Derivation on Vinberg Rings

A nonassociative ring which contains a well-known associative ring or left symmetric ring also known as Vinberg ring is of great interest. A method to construct Vinberg nonassociative ring is given; Vinberg nonassociative ring V Nn,m,s is shown as simple; all the derivations of nonassociative simple Vinberg V N0,0,1 algebra defined are determined; and finally in solid algebra it is shown that if θ is a nonzero endomorphism of V N0,0,1 , then θ is an epimorphism.


Preliminaries
Let (A, * , +) be a nonassociative algebra then the antisymmetrized algebra (A − , [, ], +) with the same set A and the Lie bracket [,] is defined as follows: [x, y] = x * y − y * x for any x, y ∈ A − .Choi proposed an interesting problem [9]: Does the equality Aut F (A) = Aut Lie (A − ) hold?The answer is no generally.Any derivation of an algebra A is a derivation of the antisymmetrized algebra A − .He also proposed an interesting problem: Is Der(A) = Der Lie (A − )?If θ is an automorphism of Vinberg ring V N then the Der(V N ) is also an automorphism.For a p-torsion free Vinberg algebra, we do not know Der(A) generally.Our method of finding Der(V N 0,0,1 ) will give a good modification to find Der(A) of an algebra A. The authors have given the description of a 2-torsion free Vinberg (-1,1) ring R in [2].They have shown that if every nonzero root space of R − for S is one-dimensional where S is a split abelian Cartan subring of R − which is nil on R then R is a Lie ring isomorphic to R − .In this paper we extend the results of [2] 10 G. Lakshmi Devi and K. Jayalakshmi to V N 0,0,1 algebra.A nonzero endomorphism of V N 0,0,1 is an epimorphism.A nonassociative ring R is called a Vinberg ring if it satisfies the identity where (x, y, z) = (xy)z − x(yz) for x, y, z ∈ R. Throughout this paper Z and N are the sets of integers and non-negative integers respectively.Let (R, +, •) be a Vinberg ring and ∂ a derivation of R.Let F [x 1 , ..., x m+s ] be the polynomial ring on the variables x 1 , ..., x m+s .Let g 1 , ..., g n be given polynomials in F [x 1 , ..., x m+s ].For n, m, s ∈ N , we define the F -algebra F n,m,s = F [e ±g1 , ..., e ±gn , x ±1 1 , ..., x ±1 m , x m+1 , ..., x m+s ] with the standard basis [3] and with the obvious addition and the multiplication [3,4,6,7].We define the F -Vector space V N (n,m,s) with the standard basis where ∂ w is the usual partial derivative with respect to x w .We define the multiplication * on V N n,m,s as for f ∂ w and h∂ u ∈ V N n,m,s .Thus we can define the Vinberg-type nonassociative ring V N n,m,s with the multiplication in (1.4) and with the set V N (n,m,s) .The nonassociative ring V N n,m,s (s ≥ 2) is not a Vinberg ring as it does not satisfy (1.1).But V N 1,0,1 is a Vinberg ring.For any element l = e a1g1 •••e angn x i1 1 •••x im+s m+s ∂ w (1 ≤ w ≤ m+s), let us call i 1 , ..., i m+s the powers of l.An ideal in a nonassociative ring is a two sided ideal of it.In this paper, we prove that the ring V N n,m,s is simple.The ring V N n,m,s is not a Jordan ring.The right annihilators of V N n,m,s is the sub ring , we know that the algebra V N n,m,s is not power associative.

Main results
Theorem 2.1.The algebra V N n,m,s is simple.
Proof: First we show that the ideal ∂ w generated by ∂ w , where 1 Let us prove the theorem by induction on the number of distinct homogeneous components of any non -zero element l in I. Assume that l has only one (0, ..., 0) -homogeneous component.We may assume that l has positive powers from l 2 = l 1 * l ∈ I by taking an appropriate element l 1 ∈ V N n,m,s .We can get the element by taking appropriate q 1 , ..., q t , 1 ≤ q 1 , ..., q t ≤ m + s, and applying ∂ q1 , ..., ∂ qt in (2.1) with appropriate times, where c is a non-zero scalar.This implies that V N n,m,s = ∂ w ⊂ I .Therefore, we have the theorem.Assume that l is in the (a 1 , ..., a n ) -homogeneous component, then 0 = e −a1g1 • • • e −angn ∂ t * l ∈ V N (0,...,0) by taking an appropriate t, 1 ≤ t ≤ m + s, where atleast one of a 1 , ..., a n is not zero.In this case, we have the theorem already.We may assume that l has n homogeneous components by induction.Let us assume that l has the (0, ..., 0, a w , ..., a n )-homogeneous component such that a w = 0.By taking where i 1 , ..., i m+s are sufficiently large positive integers so that l 1 * l ∈ I has positive powers.By taking an appropriate Therefore, we have the theorem by induction.✷ 3. Derivations of V N 0,0,1 The right annihilator of l in V N n,m,s is the set {l 1 ∈ V N n,m,s |l * l 1 = 0} and similarly the left annihilator is the set Lemma 3.2.The left annihilator of ∂ is V N 0,0,1 , and the right annihilator of ∂ is {c∂|c ∈ F }.
Proof: The proof is straightforward by the definitions of the right and left anni- where D 1 and D 2 are the derivations of V N 0,0,1 in Remark 3.1.
Proof: Let D be any derivation of V N 0,0,1 .Then By Lemma 3.1, we have Also, we have By (3.3) and (3.4), we have On the other hand, we have This implies that C(n, 0) = 0. Therefore, we have proved that This shows that D = C(0)D 2 + C(1)D 1 and completes the proof of the theorem.✷

Solid Algebras
Let A be an F -algebra.Let End F (A) be the set of all F -endomorphisms of A, and Aut F (A) the set of all automorphisms of A. An F -algebra A is solid if every non-zero endomorphism of A is surjective.Proof: It is straightforward by the fact that A is a simple algebra and the definition of the solid algebra.✷ Lemma 4.2.For any θ ∈ End F (V N 0,0,1 ) , if θ(∂) = 0, then θ is the zero map of V N 0,0,1 .
Proof: It is straightforward by Corollary 4.5.✷ By Corollary 4.6, we know that V N 0,0,1 is solid.
Proof: Since T s1 is the unique maximal right annihilator of V N n,m,s , θ(T s1 ) = T s1 holds for any θ ∈ Aut(V N n,m,s ).✷

Proposition 4 . 1 .
A simple algebra A is solid if and only if End F (A) = {0} ∪ Aut F (A).