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

Authors: Mustafa Akkurt; Zainab Alshair; Lena Lamar; Jingjing Ma
Journal: ISRN Algebra
 (2011)

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## Abstract

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


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## Body

## 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-dimensionalℓ-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.LetR be an ℓ-ring. Thepositive cone of R is R+={r∈R:r≥0}. A vector latticeVoverF is an ℓ-group and a vector space over F such that F+V+⊆V+. An ℓ-algebraA over F is an algebra over F, that is, an ℓ-ring and a vector lattice over F. An ℓ-algebra is calledunital 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 {xi:i∈I} is calleddisjoint if each xi>0 and xi∧xj=0 for any two different elements xi and xj. 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.LetA 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+⋯+αpsp, where p≥1, α1,…,αp∈F, and s1,…,sp∈S are distinct, and a≥0 if and only if each αi≥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+⋯+en, where e1,…,en∈S are distinct.In the following theorem, we record some results on unitalℓ-algebras with a d-basis from [5, 6].Theorem 1.1.
LetA be a unital finite-dimensional  ℓ-algebra over  F with a  d-basis  S such that  1=e1+⋯+en, where  n≥1 and  e1,…,en∈S are distinct.   (1)
Eachei is an idempotent f-element and eiej=0 for i≠j.(2)
Leta be a basic element. Then there exists exactly one ei such that eia=a, and for any j≠i, eja=0. Similarly, there exists exactly one ek such that aek=a, and for any j≠k, aej=0.(3)
For each elementa∈S, if eia=a or aei=a, 1≤i≤n, then either a is nilpotent or ana=αaei for some positive integer na and 0<αa∈F.(4)
Leta,b∈S. If either a or b is not nilpotent and ab≠0, then ab is basic.


## 2. The Classification of Unital Four-Dimensionalℓ-Algebras with a d-Basis

We first list someℓ-algebras that we are going to use for the classification of unital four-dimensional ℓ-algebras over F with a d-basis.(i)
LetG be a finite group, and let F[G] be the group algebra over F. Define an element ∑iαigi≥0 if each αi≥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)
Let0<α∈F, and let f(x)=xn-α∈F[x] be irreducible. We use a=αn to denote a fixed root of f(x). Consider the extension field F[αn] of F. Then
(2.1)F[αn]={αn-1an-1+⋯+α1a+α0∣αi∈F},
and {an-1,…,a,1} is a basis of F[αn] over F as a vector space over F. Now define αn-1an-1+⋯+α1a+α0≥0 if each αi≥0, i=0,1,…,n-1. Since an=α>0, F[αn] becomes an ℓ-algebra over F with respect to the above lattice order, and a is a d-element. Thus {an-1,…,a,1} is a d-basis of the ℓ-algebra F[αn] over F. In this paper, n≤4 and F[αn] always denotes the ℓ-algebra defined above.(iii)
Forn≥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 ℓ-algebras. Let eij denote the matrix with ijth entry equal to 1 and other entries equal to 0. Then {eij∣1≤i,j≤n} is a d-basis for Mn(F) over F and {eij∣1≤i≤j≤n} is a d-basis for Tn(F) over F. In the following, Mn(F) and Tn(F) always denote the ℓ-algebras with the above lattice order.All nonisomorphic unital four-dimensionalℓ-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 thatA 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 ei is an idempotent f-element, and for 1≤i,j≤4, i≠j, eiej=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 thed-basis S, respectively.Theorem 2.1.
LetA 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[α3]⊕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[α4], 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

LetA 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 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 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

Ifae1=0, then aei=a for i=2 or 3 by Theorem 1.1(2), and hence a2=(aei)(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 exists0<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:
(3.1)e1be2e3e1e1b00bbe100e200e20e3000e3
Thus A≅F[G]⊕F⊕F, where G is a group of order two. This is (2) in Theorem 2.1.(2)
The polynomialf(x)=x2-α∈F[x] has no root in F. Then we have the following multiplication table for S:
(3.2)e1ae2e3e1e1a00aaαe100e200e20e3000e3
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:
(3.3)e1ae2e3e1e1a00aa000e200e20e3000e3
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:
(3.4)e1e2ae3e1e10a0e20e200a0a00e3000e3
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:
(3.5)e11e22e12e11e110e12e220e220e120e120
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.


## 3.1.A Is ℓ-Reduced

Ifae1=0, then aei=a for i=2 or 3 by Theorem 1.1(2), and hence a2=(aei)(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 exists0<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:
(3.1)e1be2e3e1e1b00bbe100e200e20e3000e3
Thus A≅F[G]⊕F⊕F, where G is a group of order two. This is (2) in Theorem 2.1.(2)
The polynomialf(x)=x2-α∈F[x] has no root in F. Then we have the following multiplication table for S:
(3.2)e1ae2e3e1e1a00aaαe100e200e20e3000e3
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:
(3.3)e1ae2e3e1e1a00aa000e200e20e3000e3
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:
(3.4)e1e2ae3e1e10a0e20e200a0a00e3000e3
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:
(3.5)e11e22e12e11e110e12e220e220e120e120
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

LetA be a unital ℓ-algebra with a d-basis S={e1,e2,a,b} such that 1=e1+e2. By Theorem 1.1, aei=a for some i=1 or 2. Then ei∧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 thate1a=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.SinceA 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:
(4.1)e1ae2be1e1a00aaσe100e200e2bb000λe2
Depending on σ∈F or σ∉F and λ∈F or λ∉F, we have the following cases:
(4.2)A≅F[G]⊕F[G],A≅F[G]⊕F[σ],orA≅F[σ]⊕F[λ],
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:
(4.3)e1aa2e2e1e1aa20aaa2μe10a2a2μe1μabe2000e2
Now it is clear that if μ3∈F, then A≅F[G]⊕F, where G is a group of order three, and if μ3∉F, then A≅F[μ3]⊕F, where F[μ3] 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

SinceA 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 (4.4)a2=α1e1+α2e2+α3a+α4b,αi∈F+.
Let k be the smallest positive integer such that ak=0. Then k≥3 and (4.5)0=ak=a2ak-2=α1e1ak-2+α2e2ak-2+α3ak-1+α4bak-2,
so α1ak-2=α3ak-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

Fromba=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:
(4.6)e1e2a′e2e1e1aa′0e20e20ba′0a′0e1bb0e20
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)
Ifae1=a, e1b=b, and be1=b, then A≅(F1+Fa+Fb)⊕F, where Fa+Fb is the zero ℓ-ring.(2b)
Ifae1=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:
(4.7)e1e2ae1e10ae20e20a0a0
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)
Ife1b=b and be1=b, then A≅T2(F)+Fb with e11b=be11=b, be12=0, and b2=0.(2d)
Ife1b=b and be2=b, then A≅T2(F)+Fb with e11b=b, be22=b, and b2=0.(2e)
Ife2b=b and be1=b, then A≅T2(F)+Fb with e22b=b, be11=b, e12b=be12=0, and b2=0.(2f)
Ife2b=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 caseT={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

Sinceb 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: (4.8)e1ae2be1e1a00aa000e200e2bb00bϵe2From the above table, it is clear that ifϵ∈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 onA such that 1 is a sum of two disjoint basic elements.


## 4.1.A Is ℓ-Reduced

We may assume thate1a=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.SinceA 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:
(4.1)e1ae2be1e1a00aaσe100e200e2bb000λe2
Depending on σ∈F or σ∉F and λ∈F or λ∉F, we have the following cases:
(4.2)A≅F[G]⊕F[G],A≅F[G]⊕F[σ],orA≅F[σ]⊕F[λ],
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:
(4.3)e1aa2e2e1e1aa20aaa2μe10a2a2μe1μabe2000e2
Now it is clear that if μ3∈F, then A≅F[G]⊕F, where G is a group of order three, and if μ3∉F, then A≅F[μ3]⊕F, where F[μ3] 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

SinceA 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 (4.4)a2=α1e1+α2e2+α3a+α4b,αi∈F+.
Let k be the smallest positive integer such that ak=0. Then k≥3 and (4.5)0=ak=a2ak-2=α1e1ak-2+α2e2ak-2+α3ak-1+α4bak-2,
so α1ak-2=α3ak-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

Fromba=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:
(4.6)e1e2a′e2e1e1aa′0e20e20ba′0a′0e1bb0e20
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)
Ifae1=a, e1b=b, and be1=b, then A≅(F1+Fa+Fb)⊕F, where Fa+Fb is the zero ℓ-ring.(2b)
Ifae1=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:
(4.7)e1e2ae1e10ae20e20a0a0
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)
Ife1b=b and be1=b, then A≅T2(F)+Fb with e11b=be11=b, be12=0, and b2=0.(2d)
Ife1b=b and be2=b, then A≅T2(F)+Fb with e11b=b, be22=b, and b2=0.(2e)
Ife2b=b and be1=b, then A≅T2(F)+Fb with e22b=b, be11=b, e12b=be12=0, and b2=0.(2f)
Ife2b=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 caseT={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

Sinceb 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: (4.8)e1ae2be1e1a00aa000e200e2bb00bϵe2From the above table, it is clear that ifϵ∈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 onA such that 1 is a sum of two disjoint basic elements.


## 4.2.1.a2=b2=0

Fromba=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:
(4.6)e1e2a′e2e1e1aa′0e20e20ba′0a′0e1bb0e20
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)
Ifae1=a, e1b=b, and be1=b, then A≅(F1+Fa+Fb)⊕F, where Fa+Fb is the zero ℓ-ring.(2b)
Ifae1=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:
(4.7)e1e2ae1e10ae20e20a0a0
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)
Ife1b=b and be1=b, then A≅T2(F)+Fb with e11b=be11=b, be12=0, and b2=0.(2d)
Ife1b=b and be2=b, then A≅T2(F)+Fb with e11b=b, be22=b, and b2=0.(2e)
Ife2b=b and be1=b, then A≅T2(F)+Fb with e22b=b, be11=b, e12b=be12=0, and b2=0.(2f)
Ife2b=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 caseT={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

Sinceb 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: (4.8)e1ae2be1e1a00aa000e200e2bb00bϵe2From the above table, it is clear that ifϵ∈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 onA 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

Sincea is not nilpotent, by Theorem 1.1, there exist positive integer na and 0<αa∈F such that ana=αa1. We also assume that na is the smallest positive integer such that ana=αa1. Similarly bnb=αb1 and cnc=αc1, for some positive integers nb, nc, and 0<αb,αc∈F. By Theorem 1.1 again, each ak is a basic element for k≥1.Lemma 5.1.
The subset{1,a,a2,…,ana-1} of  A is a disjoint set of  A.Proof.
Let1≤i<na. Since ai is a basic element, ai=αs for some s∈S and 0<α∈F. Since i<na, s≠1, so 1∧s=0 implies that 1∧ai=0. Now let 0≤u<v<na, then au∧av=au(1∧av-u)=0 by the above argument.SinceA is four-dimensional, na≤4 by Lemma 5.1. We claim that na≠3. Supposing that na=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 αab=αa2, which is a contradiction. Hence na≠3, so na=4 or na=2.By the above analysis, we need to consider two cases: (i) one of the following:na, nb, nc, is 4, and we may assume that na=4; (ii) na=nb=nc=2.


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

In this caseT={1,a,a2,a3} is a d-basis of A over F, and we have the following multiplication table for the elements in T: (5.1)1aa2a311aa2a3aaa2a3αaa2a2a3αa1αaaa3a3αa1αaaαaa2We have the following three cases to be considered.(1)
Ifαa4∈F, then A≅F[G], where G is a cyclic group of order four.(2)
If the polynomialf(x)=x4-αa∈F[x] is irreducible, then A≅F[αa4], which is an ℓ-field extension of F of order four.(3)
The polynomialf(x)=x4-αa∈F[G] is not irreducible and αa4∉F.Suppose thatx4-αa=(x2+a1x+b1)(x2+a2x+b2). Then we have a1+a2=0, b1+b2+a1a2=0, b1a2+b2a1=0, and b1b2=-αa<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=b12=-αa<0, a contradiction. So we must have a1=0, and hence a2=0. Then b1+b2=0 and b1b2=-αa implies that b2=-b1 and b12=b22=αa. We may assume that b1>0. It follows from αa4∉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: (5.2)1caac11caaccc1acaaaacb1cb11acacab11b1cThenA≅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.na=nb=nc=2

ThenS={1,a,b,c} is a d-basis and a2=αa1, b2=αb1, and c2=αc1. 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 αba=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}.Sinceba∧a=ba∧b=0 and ba is basic, ba=β1 or ba=β(ab) for some 0<β∈F. If ba=β1, then αab=βa, which is a contradiction. Thus ba=β(ab), for some 0<β∈F, and hence (5.3)αbαa1=b(ba)a=βb(ab)a=β2(ab)(ba)=β2αbαa1.
So β=1 and the multiplication table for the elements of T is given below: (5.4)1abab11ababaaαa1abαabbbabαb1αbaababαabαbaαaαb1There are the following cases.(1)
αa and αb∈F. Replacing a and b by a/αa and b/αb, respectively, we have a2=b2=1. Therefore A≅F[G], where G is the Klein four group.(2)
αa∉F and αb∈F (or αa∈F and αb∉F). Replacing b by b/αb, we may assume that b2=1. Therefore A≅F[αa][G], where F[αa] is the quadratic extension ℓ-field of F and G is a group of order two.(3)
αa∉F and αb∉F. If αb∈F[αa], then A≅F[αa][G], where G is a group of order two. We omit the checking and leave it to the reader. If αb∉F[αa], then A≅F[αa][αb], where F[αa] is a quadratic extension ℓ-field of F and F[αa][αb] is a quadratic extension ℓ-field of F[αa]. Those are (24), (25), and (26) in Theorem 2.1.


### 5.2.A Is Not ℓ-Reduced

LetS={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.
Letx,y∈{1,a,b,c}. If either  x or  y is nilpotent and the other one is not nilpotent, then  xy is nilpotent.Proof.
Letx be nilpotent and let y be not. By Theorem 1.1, there exist a positive integer ny and 0<αy∈F such that yny=αy1. If xy=0, then xyny=0 implies that αyx=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 nz, 0<αz∈F such that znz=αz1. Let n>1 be the smallest positive integer such that xn=0. Then 0=xny=xn-1(xy)=γxn-1z, and hence 0=γxn-1z implies that 0=γxn-1znz=αzγxn-1. Thus xn-1=0, which is a contradiction. Therefore xy is nilpotent.By Lemma5.2, {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)
Eithera, 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 ak=0. Then 0=ak=ak-2a2=αak-2+βak-2b+γak-2c implies αak-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=akb=δak-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:
(5.5)1abc11abcaab+c00bb000cc000
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. Consider4×4 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:
(5.6)1abab11ababaa0ab0bb000abab000
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:
(5.7)1abab11ababaa0ab0bbβ(ab)00abab000
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 thata 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: (5.8)1caac11caacccα1acαaaaac00acacαa00Ifα∈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.Ifα∉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 Theorem2.1.


## 5.1.A Is ℓ-Reduced

Sincea is not nilpotent, by Theorem 1.1, there exist positive integer na and 0<αa∈F such that ana=αa1. We also assume that na is the smallest positive integer such that ana=αa1. Similarly bnb=αb1 and cnc=αc1, for some positive integers nb, nc, and 0<αb,αc∈F. By Theorem 1.1 again, each ak is a basic element for k≥1.Lemma 5.1.
The subset{1,a,a2,…,ana-1} of  A is a disjoint set of  A.Proof.
Let1≤i<na. Since ai is a basic element, ai=αs for some s∈S and 0<α∈F. Since i<na, s≠1, so 1∧s=0 implies that 1∧ai=0. Now let 0≤u<v<na, then au∧av=au(1∧av-u)=0 by the above argument.SinceA is four-dimensional, na≤4 by Lemma 5.1. We claim that na≠3. Supposing that na=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 αab=αa2, which is a contradiction. Hence na≠3, so na=4 or na=2.By the above analysis, we need to consider two cases: (i) one of the following:na, nb, nc, is 4, and we may assume that na=4; (ii) na=nb=nc=2.


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

In this caseT={1,a,a2,a3} is a d-basis of A over F, and we have the following multiplication table for the elements in T: (5.1)1aa2a311aa2a3aaa2a3αaa2a2a3αa1αaaa3a3αa1αaaαaa2We have the following three cases to be considered.(1)
Ifαa4∈F, then A≅F[G], where G is a cyclic group of order four.(2)
If the polynomialf(x)=x4-αa∈F[x] is irreducible, then A≅F[αa4], which is an ℓ-field extension of F of order four.(3)
The polynomialf(x)=x4-αa∈F[G] is not irreducible and αa4∉F.Suppose thatx4-αa=(x2+a1x+b1)(x2+a2x+b2). Then we have a1+a2=0, b1+b2+a1a2=0, b1a2+b2a1=0, and b1b2=-αa<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=b12=-αa<0, a contradiction. So we must have a1=0, and hence a2=0. Then b1+b2=0 and b1b2=-αa implies that b2=-b1 and b12=b22=αa. We may assume that b1>0. It follows from αa4∉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: (5.2)1caac11caaccc1acaaaacb1cb11acacab11b1cThenA≅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.na=nb=nc=2

ThenS={1,a,b,c} is a d-basis and a2=αa1, b2=αb1, and c2=αc1. 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 αba=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}.Sinceba∧a=ba∧b=0 and ba is basic, ba=β1 or ba=β(ab) for some 0<β∈F. If ba=β1, then αab=βa, which is a contradiction. Thus ba=β(ab), for some 0<β∈F, and hence (5.3)αbαa1=b(ba)a=βb(ab)a=β2(ab)(ba)=β2αbαa1.
So β=1 and the multiplication table for the elements of T is given below: (5.4)1abab11ababaaαa1abαabbbabαb1αbaababαabαbaαaαb1There are the following cases.(1)
αa and αb∈F. Replacing a and b by a/αa and b/αb, respectively, we have a2=b2=1. Therefore A≅F[G], where G is the Klein four group.(2)
αa∉F and αb∈F (or αa∈F and αb∉F). Replacing b by b/αb, we may assume that b2=1. Therefore A≅F[αa][G], where F[αa] is the quadratic extension ℓ-field of F and G is a group of order two.(3)
αa∉F and αb∉F. If αb∈F[αa], then A≅F[αa][G], where G is a group of order two. We omit the checking and leave it to the reader. If αb∉F[αa], then A≅F[αa][αb], where F[αa] is a quadratic extension ℓ-field of F and F[αa][αb] is a quadratic extension ℓ-field of F[αa]. Those are (24), (25), and (26) in Theorem 2.1.


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

In this caseT={1,a,a2,a3} is a d-basis of A over F, and we have the following multiplication table for the elements in T: (5.1)1aa2a311aa2a3aaa2a3αaa2a2a3αa1αaaa3a3αa1αaaαaa2We have the following three cases to be considered.(1)
Ifαa4∈F, then A≅F[G], where G is a cyclic group of order four.(2)
If the polynomialf(x)=x4-αa∈F[x] is irreducible, then A≅F[αa4], which is an ℓ-field extension of F of order four.(3)
The polynomialf(x)=x4-αa∈F[G] is not irreducible and αa4∉F.Suppose thatx4-αa=(x2+a1x+b1)(x2+a2x+b2). Then we have a1+a2=0, b1+b2+a1a2=0, b1a2+b2a1=0, and b1b2=-αa<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=b12=-αa<0, a contradiction. So we must have a1=0, and hence a2=0. Then b1+b2=0 and b1b2=-αa implies that b2=-b1 and b12=b22=αa. We may assume that b1>0. It follows from αa4∉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: (5.2)1caac11caaccc1acaaaacb1cb11acacab11b1cThenA≅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.na=nb=nc=2

ThenS={1,a,b,c} is a d-basis and a2=αa1, b2=αb1, and c2=αc1. 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 αba=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}.Sinceba∧a=ba∧b=0 and ba is basic, ba=β1 or ba=β(ab) for some 0<β∈F. If ba=β1, then αab=βa, which is a contradiction. Thus ba=β(ab), for some 0<β∈F, and hence (5.3)αbαa1=b(ba)a=βb(ab)a=β2(ab)(ba)=β2αbαa1.
So β=1 and the multiplication table for the elements of T is given below: (5.4)1abab11ababaaαa1abαabbbabαb1αbaababαabαbaαaαb1There are the following cases.(1)
αa and αb∈F. Replacing a and b by a/αa and b/αb, respectively, we have a2=b2=1. Therefore A≅F[G], where G is the Klein four group.(2)
αa∉F and αb∈F (or αa∈F and αb∉F). Replacing b by b/αb, we may assume that b2=1. Therefore A≅F[αa][G], where F[αa] is the quadratic extension ℓ-field of F and G is a group of order two.(3)
αa∉F and αb∉F. If αb∈F[αa], then A≅F[αa][G], where G is a group of order two. We omit the checking and leave it to the reader. If αb∉F[αa], then A≅F[αa][αb], where F[αa] is a quadratic extension ℓ-field of F and F[αa][αb] is a quadratic extension ℓ-field of F[αa]. Those are (24), (25), and (26) in Theorem 2.1.


## 5.2.A Is Not ℓ-Reduced

LetS={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.
Letx,y∈{1,a,b,c}. If either  x or  y is nilpotent and the other one is not nilpotent, then  xy is nilpotent.Proof.
Letx be nilpotent and let y be not. By Theorem 1.1, there exist a positive integer ny and 0<αy∈F such that yny=αy1. If xy=0, then xyny=0 implies that αyx=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 nz, 0<αz∈F such that znz=αz1. Let n>1 be the smallest positive integer such that xn=0. Then 0=xny=xn-1(xy)=γxn-1z, and hence 0=γxn-1z implies that 0=γxn-1znz=αzγxn-1. Thus xn-1=0, which is a contradiction. Therefore xy is nilpotent.By Lemma5.2, {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)
Eithera, 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 ak=0. Then 0=ak=ak-2a2=αak-2+βak-2b+γak-2c implies αak-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=akb=δak-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:
(5.5)1abc11abcaab+c00bb000cc000
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. Consider4×4 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:
(5.6)1abab11ababaa0ab0bb000abab000
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:
(5.7)1abab11ababaa0ab0bbβ(ab)00abab000
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 thata 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: (5.8)1caac11caacccα1acαaaaac00acacαa00Ifα∈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.Ifα∉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 Theorem2.1.


## 5.2.1.a, b, and c Are All Nilpotent

(1)
Eithera, 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 ak=0. Then 0=ak=ak-2a2=αak-2+βak-2b+γak-2c implies αak-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=akb=δak-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:
(5.5)1abc11abcaab+c00bb000cc000
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. Consider4×4 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:
(5.6)1abab11ababaa0ab0bb000abab000
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:
(5.7)1abab11ababaa0ab0bbβ(ab)00abab000
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 thata 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: (5.8)1caac11caacccα1acαaaaac00acacαa00Ifα∈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.Ifα∉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 Theorem2.1.


## 6. An Application

For anℓ-ring R, its ℓ-radical ℓ-N(R) is defined as (6.1)l-N(R)={a∈R:∃n=n(a)suchthatx0|a|x1|a|⋯|a|xn=0,∀x0,…,xn∈R},
(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.Apositive 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.
LetA 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 since1>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+⋯+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)+⋯+δ(en). Then δ is positive implying that δ(ei)=0 for each i=1,.…,n. Let a be an element in the d-basis, by Theorem 1.1, either a is nilpotent or ana=αaei for some na≥1, 0<αa∈F, and ei. If a is nilpotent, then δ(a) is also nilpotent [4]. If ana=αaei, then δ(ana)=δ(αaei)=αaδ(ei)=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 Theorem2.1 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 whichℓ-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,
(6.2)(e12+e21)δ(e12+e21)+δ(e12+e21)(e12+e21)=0,
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.



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*Source: 650969-2011-06-29.xml*