Unital Four-Dimensional Lattice-Ordered Algebras with a Distributive Basis

Abstract

All unital four-dimensional lattice-ordered algebras with a 𝑑-basis are constructed.

1. Introduction

Lattice-ordered algebras (ℓ-algebras) with a distributive basis (𝑑-basis) were studied in [5] in an effort to obtain general structure theory for a class of ℓ-algebras that includes some well-known and important examples of ℓ-algebras, such as matrix and triangular matrix ℓ-algebras over a totally ordered field with entrywise lattice order, polynomial ℓ-algebra over a totally ordered field with coordinatewise lattice order, and so forth.

All unital two-dimensional and three-dimensional ℓ-algebras with a 𝑑-basis have been constructed [6]. In this paper, we construct all unital four-dimensional ℓ-algebras with a 𝑑-basis. We will first review some definitions and results for later use. The reader is referred to [1, 5, 6] for the general theory of lattice-ordered rings (ℓ-rings) and undefined terminologies in this paper. Throughout 𝐹 always denotes a totally ordered field.

Let 𝑅 be an ℓ-ring. The positive cone of 𝑅 is 𝑅⁺={𝑟∈𝑅∶𝑟≥0}. A vector lattice 𝑉 over 𝐹 is an ℓ-group and a vector space over 𝐹 such that 𝐹⁺𝑉⁺⊆𝑉⁺. An ℓ-algebra 𝐴 over 𝐹 is an algebra over 𝐹, that is, an ℓ-ring and a vector lattice over 𝐹. An ℓ-algebra is called unital if it has an identity element 1 and called ℓ-unital if 1≥0. An element 𝑎 in an ℓ-ring is called basic if 𝑎>0 and, for any 𝑏,𝑐≥0, 𝑎≥𝑏,𝑐 implies that 𝑏 and 𝑐 are comparable, that is, either 𝑏≥𝑐 or 𝑐≥𝑏. A set of elements {𝑥_𝑖∶𝑖∈𝐼} is called disjoint if each 𝑥_𝑖>0 and 𝑥_𝑖∧𝑥_𝑗=0 for any two different elements 𝑥_𝑖 and 𝑥_𝑗. An element 𝑥≥0 in an ℓ-ring 𝑅 is called an 𝑓-element if 𝑦∧𝑧=0 implies that 𝑥𝑦∧𝑧=𝑦𝑥∧𝑧=0, and 𝑥 is called a 𝑑-element if 𝑦∧𝑧=0 implies that 𝑥𝑦∧𝑥𝑧=𝑦𝑥∧𝑧𝑥=0. An ℓ-ring 𝑅 is called ℓ-reduced if for any 𝑎∈𝑅⁺, 𝑎²=0 implies that 𝑎=0.

Let 𝐴 be an ℓ-algebra over 𝐹. A nonempty subset 𝑆 of 𝐴 is called a 𝑑-basis of 𝐴 if 𝑆 is a disjoint subset of 𝐴 that spans 𝐴 as a vector space over 𝐹 and each element in 𝑆 is a 𝑑-element [6, Definition  2.1]. Then for each 𝑎∈𝐴, 𝑎=𝛼₁𝑠₁+⋯+𝛼_𝑝𝑠_𝑝, where 𝑝≥1, 𝛼₁,…,𝛼_𝑝∈𝐹, and 𝑠₁,…,𝑠_𝑝∈𝑆 are distinct, and 𝑎≥0 if and only if each 𝛼_𝑖≥0, 𝑖=1,…,𝑝. Thus in a 𝑑-basis, each element is basic. Let 𝐴 be a unital finite-dimensional ℓ-algebra with a 𝑑-basis. Then 𝐴 is ℓ-unital [1], and there exists a 𝑑-basis 𝑆 such that 1=𝑒₁+⋯+𝑒_𝑛, where 𝑒₁,…,𝑒_𝑛∈𝑆 are distinct.

In the following theorem, we record some results on unital ℓ-algebras with a 𝑑-basis from [5, 6].

Theorem 1.1. Let 𝐴 be a unital finite-dimensional ℓ-algebra over 𝐹 with a 𝑑-basis 𝑆 such that 1=𝑒₁+⋯+𝑒_𝑛, where 𝑛≥1 and 𝑒₁,…,𝑒_𝑛∈𝑆 are distinct. (1)Each 𝑒_𝑖 is an idempotent 𝑓-element and 𝑒_𝑖𝑒_𝑗=0 for 𝑖≠𝑗. (2)Let 𝑎 be a basic element. Then there exists exactly one 𝑒_𝑖 such that 𝑒_𝑖𝑎=𝑎, and for any 𝑗≠𝑖, 𝑒_𝑗𝑎=0. Similarly, there exists exactly one 𝑒_𝑘 such that 𝑎𝑒_𝑘=𝑎, and for any 𝑗≠𝑘, 𝑎𝑒_𝑗=0. (3)For each element 𝑎∈𝑆, if 𝑒_𝑖𝑎=𝑎 or 𝑎𝑒_𝑖=𝑎, 1≤𝑖≤𝑛, then either 𝑎 is nilpotent or 𝑎^𝑛𝑎=𝛼_𝑎𝑒_𝑖 for some positive integer 𝑛_𝑎 and 0<𝛼_𝑎∈𝐹. (4)Let 𝑎,𝑏∈𝑆. If either 𝑎 or 𝑏 is not nilpotent and 𝑎𝑏≠0, then 𝑎𝑏 is basic.

2. The Classification of Unital Four-Dimensional ℓ-Algebras with a 𝑑-Basis

We first list some ℓ-algebras that we are going to use for the classification of unital four-dimensional ℓ-algebras over 𝐹 with a 𝑑-basis.(i)Let 𝐺 be a finite group, and let 𝐹[𝐺] be the group algebra over 𝐹. Define an element ∑_𝑖𝛼_𝑖𝑔_𝑖≥0 if each 𝛼_𝑖≥0. Then 𝐹[𝐺] is an ℓ-algebra over 𝐹 and 𝐺 is a 𝑑-basis. In the following, 𝐹[𝐺] always denotes the ℓ-algebra with the lattice order defined above.(ii)Let 0<𝛼∈𝐹, and let 𝑓(𝑥)=𝑥^𝑛−𝛼∈𝐹[𝑥] be irreducible. We use 𝑎=√^𝑛𝛼 to denote a fixed root of 𝑓(𝑥). Consider the extension field 𝐹[√^𝑛𝛼] of 𝐹. Then 𝐹√^𝑛𝛼=𝛼_𝑛−1𝑎^𝑛−1+⋯+𝛼₁𝑎+𝛼₀∣𝛼_𝑖∈𝐹,(2.1) and {𝑎^𝑛−1,…,𝑎,1} is a basis of 𝐹[√^𝑛𝛼] over 𝐹 as a vector space over 𝐹. Now define 𝛼_𝑛−1𝑎^𝑛−1+⋯+𝛼₁𝑎+𝛼₀≥0 if each 𝛼_𝑖≥0, 𝑖=0,1,…,𝑛−1. Since 𝑎^𝑛=𝛼>0, 𝐹[√^𝑛𝛼] becomes an ℓ-algebra over 𝐹 with respect to the above lattice order, and 𝑎 is a 𝑑-element. Thus {𝑎^𝑛−1,…,𝑎,1} is a 𝑑-basis of the ℓ-algebra 𝐹[√^𝑛𝛼] over 𝐹. In this paper, 𝑛≤4 and 𝐹[√^𝑛𝛼] always denotes the ℓ-algebra defined above.(iii)For 𝑛≥1, let 𝑀_𝑛(𝐹) and 𝑇_𝑛(𝐹) be 𝑛×𝑛 matrix algebra and upper triangular matrix algebra, respectively. Define a matrix positive if each of its entries is positive. Then 𝑀_𝑛(𝐹) and 𝑇_𝑛(𝐹) become ℓ-algebras. Let 𝑒_𝑖𝑗 denote the matrix with 𝑖𝑗th entry equal to 1 and other entries equal to 0. Then {𝑒_𝑖𝑗∣1≤𝑖,𝑗≤𝑛} is a 𝑑-basis for 𝑀_𝑛(𝐹) over 𝐹 and {𝑒_𝑖𝑗∣1≤𝑖≤𝑗≤𝑛} is a 𝑑-basis for 𝑇_𝑛(𝐹) over 𝐹. In the following, 𝑀_𝑛(𝐹) and 𝑇_𝑛(𝐹) always denote the ℓ-algebras with the above lattice order.

All nonisomorphic unital four-dimensional ℓ-algebras with a 𝑑-basis are listed in Theorem 2.1. Suppose that 𝐴 is a unital four-dimensional ℓ-algebra with a 𝑑-basis 𝑆 such that 1 is a sum of some distinct elements in 𝑆. In the proof of Theorem 2.1, we consider four different cases according to the number of elements in this sum, and in each case we consider two cases in which 𝐴 is ℓ-reduced and 𝐴 is not ℓ-reduced, respectively. We use ⊕ to denote a direct sum of ℓ-rings.

Suppose that 𝐴 is a four-dimensional ℓ-algebra with a 𝑑-basis 𝑆={𝑒₁,𝑒₂,𝑒₃,𝑒₄} such that 1=𝑒₁+𝑒₂+𝑒₃+𝑒₄. By Theorem 1.1, each 𝑒_𝑖 is an idempotent 𝑓-element, and for 1≤𝑖,𝑗≤4, 𝑖≠𝑗, 𝑒_𝑖𝑒_𝑗=0. Thus 𝐴≅𝐹⊕𝐹⊕𝐹⊕𝐹 is a four-dimensional 𝑓-algebra. This is (1) in Theorem 2.1.

In the following sections, we consider the remaining three cases where 1 is a sum of three, two, or one element from the 𝑑-basis 𝑆, respectively.

Theorem 2.1. Let 𝐴 be a unital four-dimensional ℓ-algebra over 𝐹 with a 𝑑-basis. Then 𝐴 is isomorphic to one of following ℓ-algebras over 𝐹: (1)𝐹⊕𝐹⊕𝐹⊕𝐹, (2)𝐹[𝐺]⊕𝐹⊕𝐹, where 𝐺 is a group of order two, (3)𝐹[√𝛼]⊕𝐹⊕𝐹, where 0<𝛼∈𝐹, (4)(𝐹1+𝐹𝑎)⊕𝐹⊕𝐹, where 𝑎²=0, (5)𝑇₂(𝐹)⊕𝐹, (6)𝐹[𝐺]⊕𝐹[𝐺], where 𝐺 is a group of order two, (7)𝐹[𝐺]⊕𝐹[√𝛼], where 0<𝛼∈𝐹 and 𝐺 is a group of order two, (8)𝐹[√𝛼]⊕𝐹[√𝛽], where 0<𝛼,𝛽∈𝐹, (9)𝐹[𝐺]⊕𝐹, where 𝐺 is a group of order three, (10)𝐹[√³𝛼]⊕𝐹, where 0<𝛼∈𝐹, (11)𝑀₂(𝐹), (12)(𝐹1+𝐹𝑎+𝐹𝑏)⊕𝐹, where 𝑎²=𝑏²=𝑎𝑏=𝑏𝑎=0, (13)(𝐹1+𝐹𝑎)⊕(𝐹1+𝐹𝑏), where 𝑎²=𝑏²=0, (14)𝑇₂(𝐹)+𝐹𝑎, where 𝑒₁₁𝑎=𝑎𝑒₁₁=𝑎, 𝑎𝑒₁₂=0, and 𝑎²=0, (15)𝑇₂(𝐹)+𝐹𝑎, where 𝑒₁₁𝑎=𝑎𝑒₂₂=𝑎, and 𝑎²=0, (16)𝑇₂(𝐹)+𝐹𝑎, where 𝑒₂₂𝑎=𝑎𝑒₁₁=𝑎, 𝑒₁₂𝑎=𝑎𝑒₁₂=0, and 𝑎²=0, (17)𝑇₂(𝐹)+𝐹𝑎, where 𝑒₂₂𝑎=𝑎𝑒₂₂=𝑎, 𝑒₁₂𝑎=0, and 𝑎²=0, (18)(𝐹1+𝐹𝑎+𝐹𝑎²)⊕𝐹, where 𝑎³=0, (19)𝐹[𝐺]⊕(𝐹1+𝐹𝑎), where 𝑎²=0, (20)𝐹[√𝛼]⊕(𝐹1+𝐹𝑎), where 𝑎²=0 and 0<𝛼∈𝐹, (21)𝐹[𝐺], where 𝐺 is a cyclic group of order four, (22)𝐹[√⁴𝛼], where 0<𝛼∈𝐹, (23)𝐹[𝐺]+𝐹[𝐺]𝑏, where 𝐺={𝑒,𝑎} is a group of order two and 𝑏²=𝛼𝑎 with 0<𝛼∈𝐹 and √𝛼∉𝐹, (24)𝐹[𝐺], where 𝐺 is the Klein four group, (25)𝐹[√𝛼][𝐺], where 0<𝛼∈𝐹 and 𝐺 is a group of order two, (26)𝐹[√𝛼][√𝛽], where 0<𝛼,𝛽∈𝐹, √𝛽∉𝐹[√𝛼], (27)𝐹1+𝐹𝑎+𝐹𝑏+𝐹𝑐, where 𝑎²=𝑏+𝑐 and the product of any other two elements of 𝑎,𝑏,𝑐 is zero, (28)𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑎³, where 𝑎⁴=0, (29)𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑏, where 𝑏²=𝛼𝑎², 0<𝛼∈𝐹, and the product of any two elements not both 𝑎 or 𝑏 is zero, (30)𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑏, where the product of any two elements not both 𝑎 is zero, (31)𝐹1+𝐹𝑎+𝐹𝑏+𝐹𝑐, where 𝐹𝑎+𝐹𝑏+𝐹𝑐 is a zero ring, (32)𝐹1+𝐹𝑎+𝐹𝑏+𝐹(𝑎𝑏), where 𝑎²=𝑏²=𝑏𝑎=0, (33)𝐹1+𝐹𝑎+𝐹𝑏+𝐹(𝑎𝑏), where 𝑎²=𝑏²=0 and 𝑏𝑎=𝛼(𝑎𝑏) with 0<𝛼∈𝐹, (34)𝐹[𝐺]+𝐹[𝐺]𝑎, where 𝐺 is a group of order two, 𝑎²=0, and 𝑎 commutes with each element in 𝐹[𝐺], (35)𝐹[√𝛼]+𝐹[√𝛼]𝑎, where 𝑎²=0 and 𝑎 commutes with each element in 𝐹[√𝛼].

3. 1 Is a Sum of Three Disjoint Basic Elements

Let 𝐴 be a unital four-dimensional ℓ-algebra with a 𝑑-basis 𝑆={𝑒₁,𝑒₂,𝑒₃,𝑎} such that 1=𝑒₁+𝑒₂+𝑒₃. Then by Theorem 1.1, each 𝑒_𝑖 is an idempotent 𝑓-element, 𝑖=1,2,3, and 𝑒_𝑖𝑒_𝑗=0 for any 1≤𝑖,𝑗≤3, 𝑖≠𝑗. By Theorem 1.1, we may assume that 𝑒₁𝑎=𝑎 and 𝑒₂𝑎=𝑒₃𝑎=0. We consider two cases according to whether 𝐴 is ℓ-reduced or not.

3.1. 𝐴 Is ℓ-Reduced

If 𝑎𝑒₁=0, then 𝑎𝑒_𝑖=𝑎 for 𝑖=2 or 3 by Theorem 1.1(2), and hence 𝑎²=(𝑎𝑒_𝑖)(𝑒₁𝑎)=0, which contradicts the fact that 𝐴 is ℓ-reduced. Thus 𝑎𝑒₁=𝑎 and 𝑎𝑒₂=𝑎𝑒₃=0. Since 𝑎 is not nilpotent, by Theorem 1.1(4), 𝑎² is a basic element, so 𝑎²=𝛼𝑠 for some 0<𝛼∈𝐹 and 𝑠∈𝑆. Since 𝑒₁∧𝑎=0 and 𝑒₁𝑎=𝑎, 𝑠≠𝑎. If 𝑠=𝑒₂ or 𝑒₃, then 𝑎³=(𝛼𝑠)𝑎=𝛼(𝑠𝑎)=0, which is a contradiction. Thus 𝑎²=𝛼𝑒₁ for some 0<𝛼∈𝐹. There are two cases now. (1)There exists 0<𝛽∈𝐹 such that 𝛽²=𝛼. Let 𝑏=𝑎/𝛽. Then {𝑒₁,𝑒₂,𝑒₃,𝑏} is also a 𝑑-basis with the following multiplication table: 𝑒₁𝑏𝑒₂𝑒₃𝑒𝑏₁𝑒₁𝑏00𝑒𝑏𝑒₁00₂𝑒₃00𝑒₂0000𝑒₃(3.1) Thus 𝐴≅𝐹[𝐺]⊕𝐹⊕𝐹, where 𝐺 is a group of order two. This is (2) in Theorem 2.1. (2)The polynomial 𝑓(𝑥)=𝑥²−𝛼∈𝐹[𝑥] has no root in 𝐹. Then we have the following multiplication table for 𝑆: 𝑒₁𝑎𝑒₂𝑒₃𝑒𝑎₁𝑒₁𝑎00𝑒𝑎𝛼𝑒₁00₂𝑒₃00𝑒₂0000𝑒₃(3.2) Thus 𝐴≅𝐹[√𝛼]⊕𝐹⊕𝐹. This is (3) in Theorem 2.1.

3.2. 𝐴 Is Not ℓ-Reduced

In this case, 𝑎 must be nilpotent. If 𝑎²≠0, then 𝑎² must be a positive linear combination of elements in 𝑆. Since each of 𝑒₁,𝑒₂,𝑒₃ is not nilpotent, we must have 𝑎²=𝛽𝑎 for some 0<𝛽∈𝐹. Then 𝑎 is nilpotent implying that 𝑎²=𝛽𝑎=0, which is a contradiction. Thus we must have 𝑎²=0. There are two cases that we need to consider. (1)𝑎𝑒₁=𝑎. Then 𝑎𝑒₂=𝑎𝑒₃=0, and the multiplication table for 𝑆={𝑒₁,𝑒₂,𝑒₃,𝑎} is given below: 𝑒₁𝑎𝑒₂𝑒₃𝑒𝑎₁𝑒₁𝑎00𝑒𝑎000₂𝑒₃00𝑒₂0000𝑒₃(3.3) Thus 𝐴≅(𝐹1+𝐹𝑎)⊕𝐹⊕𝐹, where 𝑎²=0 and 1𝑎=𝑎1=𝑎. This is (4) in Theorem 2.1. (2)𝑎𝑒₁=0. Without loss of generality, we may assume that 𝑎𝑒₂=𝑎 and 𝑎𝑒₃=0. Then we have the following multiplication table for elements in 𝑆: 𝑒₁𝑒₂𝑎𝑒₃𝑒𝑒₁𝑒₁0𝑎0₂𝑎0𝑒₂00𝑒0𝑎00₃000𝑒₃(3.4) Let 𝑇₂(𝐹) be the 2×2 upper triangular matrix algebra over 𝐹 with the entrywise lattice order. Then 𝑇₂(𝐹) becomes an ℓ-algebra over 𝐹 with a 𝑑-basis {𝑒₁₁,𝑒₂₂,𝑒₁₂}. The multiplication table for {𝑒₁₁,𝑒₂₂,𝑒₁₂} is given below: 𝑒₁₁𝑒₂₂𝑒₁₂𝑒₁₁𝑒₁₁0𝑒₁₂𝑒𝑒₂₂₁₂0𝑒₂₂00𝑒₁₂0(3.5) Since {𝑒₁,𝑒₂,𝑎} and {𝑒₁₁,𝑒₂₂,𝑒₁₂} have the same multiplication tables, we have 𝐴≅𝑇₂(𝐹)⊕𝐹. This is (5) in Theorem 2.1.

4. 1 Is a Sum of Two Disjoint Basic Elements

Let 𝐴 be a unital ℓ-algebra with a 𝑑-basis 𝑆={𝑒₁,𝑒₂,𝑎,𝑏} such that 1=𝑒₁+𝑒₂. By Theorem 1.1, 𝑎𝑒_𝑖=𝑎 for some 𝑖=1 or 2. Then 𝑒_𝑖∧𝑏=0 implies that 𝑎∧𝑎𝑏=0. Similarly 𝑏∧𝑎𝑏=0. Thus 𝑎𝑏=𝛼𝑒₁+𝛽𝑒₂ for some 𝛼,𝛽∈𝐹⁺. By a similar argument, 𝑏𝑎=𝛾𝑒₁+𝛿𝑒₂, for some 𝛾,𝛿∈𝐹⁺.

4.1. 𝐴 Is ℓ-Reduced

We may assume that 𝑒₁𝑎=𝑎 and 𝑒₂𝑎=0. Since 𝐴 contains no nonzero positive nilpotent elements, we also have 𝑎𝑒₁=𝑎 and 𝑎𝑒₂=0. Then 𝑎𝑏=𝛼𝑒₁+𝛽𝑒₂ implies that 𝑎𝑏=𝑒₁𝑎𝑏=𝛼𝑒₁. Similarly, 𝑏𝑎=𝛾𝑒₁.

Since 𝐴 is ℓ-reduced, 𝑎 is not a nilpotent element, so 𝑎² is basic by Theorem 1.1. Thus 𝑎²=𝜎𝑠 for some 𝑠∈𝑆 and 0<𝜎∈𝐹. Since 𝑒₁∧𝑎=0 and 𝑒₁𝑎=𝑎, 𝑠≠𝑎. If 𝑠=𝑒₂, then 𝑎³=𝜎(𝑎𝑒₂)=0, which is a contradiction. Thus 𝑠=𝑒₁ or 𝑠=𝑏. We consider these two cases separately. (1)𝑎²=𝜎𝑒₁. Suppose that 𝑎𝑏≠0. Since 𝑎𝑏=𝛼𝑒₁, 𝑎²𝑏=𝛼𝑎, so 𝜎(𝑒₁𝑏)=𝛼𝑎, which is a contradiction. Thus 𝑎𝑏=0, and hence 𝑏𝑎=0 since (𝑏𝑎)²=0. Since 0=𝑏𝑎²=𝜎𝑏𝑒₁, 𝑏𝑒₁=0. Thus 𝑏𝑒₂=𝑏, and hence 𝑒₁𝑏=0 and 𝑒₂𝑏=𝑏. Now 𝑏 is not nilpotent implying that 𝑏² is a basic element, so 𝑏²=𝜆𝑠 for some 𝑠∈𝑆 and 0<𝜆∈𝐹. Since 𝑒₂∧𝑏=0 and 𝑒₂𝑏=𝑏, 𝑠≠𝑏. If 𝑏²=𝜆𝑒₁ or 𝑏²=𝜆𝑎, then 𝜆𝑎=0 or 𝜆𝑎²=0, which are contradictions. Thus we must have 𝑏²=𝜆𝑒₂. Then we have the following multiplication table for the elements of 𝑆: 𝑒₁𝑎𝑒₂𝑏𝑒𝑎₁𝑒₁𝑎00𝑒𝑎𝜎𝑒₁00₂𝑏00𝑒₂𝑏000𝜆𝑒₂(4.1) Depending on √𝜎∈𝐹 or √𝜎∉𝐹 and √𝜆∈𝐹 or √𝜆∉𝐹, we have the following cases: 𝐴≅𝐹[𝐺]⊕𝐹[𝐺],𝐴≅𝐹[𝐺]⊕𝐹√𝜎,or𝐴≅𝐹√𝜎⊕𝐹√𝜆,(4.2) where 𝐺 is a group of order two. These are (6), (7), and (8) in Theorem 2.1. (2)𝑎²=𝜎𝑏. Since 𝑎²=𝜎𝑏 with 0<𝜎∈𝐹, 𝑇={𝑒₁,𝑒₂,𝑎,𝑎²} is also a 𝑑-basis for the ℓ-algebra 𝐴 over 𝐹 and 𝑎³=𝜎(𝑎𝑏)=(𝛼𝜎)𝑒₁. Let 𝜇=𝛼𝜎. Then 𝜇>0 since 𝑎³≠0, so we have the following multiplication table for elements in 𝑇: 𝑒₁𝑎𝑎²𝑒₂𝑒𝑎₁𝑒₁𝑎𝑎²0𝑎₂𝑎𝑎²𝜇𝑒₁0𝑒𝑎²𝜇𝑒₁𝜇𝑎𝑏₂000𝑒₂(4.3) Now it is clear that if √³𝜇∈𝐹, then 𝐴≅𝐹[𝐺]⊕𝐹, where 𝐺 is a group of order three, and if √³𝜇∉𝐹, then 𝐴≅𝐹[√³𝜇]⊕𝐹, where 𝐹[√³𝜇] is the ℓ-field extension of 𝐹 of order three. Those are (9) and (10) in Theorem 2.1.

4.2. 𝐴 Is Not ℓ-Reduced

Since 𝐴 is not ℓ-reduced, 𝑎 or 𝑏 is nilpotent. Let's assume that 𝑒₁𝑎=𝑎, and hence 𝑒₂𝑎=0. Suppose that 𝑎 is nilpotent and 𝑎²≠0. Then 𝑎²=𝛼₁𝑒₁+𝛼₂𝑒₂+𝛼₃𝑎+𝛼₄𝑏,𝛼_𝑖∈𝐹⁺.(4.4) Let 𝑘 be the smallest positive integer such that 𝑎^𝑘=0. Then 𝑘≥3 and 0=𝑎^𝑘=𝑎²𝑎^𝑘−2=𝛼₁𝑒₁𝑎^𝑘−2+𝛼₂𝑒₂𝑎^𝑘−2+𝛼₃𝑎^𝑘−1+𝛼₄𝑏𝑎^𝑘−2,(4.5) so 𝛼₁𝑎^𝑘−2=𝛼₃𝑎^𝑘−1=0. Thus 𝛼₁=𝛼₃=0, and hence 𝑎²=𝛼₂𝑒₂+𝛼₄𝑏. Then 𝑎²=𝑒₁𝑎²=𝛼₄(𝑒₁𝑏), which implies that 𝛼₂=0 and 𝑎²=𝛼₄𝑏. Then 𝑎²≠0 implies that 𝛼₄>0. So {𝑒₁,𝑒₂,𝑎,𝑎²} is a 𝑑-basis of 𝐴 over 𝐹. Similarly if 𝑏 is nilpotent and 𝑏²≠0, then {𝑒₁,𝑒₂,𝑏,𝑏²} is a 𝑑-basis of 𝐴 over 𝐹. Therefore we only need to consider the following cases.(i)𝑎²=𝑏²=0.(ii)𝑎 is nilpotent, but 𝑎²≠0. By the above argument, 𝑎²=𝛼₄𝑏 for some 0<𝛼₄∈𝐹. Thus 𝑇={𝑒₁,𝑒₂,𝑎,𝑎²} is also a 𝑑-basis for 𝐴 over 𝐹.(iii)𝑎²=0 and 𝑏 is not nilpotent.

Below, we consider each case in detail.

4.2.1. 𝑎²=𝑏²=0

From 𝑏𝑎=𝛾𝑒₁+𝛿𝑒₂, we have 0=𝑏𝑎²=𝛾𝑒₁𝑎+𝛿𝑒₂𝑎=𝛾𝑒₁𝑎, so 𝛾=0, and hence 𝑏𝑎=𝛿𝑒₂. (1)𝛿>0. Then 𝑏𝑎=𝛿𝑒₂≠0 implies that 𝑒₂𝑏=𝑏, 𝑒₁𝑏=0, 𝑎𝑒₂=𝑎, and 𝑎𝑒₁=0. Then 𝑎𝑏=𝛼𝑒₁+𝛽𝑒₂ implies that 0=𝑎²𝑏=𝛼𝑎𝑒₁+𝛽𝑎𝑒₂=𝛽𝑎, so 𝛽=0 and 𝑎𝑏=𝛼𝑒₁. Thus 𝑎𝑏𝑎=𝛼𝑒₁𝑎=𝛿𝑎𝑒₂ implies that 𝛼=𝛿. Since 𝑎𝑏=𝛼𝑒₁≠0, 𝑏𝑒₁=𝑏 and 𝑏𝑒₂=0. Let 𝑎^=𝑎/𝛼. Then {𝑒₁,𝑒₂,𝑎′,𝑏} is also a 𝑑-basis of 𝐴 with the following multiplication table: 𝑒₁𝑒₂𝑎^𝑒₂𝑒𝑒₁₂𝑒₁𝑎𝑎^0𝑎0𝑒₂0𝑏_0𝑎^0𝑒₁𝑏𝑏0𝑒₂0(4.6) It is clear now that 𝐴≅𝑀₂(𝐹), the 2×2 matrix ℓ-algebra over 𝐹. This is (11) in Theorem 2.1. (2)𝛿=0. Then 𝑏𝑎=0. From 𝑎𝑏=𝛼𝑒₁+𝛽𝑒₂, we have 0=𝑎(𝑏𝑎)=(𝑎𝑏)𝑎=𝛼𝑒₁𝑎+𝛽𝑒₂𝑎=𝛼𝑎, so 𝛼=0. Thus 𝑎𝑏=𝛽𝑒₂, and hence 0=𝑒₂𝑎𝑏=𝛽𝑒₂, which implies that 𝛽=0, so 𝑎𝑏=0. Therefore, 𝐹𝑎+𝐹𝑏 is a zero ℓ-ring. There are the following cases.(2a)If 𝑎𝑒₁=𝑎, 𝑒₁𝑏=𝑏, and 𝑏𝑒₁=𝑏, then 𝐴≅(𝐹1+𝐹𝑎+𝐹𝑏)⊕𝐹, where 𝐹𝑎+𝐹𝑏 is the zero ℓ-ring.(2b)If 𝑎𝑒₁=𝑎, 𝑒₂𝑏=𝑏, and 𝑏𝑒₂=𝑏, then 𝐴≅(𝐹1+𝐹𝑎)⊕(𝐹1+𝐹𝑏), where 𝑎²=𝑏²=0. If 𝑎𝑒₂=𝑎, then the multiplication table for {𝑒₁,𝑒₂,𝑎} is given below: 𝑒₁𝑒₂𝑎𝑒𝑒₁𝑒₁0𝑎₂𝑎0𝑒₂00𝑎0(4.7) Clearly 𝐹𝑒₁+𝐹𝑒₂+𝐹𝑎 is isomorphic to 𝑇₂(𝐹), and hence there are the following cases depending on the products between 𝑒₁, 𝑒₂, and 𝑏.(2c)If 𝑒₁𝑏=𝑏 and 𝑏𝑒₁=𝑏, then 𝐴≅𝑇₂(𝐹)+𝐹𝑏 with 𝑒₁₁𝑏=𝑏𝑒₁₁=𝑏, 𝑏𝑒₁₂=0, and 𝑏²=0.(2d)If 𝑒₁𝑏=𝑏 and 𝑏𝑒₂=𝑏, then 𝐴≅𝑇₂(𝐹)+𝐹𝑏 with 𝑒₁₁𝑏=𝑏, 𝑏𝑒₂₂=𝑏, and 𝑏²=0.(2e)If 𝑒₂𝑏=𝑏 and 𝑏𝑒₁=𝑏, then 𝐴≅𝑇₂(𝐹)+𝐹𝑏 with 𝑒₂₂𝑏=𝑏, 𝑏𝑒₁₁=𝑏, 𝑒₁₂𝑏=𝑏𝑒₁₂=0, and 𝑏²=0.(2f)If 𝑒₂𝑏=𝑏 and 𝑏𝑒₂=𝑏, then 𝐴≅𝑇₂(𝐹)+𝐹𝑏 with 𝑒₂₂𝑏=𝑏𝑒₂₂=𝑏, 𝑒₁₂𝑏=0, and 𝑏²=0.

Two more cases we need to consider are (i) 𝑎𝑒₁=𝑎, 𝑒₁𝑏=𝑏, 𝑏𝑒₁=0 and (ii) 𝑎𝑒₁=𝑎, 𝑒₂𝑏=𝑏, 𝑏𝑒₂=0. But when we switch 𝑎 and 𝑏, the case (i) becomes (2c), and when we switch 𝑎 and 𝑏, and also switch 𝑒₁ and 𝑒₂, the case (ii) becomes (2f). Thus every possible case when 𝛿=0 is covered. The above discussion has produced (12)–(17) in Theorem 2.1.

4.2.2. 𝑎 Is Nilpotent and 𝑎²≠0

As we have noticed before, in this case 𝑇={𝑒₁,𝑒₂,𝑎,𝑎²} is a 𝑑-basis for the ℓ-algebra 𝐴 over 𝐹. Since 𝑎∧𝑒₁=0 implies that 𝑎³∧𝑎²=0 and 𝑎²∧𝑒₁=0 implies that 𝑎³∧𝑎=0, 𝑎³=𝛽₁𝑒₁, for some 𝛽₁∈𝐹⁺. If 𝛽₁≠0, then 𝑎 cannot be nilpotent, so we must have 𝛽₁=0. Therefore 𝑎³=0, and 𝐴≅(𝐹1+𝐹𝑎+𝐹𝑎²)⊕𝐹 with 𝑎³=0. This is (18) in Theorem 2.1.

4.2.3. 𝑎²=0 and 𝑏 Is Not Nilpotent

Since 𝑏 is not nilpotent, 𝑏² is basic, so 𝑏²=𝜖𝑠 for some 𝑠∈𝑆 and 0<𝜖∈𝐹. Clearly 𝑠≠𝑎 or 𝑏. Thus either 𝑏²=𝜖𝑒₁ or 𝑏²=𝜖𝑒₂. In the first case, we have 𝑏²𝑎=𝜖𝑒₁𝑎=𝑏(𝛾𝑒₁+𝛿𝑒₂)=𝛾𝑏𝑒₁, so 𝜖𝑎=𝛾𝑏, which is a contradiction. Thus we must have 𝑏²=𝜖𝑒₂. If 𝑎𝑒₂=𝑎, then 𝑎𝑏²=𝜖𝑎, so (𝛼𝑒₁+𝛽𝑒₂)𝑏=𝜖𝑎, which implies that 𝛽𝑏=𝜖𝑎, a contradiction. Thus 𝑎𝑒₁=𝑎, and hence 𝑎𝑒₂=0. Then 𝑎𝑏=𝑏𝑎=0. Therefore we have the following multiplication table for the elements of 𝑆: 𝑒₁𝑎𝑒₂𝑏𝑒𝑎₁𝑒₁𝑎00𝑒𝑎000₂𝑏00𝑒₂𝑏00𝑏𝜖𝑒₂(4.8)

From the above table, it is clear that if √𝜖∈𝐹, then 𝐴≅𝐹[𝐺]⊕(𝐹1+𝐹𝑎) with 𝑎²=0, where 𝐺 is a group of order two, and if √𝜖∉𝐹, then 𝐴≅𝐹[√𝜖]⊕(𝐹1+𝐹𝑎) with 𝑎²=0, where 𝐹[√𝜖] is a quadratic extension ℓ-field of 𝐹. Those are (19) and (20) in Theorem 2.1.

Now we have constructed all lattice orders on 𝐴 such that 1 is a sum of two disjoint basic elements.

5. 1 Is a Basic Element

In this section, 𝐴 has a 𝑑-basis 𝑆={1,𝑎,𝑏,𝑐}. We still first consider the ℓ-reduced case.

5.1. 𝐴 Is ℓ-Reduced

Since 𝑎 is not nilpotent, by Theorem 1.1, there exist positive integer 𝑛_𝑎 and 0<𝛼_𝑎∈𝐹 such that 𝑎^𝑛𝑎=𝛼_𝑎1. We also assume that 𝑛_𝑎 is the smallest positive integer such that 𝑎^𝑛𝑎=𝛼_𝑎1. Similarly 𝑏^𝑛𝑏=𝛼_𝑏1 and 𝑐^𝑛𝑐=𝛼_𝑐1, for some positive integers 𝑛_𝑏, 𝑛_𝑐, and 0<𝛼_𝑏,𝛼_𝑐∈𝐹. By Theorem 1.1 again, each 𝑎^𝑘 is a basic element for 𝑘≥1.

Lemma 5.1. The subset {1,𝑎,𝑎²,…,𝑎^𝑛𝑎−1} of 𝐴 is a disjoint set of 𝐴.

Proof. Let 1≤𝑖<𝑛_𝑎. Since 𝑎^𝑖 is a basic element, 𝑎^𝑖=𝛼𝑠 for some 𝑠∈𝑆 and 0<𝛼∈𝐹. Since 𝑖<𝑛_𝑎, 𝑠≠1, so 1∧𝑠=0 implies that 1∧𝑎^𝑖=0. Now let 0≤𝑢<𝑣<𝑛_𝑎, then 𝑎^𝑢∧𝑎^𝑣=𝑎^𝑢(1∧𝑎^𝑣−𝑢)=0 by the above argument.

Since 𝐴 is four-dimensional, 𝑛_𝑎≤4 by Lemma 5.1. We claim that 𝑛_𝑎≠3. Supposing that 𝑛_𝑎=3, then we may assume that {1,𝑎,𝑎²,𝑏} is a 𝑑-basis. By Theorem 1.1, 𝑎𝑏 is a basic element and 𝑎𝑏∧𝑎=𝑎(𝑏∧1)=0, 𝑎𝑏∧𝑏=(𝑎∧1)𝑏=0, 𝑎𝑏∧𝑎²=𝑎(𝑏∧𝑎)=0, so 𝑎𝑏=𝛼1 for some 0<𝛼∈𝐹. Thus 𝑎³𝑏=𝛼𝑎², and hence 𝛼_𝑎𝑏=𝛼𝑎², which is a contradiction. Hence 𝑛_𝑎≠3, so 𝑛_𝑎=4 or 𝑛_𝑎=2.

By the above analysis, we need to consider two cases: (i) one of the following: 𝑛_𝑎, 𝑛_𝑏, 𝑛_𝑐, is 4, and we may assume that 𝑛_𝑎=4; (ii) 𝑛_𝑎=𝑛_𝑏=𝑛_𝑐=2.

5.1.1. 𝑛_𝑎=4 and Hence {1,𝑎,𝑎²,𝑎³} Is a Disjoint Subset of Basic Elements of 𝐴

In this case 𝑇={1,𝑎,𝑎²,𝑎³} is a 𝑑-basis of 𝐴 over 𝐹, and we have the following multiplication table for the elements in 𝑇: 1𝑎𝑎²𝑎³11𝑎𝑎²𝑎³𝑎𝑎𝑎²𝑎³𝛼_𝑎𝑎𝑎²₃𝑎²𝑎³𝛼_𝑎1𝛼_𝑎𝑎𝑎³𝛼_𝑎1𝛼_𝑎𝑎𝛼_𝑎𝑎²(5.1)

We have the following three cases to be considered.(1)If √⁴𝛼_𝑎∈𝐹, then 𝐴≅𝐹[𝐺], where 𝐺 is a cyclic group of order four.(2)If the polynomial 𝑓(𝑥)=𝑥⁴−𝛼_𝑎∈𝐹[𝑥] is irreducible, then 𝐴≅𝐹[√⁴𝛼_𝑎], which is an ℓ-field extension of 𝐹 of order four.(3)The polynomial 𝑓(𝑥)=𝑥⁴−𝛼_𝑎∈𝐹[𝐺] is not irreducible and √⁴𝛼_𝑎∉𝐹.

Suppose that 𝑥⁴−𝛼_𝑎=(𝑥²+𝑎₁𝑥+𝑏₁)(𝑥²+𝑎₂𝑥+𝑏₂). Then we have 𝑎₁+𝑎₂=0, 𝑏₁+𝑏₂+𝑎₁𝑎₂=0, 𝑏₁𝑎₂+𝑏₂𝑎₁=0, and 𝑏₁𝑏₂=−𝛼_𝑎<0. Thus 𝑎₂=−𝑎₁ and 0=𝑏₁𝑎₂+𝑏₂𝑎₁=𝑎₁(−𝑏₁+𝑏₂). If 𝑎₁≠0, then −𝑏₁+𝑏₂=0, and hence 𝑏₁=𝑏₂, which implies that 𝑏₁𝑏₂=𝑏²¹=−𝛼_𝑎<0, a contradiction. So we must have 𝑎₁=0, and hence 𝑎₂=0. Then 𝑏₁+𝑏₂=0 and 𝑏₁𝑏₂=−𝛼_𝑎 implies that 𝑏₂=−𝑏₁ and 𝑏²¹=𝑏²²=𝛼_𝑎. We may assume that 𝑏₁>0. It follows from √⁴𝛼_𝑎∉𝐹 that √𝑏₁∉𝐹. Let 𝑐=𝑎²/𝑏₁. Then 𝑉={1,𝑐,𝑎,𝑎𝑐} is also a 𝑑-basis of 𝐴 over 𝐹 and the multiplication table for the elements in 𝑉 is given below: 1𝑐𝑎𝑎𝑐1𝑐1𝑐𝑎𝑎𝑐𝑎𝑐1𝑎𝑐𝑎𝑎𝑐𝑎𝑎𝑐𝑏₁𝑐𝑏₁1𝑎𝑐𝑎𝑏₁1𝑏₁𝑐(5.2)

Then 𝐴≅𝐹[𝐺]+𝐹[𝐺]𝑎, where 𝐺={𝑒,𝑐} is a group of order two and 𝑎²=𝛼𝑐, 0<𝛼∈𝐹, and √𝛼∉𝐹. These are (21), (22), and (23) in Theorem 2.1.

5.1.2. 𝑛_𝑎=𝑛_𝑏=𝑛_𝑐=2

Then 𝑆={1,𝑎,𝑏,𝑐} is a 𝑑-basis and 𝑎²=𝛼_𝑎1, 𝑏²=𝛼_𝑏1, and 𝑐²=𝛼_𝑐1. Since 𝑎𝑏∧𝑎=𝑎𝑏∧𝑏=0 and 𝑎𝑏 is basic, 𝑎𝑏=𝛼1 or 𝑎𝑏=𝛼𝑐 for some 0<𝛼∈𝐹. In the first case, 𝑎𝑏²=𝛼𝑏 implies that 𝛼_𝑏𝑎=𝛼𝑏, which is a contradiction. Thus 𝑎𝑏=𝛼𝑐 for some 0<𝛼∈𝐹. Then we may replace 𝑐 by 𝑎𝑏 to get a 𝑑-basis 𝑇={1,𝑎,𝑏,𝑎𝑏}.

Since 𝑏𝑎∧𝑎=𝑏𝑎∧𝑏=0 and 𝑏𝑎 is basic, 𝑏𝑎=𝛽1 or 𝑏𝑎=𝛽(𝑎𝑏) for some 0<𝛽∈𝐹. If 𝑏𝑎=𝛽1, then 𝛼_𝑎𝑏=𝛽𝑎, which is a contradiction. Thus 𝑏𝑎=𝛽(𝑎𝑏), for some 0<𝛽∈𝐹, and hence 𝛼_𝑏𝛼_𝑎1=𝑏(𝑏𝑎)𝑎=𝛽𝑏(𝑎𝑏)𝑎=𝛽²(𝑎𝑏)(𝑏𝑎)=𝛽²𝛼_𝑏𝛼_𝑎1.(5.3) So 𝛽=1 and the multiplication table for the elements of 𝑇 is given below: 1𝑎𝑏𝑎𝑏𝑎11𝑎𝑏𝑎𝑏𝑏𝑎𝛼_𝑎1𝑎𝑏𝛼_𝑎𝑏𝑎𝑏𝑏𝑎𝑏𝛼_𝑏1𝛼_𝑏𝑎𝑎𝑏𝛼_𝑎𝑏𝛼_𝑏𝑎𝛼_𝑎𝛼_𝑏1(5.4)

There are the following cases. (1)√𝛼_𝑎 and √𝛼_𝑏∈𝐹. Replacing 𝑎 and 𝑏 by 𝑎/√𝛼_𝑎 and 𝑏/√𝛼_𝑏, respectively, we have 𝑎²=𝑏²=1. Therefore 𝐴≅𝐹[𝐺], where 𝐺 is the Klein four group. (2)√𝛼_𝑎∉𝐹 and √𝛼_𝑏∈𝐹 (or √𝛼_𝑎∈𝐹 and √𝛼_𝑏∉𝐹). Replacing 𝑏 by 𝑏/√𝛼_𝑏, we may assume that 𝑏²=1. Therefore 𝐴≅𝐹[√𝛼_𝑎][𝐺], where 𝐹[√𝛼_𝑎] is the quadratic extension ℓ-field of 𝐹 and 𝐺 is a group of order two.(3)√𝛼_𝑎∉𝐹 and √𝛼_𝑏∉𝐹. If √𝛼_𝑏∈𝐹[√𝛼_𝑎], then 𝐴≅𝐹[√𝛼_𝑎][𝐺], where 𝐺 is a group of order two. We omit the checking and leave it to the reader. If √𝛼_𝑏∉𝐹[√𝛼_𝑎], then 𝐴≅𝐹[√𝛼_𝑎][√𝛼_𝑏], where 𝐹[√𝛼_𝑎] is a quadratic extension ℓ-field of 𝐹 and 𝐹[√𝛼_𝑎][√𝛼_𝑏] is a quadratic extension ℓ-field of 𝐹[√𝛼_𝑎]. Those are (24), (25), and (26) in Theorem 2.1.

5.2. 𝐴 Is Not ℓ-Reduced

Let 𝑆={1,𝑎,𝑏,𝑐} be a 𝑑-basis of 𝐴 over 𝐹. Then some elements of 𝑎,𝑏,𝑐 must be nilpotent. We first observe the following result.

Lemma 5.2. Let 𝑥,𝑦∈{1,𝑎,𝑏,𝑐}. If either 𝑥 or 𝑦 is nilpotent and the other one is not nilpotent, then 𝑥𝑦 is nilpotent.

Proof. Let 𝑥 be nilpotent and let 𝑦 be not. By Theorem 1.1, there exist a positive integer 𝑛_𝑦 and 0<𝛼_𝑦∈𝐹 such that 𝑦^𝑛𝑦=𝛼_𝑦1. If 𝑥𝑦=0, then 𝑥𝑦^𝑛𝑦=0 implies that 𝛼_𝑦𝑥=0, which is a contradiction. Thus 𝑥𝑦≠0, so 𝑥𝑦 is basic by Theorem 1.1. If 𝑥𝑦 is not nilpotent, then 𝑥𝑦=𝛾𝑧 for some 0<𝛾∈𝐹 and 𝑧∈{1,𝑎,𝑏,𝑐} and 𝑧 is not nilpotent. By Theorem 1.1, there exist a positive integer 𝑛_𝑧, 0<𝛼_𝑧∈𝐹 such that 𝑧^𝑛𝑧=𝛼_𝑧1. Let 𝑛>1 be the smallest positive integer such that 𝑥^𝑛=0. Then 0=𝑥^𝑛𝑦=𝑥^𝑛−1(𝑥𝑦)=𝛾𝑥^𝑛−1𝑧, and hence 0=𝛾𝑥^𝑛−1𝑧 implies that 0=𝛾𝑥^𝑛−1𝑧^𝑛𝑧=𝛼_𝑧𝛾𝑥^𝑛−1. Thus 𝑥^𝑛−1=0, which is a contradiction. Therefore 𝑥𝑦 is nilpotent.

By Lemma 5.2, {𝑎,𝑏,𝑐} cannot have exactly one nilpotent element. Suppose not, let 𝑎 be the unique nilpotent element. Since 𝑎𝑏∧𝑎=0, 𝑎𝑏 cannot be a positive scalar multiple of 𝑎, so 𝑎𝑏 cannot be a nilpotent element, which is a contradiction by Lemma 5.2. So there are following two cases.

5.2.1. 𝑎, 𝑏, and 𝑐 Are All Nilpotent

(1)Either 𝑎, 𝑏, or 𝑐 has nonzero square. We may assume that 𝑎²≠0. Since 𝑎²∧𝑎=𝑎(𝑎∧1)=0, 𝑎²=𝛼1+𝛽𝑏+𝛾𝑐, where 𝛼,𝛽,𝛾∈𝐹⁺. Let 𝑘 be the smallest positive integer such that 𝑎^𝑘=0. Then 0=𝑎^𝑘=𝑎^𝑘−2𝑎²=𝛼𝑎^𝑘−2+𝛽𝑎^𝑘−2𝑏+𝛾𝑎^𝑘−2𝑐 implies 𝛼𝑎^𝑘−2=0, so 𝛼=0. Thus 𝑎²=𝛽𝑏+𝛾𝑐.(1a)𝛽>0 and 𝛾>0. Since 𝑎²∧𝑎𝑏=𝑎(𝑎∧𝑏)=0, 𝑏∧𝑎𝑏=𝑐∧𝑎𝑏=0. Also 𝑎∧𝑎𝑏=𝑎(1∧𝑏)=0. Thus 𝑎𝑏=𝛿1. Then we have 0=𝑎^𝑘𝑏=𝛿𝑎^𝑘−1, so 𝛿=0 and 𝑎𝑏=0. Similarly 𝑏𝑎=0, 𝑎𝑐=0, and 𝑐𝑎=0. Since 𝑎²=𝛽𝑏+𝛾𝑐, 0=𝑎²𝑏=𝛽𝑏²+𝛾𝑐𝑏 implies that 𝑏²=𝑐𝑏=0. Similarly 𝑐²=𝑏𝑐=0. Replacing 𝑏 and 𝑐 by 𝛽𝑏 and 𝛾𝑐, respectively, we may assume that 𝑎²=𝑏+𝑐. Therefore we have the following multiplication table: 1𝑎𝑏𝑐𝑎11𝑎𝑏𝑐𝑏𝑎𝑏+𝑐00𝑐𝑏000𝑐000(5.5) Then 𝐴≅𝐹1+𝐹𝑎+𝐹𝑏+𝐹𝑐, where 𝑎²=𝑏+𝑐 and the product of any other two elements of 𝑎, 𝑏, 𝑐 is zero. This is (27) in Theorem 2.1. We want to point out that this case provides an example in which the product of two basic elements may not be basic. (1b)𝛽=0 or 𝛾=0. We may assume that 𝛾=0, so 𝛽>0. Then 𝑎²=𝛽𝑏, and 𝑎(𝑎𝑎)=(𝑎𝑎)𝑎 implies that 𝛽(𝑎𝑏)=𝛽(𝑏𝑎), so 𝑎𝑏=𝑏𝑎.(i)𝑎𝑏=𝑏𝑎≠0. Then 𝑎𝑏=𝛿𝑐 for some 0<𝛿∈𝐹. Thus 𝑎³=𝑎𝑎²=𝛽(𝑎𝑏)=𝛽𝛿𝑐, and hence we may replace 𝑏 by 𝑎² and 𝑐 by 𝑎³ to get another 𝑑-basis 𝑇={1,𝑎,𝑎²,𝑎³}. Therefore 𝐴≅𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑎³ with 𝑎⁴=0. This is (28) in Theorem 2.1.(ii)𝑎𝑏=𝑏𝑎=0. Then 𝛽𝑏²=𝑎²𝑏=0, so 𝑏²=0. Since 𝑎𝑐=𝜇𝑏 for some 0≤𝜇∈𝐹, 𝑎²∧𝑎𝑐=0 and 𝑎²=𝛽𝑏 implies that 𝜇=0, so 𝑎𝑐=0. Similarly 𝑐𝑎=0. 0=𝑎²𝑐=𝛽(𝑏𝑐), so 𝑏𝑐=0. Similarly, 𝑐𝑏=0. Now replacing 𝑏 by 𝑎², we obtain another 𝑑-basis 𝑇={1,𝑎,𝑎²,𝑐}.If 𝑐²≠0, then 𝑐²∧𝑐=𝑐²∧1=0 implies that 𝑐²=𝛼₁𝑎+𝛼₂𝑎². Then it follows from 𝑎𝑐=0 and 𝑎³=0 that 𝛼₁𝑎²=0, so 𝛼₁=0. Thus 𝑐²=𝛼₂𝑎² with 0<𝛼₂∈𝐹. Hence 𝐴≅𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑐 with 𝑐²=𝛼𝑎² for some 0<𝛼∈𝐹 and 𝑥𝑦=0 for any two elements 𝑥 and 𝑦 in {𝑎,𝑎²,𝑐} not both 𝑎 or 𝑐. This is (29) in Theorem 2.1. If 𝑐²=0, then 𝐴≅𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑐 with 𝑥𝑦=0 for any two elements 𝑥 and 𝑦 in {𝑎,𝑎²,𝑐} which are not both 𝑎. This is (30) in Theorem 2.1.We provide some familiar examples for the above two cases. Consider 4×4 matrix algebra over 𝐹. Let 1=𝑒₁₁+𝑒₂₂+𝑒₃₃+𝑒₄₄, 𝑎=𝑒₁₂+𝑒₂₃, and 𝑐=𝑒₁₄+2𝑒₄₃. Then 𝑎²=𝑒₁₃ and 𝑐²=2𝑎². Clearly 𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑐 is a four-dimensional ℓ-algebra over 𝐹 with coordinatewise order. If 𝑐=𝑒₁₃+𝑒₄₃, then 𝑐²=0 and 𝐹1+𝐹𝑎+𝐹𝑎²+𝐹𝑐 is a four-dimensional ℓ-algebra over 𝐹 with 𝑐²=0. (2)𝑎²=𝑏²=𝑐²=0. (2a)𝑥𝑦=0 for any two elements in {𝑎,𝑏,𝑐}.Then 𝐴≅𝐹1+𝐹𝑎+𝐹𝑏+𝐹𝑐, where 𝐹𝑎+𝐹𝑏+𝐹𝑐 is zero ring. This is (31) in Theorem 2.1. (2b)At least one product of two different elements in {𝑎,𝑏,𝑐} is nonzero. Without loss of generality, we may assume that 𝑎𝑏≠0. Then 𝑎𝑏=𝛼𝑐 for some 0<𝛼∈𝐹, and we may replace 𝑐 by 𝑎𝑏 to form another 𝑑-basis 𝑇={1,𝑎,𝑏,𝑎𝑏}.If 𝑏𝑎=0, then we have the following multiplication table: 1𝑎𝑏𝑎𝑏𝑎11𝑎𝑏𝑎𝑏𝑏𝑎0𝑎𝑏0𝑎𝑏𝑏000𝑎𝑏000(5.6) Thus 𝐴≅𝐹1+𝐹𝑎+𝐹𝑏+𝐹(𝑎𝑏) with 𝑎²=𝑏²=𝑏𝑎=0. This is (32) in Theorem 2.1. If 𝑏𝑎≠0, then 𝑏𝑎=𝛽(𝑎𝑏) for some 0<𝛽∈𝐹 and we have the following multiplication table: 1𝑎𝑏𝑎𝑏𝑎11𝑎𝑏𝑎𝑏𝑏𝑎0𝑎𝑏0𝑎𝑏𝑏𝛽(𝑎𝑏)00𝑎𝑏000(5.7) Thus 𝐴≅𝐹1+𝐹𝑎+𝐹𝑏+𝐹(𝑎𝑏) with 𝑎²=𝑏²=0 and 𝑏𝑎=𝛽(𝑎𝑏) for some 0<𝛽∈𝐹. This is (33) in Theorem 2.1.

5.2.2. Two Elements in {𝑎,𝑏,𝑐} Are Nilpotent

We may assume that 𝑎 and 𝑏 are nilpotent elements. Then we must have 𝑐²=𝛼1 for some 0<𝛼∈𝐹. By Lemma 5.2, 𝑎𝑐=𝛽𝑏 for some 0<𝛽∈𝐹. If 𝑎²≠0, then 𝑎²∧𝑎=𝑎(𝑎∧1)=0 and that 1 and 𝑐 are not nilpotent implies that 𝑎²=𝛾𝑏 for some 0<𝛾∈𝐹. Thus 𝑎𝑐∧𝑎²≠0, which is a contradiction, so 𝑎²=0. Similarly, 𝑏²=0. Replacing 𝑎 by 𝑎/𝛽 and then replacing 𝑏 by 𝑎𝑐, we have another 𝑑-basis {1,𝑐,𝑎,𝑎𝑐} with the following multiplication table: 1𝑐𝑎𝑎𝑐1𝑐1𝑐𝑎𝑎𝑐𝑎𝑐𝛼1𝑎𝑐𝛼𝑎𝑎𝑐𝑎𝑎𝑐00𝑎𝑐𝛼𝑎00(5.8)

If √𝛼∈𝐹, then replacing 𝑐 by 𝑐/√𝛼, we have that 𝐴≅𝐹[𝐺]+𝐹[𝐺]𝑎 is a commutative ℓ-algebra, where 𝐺 is a group of order two and 𝑎²=0.

If √𝛼∉𝐹, then 𝐴≅𝐹[√𝛼]+𝐹[√𝛼]𝑎 is a commutative ℓ-algebra over 𝐹, where 𝐹[√𝛼] is a quadratic extension ℓ-field of 𝐹 and 𝑎²=0. Those are (34) and (35) in Theorem 2.1.

This completes the proof of Theorem 2.1.

6. An Application

For an ℓ-ring 𝑅, its ℓ-radical ℓ-𝑁(𝑅) is defined as ℓ-𝑁(𝑅)=𝑎∈𝑅∶∃𝑛=𝑛(𝑎)suchthat𝑥₀|𝑎|𝑥₁|𝑎|⋯|𝑎|𝑥_𝑛=0,∀𝑥₀,…,𝑥_𝑛∈𝑅,(6.1) (see [1, Definition, page 45]). ℓ-𝑁(𝑅) is the sum of all of the nilpotent ℓ-ideals of 𝑅 [1, Theorem  5]. It is clear that for an ℓ-algebra 𝐴, ℓ-𝑁(𝐴) is an ℓ-ideal of 𝐴.

A positive derivation on an ℓ-ring 𝑅 is a function 𝛿∶𝑅→𝑅 such that for any 𝑎,𝑏∈𝑅, 𝛿(𝑎+𝑏)=𝛿(𝑎)+𝛿(𝑏), 𝛿(𝑎𝑏)=𝑎𝛿(𝑏)+𝛿(𝑎)𝑏, and 𝛿(𝑅⁺)⊆𝑅⁺. For an Archimedean 𝑓-ring 𝑅, if 𝛿 is a positive derivation on 𝑅, then 𝛿(𝑅)⊆ℓ-𝑁(𝑅) [2, 3]. Although we believe this result is true for a unital ℓ-algebra with a 𝑑-basis, we lack ability to prove it in general. In the following, we prove this result for unital four-dimensional ℓ-algebras with a 𝑑-basis based on the classifications obtained in Theorem 2.1.

Theorem 6.1. Let 𝐴 be a unital four-dimensional ℓ-algebra with a 𝑑-basis. If 𝛿 is a positive derivation on 𝐴, then 𝛿(𝐴)⊆𝑙-𝑁(𝐴).

Proof. We notice first that since 1>0, for any 𝑎∈𝐴 and 𝛼∈𝐹, 𝛿(𝛼𝑎)=𝛼𝛿(𝑎) [4]; that is, 𝛿 is actually a linear transformation of 𝐴. Let 1=𝑒₁+⋯+𝑒_𝑛, where 𝑒₁,…,𝑒_𝑛 are distinct elements from the 𝑑-basis and 𝑛≤4. Since 𝛿(1)=𝛿(1)+𝛿(1) implies 𝛿(1)=0, 0=𝛿(𝑒₁)+⋯+𝛿(𝑒_𝑛). Then 𝛿 is positive implying that 𝛿(𝑒_𝑖)=0 for each 𝑖=1,.…,𝑛. Let 𝑎 be an element in the 𝑑-basis, by Theorem 1.1, either 𝑎 is nilpotent or 𝑎^𝑛𝑎=𝛼_𝑎𝑒_𝑖 for some 𝑛_𝑎≥1, 0<𝛼_𝑎∈𝐹, and 𝑒_𝑖. If 𝑎 is nilpotent, then 𝛿(𝑎) is also nilpotent [4]. If 𝑎^𝑛𝑎=𝛼_𝑎𝑒_𝑖, then 𝛿(𝑎^𝑛𝑎)=𝛿(𝛼_𝑎𝑒_𝑖)=𝛼_𝑎𝛿(𝑒_𝑖)=0 and 𝛿 is positive also implying that 𝛿(𝑎) is nilpotent. Thus for each element 𝑎 in the 𝑑-basis, 𝛿(𝑎) is nilpotent, so if 𝐴 is ℓ-reduced, then 𝛿(𝑎)=0 for each 𝑎 in the 𝑑-basis. Therefore 𝛿(𝐴)=0.
We only need to consider those cases in Theorem 2.1 in which 𝐴 is not ℓ-reduced. In cases (4)-(5), (12)–(20), and (27)–(35), the ℓ-radical consists of all nilpotent elements of the ℓ-algebra 𝐴. From the argument in the previous paragraph, for each element 𝑥 in the 𝑑-basis, 𝛿(𝑥) is nilpotent, so 𝛿(𝑥)∈ℓ-𝑁(𝐴). Thus 𝛿(𝐴)⊆ℓ-𝑁(𝐴) for those cases.
The only case left is (11) in which ℓ-𝑁(𝑀₂(𝐹))=0, but 𝑀₂(𝐹) contains nonzero nilpotent elements 𝑒₁₂ and 𝑒₂₁. We show that 𝛿(𝑒₁₂)=𝛿(𝑒₂₁)=0. Then together with 𝛿(𝑒₁₁)=𝛿(𝑒₂₂)=0, we have 𝛿(𝑀₂(𝐹))=0. Since (𝑒₁₂+𝑒₂₁)²=1, 𝑒₁₂+𝑒₂₁𝛿𝑒₁₂+𝑒₂₁+𝛿𝑒₁₂+𝑒₂₁𝑒₁₂+𝑒₂₁=0,(6.2) so 𝛿 is positive implying that (𝑒₁₂+𝑒₂₁)𝛿(𝑒₁₂+𝑒₂₁)=0. Multiplying this last equation by (𝑒₁₂+𝑒₂₁) from the left, we have 𝛿(𝑒₁₂+𝑒₂₁)=0. Hence 𝛿(𝑒₁₂)=𝛿(𝑒₂₁)=0. Therefore 𝛿(𝑀₂(𝐹))=0.
This completes the proof.