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Flexible Lie-admissible Superalgebras of Vector Type



G. Lakshmi Devi and K. Jayalakshmi
 
ABSTRACT

Background: First examples of simple nonassociative superalgebras were constructed by Shestakov in (1991 and 1992). Since then many researchers showed interest towards the study of superalgebras and superalgebras of vector type. Materials and Methods: Multiplication in M is uniquely defined by a fixed finite set of derivations and by elements of A. The types of derivations used in this article to obtain the results are the near derivation δx,y : a ↦ (a, x, y) the derivation D : a ↦ (x, a, x) and the derivation Dij : a ↦ (xi ,a, xj) Results: The flexible Lie-admissible superalgebra FFLSA[φ; x] over a 2, 3-torsion free field Φ on one odd generator e is isomorphic to the twisted superalgebra B0 (Φ[Γ], D, γ0) with the free generator . In a 2, 3-torsion free flexible Lie-admissible superalgebras of vector type F, the even part A is differentiably simple, associative and commutative algebra and the odd part M is a finitely generated associative and commutative A-bimodule. Conclusion: A connection between the integral domains, the finitely generated projective modules over them, the derivations of an integral domain and the flexible Lie-admissible superalgebras of vector type has been established. If A is an integral domain and M = Ax1+…+Axn be a finitely generated projective A-module of rank 1, then F (A, Δ, Γ) is a flexible Lie-admissible superalgebra with even part A and odd part M provided that the mapping M = Axi+…+Axn is a nonzero derivation of A into the A-module (M⊗A M)*, Δ = {Dij |i, j = 1,…, n} is a set of derivations of A where Dij (a) = ā (x⊗xj).

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  How to cite this article:

G. Lakshmi Devi and K. Jayalakshmi, 2017. Flexible Lie-admissible Superalgebras of Vector Type. Asian Journal of Algebra, 10: 1-9.

DOI: 10.3923/aja.2017.1.9

URL: https://scialert.net/abstract/?doi=aja.2017.1.9
 
Received: July 16, 2016; Accepted: October 19, 2016; Published: December 15, 2016


Copyright: © 2017. This is an open access article distributed under the terms of the creative commons attribution License, which permits unrestricted use, distribution and reproduction in any medium, provided the original author and source are credited.

INTRODUCTION

Okubo and Myung1 gave the classification of simple flexible Lie-admissible algebra A such that A‾ is reductive and the Levi-factor of A‾ is simple. Later using the results of Okubo and Myung1 and Benkart and Osborn2 determined flexible Lie-admissible algebras A by module approach for which the radical A‾ is abelian. Kamiya and Okubo3 constructed some class of Lie superalgebras from non-super flexible Lie-admissible algebras. They concentrated especially on the case of any associative algebra giving rise to Lie superalgebras.

In Shestakov4, the concept of a (-1, 1)-superalgebra of vector type was introduced and it was proved that a Jordan superalgebra of vector type may be obtained as the (super) symmetrized algebra A(+) of a (-1, 1)-superalgebra of vector type A. Furthermore, the (-1, 1)-superalgebras AVF: = A(V, Φ, τ, λ) and the Jordan superalgebras JVF: = J(V, Φ, τ, λ) of vector fields on a line associated with an additive homomorphism τ: V→Φ of abelian groups and partial map λ: V→V were introduced in Shestakov4. It was proved that if V ≠ 0 and the map τ is injective, then the superalgebras AVF and JVF are prime. These superalgebras were then used to construct prime degenerate (-1, 1) and Jordan algebras.

Simple (-1, 1)-superalgebras were obtained from the associative-commutative algebras which are differentiably simple with respect to a derivation. The 2, 3-torsion free simple (-1, 1)-superalgebras were described in Shestakov5. The (-1, 1)-superalgebras obtained in Shestakov5 are also called the (-1, 1)-superalgebras of vector type. Zhelyabin6 and Zhelyabin and Shestakov7 described the product in M by some fixed finite sets of derivations and the elements of A for (-1, 1)-superalgebras.

In terms of the Lie algebra of derivations of an integral domain, some necessary and sufficient conditions were found by Zhelyabin8, which allowed him to construct the Jordan superalgebras of vector type whose even part is the initial integral domain. Zhelyabin9 studied analogous questions for the (-1, 1)-superalgebras. In particular, he gave some new examples of simple and prime (-1, 1)-superalgebras of vector type whose odd part is generated as a module by two elements in the case of simple superalgebras and by an arbitrary number of elements in the case of prime superalgebras. Some properties of the universal envelopings of the simple Jordan superalgebras of vector type were described as well. In particular, he proved that the even part of the universal enveloping contains two central orthogonal idempotents such that their sum is equal to unity.

Zhelyabin9 described simple nonassociative 2, 3-torsion free (-1, 1)-superalgebras. If B has a positive characteristic, which is the only possibility in the finite dimensional case, then the even part A is local and B is isomorphic to B (Γ, D, γ). He posed an open question that whether M is always generated by one element as an A-bimodule. He also noticed that if there exists a simple (-1, 1)-superalgebra B which does not satisfy this condition and is therefore not isomorphic to B (Γ, D, γ) the attached superalgebra B+ will infact exemplify a new simple Jordan superalgebra.

In the present article using the constraints of Shestakov5 and Zhelyabin9, a description of a simple 2, 3-torsion free flexible Lie-admissible superalgebra is obtained. It is proved that the flexible Lie-admissible superalgebra FFLSA [φ; x] over a 2, 3-torsion free field Φ on one odd generator e is isomorphic to the twisted superalgebra B0 (Φ [Γ], D, γ0) with free generator . The even part A in such a superalgebra F is necessarily an associative and commutative algebra and the odd part M is finitely generated associative and commutative A-bimodule over A, which is a projective module of rank 1. Multiplication in M is defined by a fixed set of derivations and by elements of A. If the bimodule M is one-generated, i.e., M = Am for a certain m∈M, then F is isomorphic to a twisted superalgebra of vector type B (Γ, D, γ).

MATERIALS AND METHODS

A superalgebra R = R0⊕R1 is a Z2-graded algebra. A typical example of a superalgebra is the associative Grassmann algebra G = G0⊕G1 on a countable set of generators with a natural Z2-grading. If is an arbitrary homogeneous variety of algebras then R = R0⊕R1 is called an if its Grassmann envelope G (R) = G0⊗R0⊕G1⊗R1 belongs to . In particular R = R0⊕R1 is a flexible Lie-admissible superalgebra if and only if it satisfies the (super) identities:

(1)

(2)

where, x, y, z∈R0∪R1 and |r|∈{0, 1} is the parity index of an homogeneous element r: |r| = i for r∈Ri; [x, y]s = xy-(-1)|x||y| yx is the supercommutator of homogeneous elements x, y.

Let flexible Lie-admissible superalgebra satisfy the following two identities:

(*)

(**)

In a flexible Lie-admissible algebra A, the center defined by U (A) = {u∈A|[u, x] = 0, for all x∈A} is the commutative center. Throughout this study A is an associative bimodule of U (A).

Any algebra satisfies Teichmuller and semi Jacobi identities:

(3)

(4)

For r, s∈A and x, y∈M, Eq. 1 and (*) yield:

(5)

One can easily check that Eq. 1 and (*) gives rise to:

(6)

Anderson and Outcult10 have derived an important identity for r, s, x, y∈A∪M, (r, [x, y], s) = 0 for antiflexible rings. Applying the same argument for flexible Lie-admissible superalgebras, one can obtain:

(7)

One can easily check that flexible ring with (*) satisfies:

(8)

Let u∈U(A), a, x, y, z∈A. The following identities are valid in flexible Lie-admissible algebras. From Eq. 4 and 5:

(9)

From Eq. 5 and 1:

(10)

From Eq. 5:

(11)

From Eq. 6 and 8:

(12)

RESULTS

Free flexible Lie-admissible superalgebra FFLSA [φ; x]: Let Φ[Γ] be the algebra of polynomials on a countable set of variables Γ = {γ0,…, γn,…} and let D be the derivation of Φ[Γ] defined by the condition D(γi) = γi+1, i = 0, 1,… consider the twisted superalgebra of vector type B(Φ[Γ], D, γ0). Let Φ0[Γ] be a subalgebra of polynomials without constant terms, then the subspace is a subsuperalgebra of B(Φ[Γ], D, γ0) and is denoted by B0(Φ[Γ], D, γ0). It can be checked that it is generated by the odd element .

Preliminary lemmas
Lemma 1: e2∈U (F), where, U (F) is the commutative center.

Proof: Identity Eq. 2 implies:

3(e, e, e) = 0, (e2, e, e)-(e, e2, e)+(e, e, e2) = 0

that is:

(13)

(14)

Identity Eq. 3 in the form of generators are:

(e3, e, e) = -(e2, e2, e)+(e2, e, e2) = e2(e, e, e)+(e2, e, e)e,
(e2, e2, e)-(e, e3, e)+(e, e, e3) = e(e, e2, e)+(e, e, e2)e,
(e2, e, e2)-(e, e2, e2)+(e, e, e3) = e(e, e, e2)+(e, e, e)e2

which from Eq. 5 implies:

(15)

(16)

(17)

(18)

But the Jacobi identity Eq. 2 implies:

(19)

Hence by Eq. 15 one can see that 2(e2, e2, e)+2(e2, e, e)e+ (e, e3, e) = 0 and eventually using Eq. 17, 16, 14, 5 and 18, one obtains:

(20)

In particular:

[e3, e2] = 0 = (e, e2, e2)

Let p(x) be a monomial of degree n>3 in x. And suppose that [q(x), x2] = 0 for every monomial q(x) of degree less than n and flexible Lie-admissible superalgebra A satisfies the identity [[r, s], D(A)] = 0 where, D(A) is the associator ideal of A. Then in F, [[x, x]s, D(F)] = [2x2, D(F)] = 0 holds. Now, [p(x), x2] = [x2(x2 q(x)), x2], where, deg. q(x)≥0. Replacing x2 by r and q(x) by q, applying Eq. 4, induction assumption, (*) and Eq. 5 yields:

[p(x), x2] = [r(rq), r]
= r[rq, r]+[r, r]rq+(r, rq, r)-(r, r, rq)+(r, r, rq)
= (r, r, rq) = (r, r, q)r

The above arguments show that (r, r, q) = [rq, r] = 0, which proves the lemma.

Lemma 2: Let F be a flexible Lie-admissible superalgebra. Then [[F, F]s, F]s = 0.

Proof: It is sufficient to show that p(x)∈U(F) for every monomial p of even degree. This can be done by using induction on degree of p. One can check that the square of an ideal is an ideal in a flexible Lie-admissible algebra satisfying (*), it may be assumed that p = qx or p = xq, where, q is a monomial of odd degree. On the same lines again assume that q = tx or q = xt for some even t∈F. By lemma 1, x2∈U(F). By theorem 13.10 in Zhevlakov et al.11, x2∈U(F) implies that:

(21)

Applying this to q, we have (q∘x)∈U(F). So it suffices to prove that (tx)x and x(xt)∈U(F). By induction t∈U(F), thus tx2, x2t∈U(F) and hence Eq. 8 gives:

So, (tx)x∈U(F) as (tx)x = tx2+(t, x, x).

Applying the same argument one obtains x(xt) = x2t-(x, x, t) = x2t+(t, x, x)∈U(F).

Theorem 1: A 2, 3-torsion free flexible Lie-admissible superalgebra FFLSA[φ; x] over a field Φ on one odd generator e is isomorphic to the superalgebra B0 (Φ[Γ], D, γ0) with the free generator .

Proof: Let A = F0, then F1 = M1x and F = A⊕M1x, where by assumption A is an associative and commutative Φ-algebra and M1x is a commutative and associative A-module generated by x. If then D is a derivation of A. But and since it is sufficient to prove that is sub superalgebra of F. Since A0⊆U(F), for all r, s∈A, it can be seen that r⋅sx = sx⋅r = (rs)x. Now the product of the elements from M1x can be calculated:

(22)

Hence, is a subsuperalgebra of F and , A0 = A = F0. Also Eq. 21 shows that the superalgebra F is a homomorphic image of the superalgebra B0 (M1, D, x2) under the homomorphism Consider the homomorphism Obviously, this is a homomorphism of differential algebras and it can be extended to a homomorphism of superalgebras : B0(Φ[Γ], D, γ0)→B0(M1, D, x2). The composition map π∘ is onto map from B0 (Φ[Γ], D, γ0) to F. As F is a free superalgebra generated by x, this map is an isomorphism.

Lemma 3: Let F = A+M be a flexible Lie-admissible superalgebra. Then for any x, y∈M the mapping is a near derivation of A.

Proof: Equation 5 and 7 yield:

δx, y(rs) = (rs, x, y) = r(s, x, y)+(r, x, y)s+(r, x∘y, s)
= r⋅δx, y(s)+δx, y(r)⋅s+(r(x∘y))s-r((s∘y)s)
= r⋅(δx, y-Lx∘y)(s)+(δx, y+Rx∘y)(r)⋅s

Hence, the lemma is proved.

Lemma 4: If F is simple then A is Δ-simple, that is A does not contain proper Δ-invariant ideals.

Proof: Assume that I ≠ 0 is an ideal of A such that (I, M, M)⊆I. Since F is simple, = F where, is the ideal generated by I in F. Hence, M = IM+MI. The M+M2 is a graded ideal of B; therefore M2 = A. To prove that M2 = (IM+MI)2⊆I, from which I = A consider:

IM⋅M⊆I⋅MM+(I, M, M)⊆I

Let x, y∈M and i, j∈I; xi∘jy ≡ [x, i]∘jy = -jy∘[x, i] ≡ 0 by (**). The super-semi Jacobi identity Eq. 4 along with (**) gives:

Thus, M2⊆I and I = A.

Remark 1: Suppose that F = A+M is a flexible Lie-admissible superalgebra with its even part A not containing non zero Δ-invariant nil ideals. Then (A, B, A) = [A, B] = 0.

Theorem 2: Let F = A+M be a flexible Lie-admissible superalgebra. Then (A, B, A) = [A, B] = 0 there exists x1, x2,…, xn∈M such that M = Ax1+Ax2+…+Axn and the product in M is defined by rxi⋅sxj = γij⋅rs+2Dij (r)s+Dij (s)r, i, j = 1,.., n, where, γij∈A, Dij = Dji∈DerA satisfy the conditions:

(23)

(24)

for any i, j, k = 1,…, n, r∈A.

Proof: First, to prove that A satisfies the assumptions of the above remark, by lemma 4, A does not contain proper Δ-ideals and so it is sufficient to show that A itself is not nil. Since, A satisfies the identity (**), if A were nil, it would be locally nilpotent by Eq. 4 which is impossible by Shestakov12, lemma 5. Thus A is associative and commutative and M is an associative and commutative A-bimodule.

Next to show that (A, M, M) = A, for any x, y∈M and r∈A, (*) and (**) gives:

(xr)y = (rx)y = (r, x, y)+r(xy) = (r, x, y)+(xy)r = (r, x, y)+2x(ry)-(xr)y

which yields:

(25)

Assume (A, M, M) = 0; then (M, A, M) = 0 by Eq. 25. This yields A⊆Z0, where Z = Z(F) is the center of F. Since F is simple, Z0 is a field and F may be treated as a superalgebra over it. Let x, y∈M and xy ≠ 0. Then, by Eq. 1:

(xy)z = x(yz)+(zy)x-z(yx), for any z∈M

Which implies that If M = Z0⋅x then (F, F, F)⊆Z0⋅ (x, x, x) = Z0⋅[x2, x] = 0, a contradiction. Now, let M = Z0⋅ x+Z0⋅y, x2 = α, y2 = β, xy = 0, yx = η. Hence, α = 0, η = 0 since (x, x, y) = (y, x, x,). Similarly, (x, y, y) = (y, y, x) implies that β = 0, η = 0. Hence, α = β = γ = 0 and M2 = 0, a contradiction. Thus (A, M, M) is a nonzero Δ-subspace of A. From super-linearized Eq. 6, it follows that (A, M, M) is an ideal of A and hence (A, M, M) = A.

It is known that for any x, y∈M, the mapping is a derivation of A. Thus A is differentially simple and by Posner13 it contains unity 1. For any derivation δ∈Δ, δ(1) = 0. Moreover, (x, y, 1) = (1, y, x) = 0 for any x, y∈M. Hence, 1∈Z(F) and so 1 is unity in F. Since 1∈(A, M, M), there exist xi, yi∈M and ri∈A, i = 1,…, n, such that:

(26)

For any r∈A and x, y, z∈M, by super Eq. 6 and the associativity of the A-bimodule M, one has:

(27)

Substituting r = ri, x = xi, y = yi and then summing the resulting equations over i, by Eq. 26 one can obtain:

Thus M = Ax1+…+Axn and hence F = A+Ax1+…+Axn. Let Dij be the mapping By Eq. 25, is a derivation of A. For any r, s∈A, super Eq. 5 and 25 yields:

rxi⋅sxj = (rxi⋅s)xj-(rxi, s, xj)
                   = (rs⋅xi)xj-r(xi, s, xj)+(r, s, xj)xi
                    = rs⋅xixj+2 (xi, rs, xj)-r (xi, s, xj)
           = rs⋅xixj+2Dij (rs)-rDij (s)
          = γij⋅rs+2Dij (r)s+rDij (s)

where, γij = xixj∈A.

Equalities 23 and 24 are consequences of Eq. 1 and (*).

Remark 2: If there exists x∈M such that (A, x, M) = A or A is a local algebra, then B is isomorphic to a superalgebra B (Γ, D, γ), where Γ = Γ0 is a commutative and associative D-simple algebra with 0 ≠ D∈Der Γ and γ∈Γ.

Remark 3: If A is a local algebra with a maximal ideal J, the ideal cannot contain (A, M, M) since otherwise it would be a nonzero Δ-invariant ideal in A. Hence, (A, M, M)⊈J and there exists x∈M such that (A, x, M)⊈J. By super Eq. 5 and (*), for any a, b∈A and y∈M we have (a, x, y)b = (a, b, y)x+(a, x, by)+(a, b, xy) = (a, x, by). Consequently, (A, x, M) is an ideal in A and (A, x, M) = A.

Flexible Lie-admissible superalgebra F(A, Δ, Γ): As shown in theorem 2, given x, y∈M, the mapping, defined by the rule is a derivation. For all r, s∈A and x, y∈M:

sDx, y (r) = Dsx, y (r) = Dx, syy (r), Dx, y (r) Du, v (r) = Du, y (r) Dx, v (r)

Since F is a flexible superalgebra, Dx, y = Dy, x. Also, Dij = Dxi xj for i, j = 1,…, n.

From Dij (r)xk = Dik (r)xj one can deduce that:

Dij (r)xk⋅xl = Dik (r)xj⋅xl

and:

xl⋅Dij (r)xk = xl⋅Dik (r)xj for every r∈A

Let M* = HomA(M, A). Given r, define by setting Similarly the linear mapping given by is a derivation of A into the A-module (M⊗AM)* Zhelyabin8.

In what follows, all rings are considered as algebras over a field of characteristic 0. Now, let A be an associative-commutative ring with unity and without zero divisors (i.e., an integral domain) and let M be a finitely generated projective A-module of rank 1. Let be a nonzero mapping. Assume that:

for all r, s∈A. By the definition of ‾, each pair x, y∈M gives a derivation Then Zhelyabin8:

(28)

for r∈A and x, y, z∈M.

Let x1,…, xn be some generators of the A-module M, i,e., M = Axn+…+Axn. Put Then Dij = Dji.

Lemma 5: Dij (r) Dkl = Dkl (r) Dij and Dij (r) Dkl = Dkj (r) Dil for every r∈A.

Proof: Let r, s∈A. Then:

Similarly:

Dij (r) Dkl = Dkj (r)Dil

Fix some elements γij of A, where i, j = 1,…, n.

Lemma 6: Assume that:

(29)

(30)

for all i, j, k, l = 1,…, n. Then Eq. 23 holds, i.e., γij xkik xj = (γjkkj)xi for all i, j, k, = 1,…, n.

Proof: From identities Eq. 29 and 30 it can be seen that:

kl-2γlk) Dij = (γjl-2γlj) Dik

This implies:

for every r∈A. Therefore:

By the conditions on A and M Zhelyabin8:

kl-2γlk)xj = (γjl-2γlj)xk

By Eq. 30 and 28:

2DklDij = -γkl Dijjl Dik+2Djl Dik = -γkl Dijjl Dik+2Dlj Dki
= -γkl Dijjl Diklj Dkiij Dkl+2Dij Dkl

Then:

2[Dkl, Dij] = -γklDij+(γjllj)DikijDkl

for all i, j, k, l = 1, …, n. Hence:

2[Dlk, Dij] = -γlkDij+(γjkkj)DliijDlk

Thus:

2[Dkl, Dij]-2[Dlk, Dij] = (γlkkl) Dij+(γjllj)Dki+(γkjjk)Dli = 0

Similarly:

lkkl)xj+(γjllj)xk+(γkjjk)xl = 0

Then by the above:

lkxjljxk+(γkjjk)xl = 0

Hence:

γljxklkxj = (γjkkj)xl

Therefore:

γljxkikxj = (γjkkj)xi

Lemma 7: Assume that:

γlkDij+DlkDij = γljDik+DljDik, γijxkikxj = (γjkkj)xi

for all i, j, k, l = 1,…, n. Then Eq. 30 holds, i.e.,

γklDij+2DklDij = γjlDik+2DjlDik

for all i, j, k = 1,…, n.

Proof: By Eq. 28 and 29:

γlkDij+DlkDij = γljDik+DljDik, γjiDkl+DjiDkl = γjlDik+DjlDik

Therefore by Eq. 28:

[Dij, Dkl] = γlkDijjiDkl+(γjllj) Dik

Then with l = j:

[Dij, Dkj] = γjkDijjiDkj

By the hypothesis of the lemma:

By proposition Eq. 5 of Zhelyabin8:

ijji)Dlk+Dij Dlk = (γljjl) Dik+Dlj Dik

Since γjiDkl+DjiDkl = γjlDik+DjlDik, by summing up the last two equalities and by replacing i by k and j by l, it is obvious that:

γkl Dij+2Dkl Dij = γjl Dik+2Djl Dik

Similarly the following lemma may be proved.

Lemma 8: Assume that:

for all i, j, k, l = 1,…, n. Then Eq. 29 holds, i.e., for all i, j, k = 1,…, n:

γlk Dij+Dlk Dij = γlj Dik+Dlj Dik

Lemma 9: Assume that:

(31)

(32)

for all i, j, k, l = 1,…, n. Then Eq. 29 and 30 hold.

Proof: For every r∈A., from Eq. 31 and 32:

Then:

Hence:

By Eq. 31 and corollary of Zhelyabin8:

Therefore:

Let G (l.k.i, j) = Dlk Dij-Dlj Dik. Now for every r∈A, lemma 5 gives:

Since, A is an integral domain:

Hence, γlkDij+DlkDij = γljDik+DjDik. Analogously, Eq. 30 may be proved.

Corollary 1: Equation 29 and 30 are equivalent to Eq. 31 and 32.

Proof: By lemma 9 it is sufficient to show that Eq. 29 and 30 imply Eq. 31 and 32.

Assume Eq. 29 and 30. Then lemma 6 implies:

As it was shown in lemma 7:

So, Eq. 31 is valid.

As it was shown in lemma 6:

Now replacing j by k and l by j:

This implies:

i.e., Eq. 32 is valid.

Theorem 3: Let A be an integral domain. Let M = Ax1+…+Axn be a finitely generated projective A-module of rank 1. Assume that the mapping is a nonzero derivation of A into the A-module (M⊗AM)* and Δ = {Dij|i, j = 1,…, n} is a set of derivations of A, where Dij (a) =r̄(xi⊗xj). Fix the subset Γ = {γij∈A, i, j = 1,…, n} of A. Consider the vector space F(A, Δ, Γ) = A⊕M and equip it with the structure of a Z2-graded algebra putting:

where, r, s∈A and rs is the product of r and s in A. Assume that the derivations Dij and the elements of Γ satisfy Eq. 31 and 32 then F(A, Δ, Γ) is a flexible, Lie-admissible Superalgebra with even part A and odd part M.

Proof: First to show that the product in F(A, Δ, Γ) is correctly defined, let r1x1+…+rnxn = 0. Then for every fixed k = 1,…, n.

Then by lemma 9, Eq. 29 holds. So:

On the other hand, by Eq. 30:

Hence:

Therefore:

Since A is an integral domain:

Hence:

Therefore, the product in F(A, Δ, Γ) is correctly defined.

Hence F(A, Δ, Γ) is a flexible Lie-admissible superalgebra.

DISCUSSION

Kamiya and Okubo3 constructed some class of Lie superalgebras from nonsuper flexible Lie-admissible algebras without resorting to (-1, -1)-freudenthal-kantor tiple systems. Their motivation is a base for constructing the flexible Lie-admissible Grassmann algebra and hence the free unital flexible Lie-admissible superalgebra with one odd generator is isomorphic to the flexible Lie-admissible superalgebra of vector type B(Φ(Γ), D, γ0) where, D(γi) = γi+1 for i = 0, 1,… which was analogously proved for (-1, 1) superalgebras of vector type in Zhelyabin9.

If the superalgebra is not a superalgebra of nondegenerate bilinear superform then it’s even part A is differentiably simple algebra and the odd part M is a finitely generated projective A-module of rank 1. Yuan14 restricted his studies on derivations of differentiably simple associative commutative algebra of n-torsion free (n>0). Whereas in this study remarks 2 and 3 to theorem 2 state that every simple n-torsion free (n>3) nonassociative flexible Lie-admissible superalgebra is local algebra and hence isomorphic to the superalgebra B(Γ, D, γ). Pchelintsev and Shestakov15 proved that (-1, 1) monster constructed by Pchelintsev16 generates the same variety of algebras as the Grassmann (-1, 1) algebra. Following Pchelintsev16 flexible Lie-admissible monster may be constructed.

CONCLUSION

•  A connection between the integral domains, finitely generated projective modules over them, derivations of an integral domain and the flexible Lie-admissible superalgebras of vector type has been established
If A is an integral domain and M = Ax1+…+Axn be a finitely generated projective A-module of rank 1, then F(A, ΔΓ) is a flexible Lie-admissible superalgebra with even part A and odd part M provided that the mapping is a nonzero derivation of A into the A-module (M⊗AM)*, Δ = {Dij|i, j = 1,…, n} is a set of derivations of A where, Dij(a) = ā (xi⊗xj)
For F = A+M, a flexible Lie-admissible superalgebra the derivation Dij is half of δi,j
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