1

2
)0) 1

2
−16λ̄(xλ̄(zλy 1

2
)0) 1

2
=

0;

30) (zλ(xλ̄y 1
2
) 1
2
)λ̄ = (xλ̄(zλy 1

2
) 1
2
)λ = 0.



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1484

Proof. Consider the identity of Proposition 1. Setting x = e, y ∈ Aα, z ∈ Aβ, t ∈ Aγ ,
we have respectively ey = αy, ez = βz, et = γt and:

4αe(z(ty) + 4e(z(e(ty))) + 4αe(t(zy)) + 4e(e(z(ty))) + 4e(t(e(yz))) + 4e(e(t(yz)))+

4γe(z(yt)) + 4βe(t(yz)) + 4e(e(y(tz))) + 4γe(y(zt)) + 4βey(tz)) + 4e(y(e(tz))+

8γα2t(zy) + 8γαt(e(zy)) + 8γt(e(e(yz))) + 8γβt(e(yz)) + 8γβ2t(yz) + 4γαt(zy)+

4γβt(yz) + 4βz(ty) + 4β2z(ty) + 8β3z(ty) + 4αγy(zt) + 8αγ2y(zt) + 4αβy(tz)+

8αγy(e(zt)) + 8αβ2y(tz) + 8αβy(e(tz)) + 8αe(e(tz)) + 8β3y(tz) + 4β2y(tz) + 4βy(tz)+

4γt(yz) + 4γ2t(yz) + 8γ3t(yz) + 8β3z(ty) + 8β2z(e(ty)) + 8βz(e(e(ty))) + 4β2z(ty)+

8βγz(e(yt)) + 4βγz(yt) + 8βγ2z(yt) = 2βz(ty) + 2z(e(ty)) + 2αt(zy) + 2e(z(ty))+

2t(e(yz) + 2e(t(yz)) + 2γz(yt) + 2βt(yz) + 2e(y(tz)) + 2γy(zt) + 2βy(tz) + 2y(e(tz))
(2)

By setting α = β = γ = 0, the relation (2) becomes

4e(z(e(ty))) + 4e(e(z(ty))) + 4e(t(e(yz))) + 4e(e(t(yz))) + 4e(e(y(tz))) + 4e(y(e(tz))

= 2z(e(ty)) + 2e(z(ty)) + 2e(t(yz)) + 2e(y(tz)) + 2y(e(tz)) + 2t(e(yz) (3)

Since A2
0 ⊂ A 1

2
according to Theorem 2, therefore the relation (3) becomes

4e(e(z(ty))) + 4e(e(t(yz))) + 4e(e(y(tz))) = z(ty) + y(tz) + t(yz) (4)

Using the relations i) and vi) of Theorem 2, relation (4) gives [z(ty)] 1
2
+ 4λ2[z(ty)]λ +

4λ̄2[z(ty)]λ̄ + [t(yz)] 1
2
+ 4λ2[t(yz)]λ + 4λ̄2[t(yz)]λ̄ + [y(tz)] 1

2
+ 4λ2[y(tz)]λ + 4λ̄2[y(tz)]λ̄ =

[z(ty)] 1
2
+ [z(ty)]λ + [z(ty)]λ̄ + [t(yz)] 1

2
+ [t(yz)]λ + [t(yz)]λ̄ + [y(tz)] 1

2
+ [y(tz)]λ + [y(tz)]λ̄

which implies that{
(4λ2 − 1)([z(ty)]λ + [t(yz)]λ + [y(tz)]λ) = 0

(4λ̄2 − 1)([z(ty)]λ̄ + [t(yz)]λ̄ + [y(tz)]λ̄) = 0

As 4λ2 − 1 ̸= 0 and 4λ̄2 − 1 ̸= 0, then [z(ty)]λ + [t(yz)]λ + [y(tz)]λ = [z(ty)]λ̄ + [t(yz)]λ̄ +
[y(tz)]λ̄ = 0; posing t = x0; y = y0; z = z0, we have (1).
By proceeding in a similar way, we find the other identities.

the following two results are immediate consequences of Lemma 1 and Theorem2.

Corollary 1. If A0 = 0, then A2
1
2

⊂ Aλ ⊕ Aλ̄, A 1
2
Aλ ⊂ A 1

2
⊕ Aλ̄, A 1

2
Aλ̄ ⊂ A 1

2
⊕ Aλ,

A2
λ = A2

λ̄
= AλAλ̄ = 0 and for all x 1

2
∈ A 1

2
; xλ ∈ Aλ; xλ̄ ∈ Aλ̄ the following identities are

verified:

i) [(λ+ 1)x 1
2
(x21

2

)λ + (λ̄+ 1)x 1
2
(x21

2

)λ̄]1/2 = 0;

ii) (2λ+ 3)(x 1
2
(x 1

2
xλ)λ̄)λ + (λ+ 6)(x 1

2
(x 1

2
xλ) 1

2
)λ = 0;



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1485

iii) (2λ̄+ 3)(x 1
2
(x 1

2
xλ̄)λ)λ̄ + (λ̄+ 6)(x 1

2
(x 1

2
xλ̄) 1

2
)λ̄ = 0;

iv) xλ(xλx 1
2
) = 0;

v) xλ̄(xλ̄x 1
2
) = 0;

vi) (xλ(xλ̄x 1
2
))λ̄ = 0;

vii) (xλ̄(xλx 1
2
))λ = 0;

viii) (2λ− 1)xλ(xλ̄x 1
2
) + (2λ̄− 1)xλ̄(xλx 1

2
) = 0.

Corollary 2. If Aᾱ = 0 with α ∈ {λ, λ̄}, then the following identities are verified:

i) 2ex30 = x30;

ii) [12(x 1
2
(x21

2

)0) + 3(α+ 1)(x 1
2
(x21

2

)α)] 1
2
= 0;

iii) [(α+ 1)(x 1
2
(x20) 1

2
) + 8(x0(x0x 1

2
) 1
2
)]α = 0;

iV ) [(−4α− 2)(x 1
2
(x0x 1

2
)α) + 12(x 1

2
(x0x 1

2
) 1
2
)]0 = 0;

V ) [(6α+ 13)(x 1
2
(xαx 1

2
)0) + (2α+ 12)(x 1

2
(xαx 1

2
) 1
2
)]α = 0;

V i) [(7α+ 3)(xα(xαx 1
2
) 1
2
) + 8α(xα(xαx 1

2
)0)] 1

2
= 0;

V ii) [x 1
2
(x20) 1

2
]0 = [xα(x

2
0) 1

2
]0 = [xα(x0xα) 1

2
]0 = [xα(x 1

2
xα) 1

2
]0 = [xα(x 1

2
x0) 1

2
]0 = 0;

V iii) [xα(x
2
0) 1

2
] 1
2
= [xα(x0xα) 1

2
] 1
2
= [xα(x

2
1
2

)0] 1
2
= 0;

iX) [x0(x0xα) 1
2
]α = [x 1

2
(x0xα) 1

2
]α = 0.

In [4], the author give Peirce decomposition of a principal train algebra

Theorem 3. (see Theorem 1, [4]) Let (A,ω) be a principal train algebra with an idempo-
tent e and with principal train polynomial P (X) = (X − 1)(X −λ1) · · · (X −λr−1) (the λi

are two by two distinct). Then A splits into the direct sum A = Ke ⊕ V1 ⊕ V2 ⊕ · · · ⊕ Vs

where Vi = N ∩ (Le − γiid), Le : A → A, x 7→ ex, id : A → A, x 7→ x.

In [7] the authors gave a characterization of principal train algebras of rank 4.

Theorem 4. (Theorem 5, [7]) Let (A,ω) be a baric algebra. The algebra A is a principal
train algebra of rank 4, with principal train polynomial X(X − 1)(X − λ1)(X − λ2) where
λ1 and λ2 different and different from 1

2 , if and only if:

1) A possesses an idempotent e and, with respect to e, it has the Peirce decomposition
A = Ke⊕ U1/2 ⊕ Uλ1 ⊕ Uλ2 where Ui = {x ∈ kerω, ex = ix} (i ∈ {1/2, λ1, λ2})



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1486

2) U2
1/2 ⊂ Uλ1⊕Uλ2, U1/2Uλ1 ⊂ U1/2⊕Uλ2, U1/2Uλ2 ⊂ U1/2⊕Uλ1, U

2
λ1

⊂ Uλ2, U
2
λ2

⊂ Uλ1,

Uλ1Uλ2 = 0,(Uλ1 ⊕ Uλ2)
3 = 0.

3) For all x ∈ kerω, u ∈ U1/2, v ∈ Uλ1 and w ∈ Uλ2, the following relations are verified:

(i) (12 − λ2)(u(u
2)λ1)1/2 + (12 − λ1)(u(u

2)λ2)1/2 ;

(ii) (λ1 − λ2)(u(uv)1/2)λ1 + (λ1 − 1
2)(u(uv)λ2)λ1 ;

(iii) (λ2 − λ1)(u(uw)1/2)λ2 + (λ2 − 1
2)(u(uw)λ1)λ2 ;

(iv) (1− 2λ2)(v(uv))1/2 + (λ1 − 1
2)uv

2 ;

(v) (1− 2λ1)(w(uw))1/2 + (λ2 − 1
2)uw

2 ;

(vi) (12 − λ1)(v(uw)1/2)1/2 + (12 − λ2)(w(uv)1/2)1/2 ;

(vii) (λ1 − λ2)(w(uv)1/2)λ1 + (λ1 − 1
2)w(uv)λ2 ;

(viii) (λ2 − λ1)(w(uv)1/2)λ2 + (λ2 − 1
2)w(uv)λ1 ; where (x)i denotes the projection of

x ∈ kerω onto the subspace Ui, i ∈ {1
2 , λ1, λ2} ;

(ix) x4 = 0.

3. Relation with principal train algebras

Proposition 2. Let A = Ke ⊕ A0 ⊕ A 1
2
⊕ Aλ ⊕ Aλ̄ an algebra satisfying the identity

2x2x4 = ω(x)2x4 + ω(x)4x2. If A 1
2
= 0 then A is a principal train algebra satisfying the

equation x5 − 1
2ω(x)x

4 + ω(x)2x3 − 3
2ω(x)

3x2 = 0.

Proof. A 1
2
being zero, we have A2

0 = A2
λ = A2

λ̄
= AλA0 = Aλ̄A0 = Aλ̄Aλ = 0.

For x = e + x0 + xλ + xλ̄, we have x2 = e + 2λxλ + 2λ̄xλ̄, x
3 = e − 3xλ − 3xλ̄, x

4 =
e− 2λxλ − 2λ̄xλ̄, x

5 = e+ (2λ+ 3)xλ + (2λ̄+ 3)xλ̄. So, x
5 − 1

2x
4 + x3 − 3

2x
2 = 0 and we

obtain x5 − 1
2ω(x)x

4 + ω(x)2x3 − 3
2ω(x)

3x2 = 0 because the set of elements of weight 1 is
dense in A according to Zariski’s topology.

Proposition 3. Let A = Ke⊕A0⊕A 1
2
⊕Aλ⊕Aλ̄ be a Peirce decomposition of an algebra

satisfying the identity 2x2x4 = ω(x)2x4 + ω(x)4x2. If A0 = Aᾱ = 0 with α ∈ {λ̄, ¯̄λ}, then
A checks the train equation

x3 − (1 + α)ω(x)x2 + αω(x)2x = 0.

Proof. Let A = Ke ⊕ A0 ⊕ A 1
2
⊕ Aλ ⊕ Aλ̄ be an algebra satisfying the identity

2x2x4 = ω(x)2x4 + ω(x)4x2. Suppose A0 = Aλ̄ = 0.
Let x = e + x 1

2
+ xλ be an element of weight 1 of A. By exploiting the relations of the

Corrolary 2, we have: x2 = e+x 1
2
+2λxλ+x21

2

+2x 1
2
xλ, x

2−x = (2λ−1)xλ+x21
2

+2x 1
2
xλ,

x(x2 − x) = λx21
2

+ (2λ2 − λ)xλ + 2λxλx 1
2
= λ(x2 − x), so x(x2 − x) = λ(x2 − x) and



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1487

x3− (1+λ)x2+λx = 0. Since the set of elements of weight 1 is dense in A by the Zariski’s
topology, then for any x in A, we have x3 − (1 + λ)ω(x)x2 + λω(x)2x = 0. The proof is
similar when we assume A0 = Aλ = 0.

Theorem 5. Let A be an algebra satisfying the identity 2x2x4 = ω(x)2x4 + ω(x)4x2;
then A is a principal train algebra of rank 3 if and only its train equation is of the form
x3 − (1 + γ)ω(x)x2 + γω(x)2x = 0, where γ ∈ {0, λ, λ̄}

Proof. Let A be an algebra satisfying the identity 2x2x4 = ω(x)2x4 + ω(x)4x2.
Suppose A is a principal train algebra of rank 3, its equation is

x3 − (1 + α)ω(x)x2 + αω(x)2x = 0 with α ∈ K (5)

And a partial linearisation of (5) gives us

x2y + 2x(xy)− (1 + α)[ω(y)x2 + 2ω(x)xy] + α[2ω(xy)x+ ω(x)2y] = 0 (6)

setting y = x4 in (6), we have

x2x4 + 2x6 − (1 + α)ω(x)4x2 − 2(1 + α)ω(x)x5 + 2αω(x)5x+ αω(x)2x4 = 0 (7)

or 2x6 = 2(1 + α)ω(x)x5 − 2αω(x)2x4, we also know that x2x4 = 1
2ω(x)

2x4 + 1
2ω(x)

4x2;
substituting 2x6 and x2x4 by their expressions in (7), we get

(
1

2
− α)ω(x)2x4 − (

1

2
+ α)ω(x)4x2 + 2αω(x)5x = 0 (8)

We can notice that x3 = (1+α)ω(x)x2−αω(x)2x; which implies that x4 = (1+α)ω(x)x3−
αω(x)2x2 = (1+ α+ α2)ω(x)2x2 + (−α2 − α)ω(x)3x. Substituting x4 by its expression in
(8), we get
−1

2α(2α
2+α+3)ω(x)4(x2−ω(x)x) = 0. The algebra A being of rank 3, then x2−ω(x)x ̸= 0

which implies that −1
2α(2α

2 + α+ 3) = 0 hence α = 0, α = λ or α = λ̄.
Conversely, suppose that A is a principal train algebra of train equation

x3 − (1 + α)ω(x)x2 + αω(x)2x = 0 with α ∈ {0, λ, λ̄} (9)

If α = 0, A is a Bernstein Jordan algebra (see [10]) and therfore satisfies the identity
2x2x4 = ω(x)2x4 + ω(x)4x2 (see [11]).

For α = λ, the partial linearisation of (9) gives us

x2y + 2x(xy)− (1 + λ)[ω(y)x2 + 2ω(x)xy] + λ[2ω(xy)x+ ω(x)2y] = 0 (10)

By setting y = x4, we have x2x4 = −2x6+(1+λ)[ω(x)4x2+2ω(x)x5]−λ[2ω(x)5x+ω(x)2x4],
so x2x4 = [−2x6+2(1+λ)ω(x)x5−2λω(x)2x4]+λω(x)2x4+(1+λ)ω(x)4x2−2λω(x)5x =
(1 + λ)ω(x)4x2 − 2λω(x)5x+ λω(x)2x4, hence

x2x4− 1
2ω(x)

4x2− 1
2ω(x)

2x4 = (λ2+ λ
2−

1
2)ω(x)

3x3+(−λ2+ 3λ
2 + 1

2)ω(x)
4x2−2λω(x)5x,

so x2x4 − 1
2ω(x)

4x2 − 1
2ω(x)

2x4 = −2ω(x)3x3 + (2λ + 2)ω(x)4x2 − 2λω(x)5x. Therefore
x2x4− 1

2ω(x)
4x2− 1

2ω(x)
2x4 = −2ω(x)3(x3− (λ+1)ω(x)x2+λω(x)2x) = 0 and A satisfies

the identity 2x2x4 = ω(x)2x4 + ω(x)4x2. The proof is similar for α = λ̄.



D. Kabre, A. Conseibo / Eur. J. Pure Appl. Math, 16 (3) (2023), 1480-1490 1488

Theorem 6. Let A = Ke ⊕ A0 ⊕ A 1
2
⊕ Aλ ⊕ Aλ̄ be an algebra satisfying the identity

2x2x4 = ω(x)2x4 + ω(x)4x2; if A is a principal train algebra of rank 4, its train equation
is one of the following forms:

i) x4 − (1 + γ)ω(x)x3 + γω(x)2x2 = 0, γ ∈ {0, λ, λ̄};

ii) x4 − 1
2ω(x)x

3 + ω(x)2x2 − 3
2ω(x)

3x = 0;

iii) x4 − (32 + γ)ω(x)x3 + (12 + 3
2γ)ω(x)

2x2 − 1
2γω(x)

3x = 0 ; γ ∈ {1
2 , λ, λ̄};

iV) x4 − (1 + 2γ)ω(x)x3 + γ(γ + 2)ω(x)2x2 − γ2ω(x)3x = 0 ; γ ∈ {λ, λ̄}.

Proof. Let A = Ke ⊕ A0 ⊕ A 1
2
⊕ Aλ ⊕ Aλ̄ be an algebra satisfying the identity

2x2x4 = ω(x)2x4 + ω(x)4x2. Assuming that A is a principal train algebra of rank 4,
its train equation is of the form x4 − (1 + α + β)ω(x)x3 + αω(x)2x2 + βω(x)3x = 0
with α, β ∈ K so its minimal train polynomial is P (X) = X(X − 1)(X − α1)(X − α2) =
X4−(1+α1+α2)X

3+(α1+α2+α1α2)X
2−α1α2X; we then notice that α = α1+α2+α1α2

and β = −α1α2 therefore

x4 − (1 + α1 + α2)ω(x)x
3 + (α1 + α2 + α1α2)ω(x)

2x2 − α1α2ω(x)
3x = 0 (11)

Now let us look at the different cases related to the train roots α1 and α2:

1st Case: α1 ̸= α2, α1 ̸= 1
2 and α2 ̸= 1

2
By exploiting the theorem 5 of [7] and the theorem 1, we observe that A admits relatively
to an idempotent e, the following Peirce decomposition: A = Ke ⊕ A 1

2
⊕ Aα1 ⊕ Aα2 =

Ke⊕A0 ⊕A 1
2
⊕Aλ ⊕Aλ̄ then we have by identification α1, α2 ∈ {0, λ, λ̄}. Indeed:

for α1 = 0 and α2 = λ, (11) becomes x4 − (1 + λ)ω(x)x3 + λω(x)2x2 = 0,
for α1 = 0 and α2 = λ̄, the equation (11) becomes
x4 − (1 + λ̄)ω(x)x3 + λ̄ω(x)2x2 = 0, and
if α1 = λ and α2 = λ̄, (11) becomes x4 − 1

2ω(x)x
3 + ω(x)2x2 − 3

2ω(x)
3x = 0

2nd Case: α1 ̸= α2 and α1 =
1
2

Considering the theorem 1 of [4] and the theorem (1), it follows that A admits the follow-
ing Peirce decomposition: A = Ke⊕A 1

2
⊕Aα2 with α2 ∈ {0, λ, λ̄}. The train equation is

therefore one of the following forms:
For α1 =

1
2 and α2 = 0, (11) becomes x4 − 3

2ω(x)x
3 + 1

2ω(x)
2x2 = 0;

For α1 =
1
2 and α2 = λ, the equation (11) becomes

x4 − (32 + λ)ω(x)x3 + (12 + 3
2λ)ω(x)

2x2 − 1
2λω(x)

3x = 0;
For α1 =

1
2 and α2 = λ̄, (11) becomes

x4 − (32 + λ̄)ω(x)x3 + (12 + 3
2 λ̄)ω(x)

2x2 − 1
2 λ̄ω(x)

3x = 0.

3rd Case: α1 = α2, α1 ̸= 1
2 and α2 ̸= 1

2
According to the theorem 1 of [4] and as A admits nonzero idempotents, the Peirce de-
composition of A with respect to an idempotent e is



REFERENCES 1489

A = Ke ⊕ A 1
2
⊕ B with B = N ∩ Ker(ℓe − α1I)

2. If B = 0, we have x = e + x 1
2
and

x2 = e+ x 1
2
so x2 = ω(x)x which is an elementary Bernstein algebra and this contradicts

the fact that A is a train algebra of rank 4. Otherwise, there are three possibilities. Indeed:

i) α1 = α2 = 0 implies that the train equation of A is x4 − ω(x)x3 = 0;

ii) α1 = α2 = λ implies that the train equation of A is
x4 − (1 + 2λ)ω(x)x3 + λ(λ+ 2)ω(x)2x2 − λ2ω(x)3x = 0;

iii) α1 = α2 = λ̄ implies that the train equation of A is
x4 − (1 + 2λ̄)ω(x)x3 + λ̄(λ̄+ 2)ω(x)2x2 − λ̄2ω(x)3x = 0.

Definition 4. For any fixed α in K, we consider the map
φα : K[X] → K[X], P 7→ (X − α)P

We easily establish the following lemma.

Lemma 2. For α ∈ C, we have φα ◦ φᾱ = φᾱ ◦ φα

Theorem 7. Let A = Ke ⊕ A0 ⊕ A 1
2
⊕ Aλ ⊕ Aλ̄ be an algebra satisfying the identity

2x2x4 = ω(x)2x4 + ω(x)4x2 such that A0 = 0. Let setting µ = X2 −X. If A is principal
train algebra of rank n ≥ 5, its train equation is of the following form: ω(x)n((φt

λ̄
◦ φs

λ ◦
φr
1/2)(µ)(

x
ω(x)) = 0, r ≥ 0, s ≥ 0, t ≥ 0 are integers and r + t+ s = n− 2.

Proof. Let x = e + x 1
2
+ xλ + xλ̄ an element of weight 1 in A. We have x2 − x =

x21
2

+ (2λ − 1)xλ + (2λ̄ − 1)xλ̄ + 2x 1
2
xλ + 2x 1

2
xλ̄. By setting x2 − x = a 1

2
+ aλ + aλ̄ with

aα ∈ Aα, α ∈ {λ, λ̄}, we show using Corollary 1 that there exists an integer r ≥ 0 such
that φr

1/2(µ)(x) = ar,λ + ar,λ̄, ar,λ ∈ Aλ, and ar,λ̄ ∈ Aλ̄. Similarly, there exists an integer

s ≥ 0 such that (φλ ◦ φr
1/2)(µ)(x) = bλ̄ with bλ̄ ∈ Aλ̄. Finally, for some integer t ≥ 0, we

have (φλ̄ ◦φλ ◦φr
1/2)(µ)(x) = 0. The set of element of weight 1 being dense in A according

to Zariski topology, for any x in A, we have ω(x)n((φt
λ̄
◦ φs

λ ◦ φr
1/2)(µ)(

x
ω(x)) = 0.

References

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[2] J. Bayara, A. Conseibo, M. Ouattara, and F. Zitan. Power-associative algebras that
are train algebras. J. Algebra, 324:1159–1176, 2010.

[3] S. Bernstein. Solution of a mathematical problem connected with the theory of hered-
ity. Ann. Math. Stat, 13:1159–1176, 1942.



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