International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 650969 | 15 pages | https://doi.org/10.5402/2011/650969

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

Academic Editor: T. Nakatsu
Received05 May 2011
Accepted02 Jun 2011
Published29 Jun 2011

Abstract

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

1. Introduction

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

All unital two-dimensional and three-dimensional β„“-algebras with a 𝑑-basis have been constructed [6]. In this paper, we construct all unital four-dimensional β„“-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 (β„“-rings) and undefined terminologies in this paper. Throughout 𝐹 always denotes a totally ordered field.

Let 𝑅 be an β„“-ring. The positive cone of 𝑅 is 𝑅+={π‘Ÿβˆˆπ‘…βˆΆπ‘Ÿβ‰₯0}. A vector lattice 𝑉 over 𝐹 is an β„“-group and a vector space over 𝐹 such that 𝐹+𝑉+βŠ†π‘‰+. An β„“-algebra 𝐴 over 𝐹 is an algebra over 𝐹, that is, an β„“-ring and a vector lattice over 𝐹. An β„“-algebra is called unital if it has an identity element 1 and called β„“-unital if 1β‰₯0. An element π‘Ž in an β„“-ring is called basic if π‘Ž>0 and, for any 𝑏,𝑐β‰₯0, π‘Žβ‰₯𝑏,𝑐 implies that 𝑏 and 𝑐 are comparable, that is, either 𝑏β‰₯𝑐 or 𝑐β‰₯𝑏. A set of elements {π‘₯π‘–βˆΆπ‘–βˆˆπΌ} is called disjoint if each π‘₯𝑖>0 and π‘₯π‘–βˆ§π‘₯𝑗=0 for any two different elements π‘₯𝑖 and π‘₯𝑗. An element π‘₯β‰₯0 in an β„“-ring 𝑅 is called an 𝑓-element if π‘¦βˆ§π‘§=0 implies that π‘₯π‘¦βˆ§π‘§=𝑦π‘₯βˆ§π‘§=0, and π‘₯ is called a 𝑑-element if π‘¦βˆ§π‘§=0 implies that π‘₯π‘¦βˆ§π‘₯𝑧=𝑦π‘₯βˆ§π‘§π‘₯=0. An β„“-ring 𝑅 is called β„“-reduced if for any π‘Žβˆˆπ‘…+, π‘Ž2=0 implies that π‘Ž=0.

Let 𝐴 be an β„“-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 π‘Žβˆˆπ΄, π‘Ž=𝛼1𝑠1+β‹―+𝛼𝑝𝑠𝑝, where 𝑝β‰₯1, 𝛼1,…,π›Όπ‘βˆˆπΉ, and 𝑠1,…,π‘ π‘βˆˆπ‘† are distinct, and π‘Žβ‰₯0 if and only if each 𝛼𝑖β‰₯0, 𝑖=1,…,𝑝. Thus in a 𝑑-basis, each element is basic. Let 𝐴 be a unital finite-dimensional β„“-algebra with a 𝑑-basis. Then 𝐴 is β„“-unital [1], and there exists a 𝑑-basis 𝑆 such that 1=𝑒1+β‹―+𝑒𝑛, where 𝑒1,…,π‘’π‘›βˆˆπ‘† are distinct.

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

Theorem 1.1. Let 𝐴 be a unital finite-dimensional β„“-algebra over 𝐹 with a 𝑑-basis 𝑆 such that 1=𝑒1+β‹―+𝑒𝑛, where 𝑛β‰₯1 and 𝑒1,…,π‘’π‘›βˆˆπ‘† are distinct. (1)Each 𝑒𝑖 is an idempotent 𝑓-element and 𝑒𝑖𝑒𝑗=0 for 𝑖≠𝑗. (2)Let π‘Ž be a basic element. Then there exists exactly one 𝑒𝑖 such that π‘’π‘–π‘Ž=π‘Ž, and for any 𝑗≠𝑖, π‘’π‘—π‘Ž=0. Similarly, there exists exactly one π‘’π‘˜ such that π‘Žπ‘’π‘˜=π‘Ž, and for any π‘—β‰ π‘˜, π‘Žπ‘’π‘—=0. (3)For each element π‘Žβˆˆπ‘†, if π‘’π‘–π‘Ž=π‘Ž or π‘Žπ‘’π‘–=π‘Ž, 1≀𝑖≀𝑛, then either π‘Ž is nilpotent or π‘Žπ‘›π‘Ž=π›Όπ‘Žπ‘’π‘– for some positive integer π‘›π‘Ž and 0<π›Όπ‘ŽβˆˆπΉ. (4)Let π‘Ž,π‘βˆˆπ‘†. If either π‘Ž or 𝑏 is not nilpotent and π‘Žπ‘β‰ 0, then π‘Žπ‘ is basic.

2. The Classification of Unital Four-Dimensional β„“-Algebras with a 𝑑-Basis

We first list some β„“-algebras that we are going to use for the classification of unital four-dimensional β„“-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 βˆ‘π‘–π›Όπ‘–π‘”π‘–β‰₯0 if each 𝛼𝑖β‰₯0. Then 𝐹[𝐺] is an β„“-algebra over 𝐹 and 𝐺 is a 𝑑-basis. In the following, 𝐹[𝐺] always denotes the β„“-algebra with the lattice order defined above.(ii)Let 0<π›ΌβˆˆπΉ, and let 𝑓(π‘₯)=π‘₯π‘›βˆ’π›ΌβˆˆπΉ[π‘₯] be irreducible. We use π‘Ž=π‘›βˆšπ›Ό to denote a fixed root of 𝑓(π‘₯). Consider the extension field 𝐹[π‘›βˆšπ›Ό] of 𝐹. Then πΉξ‚ƒπ‘›βˆšπ›Όξ‚„=ξ€½π›Όπ‘›βˆ’1π‘Žπ‘›βˆ’1+β‹―+𝛼1π‘Ž+𝛼0βˆ£π›Όπ‘–ξ€ΎβˆˆπΉ,(2.1) and {π‘Žπ‘›βˆ’1,…,π‘Ž,1} is a basis of 𝐹[π‘›βˆšπ›Ό] over 𝐹 as a vector space over 𝐹. Now define π›Όπ‘›βˆ’1π‘Žπ‘›βˆ’1+β‹―+𝛼1π‘Ž+𝛼0β‰₯0 if each 𝛼𝑖β‰₯0, 𝑖=0,1,…,π‘›βˆ’1. Since π‘Žπ‘›=𝛼>0, 𝐹[π‘›βˆšπ›Ό] becomes an β„“-algebra over 𝐹 with respect to the above lattice order, and π‘Ž is a 𝑑-element. Thus {π‘Žπ‘›βˆ’1,…,π‘Ž,1} is a 𝑑-basis of the β„“-algebra 𝐹[π‘›βˆšπ›Ό] over 𝐹. In this paper, 𝑛≀4 and 𝐹[π‘›βˆšπ›Ό] always denotes the β„“-algebra defined above.(iii)For 𝑛β‰₯1, let 𝑀𝑛(𝐹) and 𝑇𝑛(𝐹) be 𝑛×𝑛 matrix algebra and upper triangular matrix algebra, respectively. Define a matrix positive if each of its entries is positive. Then 𝑀𝑛(𝐹) and 𝑇𝑛(𝐹) become β„“-algebras. Let 𝑒𝑖𝑗 denote the matrix with 𝑖𝑗th entry equal to 1 and other entries equal to 0. Then {π‘’π‘–π‘—βˆ£1≀𝑖,𝑗≀𝑛} is a 𝑑-basis for 𝑀𝑛(𝐹) over 𝐹 and {π‘’π‘–π‘—βˆ£1≀𝑖≀𝑗≀𝑛} is a 𝑑-basis for 𝑇𝑛(𝐹) over 𝐹. In the following, 𝑀𝑛(𝐹) and 𝑇𝑛(𝐹) always denote the β„“-algebras with the above lattice order.

All nonisomorphic unital four-dimensional β„“-algebras with a 𝑑-basis are listed in Theorem 2.1. Suppose that 𝐴 is a unital four-dimensional β„“-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 β„“-reduced and 𝐴 is not β„“-reduced, respectively. We use βŠ• to denote a direct sum of β„“-rings.

Suppose that 𝐴 is a four-dimensional β„“-algebra with a 𝑑-basis 𝑆={𝑒1,𝑒2,𝑒3,𝑒4} such that 1=𝑒1+𝑒2+𝑒3+𝑒4. By Theorem 1.1, each 𝑒𝑖 is an idempotent 𝑓-element, and for 1≀𝑖,𝑗≀4, 𝑖≠𝑗, 𝑒𝑖𝑒𝑗=0. Thus π΄β‰…πΉβŠ•πΉβŠ•πΉβŠ•πΉ 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 β„“-algebra over 𝐹 with a 𝑑-basis. Then 𝐴 is isomorphic to one of following β„“-algebras over 𝐹: (1)πΉβŠ•πΉβŠ•πΉβŠ•πΉ, (2)𝐹[𝐺]βŠ•πΉβŠ•πΉ, where 𝐺 is a group of order two, (3)√𝐹[𝛼]βŠ•πΉβŠ•πΉ, where 0<π›ΌβˆˆπΉ, (4)(𝐹1+πΉπ‘Ž)βŠ•πΉβŠ•πΉ, where π‘Ž2=0, (5)𝑇2(𝐹)βŠ•πΉ, (6)𝐹[𝐺]βŠ•πΉ[𝐺], where 𝐺 is a group of order two, (7)√𝐹[𝐺]βŠ•πΉ[𝛼], where 0<π›ΌβˆˆπΉ and 𝐺 is a group of order two, (8)√𝐹[βˆšπ›Ό]βŠ•πΉ[𝛽], where 0<𝛼,π›½βˆˆπΉ, (9)𝐹[𝐺]βŠ•πΉ, where 𝐺 is a group of order three, (10)𝐹[3βˆšπ›Ό]βŠ•πΉ, where 0<π›ΌβˆˆπΉ, (11)𝑀2(𝐹), (12)(𝐹1+πΉπ‘Ž+𝐹𝑏)βŠ•πΉ, where π‘Ž2=𝑏2=π‘Žπ‘=π‘π‘Ž=0, (13)(𝐹1+πΉπ‘Ž)βŠ•(𝐹1+𝐹𝑏), where π‘Ž2=𝑏2=0, (14)𝑇2(𝐹)+πΉπ‘Ž, where 𝑒11π‘Ž=π‘Žπ‘’11=π‘Ž, π‘Žπ‘’12=0, and π‘Ž2=0, (15)𝑇2(𝐹)+πΉπ‘Ž, where 𝑒11π‘Ž=π‘Žπ‘’22=π‘Ž, and π‘Ž2=0, (16)𝑇2(𝐹)+πΉπ‘Ž, where 𝑒22π‘Ž=π‘Žπ‘’11=π‘Ž, 𝑒12π‘Ž=π‘Žπ‘’12=0, and π‘Ž2=0, (17)𝑇2(𝐹)+πΉπ‘Ž, where 𝑒22π‘Ž=π‘Žπ‘’22=π‘Ž, 𝑒12π‘Ž=0, and π‘Ž2=0, (18)(𝐹1+πΉπ‘Ž+πΉπ‘Ž2)βŠ•πΉ, where π‘Ž3=0, (19)𝐹[𝐺]βŠ•(𝐹1+πΉπ‘Ž), where π‘Ž2=0, (20)√𝐹[𝛼]βŠ•(𝐹1+πΉπ‘Ž), where π‘Ž2=0 and 0<π›ΌβˆˆπΉ, (21)𝐹[𝐺], where 𝐺 is a cyclic group of order four, (22)𝐹[4βˆšπ›Ό], where 0<π›ΌβˆˆπΉ, (23)𝐹[𝐺]+𝐹[𝐺]𝑏, where 𝐺={𝑒,π‘Ž} is a group of order two and 𝑏2=π›Όπ‘Ž with 0<π›ΌβˆˆπΉ and βˆšπ›Όβˆ‰πΉ, (24)𝐹[𝐺], where 𝐺 is the Klein four group, (25)√𝐹[𝛼][𝐺], where 0<π›ΌβˆˆπΉ and 𝐺 is a group of order two, (26)√𝐹[βˆšπ›Ό][𝛽], where 0<𝛼,π›½βˆˆπΉ, βˆšβˆšπ›½βˆ‰πΉ[𝛼], (27)𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹𝑐, where π‘Ž2=𝑏+𝑐 and the product of any other two elements of π‘Ž,𝑏,𝑐 is zero, (28)𝐹1+πΉπ‘Ž+πΉπ‘Ž2+πΉπ‘Ž3, where π‘Ž4=0, (29)𝐹1+πΉπ‘Ž+πΉπ‘Ž2+𝐹𝑏, where 𝑏2=π›Όπ‘Ž2, 0<π›ΌβˆˆπΉ, and the product of any two elements not both π‘Ž or 𝑏 is zero, (30)𝐹1+πΉπ‘Ž+πΉπ‘Ž2+𝐹𝑏, where the product of any two elements not both π‘Ž is zero, (31)𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹𝑐, where πΉπ‘Ž+𝐹𝑏+𝐹𝑐 is a zero ring, (32)𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹(π‘Žπ‘), where π‘Ž2=𝑏2=π‘π‘Ž=0, (33)𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹(π‘Žπ‘), where π‘Ž2=𝑏2=0 and π‘π‘Ž=𝛼(π‘Žπ‘) with 0<π›ΌβˆˆπΉ, (34)𝐹[𝐺]+𝐹[𝐺]π‘Ž, where 𝐺 is a group of order two, π‘Ž2=0, and π‘Ž commutes with each element in 𝐹[𝐺], (35)√𝐹[βˆšπ›Ό]+𝐹[𝛼]π‘Ž, where π‘Ž2=0 and π‘Ž commutes with each element in √𝐹[𝛼].

3. 1 Is a Sum of Three Disjoint Basic Elements

Let 𝐴 be a unital four-dimensional β„“-algebra with a 𝑑-basis 𝑆={𝑒1,𝑒2,𝑒3,π‘Ž} such that 1=𝑒1+𝑒2+𝑒3. Then by Theorem 1.1, each 𝑒𝑖 is an idempotent 𝑓-element, 𝑖=1,2,3, and 𝑒𝑖𝑒𝑗=0 for any 1≀𝑖,𝑗≀3, 𝑖≠𝑗. By Theorem 1.1, we may assume that 𝑒1π‘Ž=π‘Ž and 𝑒2π‘Ž=𝑒3π‘Ž=0. We consider two cases according to whether 𝐴 is β„“-reduced or not.

3.1. 𝐴 Is β„“-Reduced

If π‘Žπ‘’1=0, then π‘Žπ‘’π‘–=π‘Ž for 𝑖=2 or 3 by Theorem 1.1(2), and hence π‘Ž2=(π‘Žπ‘’π‘–)(𝑒1π‘Ž)=0, which contradicts the fact that 𝐴 is β„“-reduced. Thus π‘Žπ‘’1=π‘Ž and π‘Žπ‘’2=π‘Žπ‘’3=0. Since π‘Ž is not nilpotent, by Theorem 1.1(4), π‘Ž2 is a basic element, so π‘Ž2=𝛼𝑠 for some 0<π›ΌβˆˆπΉ and π‘ βˆˆπ‘†. Since 𝑒1βˆ§π‘Ž=0 and 𝑒1π‘Ž=π‘Ž, π‘ β‰ π‘Ž. If 𝑠=𝑒2 or 𝑒3, then π‘Ž3=(𝛼𝑠)π‘Ž=𝛼(π‘ π‘Ž)=0, which is a contradiction. Thus π‘Ž2=𝛼𝑒1 for some 0<π›ΌβˆˆπΉ. There are two cases now. (1)There exists 0<π›½βˆˆπΉ such that 𝛽2=𝛼. Let 𝑏=π‘Ž/𝛽. Then {𝑒1,𝑒2,𝑒3,𝑏} is also a 𝑑-basis with the following multiplication table: 𝑒1𝑏𝑒2𝑒3𝑒1𝑒1𝑏𝑏00𝑏𝑒1𝑒00200𝑒20𝑒3000𝑒3(3.1) Thus 𝐴≅𝐹[𝐺]βŠ•πΉβŠ•πΉ, where 𝐺 is a group of order two. This is (2) in Theorem 2.1. (2)The polynomial 𝑓(π‘₯)=π‘₯2βˆ’π›ΌβˆˆπΉ[π‘₯] has no root in 𝐹. Then we have the following multiplication table for 𝑆: 𝑒1π‘Žπ‘’2𝑒3𝑒1𝑒1π‘Žπ‘Ž00π‘Žπ›Όπ‘’1𝑒00200𝑒20𝑒3000𝑒3(3.2) Thus βˆšπ΄β‰…πΉ[𝛼]βŠ•πΉβŠ•πΉ. This is (3) in Theorem 2.1.

3.2. 𝐴 Is Not β„“-Reduced

In this case, π‘Ž must be nilpotent. If π‘Ž2β‰ 0, then π‘Ž2 must be a positive linear combination of elements in 𝑆. Since each of 𝑒1,𝑒2,𝑒3 is not nilpotent, we must have π‘Ž2=π›½π‘Ž for some 0<π›½βˆˆπΉ. Then π‘Ž is nilpotent implying that π‘Ž2=π›½π‘Ž=0, which is a contradiction. Thus we must have π‘Ž2=0. There are two cases that we need to consider. (1)π‘Žπ‘’1=π‘Ž. Then π‘Žπ‘’2=π‘Žπ‘’3=0, and the multiplication table for 𝑆={𝑒1,𝑒2,𝑒3,π‘Ž} is given below: 𝑒1π‘Žπ‘’2𝑒3𝑒1𝑒1π‘Žπ‘Ž00π‘’π‘Ž000200𝑒20𝑒3000𝑒3(3.3) Thus 𝐴≅(𝐹1+πΉπ‘Ž)βŠ•πΉβŠ•πΉ, where π‘Ž2=0 and 1π‘Ž=π‘Ž1=π‘Ž. This is (4) in Theorem 2.1. (2)π‘Žπ‘’1=0. Without loss of generality, we may assume that π‘Žπ‘’2=π‘Ž and π‘Žπ‘’3=0. Then we have the following multiplication table for elements in 𝑆: 𝑒1𝑒2π‘Žπ‘’3𝑒1𝑒1𝑒0π‘Ž020𝑒2π‘Ž00𝑒0π‘Ž003000𝑒3(3.4) Let 𝑇2(𝐹) be the 2Γ—2 upper triangular matrix algebra over 𝐹 with the entrywise lattice order. Then 𝑇2(𝐹) becomes an β„“-algebra over 𝐹 with a 𝑑-basis {𝑒11,𝑒22,𝑒12}. The multiplication table for {𝑒11,𝑒22,𝑒12} is given below: 𝑒11𝑒22𝑒12𝑒11𝑒110𝑒12𝑒220𝑒220𝑒120𝑒120(3.5) Since {𝑒1,𝑒2,π‘Ž} and {𝑒11,𝑒22,𝑒12} have the same multiplication tables, we have 𝐴≅𝑇2(𝐹)βŠ•πΉ. This is (5) in Theorem 2.1.

4. 1 Is a Sum of Two Disjoint Basic Elements

Let 𝐴 be a unital β„“-algebra with a 𝑑-basis 𝑆={𝑒1,𝑒2,π‘Ž,𝑏} such that 1=𝑒1+𝑒2. By Theorem 1.1, π‘Žπ‘’π‘–=π‘Ž for some 𝑖=1 or 2. Then π‘’π‘–βˆ§π‘=0 implies that π‘Žβˆ§π‘Žπ‘=0. Similarly π‘βˆ§π‘Žπ‘=0. Thus π‘Žπ‘=𝛼𝑒1+𝛽𝑒2 for some 𝛼,π›½βˆˆπΉ+. By a similar argument, π‘π‘Ž=𝛾𝑒1+𝛿𝑒2, for some 𝛾,π›ΏβˆˆπΉ+.

4.1. 𝐴 Is β„“-Reduced

We may assume that 𝑒1π‘Ž=π‘Ž and 𝑒2π‘Ž=0. Since 𝐴 contains no nonzero positive nilpotent elements, we also have π‘Žπ‘’1=π‘Ž and π‘Žπ‘’2=0. Then π‘Žπ‘=𝛼𝑒1+𝛽𝑒2 implies that π‘Žπ‘=𝑒1π‘Žπ‘=𝛼𝑒1. Similarly, π‘π‘Ž=𝛾𝑒1.

Since 𝐴 is β„“-reduced, π‘Ž is not a nilpotent element, so π‘Ž2 is basic by Theorem 1.1. Thus π‘Ž2=πœŽπ‘  for some π‘ βˆˆπ‘† and 0<𝜎∈𝐹. Since 𝑒1βˆ§π‘Ž=0 and 𝑒1π‘Ž=π‘Ž, π‘ β‰ π‘Ž. If 𝑠=𝑒2, then π‘Ž3=𝜎(π‘Žπ‘’2)=0, which is a contradiction. Thus 𝑠=𝑒1 or 𝑠=𝑏. We consider these two cases separately. (1)π‘Ž2=πœŽπ‘’1. Suppose that π‘Žπ‘β‰ 0. Since π‘Žπ‘=𝛼𝑒1, π‘Ž2𝑏=π›Όπ‘Ž, so 𝜎(𝑒1𝑏)=π›Όπ‘Ž, which is a contradiction. Thus π‘Žπ‘=0, and hence π‘π‘Ž=0 since (π‘π‘Ž)2=0. Since 0=π‘π‘Ž2=πœŽπ‘π‘’1, 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=πœ†π‘  for some π‘ βˆˆπ‘† and 0<πœ†βˆˆπΉ. Since 𝑒2βˆ§π‘=0 and 𝑒2𝑏=𝑏, 𝑠≠𝑏. If 𝑏2=πœ†π‘’1 or 𝑏2=πœ†π‘Ž, then πœ†π‘Ž=0 or πœ†π‘Ž2=0, which are contradictions. Thus we must have 𝑏2=πœ†π‘’2. Then we have the following multiplication table for the elements of 𝑆: 𝑒1π‘Žπ‘’2𝑏𝑒1𝑒1π‘Žπ‘Ž00π‘ŽπœŽπ‘’1𝑒00200𝑒2𝑏𝑏000πœ†π‘’2(4.1) Depending on √𝜎∈𝐹 or βˆšπœŽβˆ‰πΉ and βˆšπœ†βˆˆπΉ or βˆšπœ†βˆ‰πΉ, we have the following cases: [𝐺][𝐺][𝐺]ξ‚ƒβˆšπ΄β‰…πΉβŠ•πΉ,π΄β‰…πΉβŠ•πΉπœŽξ‚„ξ‚ƒβˆš,orπ΄β‰…πΉπœŽξ‚„ξ‚ƒβˆšβŠ•πΉπœ†ξ‚„,(4.2) where 𝐺 is a group of order two. These are (6), (7), and (8) in Theorem 2.1. (2)π‘Ž2=πœŽπ‘. Since π‘Ž2=πœŽπ‘ with 0<𝜎∈𝐹, 𝑇={𝑒1,𝑒2,π‘Ž,π‘Ž2} is also a 𝑑-basis for the β„“-algebra 𝐴 over 𝐹 and π‘Ž3=𝜎(π‘Žπ‘)=(π›ΌπœŽ)𝑒1. Let πœ‡=π›ΌπœŽ. Then πœ‡>0 since π‘Ž3β‰ 0, so we have the following multiplication table for elements in 𝑇: 𝑒1π‘Žπ‘Ž2𝑒2𝑒1𝑒1π‘Žπ‘Ž20π‘Žπ‘Žπ‘Ž2πœ‡π‘’10π‘Ž2π‘Ž2πœ‡π‘’1π‘’πœ‡π‘Žπ‘2000𝑒2(4.3) Now it is clear that if 3βˆšπœ‡βˆˆπΉ, then 𝐴≅𝐹[𝐺]βŠ•πΉ, where 𝐺 is a group of order three, and if 3βˆšπœ‡βˆ‰πΉ, then 𝐴≅𝐹[3βˆšπœ‡]βŠ•πΉ, where 𝐹[3βˆšπœ‡] is the β„“-field extension of 𝐹 of order three. Those are (9) and (10) in Theorem 2.1.

4.2. 𝐴 Is Not β„“-Reduced

Since 𝐴 is not β„“-reduced, π‘Ž or 𝑏 is nilpotent. Let's assume that 𝑒1π‘Ž=π‘Ž, and hence 𝑒2π‘Ž=0. Suppose that π‘Ž is nilpotent and π‘Ž2β‰ 0. Then π‘Ž2=𝛼1𝑒1+𝛼2𝑒2+𝛼3π‘Ž+𝛼4𝑏,π›Όπ‘–βˆˆπΉ+.(4.4) Let π‘˜ be the smallest positive integer such that π‘Žπ‘˜=0. Then π‘˜β‰₯3 and 0=π‘Žπ‘˜=π‘Ž2π‘Žπ‘˜βˆ’2=𝛼1𝑒1π‘Žπ‘˜βˆ’2+𝛼2𝑒2π‘Žπ‘˜βˆ’2+𝛼3π‘Žπ‘˜βˆ’1+𝛼4π‘π‘Žπ‘˜βˆ’2,(4.5) so 𝛼1π‘Žπ‘˜βˆ’2=𝛼3π‘Žπ‘˜βˆ’1=0. Thus 𝛼1=𝛼3=0, and hence π‘Ž2=𝛼2𝑒2+𝛼4𝑏. Then π‘Ž2=𝑒1π‘Ž2=𝛼4(𝑒1𝑏), which implies that 𝛼2=0 and π‘Ž2=𝛼4𝑏. Then π‘Ž2β‰ 0 implies that 𝛼4>0. So {𝑒1,𝑒2,π‘Ž,π‘Ž2} is a 𝑑-basis of 𝐴 over 𝐹. Similarly if 𝑏 is nilpotent and 𝑏2β‰ 0, then {𝑒1,𝑒2,𝑏,𝑏2} is a 𝑑-basis of 𝐴 over 𝐹. Therefore we only need to consider the following cases.(i)π‘Ž2=𝑏2=0.(ii)π‘Ž is nilpotent, but π‘Ž2β‰ 0. By the above argument, π‘Ž2=𝛼4𝑏 for some 0<𝛼4∈𝐹. Thus 𝑇={𝑒1,𝑒2,π‘Ž,π‘Ž2} is also a 𝑑-basis for 𝐴 over 𝐹.(iii)π‘Ž2=0 and 𝑏 is not nilpotent.

Below, we consider each case in detail.

4.2.1. π‘Ž2=𝑏2=0

From π‘π‘Ž=𝛾𝑒1+𝛿𝑒2, we have 0=π‘π‘Ž2=𝛾𝑒1π‘Ž+𝛿𝑒2π‘Ž=𝛾𝑒1π‘Ž, so 𝛾=0, and hence π‘π‘Ž=𝛿𝑒2. (1)𝛿>0. Then π‘π‘Ž=𝛿𝑒2β‰ 0 implies that 𝑒2𝑏=𝑏, 𝑒1𝑏=0, π‘Žπ‘’2=π‘Ž, and π‘Žπ‘’1=0. Then π‘Žπ‘=𝛼𝑒1+𝛽𝑒2 implies that 0=π‘Ž2𝑏=π›Όπ‘Žπ‘’1+π›½π‘Žπ‘’2=π›½π‘Ž, so 𝛽=0 and π‘Žπ‘=𝛼𝑒1. Thus π‘Žπ‘π‘Ž=𝛼𝑒1π‘Ž=π›Ώπ‘Žπ‘’2 implies that 𝛼=𝛿. Since π‘Žπ‘=𝛼𝑒1β‰ 0, 𝑏𝑒1=𝑏 and 𝑏𝑒2=0. Let π‘Žξ…ž=π‘Ž/𝛼. Then {𝑒1,𝑒2,π‘Žβ€²,𝑏} is also a 𝑑-basis of 𝐴 with the following multiplication table: 𝑒1𝑒2π‘Žξ…žπ‘’2𝑒1𝑒1π‘Žπ‘Žξ…ž0𝑒20𝑒2π‘Ž0π‘ξ…ž0π‘Žξ…ž0𝑒1𝑏𝑏0𝑒20(4.6) It is clear now that 𝐴≅𝑀2(𝐹), the 2Γ—2 matrix β„“-algebra over 𝐹. This is (11) in Theorem 2.1. (2)𝛿=0. Then π‘π‘Ž=0. From π‘Žπ‘=𝛼𝑒1+𝛽𝑒2, we have 0=π‘Ž(π‘π‘Ž)=(π‘Žπ‘)π‘Ž=𝛼𝑒1π‘Ž+𝛽𝑒2π‘Ž=π›Όπ‘Ž, so 𝛼=0. Thus π‘Žπ‘=𝛽𝑒2, and hence 0=𝑒2π‘Žπ‘=𝛽𝑒2, which implies that 𝛽=0, so π‘Žπ‘=0. Therefore, πΉπ‘Ž+𝐹𝑏 is a zero β„“-ring. There are the following cases.(2a)If π‘Žπ‘’1=π‘Ž, 𝑒1𝑏=𝑏, and 𝑏𝑒1=𝑏, then 𝐴≅(𝐹1+πΉπ‘Ž+𝐹𝑏)βŠ•πΉ, where πΉπ‘Ž+𝐹𝑏 is the zero β„“-ring.(2b)If π‘Žπ‘’1=π‘Ž, 𝑒2𝑏=𝑏, and 𝑏𝑒2=𝑏, then 𝐴≅(𝐹1+πΉπ‘Ž)βŠ•(𝐹1+𝐹𝑏), where π‘Ž2=𝑏2=0. If π‘Žπ‘’2=π‘Ž, then the multiplication table for {𝑒1,𝑒2,π‘Ž} is given below: 𝑒1𝑒2π‘Žπ‘’1𝑒1𝑒0π‘Ž20𝑒20π‘Ž0π‘Ž0(4.7) Clearly 𝐹𝑒1+𝐹𝑒2+πΉπ‘Ž 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) π‘Žπ‘’1=π‘Ž, 𝑒1𝑏=𝑏, 𝑏𝑒1=0 and (ii) π‘Žπ‘’1=π‘Ž, 𝑒2𝑏=𝑏, 𝑏𝑒2=0. But when we switch π‘Ž and 𝑏, the case (i) becomes (2c), and when we switch π‘Ž 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. π‘Ž Is Nilpotent and π‘Ž2β‰ 0

As we have noticed before, in this case 𝑇={𝑒1,𝑒2,π‘Ž,π‘Ž2} is a 𝑑-basis for the β„“-algebra 𝐴 over 𝐹. Since π‘Žβˆ§π‘’1=0 implies that π‘Ž3βˆ§π‘Ž2=0 and π‘Ž2βˆ§π‘’1=0 implies that π‘Ž3βˆ§π‘Ž=0, π‘Ž3=𝛽1𝑒1, for some 𝛽1∈𝐹+. If 𝛽1β‰ 0, then π‘Ž cannot be nilpotent, so we must have 𝛽1=0. Therefore π‘Ž3=0, and 𝐴≅(𝐹1+πΉπ‘Ž+πΉπ‘Ž2)βŠ•πΉ with π‘Ž3=0. This is (18) in Theorem 2.1.

4.2.3. π‘Ž2=0 and 𝑏 Is Not Nilpotent

Since 𝑏 is not nilpotent, 𝑏2 is basic, so 𝑏2=πœ–π‘  for some π‘ βˆˆπ‘† and 0<πœ–βˆˆπΉ. Clearly π‘ β‰ π‘Ž or 𝑏. Thus either 𝑏2=πœ–π‘’1 or 𝑏2=πœ–π‘’2. In the first case, we have 𝑏2π‘Ž=πœ–π‘’1π‘Ž=𝑏(𝛾𝑒1+𝛿𝑒2)=𝛾𝑏𝑒1, so πœ–π‘Ž=𝛾𝑏, which is a contradiction. Thus we must have 𝑏2=πœ–π‘’2. If π‘Žπ‘’2=π‘Ž, then π‘Žπ‘2=πœ–π‘Ž, so (𝛼𝑒1+𝛽𝑒2)𝑏=πœ–π‘Ž, which implies that 𝛽𝑏=πœ–π‘Ž, a contradiction. Thus π‘Žπ‘’1=π‘Ž, and hence π‘Žπ‘’2=0. Then π‘Žπ‘=π‘π‘Ž=0. Therefore we have the following multiplication table for the elements of 𝑆: 𝑒1π‘Žπ‘’2𝑏𝑒1𝑒1π‘Žπ‘Ž00π‘’π‘Ž000200𝑒2𝑏𝑏00π‘πœ–π‘’2(4.8)

From the above table, it is clear that if βˆšπœ–βˆˆπΉ, then 𝐴≅𝐹[𝐺]βŠ•(𝐹1+πΉπ‘Ž) with π‘Ž2=0, where 𝐺 is a group of order two, and if βˆšπœ–βˆ‰πΉ, then βˆšπ΄β‰…πΉ[πœ–]βŠ•(𝐹1+πΉπ‘Ž) with π‘Ž2=0, where √𝐹[πœ–] is a quadratic extension β„“-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,π‘Ž,𝑏,𝑐}. We still first consider the β„“-reduced case.

5.1. 𝐴 Is β„“-Reduced

Since π‘Ž is not nilpotent, by Theorem 1.1, there exist positive integer π‘›π‘Ž and 0<π›Όπ‘ŽβˆˆπΉ such that π‘Žπ‘›π‘Ž=π›Όπ‘Ž1. We also assume that π‘›π‘Ž is the smallest positive integer such that π‘Žπ‘›π‘Ž=π›Όπ‘Ž1. Similarly 𝑏𝑛𝑏=𝛼𝑏1 and 𝑐𝑛𝑐=𝛼𝑐1, for some positive integers 𝑛𝑏, 𝑛𝑐, and 0<𝛼𝑏,π›Όπ‘βˆˆπΉ. By Theorem 1.1 again, each π‘Žπ‘˜ is a basic element for π‘˜β‰₯1.

Lemma 5.1. The subset {1,π‘Ž,π‘Ž2,…,π‘Žπ‘›π‘Žβˆ’1} of 𝐴 is a disjoint set of 𝐴.

Proof. Let 1≀𝑖<π‘›π‘Ž. Since π‘Žπ‘– is a basic element, π‘Žπ‘–=𝛼𝑠 for some π‘ βˆˆπ‘† and 0<π›ΌβˆˆπΉ. Since 𝑖<π‘›π‘Ž, 𝑠≠1, so 1βˆ§π‘ =0 implies that 1βˆ§π‘Žπ‘–=0. Now let 0≀𝑒<𝑣<π‘›π‘Ž, then π‘Žπ‘’βˆ§π‘Žπ‘£=π‘Žπ‘’(1βˆ§π‘Žπ‘£βˆ’π‘’)=0 by the above argument.

Since 𝐴 is four-dimensional, π‘›π‘Žβ‰€4 by Lemma 5.1. We claim that π‘›π‘Žβ‰ 3. Supposing that π‘›π‘Ž=3, then we may assume that {1,π‘Ž,π‘Ž2,𝑏} is a 𝑑-basis. By Theorem 1.1, π‘Žπ‘ is a basic element and π‘Žπ‘βˆ§π‘Ž=π‘Ž(π‘βˆ§1)=0, π‘Žπ‘βˆ§π‘=(π‘Žβˆ§1)𝑏=0, π‘Žπ‘βˆ§π‘Ž2=π‘Ž(π‘βˆ§π‘Ž)=0, so π‘Žπ‘=𝛼1 for some 0<π›ΌβˆˆπΉ. Thus π‘Ž3𝑏=π›Όπ‘Ž2, and hence π›Όπ‘Žπ‘=π›Όπ‘Ž2, which is a contradiction. Hence π‘›π‘Žβ‰ 3, so π‘›π‘Ž=4 or π‘›π‘Ž=2.

By the above analysis, we need to consider two cases: (i) one of the following: π‘›π‘Ž, 𝑛𝑏, 𝑛𝑐, is 4, and we may assume that π‘›π‘Ž=4; (ii) π‘›π‘Ž=𝑛𝑏=𝑛𝑐=2.

5.1.1. π‘›π‘Ž=4 and Hence {1,π‘Ž,π‘Ž2,π‘Ž3} Is a Disjoint Subset of Basic Elements of 𝐴

In this case 𝑇={1,π‘Ž,π‘Ž2,π‘Ž3} is a 𝑑-basis of 𝐴 over 𝐹, and we have the following multiplication table for the elements in 𝑇: 1π‘Žπ‘Ž2π‘Ž311π‘Žπ‘Ž2π‘Ž3π‘Žπ‘Žπ‘Ž2π‘Ž3π›Όπ‘Žπ‘Ž2π‘Ž2π‘Ž3π›Όπ‘Ž1π›Όπ‘Žπ‘Žπ‘Ž3π‘Ž3π›Όπ‘Ž1π›Όπ‘Žπ‘Žπ›Όπ‘Žπ‘Ž2(5.1)

We have the following three cases to be considered.(1)If 4βˆšπ›Όπ‘ŽβˆˆπΉ, then 𝐴≅𝐹[𝐺], where 𝐺 is a cyclic group of order four.(2)If the polynomial 𝑓(π‘₯)=π‘₯4βˆ’π›Όπ‘ŽβˆˆπΉ[π‘₯] is irreducible, then 𝐴≅𝐹[4βˆšπ›Όπ‘Ž], which is an β„“-field extension of 𝐹 of order four.(3)The polynomial 𝑓(π‘₯)=π‘₯4βˆ’π›Όπ‘ŽβˆˆπΉ[𝐺] is not irreducible and 4βˆšπ›Όπ‘Žβˆ‰πΉ.

Suppose that π‘₯4βˆ’π›Όπ‘Ž=(π‘₯2+π‘Ž1π‘₯+𝑏1)(π‘₯2+π‘Ž2π‘₯+𝑏2). Then we have π‘Ž1+π‘Ž2=0, 𝑏1+𝑏2+π‘Ž1π‘Ž2=0, 𝑏1π‘Ž2+𝑏2π‘Ž1=0, and 𝑏1𝑏2=βˆ’π›Όπ‘Ž<0. Thus π‘Ž2=βˆ’π‘Ž1 and 0=𝑏1π‘Ž2+𝑏2π‘Ž1=π‘Ž1(βˆ’π‘1+𝑏2). If π‘Ž1β‰ 0, then βˆ’π‘1+𝑏2=0, and hence 𝑏1=𝑏2, which implies that 𝑏1𝑏2=𝑏21=βˆ’π›Όπ‘Ž<0, a contradiction. So we must have π‘Ž1=0, and hence π‘Ž2=0. Then 𝑏1+𝑏2=0 and 𝑏1𝑏2=βˆ’π›Όπ‘Ž implies that 𝑏2=βˆ’π‘1 and 𝑏21=𝑏22=π›Όπ‘Ž. We may assume that 𝑏1>0. It follows from 4βˆšπ›Όπ‘Žβˆ‰πΉ that βˆšπ‘1βˆ‰πΉ. Let 𝑐=π‘Ž2/𝑏1. Then 𝑉={1,𝑐,π‘Ž,π‘Žπ‘} is also a 𝑑-basis of 𝐴 over 𝐹 and the multiplication table for the elements in 𝑉 is given below: 1π‘π‘Žπ‘Žπ‘1𝑐1π‘π‘Žπ‘Žπ‘π‘Žπ‘1π‘Žπ‘π‘Žπ‘Žπ‘Žπ‘π‘1𝑐𝑏11π‘Žπ‘π‘Žπ‘π‘Žπ‘11𝑏1𝑐(5.2)

Then 𝐴≅𝐹[𝐺]+𝐹[𝐺]π‘Ž, where 𝐺={𝑒,𝑐} is a group of order two and π‘Ž2=𝛼𝑐, 0<π›ΌβˆˆπΉ, and βˆšπ›Όβˆ‰πΉ. These are (21), (22), and (23) in Theorem 2.1.

5.1.2. π‘›π‘Ž=𝑛𝑏=𝑛𝑐=2

Then 𝑆={1,π‘Ž,𝑏,𝑐} is a 𝑑-basis and π‘Ž2=π›Όπ‘Ž1, 𝑏2=𝛼𝑏1, and 𝑐2=𝛼𝑐1. Since π‘Žπ‘βˆ§π‘Ž=π‘Žπ‘βˆ§π‘=0 and π‘Žπ‘ is basic, π‘Žπ‘=𝛼1 or π‘Žπ‘=𝛼𝑐 for some 0<π›ΌβˆˆπΉ. In the first case, π‘Žπ‘2=𝛼𝑏 implies that π›Όπ‘π‘Ž=𝛼𝑏, which is a contradiction. Thus π‘Žπ‘=𝛼𝑐 for some 0<π›ΌβˆˆπΉ. Then we may replace 𝑐 by π‘Žπ‘ to get a 𝑑-basis 𝑇={1,π‘Ž,𝑏,π‘Žπ‘}.

Since π‘π‘Žβˆ§π‘Ž=π‘π‘Žβˆ§π‘=0 and π‘π‘Ž is basic, π‘π‘Ž=𝛽1 or π‘π‘Ž=𝛽(π‘Žπ‘) for some 0<π›½βˆˆπΉ. If π‘π‘Ž=𝛽1, then π›Όπ‘Žπ‘=π›½π‘Ž, which is a contradiction. Thus π‘π‘Ž=𝛽(π‘Žπ‘), for some 0<π›½βˆˆπΉ, and hence π›Όπ‘π›Όπ‘Ž1=𝑏(π‘π‘Ž)π‘Ž=𝛽𝑏(π‘Žπ‘)π‘Ž=𝛽2(π‘Žπ‘)(π‘π‘Ž)=𝛽2π›Όπ‘π›Όπ‘Ž1.(5.3) So 𝛽=1 and the multiplication table for the elements of 𝑇 is given below: 1π‘Žπ‘π‘Žπ‘1π‘Ž1π‘Žπ‘π‘Žπ‘π‘Žπ›Όπ‘Ž1π‘Žπ‘π›Όπ‘Žπ‘π‘π‘π‘Žπ‘π›Όπ‘1π›Όπ‘π‘Žπ‘Žπ‘π‘Žπ‘π›Όπ‘Žπ‘π›Όπ‘π‘Žπ›Όπ‘Žπ›Όπ‘1(5.4)

There are the following cases. (1)βˆšπ›Όπ‘Ž and βˆšπ›Όπ‘βˆˆπΉ. Replacing π‘Ž and 𝑏 by βˆšπ‘Ž/π›Όπ‘Ž and βˆšπ‘/𝛼𝑏, respectively, we have π‘Ž2=𝑏2=1. Therefore 𝐴≅𝐹[𝐺], where 𝐺 is the Klein four group. (2)βˆšπ›Όπ‘Žβˆ‰πΉ and βˆšπ›Όπ‘βˆˆπΉ (or βˆšπ›Όπ‘ŽβˆˆπΉ and βˆšπ›Όπ‘βˆ‰πΉ). Replacing 𝑏 by βˆšπ‘/𝛼𝑏, we may assume that 𝑏2=1. Therefore βˆšπ΄β‰…πΉ[π›Όπ‘Ž][𝐺], where √𝐹[π›Όπ‘Ž] is the quadratic extension β„“-field of 𝐹 and 𝐺 is a group of order two.(3)βˆšπ›Όπ‘Žβˆ‰πΉ and βˆšπ›Όπ‘βˆ‰πΉ. If βˆšπ›Όπ‘βˆšβˆˆπΉ[π›Όπ‘Ž], then βˆšπ΄β‰…πΉ[π›Όπ‘Ž][𝐺], where 𝐺 is a group of order two. We omit the checking and leave it to the reader. If βˆšπ›Όπ‘βˆšβˆ‰πΉ[π›Όπ‘Ž], then βˆšπ΄β‰…πΉ[π›Όπ‘Žβˆš][𝛼𝑏], where √𝐹[π›Όπ‘Ž] is a quadratic extension β„“-field of 𝐹 and √𝐹[π›Όπ‘Žβˆš][𝛼𝑏] is a quadratic extension β„“-field of √𝐹[π›Όπ‘Ž]. Those are (24), (25), and (26) in Theorem 2.1.

5.2. 𝐴 Is Not β„“-Reduced

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

Lemma 5.2. Let π‘₯,π‘¦βˆˆ{1,π‘Ž,𝑏,𝑐}. If either π‘₯ or 𝑦 is nilpotent and the other one is not nilpotent, then π‘₯𝑦 is nilpotent.

Proof. Let π‘₯ be nilpotent and let 𝑦 be not. By Theorem 1.1, there exist a positive integer 𝑛𝑦 and 0<π›Όπ‘¦βˆˆπΉ such that 𝑦𝑛𝑦=𝛼𝑦1. If π‘₯𝑦=0, then π‘₯𝑦𝑛𝑦=0 implies that 𝛼𝑦π‘₯=0, which is a contradiction. Thus π‘₯𝑦≠0, so π‘₯𝑦 is basic by Theorem 1.1. If π‘₯𝑦 is not nilpotent, then π‘₯𝑦=𝛾𝑧 for some 0<π›ΎβˆˆπΉ and π‘§βˆˆ{1,π‘Ž,𝑏,𝑐} and 𝑧 is not nilpotent. By Theorem 1.1, there exist a positive integer 𝑛𝑧, 0<π›Όπ‘§βˆˆπΉ such that 𝑧𝑛𝑧=𝛼𝑧1. Let 𝑛>1 be the smallest positive integer such that π‘₯𝑛=0. Then 0=π‘₯𝑛𝑦=π‘₯π‘›βˆ’1(π‘₯𝑦)=𝛾π‘₯π‘›βˆ’1𝑧, and hence 0=𝛾π‘₯π‘›βˆ’1𝑧 implies that 0=𝛾π‘₯π‘›βˆ’1𝑧𝑛𝑧=𝛼𝑧𝛾π‘₯π‘›βˆ’1. Thus π‘₯π‘›βˆ’1=0, which is a contradiction. Therefore π‘₯𝑦 is nilpotent.

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

5.2.1. π‘Ž, 𝑏, and 𝑐 Are All Nilpotent
(1)Either π‘Ž, 𝑏, or 𝑐 has nonzero square. We may assume that π‘Ž2β‰ 0. Since π‘Ž2βˆ§π‘Ž=π‘Ž(π‘Žβˆ§1)=0, π‘Ž2=𝛼1+𝛽𝑏+𝛾𝑐, where 𝛼,𝛽,π›ΎβˆˆπΉ+. Let π‘˜ be the smallest positive integer such that π‘Žπ‘˜=0. Then 0=π‘Žπ‘˜=π‘Žπ‘˜βˆ’2π‘Ž2=π›Όπ‘Žπ‘˜βˆ’2+π›½π‘Žπ‘˜βˆ’2𝑏+π›Ύπ‘Žπ‘˜βˆ’2𝑐 implies π›Όπ‘Žπ‘˜βˆ’2=0, so 𝛼=0. Thus π‘Ž2=𝛽𝑏+𝛾𝑐.(1a)𝛽>0 and 𝛾>0. Since π‘Ž2βˆ§π‘Žπ‘=π‘Ž(π‘Žβˆ§π‘)=0, π‘βˆ§π‘Žπ‘=π‘βˆ§π‘Žπ‘=0. Also π‘Žβˆ§π‘Žπ‘=π‘Ž(1βˆ§π‘)=0. Thus π‘Žπ‘=𝛿1. Then we have 0=π‘Žπ‘˜π‘=π›Ώπ‘Žπ‘˜βˆ’1, so 𝛿=0 and π‘Žπ‘=0. Similarly π‘π‘Ž=0, π‘Žπ‘=0, and π‘π‘Ž=0. Since π‘Ž2=𝛽𝑏+𝛾𝑐, 0=π‘Ž2𝑏=𝛽𝑏2+𝛾𝑐𝑏 implies that 𝑏2=𝑐𝑏=0. Similarly 𝑐2=𝑏𝑐=0. Replacing 𝑏 and 𝑐 by 𝛽𝑏 and 𝛾𝑐, respectively, we may assume that π‘Ž2=𝑏+𝑐. Therefore we have the following multiplication table: 1π‘Žπ‘π‘1π‘Ž1π‘Žπ‘π‘π‘π‘Žπ‘+𝑐00𝑐𝑏000𝑐000(5.5) Then 𝐴≅𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹𝑐, where π‘Ž2=𝑏+𝑐 and the product of any other two elements of π‘Ž, 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 π‘Ž2=𝛽𝑏, and π‘Ž(π‘Žπ‘Ž)=(π‘Žπ‘Ž)π‘Ž implies that 𝛽(π‘Žπ‘)=𝛽(π‘π‘Ž), so π‘Žπ‘=π‘π‘Ž.(i)π‘Žπ‘=π‘π‘Žβ‰ 0. Then π‘Žπ‘=𝛿𝑐 for some 0<π›ΏβˆˆπΉ. Thus π‘Ž3=π‘Žπ‘Ž2=𝛽(π‘Žπ‘)=𝛽𝛿𝑐, and hence we may replace 𝑏 by π‘Ž2 and 𝑐 by π‘Ž3 to get another 𝑑-basis 𝑇={1,π‘Ž,π‘Ž2,π‘Ž3}. Therefore 𝐴≅𝐹1+πΉπ‘Ž+πΉπ‘Ž2+πΉπ‘Ž3 with π‘Ž4=0. This is (28) in Theorem 2.1.(ii)π‘Žπ‘=π‘π‘Ž=0. Then 𝛽𝑏2=π‘Ž2𝑏=0, so 𝑏2=0. Since π‘Žπ‘=πœ‡π‘ for some 0β‰€πœ‡βˆˆπΉ, π‘Ž2βˆ§π‘Žπ‘=0 and π‘Ž2=𝛽𝑏 implies that πœ‡=0, so π‘Žπ‘=0. Similarly π‘π‘Ž=0. 0=π‘Ž2𝑐=𝛽(𝑏𝑐), so 𝑏𝑐=0. Similarly, 𝑐𝑏=0. Now replacing 𝑏 by π‘Ž2, we obtain another 𝑑-basis 𝑇={1,π‘Ž,π‘Ž2,𝑐}.If 𝑐2β‰ 0, then 𝑐2βˆ§π‘=𝑐2∧1=0 implies that 𝑐2=𝛼1π‘Ž+𝛼2π‘Ž2. Then it follows from π‘Žπ‘=0 and π‘Ž3=0 that 𝛼1π‘Ž2=0, so 𝛼1=0. Thus 𝑐2=𝛼2π‘Ž2 with 0<𝛼2∈𝐹. Hence 𝐴≅𝐹1+πΉπ‘Ž+πΉπ‘Ž2+𝐹𝑐 with 𝑐2=π›Όπ‘Ž2 for some 0<π›ΌβˆˆπΉ and π‘₯𝑦=0 for any two elements π‘₯ and 𝑦 in {π‘Ž,π‘Ž2,𝑐} not both π‘Ž or 𝑐. This is (29) in Theorem 2.1. If 𝑐2=0, then 𝐴≅𝐹1+πΉπ‘Ž+πΉπ‘Ž2+𝐹𝑐 with π‘₯𝑦=0 for any two elements π‘₯ and 𝑦 in {π‘Ž,π‘Ž2,𝑐} which are not both π‘Ž. This is (30) in Theorem 2.1.We provide some familiar examples for the above two cases. Consider 4Γ—4 matrix algebra over 𝐹. Let 1=𝑒11+𝑒22+𝑒33+𝑒44, π‘Ž=𝑒12+𝑒23, and 𝑐=𝑒14+2𝑒43. Then π‘Ž2=𝑒13 and 𝑐2=2π‘Ž2. Clearly 𝐹1+πΉπ‘Ž+πΉπ‘Ž2+𝐹𝑐 is a four-dimensional β„“-algebra over 𝐹 with coordinatewise order. If 𝑐=𝑒13+𝑒43, then 𝑐2=0 and 𝐹1+πΉπ‘Ž+πΉπ‘Ž2+𝐹𝑐 is a four-dimensional β„“-algebra over 𝐹 with 𝑐2=0. (2)π‘Ž2=𝑏2=𝑐2=0. (2a)π‘₯𝑦=0 for any two elements in {π‘Ž,𝑏,𝑐}.Then 𝐴≅𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹𝑐, where πΉπ‘Ž+𝐹𝑏+𝐹𝑐 is zero ring. This is (31) in Theorem 2.1. (2b)At least one product of two different elements in {π‘Ž,𝑏,𝑐} is nonzero. Without loss of generality, we may assume that π‘Žπ‘β‰ 0. Then π‘Žπ‘=𝛼𝑐 for some 0<π›ΌβˆˆπΉ, and we may replace 𝑐 by π‘Žπ‘ to form another 𝑑-basis 𝑇={1,π‘Ž,𝑏,π‘Žπ‘}.If π‘π‘Ž=0, then we have the following multiplication table: 1π‘Žπ‘π‘Žπ‘1π‘Ž1π‘Žπ‘π‘Žπ‘π‘π‘Ž0π‘Žπ‘0𝑏000π‘Žπ‘π‘Žπ‘000(5.6) Thus 𝐴≅𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹(π‘Žπ‘) with π‘Ž2=𝑏2=π‘π‘Ž=0. This is (32) in Theorem 2.1. If π‘π‘Žβ‰ 0, then π‘π‘Ž=𝛽(π‘Žπ‘) for some 0<π›½βˆˆπΉ and we have the following multiplication table: 1π‘Žπ‘π‘Žπ‘1π‘Ž1π‘Žπ‘π‘Žπ‘π‘π‘Ž0π‘Žπ‘0𝑏𝛽(π‘Žπ‘)00π‘Žπ‘π‘Žπ‘000(5.7) Thus 𝐴≅𝐹1+πΉπ‘Ž+𝐹𝑏+𝐹(π‘Žπ‘) with π‘Ž2=𝑏2=0 and π‘π‘Ž=𝛽(π‘Žπ‘) for some 0<π›½βˆˆπΉ. This is (33) in Theorem 2.1.
5.2.2. Two Elements in {π‘Ž,𝑏,𝑐} Are Nilpotent

We may assume that π‘Ž and 𝑏 are nilpotent elements. Then we must have 𝑐2=𝛼1 for some 0<π›ΌβˆˆπΉ. By Lemma 5.2, π‘Žπ‘=𝛽𝑏 for some 0<π›½βˆˆπΉ. If π‘Ž2β‰ 0, then π‘Ž2βˆ§π‘Ž=π‘Ž(π‘Žβˆ§1)=0 and that 1 and 𝑐 are not nilpotent implies that π‘Ž2=𝛾𝑏 for some 0<π›ΎβˆˆπΉ. Thus π‘Žπ‘βˆ§π‘Ž2β‰ 0, which is a contradiction, so π‘Ž2=0. Similarly, 𝑏2=0. Replacing π‘Ž by π‘Ž/𝛽 and then replacing 𝑏 by π‘Žπ‘, we have another 𝑑-basis {1,𝑐,π‘Ž,π‘Žπ‘} with the following multiplication table: 1π‘π‘Žπ‘Žπ‘1𝑐1π‘π‘Žπ‘Žπ‘π‘Žπ‘π›Ό1π‘Žπ‘π›Όπ‘Žπ‘Žπ‘Žπ‘00π‘Žπ‘π‘Žπ‘π›Όπ‘Ž00(5.8)

If βˆšπ›ΌβˆˆπΉ, then replacing 𝑐 by βˆšπ‘/𝛼, we have that 𝐴≅𝐹[𝐺]+𝐹[𝐺]π‘Ž is a commutative β„“-algebra, where 𝐺 is a group of order two and π‘Ž2=0.

If βˆšπ›Όβˆ‰πΉ, then βˆšπ΄β‰…πΉ[βˆšπ›Ό]+𝐹[𝛼]π‘Ž is a commutative β„“-algebra over 𝐹, where √𝐹[𝛼] is a quadratic extension β„“-field of 𝐹 and π‘Ž2=0. Those are (34) and (35) in Theorem 2.1.

This completes the proof of Theorem 2.1.

6. An Application

For an β„“-ring 𝑅, its β„“-radical β„“-𝑁(𝑅) is defined as ξ€½β„“-𝑁(𝑅)=π‘Žβˆˆπ‘…βˆΆβˆƒπ‘›=𝑛(π‘Ž)suchthatπ‘₯0|π‘Ž|π‘₯1|π‘Ž|β‹―|π‘Ž|π‘₯𝑛=0,βˆ€π‘₯0,…,π‘₯π‘›ξ€Ύβˆˆπ‘…,(6.1) (see [1, Definition, page 45]). β„“-𝑁(𝑅) is the sum of all of the nilpotent β„“-ideals of 𝑅 [1, Theorem  5]. It is clear that for an β„“-algebra 𝐴, β„“-𝑁(𝐴) is an β„“-ideal of 𝐴.

A positive derivation on an β„“-ring 𝑅 is a function π›ΏβˆΆπ‘…β†’π‘… such that for any π‘Ž,π‘βˆˆπ‘…, 𝛿(π‘Ž+𝑏)=𝛿(π‘Ž)+𝛿(𝑏), 𝛿(π‘Žπ‘)=π‘Žπ›Ώ(𝑏)+𝛿(π‘Ž)𝑏, and 𝛿(𝑅+)βŠ†π‘…+. For an Archimedean 𝑓-ring 𝑅, if 𝛿 is a positive derivation on 𝑅, then 𝛿(𝑅)βŠ†β„“-𝑁(𝑅) [2, 3]. Although we believe this result is true for a unital β„“-algebra with a 𝑑-basis, we lack ability to prove it in general. In the following, we prove this result for unital four-dimensional β„“-algebras with a 𝑑-basis based on the classifications obtained in Theorem 2.1.

Theorem 6.1. Let 𝐴 be a unital four-dimensional β„“-algebra with a 𝑑-basis. If 𝛿 is a positive derivation on 𝐴, then 𝛿(𝐴)βŠ†π‘™-𝑁(𝐴).

Proof. We notice first that since 1>0, for any π‘Žβˆˆπ΄ and π›ΌβˆˆπΉ, 𝛿(π›Όπ‘Ž)=𝛼𝛿(π‘Ž) [4]; that is, 𝛿 is actually a linear transformation of 𝐴. Let 1=𝑒1+β‹―+𝑒𝑛, where 𝑒1,…,𝑒𝑛 are distinct elements from the 𝑑-basis and 𝑛≀4. Since 𝛿(1)=𝛿(1)+𝛿(1) implies 𝛿(1)=0, 0=𝛿(𝑒1)+β‹―+𝛿(𝑒𝑛). Then 𝛿 is positive implying that 𝛿(𝑒𝑖)=0 for each 𝑖=1,.…,𝑛. Let π‘Ž be an element in the 𝑑-basis, by Theorem 1.1, either π‘Ž is nilpotent or π‘Žπ‘›π‘Ž=π›Όπ‘Žπ‘’π‘– for some π‘›π‘Žβ‰₯1, 0<π›Όπ‘ŽβˆˆπΉ, and 𝑒𝑖. If π‘Ž is nilpotent, then 𝛿(π‘Ž) is also nilpotent [4]. If π‘Žπ‘›π‘Ž=π›Όπ‘Žπ‘’π‘–, then 𝛿(π‘Žπ‘›π‘Ž)=𝛿(π›Όπ‘Žπ‘’π‘–)=π›Όπ‘Žπ›Ώ(𝑒𝑖)=0 and 𝛿 is positive also implying that 𝛿(π‘Ž) is nilpotent. Thus for each element π‘Ž in the 𝑑-basis, 𝛿(π‘Ž) is nilpotent, so if 𝐴 is β„“-reduced, then 𝛿(π‘Ž)=0 for each π‘Ž in the 𝑑-basis. Therefore 𝛿(𝐴)=0.
We only need to consider those cases in Theorem 2.1 in which 𝐴 is not β„“-reduced. In cases (4)-(5), (12)–(20), and (27)–(35), the β„“-radical consists of all nilpotent elements of the β„“-algebra 𝐴. From the argument in the previous paragraph, for each element π‘₯ in the 𝑑-basis, 𝛿(π‘₯) is nilpotent, so