Characterization of Multiplicative Lie Triple Derivations on Rings

Abstract

Let ℛ be a ring having unit 1. Denote by 𝒵 (ℛ) the center of ℛ. Assume that the characteristic of ℛ is not 2 and there is an idempotent element 𝑒∈ ℛ such that 𝑎ℛ𝑒 ={0}⇒𝑎=0and𝑎ℛ(1−𝑒)={0}⇒𝑎=0. It is shown that, under some mild conditions, a map 𝐿∶ℛ→ℛ is a multiplicative Lie triple derivation if and only if 𝐿(𝑥)=𝛿(𝑥)+ℎ(𝑥) for all 𝑥∈ℛ, where 𝛿∶ℛ→ℛ is an additive derivation and ℎ∶ℛ→𝒵(ℛ) is a map satisfying ℎ([[𝑎,𝑏],𝑐])=0 for all 𝑎,𝑏,𝑐∈ℛ. As applications, all Lie (triple) derivations on prime rings and von Neumann algebras are characterized, which generalize some known results.

1. Introduction

Let ℛ be an associative ring with the center 𝒵(ℛ). For any element 𝑎,𝑏∈ℛ, we set [𝑎,𝑏]=𝑎𝑏−𝑏𝑎. Recall that a map 𝐿∶ℛ→ℛ is a multiplicative derivation or nonlinear derivation if 𝐿(𝑎𝑏)=𝐿(𝑎)𝑏+𝑎𝐿(𝑏), for all 𝑎,𝑏∈ℛ, is a multiplicative Lie derivation, if 𝐿([𝑎,𝑏])=[𝐿(𝑎),𝑏]+[𝑎,𝐿(𝑏)], for all 𝑎,𝑏∈ℛ, and is a multiplicative Lie triple derivation, if 𝐿([[𝑎,𝑏],𝑐])=[[𝐿(𝑎),𝑏],𝑐]+[[𝑎,𝐿(𝑏)],𝑐]+[[𝑎,𝑏],𝐿(𝑐)], for all 𝑎,𝑏,𝑐∈ℛ. Particularly, if 𝐿 is additive (linear), then above maps are, respectively, additive (linear) derivations, additive (linear) Lie derivations, and additive (linear) Lie triple derivations. We often omit “linear” for “linear derivations.”

The structure of additive (linear) derivations and additive (linear) Lie (triple) derivations on rings or algebras has been studied by many authors. Bre_̂‌sar in [1] proved that every additive Lie derivation on a prime ring ℛ with characteristic not 2 can be decomposed as 𝜏+𝜁, where 𝜏 is an additive derivation from ℛ into its central closure and 𝜁 is an additive map of ℛ into the extended centroid 𝒞 sending commutators to zero. Mathieu and Villena [2] showed that every linear Lie derivation on a 𝐶^∗-algebra is standard, that is, can be decomposed as the form 𝜏+ℎ, where 𝜏 is a derivation and ℎ is a central valued linear map vanishing at each commutator. In [3] Qi and Hou proved that the same is true for additive Lie derivations of nest algebras on Banach spaces. Miers [4] showed that every linear Lie triple derivation on ℳ, a von Neumann algebra with no central summands of type 𝐼₁, is of the form 𝜏+ℎ, where 𝜏 is a derivation and ℎ is a central valued linear map vanishing at every Lie triple products [[𝐴,𝐵],𝐶]. Recently, Wang and Lu [5] described the structure of linear Lie triple derivations on 𝒥-subspace lattice algebras. For other results, see [6–10] and the references therein.

For the study of multiplicative derivations and multiplicative Lie (triple) derivations, Daif [11] initially proved that each multiplicative derivation on a 2-torsion free prime ring containing a nontrivial idempotent is additive. Yu and Zhang [12] showed that every multiplicative Lie derivation on triangular algebras is the sum of an additive derivation and a map into its center sending commutators to zero, and, later, Ji et al. [13] generalized this result to the case of multiplicative Lie triple derivations. Assume that 𝒩 is a nontrivial nest on a Banach space 𝑋 over the complex field which contains a nontrivial complemented element, and Alg𝒩 is the associated nest algebra. Li and Fang in [14] obtained the same result as the above for multiplicative Lie triple derivations on Alg𝒩.

The purpose of the present paper is to consider the problem of characterizing nonlinear Lie triple derivations on general rings.

Let ℛ be a ring having unit 1 and an idempotent element 𝑒, and let 𝒵(ℛ) denote the center of ℛ. Assume that the characteristic of ℛ is not 2 and satisfies that 𝑎ℛ𝑒={0}⇒𝑎=0 and 𝑎ℛ(1−𝑒)={0}⇒𝑎=0. Let 𝐿∶ℛ→ℛ be a multiplicative Lie triple derivation. We show that if 𝑒ℛ𝑒 and (1−𝑒)ℛ(1−𝑒) do not contain nonzero central ideals, then 𝐿(𝑎+𝑏)=𝐿(𝑎)+𝐿(𝑏)+𝜆_𝑎,𝑏, for all 𝑎,𝑏∈ℛ, where 𝜆_𝑎,𝑏∈𝒵(ℛ) is a central element depending on 𝑎 and 𝑏 (Theorem 1); furthermore, if ℛ also satisfies that, for 𝑎∈ℛ, [𝑒𝑎𝑒,𝑒ℛ𝑒]⊆𝒵(𝑒ℛ𝑒)⇒𝑒𝑎𝑒∈𝒵(𝑒ℛ𝑒), and [(1−𝑒)𝑎(1−𝑒),(1−𝑒)ℛ(1−𝑒)]⊆𝒵((1−𝑒)ℛ(1−𝑒))⇒(1−𝑒)𝑎(1−𝑒)∈𝒵((1−𝑒)ℛ(1−𝑒)), then 𝐿(𝑎)=𝛿(𝑎)+ℎ(𝑎), for all 𝑎∈ℛ, where 𝛿∶ℛ→ℛ is an additive derivation and ℎ∶ℛ→𝒵(ℛ) is a map satisfying ℎ([[𝑎,𝑏],𝑐])=0, for all 𝑎,𝑏,𝑐∈ℛ (Theorem 2). As applications, some characterizations of multiplicative (additive) Lie (triple) derivations on prime rings and von Neumann algebras are obtained, respectively (Corollaries 3–7).

2. Main Results and Corollaries

The following are our main results in this paper.

Theorem 1. Let ℛ be a ring having unit 1 and an idempotent element 𝑒. Assume that the characteristic of ℛ is not 2 and ℛ satisfies the following two conditions:(i)𝑎ℛ𝑒={0}⇒𝑎=0 and 𝑎ℛ(1−𝑒)={0}⇒𝑎=0;(ii)𝑒ℛ𝑒 and (1−𝑒)ℛ(1−𝑒) do not contain nonzero central ideals.
Assume that 𝐿∶ℛ→ℛ is a multiplicative Lie triple derivation. Then 𝐿(𝑎+𝑏)=𝐿(𝑎)+𝐿(𝑏)+𝜆_𝑎,𝑏,  for all 𝑎,𝑏∈ℛ, where 𝜆_𝑎,𝑏∈𝒵(ℛ) is a central element depending on 𝑎 and 𝑏.

Moreover, if the ring ℛ in Theorem 1 also satisfies that, for 𝑎∈ℛ, [𝑒𝑎𝑒,𝑒ℛ𝑒]⊆𝒵(𝑒ℛ𝑒)⇒𝑒𝑎𝑒∈𝒵(𝑒ℛ𝑒), and [(1−𝑒)𝑎(1−𝑒),(1−𝑒)ℛ(1−𝑒)]⊆𝒵((1−𝑒)ℛ(1−𝑒))⇒(1−𝑒)𝑎(1−𝑒)∈𝒵((1−𝑒)ℛ(1−𝑒)), then 𝐿 has more concrete form.

Theorem 2. Let ℛ be a ring having unit 1 and an idempotent element 𝑒. Assume that the characteristic of ℛ is not 2 and ℛ satisfies the following two conditions:(i)𝑎ℛ𝑒={0}⇒𝑎=0 and 𝑎ℛ(1−𝑒)={0}⇒𝑎=0;(ii)𝑒ℛ𝑒 and (1−𝑒)ℛ(1−𝑒) do not contain nonzero central ideals;(iii)for 𝑎∈ℛ, [𝑒𝑎𝑒,𝑒ℛ𝑒]⊆𝒵(𝑒ℛ𝑒)⇒𝑒𝑎𝑒∈𝒵(𝑒ℛ𝑒), and [(1−𝑒)𝑎(1−𝑒),(1−𝑒)ℛ(1−𝑒)]⊆𝒵((1−𝑒)ℛ(1−𝑒))⇒(1−𝑒)𝑎(1−𝑒)∈𝒵((1−𝑒)ℛ(1−𝑒)).
Then a map 𝐿∶ℛ→ℛ is a multiplicative Lie triple derivation if and only if 𝐿(𝑎)=𝛿(𝑎)+ℎ(𝑎), for all 𝑎∈ℛ, where 𝛿∶ℛ→ℛ is an additive derivation and ℎ∶ℛ→𝒵(ℛ) is a map satisfying ℎ([[𝑎,𝑏],𝑐])=0, for all 𝑎,𝑏,𝑐∈ℛ.

Recall that a ring ℛ is prime if, for any 𝑎,𝑏∈ℛ, 𝑎ℛ𝑏={0} implies 𝑎=0 or 𝑏=0.

Applying Theorem 2 to prime rings, we have the following result.

Corollary 3. Let ℛ be a prime ring having unit 1 and a nontrivial idempotent, and let 𝐿∶ℛ→ℛ be a map. If the characteristic of ℛ is not 2 and 𝑒ℛ𝑒, (1−𝑒)ℛ(1−𝑒) are noncommutative, then the following two statements are equivalent.(1)𝐿 is a multiplicative Lie triple derivation.(2)There exist an additive derivation 𝛿∶ℛ→ℛ and a map ℎ∶ℛ→𝒵(ℛ) satisfying ℎ([[𝑎,𝑏],𝑐])=0 for all 𝑎,𝑏,𝑐∈ℛ such that 𝐿(𝑥)=𝛿(𝑥)+ℎ(𝑥) for all 𝑥∈ℛ.

Proof. Let 𝑒∈ℛ be a nontrivial idempotent. It is obvious that ℛ satisfies the condition (i) in Theorem 2.
Claim. If 𝒰 is a central ideal of a noncommutative prime ring ℛ^′, then 𝒰={0}.
Take any 𝑢∈𝒰. Since 𝒰⊆ℛ^′ is central, we have 𝑢𝑎=𝑎𝑢∈𝒰 for all 𝑎∈ℛ^′. Thus, for any 𝑎,𝑏∈ℛ^′, one gets 𝑢[]()()()𝑎,𝑏=𝑢𝑎𝑏−𝑢𝑏𝑎=𝑢𝑎𝑏−𝑎𝑢𝑏=𝑢𝑎𝑏−𝑎𝑢𝑏=𝑢𝑎𝑏−𝑢𝑎𝑏=0, (1) and so 𝑢𝑐[𝑎,𝑏]=𝑢[𝑐𝑎,𝑏]−𝑢[𝑐,𝑏]𝑎=0∀𝑎,𝑏,𝑐∈ℛ^′. (2) Since ℛ^′ is noncommutative, there exist two elements 𝑎₀,𝑏₀∈ℛ^′ such that [𝑎₀,𝑏₀]≠0. It follows from the primeness of ℛ^′ that 𝑢=0. The claim holds.
Since ℛ is prime, both 𝑒ℛ𝑒 and (1−𝑒)ℛ(1−𝑒) are prime. Thus, by the above claim, the condition (ii) in Theorem 2 is satisfied.
Now, for any fixed 𝑎∈ℛ, define two maps 𝛿₁∶𝑒ℛ𝑒→𝑒ℛ𝑒 and 𝛿₂∶(1−𝑒)ℛ(1−𝑒)→(1−𝑒)ℛ(1−𝑒), respectively, by 𝛿₁()=[],𝛿𝑒𝑥𝑒𝑒𝑎𝑒,𝑒𝑥𝑒₂((1−𝑒)𝑥(1−𝑒))=[(1−𝑒)𝑎(1−𝑒),(1−𝑒)𝑥(1−𝑒)], (3) for all 𝑥∈ℛ. It is clear that both 𝛿₁ and 𝛿₂ are derivations. Posner in [9, Theorem 2] proved that, if 𝑑 is a derivation of a noncommutative prime ℛ^′ such that, for all 𝑎∈ℛ^′, 𝑎𝑑(𝑎)−𝑑(𝑎)𝑎 is in the center of ℛ^′, then 𝑑 is the zero derivation. Thus, by [9, Theorem 2], for 𝑎∈ℛ, if 𝑒ℛ𝑒 is noncommutative and [𝑒𝑎𝑒,𝑒ℛ𝑒]⊆𝒵(𝑒ℛ𝑒), then [𝑒𝑎𝑒,𝑒ℛ𝑒]=0; that is, 𝑒𝑎𝑒∈𝒵(𝑒ℛ𝑒). Similarly, if (1−𝑒)ℛ(1−𝑒) is noncommutative and [(1−𝑒)𝑎(1−𝑒),(1−𝑒)ℛ(1−𝑒)]⊆𝒵((1−𝑒)ℛ(1−𝑒)), then (1−𝑒)𝑎(1−𝑒)∈𝒵((1−𝑒)ℛ(1−𝑒)). Hence the condition (iii) in Theorem 2 is also satisfied.
Now, by Theorem 2, the corollary is true.

Let ℬ(𝐻) be the algebra of all bounded linear operators acting on a complex Hilbert space 𝐻. Recall that a von Neumann algebra ℳ is a subalgebra of some ℬ(𝐻) satisfying ℳ^′′={ℳ^′}^′=ℳ, where ℳ^′={𝑇∶𝑇∈ℬ(𝐻)and𝑇𝐴=𝐴𝑇∀𝐴∈ℳ} ([15]). For 𝐴∈ℳ, the central carrier of 𝐴, denoted by 𝐴, is the intersection of all central projections 𝑃 such that 𝑃𝐴=0. If 𝐴 is self-adjoint, then the core of 𝐴, denoted by 𝐴, is sup{𝑆∈𝒵(ℳ)∶𝑆=𝑆^∗,𝑆≤𝐴}. Particularly, if 𝐴=𝑃 is a projection, it is clear that 𝑃 is the largest central projection ≤𝑃. A projection 𝑃 is core-free if 𝑃=0. It is easy to see that 𝑃=0 if and only if 𝐼−𝑃=𝐼 ([16]).

Applying Theorem 2 to von Neumann algebras, we have the following corollary.

Corollary 4. Let ℳ be a von Neumann algebra without central summands of type 𝐼₁ and 𝐿∶ℳ→ℳ a map. Then the following statements are equivalent.(1)𝐿 is a multiplicative Lie triple derivation.(2)There exist an additive derivation 𝛿∶ℳ→ℳ and a map ℎ∶ℳ→𝒵(ℳ) vanishing at each Lie triple product [[𝐴,𝐵],𝐶] such that 𝐿(𝐴)=𝛿(𝐴)+ℎ(𝐴), for all 𝐴∈ℳ.

Proof. Assume that ℳ is a von Neumann algebra without central summands of type 𝐼₁. Then, by [16], there exists a nonzero core-free projection 𝑃∈ℳ with 𝑃=𝐼. Fix such 𝑃 and note that 𝑃=𝐼−𝑃=𝐼. It follows from the definition of the central carrier that both span{𝑇𝑃(𝑥)∶𝑇∈ℳ,𝑥∈𝐻} and span{𝑇(𝐼−𝑃)(𝑥)∶𝑇∈ℳ,𝑥∈𝐻} are dense in 𝐻. So 𝐴ℳ𝑃={0}⇒𝐴=0 and 𝐴ℳ(𝐼−𝑃)={0}⇒𝐴=0.
Bre_̂‌sar and Miers [17] proved that if 𝑍∈𝒵(ℳ) such that 𝑍ℳ⊆𝒵(ℳ), then 𝑍=0. This implies that ℳ has no nonzero central ideals. Note that 𝑃ℳ𝑃 and (𝐼−𝑃)ℳ(𝐼−𝑃) are also von Neumann algebras without central summands of type 𝐼₁. So both 𝑃ℳ𝑃 and (𝐼−𝑃)ℳ(𝐼−𝑃) have no nonzero central ideals.
Finally, for 𝐴∈ℳ, if [𝑃𝐴𝑃,𝑃ℳ𝑃]⊆𝒵(𝑃ℳ𝑃) and [(𝐼−𝑃)𝐴(𝐼−𝑃),(𝐼−𝑃)ℳ(𝐼−𝑃)]⊆𝒵((𝐼−𝑃)ℳ(𝐼−𝑃)), by the Kleinecke-Shirokov theorem ([18]), both [𝑃𝐴𝑃,𝑃𝑇𝑃] and [(𝐼−𝑃)𝐴(𝐼−𝑃),(𝐼−𝑃)𝑇(𝐼−𝑃)] are central quasinilpotent for all 𝑇∈ℳ. Hence [𝑃𝐴𝑃,𝑃𝑇𝑃]=[(𝐼−𝑃)𝐴(𝐼−𝑃),(𝐼−𝑃)𝑇(𝐼−𝑃)]=0 for all 𝑇∈ℳ; that is, 𝑃𝐴𝑃∈𝒵(𝑃ℳ𝑃) and (𝐼−𝑃)𝐴(𝐼−𝑃)∈𝒵((𝐼−𝑃)ℳ(𝐼−𝑃)).
Thus, if ℳ has no central summands of type 𝐼₁, by what the above stated, ℳ satisfies the corresponding assumptions (i)–(iii) in Theorem 2. By Theorem 2, the corollary is true.

Note that a multiplicative Lie derivation must be a multiplicative Lie triple derivation. So the following corollary is immediate.

Corollary 5. Let ℳ be a von Neumann algebra without central summands of type 𝐼₁ and 𝐿∶ℳ→ℳ a map. Then the following statements are equivalent.(1)𝐿 is a multiplicative Lie derivation.(2)There exist an additive derivation 𝛿∶ℳ→ℳ and a map ℎ∶ℳ→𝒵(ℳ) vanishing at each commutator [𝐴,𝐵] such that 𝐿(𝐴)=𝛿(𝐴)+ℎ(𝐴), for all 𝐴∈ℳ.

If 𝐿 in Corollaries 4 and 5 is additive, more can be said. In fact, a complete characterization of additive Lie (triple) derivations on any von Neumann algebras can be obtained, which is a slight generalization of the corresponding result in [4].

Corollary 6. Let ℳ be a von Neumann algebra and 𝐿∶ℳ→ℳ an additive map. Then the following statements are equivalent.(1)𝐿 is a Lie triple derivation.(2)There exist an additive derivation 𝛿∶ℳ→ℳ and an additive map ℎ∶ℳ→𝒵(ℳ) vanishing at each Lie triple product [[𝐴,𝐵],𝐶] such that 𝐿(𝐴)=𝛿(𝐴)+ℎ(𝐴), for all 𝐴∈ℳ.

Proof. Clearly, one only needs to check (1)⇒(2). In the following assume that 𝐿 is an additive Lie triple derivation.
Take the central projection 𝑄∈ℳ⊆ℬ(𝐻), so that, with respect to the space decomposition 𝐻=𝑄𝐻⊕(𝐼−𝑄)𝐻, ℳ=ℳ₁⊕ℳ₂, where ℳ₁ is of type 𝐼₁ and ℳ₂ has no central summands of type 𝐼₁. Note that we may have 𝑄=0. Then 𝐿(𝐴) can be decomposed as 𝐿(𝐴)=𝑄𝐿(𝐴)+(𝐼−𝑄)𝐿(𝑄𝐴)+(𝐼−𝑄)𝐿((𝐼−𝑄)𝐴)∀𝐴∈ℳ. (4)
Claim. 𝐿(𝒵(ℳ))⊆𝒵(ℳ).
For any 𝑍∈𝒵(ℳ) and any 𝐴,𝐵∈ℳ, we have 0=𝐿(0)=𝐿([[𝑍,𝐴],𝐵])=[[𝐿(𝑍),𝐴],𝐵], and so [𝐿(𝑍),𝐴]∈𝒵(ℳ) for all 𝐴∈ℳ. Hence [𝐿(𝑍),𝐴] is central quasinilpotent, which implies [𝐿(𝑍),𝐴]=0 for all 𝐴∈ℳ. It follows that 𝐿(𝑍)∈𝒵(ℳ). The claim holds.
Since 𝑄 is a central projection and 𝑄ℳ is of type 𝐼₁, we have 𝑄ℳ⊆𝒵(ℳ). (5) By the above claim, 𝐿(𝑄𝐴)∈𝒵(ℳ), for all 𝐴∈ℳ, and so (𝐼−𝑄)𝐿(𝑄𝐴)=𝐿(𝑄𝐴)−𝑄𝐿(𝑄𝐴)∈𝒵(ℳ)∀𝐴∈ℳ. (6)
Define a map Ψ∶ℳ₂→ℳ₂ by Ψ(𝐵)=𝐿(0⊕𝐵)|_{(𝐼−𝑄)𝐻}=𝐿((𝐼−𝑄)(0⊕𝐵))|_{(𝐼−𝑄)𝐻}∀𝐵∈ℳ₂. (7) Then, for any 𝐴=𝐴₁⊕𝐴₂∈ℳ=ℳ₁⊕ℳ₂, we have 𝐿(𝐴)=𝑄𝐿(𝐴)+(𝐼−𝑄)𝐿(𝑄𝐴)+0⊕Ψ𝐴₂. (8) It is easy to prove that Ψ is an additive Lie triple derivation on ℳ₂. So, by Corollary 4, there exist an additive derivation 𝛿₂∶ℳ₂→ℳ₂ and an additive map ℎ₂∶ℳ₂→𝒵(ℳ₂) vanishing at each Lie triple product such that Ψ(𝐵)=𝛿₂(𝐵)+ℎ₂(𝐵)∀𝐵∈ℳ₂. (9) Let 𝛿=0⊕𝛿₂ and ℎ(𝐴)=𝑄𝐿(𝐴)+(𝐼−𝑄)𝐿(𝑄𝐴)+(0⊕ℎ₂(𝐴|_{(𝐼−𝑄)𝐻})). By (4)–(9), one gets that 𝛿 is an additive derivation on ℳ and ℎ∶ℳ→𝒵(ℳ) is an additive map satisfying ℎ([[𝐴,𝐵],𝐶])=0 for all 𝐴,𝐵,𝐶 such that 𝐿(𝐴)=𝛿(𝐴)+ℎ(𝐴) holds for all 𝐴∈ℳ. Hence (2) is true and the proof is finished.

Particularly, for additive Lie derivations, we have the following corollary.

Corollary 7. Let ℳ be a von Neumann algebra and 𝐿∶ℳ→ℳ an additive map. Then the following statements are equivalent.(1)𝐿 is a Lie derivation.(2)There exist an additive derivation 𝛿∶ℳ→ℳ and an additive map ℎ∶ℳ→𝒵(ℳ) vanishing at each commutator such that 𝐿(𝐴)=𝛿(𝐴)+ℎ(𝐴) for all 𝐴∈ℳ.

3. The Proof of Main Results

In this section, we will give proofs of our main results, Theorems 1 and 2.

In the sequel, assume that ℛ is a unital ring and containing an idempotent 𝑒 satisfying 𝑎ℛ𝑒={0}⇒𝑎=0 and 𝑎ℛ(1−𝑒)={0}⇒𝑎=0. It is clear that 𝑒≠0,1. Write 𝑒₁=𝑒 and 𝑒₂=1−𝑒. Then ℛ can be written as ℛ=ℛ₁₁+ℛ₁₂+ℛ₂₁+ℛ₂₂, where ℛ_𝑖𝑗=𝑒_𝑖ℛ𝑒_𝑗 (𝑖,𝑗∈{1,2}).

We first give several lemmas, which are needed to prove the main results.

Lemma 8 (see [3, Lemma 3.1]). The center of ℛ is 𝒵(ℛ)=𝑧₁₁+𝑧₂₂∶𝑧₁₁∈ℛ₁₁,𝑧₂₂∈ℛ₂₂,𝑧₁₁𝑎₁₂=𝑎₁₂𝑧₂₂,𝑧₂₂𝑎₂₁=𝑎₂₁𝑧₁₁∀𝑎₁₂∈ℛ₁₂,𝑎₂₁∈ℛ₂₁. (10)

By Lemma 8, it is easily seen that if 𝑧₁₁+𝑧₂₂∈𝒵(ℛ), then 𝑧₁₁∈𝒵(ℛ₁₁) and 𝑧₂₂∈𝒵(ℛ₂₂).

Lemma 9. Let 𝑎∈ℛ. If 𝑎𝑏₁₂=𝑏₁₂𝑎 and 𝑎𝑏₂₁=𝑏₂₁𝑎 hold for all 𝑏₁₂∈ℛ₁₂ and all 𝑏₂₁∈ℛ₂₁, then 𝑎∈𝒵(ℛ).

Proof. Write 𝑎=𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂. Since 𝑎𝑏₁₂=𝑏₁₂𝑎 for all 𝑏₁₂∈ℛ₁₂, we have 𝑎₁₁𝑏₁₂+𝑎₂₁𝑏₁₂=𝑏₁₂𝑎₂₁+𝑏₁₂𝑎₂₂. (11) Multiplying 𝑒₁ and 𝑒₂ from the left and right in the equation, one gets 𝑎₁₁𝑏₁₂=𝑏₁₂𝑎₂₂. Multiplying 𝑒₂ from both sides in the equation, one has 𝑎₂₁𝑏₁₂=0; that is, 𝑎₂₁𝑏𝑒₂=0 for all 𝑏∈ℛ. This implies 𝑎₂₁=0.
Similarly, from the relation 𝑎𝑏₂₁=𝑏₂₁𝑎 for all 𝑏₂₁∈ℛ₂₁, one can prove 𝑎₂₂𝑏₂₁=𝑏₂₁𝑎₁₁ and 𝑎₁₂=0. It follows from Lemma 8 that 𝑎=𝑎₁₁+𝑎₂₂∈𝒵(ℛ).

Lemma 10. Let 𝑎∈ℛ. If  [[𝑎,𝑡_𝑖𝑗],𝑒_𝑗]=0 holds for all 𝑡_𝑖𝑗∈ℛ_𝑖𝑗 (1≤𝑖≠𝑗≤2), then 𝑒₁𝑎𝑒₁+𝑒₂𝑎𝑒₂∈𝒵(ℛ).

Proof. For 1≤𝑖≠𝑗≤2, since [[𝑎,𝑡_𝑖𝑗],𝑒_𝑗]=0, we have 𝑒_𝑖𝑎𝑒_𝑖𝑡_𝑖𝑗=𝑡_𝑖𝑗𝑒_𝑗𝑎𝑒_𝑗 for all 𝑡_𝑖𝑗∈ℛ_𝑖𝑗. It follows from Lemma 8 that 𝑒₁𝑎𝑒₁+𝑒₂𝑎𝑒₂∈𝒵(ℛ).

Lemma 11. For 𝑎∈ℛ, if 𝑎∈𝒵(ℛ)∩ℛ_𝑖𝑖 (𝑖=1,2), then 𝑎=0.

Proof. For 𝑖∈{1,2}, assume that 𝑎∈𝒵(ℛ)∩ℛ_𝑖𝑖. So 𝑎𝑒_𝑖𝑏𝑒_𝑗=𝑒_𝑖𝑏𝑒_𝑗𝑎=0 for all 𝑏∈ℛ, where 𝑗≠𝑖. It follows from the assumption on ℛ that 𝑎=𝑎𝑒_𝑖=0.

Lemma 12 (see [19, Lemma 4]). Let ℛ^′ be a ring having unit 1 and an idempotent element 𝑒. Assume that ℛ^′ satisfies that 𝑎ℛ^′𝑒={0}, which implies that 𝑎=0. For 𝑎∈ℛ^′, if 𝑒𝑥𝑒𝑎𝑒=𝑒𝑎𝑒𝑥𝑒 for all 𝑥∈ℛ^′, then there exists an element 𝜆∈𝒵(ℛ^′) such that 𝑒𝑎𝑒=𝜆𝑒.

Applying Lemma 12 to our ring ℛ, we get that, for any 𝑎_𝑖𝑖∈ℛ_𝑖𝑖, if 𝑒_𝑖𝑥𝑎_𝑖𝑖=𝑎_𝑖𝑖𝑥𝑒_𝑖 holds for all 𝑥∈ℛ, then there exists 𝜆_𝑖∈𝒵(ℛ) such that 𝑎_𝑖𝑖=𝜆_𝑖𝑒_𝑖, 𝑖=1,2.

Now, we are in the position to give the proofs of Theorems 1 and 2.

Proof of Theorem 1. We will prove the theorem by a series of claims.
Claim 1. 𝐿(0)=0.
By the definition of 𝐿, we have 𝐿(0)]=𝐿(0)]]+(0]+](0=𝐿([[0,0,0])[[,0,0[[0,𝐿)],0[[0,0,𝐿)]=0. (12)
Claim 2. 𝑒₁𝐿(𝑒_𝑖)𝑒₁+𝑒₂𝐿(𝑒_𝑖)𝑒₂∈𝒵(ℛ),  𝑖=1,2.
Let 1≤𝑖≠𝑗≤2. For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗 and 𝑒_𝑖, we have 𝐿𝑎_𝑖𝑗=𝐿𝑎_𝑖𝑗,𝑒_𝑖,𝑒_𝑖=𝐿𝑎_𝑖𝑗,𝑒_𝑖,𝑒_𝑖+𝑎_𝑖𝑗,𝐿𝑒_𝑖,𝑒_𝑖+𝑎_𝑖𝑗,𝑒_𝑖,𝐿𝑒_𝑖=𝐿𝑎_𝑖𝑗𝑒_𝑖−2𝑒_𝑖𝐿𝑎_𝑖𝑗𝑒_𝑖+𝑒_𝑖𝐿𝑎_𝑖𝑗+𝑎_𝑖𝑗𝐿𝑒_𝑖𝑒_𝑖−𝑎_𝑖𝑗𝐿𝑒_𝑖+𝑒_𝑖𝐿𝑒_𝑖𝑎_𝑖𝑗−𝑎_𝑖𝑗𝐿𝑒_𝑖+𝐿𝑒_𝑖𝑎_𝑖𝑗. (13) Multiplying by 𝑒_𝑖 and 𝑒_𝑗 from the left and the right in the above equation, respectively, one gets 2𝑒_𝑖𝐿(𝑒_𝑖)𝑒_𝑖𝑎_𝑖𝑗=2𝑎_𝑖𝑗𝑒_𝑗𝐿(𝑒_𝑖)𝑒_𝑗 for all 𝑎_𝑖𝑗∈ℛ_𝑖𝑗, which implies 𝑒_𝑖𝐿𝑒_𝑖𝑒_𝑖𝑎_𝑖𝑗=𝑎_𝑖𝑗𝑒_𝑗𝐿𝑒_𝑖𝑒_𝑗∀𝑎_𝑖𝑗∈ℛ_𝑖𝑗, (14) since the characteristic of ℛ is not 2.
Similarly, for any 𝑎_𝑗𝑖∈ℛ_𝑗𝑖, by the relation 𝐿(𝑎_𝑗𝑖)=𝐿([[𝑎_𝑗𝑖,𝑒_𝑖],𝑒_𝑖]), one can check that 𝑎_𝑗𝑖𝑒_𝑖𝐿𝑒_𝑖𝑒_𝑖=𝑒_𝑗𝐿𝑒_𝑖𝑒_𝑗𝑎_𝑗𝑖∀𝑎_𝑗𝑖∈ℛ_𝑗𝑖. (15) Applying Lemma 8 to (14) and (15), one obtains 𝑒_𝑖𝐿(𝑒_𝑖)𝑒_𝑖+𝑒_𝑗𝐿(𝑒_𝑖)𝑒_𝑗∈𝒵(ℛ) for 1≤𝑖≠𝑗≤2. The claim holds.
Now let 𝛿(𝑎)=𝐿(𝑎)−[𝑎,𝑒₁𝐿(𝑒₁)𝑒₂−𝑒₂𝐿(𝑒₁)𝑒₁] for all 𝑎∈ℛ. Then 𝛿∶ℛ→ℛ is also a multiplicative Lie triple derivation and satisfies 𝛿(𝑒₁)∈𝒵(ℛ) as Claim 2. Thus, without loss of generality, we may assume 𝐿𝑒₁∈𝒵(ℛ). (16)
Claim 3. 𝐿(𝑒₂)∈𝒵(ℛ).
By Claim 2, 𝑒₁𝐿(𝑒₂)𝑒₁+𝑒₂𝐿(𝑒₂)𝑒₂∈𝒵(ℛ). So we only need to check 𝑒₁𝐿(𝑒₂)𝑒₂=𝑒₂𝐿(𝑒₂)𝑒₁=0. In fact, by (16), we have 0=𝐿𝑒₂,𝑒₁,𝑒₁=𝐿𝑒₂,𝑒₁,𝑒₁+𝑒₂,𝐿𝑒₁,𝑒₁+𝑒₂,𝑒₁,𝐿𝑒₁=𝐿𝑒₂,𝑒₁,𝑒₁=𝑒₂𝐿𝑒₂𝑒₁+𝑒₁𝐿𝑒₂𝑒₂, (17) which implies 𝑒₁𝐿(𝑒₂)𝑒₂=𝑒₂𝐿(𝑒₂)𝑒₁=0. Hence 𝐿(𝑒₂)=𝑒₁𝐿(𝑒₂)𝑒₁+𝑒₂𝐿(𝑒₂)𝑒₂∈𝒵(ℛ).
Claim 4. 𝐿(ℛ_𝑖𝑗)⊆ℛ_𝑖𝑗, 1≤𝑖≠𝑗≤2.
Here, we only give the proof for 𝐿(ℛ₁₂)⊆ℛ₁₂. The proof for the other inclusion 𝐿(ℛ₂₁)⊆ℛ₂₁ is similar.
For any 𝑎₁₂∈ℛ₁₂, by (16), (13) can be reduced to 𝐿𝑎₁₂=𝐿𝑎₁₂𝑒₁−2𝑒₁𝐿𝑎₁₂𝑒₁+𝑒₁𝐿𝑎₁₂=𝑒₁𝐿𝑎₁₂𝑒₂+𝑒₂𝐿𝑎₁₂𝑒₁. (18) Taking any 𝑏₁₂∈ℛ₁₂ and any 𝑐∈ℛ, one obtains 0=𝐿𝑎₁₂,𝑏₁₂=𝐿𝑎,𝑐₁₂,𝑏₁₂,𝑐+𝑎₁₂,𝐿𝑏₁₂+𝑎,𝑐₁₂,𝑏₁₂(𝑐=𝐿𝑎,𝐿)₁₂,𝑏₁₂+𝑎₁₂,𝐿𝑏₁₂,𝑐, (19) which implies [𝐿(𝑎₁₂),𝑏₁₂]+[𝑎₁₂,𝐿(𝑏₁₂)]∈𝒵(ℛ). Note that 𝑎₁₂,𝐿𝑏₁₂=𝑒₁,𝑎₁₂,𝐿𝑏₁₂=𝐿𝑒₁,𝑎₁₂,𝑏₁₂−𝐿𝑒₁,𝑎₁₂,𝑏₁₂−𝑒₁,𝐿𝑎₁₂,𝑏₁₂=−𝑒₁,𝐿𝑎₁₂,𝑏₁₂. (20) So, by (18), one achieves 𝐿𝑎₁₂,𝑏₁₂+𝑎₁₂,𝐿𝑏₁₂=𝑒₂𝐿𝑎₁₂𝑒₁,𝑏₁₂−𝑒₁,𝐿𝑎₁₂,𝑏₁₂=2𝑒₂𝐿𝑎₁₂𝑒₁,𝑏₁₂∈𝒵(ℛ). (21) It follows from char  ℛ≠2 that [𝑒₂𝐿(𝑎₁₂)𝑒₁,𝑏₁₂]=𝑒₂𝐿(𝑎₁₂)𝑒₁𝑏₁₂−𝑏₁₂𝑒₂𝐿(𝑎₁₂)𝑒₁∈𝒵(ℛ) for all 𝑏₁₂∈ℛ₁₂. By Lemma 8, one can get 𝑒₂𝐿(𝑎₁₂)𝑒₁𝑏₁₂∈𝒵(ℛ₂₂) for all 𝑏₁₂∈ℛ₁₂. It is easily checked that 𝑒₂𝐿(𝑎₁₂)𝑒₁ℛ𝑒₂ is an ideal of ℛ₂₂. So, by the assumption (ii), 𝑒₂𝐿(𝑎₁₂)𝑒₁ℛ𝑒₂={0}, and so 𝑒₂𝐿(𝑎₁₂)𝑒₁=0 by the condition (i). This and (18) imply that 𝐿(𝑎₁₂)=𝑒₁𝐿(𝑎₁₂)𝑒₂∈ℛ₁₂, as desired.
Claim 5. For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗, we have 𝐿(−𝑎_𝑖𝑗)=−𝐿(𝑎_𝑖𝑗), 1≤𝑖≠𝑗≤2.
By (16) and Claims 3 and 4, it is clear that 𝐿−𝑎_𝑖𝑗=𝐿𝑎_𝑖𝑗,𝑒_𝑖,𝑒_𝑗=𝐿𝑎_𝑖𝑗,𝑒_𝑖,𝑒_𝑗=−𝐿𝑎_𝑖𝑗. (22)
Claim 6. For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗, we have 𝐿(𝑒_𝑖+𝑎_𝑖𝑗)−𝐿(𝑎_𝑖𝑗)∈𝒵(ℛ), 1≤𝑖≠𝑗≤2.
Let 1≤𝑖≠𝑗≤2 and 𝑎_𝑖𝑗∈ℛ_𝑖𝑗. For any 𝑏_𝑖𝑗∈ℛ_𝑖𝑗, by (16) and Claims 3 and 4, one has 𝐿𝑏_𝑖𝑗=𝐿𝑒_𝑖+𝑎_𝑖𝑗,𝑏_𝑖𝑗,𝑒_𝑗=𝐿𝑒_𝑖+𝑎_𝑖𝑗,𝑏_𝑖𝑗,𝑒_𝑗+𝐿𝑏_𝑖𝑗; (23) that is, [[𝐿(𝑒_𝑖+𝑎_𝑖𝑗),𝑏_𝑖𝑗],𝑒_𝑗]=0. Note that [[𝐿(𝑎_𝑖𝑗),𝑏_𝑖𝑗],𝑒_𝑗]=0 by Claim 4. So 𝐿𝑒_𝑖+𝑎_𝑖𝑗−𝐿𝑎_𝑖𝑗,𝑏_𝑖𝑗,𝑒_𝑗=0. (24) Similarly, for any 𝑏_𝑗𝑖∈ℛ_𝑗𝑖, by the relations −𝐿(𝑏_𝑗𝑖)=𝐿(−𝑏_𝑗𝑖)=𝐿([[𝑒_𝑖+𝑎_𝑖𝑗,𝑏_𝑗𝑖],𝑒_𝑖]) and [[𝐿(𝑎_𝑖𝑗),𝑏_𝑗𝑖],𝑒_𝑖]=0, one can check 𝐿𝑒_𝑖+𝑎_𝑖𝑗−𝐿𝑎_𝑖𝑗,𝑏_𝑗𝑖,𝑒_𝑖=0. (25) Combining (24) and (25) and by Lemma 10, we achieve 𝑒₁𝐿𝑒_𝑖+𝑎_𝑖𝑗−𝐿𝑎_𝑖𝑗𝑒₁+𝑒₂𝐿𝑒_𝑖+𝑎_𝑖𝑗−𝐿𝑎_𝑖𝑗𝑒₂∈𝒵(ℛ). (26) Thus, to complete the proof of the claim, we still need to check 𝑒₁(𝐿(𝑒_𝑖+𝑎_𝑖𝑗)−𝐿(𝑎_𝑖𝑗))𝑒₂=𝑒₂(𝐿(𝑒_𝑖+𝑎_𝑖𝑗)−𝐿(𝑎_𝑖𝑗))𝑒₁=0. In fact, by (16), we have 𝐿𝑎_𝑖𝑗=𝐿𝑒_𝑖+𝑎_𝑖𝑗,𝑒₁,𝑒₁=𝐿𝑒_𝑖+𝑎_𝑖𝑗,𝑒₁,𝑒₁,𝐿𝑎_𝑖𝑗=𝐿𝑎_𝑖𝑗,𝑒₁,𝑒₁=𝐿𝑎_𝑖𝑗,𝑒₁,𝑒₁. (27) It follows that [[𝐿(𝑒_𝑖+𝑎_𝑖𝑗)−𝐿(𝑎_𝑖𝑗),𝑒₁],𝑒₁]=0, which implies 𝑒₁(𝐿(𝑒_𝑖+𝑎_𝑖𝑗)−𝐿(𝑎_𝑖𝑗))𝑒₂=𝑒₂(𝐿(𝑒_𝑖+𝑎_𝑖𝑗)−𝐿(𝑎_𝑖𝑗))𝑒₁=0.
Claim 7. 𝐿 is additive on ℛ_𝑖𝑗, 1≤𝑖≠𝑗≤2.
Take any 𝑎_𝑖𝑗,𝑏_𝑖𝑗∈ℛ_𝑖𝑗 (1≤𝑖≠𝑗≤2). By (16) and Claims 3–6, one obtains 𝐿𝑎_𝑖𝑗+𝑏_𝑖𝑗=𝐿𝑒_𝑖−𝑎_𝑖𝑗,𝑒_𝑖+𝑏_𝑖𝑗,𝑒_𝑗=𝐿𝑒_𝑖−𝑎_𝑖𝑗,𝑒_𝑖+𝑏_𝑖𝑗,𝑒_𝑗+𝑒_𝑖−𝑎_𝑖𝑗,𝐿𝑒_𝑖+𝑏_𝑖𝑗,𝑒_𝑗=−𝐿𝑎_𝑖𝑗,𝑒_𝑖+𝑏_𝑖𝑗,𝑒_𝑗+𝑒_𝑖−𝑎_𝑖𝑗,𝐿𝑏_𝑖𝑗,𝑒_𝑗=𝐿𝑎_𝑖𝑗+𝐿𝑏_𝑖𝑗. (28)
Claim 8. For 1≤𝑖≠𝑗≤2, we have 𝐿(ℛ_𝑖𝑖)⊆(ℛ_𝑖𝑖+ℛ_𝑗𝑗), 𝐿(ℛ_𝑖𝑖+ℛ_𝑖𝑖)⊆(ℛ_𝑖𝑖+ℛ_𝑗𝑗), and 𝐿(ℛ_𝑖𝑖+ℛ_𝑗𝑗)⊆(ℛ_𝑖𝑖+ℛ_𝑗𝑗).
Let 1≤𝑖≠𝑗≤2. For any 𝑎_𝑖𝑖,𝑏_𝑖𝑖∈ℛ_𝑖𝑖 and 𝑎_𝑗𝑗∈ℛ_𝑗𝑗, by (16) and Claim 3, one has 0=𝐿𝑎_𝑖𝑖+𝑎_𝑗𝑗,𝑒_𝑗,𝑒_𝑗=𝐿𝑎_𝑖𝑖+𝑎_𝑗𝑗,𝑒_𝑗,𝑒_𝑗,0=𝐿𝑎_𝑖𝑖,𝑒_𝑗,𝑒_𝑗=𝐿𝑎_𝑖𝑖,𝑒_𝑗,𝑒_𝑗,0=𝐿𝑎_𝑖𝑖+𝑏_𝑖𝑖,𝑒_𝑗,𝑒_𝑗=𝐿𝑎_𝑖𝑖+𝑏_𝑖𝑖,𝑒_𝑗,𝑒_𝑗. (29) A simple calculation yields 𝑒₁𝐿(𝑎_𝑖𝑖+𝑎_𝑗𝑗)𝑒₂=𝑒₂𝐿(𝑎_𝑖𝑖+𝑎_𝑗𝑗)𝑒₁=0, 𝑒₁𝐿(𝑎_𝑖𝑖)𝑒₂=𝑒₂𝐿(𝑎_𝑖𝑖)𝑒₁=0, and 𝑒₁𝐿(𝑎_𝑖𝑖+𝑏_𝑖𝑖)𝑒₂=𝑒₂𝐿(𝑎_𝑖𝑖+𝑏_𝑖𝑖)𝑒₁=0. So the claim is true.
Claim 9. For any 𝑎₁₁∈ℛ₁₁ and 𝑎₂₂∈ℛ₂₂, we have 𝐿(𝑎₁₁+𝑎₂₂)−𝐿(𝑎₁₁)−𝐿(𝑎₂₂)∈𝒵(ℛ).
Take any 𝑎₁₁∈ℛ₁₁and𝑎₂₂∈ℛ₂₂ and write 𝐿(𝑎₁₁+𝑎₂₂)−𝐿(𝑎₁₁)−𝐿(𝑎₂₂)=𝑠₁₁+𝑠₁₂+𝑠₂₁+𝑠₂₂. Note that, by Claim 8, one has 𝐿(𝑎₁₁+𝑎₂₂)−𝐿(𝑎₁₁)−𝐿(𝑎₂₂)=𝑠₁₁+𝑠₂₂. So, to prove the claim, one only needs to check 𝑠₁₁+𝑠₂₂∈𝒵(ℛ).
To do this, taking any 𝑏₁₂∈ℛ₁₂, by Claims 3--5 and 7, we have 𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂=𝐿𝑎₁₁𝑏₁₂−𝑏₁₂𝑎₂₂=𝐿𝑎₁₁𝑏₁₂−𝐿𝑏₁₂𝑎₂₂=𝐿𝑎₁₁,𝑏₁₂,𝑒₂−𝐿𝑏₁₂,𝑎₂₂,𝑒₂=𝐿𝑎₁₁,𝑏₁₂,𝑒₂+𝑎₁₁,𝐿𝑏₁₂,𝑒₂−𝐿𝑏₁₂,𝑎₂₂,𝑒₂−𝑏₁₂,𝐿𝑎₂₂,𝑒₂=𝐿𝑎₁₁,𝑏₁₂,𝑒₂+𝑎₁₁,𝐿𝑏₁₂+𝑎₂₂,𝐿𝑏₁₂+𝐿𝑎₂₂,𝑏₁₂,𝑒₂,𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂=𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂,𝑒₂=𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂,𝑒₂+𝑎₁₁+𝑎₂₂,𝐿𝑏₁₂,𝑒₂=𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂,𝑒₂+𝑎₁₁,𝐿𝑏₁₂+𝑎₂₂,𝐿𝑏₁₂. (30) Combining the above two equations yields 𝐿𝑎₁₁+𝑎₂₂−𝐿𝑎₁₁−𝐿𝑎₂₂,𝑏₁₂,𝑒₂=0∀𝑏₁₂∈ℛ₁₂. (31)
Similar to the above discussion, one can also prove that 𝐿𝑎₁₁+𝑎₂₂−𝐿𝑎₁₁−𝐿𝑎₂₂,𝑏₂₁,𝑒₁=0∀𝑏₂₁∈ℛ₂₁. (32) Now, by Lemma 10, it follows that 𝑠₁₁+𝑠₂₂∈𝒵(ℛ), and the claim holds.
Claim 10. For any 𝑎∈ℛ, any 𝑎₁₂∈ℛ₁₂, and any 𝑎₂₁∈ℛ₂₁, we have(i)𝐿(𝑎+𝑎₁₂)−𝐿(𝑎)−𝑒₁(𝐿(𝑎+𝑎₁₂)−𝐿(𝑎))𝑒₂∈𝒵(ℛ);(ii)𝐿(𝑎+𝑎₂₁)−𝐿(𝑎)−𝑒₂(𝐿(𝑎+𝑎₂₁)−𝐿(𝑎))𝑒₁∈𝒵(ℛ).
For any 𝑎,𝑐∈ℛ and any 𝑎₁₂,𝑏₁₂∈ℛ₁₂, since [[𝑎+𝑎₁₂,𝑏₁₂],𝑐]=[[𝑎,𝑏₁₂],𝑐], by Claim 4, one can easily get [[𝐿(𝑎+𝑎₁₂),𝑏₁₂],𝑐]=[[𝐿(𝑎),𝑏₁₂],𝑐]; that is, 𝐿𝑎+𝑎₁₂−𝐿(𝑎),𝑏₁₂,𝑐=0∀𝑐∈ℛ. (33) Write 𝑡=𝐿(𝑎+𝑎₁₂)−𝐿(𝑎)=𝑡₁₁+𝑡₁₂+𝑡₂₁+𝑡₂₂. Then (33) implies that 𝑡,𝑏₁₂=𝑡₁₁𝑏₁₂+𝑡₂₁𝑏₁₂−𝑏₁₂𝑡₂₁−𝑏₁₂𝑡₂₂∈𝒵(ℛ). (34) Letting 𝑐=𝑒₂ in (33), a simple calculation yields 𝑡₁₁𝑏₁₂=𝑏₁₂𝑡₂₂∀𝑏₁₂∈ℛ₁₂. (35) So (34) is reduced to 𝑡₂₁𝑏₁₂−𝑏₁₂𝑡₂₁∈𝒵(ℛ), which and Lemma 8 imply that 𝑡₂₁𝑏₁₂∈𝒵(ℛ₂₂) holds for all 𝑏₁₂∈ℛ₁₂. Note that 𝑡₂₁𝑒₁ℛ𝑒₂ is an ideal of ℛ₂₂. It follows from the assumption (ii) that 𝑡₂₁𝑒₁ℛ𝑒₂={0}, and so 𝑡₂₁=0. (36) On the other hand, by using the equation [[𝑎+𝑎₁₂,𝑏₂₁],𝑒₂]=[[𝑎,𝑏₂₁],𝑒₂], one can also show 𝑡₂₂𝑏₂₁=𝑏₂₁𝑡₁₁∀𝑏₂₁∈ℛ₂₁,𝑡₁₂=0. (37) Combining (35)–(37) and Lemma 8, we get that (i) holds.
The proof of (ii) is similar and we omit it here.
Claim 11. For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗 (1≤𝑖,𝑗≤2), we have(i)𝐿(𝑎₁₂)=𝑒₁𝐿(𝑎₁₁+𝑎₁₂+𝑎₂₂)𝑒₂ and 𝑒₂𝐿(𝑎₁₁+𝑎₁₂+𝑎₂₂)𝑒₁=0;(ii)𝐿(𝑎₂₁)=𝑒₂𝐿(𝑎₁₁+𝑎₂₁+𝑎₂₂)𝑒₁ and 𝑒₁𝐿(𝑎₁₁+𝑎₂₁+𝑎₂₂)𝑒₂=0.
Here, we only give the proof of (i). The proof of (ii) is similar.
For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗 (1≤𝑖,𝑗≤2), by Claim 3, one gets 𝐿𝑎₁₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑒₂,𝑒₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑒₂,𝑒₂=𝑒₁𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂𝑒₂+𝑒₂𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂𝑒₁, (38) which, together with Claim 4, implies 𝐿(𝑎₁₂)=𝑒₁𝐿(𝑎₁₁+𝑎₁₂+𝑎₂₂)𝑒₂ and 𝑒₂𝐿(𝑎₁₁+𝑎₁₂+𝑎₂₂)𝑒₁=0, as desired.
Claim 12. For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗 (1≤𝑖,𝑗≤2), we have(i)𝐿(𝑎₁₁+𝑎₁₂+𝑎₂₂)−𝐿(𝑎₁₁)−𝐿(𝑎₁₂)−𝐿(𝑎₂₂)∈𝒵(ℛ);(ii)𝐿(𝑎₁₁+𝑎₂₁+𝑎₂₂)−𝐿(𝑎₁₁)−𝐿(𝑎₂₁)−𝐿(𝑎₂₂)∈𝒵(ℛ).
Still, we only give the proof of (i).
Take any 𝑎₁₁∈ℛ₁₁, 𝑎₁₂∈ℛ₁₂, and 𝑎₂₂∈ℛ₂₂. Firstly, by Claims 8, 9, and 11, there exists some 𝑧∈𝒵(ℛ) such that 𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂−𝐿𝑎₁₁−𝐿𝑎₁₂−𝐿𝑎₂₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂−𝐿𝑎₁₁+𝑎₂₂−𝐿𝑎₁₂+𝑧=𝑒₁𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂−𝐿𝑎₁₁+𝑎₂₂𝑒₁+𝑒₂𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂−𝐿𝑎₁₁+𝑎₂₂𝑒₂+𝑧. (39)
Next, for any 𝑏₁₂∈ℛ₁₂, by Claims 3 and 4, one has 𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑏₁₂,𝑒₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑏₁₂,𝑒₂+𝑎₁₁+𝑎₁₂+𝑎₂₂,𝐿𝑏₁₂,𝑒₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑏₁₂,𝑒₂+𝑎₁₁+𝑎₂₂,𝐿𝑏₁₂,𝑒₂,𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑏₁₂,𝑒₂=𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂,𝑒₂=𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂,𝑒₂+𝑎₁₁+𝑎₂₂,𝐿𝑏₁₂,𝑒₂. (40) Comparing the above equations, we get 𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂−𝐿𝑎₁₁+𝑎₂₂,𝑏₁₂,𝑒₂=0. (41)
Finally, for any 𝑏₂₁∈ℛ₂₁, a similar argument to the above achieves 𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂−𝐿𝑎₁₁+𝑎₂₂,𝑏₂₁,𝑒₁=0. (42)
Now, combining (39)–(42) and Lemma 10, the claim holds.
Claim 13. For any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗(1≤𝑖,𝑗≤2), we have 𝐿(𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂)−𝐿(𝑎₁₁)−𝐿(𝑎₁₂)−𝐿(𝑎₂₁)−𝐿(𝑎₂₂)∈𝒵(ℛ).
Take any 𝑎_𝑖𝑗∈ℛ_𝑖𝑗(1≤𝑖,𝑗≤2) and write 𝑡=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂−𝐿𝑎₁₁−𝐿𝑎₁₂−𝐿𝑎₂₁−𝐿𝑎₂₂. (43) For any 𝑏₁₂∈ℛ₁₂ and any 𝑏₂₁∈ℛ₂₁, by Claims 4, 10, and 12, one has 𝑡,𝑏₁₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂−𝐿𝑎₁₁−𝐿𝑎₂₁−𝐿𝑎₂₂,𝑏₁₂=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂−𝐿𝑎₁₁+𝑎₂₁+𝑎₂₂,𝑏₁₂=0,𝑡,𝑏₂₁=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂−𝐿𝑎₁₁−𝐿𝑎₁₂−𝐿𝑎₂₂,𝑏₂₁=𝐿𝑎₁₁+𝑎₁₂+𝑎₂₁+𝑎₂₂−𝐿𝑎₁₁+𝑎₁₂+𝑎₂₂,𝑏₂₁=0. (44) Now the above two equations and Lemma 9 imply 𝑡∈𝒵(ℛ).
Claim 14. For any 𝑎_𝑖𝑖,𝑏_𝑖𝑖