aa r X i v : . [ m a t h . R A ] M a y Algebras of quotients of Hom-Lie algebras
Chenrui Yao, Liangyun Chen
School of Mathematics and Statistics, Northeast Normal University,Changchun 130024, CHINA
Abstract
In this paper, we introduce the notion of algebras of quotients of Hom-Lie algebrasand investigate some properties which can be lifted from a Hom-Lie algebra to its algebraof quotients. We also give some necessary and sufficient conditions for Hom-Lie algebrashaving algebras of quotients. We also examine the relationship between a Hom-Lie algebraand the associative algebra generated by inner derivations of the corresponding Hom-Liealgebra of quotients. Moreover, we introduce the notion of dense extensions and get aproposition about Hom-Lie algebras of quotients via dense extensions.
Keywords:
Hom-Lie algebras; Algebras of quotients; dense extensions.
Algebras where the identities defining the structure are twisted by a homomorphismare called Hom-algebras. The notion of Hom-Lie algebras was introduced by Hartwig,Larsson and Silvestrov to describe the structures on certain deformations of the Witt alge-bras and the Virasoro algebras in [9]. Hom-Lie algebras are also related to deformed vectorfields, the various versions of the Yang-Baxter equations, braid group representations, andquantum groups in [9, 19, 20]. More research on Hom-Lie algebras see references [7, 17].The notion of ring of quotients was introduced by Utumi in [18], which played animportant role in the development of the theories of associative and commutative rings.He showed that every associative ring without right zero divisors has a maximal leftquotients ring and constructed it. Inspired by [18], Siles Molina studied the algebrasof quotients of Lie algebras in [16]. Martindale rings of quotients were introduced byMartindale in 1969 for prime rings in [12], which was designed for applications to ringssatisfying a generalized polynomial identity. In [6], Garc´ıa and G´omez defined Martindale-like quotients for Lie triple systems with respect to power filters of sturdy ideals and
Corresponding author(L. Chen): [email protected] by NNSF of China (No. 11771069). Q, [ · , · ] , α ) is an algebra of quotients of a Hom-Lie algebra ( L, [ · , · ] , α ) if and onlyif ( Q, [ · , · ] , α ) is ideally absorbed into ( L, [ · , · ] , α ), presented as Theorem 3.10. In Section 4,we focus on semiprime Hom-Lie algebras and construct the maximal algebras of quotientsof them and characterize them. In Section 5, we mainly study the relationship betweenalgebras of quotients of Hom-Lie algebras and associative algebras generated by innerderivations of Hom-Lie algebras, and get the main result that A ( Q ), the associative algebragenerated by inner derivations of ( Q, [ · , · ] , α ), is a left quotients of A if ( Q, [ · , · ] , α ) is analgebra of quotients of ( L, [ · , · ] , α ), where A denotes the subalgebra of A ( Q ) satisfying A ( L ) ⊆ L , written as Theorem 5.11. In Section 6, we introduce the concept of denseextension first of all. Next, we get a proposition about dense extension, Proposition 6.7,that every I ⊆ Q is also dense for every essential Hom-ideal I of ( L, [ · , · ] , α ) if L ⊆ Q is adense extension and ( Q, [ · , · ] , α ) is an algebra of quotients of ( L, [ · , · ] , α ). Definition 2.1. [9] Let L be a vector space over a field F . Define a bilinear map [ · , · ] : L × L → L and a linear map α : L → L . If the triple ( L, [ · , · ] , α ) satisfies the followingconditions:(1) [ x, y ] = − [ y, x ] , ∀ x, y ∈ L ;(2) [ α ( x ) , [ y, z ]] + [ α ( y ) , [ z, x ]] + [ α ( z ) , [ x, y ]] = 0 , ∀ x, y, z ∈ L .then ( L, [ · , · ] , α ) is called a Hom-Lie algebra. efinition 2.2. [9] A subspace W of a Hom-Lie algebra ( L, [ · , · ] , α ) is called a Hom-subalgebra of L if it satisfies α ( W ) ⊆ W and [ W, W ] ⊆ W . Definition 2.3. [9] A subspace I of a Hom-Lie algebra ( L, [ · , · ] , α ) is called a Hom-idealof L if it satisfies α ( I ) ⊆ I and [ I, L ] ⊆ I . Remark 2.4.
Suppose that ( L, [ · , · ] , α ) is a Hom-Lie algebra. For any x ∈ L , define ad x : L → L to be a linear map by ad x ( y ) = [ α ( x ) , y ] for any y ∈ L . Then ad x is aderivation if α satisfies α ([ x, y ]) = [ α ( x ) , y ] = [ x, α ( y )] , ∀ x, y ∈ L, (2.1) which is to say that α belongs to α -centroid. In this sense, ad x is called an inner deriva-tion. Example 2.5.
Suppose that L is a -dimensional vector over F with a basis { e , e , e , e } .The multiplication is [ e , e ] = [ e , e ] = e , [ e , e ] = e , others are all zero . One can see that ( L, [ · , · ]) isn’t a Lie algebra since [[ e , e ] , e ] + [[ e , e ] , e ] + [[ e , e ] , e ] = − e = 0 . Define α : L → L to be a linear map by α ( e ) = ae , α ( e ) = α ( e ) = 0 , α ( e ) = be , where a, b ∈ F . Then ( L, [ · , · ] , α ) is a non-trivial Hom-Lie algebra satisfying (2.1). Definition 2.6.
For a Hom-subalgebra H of a Hom-Lie algebra ( L, [ · , · ] , α ) , define Ann L ( H ) = { x ∈ L | [ x , α ( y )] = 0 , ∀ y ∈ H } to be the α -annihilator of H in L . If H is a Hom-ideal of L and α satisfies (2.1), then sois Ann L ( H ) . In particular, if H = L , write Ann( L ) = Ann L ( L ) . Definition 2.7.
Let ( L, [ · , · ] , α ) be a Hom-Lie algebra. An element x ∈ L is an absolutezero divisor if (ad x ) = 0 . The Hom-Lie algebra L is said to be nondegenerate if it doesn’tcontain any nonzero absolute zero divisor. Definition 2.8.
Suppose that ( L, [ · , · ] , α ) is a Hom-Lie algebra. Then L is said to besemiprime if for any nonzero Hom-ideal I of L , [ I, α ( I )] = { } and prime if for any twononzero Hom-ideals I , J of L , [ I, α ( J )] = { } . In the following, we suppose that all Hom-Lie algebras satisfy (2 . Remark 2.9.
For a Hom-Lie algebra ( L, [ · , · ] , α ) , we have L nondegenerate ⇒ L semiprime ⇒ Ann( L ) = { } . roof. (1). Suppose that L is nondegenerate. If L isn’t semiprime, there exists anonzero Hom-ideal I of L satisfying [ I, α ( I )] = { } . Take 0 = x ∈ I , then we have(ad x ) ( L ) = [ α ( x ) , [ α ( x ) , L ]] ⊆ [ α ( I ) , I ] = { } , which implies that x is a nonzero absolutezero divisor. Contradict with L is nondegenerate.(2). Suppose that L is semiprime. If Ann( L ) = { } , then Ann( L ) is a nonzero Hom-ideal of L satisfying [Ann( L ) , α (Ann( L ))] ⊆ [Ann( L ) , α ( L )] = { } . Contradiction.A Hom-ideal I of a Hom-Lie algebra ( L, [ · , · ] , α ) is said to be essential if every nonzeroHom-ideal of L hits I ( I ∩ J = { } for every nonzero Hom-ideal J of L ). Proposition 2.10.
Suppose that ( L, [ · , · ] , α ) is a Hom-Lie algebra and I a Hom-ideal of L . (1) If Ann L ( I ) = { } , then I is essential.(2) If L is semiprime, then I ∩ Ann L ( I ) = { } and I is essential if and only if Ann L ( I ) = { } .Proof. (1). Otherwise, there exists a nonzero Hom-ideal J of L such that I ∩ J = { } .Take 0 = v ∈ J , we have [ v , α ( I )] ⊆ I ∩ J = { } . So v ∈ Ann L ( I ), which is acontradiction. Hence I is essential.(2). If not, [ I ∩ Ann L ( I ) , α ( I ∩ Ann L ( I ))] ⊆ [Ann L ( I ) , α ( I )] = { } . Note that I ∩ Ann L ( I ) is a nonzero Hom-ideal of L , this contradicts with L being semiprime.Suppose that I is essential. If Ann L ( I ) = { } , then I ∩ Ann L ( I ) = { } . Contradiction.Hence, Ann L ( I ) = { } . Combining with (1), we get that I is essential if and only ifAnn L ( I ) = { } . Proposition 2.11.
Suppose that ( L, [ · , · ] , α ) is a Hom-Lie algebra. Then L is prime ifand only if for every nonzero Hom-ideal I of L , Ann L ( I ) = { } .Proof. Suppose that L is prime. If there exists a nonzero Hom-ideal J of L suchthat Ann L ( J ) = { } . Then [Ann L ( J ) , α ( J )] = { } according to the definition of Ann L ( J ),which contradicts with L being prime.On the other hand, suppose that for every nonzero Hom-ideal I of L , Ann L ( I ) = { } .If L isn’t prime, there exist two nonzero Hom-ideals I and J of L such that [ I, α ( J )] = { } ,which implies that I ⊆ Ann L ( J ). Contradiction. Inspired by the notion of algebras of quotients of Lie algebras in [16], we introducethe notion of algebras of quotients of Hom-Lie algebras.4uppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of Hom-Lie algebra ( Q, [ · , · ] , α ). Forevery q ∈ Q , set L ( q ) = F q + ( n X i =1 [[[[ q, x i ] , x i ] , · · · ] , x ik i ] | x ij ∈ L with n , k i ∈ N ) , That is, L ( q ) is the linear span in Q of the elements of the form [[[[ q, x ] , x ] , · · · ] , x n ]and q , where n ∈ N and x , · · · , x n ∈ L . It is clear that [ L ( q ) , L ] ⊆ L ( q ). Now we define( L : q ) = { x ∈ L | [ x, α ( L ( q ))] ⊆ L } . Obviously, if q ∈ L , then ( L : q ) = L . Proposition 3.1.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) . Take q ∈ Q . Then ( L : q ) is a Hom-ideal of L . Moreover, it’s maximal among the Hom-ideals I of L such that [ I, α ( q )] ⊆ L .Proof. For any x ∈ ( L : q ), [ α ( x ) , α ( L ( q ))] = α ([ x, α ( L ( q ))]) ⊆ α ( L ) ⊆ L , whichimplies that α ( x ) ∈ ( L : q ), i.e., α (( L : q )) ⊆ ( L : q ).For any x ∈ ( L : q ), y ∈ L , [[ x, y ] , α ( L ( q ))] = [ α ( x ) , [ y, L ( q )]] + [[ x, L ( q )] , α ( y )] =[ x, α ([ y, L ( q )])] + [[ x, α ( L ( q ))] , y ] ⊆ [ x, α ( L ( q ))] + [ L, y ] ⊆ L , which implies that [ x, y ] ∈ ( L : q ), i.e., [( L : q ) , L ] ⊆ ( L : q ). Hence, ( L : q ) is a Hom-ideal of L .Suppose that I is a Hom-ideal of L such that [ I, α ( q )] ⊆ L . We’ll show that for any x , · · · , x n ∈ L and n ∈ N , [ I, α ([[[[ q, x ] , x ] , · · · ] , x n ])] ⊆ L . We’ll prove it by inductionon n .When n = 1, for any x ∈ L , [ I, α ([ q, x ])] = [ α ( I ) , [ q, x ]] = [ α ( q ) , [ I, x ]] +[ α ( x ) , [ q, I ]] = [ α ( q ) , [ I, x ]] + [ x , [ α ( q ) , I ]] ⊆ [ α ( q ) , I ] + [ x , L ] ⊆ L .Suppose that when n = k , [ I, α ([[[[ q, x ] , x ] , · · · ] , x k ])] ⊆ L for any x , · · · , x k ∈ L .When n = k + 1, for any x , · · · , x k , x k +1 ∈ L , we denote [[[[ q, x ] , x ] , · · · ] , x k ] by z .By hypothesis, we have [ I, α ( z )] ⊆ L .Therefore, [ I, α ([[[[[ q, x ] , x ] , · · · ] , x k ] , x k +1 ])] = [ I, α ([ z, x k +1 ])] = [ α ( I ) , [ z, x k +1 ]] =[ α ( z ) , [ I, x k +1 ]]+[ α ( x k +1 ) , [ z, I ]] = [ α ( z ) , [ I, x k +1 ]]+[ x k +1 , [ α ( z ) , I ]] ⊆ [ α ( z ) , I ]+[ x k +1 , L ] ⊆ L . Hence, we have [ I, α ( L ( q ))] ⊆ L , i.e., I ⊆ ( L : q ). Definition 3.2.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) . Then Q iscalled an algebra of quotients of L , if given p, q ∈ Q with p = 0 , there exists x ∈ L suchthat [ x, α ( p )] = 0 and x ∈ ( L : q ) . Proposition 3.3.
Let ( L, [ · , · ] , α ) be a Hom-subalgebra of ( Q, [ · , · ] , α ) .(1) If Ann( L ) = { } , then L is an algebra of quotients of itself.(2) If Q is an algebra of quotients of L , then Ann Q ( L ) = Ann( L ) = { } . roof. (1). For any p, q ∈ L with p = 0, we have p / ∈ Ann( L ). There exists x ∈ L such that [ p, α ( x )] = 0, so [ x, α ( p )] = 0. It’s obvious that x ∈ ( L : q ). Therefore L is analgebra of quotients of itself.(2). For any 0 = q ∈ Q , there exists x ∈ L such that [ x, α ( q )] = 0, so [ q, α ( x )] = 0.Hence q / ∈ Ann Q ( L ). Therefore Ann Q ( L ) = { } . Similarly, we have Ann( L ) = { } .The above proposition says for a Hom-Lie algebra ( L, [ · , · ] , α ), that Ann( L ) = { } isa sufficient and necessary condition such that L has algebras of quotients.We will prove that some properties of a Hom-Lie algebra ( L, [ · , · ] , α ) can be inheritedby its algebras of quotients Q . Actually, Q just needs a weaker condition. Definition 3.4.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) . Then Q iscalled a weak algebra of quotients of L , if given = q ∈ Q , there exists x ∈ L such that = [ x, α ( q )] ∈ L . Remark 3.5.
Every algebra of quotients of a Hom-Lie algebra ( L, [ · , · ] , α ) is a weak algebraof quotients. Proposition 3.6.
Let ( Q, [ · , · ] , α ) be a weak algebra of quotients of the Hom-Lie algebra ( L, [ · , · ] , α ) .(1) If I is a nonzero Hom-ideal of Q , then I ∩ L is a nonzero Hom-ideal of L .(2) If L is semiprime(prime), so is Q .Proof. (1). Since I is a Hom-ideal of Q , α ( I ) ⊆ I . Hence α ( I ∩ L ) ⊆ α ( I ) ∩ α ( L ) ⊆ I ∩ L .Note that I = { } , we can take 0 = x ∈ I . There exists x ∈ L such that 0 =[ x, α ( x )] ∈ L . [ x, α ( x )] ∈ I since I is a Hom-ideal of Q . So we have 0 = [ x, α ( x )] ∈ I ∩ L .Moreover, [ I ∩ L, L ] ⊆ [ I, Q ] ⊆ I and [ I ∩ L, L ] ⊆ [ L, L ] ⊆ L . Hence, [ I ∩ L, L ] ⊆ I ∩ L .Therefore, I ∩ L is a nonzero Hom-ideal of L .(2). Suppose that L is semiprime. If Q isn’t semiprime, there exists a nonzero Hom-ideal I of Q such that [ I, α ( I )] = { } . By (1), I ∩ L is a nonzero Hom-ideal of L and[ I ∩ L, α ( I ∩ L )] ⊆ [ I, α ( I )] = { } . Contradict with L being semiprime.Similarly, we have Q is prime if L is prime. Definition 3.7.
Let ( L, [ · , · ] , α ) be a Hom-subalgebra of ( Q, [ · , · ] , α ) . Then Q is calledideally absorbed into L if, for every = q ∈ Q , there exists a Hom-ideal I of L with Ann L ( I ) = { } such that { } 6 = [ I, α ( q )] ∈ L . Proposition 3.8.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) . Take q ∈ Q .(1) If Q is an algebra of quotients of L , then ( L : q ) is an essential Hom-ideal of L .Moreover, Ann L (( L : q )) = { } .
2) If Q is ideally absorbed into L , then ( L : q ) is an essential Hom-ideal of L .Moreover, Ann L (( L : q )) = { } .Proof. (1). Let I be a nonzero Hom-ideal of L . Take 0 = y ∈ I . Apply that Q isan algebra of quotients of L to find x ∈ L such that 0 = [ x, α ( y )] and x ∈ ( L : q ). Since( L : q ) is a Hom-ideal of L , [ x, α ( y )] ∈ I ∩ ( L : q ). This implies that I ∩ ( L : q ) = { } .Therefore ( L : q ) is an essential Hom-ideal of L .Suppose that Ann L (( L : q )) = { } . Then Ann L (( L : q )) ∩ ( L : q ) = { } since ( L : q ) isessential. Take 0 = u ∈ Ann L (( L : q )) ∩ ( L : q ). Apply that Q is an algebra of quotients of L to find x ∈ L such that 0 = [ x, α ( u )] and x ∈ ( L : q ). However, [ x, α ( u )] = − [ u, α ( x )] =0 since u ∈ Ann L (( L : q )). Contradiction.(2). Since Q is ideally absorbed into L , there exists a nonzero Hom-ideal I of L withAnn L ( I ) = { } such that { } 6 = [ I, α ( q )] ⊆ L . According to the proof of Proposition 3.1,[ I, α ( L ( q ))] ⊆ L . So I ⊆ ( L : q ). Hence, Ann L (( L : q )) ⊆ Ann L ( I ) = { } . According toProposition 2.10 (1), ( L : q ) is essential. Lemma 3.9.
Let ( Q, [ · , · ] , α ) be a weak algebra of quotients of Hom-Lie algebra ( L, [ · , · ] , α ) .Let I be a Hom-ideal of L with Ann L ( I ) = { } . Then there is no nonzero element x in Q such that [ α ( x ) , I ] = { } .Proof. Suppose that there exists 0 = x ∈ Q such that [ α ( x ) , I ] = { } . Apply that Q is a weak algebra of quotients of L to find y ∈ L such that 0 = [ y, α ( x )] ∈ L . SinceAnn L ( I ) = { } , [[ y, α ( x )] , α ( I )] = { } . However,[[ y, α ( x )] , α ( I )] = α ([[ y, x ] , α ( I )]) = α ([ α ( x ) , [ I, y ]] + [ α ( y ) , [ x, I ]])= α ([ α ( x ) , [ I, y ]] + [ y, [ α ( x ) , I ]]) = { } . This is a contradiction.
Theorem 3.10.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) . Then Q isan algebra of quotients of L if and only if Q is ideally absorbed into L .Proof. Suppose that Q is an algebra of quotients of L . For any 0 = q ∈ Q , ( L : q ) isa Hom-ideal of L with Ann L (( L : q )) = { } according to Proposition 3.8 (1). By Lemma3.9, [ α ( q ) , ( L : q )] = { } . According to the definition of ( L : q ), [( L : q ) , α ( q )] ⊆ L . Hence, Q is ideally absorbed into L .Now suppose that Q is ideally absorbed into L . Then Q is a weak algebra of quotientsof L . For any p, q ∈ Q with p = 0, we have Ann L (( L : q )) = { } . Hence, [ α ( p ) , ( L : q )] =[ p, α (( L : q ))] = { } . Therefore, there exists x ∈ ( L : q ) such that [ α ( p ) , x ] = 0. Hence, Q is an algebra of quotients of L .In the proof of Theorem 3.10, we can conclude that Q is an algebra of quotients of L if and only if Q is a weak algebra of quotients of L satisfying Ann L (( L : q )) = { } forevery q ∈ Q . 7 roposition 3.11. Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and Q aweak algebra of quotients of L . Then for every Hom-ideal I of L , Ann L ( I ) = { } implies Ann Q ( I ) = { } .Proof. Suppose that 0 = q ∈ Ann Q ( I ). By hypothesis, there exists an element x in L such that 0 = [ x, α ( q )] ∈ L . Since Ann L ( I ) = { } , there exists y ∈ I such that[[ x, α ( q )] , α ( y )] = 0. However,[[ x, α ( q )] , α ( y )] = α ([[ x, q ] , α ( y )]) = α ([[ y, q ] , α ( x )] + [[ x, y ] , α ( q )])= α ([[ α ( y ) , q ] , x ] + [ α ([ x, y ]) , q ]) ⊆ α ([[ α ( y ) , q ] , x ] + [ α ( I ) , q ]) = 0 . This is a contradiction.
Notation 3.12.
Denote by J e ( L ) the set of all essential Hom-ideals of a Hom-Lie algebra ( L, [ · , · ] , α ) . Clearly, given
I, J ∈ J e ( L ), we have I ∩ J ∈ J e ( L ). If L is semiprime, then I ∈ J e ( L ), where I = [ I, α ( I )]. If L is prime, then [ I, α ( J )] ∈ J e ( L ) for any I, J ∈ J e ( L ). Proposition 3.13.
Let ( L, [ · , · ] , α ) be a semiprime Hom-Lie algebra and ( Q, [ · , · ] , α ) analgebra of quotients of L . Then for every essential Hom-ideal I of L , we have that Q isan algebra of quotients of I .Proof. For any p, q ∈ Q with p = 0, we have ( L : q ) ∈ J e ( L ). So that ( L : q ) ∩ I ∈ J e ( L ) and consequently, (( L : q ) ∩ I ) ∈ J e ( L ). Then we have Ann Q ((( L : q ) ∩ I ) ) = { } . So there exist y, z ∈ ( L : q ) ∩ I such that [ α ([ y, α ( z )]) , p ] = 0, i.e., [[ y, α ( z )] , α ( p )] = 0.Moreover,[[ y, α ( z )] , α ( I ( q ))] = α ([[ I ( q ) , z ] , α ( y )] + [[ y, I ( q )] , α ( z )]) ⊆ α ([[ L ( q ) , z ] , α ( y )] + [[ y, L ( q )] , α ( z )])= α ([[ α ( L ( q )) , z ] , y ] + [[ y, α ( L ( q ))] , z ]) ⊆ α ([ L, I ]) ⊆ α ( I ) ⊆ I, which implies that [ y, α ( z )] ∈ ( I : q ). Therefore, Q is an algebra of quotients of I . In this section, we will construct a maximal algebra of quotients for every semiprimeHom-Lie algebra ( L, [ · , · ] , α ) with α ( L ) is an essential ideal. Maximal in the sense thatevery algebra of quotients of L can be considered inside this maximal algebra of quotientsvia an injective map which restricted in L is the identity. Definition 4.1.
Given a Hom-ideal I of a Hom-Lie algebra ( L, [ · , · ] , α ) , we say that alinear map δ : I → L is an α k -partial derivation for k ∈ N if for any x, y ∈ I it issatisfied: δ ◦ α = α ◦ δ ;(2) δ ([ x, y ]) = [ δ ( x ) , α k ( y )] + [ α k ( x ) , δ ( y )] . Denote by PDer k ( I , L ) the set of all α k -partial derivations of I in L . Lemma 4.2.
Let ( L, [ · , · ] , α ) be a semiprime Hom-Lie algebra and consider the set D k := { ( δ, I ) | I ∈ J e ( L ) , δ ∈ PDer k ( I , L ) } . Define on D k the following relation: ( δ, I ) ≡ ( µ, J ) if and only if there exists K ∈ J e ( L ) , K ⊆ I ∩ J such that δ | K = µ | K . Then ≡ is an equivalence relation.Proof. (1). For any ( δ, I ) ∈ D k , δ | I = δ | I , so ( δ, I ) ≡ ( δ, I ).(2). For any ( δ, I ) , ( µ, J ) ∈ D k and ( δ, I ) ≡ ( µ, J ), there exists K ∈ J e ( L ), K ⊆ I ∩ J such that δ | K = µ | K . It’s obvious that µ | K = δ | K . Hence ( µ, J ) ≡ ( δ, I ).(3). For any ( δ, I ) , ( ξ, J ) , ( β, K ) ∈ D k and ( δ, I ) ≡ ( ξ, J ) and ( ξ, J ) ≡ ( β, K ), thereexist P, Q ∈ J e ( L ), P ⊆ I ∩ J , Q ⊆ J ∩ K such that δ | P = ξ | P and ξ | Q = β | Q . Since P, Q ∈ J e ( L ), P ∩ Q = ∅ . Take T ⊆ P ∩ Q . We also get T ∈ J e ( L ). Moreover, δ | T = ξ | T = β | T . Therefore ( δ, I ) ≡ ( β, K ).The proof is completed. Notation 4.3.
Denote by Q k ( L ) the quotient set D k / ≡ . Set Q ( L ) = Q ( L ) . Let δ I denote the equivalence class of ( δ, I ) in Q ( L ) . Theorem 4.4.
Let ( L, [ · , · ] , α ) be a semiprime Hom-Lie algebra over F , and let Q := Q ( L ) be as in Notation 4.3. Define the following maps: · : F × Q → Q, ( p, δ I ) ( pδ ) I where p δ : I → L , y δ ( py ) , + : Q × Q → Q, ( δ I , µ J ) ( δ + µ ) I ∩ J where δ + µ : I ∩ J → L , x δ ( x ) + µ ( x ) , [ · , · ] : Q × Q → Q, ( δ I , µ J ) [ δ, µ ] [ I ∩ J,I ∩ J ] where [ δ, µ ] : [ I ∩ J , I ∩ J ] → L , x δµ ( x ) − µδ ( x ) , ˜ α : Q → Q, δ I ( α ◦ δ ) I where α ◦ δ : I → L , x αδ ( x ) . Then Q , with these operations, is a Hom-Lie algebra satisfying ˜ α ([ δ I , µ J ]) = [ ˜ α ( δ I ) , µ J ] = [ δ I , ˜ α ( µ J )] for any δ I , µ J ∈ Q and containing L as a subalgebra, via the injection ϕ : L → Q, x (ad x ) L . roof. (1). For any δ I , µ J ∈ Q and x, y ∈ I ∩ J ,[ δ, µ ]([ x, y ]) = δµ ([ x, y ]) − µδ ([ x, y ]) = δ ([ µ ( x ) , y ] + [ x, µ ( y )]) − µ ([ δ ( x ) , y ] + [ x, δ ( y )])= [ δµ ( x ) , y ] + [ µ ( x ) , δ ( y )] + [ δ ( x ) , µ ( y )] + [ x, δµ ( y )] − [ µδ ( x ) , y ] − [ δ ( x ) , µ ( y )] − [ µ ( x ) , δ ( y )] − [ x, µδ ( y )]= [[ δ, µ ]( x ) , y ] + [ x, [ δ, µ ]( y )] , which implies that [ δ, µ ] ∈ PDer ([ I ∩ J , I ∩ J ] , L ).For any δ I ∈ Q and x, y ∈ I , α ◦ δ ([ x, y ]) = α ([ δ ( x ) , y ] + [ x, δ ( y )]) = [ α ◦ δ ( x ) , y ] + [ x, α ◦ δ ( y )] , which implies that α ◦ δ ∈ PDer ( I , L ).Therefore, [ · , · ], ˜ α are well-defined.(2). It’s obvious that ( Q, · , +) is a vector space over F and [ · , · ] is skew-symmetric.Next we’ll show that Q satisfies the Jacobi identity. For any δ I , µ J , β K ∈ Q , we have[[ δ I , µ J ] , ˜ α ( β K )] + [[ µ J , β K ] , ˜ α ( δ I )] + [[ β K , δ I ] , ˜ α ( µ J )] = 0since α commutes with δ, µ, β . Hence, Q is a Hom-Lie algebra.(3). For any δ I , µ J ∈ Q ,˜ α ([ δ I , µ J ]) = ˜ α ([ δ, µ ] [ I ∩ J,I ∩ J ] ) = ( α ◦ [ δ, µ ]) [ I ∩ J,I ∩ J ] = ( αδµ − αµδ ) [ I ∩ J,I ∩ J ] = ( αδµ − µαδ ) [ I ∩ J,I ∩ J ] = [ αδ, µ ] [ I ∩ J,I ∩ J ] = [( αδ ) I , µ J ] = [ ˜ α ( δ I ) , µ J ] . Similarly, we get ˜ α ([ δ I , µ J ]) = [ δ I , ˜ α ( µ J )]. Hence, we have ˜ α ([ δ I , µ J ]) = [ ˜ α ( δ I ) , µ J ] =[ δ I , ˜ α ( µ J )] for any δ I , µ J ∈ Q .(4). According to Remark 2.4, we get for any x ∈ L , ad x ∈ PDer ( L , L ). So ϕ iswell-defined. If ϕ ( x ) = 0, we have (ad x ) L = 0. Then we have [ x, α ( L )] = [ α ( x ) , L ] = { } ,which implies that x ∈ Ann( L ). Note that L is semiprime, we have Ann( L ) = { } byProposition 2.10 (2). Hence x = 0, i,e, ϕ is injective.For any X ⊆ L , write X ϕ to denote the image of X inside Q ( L ) via the map definedin Theorem 4.4. Lemma 4.5.
For every δ I ∈ Q ( L ) , and (ad x ) L ∈ I ϕ ( x ∈ I ) we have [ δ I , (ad x ) L ] =(ad δ ( x ) ) L ∈ L ϕ .Proof. For any y ∈ I ,[ δ, ad x ]( y ) = δ ad x ( y ) − ad x δ ( y ) = δ ([ α ( x ) , y ]) − [ α ( x ) , δ ( y )]= [ α ( x ) , δ ( y )] + [ δα ( x ) , y ] − [ α ( x ) , δ ( y )] = [ δα ( x ) , y ] = [ αδ ( x ) , y ] = ad δ ( x ) ( y ) , and so [ δ, ad x ] I = (ad δ ( x ) ) I = (ad δ ( x ) ) L ∈ L ϕ .10 roposition 4.6. Let ( L, [ · , · ] , α ) be a semiprime Hom-Lie algebra with α ( L ) is essential.Then Q ( L ) is semiprime and an algebra of quotients of L . Moreover, Q ( L ) is maximalamong the algebras of quotients of L , in the sense that if S is an algebra of quotients of L , then there exists an injection ψ : S → Q ( L ) which is the identity in L . In particular,the map ψ : S → Q ( L ) , s (ad s ) ( L : s ) is an injective map which is the identity when restricted to L .Proof. Take δ I , µ I ∈ Q ( L ) with δ I = 0(we can consider the same I for δ and µ because if J, K ∈ J e ( L ) are such that δ : J → L , µ : K → L , then I := J ∩ K ∈ J e ( L )and so δ J = δ I and µ K = µ I ). Choose a ∈ I such that δ ( a ) = 0. Then (ad a ) L ∈ L ϕ satisfies [ ˜ α ( δ I ) , (ad a ) L ] = (ad αδ ( a ) ) L = 0. Otherwise, [ δ ( a ) , α ( L )] = [ α δ ( a ) , L ] = { } ,which implies that δ ( a ) ∈ Ann L ( α ( L )) = { } . And for every λ I ∈ ϕL ( µ I ), [ ˜ α ( λ I ) , (ad a ) L ] =(ad αλ ( a ) ) L ∈ L ϕ . This implies that (ad a ) L ∈ ( L ϕ : µ I ). Hence, Q ( L ) is an algebra ofquotients of L ϕ . The semiprimeness of Q ( L ) follows by Proposition 3.6 (2).Now suppose that S is an algebra of quotients of L and consider the map ψ : S → Q ( L ) , s (ad s ) ( L : s ) . According to Proposition 3.8 (1), ψ is well-defined. To prove the injectivity, suppose that s ∈ S such that ψ ( s ) = 0, that is [ α ( s ) , K ] = { } for some Hom-ideal K ∈ J e ( L ), K ⊆ ( L : s ). This implies that s ∈ Ann L ( K ). Note that K is essential, Ann L ( K ) = { } ,so s = 0. Definition 4.7.
For a semiprime Hom-Lie algebra ( L, [ · , · ] , α ) with α ( L ) is essential, theHom-Lie algebra constructed in Theorem 4.4 will be called the maximal algebra of quotientsof L . Denote it by Q m ( L ) . The axiomatic characterization of the Martindale ring of quotients given by D. Pass-man in [14] has inspired us to give the following description of the maximal algebra ofquotients of a semiprime Hom-Lie algebra.
Proposition 4.8.
Let ( L, [ · , · ] , α ) be a semiprime Hom-Lie algebra with α ( L ) is essentialand consider an overalgebra S of L . Then there exists an injection between S and Q m ( L ) which is the identity on L , if and only if S satisfies the following properties:(1) For any s ∈ S , there exists I ∈ J e ( L ) such that [ α ( I ) , s ] ⊆ L .(2) For s ∈ S and I ∈ J e ( L ) , [ α ( I ) , s ] = { } implies that s = 0 .Proof. Define ψ : S → Q m ( L ), s (ad s ) I where I is a nonzero essential Hom-idealof L satisfying [ α ( I ) , s ] ⊆ L ; this exists by (1), and so the map is well-defined.If ψ ( s ) = 0, i.e., (ad s ) I = 0, then [ α ( s ) , I ] = { } . By (2), we get s = 0. This impliesthat ψ is an injective map. 11inally, ψ is the identity on L , by identifying L with L ϕ , where ϕ is the map definedin Theorem 4.4.Conversely, we’ll prove that Q ( L ) satisfies the two conditions.(1). Consider q ∈ Q ( L ). According to Propositions 4.6 and 3.8 (1), ( L : q ) ∈ J e ( L ),and by the definition, [ α (( L : q )) , q ] = [( L : q ) , α ( q )] ⊆ L .(2). Take q ∈ Q ( L ) and I ∈ J e ( L ) such that [ α ( I ) , q ] = { } . We have q ∈ Ann Q ( L ) ( I ) = 0 according to Propositions 4.6 and 3.11. In [15], authors examined how the notion of algebras of quotients for Lie algebrastied up with the corresponding well-known concept in the associative case. Inspired bythe method in [15], we mainly study the relationship between Hom-Lie algebras and theassociative algebras generated by inner derivations of the corresponding Hom-Lie algebrasof quotients. First of all, we will give some definitions and basic notations.We shall denote by A ( L ) the associative subalgebra (possibly without identity) ofEnd( L ) generated by the elements ad x for x in ( L, [ · , · ] , α ).By an extension of Hom-Lie algebras L ⊆ Q we will mean that ( L, [ · , · ] , α ) is a Hom-subalgebra of the Hom-Lie algebra ( Q, [ · , · ] , α ).Let L ⊆ Q be an extension of Hom-Lie algebras and let A Q ( L ) be the associativesubalgebra of A ( Q ) generated by { ad x : x ∈ L } .Recall that, given an associative algebra A and a subset X of A , we define the rightannihilator of X in A as rann A ( X ) = { a ∈ A | Xa = 0 } , which is always a right ideal of A (and two-sided if X is a right ideal). One similarlydefines the left annihilator, which shall be denoted by lann A ( X ). Lemma 5.1.
Let I be a Hom-ideal of a Hom-Lie algebra ( L, [ · , · ] , α ) with Ann( L ) = { } .Then Ann L ( I ) = { } if and only if rann A ( L ) ( A L ( I )) = { } .Proof. Suppose that Ann L ( I ) = { } . For any µ in rann A ( L ) ( A L ( I )), we have ad y µ =0 for any y ∈ I . In particular if x ∈ L , we get 0 = ad y µ ( x ) = [ α ( y ) , µ ( x )], and this impliesthat µ ( x ) ∈ Ann L ( I ) = 0. Hence, µ = 0.Conversely, suppose that rann A ( L ) ( A L ( I )) = { } . For any x in Ann L ( I ), y in I and z in L , we havead y ad x ( z ) = [ α ( y ) , [ α ( x ) , z ]] = α ([ α ( y ) , [ x , z ]]) = α ([[ z , y ] , α ( x )] + [[ y , x ] , α ( z )]) = 0 , which implies that ad x ∈ rann A ( L ) ( A L ( I )) = 0. Since by assumption Ann( L ) = { } , weobtain x = 0. 12or a subset X of an associative algebra A , denote by h X i lA , h X i rA and h X i A the left,right and two sided ideal of A , respectively, generated by X . Lemma 5.2.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and I a Hom-ideal of ( L, [ · , · ] , α ) . Then h A Q ( I ) i lA Q ( L ) = h A Q ( I ) i rA Q ( L ) = h A Q ( I ) i A Q ( L ) . Proof.
Obviously, we have h A Q ( I ) i lA Q ( L ) = A Q ( L ) A Q ( I )+ A Q ( I ) and h A Q ( I ) i rA Q ( L ) = A Q ( I ) A Q ( L ) + A Q ( I ). Notice that given x ∈ L and y ∈ I , we havead x ad y = ad α ([ x , y ]) + ad y ad x . Then we have h A Q ( I ) i lA Q ( L ) = h A Q ( I ) i rA Q ( L ) . Hence we come to the conclusion. Lemma 5.3.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and I a Hom-ideal of ( L, [ · , · ] , α ) . Write ˜ I to denote the ideal of A Q ( L ) generated by A Q ( I ) . Then(1) rann A Q ( L ) ( ˜ I ) = rann A Q ( L ) ( A Q ( I )) .(2) lann A Q ( L ) ( ˜ I ) = lann A Q ( L ) ( A Q ( I )) .Proof. (1) Since A Q ( I ) ⊆ ˜ I , we have rann A Q ( L ) ( ˜ I ) ⊆ rann A Q ( L ) ( A Q ( I )). So it’senough to show that rann A Q ( L ) ( A Q ( I )) ⊆ rann A Q ( L ) ( ˜ I ). Let λ ∈ rann A Q ( L ) ( A Q ( I )). ByLemma 5.2 we know that, if µ ∈ ˜ I there exist a natural number n , elements x ,i , · · · , x r i ,i ∈ L and y ,i , · · · , y s i ,i ∈ I with 0 ≤ r i ∈ N for all i and ∅ 6 = { s , · · · , s n } ⊆ N , such that µ = n X i ad x , i · · · ad x ri , i ad y , i · · · ad y si , i . Since ad y si , i λ = 0, we see that µλ = 0.(2) The proof is similar to (1). Lemma 5.4.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) such that Q isa weak algebra of quotients of L . Let I be a Hom-ideal of ( L, [ · , · ] , α ) . If Ann L ( I ) = { } ,then rann A ( Q ) ( A Q ( I )) = { } .Proof. According to Proposition 3.11, we have Ann Q ( I ) = { } . For any µ ∈ rann A ( Q ) ( A Q ( I )), we have ad y µ = 0 for every y ∈ I . If q ∈ Q , we then have 0 =ad y µ ( q ) = [ α ( y ) , µ ( q )]. This implies that µ ( q ) ∈ Ann Q ( I ) = 0, and so µ = 0. Lemma 5.5.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and let x , · · · , x n , y in L . Then we have, in A ( Q ) : ad x · · · ad x n ad y = ad y ad x · · · ad x n + n X i = ad x · · · ad α ([ x i , y ]) · · · ad x n . n particular, if I is a Hom-ideal of L and x , · · · , x n ∈ I , then ad x · · · ad x n ad y = ad y ad x · · · ad x n + δ where δ ∈ span { ad z · · · ad z n | z i ∈ I } .Proof. We’ll prove it by induction on n .When n = 1, we get ad x ad y = ad α ([ x , y ]) + ad y ad x for any x, y ∈ L .Suppose that ad x · · · ad x k ad y = ad y ad x · · · ad x k + P ki = ad x · · · ad α ([ x i , y ]) · · · ad x k forany x , · · · , x k , y ∈ L . Then when n = k + 1, for any x , · · · , x k , x k +1 , y ∈ L , we getad x · · · ad x k ad x k + ad y = ad x · · · ad x k (ad y ad x k + + ad α ([ x k + , y ]) )= (ad x · · · ad x k ad y )ad x k + + ad x · · · ad x k ad α ([ x k + , y ]) = (ad y ad x · · · ad x k + k X i = ad x · · · ad α ([ x i , y ]) · · · ad x k )ad x k + + ad x · · · ad x k ad α ([ x k + , y ]) = ad y ad x · · · ad x k ad x k + + k + X i = ad x · · · ad α ([ x i , y ]) · · · ad x k + . The proof is completed.Let ( L, [ · , · ] , α ) be a Hom-subalgebra of ( Q, [ · , · ] , α ). Denote by A the associativesubalgebra of A ( Q ) whose elements are those µ in A ( Q ) such that µ ( L ) ⊆ L . We obviouslyhave the containments: A Q ( L ) ⊆ A ⊆ A ( Q ) . Lemma 5.6.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and I a Hom-ideal of ( L, [ · , · ] , α ) . Let q , · · · , q n in Q such that [ α ( q i ) , I ] ⊆ L for every i = 1 , · · · , n .Then for µ = ad q · · · ad q n in A ( Q ) , we have that µ ( ˜ I ) n + ( ˜ I ) n µ ⊆ A (where ( ˜ I ) n denotesthe n -th power of ˜ I in the associative algebra A Q ( L ) ).Proof. According to Lemma 5.2, we have( ˜ I ) n = A Q ( L )( A Q ( I )) n + ( A Q ( I )) n = ( A Q ( I )) n A Q ( L ) + ( A Q ( I )) n . Thus it’s enough to prove that, for any y = ad x · · · ad x n where x i ∈ I , both µy and yµ belong to A .We’ll use induction on n . For n = 1, we have ad x ad q = ad α ([ x , q ]) + ad q ad x and since α ([ x, q ]) = [ x, α ( q )] ∈ L we see that ad α ([ x , q ]) ∈ A . On the other hand, ad q ad x ( L ) =[ α ( q ) , [ α ( x ) , L ]] ⊆ [ α ( q ) , I ] ⊆ L , and so ad q ad x ∈ A .Assume that the result is true for n −
1. Now, by Lemma 5.5 we havead x · · · ad x n ad q · · · ad q n = (ad q ad x )ad x · · · ad x n ad q · · · ad q n n X i =1 ad x · · · ad α ([ x i , q ]) · · · ad x n ad q · · · ad q n . (*)The first summand on the right side belongs to A because, as proved before, ad q ad x ∈ A and ad x · · · ad x n ad q · · · ad q n ∈ A by induction hypothesis. On the other hand,for each of the terms ad x · · · ad α ([ x i , q ]) · · · ad x n ad q · · · ad q n we have that x i ∈ I and α ([ x i , q ]) = [ x i , α ( q )] ∈ L . Using Lemma 5.5, we may write this as:ad α ([ x i , q ]) ad x · · · ad x i − ad x i + · · · ad x n ad q · · · ad q n + δ · ad x i + · · · ad x n ad q · · · ad q n , where δ ∈ span { ad z · · · ad z i − | z j ∈ I } . The induction hypothesis applies again to showthat this belongs to A . Hence, yµ ∈ A .If we continue to develop in the expression (*), we get, for some δ ∈ A ,ad q ad q ad x ad x · · · ad x n ad q · · · ad q n + n X i = ad q ad x · · · ad α ([ x i , q ]) · · · ad x n ad q · · · ad q n + δ . Using Lemma 5.5 we can write each term of the formad q ad x · · · ad α ([ x i , q ]) · · · ad x n ad q · · · ad q n as:ad q ad α ([ x i , q ]) ad x · · · x i − ad x i + · · · ad x n ad q · · · ad q n + ad q δ · ad x i + · · · ad x n ad q · · · ad q n , where δ ∈ span { ad z · · · ad z i − | z j ∈ I } . Notice thatad q ad α ([ x i , q ]) ad x = ad q ad α ([ α ([ x i , q ]) , x ]) + ad q ad x ad α ([ x i , q ]) . Hence, using α ([ x i , q ]) ∈ L and x i ∈ I , we see that the first summand above belongsto A . For the second summand, assuming that δ = ad z · · · ad z i − with z j ∈ I , wehave (ad q ad z )ad z · · · ad z i − ad x i + · · · ad x n ad q · · · ad q n , which is also an element of A .Continuing in this way, we find thatad x · · · ad x n ad q · · · ad q n − ad q · · · ad q n ad x · · · ad x n ∈ A and by what we have just proved, we see that ad q · · · ad q n ad x · · · ad x n ∈ A , as was tobe shown. Corollary 5.7.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and I a Hom-ideal of ( L, [ · , · ] , α ) . Let q , · · · , q n in Q such that [ α ( q i ) , I ] ⊆ L for every i = 1 , · · · , n .Then for µ = ad q · · · ad q n in A ( Q ) , we have that µ ˜ I n ∈ A , ˜ I n µ ⊆ A (where I = I and I k = [ I k − , α ( I )] for k ≥ ).Proof. It’s straightforward to show that I n is a Hom-ideal of L for each n ≥ e I n ⊆ ( ˜ I ) n . We’ll show it by induction on n . When n = 1, it’s obvious.Suppose that e I k ⊆ ( ˜ I ) k . When n = k + 1, for any y ∈ I k , z ∈ I ,ad [ y ,α ( z )] = ad α ([ y , z ]) = ad y ad z − ad z ad y ∈ ˜I k ˜I ⊆ ( ˜I ) k ˜I = ( ˜I ) k + . According to Lemma 5.6, we come to the conclusion.15 emma 5.8.
Let ( L, [ · , · ] , α ) be a semiprime Hom-Lie algebra. If I is a Hom-ideal of ( L, [ · , · ] , α ) with Ann L ( I ) = { } , then Ann L ( I s ) = { } for any s ≥ . Any finite intersec-tion of ideals with zero α -annihilator will also have zero α -annihilator.Proof. We’ll show it by induction on s . When s = 1, it’s obvious. Suppose thatAnn L ( I k ) = { } , i.e., I k is essential. When s = k + 1, for any nonzero Hom-ideal J of L ,we have I k ∩ J = { } since I k is essential. So { } 6 = [ I k ∩ J, α ( I k ∩ J )] ⊆ [ I k , α ( I )] ∩ J = I k +1 ∩ J, which implies that I k +1 is essential and so Ann L ( I k + ) = { } .Similarly, we can show that any finite intersection of ideals with zero α -annihilatorwill also have zero α -annihilator by induction. Proposition 5.9.
Suppose that ( L, [ · , · ] , α ) is a semiprime Hom-subalgebra of ( Q, [ · , · ] , α ) .Then the following conditions are equivalent:(1) Q is an algebra of quotients of L ;(2) Ann( Q ) = { } and, if µ ∈ A ( Q ) \ { } , there exists a Hom-ideal I of L with Ann L ( I ) = { } such that µ ˜ I ⊆ A and { } 6 = ˜ Iµ ⊆ A . If µ = ad q , then we also have µ ˜ I ( L ) = { } .Proof. (2) ⇒ (1) Let q ∈ Q \ { } . Then µ = ad q = 0 since Ann( Q ) = { } . Let ˜ I beas in (2), so it satisfies µ ˜ I ( L ) = { } and µ ˜ I ⊆ A . Set I := span { δ ( x ) | x ∈ L and δ ∈ ˜ I } . Then I is a Hom-ideal of L such that Ann L ( I ) = { } . Indeed, for any δ ∈ ˜ I , there exista natural number n , elements x ,i , · · · , x r i ,i ∈ L and y ,i , · · · , y s i ,i ∈ I with 0 ≤ r i ∈ N forall i and ∅ 6 = { s , · · · , s n } ⊆ N , such that δ = n X i ad x , i · · · ad x ri , i ad y , i · · · ad y si , i . For any x, y ∈ L , if r i = 0 and s i = 1, we get[ y, ad y , i ( x )] = [ y , [ α ( y , i ) , x ]] = [ α ( y , i ) , [ y , x ]] + [ α ([ y , y , i ]) , x ]= ad y , i ([ y , x ]) + ad [ y , y , i ] ( x ) ∈ I , if r i > s i = 1, we’ll prove[ y, ad x , i · · · ad x ri , i ad y , i ( x )] ∈ I by induction on r i . When r i = 1, we get[ y, ad x , i ad y , i ( x )] = [ y , ad y , i ad x , i ( x ) + ad α ([ x , i , y , i ]) ( x )]16 [ y, ad y , i ad x , i ( x )] + [ y , ad α ([ x , i , y , i ]) ( x )]= [ α ( y ) , ad y , i ([ x , i , x ])] + [ y , ad α ([ x , i , y , i ]) ( x )] = ad y ad y , i ([ x , i , x ]) + [ y , ad α ([ x , i , y , i ]) ( x )] ∈ I . Suppose that when r i = k , [ y, ad x , i · · · ad x k , i ad y , i ( x )] ∈ I for any x ,i , · · · , x k,i ∈ L and y ,i ∈ I . When r i = k + 1, we get[ y, ad x , i · · · ad x k + , i ad y , i ( x )]= " y, ad y , i ad x , i · · · ad x k + , i + k + X j = ad x , i · · · ad α ([ x j , i , y , i ]) · · · ad x k + , i ! ( x ) = [ y, ad y , i ad x , i · · · ad x k + , i ( x )] + " y , k + X j = ad x , i · · · ad α ([ x j , i , y , i ]) · · · ad x k + , i ( x ) = [ α ( y ) , ad y , i ad x , i · · · ad x k , i ([ x k + , i , x ])] + k X j = [ α ( y ) , ad x , i · · · ad α ([ x j , i , y , i ]) · · · ad x k , i ([ x k + , i , x ])]+ [ y, ad x , i · · · ad x k , i ad α ([ x k + , i , y , i ]) ( x )]= ad y ad y , i ad x , i · · · ad x k , i ([ x k + , i , x ]) + k X j = ad y ad x , i · · · ad α ([ x j , i , y , i ]) · · · ad x k , i ([ x k + , i , x ])+ [ y, ad x , i · · · ad x k , i ad α ([ x k + , i , y , i ]) ( x )]= ad y ad x , i · · · ad x k , i ad y , i ([ x k + , i , x ]) − k X j = ad y ad x , i · · · ad α ([ x j , i , y , i ]) · · · ad x k , i ([ x k + , i , x ])+ k X j =1 ad y ad x , i · · · ad α ([ x j , i , y , i ]) · · · ad x k , i ([ x k + , i , x ]) + [ y , ad x , i · · · ad x k , i ad α ([ x k + , i , y , i ]) ( x )]= ad y ad x , i · · · ad x k , i ad y , i ([ x k + , i , x ]) + [ y , ad x , i · · · ad x k , i ad α ([ x k + , i , y , i ]) ( x )] ∈ I , if s i >
1, we have[ y, ad x , i · · · ad x ri , i ad y , i · · · ad y si , i ( x )] = [ y , ad x , i · · · ad x ri , i ad y , i · · · ad y si − , i ([ α ( y s i , i ) , x ])]= [ α ( y ) , ad x , i · · · ad x ri , i ad y , i · · · ad y si − , i ([ y s i , i , x ])]= ad y ad x , i · · · ad x ri , i ad y , i · · · ad y si − , i ([ y s i , i , x ]) ∈ I . Therefore we get [ y, δ ( x )] ∈ I , i.e., [ L, I ] ⊆ I . It’s straightforward to show that α ( I ) ⊆ I . If now [ x, α ( I )] = { } for x ∈ L , then ad x ˜I ( L ) = { } . In particular, for y, z ∈ I , wehave ad x ad y ( z ) = 0 and so x ∈ Ann L ( I ) = { } , which is zero by Lemma 5.8.Finally, { } 6 = [ α ( q ) , I ] = ad q ˜I ( L ) ⊆ A ( L ) ⊆ L , which implies that Q is ideallyabsorbed into L . According to Theorem 3.10, Q is an algebra of quotients of L .(1) ⇒ (2) Since Q is an algebra of quotients of L , we have Ann( Q ) = { } . Next let µ = P i ≥ ad q i , · · · ad q i , ri ∈ A ( Q ) \ { } . We may of course assume that all q i,j are nonzero17lements in Q . Set s = P i ≥ r i . As Q is an algebra of quotients of L , there exists, for every i and j , a Hom-ideal J i,j of L such that Ann L ( J i , j ) = { } , and { } 6 = [ α ( q i,j ) , J i,j ] ⊆ L . ByLemma 5.8, the Hom-ideal J = ∩ i,j J i,j and hence also I = J s will have zero α -annihilatorin L . Moreover, [ α ( q i,j ) , I ] ⊆ [ α ( q i,j ) , J i,j ] ⊆ L . According to Corollary 5.7 and takinginto account that J s ⊆ J r i for every i , we have µ ˜ I ⊆ A and ˜ Iµ ⊆ A .If ˜ Iµ = { } , then µ ∈ rann A ( Q ) ( A Q ( I )) which is zero by Lemma 5.4.Finally, suppose that µ = ad q where q ∈ A ( Q ) \ { } . Note that Q is also an algebra ofquotients of I , this implies that there exist elements y, z ∈ I such that 0 = [ α ( q ) , [ y, α ( z )]].But [ α ( q ) , [ y, α ( z )]] = ad q ad y ( z ) ∈ ad q ˜I ( L ).Recall that an associative algebra S is said to be a left quotient algebra of a subalgebra A if whenever p and q ∈ S , with p = 0, there exists x in A such that xp = 0 and xq ∈ A .An associative algebra A has a left quotient algebra if and only if it has no total rightzero divisors different from zero. (Here, an element x in A is said to be a total right zerodivisor if Ax = { } .) Lemma 5.10. [16] Let A be a subalgebra of an associative algebra Q . Then Q is a leftquotient algebra of L if and only if for every nonzero element q ∈ Q there exists a leftideal I of A with rann A ( I ) = { } such that { } 6 = Iq ⊆ A . Theorem 5.11.
Suppose that ( L, [ · , · ] , α ) is a semiprime Hom-subalgebra of ( Q, [ · , · ] , α ) .Moreover, suppose that Q is an algebra of quotients of L . Then A ( Q ) is a left quotientalgebra of A .Proof. Let µ ∈ A ( Q ) \ { } and let I be a Hom-ideal of L satisfying condition (2) inProposition 5.9. Set J = A ˜ I + ˜ I , a left ideal of A that satisfies { } 6 = J µ ⊆ A (because { } 6 = ˜ Iµ ⊆ A ).Since also Ann L ( I ) = { } , we obtain from Lemma 5.4 that rann A ( Q ) ( A Q ( I )) = { } .This, together with the fact that ˜ I contains A Q ( I ), implies that rann A ( Q ) ( ˜ I ) = { } . Since rann A ( ˜ I ) ⊆ rann A ( Q ) ( ˜ I ) we get that also rann A ( ˜ I ) = { } . Hence we get rann A ( J ) = { } since ˜ I ⊆ J . This concludes the proof according to Lemma 5.10. Remark 5.12.
As established in the proof of the previous result, if ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) and if ( L, [ · , · ] , α ) is semiprime and I is a Hom-ideal of ( L, [ · , · ] , α ) with Ann L ( I ) = { } , then A ˜ I + ˜ I is a left ideal of A with zero right annihilator. In this section, we mainly study algebras of quotients of Hom-Lie algebras via theirdense extensions which is introduced by Cabrera in [3]. We show that dense extensioncan be lifted from a Hom-Lie algebra to its Hom-ideals if the extension is also an algebra18f quotients. For any Hom-Lie algebra ( L, [ · , · ] , α ), let M ( L ) be the associative algebragenerated by the identity map together with inner derivations of L . Moreover, M ( L ) isnothing but the unitization of A ( L ).Following [3], given an extension of Hom-Lie algebras L ⊆ Q , the annihilator of L in M ( Q ) is defined by L ann := { µ ∈ M ( Q ) | µ ( x ) = 0 , x ∈ L } . If L ann = { } , then L is said to be a dense Hom-subalgebra of Q , and we will say that L ⊆ Q is a dense extension of Hom-Lie algebras. Lemma 6.1.
Suppose that ( L, [ · , · ] , α ) is a Hom-subalgebra of ( Q, [ · , · ] , α ) with Ann( Q ) = { } . Then the following conditions are equivalent:(1) L is a dense Hom-subalgebra of Q .(2) If µ ( L ) = { } for some µ in A ( Q ) , then µ = 0 .Proof. Clearly, (1) implies (2). Conversely, suppose that µ ∈ M ( Q ) satisfies µ ( L ) = { } . If µ ( p ) = 0 for some p in Q , then use Ann( Q ) = { } to find a nonzero elements q in Q satisfying ad q µ ( p ) = 0. But then ad q µ ( L ) = { } and since A ( Q ) is a two-sidedideal of M ( Q ), we have that ad q µ ∈ A ( Q ). Hence, condition (2) yields ad q µ = 0, acontradiction.Following Cabrera and Mohammed in [4], we say that a Hom-Lie algebra ( L, [ · , · ] , α ) ismultiplicatively semiprime whenever L and its multiplication algebra M ( L ) are semiprime.Observe that in this situation, and if L is a Hom-Lie algebra, then A ( L ), being an idealof M ( L ), will also be a semiprime algebra. Lemma 6.2.
Suppose that ( L, [ · , · ] , α ) is a dense Hom-subalgebra of ( Q, [ · , · ] , α ) and I aHom-ideal of L . If µ is an element of M ( Q ) such that µ ( I ) = { } , then µ ( M ( Q )( I )) = { } . Proof. Consider the set H = { ϕ ∈ M ( Q ) | µωϕ ( I ) = { } for each ω ∈ M ( Q ) such that µω ( I ) = { }} . It’s clear that H is a subalgebra of M ( Q ). Indeed, for any ϕ, ψ ∈ H and each ω ∈ M ( Q ) such that µω ( I ) = { } , we get that µωϕψ ( I ) = { } since ωϕ ∈ M ( Q ) satisfying µωϕ ( I ) = { } and ψ ∈ H , which implies that ϕψ ∈ H .Furthermore, for ω ∈ M ( Q ) such that µω ( I ) = { } , x ∈ I and y ∈ L we see that0 = µω ([ α ( x ) , y ]) = µω ad x ( y ) . Therefore µω ad x lies in L ann , and so it is equal to zero. Consequently for each z ∈ Q , itcan be seen that0 = µω ad x ( z ) = µω ([ α ( x ) , z ]) = − µω ([ α ( z ) , x ]) = − µω ad z ( x ) . µω ad z ( I ) = { } , and hence ad z belongs to H . Thus H = M ( Q ).Finally the statement follows directly from the definition of H and from the fact that, byassumption, µ id Q ( I ) = { } . Corollary 6.3.
Suppose that ( L, [ · , · ] , α ) is a dense Hom-subalgebra of ( Q, [ · , · ] , α ) and I, J two Hom-ideals of L . If [ α ( I ) , J ] = { } , then [ M ( Q )( I ) , α ( M ( Q )( J ))] = { } .Proof. First, apply Lemma 6.2 to the Hom-ideal J taking µ = ad x for any x ∈ I toobtain the result [ α ( I ) , M ( Q )( J )] = { } . Now, to conclude the proof, apply once moreLemma 6.2 to the Hom-ideal I taking µ = ad z where z ∈ M ( Q )( J ). Corollary 6.4.
Suppose that ( L, [ · , · ] , α ) is a dense Hom-subalgebra of ( Q, [ · , · ] , α ) . If ( Q, [ · , · ] , α ) is semiprime, then ( L, [ · , · ] , α ) is also semiprime.Proof. Otherwise, there exists a nonzero Hom-ideal I such that [ I, α ( I )] = { } .By Corollary 6.3, [ M ( Q )( I ) , α ( M ( Q )( I ))] = { } where M ( Q )( I ) is a Hom-ideal of Q ,contradiction. Lemma 6.5.
Suppose that ( L, [ · , · ] , α ) is a dense Hom-subalgebra of ( Q, [ · , · ] , α ) . If Q ismultiplicatively semiprime, then L is also multiplicatively semiprime.Proof. It’s sufficient to show that M ( L ) is semiprime by Corollary 6.4. Suppose that µ in M ( L ) satisfies µM ( L ) µ = 0. Then µ ( I ) = { } where I denotes the Hom-ideal of L generated by µ ( L ). Now, choose an element of M ( Q ), say ˆ µ , satisfying ˆ µ | L = µ | L . Henceˆ µ ( I ) = { } . By Lemma 6.2, ˆ µ ( M ( Q )( I )) = { } , and so ˆ µM ( Q )ˆ µ ( L ) = { } . Consequentlywe have ˆ µM ( Q )ˆ µ = 0 since L ⊆ Q is dense. Note that M ( Q ) is semiprime, we get ˆ µ = 0,and so µ = 0. Remark 6.6.
Given a dense extension of Hom-Lie algebras L ⊆ Q , where Q is mul-tiplicatively semiprime, we have that L is also multiplicatively semiprime from Lemma6.5. Proposition 6.7.
Suppose that ( L, [ · , · ] , α ) is a dense Hom-subalgebra of ( Q, [ · , · ] , α ) and Q a multiplicative semiprime algebra of quotients of L . Then I ⊆ Q is a dense extensionfor every essential Hom-ideal of I of L .Proof. We first observe that Ann( Q ) = { } since Q is an algebra of quotients of L and hence Lemma 6.1 applies. Thus, let µ be in A ( Q ) such that µ ( I ) = { } , andby way of contradiction assume that µ = 0. According to Theorem 5.11, A ( Q ) is a leftquotients algebra of A and hence there exists λ in A such that 0 = λµ ∈ A . Since theextension L ⊆ Q is dense, λµ ( L ) = { } , and since Ann L ( I ) = { } , there exists a nonzeroelement y in I such that ad y λµ ( L ) = { } . Using now that A ( Q ) has no total right zerodivisor according to Lemma 5.1, we get that A ( Q )ad y λµ = 0, and this, coupled with thesemiprimeness of A ( Q ), implying that A ( Q )ad y λµ A ( Q )ad y λµ = 0. A second applicationof the fact that L ⊆ Q is a dense extension yields A ( Q )ad y λµ A ( Q )ad y λµ ( L ) = { } .20owever, µ ( I ) = { } by assumption and thus implies that µM ( Q )( I ) = { } . But this isa contradiction, because of the containments µA ( Q )ad y λµ ( L ) ⊆ µ A ( Q )([ α ( I ) , L ]) ⊆ µ A ( Q )( I ) ⊆ µ M ( Q )( I ) = { } . This completes the proof.
Corollary 6.8.
Suppose that ( L, [ · , · ] , α ) is a dense Hom-subalgebra of ( Q, [ · , · ] , α ) and Q a multiplicative semiprime algebra of quotients of L . Then for every essential Hom-ideal I of L , lann A ( Q ) ( ˜ I ) = { } .Proof. Let µ ∈ lann A ( Q ) ( ˜ I ). Then, if y ∈ I , we have µ ad y ( I ) = { } . This impliesthat µ ( I ) = { } . By Proposition 6.7 applied to the essential Hom-ideal I of L , theextension I ⊆ Q is dense, so µ = 0. References [1] Anquela, J. A., Garc´ıa, E., G´omez-Lozano, M.
Maximal algebras of Martindale-likequotients of strongly prime linear Jordan algebras.
J. Algebra 280(2004), No. 1, 367-383.[2] Anquela, J. A., McCrimmon, K.
Martindale quotients of Jordan algebras.
J. PureAppl. Algebra 213(2009), No. 3, 299-312.[3] Cabrera, M.
Ideals which memorize the extended centroid.
J. Algebras Appl. 1(2002),No. 3, 281-288.[4] Cabrera, M., Mohammed, A. A.
Extended centroid and central closure of multiplica-tively semiprime algebra.
Comm. Algebra 29(2001), No. 3, 1215-1233.[5] Garc´ıa, E., Lozano, M. G.
Jordan systems of Martindale-like quotients.
J. Pure Appl.Algebra 194(2004), No. 1-2, 127-145.[6] Garc´ıa, E., Lozano, M. G.
Quotients in graded Lie algebras. Martindale-like quotientsfor Kantor pairs and Lie triple systems.
Algebr. Represent. Theory 16(2013), No. 1,229-238.[7] Gohr, A.
On hom-algebras with surjective twisting.
J. Algebra 324(2010), No. 7,1483-1491.[8] Guo, W., Chen, L.
Algebras of quotients of Jordan-Lie algebras.
Comm. Algebra44(2016), No. 9, 3788-3795.[9] Hartwig, J., Larsson, D., Silvestrov, S.
Deformations of Lie algebras using σ -derivations. J. Algebra 295(2006), No. 2, 314-361.2110] Ma, Y., Chen, L., Lin, J.
Systems of quotients of Lie triple systems.
Comm. Algebra42(2014), No. 8, 3339-3349.[11] Mart´ınez, C.
The ring of fractions of a Jordan algebra.
J. Algebra 237(2001), No. 2,798-812.[12] Martindale, W. S. III
Prime rings satisfying a generalized polynomial identity.
J.Algebra 12(1969), 576-584.[13] Montaner, F.
Algebras of quotients of Jordan algebras.
J. Algebra 323(2010), No. 10,2638-2670.[14] Passman, D. S.
Computing the symmetric ring of quotients.
J. Algebra 105(1987),207-235.[15] Perera, F., Siles Molina, M.
Associative and Lie algebras of quotients.
Publ. Mat.52(2008), 129-149.[16] Siles Molina, M.
Algebras of quotients of Lie algebras.
J. Pure Appl. Algebra188(2004), No. 1-3, 175-188.[17] Sheng, Y.
Representations of Hom-Lie algebras.
Algebr. Represent. Theory 15(2012),No. 6, 1081-1098.[18] Utumi, Y.
On quotient rings.
Osaka Math. J. 8(1956), 1-18.[19] Yau, D.
The Hom-Yang-Baxter equation and Hom-Lie algebras.
J. Math. Phys.52(2011), No. 5, 053502, 19 pp.[20] Yau, D.