EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 4, 2022, 1887-1907 ISSN 1307-5543 – ejpam.com Published by New York Business Global Classification of four dimensional train algebras of degree 2 and exponent 4 W. Achile Zangre1, André Conseibo1,∗ 1 Département de Mathématiques et Informatique, Université Norbert ZONGO, BP 376 Koudougou, Burkina Faso Abstract. This paper is devoted to the classification of train algebras of degree 2 and exponent 4. This classification is made in dimension at most four and according to the type of the algebra. We first show that in four dimension, the type of algebra can only be of (2, 0, 1, 1), (2, 1, 0, 1) or (2, 1, 1, 0). 2020 Mathematics Subject Classifications: 17D92,17A05 Key Words and Phrases: Peirce decomposition, Train algebra of degree 2 and exponent 4, Bernstein algebra, Idempotent 1. Introduction Several nonassociative algebras verifying polynomial identities are used in algebra mod- eling of population genetics. We can cite among others, Bernstein algebras (cf [2],[6],[8]), Jordan algebras (see [10],[7]), power-associative train algebras ([5]) train algebras of de- gree 2 and exponent n ([9],[4]). In [1] the authors defined an algebra satisfying a train identity of degree 2 and exponent 4 as an algebra A such that for any x in A, we have: (x4)2 = ω(x)4x4, where K is an infinite and algebrically closed commutative field of char- acteristic different from 2 and 3. If such an algebra does not verify a polynomial identity of degree less than or equal to 7, we say that it is a train algebras of degree 2 and exponent 4. This class of algebras models populations whose genetic potential becomes stable in the fourth generation. In particular, this class contains Bernstein algebras([3]). In this paper we are interested in the classification of these algebras in dimension 4; it is made according to the type of the algebra A; this one cannot be of lower dimension. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4586 Email addresses: achilzangre@gmail.com (W.A. Zangre), andreconsebo@yahoo.fr (A. Conseibo) https://www.ejpam.com 1887 © 2022 EJPAM All rights reserved. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1888 2. Preliminaries Let K be a commutative field and A a commutative K-algebra, not necessarily asso- ciative. For any element x of A we define the principal powers of x by: x1 = x, xk+1 = xxk for any integer k ≥ 1. We will say that the algebra A is a baric algebra if there exists a non-zero homomor- phism of algebras ω : A → K called the weight function of the algebra A. The weight of an element x of A is the scalar ω(x). A baric K-algebra (A,ω) is a Bernstein algebra if (x2)2 = ω(x)2x2 for any x in A. In the following K is a commutative algebraically closed field of characteristic distinct from 2 and 3, and A is a commutative nonassociative algebra. A baric algebra (A,ω) satisfies a train identity of degree 2 and exponent 4 if for any x in A, we have: (x4)2 = ω(x)4x4. (1) An algebra A is called a train algebra of degree 2 and exponent 4 if it satisfies a train identity of degree 2 and exponent 4 and does not verify a polynomial identity of degree less than 8. For an element x of weight 1 in A (i.e. ω(x) = 1), the identity (1) gives (x4)2 = x4 and thus x4 is an non zero idempotent of A. In [1], the authors have established the following two theorems: Theorem 1. Let A be an algebra satisfying a train identity of degree 2 and exponent 4. Then A has a Peirce decomposition A = Ke⊕A1/2⊕A0⊕Aλ⊕Aλ, where e2 = e ̸= 0 and Aα = {y ∈ ker(ω) | ey = αy} for α ∈ {0; 1/2;λ = −1 + i √ 7 4 ;λ = −1 + i √ 7 4 }. Theorem 2. Let A = Ke⊕A1/2 ⊕A0 ⊕Aλ ⊕Aλ be a Peirce decomposition relative to a non-zero idempotent e of an algebra satisfying a train identity of degree 2 and exponent 4. Then: (i) A2 0 ⊂ A0 ⊕A1/2; (ii) A2 1/2 ⊂ A0 ⊕Aλ ⊕Aλ; (iii) A2 λ ⊂ A1/2; (iv) A2 λ ⊂ A1/2; (v) A1/2A0 ⊂ A1/2 ⊕Aλ ⊕Aλ; (vi) A0Aλ ⊂ A1/2; (vii) A0Aλ ⊂ A1/2; (viii) A1/2Aλ ⊂ A1/2 ⊕A0 ⊕Aλ; (ix) A1/2Aλ ⊂ A1/2 ⊕A0 ⊕Aλ; W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1889 (x) AλAλ ⊂ A1/2. Lemma 1. Let A = Ke ⊕ A1/2 ⊕ A0 ⊕ Aλ ⊕ Aλ be a Peirce decomposition relative to a non-zero idempotent e of an algebra satisfying a train identity of degree 2 and exponent 4, then for all x1/2, x0, xλ and xλ respectively in A1/2, A0, Aλ, Aλ, the following assertions are verified: (i) 2x1/2(e(ex 2 1/2)) + 2e(x1/2(ex 2 1/2)) + x1/2(ex 2 1/2) + 2e(ex31/2) + ex31/2 + x31/2 =0; (ii) 2x1/2(x1/2(ex 2 1/2)) + 2x1/2(ex 3 1/2) + 2ex41/2 + x41/2 + (e(ex21/2) + ex21/2 + x21/2) 2 = 0; (iii) 2e(x0(ex 2 0)) + 2e(ex30)− x0(ex 2 0)− ex30 = 0; (iv) 8ex40 + (ex20) 2 − 4x40 = 0; (v) 4e(ex3λ) + 8λex3λ − (4λ+ 1)x3λ = 0; (vi) 32ex4λ + (1− 32λ)(x2λ) 2 − 16x4λ = 0; (vii) 4e(ex3 λ ) + 8λex3 λ − (4λ+ 1)x3 λ = 0; (viii) 32ex4 λ + (1− 32λ)(x2 λ )2 − 16x4 λ = 0. Proof. It suffices to set x = e+ αx1/2 + βx0 + γxλ + µxλ and identify the coefficients of αiβjγkµℓ (1 ≤ i+ j+ k+ ℓ ≤ 8), equality x4 = (x4)2 allowing to obtain respectively (i), (ii), (iii), (iv), (v), (vi), (vii) and (viii). Lemma 2. Let A = Ke ⊕ A1/2 ⊕ A0 ⊕ Aλ ⊕ Aλ be a Peirce decomposition relative to a non-zero idempotent e of an algebra satisfying a train identity of degree 2 and exponent 4. If Aλ = Aλ = 0 then, for all x0 ∈ A0 and x1/2 ∈ A1/2, we have the following identities: (i) 2e(x0x 2 1/2) + 2x1/2(x0x1/2)− x0x 2 1/2 = 0; (ii) x31/2 = 0; (iii) (x21/2) 2 = 0; (iv) (ex20) 2 + 8ex40 − 4x40 = 0. (v) x1/2(ex 2 0)− x1/2x 2 0 = 0; (vi) 2e(x0(x1/2(x0x1/2)))+2e(x1/2(x1/2x 2 0))+x21/2(ex 2 0)−x0(x1/2(x0x1/2))−x1/2(x1/2x 2 0)+ 4(x0x1/2) 2 = 0; (vii) 4x1/2(x1/2(x1/2x0))+2e(x1/2(x0x 2 1/2))+2x1/2(e(x0x 2 1/2))+x1/2(x0x 2 1/2)+4x21/2(x0x1/2) = 0; W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1890 (viii) (ex20)(x0x1/2)+4x1/2(ex 3 0)+4x1/2(x0(ex 2 0))+4e(x0(x 2 0x1/2))+4e(x1/2x 3 0)−2x0(x1/2x 2 0)− 2x1/2x 3 0 = 0. Proof. The proof is similar to that of the Lemma 1. According to Propositions 3.1 and 3.2 of [1], the possible types of A in dimension 4 are: (2, 2, 0, 0), (2, 1, 0, 1), (2, 1, 1, 0) and (2, 0, 1, 1). Theorem 3. Let A = Ke ⊕ A1/2 ⊕ A0 ⊕ Aλ ⊕ Aλ be a Peirce decomposition relative to a non-zero idempotent e of an algebra satisfying a train identity of degree 2 and exponent 4. If Aλ = Aλ = 0, then A verifies a polynomial equation of degree less than eight and therefore, A is not train of degree 2 and exponent 4. Proof. The type of A being (2, 2, 0, 0), then Aλ = Aλ = 0, A2 1/2 ⊂ A0, A2 0 ⊂ A1/2 ⊕A0 and A1/2A0 ⊂ A1/2; we can then set A1/2 =< e0 >, A0 =< e1, e2 >, so A =< e, e0, e1, e2 > such that: e2 = e, ee0 = 1 2e0, ee1 = ee2 = 0, e20 = α0e1 + α1e2, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0+ γ1e1+ γ2e2, e22 = µ0e0+µ1e1+µ2e2, e0e1 = µe0, e0e2 = γe0; According to the Lemma (2) we have x31/2 = (x21/2) 2 = 0 which allows us to obtain the following relations: x31/2 = 0 ⇒ e30 = (α0µ + α1γ)e0 = 0 so : α0µ + α1γ = 0. Moreover, (x21/2) 2 = 0 ⇒ (α2 0γ0+α2 1µ0+2α0α1β0)e0+(α2 0γ1+α2 1µ1+2α0α1β1)e1+(α2 0γ2+α2 1µ2+2α0α1β2)e2 = 0 so : α2 0γ0+α2 1µ0+2α0α1β0 = 0; α2 0γ1+α2 1µ1+2α0α1β1 = 0; and α2 0γ2+α2 1µ2+2α0α1β2 = 0. Thus we have the following cases, the products not mentioned in the multiplication table of A are zero. 1st Case: A2 1/2 ̸= 0. We can set e1 = e20 and therefore α0 = 1, α1 = 0. Thus, we have µ = γ0 = γ1 = γ2 = 0 and using the others identities of the Lemma 2; µ0 = β2 = β1 = γ = µ2 = 0. Therefore, the multiplication table of A becomes e2 = e, ee0 = 1 2e0, e 2 0 = e1, e1e2 = β0e0, e22 = µ1e1. Let x = e+ ae0 + be1 + ce2 an element in A of weight 1. We have (x2)3 − (x2)2 = 0, so for any x in A, (x2)3 − ω(x)2(x2)2 = 0. 2nd Case: A2 1/2 = 0. We have e2 = e, ee0 = 1 2e0, ee1 = ee2 = 0, e1e2 = β0e0+β1e1+β2e2, e21 = γ0e0+γ1e1+γ2e2, e22 = µ0e0 + µ1e1 + µ2e2, e0e1 = µe0, e0e2 = γe0. Determine the identity verified by A according to its multiplication table. Using the identities of the Lemma 2 we have γ1µ+γ2γ = 0, µ1µ+µ2γ = 0, γ1(γ21 +γ2β1)+β1γ2(γ1+ β2) = 0, γ2(γ 2 1 + γ2β1) + β2γ2(γ1 + β2) = 0, µ1β1(β1 + µ2 2) + µ1(µ1β2 + µ3 2) = 0 and µ1β2(β1 + µ2 2) + µ2(µ1β2 + µ3 2) = 0. i) γ = 0 and µ ̸= 0. We can set µ = 1 and then γ1 = µ1 = µ2 = γ2β1 = γ2β2 = 0. The multiplication table of A becomes e2 = e, ee0 = 1 2e0, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0 + γ2e2, e22 = µ0e0, e0e1 = e0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1891 i.1) Suppose γ2 ̸= 0. Then β1 = β2 = 0, so e2 = e, ee0 = 1 2e0, e1e2 = β0e0, e21 = γ0e0 + e2, e22 = µ0e0, e0e1 = e0 and for any x in A, (x3)2 = ω(x)3x3. i.2) γ2 = 0. Then e2 = e, ee0 = 1 2e0, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0, e22 = µ0e0, e0e1 = e0. Using the identities of the Lemma 2 we have β1 = β2 = 0. Therefore, e2 = e, ee0 = 1 2e0, e1e2 = β0e0, e21 = γ0e0, e22 = µ0e0, e0e1 = e0. For any x in A, (x2)2 = ω(x)2x2. ii) µ = 0 and γ ̸= 0. Then γ1 = γ2 = µ2 = µ1β1 = µ1β2 = 0 and e2 = e, ee0 = 1 2e0, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0, e22 = µ0e0 + µ1e1, e0e1 = µe0, e0e2 = γe0. ii.1) µ1 ̸= 0, then β1 = β2 = 0, so e2 = e, ee0 = 1 2e0, e1e2 = β0e0, e21 = γ0e0, e22 = µ0e0+e1 (we can set µ1 = 1), e0e1 = µe0, e0e2 = γe0 and for any x in A, (x2)2 = ω(x)2x2. ii.2) µ1 = 0, then e2 = e, ee0 = 1 2e0, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0, e22 = µ0e0, e0e1 = µe0, e0e2 = γe0. Using the identities of the Lemma 2 we have β1 = β2 = 0. Therefore, e2 = e, ee0 = 1 2e0, e1e2 = β0e0, e21 = γ0e0, e22 = µ0e0, e0e1 = µe0, e0e2 = γe0. For any x in A, (x2)2 = ω(x)2x2. iii) γ = µ = 0 e2 = e, ee0 = 1 2e0, e1e2 = β0e0+β1e1+β2e2, e21 = γ0e0+γ1e1+γ2e2, e22 = µ0e0+µ1e1+µ2e2. iii.1) γ1 = 0 and γ2 ̸= 0, then β1 = β2 = µ2 = 0, so e2 = e, ee0 = 1 2e0, e1e2 = β0e0, e21 = γ0e0 + e2, e22 = µ0e0 + µ1e1. The identity (x4)2 = ω(x)4x4 implies that µ1 = 0 and for any x in A, we have (x3)2 = ω(x)3x3. iii.2) γ2 = 0 and γ1 ̸= 0, this case is impossible. iii.3) γ2 = γ1 = 0, then e2 = e, ee0 = 1 2e0, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0, e22 = µ0e0 + µ1e1 + µ2e2. iii.3.1) µ1 = 0, then µ2 = 0 and β1 = β2 = 0. Therefore for any x in A, we have (x2)2 = ω(x)2x2. iii.3.2) µ1 ̸= 0, then e2 = e, ee0 = 1 2e0, e1e2 = β0e0 + β1e1 + β2e2, e21 = γ0e0, e22 = µ0e0 + µ1e1 + µ2e2. If µ2 = 0, then β1 = β2 = 0 and for any x in A, we have (x2)2 = ω(x)2x2. If µ2 ̸= 0, then µ1β2 = µ2β1. For x = e0, the identity (x4)2 = 0 implies that β1 = −µ2 or β1 = −µ2 2 . We show that this is impossible. iii.4) γ2γ1 ̸= 0, then e2 = e, ee0 = 1 2e0, e1e2 = β0e0+β1e1+β2e2, e21 = γ0e0+γ1e1+γ2e2, e22 = µ0e0 + µ1e1 + µ2e2. iii.4.1) µ2 = 0. Then µ1 = 0 and β1β2 ̸= 0. without loss of generality. It suffices to make some basis transformations to prove that we can set γ2 = 1 and therefore β1 = γ1β2. We show this is impossible. iii.4.2) µ2 ̸= 0. Then µ1 ̸= 0 and β1β2 ̸= 0, so µ1β2 = µ2β1. We show that β1 = −µ2 W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1892 or β1 = −µ2 2 this is impossible. We conclude that the type of A cannot be (2, 2, 0, 0). The above result allows us to discard the type (2, 2, 0, 0) in the classification. 3. Classification of algebras of type (2, 1, 1, 0) In this section we will set x = e + αe0 + βe1 + θe2 an element of weight 1 in A. Also we will use essentially the assertion of the Lemma 1. If Type A = (2, 1, 1, 0) then Aλ = 0 and A2 1/2 ⊂ A0 ⊕Aλ, A2 0 ⊂ A1/2 ⊕A0, A2 λ ⊂ A1/2, A1/2A0 ⊂ A1/2 ⊕ Aλ, A1/2Aλ ⊂ A1/2 ⊕ A0, A0Aλ ⊂ A1/2; we can thus set A1/2 =< e0 >, A0 =< e1 >, Aλ =< e2 >. The multiplication table of A is given by: e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α0e1 + α1e2, e21 = β0e0 + β1e1, e22 = γe0, e0e1 = µ0e0 + µ1e2, e0e2 = γ0e0 + γ1e1, e1e2 = µe0. The assertion (i) of the Lemma 1 allows to obtain the equality: 2(α0µ0 − λ2α1γ0)e0 = 0 so: α0µ0 = λ2α1γ0 (2) As for (ii), it leads to: [2(1 + λ)(α1γ1µ0 + α0γ0µ1) + α2 0β0 + α0α1µ(1 + λ) + 1 8(1 + 3λ)α2 1γ]e0 + [α2 0β1 + (1 + 2λ)α0µ1γ1 − 2α2 0µ0]e1 + [(2 + 4λ)α2 1γ + (1 + 4λ)α1γ1µ1 − (1 + λ)2α2 1γ0]e2 = 0 then: 2(1 + λ)(α1γ1µ0 + α0γ0µ1) + α2 0β0 + α0α1µ(1 + λ) + 1 8 (1 + 3λ)α2 1γ = 0. (3) α2 0β1 + (1 + 2λ)α0µ1γ1 − 2α2 0µ0 = 0. (4) (2 + 4λ)α2 1γ + (1 + 4λ)α1γ1µ1 − (1 + λ)2α2 1γ0 = 0. (5) The assertion (iii) of the Lemma 1 leads to the equality: −1 2(3 + λ)β0µ1e2 = 0 or: β0µ1 = 0. (6) The assertion (iv) of the Lemma 1 allows to obtain: ( 1 16α0β 2 0 − β3 1)e1 + 1 16α1β 2 0e2 = 0 which gives us: α0β 2 0 = 16β3 1 . (7) α1β0 = 0. (8) Using the assertion (v) of the lemma 1 we have: W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1893 (2λ+ 1 2)γγ1e1 = 0 thus: γγ1 = 0. (9) The assertion (vi) of the lemma 1 makes it possible to obtain the equality: (1− 32λ)α0γ 2e1 + (1− 32λ)α1γe2 = 0, so: α0γ = 0. (10) α1γ = 0. (11) By multiplying (7) by γ we obtain γβ1 = 0 since α0γ = 0. And by reasoning in the same way but with µ1 we obtain µ1β1 = 0; and we get α1β1 = 0 with α1. Finally the product of (7) with µ0 gives µ0β1 = 0. These equalities clearly show that some scalars cannot be non-zero simultaneously; this is the case of γ and γ1, β0 and µ1 for examples and among many others. Suppose that β1 ̸= 0. Then α1 = µ1 = 0 and the equalities (3) and (7) implies that 0 = α2 0β0 = 16β3 1 ,which is absurd. Thus β1 = 0. Thus we have β0α0 = β0α1 = β0µ1 = 0, γγ1 = γα0 = γα1 = 0, α0µ0 = λ2α1γ0, (1 + 4λ)α1γ1µ1 − (1 + λ)2α2 1γ0 = 0, (1 + 2λ)α0γ1µ1 − 2α2 0µ0 = 0, 2α1γ1µ0 + 2α0γ0µ1 + α0α1µ = 0. Let us then distinguish the following cases which satisfy these equalities: 3.1. 1st Case : α0 ̸= 0 and α1 = 0 then β0 = γ = µ0 = γ0µ1 = γ1µ1 = 0. i) µ1 ̸= 0, then γ0 = γ1 = 0 and e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = e1, e21 = 0, e22 = 0, e0e1 = e2, e0e2 = 0, e1e2 = µe0. This case is impossible because A not satisfy identity (x4)2 − ω(x)4x4 = 0. ii) µ1 = µ = γ1 = 0, then e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = e1, e21 = 0, e22 = 0, e0e1 = 0, e0e2 = γ0e0, e1e2 = 0. A verifies the identity x(x2)2 − 4λω(x)x4 + (4λ2 + 2λ− 1)ω(x)2x3 + (2λ− 4λ2)ω(x)3x2 = 0. iii) µ1 = µ = 0, γ1 ̸= 0 then e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = e1, e21 = 0, e22 = 0, e0e1 = 0, e0e2 = γ0e0 + e1, e1e2 = 0. A verifies the identity x(x2)2 − 4λω(x)x4 + (4λ2 + 2λ− 1)ω(x)2x3 + (2λ− 4λ2)ω(x)3x2 = 0. iv) µ1 = 0, γ1γ0µ ̸= 0 then e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = e1, e21 = 0, e22 = 0, e0e1 = 0, e0e2 = γ0e0 + γ1e1, e1e2 = e0. A verifies the identity x(x2)2 − 4λω(x)x4 + (4λ2 + 2λ− 1)ω(x)2x3 + (2λ− 4λ2)ω(x)3x2 = 0. 3.2. 2nd Case : α0α1 ̸= 0, then β0 = β1 = γ = µ0 = γ0 = µ1γ1 = 0. The multiplication table of A becomes: e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α0e1 + α1e2, e21 = 0, e22 = 0, e0e1 = µ1e2, e0e2 = γ1e1,e1e2 = 0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1894 i) µ1 = 0. The multiplication table for A is given as follows : e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = e1 + e2, e21 = 0, e22 = 0, e0e1 = 0, e0e2 = γ1e1,e1e2 = 0. Where e1 is replaced by α−1 0 e1 and e2 by α−1 1 e2. A verifies polynomial identity (x2)3 − ω(x)x(x2)2 − ω(x)2(x2)2 + ω(x)3x3 = 0. ii) µ1 ̸= 0, then γ1 = 0. In this case, the multiplication table for A is given as follows : e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = e1 + e2, e21 = 0, e22 = 0, e0e1 = µ1e2, e0e2 = 0,e1e2 = 0, where e1 is replaced by α−1 0 e1 and e2 by α−1 1 e2. We prove that A checks the polynomial identity x(x2)2 − λω(x)(x2)2 − ω(x)2x3 + λω(x)3x2 = 0. 3.3. 3rd Case : α0 = α1 = 0 then β0µ1 = γγ1 = 0. i) β0γ ̸= 0 and µ1 ̸= 0, then γ1 = 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = β0e0, e22 = γe0, e0e1 = µ0e0, e0e2 = γ0e0, e1e2 = µe0. Then algebra A verifies polynomial identity (x2)2 − ω(x)2x2 = 0. ii) β0 ̸= 0 and γ = 0, then µ1 = 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = β0e0, e22 = 0, e0e1 = µ0e0, e0e2 = γ0e0 + γ1e1, e1e2 = µe0. This case is impossible because A not satisfy identity (x4)2 − ω(x)4x4 = 0. iii) β0 = γ = 0, then e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e0e1 = µ0e0 + µ1e2, e0e2 = γ0e0 + γ1e1, e1e2 = µe0. If µ1 = 0, the algebra A veri- fies the identity (x2 − 2λω(x))3(x2 − 2λω(x))2 − (1 − 2λ)ω(x)2((x2 − 2λω(x))2)2 − (1−2λ)2 2 ω(x)4(x2 − 2λω(x))3 + (1−2λ)3 2 ω(x)6(x2 − 2λω(x))2 = 0. Suppose now, µ1 ̸= 0. Then γ1 = 0 and we can set µ1 = 1 and the multiplication table of A is one of: iii.1) e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e0e1 = e2, e0e2 = e0, e1e2 = 0. iii.2) e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e0e1 = µ0e0 + e2, e0e2 = e0, e1e2 = 0. iii.3) e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e0e1 = e2, e0e2 = e0, e1e2 = µe0. iii.4) e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e0e1 = µ0e0 + e2, e0e2 = e0, e1e2 = µe0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1895 iii.5) e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e0e1 = µ0e0 + e2, e0e2 = 0, e1e2 = µe0. Then the algebra A verifies polynomial identity (x2)3 − (1 + λ)ω(x)2(x2)2 + λω(x)4x2 = 0. iv) β0 = 0, γ ̸= 0, then γ1 = 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = 0, e21 = 0, e22 = γe0, e0e1 = µ0e0 + µ1e2, e0e2 = γ0e0, e1e2 = µe0. If µ1 = 0, then the algebra A verifies polynomial identity 4λ2(x3)2 + 4λω(x)2x3x2 + ω(x)2(x2)2 + 2ω(x)3x3 − (2λ + 1)ω(x)4x2 = 0. If µ1 ̸= 0, then this is impossible because A not satisfy identity (x4)2 − ω(x)4x4 = 0. 3.4. 4th case : α0 = 0 and α1 ̸= 0 then β0 = γ = γ0 = γ1µ0 = γ1µ1 = 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e0e1 = µ0e0 + µ1e2, e0e2 = γ1e1, e1e2 = µe0. i) α0 = γ = γ0 = µ1 = µ0 = 0 and γ1 ̸= 0 e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e0e1 = 0, e0e2 = γ1e1, e1e2 = µe0. We can suppose α1 = γ1 = 1. If µ = 0, the algebra A verifies polynomial identity (x2)2 − 2ω(x)x3 +ω(x)2x2 = 0. If µ ̸= 0, it is impossible because A not satisfy identity (x4)2 − ω(x)4x4 = 0. ii) α0 = γ = γ0 = γ1 = µ0 = µ1 = 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e0e1 = 0, e0e2 = 0, e1e2 = µe0. If µ = 0, the algebra A verifies polynomial identity (x2)3−ω(x)2(x2)2 = 0. If µ ̸= 0, we can set α1 = µ = 1 and the algebra A verifies polynomial identity (x2)3 − (1 + λ)ω(x)2(x2)2 + λω(x)4x2 = 0. iii) α0 = γ = γ0 = γ1 = µ1 = 0, µ0 ̸= 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e0e1 = µ0e0, e0e2 = 0, e1e2 = µe0. If µ = 0, the algebra A verifies polynomial identity (x2)3 − (1 + λ)ω(x)2(x2)2 + λω(x)4x2 = 0.If µ ̸= 0, we can set α1 = µ = 1 and A verifies the same identity. iv) α0 = γ = γ0 = γ1 = µ0 = 0, µ1 ̸= 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e0e1 = µ1e2, e0e2 = 0, e1e2 = µe0. If µ = 0, the algebra A verifies polynomial identity (x2)3 − ω(x)2(x2)2 = 0. If µ ̸= 0, we can set α1 = µ = 1 and the algebra A verifies polynomial identity (x2)3 − (1 + λ)ω(x)2(x2)2 + λω(x)4x2 = 0. v) α0 = γ = γ0 = γ1 = 0, µ0µ1 ̸= 0. e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e0e1 = µ0e0 + µ1e2, e0e2 = 0, e1e2 = µe0. We can set α1 = µ1 = 1. If µ = 0, the algebra A verifies polynomial identity (x2)3 − ω(x)2(x2)2 = 0. If µ ̸= 0, the algebra A verifies polynomial identity (x2)3 − (1 + λ)ω(x)2(x2)2 + λω(x)4x2 = 0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1896 Thus we have the following theorem: Theorem 4. Let A be a train algebra of degree 2 and exponent 4. If the type of A is (2, 1, 1, 0) then we have the algebras whose multiplication tables are given as follows, the products not mentioned are zero. If one or both parameters α and β appear in the multi- plication table, the algebra will be denoted A(α) or A(α, β). 1◦) e2 = e, ee0 = 1 2e0, ee2 = λe2, e0e1 = e2, e0e2 = e0. 2◦) e2 = e, ee0 = 1 2e0, ee2 = λe2, e0e1 = αe0 + e2, e0e2 = e0. 3◦) e2 = e, ee0 = 1 2e0, ee2 = λe2, e0e1 = e2, e0e2 = e0, e1e2 = αe0. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 4◦) e2 = e, ee0 = 1 2e0, ee2 = λe2 e0e1 = αe0 + e2, e0e2 = e0, e1e2 = βe0. For α, α′, β, β′ in K⋆, A(α, β) and A(α′, β′) are isomorph if and only if it exist k ∈ K⋆ such that β′ = k2β, so β′β−1 ∈ (K⋆)2. 4. Classification of algebras of type (2, 0, 1, 1) The type of A being (2, 0, 1, 1) then A0 = 0 and we have : A2 1/2 ⊂ Aλ⊕Aλ, A 2 λ ⊂ A1/2, A2 λ ⊂ A1/2, A1/2Aλ ⊂ A1/2 ⊕Aλ, A1/2Aλ ⊂ A1/2 ⊕Aλ, AλAλ ⊂ A1/2. We can set: A1/2 =< e0 >, Aλ =< e1 >, Aλ =< e2 >, such that e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α0e1 + α1e2, e21 = γe0, e22 = µe0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, e0e2 = γ0e0 + γ1e1. According the type, we use here identities of Lemma (1) The assertion (i) leads to equality: (α0β0(1 + λ) + α1γ0(1 + λ))e0 = 0 so: α0β0(1 + λ) + α1γ0(1 + λ) = 0. (12) The assertion (ii) leads to equality: [α0β1γ0 + α1β0γ1 + 1 8(γα 2 0(1 + 3λ) + 4α0α1ρ + µα1(1+3λ))]e0+(4α0β0λ+2α2 0β0+2α0β1γ1λ+α1γ0α0)e1+(4α2 1γ0λ+2α2 1γ0+2α1β1γ1λ+ α0α1β0)e2 = 0 or: α0β1γ0 + α1β0γ1 + 1 8 (γα2 0(1 + 3λ) + 4α0α1ρ+ µα2 1(1 + 3λ)) = 0. (13) 4α2 0β0λ+ 2α2 0β0 + 2α0β1γ1λ+ α1γ0α0 = 0. (14) 4α2 1γ0λ+ 2α2 1γ0 + 2α1β1γ1λ+ α0α1β0 = 0. (15) The assertion (v) leads to: 2(1− λ)γβ1e2 = 0 so: W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1897 γβ1 = 0. (16) The assertion (vi) leads to: (1− 32λ)α0γ 2e1 + (1− 32λ)α1α1γ 2e2 = 0, so: α0γ = 0. (17) α1γ = 0. (18) The assertion (vii) leads to: 2(1− λ)µγ1e1 = 0 thus: µγ1 = 0. (19) Finally the assertion (vii) allows to obtain (1− 32λ)α0µ 2e1 + (1− 32λ)α1µ 2e2 = 0 so: α0µ = 0. (20) α1µ = 0. (21) Let us distinguish the following cases satisfying the preceding equalities: 4.1. 1st Case: α0 ̸= 0 and α1 = 0 Then γ = µ = β0 = β1γ1 = β1γ0 = 0 and the multiplication table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α0e1, e1e2 = ρe0, e0e1 = β1e2, e0e2 = γ0e0 + γ1e1. i) β1 ̸= 0,γ1 = γ0 = 0, ρ ̸= 0. e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = e1, e1e2 = ρe0 where e1 is re- placed by α−1 0 e1 and e2 is replaced by α−1 0 β−1 1 e2. A verifies the polynomial identity : (x2−2λ̄ω(x)x)3−(λ−2λ̄)ω(x)2(x2−2λ̄ω(x)x)2+(λ+2λ̄−2)ω(x)4(x2−2λ̄ω(x)) = 0. ii) β1 ̸= 0,γ1 = γ0 = ρ = 0 e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2 , e20 = e1, e1e2 = 0 and e0e1 = e2, where e1 is replaced by α−1 0 e1 and e2 by α−1 0 β−1 1 e2. A verifies the identity (x2)2 − 2ω(x)x3 + ω(x)2x2 = 0. iii) β1 = 0 e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = e1 and e1e2 = ρe0, e0e1 = 0, e0e2 = γ0e0 + γ1e1, where e1 is replaced by α−1 0 e1. A verifies the identity (x2)2 − 2ω(x)x3 + ω(x)2x2 = 0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1898 4.2. 2nd Case: α0 ̸= 0 and α1 ̸= 0, then 0 = γ = ρ = µ = β1γ1. i) β0 = β1 = γ0 = γ1 = 0. The multiplication table of A is : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = e1 + e2. Where e1 is replaced by α−1 0 e1 et e2 par α−1 1 e2. A verifies the polynomial identity: x4 − 1 2ω(x)x 3 − 1 2ω(x) 3x = 0. ii) β0 = γ0 = γ1 = 0 and β1 ̸= 0. The multiplication table of A is : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = e1 + e2, e0e1 = e2. Where e1 is replaced by α−1 0 e1 and e2 by α−1 1 e2. A verifies the polynomial identity: x4 − 1 2ω(x)x 3 − 1 2ω(x) 3x = 0 iii) β0 = γ0 = β1 = 0 and γ1 ̸= 0. The multiplication table of A is : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = e1 + e2, e0e2 = e1. Where e1 is replaced by α−1 0 e1 and e2 by α−1 1 e2, so A verifies the polynomial identity : (x2−2λ̄ω(x)x)3− (1+λ)ω(x)2(x2− 2λ̄ω(x)x)2 + λω(x)4(x2 − 2λ̄ω(x)) = 0. 4.3. 3rd Case: α0 = 0 and α1 ̸= 0. Then µ = γ = γ0 = β1γ1 = β0γ1 = 0, so the multiplication table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0+β1e2, e0e2 = γ1e1. We distinguish the algebras whose multiplication tables are: i) γ1 ̸= 0, then β0 = β1 = 0 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = 0, e0e2 = γ1e1. We can set α1 = γ1 = 1. If ρ = 0, A verifies the polynomial identity (x2)2 − 2ω(x)x3 + ω(x)2x2 = 0 ii) γ1 = β1 = β0 = 0, then e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = 0, e0e2 = 0. We can set α1 = 1 and A verifies the polynomial identity : (x2 − 2λω(x)x)3 − (λ̄− 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. iii) γ1 = 0, β1β0 ̸= 0, e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, e0e2 = 0. We can set α1 = 1 and A verifies the polynomial identity: (x2 − 2λω(x)x)3 − (λ̄ − 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄ + 2λ − 2)ω(x)4(x2 − 2λω(x)) = 0. iv) γ1 = β1 = 0, β0 ̸= 0,e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0, e0e2 = 0. We can set α1 = β0 = 1 and A verifies the polynomial identity: (x2 − 2λω(x)x)3 − (λ̄ − 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄ + 2λ − 2)ω(x)4(x2 − 2λω(x)) = 0. v) γ1 = β0 = 0, β1 ̸= 0, e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = α1e2, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β1e2, e0e2 = 0.We can set α1 = β1 = 1 and A verifies the polynomial identity: (x2 − 2λω(x)x)3 − (λ̄ − 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄ + 2λ − 2)ω(x)4(x2 − 2λω(x)) = 0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1899 4.4. 4th case: α0 = α1 = 0 i) α0 = α1 = γ1 = γ0 = β1 = 0. Then the multiplication table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = µe0, e0e1 = β0e0, e0e2 = 0, e1e2 = ρe0. The classification of algebras in this case amounts to the classification of the quadratic form q of polar forms φ defined from A1/2 ⊕ Aλ ⊕ Aλ to K by : φ(e0, e0) = q(e0) = 0, φ(e1, e1) = q(e1) = γ, φ(e2, e2) = q(e2) = µ, φ(e0, e1) = β0, φ(e0, e2) = 0, φ(e1, e2) = ρ . The matrix of q in the basis (e0, e1, e2) is given by: M =  0 β0 0 β0 γ ρ 0 ρ µ  , so det(M) = −β2 0µ. If q is degenerate, then det(M) = 0, so µβ0 = 0. Suppose β0 ̸= 0, then µ = 0 and we have the cases: i.1) γ = ρ = 0. The multiplication table of A is given by : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0. Where e1 is replaced by β−1 0 e1.A verifies the polynomial identity : (x2 − 2λω(x)x)3 − (λ̄− 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. i.2) γ = 0 and ρ ̸= 0. The multiplication table of A is given by : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0, e1e2 = e0. Where e1 is replaced by β−1 0 e1 et e2 par β0ρ −1e2.A verifies the polynomial identity: (x2 − 2λω(x)x)3 − (λ̄ − 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. i.3) ρ = 0 and γ ̸= 0. The multiplication table of A is given by : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0 and e21 = e0. Where e1 is replaced byβ−1 0 e1 et e0 par β2 0γ −1e0. A verifies the polynomial identity: (x2 − 2λω(x)x)3 − (λ̄− 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. i.4) ρ ̸= 0 and γ ̸= 0. The multiplication table of A is given by: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0, e21 = e0 et e1e2 = ρe0, where e1 is replaced by β−1 0 e1 and e0 by β2 0γ −1e0. A verifies the polynomial identity: (x2−2λω(x)x)3− (λ̄− 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. Suppose β0 = 0, then e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = µe0, e0e1 = 0, e0e2 = 0, e1e2 = ρe0. i.5) µ = 0. The multiplication table of A is given by : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = 0, e0e1 = 0, e0e2 = 0, e1e2 = ρe0. A verifies the polynomial identity : (x2 − 2λω(x)x)3 − (λ̄ − 2λ)ω(x)2(x2 − 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. i.6) µγρ ̸= 0. We can set ρ = γ = 1 and the multiplication table of A is given by: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = µe0, e0e1 = 0, e0e2 = 0, e1e2 = e0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1900 i.7) µ ̸= 0, ρ = γ = 0. We can set µ = 1. The multiplication table of A is given by: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e0e1 = 0, e0e2 = 0, e1e2 = 0. A verifies the polynomial identity: (x2−2λ̄ω(x)x)3−(λ−2λ̄)ω(x)2(x2−2λ̄ω(x)x)2+(λ+2λ̄−2)ω(x)4(x2−2λ̄ω(x)) = 0. i.8) µγ ̸= 0, ρ = 0. We can set γ = 1. The multiplication table of A is given by: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = µe0, e0e1 = 0, e0e2 = 0, e1e2 = 0. A verifies the polynomial identity: (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. i.9) µρ ̸= 0, γ = 0. We can set ρ = ρ = 1. The multiplication table of A is given by : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e0e1 = 0, e0e2 = 0, e1e2 = e0. A verifies the polynomial identity: (x2 − 2λ̄ω(x)x)3 − (λ− 2λ̄)ω(x)2(x2 − 2λ̄ω(x)x)2 + (λ+ 2λ̄− 2)ω(x)4(x2 − 2λ̄ω(x)) = 0. If q is regular then det(M) ̸= 0 therefore β0µ ̸= 0 and we have: i.10) γ = ρ = 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0, e22 = e0. Where e1 is replaced by β−1 0 e1 and e0 by µ−1e0. A verifies the polynomial identity: (x3)2+2ω(x)2x4−2ω(x)3x3+ω(x)4x2−2ω(x)5x = 0. i.11) γ = 0 and ρ ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0, e1e2 = ρe0, e22 = e0, where e1 is replaced by β−1 0 e1 and e0 by µ−1e0. A verifies the polynomial identity : (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. i.12) ρ = 0 and γ ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0, e21 = ρe0, e22 = e0, where e1 is replaced by β−1 0 e1 and e0 by µ−1e0. A verifies the polynomial identity: (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. i.13) ρ ̸= 0 and γ ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e0, e1e2 = e0, e21 = e0, e22 = ρe0, where e1 is replaced by β−1 0 e1 and e2 by β0ρ −1e2 and e0 by β2 0γ −1e0. A verifies the polynomial identity: (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. ii) α0 = α1 = γ1 = β1 = 0, γ0 ̸= 0. Then the multiplication table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = µe0, e0e1 = β0e0, e0e2 = γ0e0, e1e2 = ρe0. We can set γ0 = 1. ii.1) Suppose β0 ̸= 0, then we can set β0 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = µe0, e0e1 = e0, e0e2 = e0, e1e2 = ρe0. A verifies the polynomial identity: (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. Suppose β0 = 0, then detM = −γ. Let q be the quadratic of polar forms φ defined from A1/2 ⊕ Aλ ⊕ Aλ to K by: W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1901 φ(e0, e0) = q(e0) = 0, φ(e1, e1) = q(e1) = γ, φ(e2, e2) = q(e2) = µ, φ(e0, e1) = 0, φ(e0, e2) = 1, φ(e1, e2) = ρ. The matrix of q in the basis (e0, e1, e2) is given by: M =  0 0 1 0 γ ρ 1 ρ µ , so detM = −γ. ii.2) If detM = 0, γ = 0 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = µe0, e0e1 = 0, e0e2 = e0, e1e2 = ρe0. A verifies the polynomial identity: (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. ii.3) If detM ̸= 0, we can set γ = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = µe0, e0e1 = 0, e0e2 = e0, e1e2 = ρe0.A verifies the polynomial identity: (x3)2 + 2ω(x)2x4 − 2ω(x)3x3 + ω(x)4x2 − 2ω(x)5x = 0. iii) α1 = α0 = β1 = γ0 = µ = 0 and γ1 ̸= 0. We can set γ1 = 1. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0, e0e2 = e1. iii.1) Suppose β0 ̸= 0, then we can set β0 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = 0, e1e2 = ρe0, e0e1 = e0, e0e2 = e1. Suppose β0 = 0, then detM = −γ. Let q be the quadratic of polar forms φ defined from A1/2 ⊕ Aλ ⊕ Aλ to K by : φ(e0, e0) = q(e0) = 0, φ(e1, e1) = q(e1) = γ, φ(e2, e2) = q(e2) = 0, φ(e0, e1) = 0, φ(e0, e2) = 1, φ(e1, e2) = ρ . The matrix of q in the basis (e0, e1, e2) is given by: M =  0 0 1 0 γ ρ 1 ρ 0 , so detM = −γ. iii.2) If detM = 0, γ = 0 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = 0, e0e2 = e1. A verifies the polynomial identity: (x2−2λ̄ω(x)x)3−(λ−2λ̄)ω(x)2(x2−2λ̄ω(x)x)2+(λ+2λ̄−2)ω(x)4(x2−2λ̄ω(x)) = 0. iii.3) If detM ̸= 0, γ ̸= 0. We can set γ = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = 0, e1e2 = ρe0, e0e1 = 0, e0e2 = e1. A verifies the polynomial identity: (x2 − 2λ̄ω(x)x)3 − (λ − 2λ̄)ω(x)2(x2 − 2λ̄ω(x)x)2 + (λ + 2λ̄ − 2)ω(x)4(x2 − 2λ̄ω(x)) = 0. iv) α1 = α0 = β1 = µ = 0 and γ0γ1 ̸= 0. The table of A is : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = γe0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0, e0e2 = γ0e0 + γ1e1. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1902 iv.1) β0 = γ = 0, then A verifies the polynomial identity: (x2 − 2λ̄ω(x)x)3 − (λ − 2λ̄)ω(x)2(x2 − 2λ̄ω(x)x)2 + (λ+ 2λ̄− 2)ω(x)4(x2 − 2λ̄ω(x)) = 0. iv.2) β0 = 0, ργ ̸= 0, then we can set γ = γ1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = 0, e1e2 = ρe0, e0e1 = 0, e0e2 = γ0e0 + e1. iv.3) β0 = ρ = 0, γ ̸= 0, then we can set γ = γ1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = 0, e1e2 = 0, e0e1 = 0, e0e2 = γ0e0 + e1. iv.4) β0ρ ̸= 0, γ = 0, then we can set β0 = γ1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = e0, e0e2 = γ0e0 + e1. iv.5) β0 ̸= 0, ρ = γ = 0, then we can set β0 = γ1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = e0, e0e2 = γ0e0 + e1. iv.6) β0γ ̸= 0, then we can set β0 = γ = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = e0, e22 = 0, e1e2 = ρe0, e0e1 = e0, e0e2 = γ0e0 + e1. v) α1 = α0 = γ1 = γ0 = γ = 0 and β1 ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = µe0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, e0e2 = 0. v.1) µ = 0, A verifies the polynomial identity: (x2− 2λω(x)x)3− (λ̄− 2λ)ω(x)2(x2− 2λω(x)x)2 + (λ̄+ 2λ− 2)ω(x)4(x2 − 2λω(x)) = 0. v.2) ρ = β0 = 0, µ ̸= 0, then we can set µ = β1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = 0, e0e1 = e2, e0e2 = 0. v.3) ρ = 0, β0µ ̸= 0, then we can set µ = β1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = 0, e0e1 = β0e0 + e2, e0e2 = 0. v.4) β0 = 0, ρµ ̸= 0, then we can set µ = β1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = ρe0, e0e1 = e2, e0e2 = 0. v.5) β0ρµ ̸= 0, then we can set µ = β1 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = ρe0, e0e1 = β0e0 + e2, e0e2 = 0. vi) α1 = α0 = γ1 = γ = 0 and γ0β1 ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = µe0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, e0e2 = γ0e0. W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1903 vi.1) β0 ̸= 0, ρ = µ = 0. We can set γ0 = β1 = 1 and he table of A is : e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = β0e0 + e2, e0e2 = e0. vi.2) β0 ̸= 0, ρ = 0, µ ̸= 0. We can set µ = γ0 = β1 = 1. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = 0, e0e1 = β0e0 + e2, e0e2 = e0. vi.3) ρβ0 ̸= 0, µ = 0. We can set β1 = ρ = γ0 = 1. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = e0, e0e1 = β0e0 + e2, e0e2 = e0. vi.4) β0ρµ ̸= 0. We can set γ0 = ρ = µ = 1 and he table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = e0, e0e1 = β0e0 + β1e2, e0e2 = e0. vi.5) β0 = 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = µe0, e1e2 = ρe0, e0e1 = β1e2, e0e2 = γ0e0. Let q be the quadratic of polar forms φ defined from A1/2 ⊕ Aλ ⊕ Aλ to K by : φ(e0, e0) = q(e0) = 0, φ(e1, e1) = q(e1) = 0, φ(e2, e2) = q(e2) = µ, φ(e0, e1) = β1, φ(e0, e2) = γ1, φ(e1, e2) = ρ . The matrix of q in the basis (e0, e1, e2) is given by: M =  0 β1 γ0 β1 0 ρ γ0 ρ µ , so detM = β1(2γ0ρ− β1µ). vi.6) ρ = µ = 0, then we can set β1 = γ0 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = e2, e0e2 = e0. vi.7) ρ = 0, µ ̸= 0, then we can set β1 = γ0 = µ = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = 0, e0e1 = e2, e0e2 = e0. vi.8) ρ ̸= 0, µ = 0, then we can set β1 = γ0 = 1 and e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = e2, e0e2 = e0. vi.9) ρµ ̸= 0, then we can set β1 = γ0 = µ = 1, e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = e0, e1e2 = ρe0, e0e1 = e2, e0e2 = e0. vii) α1 = α0 = γ0 = µ = γ = 0 and γ1 ̸= 0, β1 ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1904 e0e2 = γ1e1. vii.1) β0 = ρ = 0, we can set β1 = γ1 = 1, the table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = e2, e0e2 = e1. vii.2) β0 = 0, ρ ̸= 0, we can set β1 = γ1. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = e2, e0e2 = e1. vii.3) β0 ̸= 0, ρ = 0. We can set γ1 and the table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = β0e0+β1e2, e0e2 = e1. vii.4) β0ρ ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, e0e2 = γ1e1. This case is impossible because A not verifies the identity (x4)2 − ω(x)4x4 = 0. viii) α1 = α0 = µ = γ = 0 and γ0γ1 ̸= 0, β1 ̸= 0. The table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = β0e0 + β1e2, e0e2 = γ1e1 + γ0e0. viii.1) β0 = ρ = 0, without loss of generality. It suffices to make some basis trans- formations to prove that β1 = 1 and the table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = e2, e0e2 = γ1e1 + γ0e0. viii.2) β0 = 0, ρ ̸= 0, without loss of generality. It suffices to make some basis trans- formations to prove that β1 = 1 and the table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = ρe0, e0e1 = e2, e0e2 = γ1e1+γ0e0. This case is impossible because A not verifies the identity (x4)2−ω(x)4x4 = 0. (for x = e0+e2) viii.3) β0 ̸= 0, ρ = 0, without loss of generality. It suffices to make some basis trans- formations to prove that β1 = 1 and the table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = 0, e0e1 = β0e0 + e2, e0e2 = γ1e1 + γ0e0. This case is impossible because A not verifies the identity (x4)2 − ω(x)4x4 = 0. viii.4) β0ρ ̸= 0, without loss of generality. It suffices to make some basis transfor- mations to prove that ρ = 1 and the table of A is: e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e20 = 0, e21 = 0, e22 = 0, e1e2 = e0, e0e1 = β0e0 + β1e2, e0e2 = γ1e1 + γ0e0. This case is impossible because A not verifies the identity (x4)2 − ω(x)4x4 = 0. Theorem 5. Let A be a train algebra of degree 2 and exponent 4. If the type of A is (2, 0, 1, 1), then we have algebras whose multiplication tables are given as follows, the un- mentioned products being zero, the algebra is denoted A(α) or A(α, β) if α and β are the parameters which appear in the multiplication table: W. A. Zangre, A. Conseibo / Eur. J. Pure Appl. Math, 15 (4) (2022), 1887-1907 1905 1) e2 = e, ee0 = 1 2e0, ee1 = λe1,ee2 = λe2, e20 = e2, e1e2 = αe0, e0e2 = e1. for α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 2) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e21 = e0, e22 = αe0, e1e2 = e0; for α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if α′ = α. 3) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e21 = αe0, e1e2 = βe0, e0e1 = e0, e0e2 = e1; for α, α′, β and β′ in K⋆, A(β) and A(β′) are isomorph if and only if it exist k ∈ K⋆ such that β′ = k2β, so β′β−1 ∈ (K⋆)2. 4) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e21 = e0, e1e2 = αe0, e0e2 = βe0 + e1. for α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 5) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e0e1 = αe0 + e2. 6) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e1e2 = αe0, e0e1 = e2. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 7) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e1e2 = αe0, e0e1 = βe0 + e2. for α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 8) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = αe0 + e2, e0e2 = e0. 9) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e0e1 = αe0 + e2, e0e2 = e0. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if α′ = α. 10) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e1e2 = e0, e0e1 = αe0 + e2, e0e2 = e0. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if α′ = α or α′ = −α. 11) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e1e2 = e0, e0e1 = αe0 + βe2, e0e2 = e0. For α, α′, β and β′ in K⋆, A(β) and A(β′) are isomorph if and only if it exist k ∈ K⋆ such that β′ = k2β, so β′β−1 ∈ (K⋆)2. 12) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e2, e0e2 = e0. 13) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e0e1 = e2, e0e2 = e0. 14) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e1e2 = αe0, e0e1 = e2, e0e2 = e0. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 15) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e22 = e0, e1e2 = αe0, e0e1 = e2, e0e2 = e0. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if α′ = α. REFERENCES 1906 16) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e2, e0e2 = e1. 17) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e1e2 = αe0, e0e1 = e2, e0e2 = e1. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 18) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = αe0 + βe2, e0e2 = e1. For α, α′, β and β′ in K⋆, A(β) and A(β′) are isomorph if and only if it exist k ∈ K⋆ such that β′ = k2β, so β′β−1 ∈ (K⋆)2. 19) e2 = e, ee0 = 1 2e0, ee1 = λe1, ee2 = λe2, e0e1 = e2, e0e2 = αe0 + βe1. For α, α′, β and β′ in K⋆, A(β) and A(β′) are isomorph if and only if it exist k ∈ K⋆ such that β′ = k2β, so β′β−1 ∈ (K⋆)2. 5. Classification of algebras of type (2, 1, 0, 1) The type of A being (2, 1, 0, 1) then Aλ = 0 and we have : A2 1/2 ⊂ A0 ⊕ Aλ, A2 0 ⊂ A1/2 ⊕ A0, A2 λ ⊂ A1/2, A1/2A0 ⊂ A0 ⊕ Aλ, A1/2Aλ ⊂ A1/2 ⊕ A0, A0Aλ ⊂ A1/2. In this way, it is possible to set A1/2 =< e0 >, A0 =< e1 >, Aλ < e2 > and the multiplication table of A is : e2 = e, ee0 = 1 2e0, ee1 = 0, ee2 = λe2; e20 = α0e1 + α1e2, e21 = β0e0 + β1e1, e22 = γe0, e0e1 = µ0e0 + µ1e2, e0e2 = γ0e0 + γ1e1, e1e2 = µe0. The reasoning is similar to that of the type (2, 1, 1, 0). This allows us to obtain the algebras given in the proposition below. Theorem 6. Let A be a train algebra of degree 2 and exponent 4. If the type of A is (2, 1, 0, 1) then we have the algebras whose multiplication table are given as follows, the products not mentioned being zero. If one or both parameters α and β appear in the multiplication table, the algebra will be denoted A(α) or A(α, β). 1◦) e2 = e, ee0 = 1 2e0, ee2 = λ̄e2, e0e1 = e2, e0e2 = e0. 2◦) e2 = e, ee0 = 1 2e0, ee2 = λ̄e2, e0e1 = αe0 + e2, e0e2 = e0. 3◦) e2 = e, ee0 = 1 2e0, ee2 = λ̄e2, e0e1 = e2, e0e2 = e0, e1e2 = αe0. For α and α′ in K⋆, A(α) and A(α′) are isomorph if and only if it exist k ∈ K⋆ such that α′ = k2α, so α′α−1 ∈ (K⋆)2. 4◦) e2 = e, ee0 = 1 2e0, ee2 = λ̄e2 e0e1 = αe0 + e2, e0e2 = e0, e1e2 = βe0. For α, α′, β, β′ in K⋆, A(α, β) and A(α′, β′) are isomorph if and only if it exist k ∈ K⋆ such that β′ = k2β, so β′β−1 ∈ (K⋆)2. References [1] Zangré W. Achile and Conseibo André. On train algebras of degree 2 and exponent 4. Gulf Journal of Mathematics, 13(1):41–53, July 2022. REFERENCES 1907 [2] M. T. Alcalde, Burgueno C., Labra A., and Micali A. Sur Les Algebres de Bernstein. Proceedings of the London Mathematical Society, s3-58(1):51–68, 1989. [3] Worz-Busekros Angelika. Algebras in Genetics, volume 36 of Lecture Notes in Biomathematics. Springer Berlin Heidelberg, Berlin, Heidelberg, 1980. [4] I. Basso, R. Costa, J. Carlos Gutiérrez, and H. Guzzo Jr. Cubic algebra of exponent 2: basic properties. pages 245–258, 1999. [5] Joseph Bayara, André Conseibo, Moussa Ouattara, and Fouad Zitan. Power- associative algebras that are train algebras. Journal of Algebra, 324(6):1159–1176, September 2010. [6] Serge Bernstein. Démonstration Mathématique de la loi de l’hérédité de Mendel. Comptes rendus hebdomadaires des sé, pages = 528-531, year = 1923, note = Paris, Tome 177,. [7] Schafer R. D. Structure of genetic algebras. Amer. 1. Math. vol 71, pages 121–135, 1949. [8] P. Holgate. Selfing in genetic algebras. Journal of the London Mathematical Society, s2-9(4):613–623, April 1975. [9] Bayara Joseph, Conseibo André, Moussa Ouattara, and Micali Artibano. Train al- gebras of degree 2 and exponent 3. Discrete & Continuous Dynamical Systems - S, 4(6):1371–1386, 2011. [10] Holgate P. Jordan algebras arising in population genetics. Proc. Edinburgh Math. Soc., 15:291–294, 1967.