Algebras of quotients and Martindale-like quotients of Leibniz algebras
aa r X i v : . [ m a t h . R A ] A ug Algebras of quotients and Martindale-likequotients of Leibniz algebras
Chenrui Yao , Yao Ma , Liming Tang , Liangyun Chen School of Mathematics and Statistics, Northeast Normal University,Changchun 130024, China School of Mathematical Sciences, Harbin Normal University,Harbin 150025, China
Abstract
In this paper, the definitions of algebras of quotients and Martandale-like qoutientsof Leibniz algebras are introduced and the interactions between the two quotients aredetermined. Firstly, some important properties which not only hold for a Leibniz al-gebras but also can been lifted to its algebras of quotients are investigated. Secondly,for any semiprime Leibniz algebra, its maximal algebra of quotients is constucted anda Passman-like characterization of the maximal algebra is described. Thirdly, the rela-tionship between a Leibniz algebra and the associative algebra which is generated by leftand right multiplication operators of the corresponding Leibniz algebras of quotients areexamined. Finally, the definition of dense extensions and some vital properties aboutLeibnia algebras via dense extensions are introduced.
Keywords:
Leibniz algebras; Algebras of quotients; Martindale-like quotients; denseextensions.
The notion of Leibniz algebras was first introduced by Loday in [15], which is ageneralization of Lie algebras where the skew-symmetry is omitted. This gives rise to twotypes of Leibniz algebras: left Leibniz algebra and right Leibniz algebra. A right(resp.left) Leibniz algebra is a vector space L over F with a bilinear map [ · , · ] : L × L → L satisfying for all x, y, z in L [ x, [ y, z ]] = [[ x, y ] , z ] − [[ x, z ] , y ] . (resp . [ x , [ y , z ]] = [[ x , y ] , z ] + [ y , [ x , z ]] . ) Corresponding author(L. Chen): [email protected] by NNSF of China (Nos. 11771069 and 11801066). L . If L is both a right and left Leibniz algebra, then L is called a symmetric Leibniz algebra,introduced by Geoffrey and Gaywalee in [13]. In recent years, study about Leibniz algebrasis becoming more and more popular and many results of Lie algebras have been generalizedto Leibniz algebras, such as references [4, 5]. For more references about Leibniz algebras,one can see [6–8].The notion of ring of quotients was introduced by Utumi in [23], which played animportant role in the development of the theories of associative and commutative rings. Heshowed that every associative ring without right zero divisors has a maximal left quotientsring and constructed it. Inspired by [23], Siles Molina studied the algebras of quotients ofLie algebras in [22]. Martindale rings of quotients were introduced by Martindale in 1969for prime rings in [16], which was designed for applications to rings satisfying a generalizedpolynomial identity. In [12], Garc´ıa and G´omez defined Martindale-like quotients for Lietriple systems with respect to power filters of sturdy ideals and constructed the maximalsystem of quotients in the nondegenerate cases. More research about quotients of Liesystems refer in references [14, 19]. In [17], Mart´ınez derived a necessary and sufficientOre type condition for a Jordan algebra to have a ring of fractions, which is the originof algebras of quotients of Jordan systems. In [1, 2, 11, 18], authors researched algebrasof quotients and Martindale-like quotients of Jordan systems respectively. Inspired by[22, 23], we will introduce the notion of algebras of quotients of Leibniz algebras andprove the properties, such as semiprimeness and primeness, can be lifted from a Leibnizalgebra to its algebra of quotients in this paper.As we know, one popular topic of research is the relationship between algebras ofquotients of non-associative algebras and associative algebras in recent years. In [21],authors examined the relationship between associative and Lie algebras of quotients. In-spired by them, we’ll explore whether there exists a relationship between associative andLeibniz algebras of quotients. The answer is affirmative and we will describe in detail inthe following.The paper is organised as follows: in Section 2, some basic definition are introducedand some elemental propositions are given. In Section 3, the definition of quotients andideally absorbed are introduced first, which proved to be equivalent in the following,presented as Theorem 3.10. What’s more, the concept of Martindale-like quotients areintroduced and the relationship between them are determined(See Proposition 3.16). InSection 4, we focus on semiprime Leibniz algebras and construct the maximal algebras ofquotients of them and characterize them. In Section 5, we mainly study the relationshipbetween algebras of quotients of Leibniz algebras and associative algebras generated byleft and right multiplication operators of Leibniz algebras, and get the main result that A ( Q ), the associative algebra generated by left and right multiplication operators of Q , isa left quotients of A if Q is an algebra of quotients of L , where A denotes the subalgebraof A ( Q ) satisfying A ( L ) ⊆ L , written as Theorem 5.11. In Section 6, we introduce theconcept of dense extension first of all. Next, we get a proposition about dense extension,Proposition 6.5, that every I ⊆ Q is also dense for every essential ideal I of L if L ⊆ Q is a dense extension and Q is an algebra of quotients of L . Also, we find the possible2onverse, Proposition 6.9, to Theorem 5.11 in the presence of dense extensions of Leibnizalgebras.In Sections 2, 3 and 4, we suppose that the Leibniz algebras are right Leibniz algebras.While in Sections 5 and 6, we assume that all Leibniz algebras are symmetric Leibnizalgebras. Definition 2.1.
Let L be a Leibniz algebra and H a subset of L .(1) The vector space lan L ( H ) = { x ∈ L | [ x, y ] = 0 , ∀ y ∈ H } is called the left annihilatorof H in L . We denote lan L ( L ) by lan ( L ) .(2) The vector space ran L ( H ) = { x ∈ L | [ y, x ] = 0 , ∀ y ∈ H } is called the rightannihilator of H in L . We denote ran L ( L ) by ran ( L ) .(3) The vector space Ann L ( H ) = lan L ( H ) ∩ ran L ( H ) is called the annihilator of H in L .We denote Ann L ( L ) by Ann( L ) . Remark 2.2. (1) ran L ( H ) is an ideal of L if H is a right ideal of L .(2) Ann L ( H ) is an ideal of L if H is an ideal of L .Proof. (1) For any z ∈ L , x ∈ ran L ( H ) and y ∈ H ,[ y, [ x, z ]] = [[ y, x ] , z ] − [[ y, z ] , x ] = 0 , [ y, [ z, x ]] = [[ y, z ] , x ] − [[ y, x ] , z ] = 0 , which imply that both [ x, z ] and [ z, x ] belong to ran L ( H ), i.e., ran L ( H ) is an ideal of L .(2) For any z ∈ L , x ∈ Ann L ( H ) and y ∈ H ,[ y, [ x, z ]] = [[ y, x ] , z ] − [[ y, z ] , x ] = 0 , [[ x, z ] , y ] = [[ x, y ] , z ] + [ x, [ z, y ]] = 0 , which imply that [ x, z ] ∈ Ann L ( H ), i.e., Ann L ( H ) is a right ideal of L . Similarly, we canshow that Ann L ( H ) is a left ideal of L . Thus, Ann L ( H ) is an ideal of L . Definition 2.3.
Suppose that L is a Leibniz algebra. Then L is semiprime if for anynonzero ideal I of L , [ I, I ] = { } and prime if for any two nonzero ideals I , J of L , [ I, J ] = { } . Proposition 2.4.
For a Leibniz algebra L , we have Ann( L ) = ran ( L ) = { } if L issemiprime. roof. Otherwise, ran ( L ) is a nonzero ideal of L such that [ ran ( L ) , ran ( L )] ⊆ [ L, ran ( L )] = { } according to the definition of ran ( L ), which contradicts with L be-ing semiprime. Thus ran ( L ) = { } , consequently Ann( L ) = { } . Definition 2.5.
An ideal I of a Leibniz algebra L is called essential if for any nonzeroideal J of L , I ∩ J = { } and sturdy if Ann L ( I ) = { } . Proposition 2.6.
Suppose that L is a Leibniz algebra and I an ideal of L .(1) If Ann L ( I ) = { } (resp . ran L ( I ) = { } ) , then I is essential.(2) If L is semiprime, then I ∩ Ann L ( I ) = { } (resp . I ∩ ran L ( I ) = { } ) and I is essentialif and only if Ann L ( I ) = { } (resp . ran L ( I ) = { } ) .Proof. (1) Otherwise, there exists a nonzero ideal J of L such that I ∩ J = { } .Take 0 = x ∈ J . We get[ x, I ] ⊆ I ∩ J = { } , [ I, x ] ⊆ I ∩ J = { } , which implies that 0 = x ∈ Ann L ( I )(resp . = x ∈ ran L ( I )). Contradiction.(2) Suppose that I ∩ Ann L ( I ) = { } (resp . I ∩ ran L ( I ) = { } ). Then we can see that I ∩ Ann L ( I )(resp . I ∩ ran L ( I )) is a nonzero ideal of L satisfying[ I ∩ Ann L ( I ) , I ∩ Ann L ( I )] ⊆ [ I , Ann L ( I )] = { } , (resp . [ I ∩ ran L ( I ) , I ∩ ran L ( I )] ⊆ [ I , ran L ( I )] = { } ) , which contradicts with L being semiprime.Now suppose that I is essential. If Ann L ( I ) = { } (resp . ran L ( I ) = { } ), then I ∩ Ann L ( I ) = { } (resp . I ∩ ran L ( I ) = { } ), contradiction. Combining with (1), we come tothe conclusion. Proposition 2.7.
Suppose that L is a Leibniz algebra. Then L is prime if and only if forevery nonzero ideal I of L , ran L ( I ) = { } .Proof. Suppose that L is prime. If there exists a nonzero ideal J of L with ran L ( J ) = { } , then [ J, ran L ( J )] = { } according to the definition of ran L ( J ), which contradicts with L being prime.Conversely, suppose that for every nonzero ideal I of L , ran L ( I ) = { } . If L isn’tprime, there exists two nonzero ideals I, J of L such that [ J, I ] = { } , which implies that I ⊆ ran L ( J ). Contradiction. 4 Algebras of quotients of Leibniz algebras
Inspired by the notion of algebras of quotients of associative algebras and Lie algebrasin [22, 23], we introduce the notion of algebras of quotients of Leibniz algebras.Suppose that L is a subalgebra of the Leibniz algebra Q . For any q ∈ Q , set L ( q ) = F q + ( n X i =1 ξ i ( q ) | ξ i ∈ A ( L ) with n ∈ N ) where A ( L ) denotes the associative algebra generated by { R x , L y | x, y ∈ L } .That is, L ( q ) is the linear span in Q of the elements of the form ξ ( q ) and q where ξ ∈ A ( L ). It’s obvious that for any x, y ∈ L , both R x ( L ( q )) and L y ( L ( q )) still be containedin L ( q ). Now define ( L : q ) = { x ∈ L | [ x, L ( q )] ⊆ L, [ L ( q ) , x ] ⊆ L } . Clearly, ( L : q ) is equal to L if q ∈ L . Proposition 3.1.
Suppose that L is a subalgebra of Q . Then ( L : q ) is an ideal of L .Moreover, it’s maximal among the ideals I of L such that [ I, q ] ⊆ L and [ q, I ] ⊆ L .Proof. For any x ∈ ( L : q ) and y ∈ L , we have[[ x, y ] , L ( q )] = [[ x, L ( q )] , y ] + [ x, [ y, L ( q )]] ⊆ L, [ L ( q ) , [ x, y ]] = [[ L ( q ) , x ] , y ] − [[ L ( q ) , y ] , x ] ⊆ L, which imply that [ x, y ] ∈ ( L : q ). Similarly, we have [ y, x ] ∈ ( L : q ). Hence, ( L : q ) is anideal of L .Suppose that I is an ideal of L such that [ I, q ] ⊆ L and [ q, I ] ⊆ L . Our goal is toshow that [ I, L ( q )] ⊆ L and [ L ( q ) , I ] ⊆ L .For any y ∈ L ( q ) with y = q , there exist ξ , · · · , ξ n ∈ A ( L ) such that y = P ni =1 ξ i ( q ).Moreover, for each 1 ≤ i ≤ n , there exist η i , · · · , η r i i such that ξ i = η r i i · · · η i where η ji = R u or L v for some u, v ∈ L for every 1 ≤ j ≤ r i . Indeed, we only need to show thatfor each 1 ≤ i ≤ n , [ I, ξ i ( q )] ⊆ L and [ ξ i ( q ) , I ] ⊆ L . Let’s show that by induction on r i .When r i = 1, if η i = R x for some x ∈ L , we have[ I, ξ i ( q )] = [ I, η i ( q )] = [ I, R x ( q )] = [ I, [ q, x ]] = [[ I, q ] , x ] − [[ I, x ] , q ] ⊆ L, [ ξ i ( q ) , I ] = [ η i ( q ) , I ] = [ R x ( q ) , I ] = [[ q, x ] , I ] = [[ q, I ] , x ] + [ q, [ x, I ]] ⊆ L, if η i = L y for some y ∈ L , we have[ I, ξ i ( q )] = [ I, η i ( q )] = [ I, L y ( q )] = [ I, [ y, q ]] = [[ I, y ] , q ] − [[ I, q ] , y ] ⊆ L, [ ξ i ( q ) , I ] = [ η i ( q ) , I ] = [ L y ( q ) , I ] = [[ y, q ] , I ] = [[ y, I ] , q ] + [ y, [ q, I ]] ⊆ L. r i = k . When r i = k +1, denote η ki · · · η i ( q )by z . By induction hypothesis, [ I, z ] ⊆ L and [ z, I ] ⊆ L . If η k +1 i = R x for some x ∈ L , wehave[ I, ξ i ( q )] = [ I, η k +1 i η ki · · · η i ( q )] = [ I, η k +1 i ( z )] = [ I, [ z, x ]] = [[ I, z ] , x ] − [[ I, x ] , z ] ⊆ L, [ ξ i ( q ) , I ] = [ η k +1 i η ki · · · η i ( q ) , I ] = [ η k +1 i ( z ) , I ] = [[ z, x ] , I ] = [[ z, I ] , x ] + [ z, [ x, I ]] ⊆ L. Similarly, we have both [
I, ξ i ( q )] and [ ξ i ( q ) , I ] are contained in L if η k +1 i = L y for some y ∈ L .Therefore, I ⊆ ( L : q ) and the proof is completed. Definition 3.2.
Suppose that L is a subalgebra of Q . Then Q is called an algebra ofquotients, if given p, q ∈ Q with p = 0 , there exists x ∈ L such that [ x, p ] = 0 and x ∈ ( L : q ) or there exists y ∈ L such that [ p, y ] = 0 and y ∈ ( L : q ) . Proposition 3.3.
Let L be a subalgebra of Q .(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 ) = { } .Proof. (1) For any x, y ∈ L with x = 0, x / ∈ Ann( L ) by assumption. Then thereexists u ∈ L such that [ u, x ] = 0 or v ∈ L such that [ x, v ] = 0. Obviously, u, v ∈ ( L : y ).(2) For any 0 = q ∈ Q , there exists x ∈ L such that [ q, x ] = 0 or y ∈ L such that[ y, q ] = 0, which implies that q / ∈ Ann Q ( L ). Hence Ann Q ( L ) = { } . Similarly we haveAnn( L ) = { } .The above proposition says for a Leibniz algebra L , that Ann( L ) = { } is a sufficientand necessary condition such that L has algebras of quotients.We will prove that some properties of a Leibniz algebra L can be inherited by itsalgebras of quotients Q . Actually, Q just needs a weaker condition. Definition 3.4.
Suppose that L is a subalgebra of Q . Then Q is called a weak algebra ofquotients of L , if given = q ∈ Q , there exists x ∈ L such that = [ q, x ] ∈ L or y ∈ L such that = [ y, q ] ∈ L . Remark 3.5.
Every algebra of quotients of a Leibniz algebra is a weak algebra of quotients.
Proposition 3.6.
Let Q be a weak algebra of quotients of L .(1) If I is a nonzero ideal of Q , then I ∩ L is a nonzero ideal of L .(2) If L is semiprime(prime), so is Q . roof. (1) Take 0 = x ∈ I . There exists y ∈ L such that 0 = [ x, y ] ∈ L or z ∈ L such that 0 = [ z, x ] ∈ L . Note that I is an ideal of Q , both [ x, y ] and [ z, x ] belong to I .Therefore, 0 = [ x, y ] ∈ I ∩ L or 0 = [ z, x ] ∈ I ∩ L . It’s straightforward to show that I ∩ L is an ideal of L .(2) Suppose that Q isn’t semiprime. Then there exists a nonzero ideal I of Q suchthat [ I, I ] = { } . By (1), I ∩ L is a nonzero ideal of L such that [ I ∩ L, I ∩ L ] ⊆ [ I, I ] = { } ,which contradicts with L being simiprime.Similarly, we can show that Q is prime if L is prime. Definition 3.7.
Let L be a subalgebra of Q . Then Q is called ideally absorbed into L , iffor each = q ∈ Q , there exists an ideal I of L with Ann L ( I ) = { } such that [ I, q ] = { } or [ q, I ] = { } and both [ I, q ] and [ q, I ] are contained in L . Proposition 3.8.
Suppose that L is a subalgebra of Q . Take q ∈ Q .(1) If Q is an algebra of quotients of L , then ( L : q ) is an essential ideal of L . Moreover, Ann L (( L : q )) = { } .(2) If Q is ideally absorbed into L , then ( L : q ) is an essential ideal of L . Moreover, Ann L (( L : q )) = { } .Proof. (1) Let I be a nonzero ideal of L . Take 0 = x ∈ I . Apply that Q is analgebra of quotients of L to find y ∈ L such that [ x, y ] = 0 , y ∈ ( L : q ) or z ∈ L suchthat [ z, x ] = 0 , z ∈ ( L : q ). Note that ( L : q ) is an ideal of L , we have both [ x, y ] and[ z, x ] belong to ( L : q ) ∩ I , which implies that ( L : q ) ∩ I = { } . Therefore, ( L : q ) is anessential ideal of L .Suppose that Ann L (( L : q )) = { } . Then ( L : q ) ∩ Ann L (( L : q )) = { } since ( L : q ) isessential. Take 0 = u ∈ ( L : q ) ∩ Ann L (( L : q )). Apply that Q is an algebra of quotients of L to find w ∈ L such that [ u, w ] = 0 , w ∈ ( L : q ) or v ∈ L such that [ v, u ] = 0 , v ∈ ( L : q ).However, both [ u, w ] and [ v, u ] are equal to 0 since u ∈ Ann L (( L : q )). Contradiction.(2) Since Q is ideally absorbed into L , there exists a nonzero ideal I of L withAnn L ( I ) = { } such that [ I, q ] ⊆ L and [ q, I ] ⊆ L . According to the proof of Proposition3.1, I ⊆ ( L : q ). Hence, Ann L (( L : q )) ⊆ Ann L ( I ) = { } . By Proposition 2.6 (1), we get( L : q ) is an essential ideal of L . Lemma 3.9.
Let Q be a weak algebra of quotients of L . Let I be an ideal of L with Ann L ( I ) = { } . Then there is no nonzero element x in Q such that [ x, I ] = [ I, x ] = { } .Proof. Suppose that there exists 0 = x ∈ Q such that [ x, I ] = [ I, x ] = { } . Applythat Q is a weak algebra of quotients of L to find y ∈ L such that 0 = [ x, y ] ∈ L or z ∈ L such that 0 = [ z, x ] ∈ L .If 0 = [ x, y ] ∈ L , we have [ I, [ x, y ]] = { } or [[ x, y ] , I ] = { } since Ann L ( I ) = { } .Either of the two situations leads to a contradiction. Indeed,[ I, [ x, y ]] = [[ I, x ] , y ] − [[ I, y ] , x ] = { } , x, y ] , I ] = [[ x, I ] , y ] + [ x, [ y, I ]] = { } . The situation is similarly if 0 = [ z, x ] ∈ L . Therefore the proof is completed. Theorem 3.10.
Suppose that L is a subalgebra of Q . Then Q is an algebra of quotientsof 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 )is an ideal of L with Ann L (( L : q )) = { } according to Proposition 3.8 (1). By Lemma3.9, [ q, ( L : q )] = { } or [( L : q ) , q ] = { } . Also, both [ q, ( L : q )] and [( L : q ) , q ] contain in( L : q ) according to the definition of ( L : q ). Hence, Q is ideally absorbed into L .Conversely, suppose that Q is ideally absorbed into L . Then Q is a weak algebraof quotients of L . For any p, q ∈ Q with p = 0, we have Ann L (( L : q )) = { } . Hence,[ p, ( L : q )] = { } or [( L : q ) , p ] = { } according to Lemma 3.9. Therefore, there exists x ∈ ( L : q ) such that [ p, x ] = 0 or y ∈ ( L : q ) such that [ y, p ] = 0. Hence, Q is an algebraof 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 . Proposition 3.11.
Suppose that Q is a weak algebra of quotients of L . Then for everyideal I of L , Ann L ( I ) = { } ⇒ Ann Q ( I ) = { } and ran L ( I ) = { } ⇒ ran Q ( I ) = { } .Proof. Suppose that 0 = q ∈ Ann Q ( I ). By hypothesis, there exists x ∈ L such that0 = [ q, x ] ∈ L or y ∈ L such that 0 = [ y, q ] ∈ L .If 0 = [ q, x ] ∈ L , we have [ I, [ q, x ]] = { } or [[ q, x ] , I ] = { } since Ann L ( I ) = { } .Either of the two situations leads to a contradiction. Indeed,[ I, [ q, x ]] = [[ I, q ] , x ] − [[ I, x ] , q ] = { } , [[ q, x ] , I ] = [[ q, I ] , x ] + [ q, [ x, I ]] = { } . The situation is similarly if 0 = [ y, q ] ∈ L , consequently Ann Q ( I ) = { } .Similarly, we have ran Q ( I ) = { } if ran L ( I ) = { } . Notation 3.12.
Denote by J e ( L ) the set of all essential ideals of a Leibniz algebra L . Clearly, given
I, J ∈ J e ( L ), we have I ∩ J ∈ J e ( L ). If L is semiprime, then I ∈ J e ( L ). Proposition 3.13.
Let L be a semiprime Leibniz algebra and Q an algebra of quotientsof L . Then for every essential ideal I of L , we have Q is also an algebra of quotients of I . roof. For any p, q ∈ Q with p = 0, we have ( L : q ) ∈ J e ( L ). So that I ∩ ( L : q ) ∈ J e ( L ) and consequently ( I ∩ ( L : q )) ∈ J e ( L ). Thus Ann Q (( I ∩ ( L : q )) ) = { } byProposition 3.11. So there exist y, z ∈ I ∩ ( L : q ) such that [[ y, z ] , p ] = 0 or u, v ∈ I ∩ ( L : q )such that [ p, [ u, v ]] = 0. Moreover,[[ y, z ] , I ( q )] = [[ y, I ( q )] , z ] + [ y, [ z, I ( q )]] ⊆ [[ y, L ( q )] , z ] + [ y, [ z, L ( q )]] ⊆ [ L, I ] + [
I, L ] ⊆ I, [ I ( q ) , [ y, z ]] = [[ I ( q ) , y ] , z ] − [[ I ( q ) , z ] , y ] ⊆ [[ L ( q ) , y ] , z ] − [[ L ( q ) , z ] , y ] ⊆ [ L, I ] ⊆ I, which implies that [ y, z ] ∈ ( I : q ). Similarly [ u, v ] ∈ ( I : q ).Therefore, Q is an algebra of quotients of I .Next, let’s focus on Martindale-like quotients. Definition 3.14.
A filter F on a Leibniz algebra is a nonempty family of nonzero idealssuch that for any I , I ∈ F there exists I ∈ F such that I ⊆ I ∩ I . Moreover, F is apower filter if for any I ∈ F there exists K ∈ F such that K ⊆ [ I, I ] . Definition 3.15.
A Leibniz algebra of Martindale-like quotients Q of a Leibniz algebra L with respect to a power filter of sturdy ideals F if L ⊆ Q such that for every nonzeroelement q ∈ Q there exists an ideal I q ∈ F such that [ q, I q ] = { } or [ I q , q ] = { } andboth [ q, I q ] and [ I q , q ] contain in L . Proposition 3.16.
Let L be a Leibniz algebra. Then a Martindale-like quotients of L is also an algebra of quotients of L . Moreover, if L is semiprime, then any algebra ofquotients of L is Martindale-like quotients of L .Proof. Suppose that Q is a Martindale-like quotients of L . Then Q is ideally ab-sorbed into L by definition and so an algebra of quotients of L .Conversely, suppose that Q is an algebra of quotients of L with L semiprime. Thenall sturdy ideals of L form a power filter of L . Indeed, for any sturdy ideal I of L , wehave [ I, I ] is also a sturdy ideal since L is semiprime. So it just need to take K = [ I, I ].Hence by hypothesis, Q is also a Martindale-like quotients of L . In this section, we will construct a maximal algebra of quotients for every semiprimeLeibniz algebra L . Maximal in the sense that every algebra of quotients of L can beconsidered inside this maximal algebra of quotients via a monomorphism which restrictedin L is the identity. Definition 4.1.
Given an ideal I of a Leibniz algebra L , we say that a linear map δ : I → L is a partial derivation if for any x, y ∈ I it satisfies: δ ([ x, y ]) = [ δ ( x ) , y ] + [ x, δ ( y )] . I , L ) the set of all partial derivations of I in L . Lemma 4.2.
Let L be a semiprime Leibniz algebra and consider the set D := { ( δ, I ) | I ∈ J e ( L ) , δ ∈ PDer( I , L ) } . Define on D 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 , δ | I = δ | I , so ( δ, I ) ≡ ( δ, I ).(2) For any ( δ, I ) , ( µ, J ) ∈ D 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 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 ). Notation 4.3.
Denote by Q ( L ) the quotient set D / ≡ . Let δ I denote the equivalenceclass of ( δ, I ) in Q ( L ) . Theorem 4.4.
Let L be a semiprime Leibniz algebra over F , and let Q := Q ( L ) be as inNotation 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 ) where [ δ, µ ] : ( I ∩ J ) → L , x µδ ( x ) − δµ ( x ) . Then Q , with these operations, is a Leibniz algebra containing L as a subalgebra, via themonomorphism: ϕ : L → Q, x ( R x ) L . Proof. (1) For any δ I , µ J ∈ Q and x, y ∈ ( I ∩ J ) ,[ δ, µ ]([ x, y ]) = µδ ([ x, y ]) − δµ ([ x, y ]) = µ ([ δ ( x ) , y ] + [ x, δ ( y )]) − δ ([ µ ( x ) , y ] + [ x, µ ( y )])= [ µδ ( x ) , y ] + [ δ ( x ) , µ ( y )] + [ µ ( x ) , δ ( y )] + [ x, µδ ( y )] − [ δµ ( x ) , y ] − [ µ ( x ) , δ ( y )] − [ δ ( x ) , µ ( y )] − [ x, δµ ( y )]= [[ δ, µ ]( x ) , y ] + [ x, [ δ, µ ]( y )] , which makes sense since δ ( x ), δ ( y ), µ ( x ), µ ( y ), [ δ ( x ) , y ], [ x, δ ( y )], [ µ ( x ) , y ], [ x, µ ( y )] ∈ I ∩ J .Moreover, δ ([ x, y ]), µ ([ x, y ]) ∈ I ∩ J . So [ δ, µ ] ∈ PDer(( I ∩ J ) , L ), i.e., [ · , · ] is well-defined.102) It’s obvious that ( Q, · , +) is a vector space over F . For any δ I , µ J , β K ∈ Q (we canconsider the same T for δ , µ and β because if I, J, K ∈ J e ( L ) are such that δ : I → L , µ : J → L , β : K → L , then T := I ∩ J ∩ K ∈ J e ( L ) and so δ I = δ T and µ J = µ T and β K = β T ),[[ δ T , µ T ] , β T ] − [[ δ T , β T ] , µ T ] = [[ δ, µ ] T , β T ] − [[ δ, β ] T , µ T ]= [[ δ, µ ] , β ] ( T ∩ T ) − [[ δ, β ] , µ ] ( T ∩ T ) = ([[ δ, µ ] , β ] − [[ δ, β ] , µ ]) ( T ∩ T ) = [ δ, [ µ, β ]] ( T ∩ T ) = [ δ T , [ µ, β ] T ] = [ δ T , [ µ T , β T ]] . Therefore, Q is a Leibniz algebra.(3) ϕ is well-defined and a homomorphism by the definitions of Leibniz algebra and[ · , · ]. The injectivity of ϕ follows since ( R x ) L = 0 for some x ∈ L means [ L, x ] = { } andhence x ∈ ran ( L ), which is zero by the semiprimeness of L and Proposition 2.4.For any X ⊆ L , write X ϕ to denote the image of X inside Q ( L ) via the monomorphismdefined in Theorem 4.4. Lemma 4.5.
For every δ I ∈ Q ( L ) , and ( R x ) L ∈ I ϕ ( x ∈ I ) , we have [ δ I , ( R x ) L ] =( R − δ ( x ) ) L ∈ L ϕ and [( R x ) L , δ I ] = ( R δ ( x ) ) L ∈ L ϕ .Proof. For any y ∈ I ,[ δ I , ( R x ) L ]( y ) = R x δ ( y ) − δR x ( y ) = [ δ ( y ) , x ] − δ ([ y, x ]) = [ δ ( y ) , x ] − [ δ ( y ) , x ] − [ y, δ ( x )]= − [ y, δ ( x )] = R − δ ( x ) ( y )and so [ δ I , ( R x ) L ] = ( R − δ ( x ) ) I = ( R − δ ( x ) ) L ∈ L ϕ .Similarly, we have [( R x ) L , δ I ] = ( R δ ( x ) ) L ∈ L ϕ . Proposition 4.6.
Let L be a semiprime Leibniz algebra. Then Q ( L ) is semiprime and analgebra of quotients of L . Moreover, Q ( L ) is maximal among the algebras of quotients of L , in the sense that if S is an algebra of quotients of L , then there exists a monomorphism ψ : S → Q ( L ) which is the identity in L . In particular, the map ψ : S → Q ( L ) , s ( R s ) ( L : s ) is a monomorphism 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 ( R a ) L ∈ L ϕ satisfies[( R a ) L , δ I ] = ( R δ ( a ) ) L = 0. Otherwise, [ L, δ ( a )] = { } , which implies that δ ( a ) ∈ ran ( L ) = { } . And for every λ I ∈ L ϕ ( µ I ), [ λ I , R δ ( a ) ] = ( R − λ ( a ) ) L ∈ L ϕ and [ R a , λ I ] = R λ ( a ) ∈ L ϕ .This imply that ( R a ) L ∈ ( L ϕ : µ I ). Hence, Q ( L ) is an algebra of quotients of L ϕ . Thesemiprimeness of Q ( L ) follows by Proposition 3.6 (2).11ow suppose that S is an algebra of quotients of L and consider the map ψ : S → Q ( L ) , s ( R s ) ( L : s ) . According to the definition of ( L : s ), ψ is well-defined. Moreover, ψ is a monomorphism.To prove the injectivity, suppose that s ∈ S such that ψ ( s ) = 0, that is [ K, s ] = { } forsome ideal K ∈ J e ( L ), K ⊆ ( L : s ). This implies that s ∈ ran S ( K ). Note that K isessential, ran S ( K ) = { } according to Proposition 3.11, so s = 0. Definition 4.7.
For a semiprime Leibniz algebra L , the Leibniz algebra constructed inTheorem 4.4 is called the maximal algebra of quotients of L . Denote it by Q m ( L ) . The axiomatic characterization of the Martindale ring of quotients given by D. Pass-man in [20] has inspired us to give the following description of the maximal algebra ofquotients of a semiprime Leibniz algebra.
Proposition 4.8.
Let L be a semiprime Leibniz algebra and consider an overalgebra S of L . Then there exists a monomorphism 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 [ I, s ] ⊆ L .(2) For s ∈ S and I ∈ J e ( L ) , [ I, s ] = { } implies that s = 0 .(3) For any I ∈ J e ( L ) , δ ∈ PDer( I , L ) , there exists s ∈ S such that δ ( x ) = [ x, s ] forevery x ∈ I .Proof. Define ψ : S → Q m ( L ), s ( R s ) I where I is a nonzero essential ideal of L satisfying [ I, s ] ⊆ L ; this exists by (1), and so the map is well-defined. Moreover, it’s ahomomorphism.The map ψ is injective(if ( R s ) I = 0 for some s ∈ S , by (2), s = 0) and surjec-tive(consider δ I ∈ Q ( L ), by (3), there exists s ∈ S such that δ ( x ) = R s ( x ) for each x ∈ I ,hence δ I = ψ ( s )).Finally, 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 three conditions.(1) Consider q ∈ Q ( L ). According to Propositions 3.8 (1) and 4.6, ( L : q ) ∈ J e ( L ),and by the definition, [( L : q ) , q ] ⊆ L .(2) Take q ∈ Q ( L ) and I ∈ J e ( L ) such that [ I, q ] = { } . We have q ∈ ran Q ( L ) ( I ) = 0according to Propositions 3.11 and 4.6.(3) Given I ϕ ∈ J e ( L ϕ ) and ¯ δ ∈ PDer( I ϕ , L ϕ ), we have to find q ∈ Q ( L ) such thatfor every x ϕ ∈ I ϕ , ¯ δ ( x ϕ ) = [ x ϕ , q ]. Define δ : I → L, x ϕ − ¯ δ (( R x ) L ) . q := δ I is a partial derivation of I in L . Indeed, for any x, y ∈ I , q ([ x, y ]) = δ I ([ x, y ]) = ϕ − ¯ δ (( R [ x,y ] ) L ) = ϕ − ¯ δ ([( R x ) L , ( R y ) L ])= ϕ − ([¯ δ (( R x ) L ) , ( R y ) L ] + [( R x ) L , ¯ δ (( R y ) L )])= [ ϕ − ¯ δ (( R x ) L ) , ϕ − (( R y ) L )] + [ ϕ − (( R x ) L ) , ϕ − ¯ δ (( R y ) L )]= [ δ I ( x ) , y ] + [ x, δ I ( y )] = [ q ( x ) , y ] + [ x, q ( y )] . Moreover, [ x ϕ , q ] = [( R x ) L , δ I ] = ( R δ ( x ) ) L = ( δ ( x )) ϕ = ¯ δ (( R x ) L ) = ¯ δ ( x ϕ ). In [21], authors examined how the notion of algebras of quotients for Lie algebras tiedup with the corresponding well-known concept in the associative case. Inspired by themethod in [21], we mainly study the relationship between Leibniz algebras and the asso-ciative algebras generated by right and left multiplication operators of the correspondingLeibniz algebras of quotients. First of all, we will give some definitions and basic notations.As defined above, A ( L ) denotes the associative subalgebra (possibly without identity)of End( L ) generated by the elements R x and L y for x, y in L .By an extension of Leibniz algebras L ⊆ Q we will mean that L is a subalgebra ofthe Leibniz algebra Q .Let L ⊆ Q be an extension of Leibniz algebras and let A Q ( L ) be the associativesubalgebra of A ( Q ) generated by { R x , L y : x, y ∈ L } .Recall that, given an associative algebra A and a subset X of A , we define the rightannihilator of X in A as ran 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 lan A ( X ). Lemma 5.1.
Let I be an ideal of a Leibniz algebra L with ran ( L ) = { } . Then Ann L ( I ) = { } if and only if ran A ( L ) ( A L ( I )) = { } .Proof. Suppose that Ann L ( I ) = { } . For any µ in ran A ( L ) ( A L ( I )), we have R y µ = 0for any y ∈ I and L z µ = 0 for any z ∈ I . In particular if x ∈ L , we get0 = R y µ ( x ) = [ µ ( x ) , y ] , L z µ ( x ) = [ z, µ ( x )] , and these imply that µ ( x ) ∈ Ann L ( I ) = 0. Hence, µ = 0.13onversely, suppose that ran A ( L ) ( A L ( I )) = { } . Let x be in Ann L ( I ). Then for any y, z in I and u in L , we have L y R x ( u ) = [ y, [ u, x ]] = [[ y, u ] , x ] + [ u, [ y, x ]] = 0 ,R z R x ( u ) = [[ u, x ] , z ] = [[ u, z ] , x ] + [ u, [ x, z ]] = 0 , which imply that R x ∈ ran A ( L ) ( A L ( I )) = 0. Since by assumption ran ( L ) = { } , weobtain x = 0.For 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 is a subalgebra of Q and I an ideal of L . Then h A Q ( I ) i l A Q ( L ) = h A Q ( I ) i r A Q ( L ) = h A Q ( I ) i A Q ( L ) . Proof.
Notice that given x ∈ L and y ∈ I , we have R x R y = R y R x + R [ y,x ] , R x L y = L y R x + L [ y,x ] , L x R y = R y L x + R [ x,y ] , L x L y = L y L x + L [ x,y ] . Then we have h A Q ( I ) i l A Q ( L ) = h A Q ( I ) i r A Q ( L ) . Hence we come to the conclusion. Lemma 5.3.
Suppose that L is a subalgebra of Q and I an ideal of L . Write ˜ I to denotethe ideal of A Q ( L ) generated by A Q ( I ) . Then(1) ran A Q ( L ) ( ˜ I ) = ran A Q ( L ) ( A Q ( I )) .(2) lan A Q ( L ) ( ˜ I ) = lan A Q ( L ) ( A Q ( I )) .Proof. (1) Since A Q ( I ) ⊆ ˜ I , we have ran A Q ( L ) ( ˜ I ) ⊆ ran A Q ( L ) ( A Q ( I )). So it’s enoughto show that ran A Q ( L ) ( A Q ( I )) ⊆ ran A Q ( L ) ( ˜ I ). Let λ ∈ ran A Q ( L ) ( A Q ( I )). By Lemma 5.2we 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 ξ ,i · · · ξ r i ,i η ,i · · · η s i ,i where ξ j,i = R x j,i or L x j,i for 1 ≤ j ≤ r i and η k,i = R y k,i or L y k,i for 1 ≤ k ≤ s i .Since η s i ,i λ = 0, we see that µλ = 0.(2) The proof is similar to (1). Lemma 5.4.
Suppose that L is a subalgebra of Q such that Q is a weak algebra of quotientsof L . Let I be an ideal of L . If Ann L ( I ) = { } , then ran A ( Q ) ( A Q ( I )) = { } . roof. According to Proposition 3.11, we have Ann Q ( I ) = { } . For any µ ∈ ran A ( Q ) ( A Q ( I )), we have R y µ = 0 for any y ∈ I and L z µ = 0 for any z ∈ I . If q ∈ Q , then we have 0 = R y µ ( q ) = [ µ ( q ) , y ] , L z µ ( q ) = [ z, µ ( q )] , which imply that µ ( q ) ∈ Ann Q ( I ) = 0, and so µ = 0. Lemma 5.5.
Suppose that L is a subalgebra of Q and let x , · · · , x n , y in L . Then wehave, in A ( Q ) : ξ · · · ξ n η = ηξ · · · ξ n + n X i =1 ξ · · · ξ i − δ i ξ i +1 · · · ξ n where ξ i = R x i or L x i for ≤ i ≤ n , η = R y or L y and δ i is of one of the following forms R [ x i ,y ] , R [ y,x i ] , L [ x i ,y ] , L [ y,x i ] .In particular, if I is an ideal of L and x , · · · , x n ∈ I , then ξ · · · ξ n η = ηξ · · · ξ n + α where α ∈ span { ρ · · · ρ n | ρ i = R z i or L z i with z i ∈ I , ≤ i ≤ n } .Proof. Let’s prove the conclusion by induction on n . When n = 1, in the case of ξ = R x and η = R y , we have ξ η = R x R y = R y R x + R [ y,x ] = ηξ + δ ;in the case of ξ = R x and η = L y , we have ξ η = R x L y = L y R x + L [ y,x ] = ηξ + δ ;in the case of ξ = L x and η = R y , we have ξ η = L x R y = R y L x + R [ x ,y ] = ηξ + δ ;in the case of ξ = L x and η = L y , we have ξ η = L x L y = L y L x + L [ x ,y ] = ηξ + δ . Suppose that the conclusion is valid for n = k . Then for n = k + 1, ξ · · · ξ k ξ k +1 η = ξ · · · ξ k ( ηξ k +1 + δ k +1 ) = ξ · · · ξ k ηξ k +1 + ξ · · · ξ k δ k +1 = ηξ · · · ξ k + k X i =1 ξ · · · ξ i − δ i ξ i +1 · · · ξ k ! ξ k +1 + ξ · · · ξ k δ k +1 = ηξ · · · ξ k +1 + k +1 X i =1 ξ · · · ξ i − δ i ξ i +1 · · · ξ k +1 . The proof is completed. 15et L be a subalgebra of Q . Denote by A the associative subalgebra of A ( Q ) whoseelements are those µ in A ( Q ) such that µ ( L ) ⊆ L . We obviously have the containments: A Q ( L ) ⊆ A ⊆ A ( Q ) . Lemma 5.6.
Suppose that L is a subalgebra of Q and I an ideal of L . Let q , · · · , q n in Q such that [ q i , I ] ⊆ L and [ I, q i ] ⊆ L for every i = 1 , · · · , n . Then for µ = ξ · · · ξ n in A ( Q ) where ξ i = R q i or L q i for each ≤ i ≤ n , we have that µ ( ˜ I ) n + ( ˜ I ) n µ ⊆ A (where ( ˜ I ) n denotes the 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 x , · · · , x n ∈ I , y = η · · · η n where η i = R x i or L x i for each 1 ≤ i ≤ n , both µy and yµ belong to A .Let’s prove the conclusion by induction on n . For n = 1, we have η ξ = ξ η + δ where δ is of one of the following forms R [ x ,q ] , R [ q ,x ] , L [ x ,q ] , L [ q ,x ] . Since [ x , q ] , [ q , x ] ∈ L we see that δ ∈ A . On the other hand, ξ η ( L ) ⊆ ξ ( I ) ⊆ L , and so ξ η , η ξ ∈ A .Assume that the result is true for n −
1. Now, by Lemma 5.5 we have η · · · η n ξ · · · ξ n = ( ξ η ) η · · · η n ξ · · · ξ n + n X i =1 η · · · η i − δ i η i +1 · · · η n ξ · · · ξ n (*)where δ i is of one of the following forms R [ x i ,q ] , R [ q ,x i ] , L [ x i ,q ] , L [ q ,x i ] .The first summand on the right side belongs to A since ξ η ∈ A and η · · · η n ξ · · · ξ n ∈ A by induction hypothesis.On the other hand, for each of the terms η · · · η i − δ i η i +1 · · · η n ξ · · · ξ n we have that x i ∈ I and [ x i , q ] , [ q , x i ] ∈ L . Using Lemma 5.5, we may write this as: δ i η · · · η i − η i +1 · · · η n ξ · · · ξ n + αη i +1 · · · η n ξ · · · ξ n where α ∈ span { ρ · · · ρ i − | ρ j = R z j or L z j with z j ∈ I , ≤ j ≤ i − } . The inductionhypothesis applies to show that this belongs to A again. Hence, yµ ∈ A .If we continue to develop in the expression (*), we get, for some α ∈ A , ξ ξ η · · · η n ξ · · · ξ n + n X i =1 ξ η · · · η i − δ i η i +1 · · · η n ξ · · · ξ n + α where δ i is of one of the following forms R [ x i ,q ] , R [ q ,x i ] , L [ x i ,q ] , L [ q ,x i ] .Using Lemma 5.5 we can write each term of the form ξ η · · · η i − δ i η i +1 · · · η n ξ · · · ξ n ξ δ i η η · · · η i − η i +1 · · · η n ξ · · · ξ n + ξ αη i +1 · · · η n ξ · · · ξ n , where α ∈ span { ρ · · · ρ i − | ρ j = R z j or L z j with z j ∈ I , ≤ j ≤ i − } . Notice that ξ δ i η = ξ η δ i + ξ γ where γ is of one of the following forms R [[ x i ,q ] ,x ] , R [[ q ,x i ] ,x ] , R [ x , [ x i ,q ]] , R [ x , [ q ,x i ]] , L [[ x i ,q ] ,x ] , L [[ q ,x i ] ,x ] , L [ x , [ x i ,q ]] , L [ x , [ q ,x i ]] .Hence, using [ q , x i ] , [ x i , q ] ∈ L and x i ∈ I , we see that the first summand abovebelongs to A . For the second summand, assuming that α = ρ · · · ρ i − where ρ j = R z j or L z j with z j ∈ I for 1 ≤ j ≤ i −
1, we have ( ξ ρ ) ρ · · · ρ i − η i +1 · · · η n ξ · · · ξ n , which isalso an element of A . Continuing in this way, we find that η · · · η n ξ · · · ξ n − ξ · · · ξ n η · · · η n ∈ A and by what we have just proved, we see that ξ · · · ξ n η · · · η n ∈ A .This completes the proof. Corollary 5.7.
Suppose that L is a subalgebra of Q and I an ideal of L . Let q , · · · , q n in Q such that [ q i , I ] ⊆ L and [ I, q i ] ⊆ L for every i = 1 , · · · , n . Then for µ = ξ · · · ξ n in A ( Q ) where ξ i = R q i or L q i for each ≤ i ≤ n , we have that µ e I n ⊆ A , e I n µ ⊆ A (where I n denotes the n -th power of I in the Leibniz algebra L ).Proof. It’s straightforward to show that I n is an 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 x ∈ I k +1 , there exist y ∈ I k , z ∈ I , R x = R [ y,z ] = R z R y − R y R z ∈ e I k ˜ I ⊆ ( ˜ I ) k ˜ I = ( ˜ I ) k +1 ,L x = L [ y,z ] = L y L z − L z L y ∈ e I k ˜ I ⊆ ( ˜ I ) k ˜ I = ( ˜ I ) k +1 . According to Lemma 5.6, we come to the conclusion.
Lemma 5.8.
Let L be a semiprime Leibniz algebra. If I is an ideal of L with Ann L ( I ) = { } , then Ann L ( I s ) = { } for any s ≥ . Any finite intersection of ideals with zeroannihilator also have zero annihilator.Proof. Let’s 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 ideal J of L , wehave I k ∩ J = { } since I k is essential. So { } 6 = [ I k ∩ J, I k ∩ J ] ⊆ [ I k , 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 annihilator alsohave zero annihilator by induction. 17 roposition 5.9. Suppose that L is a semiprime subalgebra of Q . Then the followingconditions are equivalent:(1) Q is an algebra of quotients of L ;(2) For any µ ∈ A ( Q ) \ { } , there exists an ideal I of L with Ann L ( I ) = { } such that µ ˜ I ⊆ A and { } 6 = ˜ Iµ ⊆ A . For any = q ∈ Q , we also have R q ˜ I ( L ) = { } or L q ˜ I ( L ) = { } .Proof. (2) ⇒ (1) Let q ∈ Q \ { } . Let ˜ I be as in (2), so it satisfies R q ˜ I ( L ) = { } or L q ˜ I ( L ) = { } and both R q ˜ I and L q ˜ I contain in A . Set I := span { α ( x ) | x ∈ L and α ∈ ˜ I } . Then I is an ideal of L such that Ann L ( I ) = { } . Indeed, if x, y ∈ L and α ∈ ˜ I , we have[ y, α ( x )] = L y α ( x ) and L y α ∈ ˜I , [ α ( x ) , y ] = R y α ( x ) and R y α ∈ ˜I . According to Proposition 2.6, we know that for a semiprime Leibniz algebra L , Ann L ( I ) = { } ⇔ ran L ( I ) = { } for any ideal I of L . If now [ I , x ] = { } for x ∈ L , then R x ˜ I ( L ) = { } . In particular, for y, z ∈ I , we have R x R y ( z ) = 0 and so x ∈ ran L ( I ), which is zeroby Lemma 5.8. So ran L ( I ) = { } , consequently Ann L ( I ) = { } .Finally, [ q, I ] = L q ˜ I ( L ) = { } or [ I , q ] = R q ˜ I ( L ) = { } , and both [ q, I ] and [ I , q ]contain in L , which implies that Q is ideally absorbed into L . According to Theorem3.10, Q is an algebra of quotients of L .(1) ⇒ (2) Let µ = P i ≥ ξ i, · · · ξ i,r i ∈ A ( Q ) \ { } where ξ i,j = R q i,j or L q i,j for1 ≤ j ≤ r i and q i,j ∈ Q . We may of course assume that all q i,j are nonzero elements in Q .Set s = P i ≥ r i . As Q is an algebra of quotients of L , there exists, for every i and j , anideal J i,j of L such that Ann L ( J i , j ) = { } , and [ q i,j , J i,j ] ⊆ L, [ J i,j , q i,j ] ⊆ L . By Lemma5.8, the ideal J = ∩ i,j J i,j and hence also I = J s have zero annihilator in L . Moreover,[ q i,j , J ] ⊆ [ q i,j , J i,j ] ⊆ L and [ J, q i,j ] ⊆ [ J i,j , q i,j ] ⊆ L . According to Corollary 5.7, we have µ ˜ I ⊆ A and ˜ Iµ ⊆ A .If ˜ Iµ = { } , then µ ∈ ran A ( Q ) ( A Q ( I )) which is zero by Lemma 5.4.Finally, for any 0 = q ∈ Q , apply that Q is an algebra of quotients of I to find y, z ∈ I such that [ q, [ y, z ]] = 0 or u, v ∈ I such that [[ u, v ] , q ] = 0. And [ q, [ y, z ]] = L q L y ( z ) ∈ L q ˜ I ( L )(resp. [[ u, v ] , q ] = R q L u ( v ) ∈ R q ˜ I ( L )).Recall that an associative algebra S is a left quotient algebra of a subalgebra A ifwhenever 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 a total right zero divisor if Ax = { } .) Lemma 5.10. [22] 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 ran A ( I ) = { } such that { } 6 = Iq ⊆ A . heorem 5.11. Suppose that L is a semiprime subalgebra of Q . Moreover, suppose that Q is an algebra of quotients of L . Then A ( Q ) is a left quotient algebra of A .Proof. Let µ ∈ A ( Q ) \ { } and let I be an 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 ) = { } , from Lemma 5.4 we obtain that ran A ( Q ) ( A Q ( I )) = { } .This, together with the fact that ˜ I contains A Q ( I ), implies that ran A ( Q ) ( ˜ I ) = { } . Since ran A ( ˜ I ) ⊆ ran A ( Q ) ( ˜ I ) we get that ran A ( ˜ I ) = { } . Hence we get ran 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 is a subalgebra of Q and if L is semiprime and I is an ideal of L with Ann L ( I ) = { } , then A ˜ I + ˜ I is aleft ideal of A with zero right annihilator. In this section, we mainly study algebras of quotients of Leibniz algebras via theirdense extensions which is introduced by Cabrera in [10]. We show that dense extensioncan be lifted from a Leibniz algebra to its essential ideals if the extension is also an algebraof quotients. What’s more, we get a conclusion, a converse to Theorem 5.11, via denseextension of Leibniz algebras. Specifically, for any algebra L (not necessary associative),let M ( L ) be the associative algebra generated by the identity map together with theoperators given by right and left multiplication by elements of L . In the case of a Leibnizalgebra L , note that M ( L ) is nothing but the unitization of A ( L ).Following [10], given an extension of (not necessarily associative) algebra L ⊆ Q , theannihilator of L in M ( Q ) is defined by L ann := { µ ∈ M ( Q ) | µ ( x ) = 0 , x ∈ L } . If L ann = { } , then L is a dense subalgebra of Q , and L ⊆ Q is a dense extension ofalgebras in this sense. Lemma 6.1.
Suppose that L is a subalgebra of the Leibniz algebra Q with Ann( Q ) = { } .Then the following conditions are equivalent:(1) L is a dense 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 element q in Q R q µ ( p ) = 0 or 0 = s ∈ Q such that L s µ ( p ) = 0. But then R q µ ( L ) = L s µ ( L ) = { } and since A ( Q ) is a two-sided ideal of M ( Q ), we have that R q µ, L s µ ∈ A ( Q ). Hence,condition (2) yields R q µ = L s µ = 0, a contradiction. Lemma 6.2. [10] Suppose that L is a dense subalgebra of Q and I an ideal of L . If µ isan element of M ( Q ) such that µ ( I ) = { } , then µ ( M ( Q )( I )) = { } . Following Cabrera and Mohammed in [9], we say that a Leibiniz algebra L is mul-tiplicatively semiprime whenever L and its multiplication algebra M ( L ) are semiprime.Observe that in this situation, and if L is a Leibniz algebra, then A ( L ), being an idealof M ( L ), will also be a semiprime algebra. Lemma 6.3.
Let L ⊆ Q be a dense extension of Leibniz algebras. If Q is multiplicativelysemiprime, then A is semiprime.Proof. Let I be an ideal of A with I = { } . As in the proof of Proposition 5.9, let I = { α ( x ) | α ∈ I, x ∈ L } , which is clearly an ideal of L . For any µ ∈ I , we evidentlyhave µ ( I ) = { } since I = { } . It then follows Lemma 6.2 that µ ( M ( Q )( I )) = { } .This implies that µM ( Q ) µ ( L ) = { } , and thus µM ( Q ) µ = { } since L is dense in Q . But M ( Q ) is semiprime by hypothesis, so µ = 0 and since µ is an arbitrary element in I , weget that I = { } , that is, A is semiprime. Lemma 6.4. [10] Suppose that L is a dense subalgebra of Q . If Q is multiplicativelysemiprime(resp. multiplicatively prime), then L is also multiplicatively semiprime(resp.multiplicatively prime). Proposition 6.5.
Suppose that L is a dense subalgebra of Q and Q a multiplicativelysemiprime algebra of quotients of L . Then I ⊆ Q is a dense extension for every essentialideal of I of L .Proof. We first observe that Ann( Q ) = { } since Q is an algebra of quotients of L .Hence Lemma 6.1 applies. Thus, let µ be in A ( Q ) such that µ ( I ) = { } , and by way ofcontradiction assume that µ = 0. According to Theorem 5.11, A ( Q ) is a left quotientsalgebra of A and hence there exists λ in A such that 0 = λµ ∈ A . Since the extension L ⊆ Q is dense, λµ ( L ) = { } , and since ran L ( I ) = { } , there exists a nonzero element y in I such that L y λµ ( L ) = { } . Using now that A ( Q ) has no total right zero divisor accordingto Lemma 5.1, we get that A ( Q ) L y λµ = { } , and this, coupled with the semiprimeness of A ( Q ), implying that A ( Q ) L y λµ A ( Q ) L y λµ = { } . A second application of the fact that L ⊆ Q is a dense extension yields A ( Q ) L y λµ A ( Q ) L y λµ ( L ) = { } . However, µ ( I ) = { } by assumption and thus implies that µM ( Q )( I ) = { } . But this is a contradiction,because of the containments µ A ( Q ) L y λµ ( L ) ⊆ µ A ( Q )([ I, L ]) ⊆ µ A ( Q )( I ) ⊆ µM ( Q )( I ) = { } . This completes the proof. 20 orollary 6.6.
Let L ⊆ Q be a dense extension of Leibniz algebras. Suppose that Q isa multiplicatively prime algebra of quotients of L . Then I ⊆ Q is a dense extension forany nonzero ideal I of L .Proof. According to Proposition 2.7, we get ran L ( I ) = { } for each nonzero ideal I of L . Since L is prime, we get I is essential where I is a nonzero ideal of L . The proof iscompleted by Proposition 6.5. Corollary 6.7.
Suppose that L is a dense subalgebra of Q and Q a multiplicative semiprimealgebra of quotients of L . Then for every essential ideal I of L , lan A ( Q ) ( ˜ I ) = { } .Proof. Let µ ∈ lan A ( Q ) ( ˜ I ). Then, if y ∈ I , we have µR y ( I ) = { } . This impliesthat µ ( I ) = { } . By Proposition 6.5 applied to the essential ideal I of L , the extension I ⊆ Q is dense, so µ = 0.We close by exploring the possible converse to Theorem 5.11 in the presence of denseextensions of Leibniz algebras. Definition 6.8.
Given an extension of associative algebra A ⊆ S , we say that S isstrong right ideally absorbed into A if for any p, q ∈ S \ { } there is an ideal I of A with lan A ( I ) = { } and such that pI = { } or qI = { } and both pI and qI are contained in A . Proposition 6.9.
Let L ⊆ Q be a dense extension of Leibniz algebras with ran ( Q ) = lan ( Q ) = { } . Suppose that A ( Q ) is strong right ideally absorbed into A . Then Q is analgebra of quotients of L .Proof. Let q ∈ Q \ { } . Since ran ( Q ) = lan ( Q ) = { } , we have R q = 0, L q = 0. Byhypothesis, there is an ideal I of A such that lan A ( I ) = { } and R q I ⊆ A , L q I ⊆ A and R q I = { } or L q I = { } . Set I = { α ( x ) | α ∈ I, x ∈ L } , which is an ideal of L .Moreover, Ann L ( I ) = { } . Indeed, suppose that an element x in L satisfies [ x, I ] = { } .By definition, this means that L x I ( L ) = { } , and since the extension is dense we havethat L x I = { } . Thus L x ∈ lan A ( I ) = 0. Note that lan ( L ) = { } , and so x = 0,consequently lan L ( I ) = { } . Thus Ann L ( I ) = { } .Finally, [ q, I ] = L q ˜ I ( L ) = { } or [ I , q ] = R q ˜ I ( L ) = { } , and both [ q, I ] and [ I , q ]contain in L , which implies that Q is ideally absorbed into L . According to Theorem3.10, Q is an algebra of quotients of L . Corollary 6.10.
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