aa r X i v : . [ m a t h . R A ] S e p A remark on n − Jordan homomorphisms
M. El Azhari
Abstract.
Let A and B be commutative algebras and n > n − Jordan homomorphism h : A → B is an n − homomorphism. Mathematics Subject Classification 2010:
Keywords: n − Jordan homomorphism, n − homomorphism.
1. Preliminaries
Let
A, B be two algebras and n > h : A → B iscalled an n − Jordan homomorphism if h ( a n ) = h ( a ) n for all a ∈ A. Also, a lin-ear map h : A → B is called an n − homomorphism if h ( Q ni =1 a i ) = Q ni =1 h ( a i )for all a , ..., a n ∈ A. In the usual sense, a 2 − Jordan homomorphism is a Jordanhomomorphism and a 2 − homomorphism is a homomorphism.It is obvious that n − homomorphisms are n − Jordan homomorphisms. Con-versely, under certain conditions, n − Jordan homomorphisms are n − homo-morphisms. For example, if A, B are commutative algebras and h : A → B is a Jordan homomorphism, then h is a homomorphism: Let x, y ∈ A, since h (( x + y ) ) = h ( x + y ) , a simple calculation shows that h ( xy ) = h ( x ) h ( y ) . Also, it was shown in [2] that if n ∈ { , } , A, B are commutative algebras and h : A → B is an n − Jordan homomorphism, then h is an n − homomorphism.This result was also proved for n = 5 in [3].
2. Result
In [1], Cheshmavar et al claimed the following result:Theorem 2.3 of [1]. Let A and B be commutative algebras, n > , and let h : A → B be an n − Jordan homomorphism. Then h is an n − homomorphism.The proof of this theorem is essentially based on a formula which is not correct:Let h : A → B be a linear map, n > , and consider the maps ψ : A n → B, ψ ( x , ..., x n ) = h ( Q ni =1 x i ) − Q ni =1 h ( x i )and for 2 k n,ϕ k − : A k → B, ϕ k − ( x , ..., x k ) = h (( x + ... + x k ) n ) − h ( x + ... + x k ) n . In [1], the authors claimed that 1 n − ( x , ..., x n ) = P ni,j =1 ,i Let A, B be two algebras, n > , and let h : A → B be an n − Jordan homomorphism. Then P σ ∈ S n ( h ( Q ni =1 x σ ( i ) ) − Q ni =1 h ( x σ ( i ) )) = 0 for all x , ..., x n ∈ A, where S n is the set of permutations on { , ..., n } . Furthermore,if A and B are commutative, then h is an n − homomorphism. Proof. Case n = 2 : Let h : A → B be a Jordan homomorphism and x , x ∈ A. Since h (( x + x ) ) = h ( x + x ) , a simple calculation showsthat h ( x x ) − h ( x ) h ( x ) + h ( x x ) − h ( x ) h ( x ) = 0 , then P σ ∈ S ( h ( x σ (1) x σ (2) ) − h ( x σ (1) ) h ( x σ (2) )) = 0 . If A and B are commutative, we deduce that h is a homomorphism.Case n > h : A → B be a linear map and x , ..., x n ∈ A. We willwrite h (( x + ... + x n ) n ) − h ( x + ... + x n ) n as the sum of expressions con-taining respectively 1 , , ..., n coordinates of ( x , ..., x n ) . Let 1 k n and i , ..., i k ∈ { , ..., n } , i < ... < i k , we denote by ψ ( n ) k ( x i , ..., x i k ) the expressionin h (( x + ... + x n ) n ) − h ( x + ... + x n ) n containing exactly the k coordinates x i , ..., x i k . Then h (( x + ... + x n ) n ) − h ( x + ... + x n ) n = P ni =1 ψ ( n )1 ( x i ) + P ni,j =1 ,i