ABSTRACT Finitary Incidence Algebras Bradley M Wagner, Ph.D

ABSTRACT

Finitary Incidence Algebras

Bradley M Wagner, Ph.D.

Advisor: Manfred Dugas, Ph.D.

Let P be an arbitrary partially ordered set and I(P ) its incidence space. Then

F (P ) is the ﬁnitary incidence algebra and I(P ) is a bimodule over it. Consequently we can form D(P ) = FI(P ) ⊕ I(P ) the idealization of I(P ). In this paper we will study the automorphisms of FI(P ) and D(P ). We will also explore suﬃcient conditions for FI(P ) to be zero product determined. Finitary Incidence Algebras

Bradley M Wagner, B.S., M.S.

A Dissertation

Approved by the Department of Mathematics

Lance L. Littlejohn, Ph.D., Chairperson

Submitted to the Graduate Faculty of Baylor University in Partial Fulﬁllment of the Requirements for the Degree of Doctor of Philosophy

Approved by the Dissertation Committee

Manfred Dugas, Ph.D., Chairperson

David Arnold, Ph.D.

Markus Hunziker, Ph.D.

Lorin S Matthews, Ph.D.

Mark R Sepanski, Ph.D.

Accepted by the Graduate School May 2014

J. Larry Lyon, Ph.D., Dean

ACKNOWLEDGMENTS v

1 Introduction 1

2 Finitary Incidence Algebras 3

2.1 Deﬁnition and Basics ...... 3

2.2 Examples ...... 5

2.3 Prior Results ...... 7

2.4 Ring Theoretic Properties ...... 8

3 Derivations of FI(P ) 15

4 Automorphisms of FI(P ) 24

5 The Idealization of FI(P ) 34

5.1 Deﬁnition ...... 34

5.2 Basics ...... 34

6 Automorphisms of D(P ) 39

7 Zero Product Determined Algebras 51

7.1 Prior Results ...... 51

7.2 Cartesian Products of Zero Product Determined Algebras ...... 52

7.3 FI(P ) as a ZPD Algebra ...... 60

8 Future Research 63

BIBLIOGRAPHY 64

iv ACKNOWLEDGMENTS

This dissertation would never have been possible without the many individuals whose encouragment and support have carried me through the last 5 years.

To begin, I need to thank my Lord and Savior. Apart from His constant sustaining grace none of this would be possible. Next, I would like to thank my adviser, Dr. Manfred Dugas. His patience and dedication towards this project were fundamental to its completion. Further, I would like to thank the remaining members of my committee - Dr. David Arnold, Dr. Markus Hunziker, Dr. Lorin

Matthews, and Dr. Mark Sepanski. I am grateful for their willingness to give up their valuable time to assist me. I would also like to thank the rest of the staﬀ and faculty of the math department. Throughout the last 5 years, their support, advice, and friendship have been integral to my graduate existence.

I would be remiss in not thanking my fellow graduate students. The cama- raderie we shared made the past 5 years easier. From going to sporting events together to simply talking in the graduate lounge during lunch, the friendships created will continue to inﬂuence me as I leave Baylor. Particularly, I would like to thank my cohort. Our weekly homework meetings were not only necessary to completing that ﬁrst year, but were also a source of strength as we struggled together.

Finally, apart from my Lord, my family has been my next greatest source of strength, not only during these last 5 years but throughout my life. I thank my parents for their support and encouragement, and for instilling in me the desire to learn. I would like to thank my wife for immeasurable support in all forms. She has made many sacriﬁces in helping me pursue this goal. I would like to thank my new son in giving me the motivation to push through this last year so that he might one day be proud of his father.

v CHAPTER ONE

Introduction

The study of incidence spaces is the study of vectors over some ring R indexed by intervals of the form [x, y] for which x, y ∈ P , where P is some partially ordered set with the condition that x ≤ y. In the case where P is locally finite, this incidence space I(P ) becomes an associative algebra and a good deal is known about these types of spaces (see [7]). However, in the case of P being not locally finite, a multiplication couldn’t be defined in any obvious way.

In [5], Khripchenko and Novikov provided a solution to this problem. They defined a finitary series. Let P be any poset with x, y ∈ P and F a field and

α(x, y): P × P → F such that α(x, y) = 0 whenever x > y or x and y are not comparable. Then a ﬁnitary series is a formal sum

X α = α(x, y)[x, y], x,y∈P with certain restrictions on subintervals. They determined that the collection of all such sums formed an F -algebra with multiplication deﬁned by convolution and they called this algebra FI(P ): the ﬁnitary incidence algebra of P over F . Note that

I(P ) is a natural FI(P )-bimodule.

Armed with this definition they went on to show that every α in FI(P ) for which α(x, x) 6= 0 for all x ∈ P , is invertable with α−1 ∈ FI(P ). They also showed that the Jacobson Radical consists of all α ∈ FI(P ) such that α(x, x) = 0 for all x ∈ P . Finally, they determined that FI(P ) ∼= FI(Q) implies P ∼= Q. The final corollary of Khripchenko and Novikov in [5] shows that the study of FI(P ) is not redundant to the study of I(P ) for P locally finite.

The dissertation is organized as follows. In Chapter One, we develop the

finitary incidence algebra. We present the definition along with known results. We 1 also provide some examples to stimulate an intuitive understanding of the structure of an incidence algebra (finitary or otherwise). Finally we discuss ring theoretic properties of FI(P ) and how the structure of P affects the relationship between regular and invertable elements.

In Chapter Three, we develop derivations of FI(P ). We give the definition of a derivation of FI(P ) and look at several types. Finally we provide a decomposition of the space of all derivations. The techniques used in this section mirror those used by Spiegel and O’Donnell in [7], the only difference being that we now allow P to be an arbitrary poset rather than locally finite, and we are working over a field F rather than a commutative ring R. We show that all major results carry over from the locally finite to the arbitrary case without change.

In Chapter Four, we look at automorphisms of FI(P ) and provide a general decomposition of an arbitrary automorphisms as a composition of three types of automorphisms. Although the techniques used in this section again follow in the spirit of [7] much more adaptation is needed. The end result shown in [7] is again achieved but not without adopting techniques used by Khripchenko in [6].

In Chapters Five and Six we study the Nagata idealization, D(P ), of FI(P ).

In Chapter Five, we provide the deﬁnition and some basic results. In Chapter Six we study isomorphisms of D(P ). We use results from Chapter Four to again provide a decomposition of an arbitrary isomorphism of D(P ). We also show that D(P ) and

D(Q) are isomorphic as F -algebras if and only if P and Q are isomorphic as posets.

In Chapter Seven, we turn our attention to zero product determined algebras.

We start with a deﬁnition and some known results. We look at cartesian products of zero product determined algebras in order to determine under what conditions the product remains zero product determined. We then apply these more general results to the case of incidence algebras.

2 CHAPTER TWO

Finitary Incidence Algebras

2.1 Deﬁnition and Basics

Let P be an arbitrary partially ordered set and F a ﬁeld.

Deﬁnition 2.1. [5] Denote by I(P ) the set of formal sums of the form

X α = α(x, y)[x, y], (2.1) x≤y where x, y ∈ P , α(x, y) ∈ F ,[x, y] = {z ∈ P : x ≤ z ≤ y}. In the general case I(P ) is not an algebra, so we call it an incidence space.

Definition 2.2. A poset P is called locally finite if [x, y] is finite for all x, y ∈ P with x ≤ y.

Given α, β ∈ I(P ) we deﬁne

X (αβ)(x, y) = α(x, z)β(z, y). x≤z≤y

If P is locally finite, this multiplication is always defined and I(P ) is an algebra. P Locally finite or otherwise it is easy to see that e := x∈P [x, x] acts as the identity on I(P ).

While incidence spaces have been studied in the case where P is locally finite, the following definition from Khripchenko and Novikov in [5] has allowed for the extension of the idea to non-locally finite posets:

Definition 2.3. A formal sum 2.1 is called a finitary series if for any x, y ∈ P , x < y, there is only a finite number of subsegments [u, v] ⊂ [x, y] such that u 6= v and

α(u, v) 6= 0. The set of all ﬁnitary series is denoted by FI(P ).

3 Using this deﬁnition Khripchenko and Novikov then proved the following:

Theorem 2.1. [5] FI(P ) is an associative algebra and I(P ) is a bimodule over it.

We now introduce some useful notation:

Notation 1. Let δxy : P × P → F be deﬁned by   1 if u = x, v = y and x ≤ y, δxy(u, v) =  0 otherwise.

Hence δxy can be thought of as the element of I(P ) consisting of the single interval

[x, y] with coeﬃcient 1. We will use the further notation δxx := ex and then the P identity e can be written e = x∈P ex.

We see then that δxy = exδxyey which we will ﬁnd useful later on.

Remark 2.1. It should be noted that neither FI(P ) nor I(P ) are commutative:

δxyey = δxy 6= 0 = eyδxy.

Our ﬁrst minor result:

Theorem 2.2. If α is in the center of FI(P ) then α = fe for some f ∈ F .

Proof. Let x ∈ P . Since α is in the center of FI(P ) then αex = exα. Clearly

X αex = α(u, x)[u, x], u≤x and X exα = α(x, v)[x, v]. x≤v Since the only overlapping terms occur at [x, x], for all x 6= y we have that α(x, y) =

0. Hence X α = α(x, x)[x, x]. x∈P 4 Next, since αδxy = δxyα we have

αδxy = α(x, x)[x, y], and

δxyα = α(y, y)[x, y], so that α(x, x) = α(y, y) for all x, y ∈ P .

Remark 2.2. Similarly if α ∈ FI(P ) such that αγ = γα for all γ ∈ I(P ) then α = fe for some f ∈ F .

2.2 Examples

In this section we will see several examples useful in developing an intuition for the structure of FI(P ). Although the theorems in this paper will often be for an arbitrary poset, many of the examples we look at will be for ﬁnite (and thus locally

ﬁnite) posets.

Example 2.1. Many examples will be for Pn being ﬁnite and linearly ordered, or ∼ equivalently Pn = {1, 2, . . . , n} with the natural order. In this case, an element of FI(P ) would take the form of the n × n upper triangular matrices with entries from

F :

  f11 f12 ··· f1n      0 f22 ··· f2n     . . . .   ......      0 ··· 0 fnn

Since the elements of a poset may not always be linearly ordered, a Hasse diagram can be useful in expressing the relationship between elements.

5 Example 2.2. Take the set {1, 2, 3, 4} along with the partial order illustrated by the

Hasse diagram:

4 .

2 3

1 Then elements of the F -algebra FI(P ) would take the form

  f11 f12 f13 f14      0 f22 0 f24       0 0 f f   33 34    0 0 0 f44

with fij ∈ F .

Although ﬁnite examples are useful for visualization, a ﬁnitary incidence algebra can be formed from any poset.

Example 2.3. Let P be the positive integers with the normal ordering. Then FI(P ) can be though of as an inﬁnite upper triangular matrix:

  f11 f22 f13 ···      0 f22 f23 ···       0 0 f ···   33   . . . .  ......

In the case where P consists of multiple connected components, P = ∪i∈I Pi, then FI(P ) may be written as a direct product over said components:

Y FI(P ) = FI(Pi). i∈I 6 Then, when visualizing FI(P ) as a matrix, said matrix takes on a block matrix form.

Example 2.4. Let P consists of two connected components P = {1, 2, 3} ∪ {4, 5} whose order relation can be visualized as

2 5

1 4 Then FI(P ) can be written as the 5 × 5 block matrix:

  f f f 0 0  11 12 13     0 f f 0 0   22 23       0 0 f33 0 0       0 0 0 f f   44 45    0 0 0 0 f55 for which the ﬁrst block represents the component {1, 2, 3} and the second block represents the component {4, 5}.

2.3 Prior Results

We would like to quickly review the more important results ﬁrst discovered by

Khripchenko and Novikov in [5]. Their ﬁrst results was obviously to show that their

ﬁnitary construction did, in fact, form an algebra.

Theorem 2.3. [5] FI(P ) is an associative algebra and I(P ) is a bimodule over it.

Next came a few basic preliminary results:

Theorem 2.4. [5] A series α ∈ FI(P ) is invertible if and only if α(x, x) 6= 0 for each x ∈ P . Moreover, α−1 ∈ FI(P ). 7 Corollary 2.1. [5] α ∈ J(FI(P )) if and only if α(x, x) = 0 for all x ∈ P .

Their ﬁnal set of results culminated in a solution to the isomorphism problem:

Theorem 2.5. [5] Each idempotent α ∈ FI(P ) is conjugate to the diagonal idempotent ε, such that ε(x, x) = α(x, x) for all x ∈ P .

Remark 2.3. In this paper, since we will be studying FI(P ) over a ﬁeld F , this will imply that α(x, x) ∈ {0, 1} for α an idempotent in FI(P ).

Deﬁnition 2.4. An idempotent α 6= 0 is primitive, if α = α = for some idempotent

implies that is equal to 0 or α.

Lemma 2.1. [5] An idempotent α is primitive if and only if it is conjugate to ex for some x ∈ P .

Remark 2.4. [5] The x ∈ P in the previous lemma is unique. That is, if ex is conjugate to ey then x = y.

Theorem 2.6. [5] Let P and Q be arbitrary partially ordered sets. Then

FI(P ) ∼= FI(Q) ⇒ P ∼= Q.

Corollary 2.2. [5] Let P be not locally ﬁnite. Then there is no locally ﬁnite partially ordered set Q, such that FI(P ) ∼= I(Q).

Throughout this paper we will make use of these results.

2.4 Ring Theoretic Properties

We ﬁnish this chapter with a discussion of the ring theoretic properties of

FI(P ) and their relationship to the structure of P .

Deﬁnition 2.5. An element s in a ring T with identity is regular if s is not a zero divisor of T ; that is, for all t ∈ T with t 6= 0 we have st 6= 0 6= ts.

8 Deﬁnition 2.6. For α ∈ FI(P ), deﬁne T0(α) = {x ∈ P : α(x, x) = 0}.

Lemma 2.2. Let α ∈ FI(P ). Suppose that there is some element v0 ∈ T0(α) such that v0 is minimal in T0(α). Then there exists some 0 6= β ∈ FI(P ) such that αβ = 0. That is, α is a left zero divisor.

Proof. Define β(v0, v0) = 1 and β(x, v) = 0 for all x ≤ v 6= v0. What remains to be defined are the entries β(x, v0) for all x < v0. For u < v ∈ P we define

Cα(u, v) = {[x, y]: u ≤ x < y ≤ v, α(x, y) 6= 0}. Note that Cα(u, v) is a ﬁnite set since α ∈ FI(P ). We will deﬁne β(u, v0) by induction over the size of Cα(u, v0) such that the equation

X α(u, x)β(x, v0) = 0 (2.2)

u≤x≤v0 holds.

To start our induction, assume that Cα(u, v0) = ∅. In this case equation 2.2 reduces to

α(u, u)β(u, v0) = 0

which we can obtain by setting β(u, v0) = 0. Let n ∈ N and assume that we have deﬁned β(u, v0) for all u < v0 with |Cα(u, v0)| < n such that 2.2 holds.

Let |Cα(u, v0)| = n. If α(u, x) = 0 for all u < x ≤ v0, then 2.2 reduces to

α(u, u)β(u, v0) = 0 and we may deﬁne β(u, v0) = 0. If not there exists u < x0 < v0 such that α(u, x0) 6= 0. Then [u, x0] ∈ Cα(u, v0) and Cα(x0, v0) ⊂ Cα(u, v0), but

[u, x0] ∈/ Cα(x0, v0). This shows that |Cα(x0, v0)| < |Cα(u, v0)| = n. By the induction hypothesis we have that the term

X α(u, x)β(x, v0),

X X α(u, x)β(x, v0) = α(u, u)β(u, v0) + α(u, x)β(x, v0) = 0.

u≤x≤v0 u

We infer that 0 6= β ∈ I(P ) and αβ = 0.

It remains to be shown that β ∈ FI(P ). Since β(x, v) = 0 for all x ≤ v 6= v0 we need only be concerned with intervals of the form [u, v0]. By way of contradiction, assume that there exists some u0 < v0 and ui ∈ [u0, v0] such that β(ui, v0) 6= 0 for all i ∈ N. Now consider 2.2 for u = ui:

X α(ui, ui)β(ui, v0) + α(ui, x)β(x, v0) = 0.

Since ui < v0 we know that α(ui, ui) 6= 0 and thus α(ui, ui)β(ui, v0) 6= 0. This implies that for all i ∈ N there is some x = xi such that α(ui, xi) 6= 0 and u0 < ui < xi ≤ v0. But this would mean α∈ / FI(P ), a contradiction. We conclude that β ∈ FI(P ).

By duality we have:

Lemma 2.3. Let α ∈ FI(P ). Suppose that there is some element v0 ∈ T0(α) such that v0 is maximal in T0(α). Then there exists some 0 6= β ∈ FI(P ) such that βα = 0. That is, α is a right zero divisor.

Remark 2.5. The previous two lemmas show that if α ∈ FI(P ) is regular, then T0(α) has neither minimal nor maximal elements.

In the case of P being locally ﬁnite and the incidence space being generated over a ﬁeld F it is known that:

Theorem 2.7. [7] All regular elements of I(P ) are units if and only if Z is not embeddable in P .

Example 2.5. Let P = Z ∪ {x∗} where x∗ is not comparable to any other element of

P . Note that P in this is case is locally ﬁnite so that I(P ) = FI(P ). Clearly Z is 10 embedabble in P . Now deﬁne φ ∈ FI(P ) by   1 if x = y = x∗,   φ(x, y) = 1 if x, y ∈ Z with x = xi and y = xi+1 for some i,    0 otherwise. Let β be some nonzero element of FI(P ). Then there exists some x ≤ y ∈ FI(P ) with β(x, y) 6= 0. If y∈ / Z then (βφ)(x, y) = β(x, y) 6= 0. If y = xi ∈ Z then

(βφ)(x, xi+1) = β(x, xi) = β(x, y) 6= 0. Similarly, if x∈ / Z then (φβ)(x, y) =

β(x, y) 6= 0 while if x = xi ∈ Z then (φβ)(xi−1, y) = β(xi, y) = β(x, y) 6= 0. Hence φβ 6= 0 6= βφ and φ is regular. However, ϕ is clearly not a unit since all its diagonal terms are zero except at x∗.

We would desire that the previous theorem extend to the FI(P ) case for non- locally ﬁnite posets.

Theorem 2.8. Let P be a partially ordered set. If Z is not embeddable in P then all regular elements of FI(P ) are units.

Proof. Let α ∈ FI(P ) be regular. If T0(α) = ∅, then for each x ∈ P , the element

α(x, x) is non-zero. Hence by Theorem 2 of [5], α is a unit in FI(P ). If T0(α) 6= ∅, then T0(α) contains neither a maximal nor minimal element by Remark 2.5. Hence

Z is embeddable in T0(α) ⊂ P , a contradiction. Thus all regular elements are units.

With stronger restrictions on P we can make this a biconditional.

Theorem 2.9. Let P be a partially ordered set in which, given any interval [a, b] then

[a, b] does not contain an inﬁnite chain of elements of P . Then all regular elements of FI(P ) are units if and only if Z is not embeddable in P .

Remark 2.6. Note that this is not equivalent to P being locally ﬁnite. P could have an inﬁnite number of non-comparable entries, for example:

zi ∈ [a, b], ∀i ∈ Z but zi 6≤ zj for i 6= j. 11 i.e.

z1 z2 z3 ······

a Proof. The forward direction follows from the previous theorem.

For the reverse direction assume that all regular elements of FI(P ) are units.

We must verify that Z is not embeddable in P . Suppose, the contrary, that for each i ∈ Z, there exists xi ∈ P with xi < xj when i < j. Write U = {xi : i ∈ Z} as our embedding of Z in P and let φ be deﬁned by   1 if x = y ∈ P \U,   φ(x, y) = 1 if x, y ∈ U with x = xi and y = xi+1 for some i,    0 otherwise.

Note that φ ∈ FI(P ) since any interval [u, v] will contain at most a ﬁnite number of subintervals of the form [xi, xi+1] since by hypothesis [u, v] does not contain an inﬁnite chain. Furthermore, note that T0(φ) = U and φ is not a unit in FI(P ). Let β be some nonzero element of FI(P ). Then there exists some x ≤ y ∈ FI(P ) with β(x, y) 6= 0. If y∈ / U then (βφ)(x, y) = β(x, y) 6= 0. If y = xi ∈ U then

(βφ)(x, xi+1) = β(x, xi) = β(x, y) 6= 0. Similarly, if x∈ / U then (φβ)(x, y) =

β(x, y) 6= 0 while if x = xi ∈ U then (φβ)(xi−1, y) = β(xi, y) = β(x, y) 6= 0. Hence φβ 6= 0 6= βφ and φ is regular and also not a unit. This is a contradiction and the result is established.

Unfortunately a generalized converse to Theorem 2.8 does not exist (and thus neither does a generalization of Theorem 2.7). We provide the following counterex- ample:

12 Example 2.6. Let P = Z ∪ {x0, x1} where x0 ≤ x ∈ P and x1 ≥ x ∈ P ; i.e. x0 is the minimum element of P and x1 is the maximum element of P :

x0 ··· −2 −1 0 1 2 ··· x1

Note that Z is clearly embeddable in P and P is not locally ﬁnite. We will show that every regular element is a unit; or rather, we will show that every non-unit is a zero divisor.

Let ϕ ∈ FI(P ) such that there exists some t ∈ P with ϕ(t, t) = 0; that is, ϕ is not a unit. If T0(ϕ) has either a minimal or maximal element then ϕ is a left or right zero divisor respectively.

Suppose T0(ϕ) 6= ∅ has neither a maximal nor minimal element. Since ϕ ∈

FI(P ) then ϕ(x, y) 6= 0 for at most a ﬁnite number of subintervals [x, y] ⊂ [x0, x1] with x 6= y. Thus {x ∈ P : ϕ(x, y) 6= 0, x 6= x0, x1, x 6= y} is ﬁnite and has a

∗ ∗ minimal element, call itx ˆ. Choose x ∈ T0(ϕ) with x < xˆ which exists since T0(ϕ) has no minimal element. Hence ϕ(x∗, y) = 0 for all y ∈ P . Deﬁne α ∈ FI(P ) by

α(x∗, x∗) = 1 and α(x, y) = 0 for all x ∈ P with x 6= x∗; i.e. α = [x∗, x∗]. Then

X (αϕ)(x, y) = α(x, z)ϕ(z, y), x≤z≤y

= α(x∗, x∗)ϕ(x∗, y),

= ϕ(x∗, y).

But ϕ(x∗, y) = 0. Thus ϕ is a right zero divisor.

Similarly it can be shown ϕ is a left zero divisor as well. In either case ϕ is a zero divisor and our example is complete.

Since we cannot ﬁnd a general converse to Theorem 2.8 we must try to determine the nature of any embedding of Z in P for which all regular elements are units in FI(P ).

13 Theorem 2.10. Let P be a partial ordered set. If all regular elements are units in

FI(P ), then any embedding, U of Z in P must be bounded either above or below.

Proof. Let U be an embedding of Z in P . Now suppose that U is neither bounded above or below. Let φ be deﬁned by   1 if x = y ∈ P \U,   φ(x, y) = 1 if x, y ∈ U with x = xi and y = xi+1 for some i,    0 otherwise.

Then, given any interval [x, y] since x is not a lower bound for U and y is not an upper bound for U,[x, y] will contain at most a ﬁnite number of subintervals of the form (xi, xi+1) with xi, xi+1 ∈ U and hence φ ∈ FI(P ). Furthermore, note that T0(φ) = U and φ is not a unit in FI(P ). Let β be some nonzero element of FI(P ). Then there exists some x ≤ y ∈ FI(P ) with β(x, y) 6= 0. If y∈ / U then

(βφ)(x, y) = β(x, y) 6= 0. If y = xi ∈ U then (βφ)(x, xi+1) = β(x, xi) = β(x, y) 6=

0. Similarly, if x∈ / U then (φβ)(x, y) = β(x, y) 6= 0 while if x = xi ∈ U then

(φβ)(xi−1, y) = β(xi, y) = β(x, y) 6= 0. Hence φβ 6= 0 6= βφ and φ is regular and also not a unit. This is a contradiction and the result is established.

Corollary 2.3. Let P be a partially ordered set. If all regular elements are units then either Z is not embeddable in P or, if Z is embeddable in P then its embedding is bounded either above or below.

14 CHAPTER THREE

Derivations of FI(P )

In [7] Spiegel and O’Donnell studied derivations of I(P ) with P being locally

finite and the incidence algebra being generated over a commutative ring with identity. As it turns out, many of their proofs can be applied to the finitary case over an arbitrary poset and a field with little to no change. Thus, all but the last lemma in this chapter is adapted from their book with changes as necessary.

Deﬁnition 3.1. D : FI(P ) −→ I(P ) is a derivation if for all α1, α2 ∈ FI(P ) and f ∈ F ,

(1) D(α1 + α2) = D(α1) + D(α2),

(2) D(f · α1) = f · D(α1); and

(3) D(α1 · α2) = D(α1) · α2 + α1 · D(α2).

We will denote the set of all derivations of FI(P ) by D(FI(P )).

Observe that if D1 and D2 are derivations of FI(P ), then so are D1 + D2 and f · D1, for all f ∈ F . Thus D(FI(P )) is a left F -module.

For β ∈ I(P ), deﬁne dβ : FI(P ) −→ I(P ) by

dβ(α) = α · β − β · α,

for each α ∈ FI(P ). If α1, α2 ∈ FI(P ) then

dβ(α1 + α2) = (α1 + α2) · β − β · (α1 + α2)

= (α1 · β − β · α1) + (α2 · β − β · α2)

= dβ(α1) + dβ(α2).

15 Furthermore, for f ∈ F we have

dβ(f · α1) = (f · α1) · β − β · (f · α1)

= f · (α1 · β − β · α1)

= f · dβ(α1).

Finally,

dβ(α1 · α2) = (α1 · α2) · β − β · (α1 · α2)

= (α1 · β · α2) − (β · α1 · α2) + (α1 · α2 · β) − (α1 · β · α2)

= dβ(α1) · α2 + α1 · dβ(α2).

Therefore, dβ is a derivation of FI(P ). Formally:

Deﬁnition 3.2. The collection of dβ, for β ∈ I(P ), is the set of inner derivations and is denoted by Inn(FI(P )).

If β1, β2 ∈ I(P ) and f ∈ F , then it is straightforward to show that dβ1+β2 =

dβ1 + dβ2 and df·β1 = f · dβ1 . Thus Inn(FI(P )) a submodule of D(FI(P )).

Deﬁnition 3.3. For α, β ∈ I(P ) we deﬁne the Hadamard product of α and β, denoted by α ? β, to be

(α ? β)(x, y) = α(x, y) · β(x, y) for each pair of elements x, y ∈ P with x ≤ y.

Thus the Hadamard product is simply entrywise multiplication. Note that if

α ∈ FI(P ) and β ∈ I(P ) then α ? β = β ? α ∈ FI(P ).

Deﬁnition 3.4. σ ∈ I(P ) is called additive if

σ(x, z) = σ(x, y) + σ(y, z) for all x ≤ y ≤ z.

16 Remark 3.1. Notice that Deﬁnition 3.4 implies that if σ is additive, then σ(x, x) = 0 for each x ∈ P .

Suppose that σ ∈ I(P ) is additive. We deﬁne Lσ : FI(P ) −→ I(P ), by

Lσ(α) = σ ? α, for all α ∈ FI(P ). In fact, since multiplication is component-wise, any zeros in α will be propagated to the image of α and hence the image is in FI(P ). If α, β ∈ FI(P ) and f ∈ F , then Lσ(α + β) = Lσ(α) + Lσ(β) and Lσ(f · α) = f · Lσ(α). We now check the last condition for Lσ to be a derivation. For α, β ∈ FI(P ) and x, y ∈ P , the left hand side yields

(Lσ(α · β))(x, y) = (σ ? (α · β))(x, y)

= σ(x, y) · (α · β)(x, y) X = σ(x, y) · α(x, z) · β(z, y). x≤z≤y We compare this to the right hand side:

(Lσ(α) · β + α · Lσ(β))(x, y) = ((σ ? α) · β)(x, y) + (α · (σ ? β))(x, y) X = (σ ? α)(x, z) · β(z, y) x≤z≤y X + α(x, z) · (σ ? β)(z, y) x≤z≤y X = σ(x, z) · α(x, z) · β(z, y) x≤z≤y X + α(x, z) · σ(z, y) · β(z, y) x≤z≤y X = α(x, z) · β(z, y) · [σ(x, z) + σ(z, y)] x≤z≤y X = σ(x, y) · α(x, z) · β(z, y). x≤z≤y

Hence, the two expressions are equal and Lσ is a derivation. 17 If σ and τ are additive functions in I(P ) and f ∈ F , then the functions σ + τ and f · σ are additive. Hence, Lσ+τ = Lσ + Lτ and Lf·σ = f · Lσ. Therefore, the collection of Lσ, such that σ is additive, is a submodule of D(FI(P )). We deﬁne this formally:

Deﬁnition 3.5. The collection of Lσ, such that σ is additive are called the additive derivations and denoted by Ad(FI(P )).

The next proposition describes a property by which we may identify an additive derivation.

Proposition 3.1. [7] A derivation D of FI(P ) is additive if and only if D(ex) = 0 for each x ∈ P .

Proof. If D is an additive derivation of FI(P ), then there is an additive function

τ ∈ I(P ) such that D(α) = τ ? α for each α ∈ FI(P ). In particular, D(ex) = τ ? ex.

By Remark 3.1 τ(x, x) = 0 for each x ∈ P . It follows that D(ex) = 0.

Conversely, suppose that D is a derivation such that D(ex) = 0 for each x ∈ P . Let x and y be elements of P with x ≤ y. Then

D(δxy) = D(ex · δxy · ey)

= (D(ex) · δxy · ey) + (ex · D(δxy · ey))

= (D(ex) · δxy · ey) + ex · [D(δxy) · ey + δxy · D(ey)]

= (D(ex) · δxy · ey) + (ex · D(δxy) · ey) + (ex · δxy · D(ey))

= ex · D(δxy) · ey.

But for each β ∈ I(P ), we have ex · β · ey = β(x, y) · δxy, and so

D(δxy) = (D(δxy))(x, y) · δxy.

Deﬁne τ ∈ I(P ) by τ(x, y) = (D(δxy))(x, y) so that D(δxy) = τ(x, y) · δxy. If x, y, z ∈ P are such that x ≤ z ≤ y, then 18 τ(x, y) · δxy = D(δxy)

= D(δxz · δzy)

= D(δxz) · δzy + δxz · D(δzy)

= (τ(x, z) · δxz · δzy) + (δxz · τ(z, y) · δzy)

= (τ(x, z) + τ(z, y)) · δxy.

Therefore τ is an additive function. To complete the proof, we show that D(α) = τ?α for each α ∈ FI(P ).

For α ∈ FI(P ), we have

D(ex · α · ey) = D(α(x, y) · δxy)

= α(x, y) · D(δxy)

= α(x, y) · τ(x, y) · δxy, and

D(ex · α · ey) = ey · D(α) · ey

= D(α)(x, y) · δxy.

Therefore, (D(α))(x, y) = α(x, y) · τ(x, y) and D(α) = τ ? α.

We can now describe the derivations of the ﬁnitary incidence algebra.

Theorem 3.1. [7] D(FI(P )) = Inn(FI(P )) + Ad(FI(P )).

Proof. It suﬃces to show that every derivation can be written as the sum of an inner derivation and an additive derivation. Let D be a derivation of FI(P ) and deﬁne

γ ∈ I(P ) by   D(ey)(x, y) if x ≤ y, γ(x, y) =  0 otherwise.

19 As   γ(u, x) if v = x, (γ · ex)(u, v) =  0 otherwise, for u ≤ v we have

(γ · ex)(u, v) = γ(u, v) = D(ex)(u, v) = (D(ex) · ex)(u, v), for v = x, and

(γ · ex)(u, v) = 0 = (D(ex) · ex)(u, v), for v 6= x, so that γ · ex = D(ex) · ex, for any x ∈ P . We write

X X γ = γ · ey = D(ey) · ey. y∈P y∈P

Since Inn(FI(P )) is a submodule of D(FI(P )), by setting A = D + dγ, it suﬃces to show that A is additive.

By deﬁnition,

D(ex · ey) = D(ex) · ey + ex · D(ey).

2 If x 6= y, then ex · ey = 0 and so ex · D(ey) = −(D(ex) · ey). If x = y, then (ex) = ex and so D(ex) = D(ex)·ex+ex·D(ex). Therefore, D(ex)·ex = D(ex)·ex+(ex·D(ex)·ex), which implies that ex · D(ex) · ex = 0. It follows that

X ex · γ = ex · D(ey) · ey y6=x X = ex · D(ey) · ey y6=x X = −(D(ex) · ey) · ey y6=x X = − D(ex) · ey. y6=x

20 We conclude that

dγ(ex) = ex · γ − γ · ex ! X = − D(ex) · ey − D(ex) · ex y6=x X = − D(ex) · ey y∈P X = −D(ex) · ey y∈P

= −D(ex),

P as y∈P ey = e. Therefore, A(ex) = 0 for each x ∈ P . By Proposition 3.1, the derivation A is additive.

Lemma 3.1. If dg(FI(P )) ⊆ FI(P ), then g ∈ FI(P ).

Proof. Suppose dg(FI(P )) ⊆ FI(P ) and g∈ / FI(P ). Then there exists a < b ∈ P and an inﬁnite set [ui, vi] for i ∈ N with a ≤ ui < vi ≤ b and g(ui, vi) 6= 0. We will prove the lemma in three cases.

Case 1 Suppose that there are an inﬁnite number of distinct ui. Then there exists an inﬁnite subset N1 ⊆ N with ui = uj, i, j ∈ N1 implies i = j. Thus the ui are unique in N1.

Case 1.1 Suppose {vi : i ∈ N1} is infinite vi 6= vj for i 6= j ∈ N1. By induction define an infinite subset N2 ⊆ N1 such that

{ui : i ∈ N2} ∩ {vi : i ∈ N2} = ∅. (3.1)

(1) Let i1 be minimal in N2 and pick ui1 . If ∃j1 ∈ N1 with ui1 = vj1 set N1 = N1 −{j1}. (1) That j1 is unique and not equal to i1 which implies i1 ∈ N1 . Pick ui2 when i2 is (1) (1) (2) (1) the successor of i1 in N1 . If ui2 = vj2 for some j2 ∈ N1 , put N1 = N1 − {j2}. (2) Again j2 is unique and not equal to i2. So for ui1 , ui2 ∈/ {vj : j ∈ N1 } let i3 be the

21 (2) (2) (3) (2) (2) successor of i2 in N1 . If ui3 = vj3 some j3 ∈ N1 , put N1 = N1 − {j3} ⊆ N1 (3) and i3 6= j3 so i3 ∈ N1 . Continue... Call the resulting N2.

X N2 := {ik : k ∈ N} and s = [ui, ui] ∈ FI(P ). i∈N2 Then

dg(s) = sg − gs   X X =  s(x, z)g(z, y) [x, y] x≤y z∈[x,y]   X X −  g(x, z)s(z, y) [x, y] x≤y z∈[x,y] X X = s(x, x)g(x, y)[x, y] − g(x, y)s(y, y)[x, y] x≤y x≤y X X = s(ui, ui)g(ui, y)[ui, y] − g(x, ui)s(ui, ui)[x, ui]

ui≤y x≤ui X X = g(ui, y)[ui, y] − g(x, ui)[x, ui]

ui≤y x≤ui j ∈ N2,(sg−gs)(uj, vj) its ﬁrst term is 0 for i 6= j and g(uj, vj) for i = j second term is 0 unless ui = vj which contradicts 3.1 so it equals 0. So (sg−gs)(uj, vj) = g(uj, vj) which implies sg − gs∈ / FI(P ) which is a contradiction.

Case 1.2 Suppose {vi : i ∈ N1} is ﬁnite. N1 is inﬁnite implies there exists

some i0 ∈ N1 such that N2 = {i ∈ N1 : vi = vi0 } is inﬁnite and ui 6= vi0 . Put v := vi0 and s = [v, v]. Then

X X sg − gs = g(v, y)[v, y] − g(x, v)[x, v]. v≤y x≤v

Then for j ∈ N2 the ﬁrst term of (sg − gs)(uj, v) is 0 since uj 6= v. In the second term however g(uj, v) 6= 0 so sg − gs∈ / FI(P ) which is a contradiction.

Case 2 Suppose there are only a ﬁnite number of distinct ui. This implies

{vi : i ∈ N} contains an inﬁnite number of distinct vi. For some minimal i0 ∈ N the 22 set {ui : ii0 = ui} is inﬁnite. Let N1 = {i : g(ui0 , vi) 6= 0}. Note that ui0 6= vi for all

i ∈ N1. Set u := ui0 and s = [u, u]. Then

dg(s) = sg − gs X X = g(u, y)[u, y] − g(x, u)[x, u]. y≥u x≤u

So (sg − gs)(u, vi) = g(u, vi) + 0 = g(u, vi), which implies that sg − gs∈ / FI(P ). A contradiction.

23 CHAPTER FOUR

Automorphisms of FI(P )

Much of the content from this chapter is adapted from [7]. However, unlike the previous chapter, many results can not be carried from I(P ) to FI(P ) without change. Hence we will also refer to results in [6] in order to reach the desired ﬁnal decomposition.

Let Aut(FI(P )) denote the set of F -algebra automorphisms of FI(P ) under the operation of composition.

Deﬁnition 4.1. Let f ∈ FI(P ) be invertible. The automorphism ψf : FI(P ) →

−1 FI(P ) deﬁned by ψf (g) = fgf will be called an inner automorphism and will be denoted by Inn(FI(P )).

It is easy to show that Inn(I(P )) is a subgroup of Aut(FI(P )).

Deﬁnition 4.2. If α is an automorphism of P , then α induces an automorphismα ˆ of FI(P ) given by (ˆα(f))(x, y) = f(α(x), α(y)), for each f ∈ FI(P ) and each pair of elements x, y ∈ P . We then denote the subgroup of such automorphisms induced by α by Aut(P ).

Note that Aut(P ) is isomorphic to the group of automorphisms of P in the natural way.

Deﬁnition 4.3. We call σ ∈ I(P ) multiplicative if whenever x ≤ z ≤ y in P then

σ(x, y) is a unit of F and σ(x, y) = σ(x, z)σ(z, y).

As a consequence of the previous deﬁnition, if σ is multiplicative and σ ∈

FI(P ) then σ(x, x) = 1 for each x ∈ P since σ(x, x)σ(x, x) = σ(x, x) and F is a

ﬁeld. Hence σ is invertible in FI(P ).

24 Recall that the Hadamard product of f, g ∈ FI(P ) is denoted by f ? g and is given by (f ? g)(x, y) = f(x, y)g ˙(x, y) for each pair of x, y ∈ P . If σ ∈ I(P ) is multiplicative, we deﬁne Mσ : FI(P ) → FI(P ) by

Mσ(f) = σ ? f

for each f ∈ FI(P ). It is routine to check that Mσ is an automorphism of FI(P ). Such an automorphism is called a multiplicative automorphism.

By Mult(FI(P )) we mean the the subgroup of Aut(FI(P )) consisting of all multiplicative automorphisms.

Theorem 4.1. [7] Let P be a partially ordered set and F a ﬁeld. Suppose that ψ ∈

Aut(FI(P )). The ψ is in Mult(FI(P )) if and only if ψ(ex) = ex for each x ∈ P .

Proof. If σ is multiplicative, then as σ(x, x) = 1 for each x ∈ P , it is straight forward to show that Mσ(ex) = σ ? ex = ex.

Conversely, suppose that ψ is an automorphism of FI(P ) such that ψ(ex) = ex for each x ∈ P . Let f be an element of the incidence algebra and x, y ∈ P . Then exfey = f(x, y)δxy and so exFI(P )ey = F δxy. It follows that

ψ(ex)ψ(FI(P ))ψ(ey) = exFI(P )ey = F δxy.

The applying ψ to both sides of the last equality yields

ψ(ex)ψ(FI(P ))ψ(ey) = ψ(F δxy) = F ψ(δxy).

If x 6≤ y, then set σ(x, y) = 0. If x ≤ y, then let σ(x, y) take the value such that

ψ(δxy) = σ(x, y)δxy. In the latter case, as F ψ(δxy) = F δxy, there is an r ∈ F such that rσ(x, y) = 1 and so σ(x, y) is a unit in F . If x ≤ z ≤ y, then we have

σ(x, y)δxy = ψ(δxy)

= ψ(δxzδzy) 25 = ψ(δxz)ψ(δzy)

= (σ(x, z)δxz)(σ(z, y)δzy)

= σ(x, y)σ(y, z)δxy,

and so σ is multiplicative. To show that ψ = Mσ, consider

ψ(exfey) = ψ(f(x, y)δxy)

= f(x, y)ψ(δxy)

= f(x, y)σ(x, y)δxy.

Hence, ψ(f) = σ ? f and ψ = Mσ.

Theorem 4.2. [6] Let ρ ∈ Aut(FI(P )). Then for

X β = ρ(eu)(u, v)[u, v], u≤v we have

βρ(ex) = exβ. for all x ∈ P .

Proof. Since

X X exβ = ρ(eu)(u, v)ex[u, v] = ρ(ex)(x, v)[x, v], u≤v x≤v we can write

X X βρ(ex) = ρ(eu)(u, v)[u, v] ρ(ex)(a, b)[a, b] u≤v a≤b ! X X = ρ(eu)(u, v)ρ(ex)(v, b) [u, b] u≤b u≤v≤b X = (ρ(ex)ρ(ex)) (v, b)[v, b], since ρ(eu)ρ(ex) = 0 for u 6= x, x≤b

26 X = ρ(ex)(x, b)[x, b] x≤b

= exβ.

Theorem 4.3. β as deﬁned in Theorem 4.2 is in FI(P ).

Proof. Assume that there is an inﬁnite set A of intervals x = [l(x), r(x)] such that, for some a, b ∈ P we have a ≤ l(x) < r(x) ≤ b for all x ∈ A and ρ(el(x))(l(x), r(x)) 6= 0; i.e. that β∈ / FI(P ).

First we show that there exists a subset A0 of A such that l, r : A0 → P are injective maps. For x ∈ A define Lx = {y ∈ A : l(y) = l(x)}. Assume that Lx is an infinite set. Then ρ(el(y))(l(y), r(y)) = ρ(el(x))(l(x), r(y)) 6= 0 for all y ∈ Lx. Since ρ(el(x)) ∈ FI(P ) we infer that {r(y): y ∈ Lx} is finite. Since

0 0 Lx is inﬁnite, there exists y 6= y ∈ Lx with r(y) = r(y ). Note that we have y0 = [l(y0), r(y0)] = [l(x), r(y)] = [l(y), r(y)] = y, a contradiction.

This shows that the sets Lx are ﬁnite for all x ∈ P . Thus we may pick an

0 0 inﬁnite subset A of A such that the map r|A0 : A → P is injective.

Now deﬁne Rx = {y ∈ A : r(y) = r(x)} and assume that there exists some x ∈ A such that Rx is inﬁnite. Since ρ(el(y))(l(y), r(y)) 6= 0 for all y ∈ A, by Lemma

−1 2 [6] we have that ρ (er(y))(l(y), r(y)) 6= 0 for all y ∈ A. Now let y ∈ Rx. Then

−1 −1 −1 0 6= ρ (eer(y))(l(y), r(y)) = ρ (er(x))(l(x), r(y)) for all y ∈ Rx. Since ρ (er(y)) ∈

0 FI(P ), we infer that {r(y): y ∈ Rx} is ﬁnite. Thus there exist y 6= y ∈ Rx such that r(y) = r(y0). It follows that y0 = [l(y0), r(y0)] = [l(x), r(y)] = [l(y), r(y)] = y, a contradiction. This shows that the sets Rx are ﬁnite for all x ∈ A. Thus there is an

00 0 00 inﬁnite subset A ofA such that r|A00 : A → P is injective. This allows us, by renaming A := A00, to assume that both maps l, r : A → P are injective. We will now prove the following: There exists an inﬁnite sequence

27 xi = [l(xi), r(xi)] ∈ A for all i < ω such that ρ(el(xi))(l(xi), r(xj)) 6= 0 if and only if i = j.

We will prove this claim by induction. Let x0 = [l(x0), r(x0)] be any element of the set A. By induction hypothesis, there exist xi ∈ A for all 0 ≤ i < n such

that ρ(er(xi))(l(xi), r(xj)) 6= 0 if and only if i = j for all 0 ≤ i, j < n. Now

consider the set B = {x ∈ A : ρ(el(xi))(l(xi), r(x)) 6= 0 for some 0 ≤ i < n}.

0 This set is ﬁnite because ρ(el(xi)) ∈ FI(P ) and r is injective. Let B = {x ∈ A :

−1 ρ (er(xi))(l(x), r(xi)) 6= 0 for some 0 ≤ i < n}. This set is ﬁnite because L is

0 injective. Now pick xnA − (B ∪ B ). This implies that ρ(el(xi))(l(xi), r(xn)) = 0 for all 0 ≤ i < n but ρ(el(xn))(l(xn), r(xn)) 6= 0. If ρ(el(xn))(l(xn), r(xi)) 6= 0 for some

−1 0 ≤ i < n, then ρ (el(xi))(l(xn), r(xi)) 6= 0 by Lemma 2 [6]. This is a contradiction

0 to xn ∈ B . This shows that ρ(el(xn))(l(xn), r(xi)) = 0 for all 0 ≤ i < n. P Now let X = {xi : r ≤ ω} and eX = i<ω el(xi). By the Lemma 2 [6] we have

0 0 ρ(eX )(l(xj), r(xj)) = ρ(eX )(l(xj), r(xj)) where X = {l(xi): ρ(el(xi))(l(xj), r(xi)) 6=

0 6= ρ(el(xi))(l(xi), r(xj))}. The sequence {xi : i < ω} has the property that

0 ρ(el(xi))(l(xi), r(xj)) 6= 0 if and only if i = j. This means that X = {l(xj)} is

a singleton and thus ρ(eX )(l(xi), r(xj)) = ρ(el(xj ))(l(xj), r(xj)) 6= 0 for all j < ω.

This is a contradiction to ρ(eX ) ∈ FI(P ). This shows that β ∈ FI(P ).

Lemma 4.1. If ρ ∈ Aut(FI(P )) with ρ(ex) conjugate to ex then ρ(ex) ∈ ex + J(FI(P )).

Proof. Since ρ is an automorphism and ex is a primitive idempotent then rho(ex) is a primitive idempotent. By Theorem 3 of [5] ρ(ex) is conjugate to the diagonal idempotent ε such that ρ(ex)(u, u) = ε(u, u) for all u ∈ P . But by Lemma 1 [5] ε must be ex. Hence

ρ(ex)(x, x) = ex(x, x) = 1,

28 and 0 for u 6= x. Hence

ρ(ex) ∈ ex + J(FI(P )).

Corollary 4.1. If ρ ∈ Aut(FI(P )) such that ρ(ex) is conjugate to ex for each x ∈ P then there exists β ∈ FI(P ) such that ρ(ex) is conjugate to ex for all x ∈ P via β.

Proof. By the previous two theorems we have the existence of a β ∈ FI(P ) such that β · ρ(ex) = ex · β. But β(u, u) = ρ(eu)(u, u) = eu(u, u) = 1 for all u ∈ P by 4.1. Thus β is invertible. Hence

−1 ρ(ex) = β · exβ .

The following is a generalization of a result in [7] and is proven in even greater abstraction in [6], although the proof is more direct than in [6].

Theorem 4.4. Let P be a partially ordered set and F a ﬁeld. Let ρ ∈ Aut(FI(P )).

Then there are ψβ ∈ Inn(FI(P )), αˆ ∈ Aut(P ), and Mσ ∈ Mult(FI(P )) such that

ρ = ψβ ◦ Mσ ◦ α.ˆ

Proof. First recall that α induces an automorphism of FI(P ) and that αd−1 = (ˆα)−1.

Since ρ ∈ Aut(FI(P )) we have that ρ(ex) is conjugate to some ey and hence (ρ ◦

−1 αd)(ey) is conjugate to ey. Hence there exists a β ∈ FI(P ) such that β · ey ·

−1 −1 −1 β = (ρ ◦ αd)(ey). Thus ψβ(ey) = (ρ ◦ αd)(ey) where ψβ ∈ InnAut(FI(P )). By

−1 −1 Theorem 4.1 we have that (ψβ) ◦ρ◦αd is a multiplicative automorphism of FI(P ).

−1 −1 Therefore, there is a multiplicative σ ∈ FI(P ) such that Mσ = (ψβ) ◦ ρ ◦ αd or

ρ = ψβ ◦ Mσ ◦ αˆ.

Deﬁnition 4.4. Let R be an F -algebra and M an R-bimodule. The F -linear map

φ : M → M is called an α-semilinear bimodule homomorphism if α ∈ Aut(R) and

φ(rms) = α(r)φ(m)α(s) for all r, s ∈ R and m ∈ M. 29 For later use, we will now extend the previously deﬁned group automorphisms to I(P ), and show that they remain group isomorphisms and are also semilinear with respect to the function they are extending.

−1 Theorem 4.5. Given ψf : FI(P ) → FI(P ) where ψf (g) = fgf for a unit f ∈

−1 FI(P ) then ψ ∈ Aut(FI(P )). Deﬁne ψff : I(P ) → I(P ) by ψff (g) = fgf . Then

ψff |FI(P ) = ψf and ψff is both a group homomorphism and ψf -semilinear.

Proof. ψff is a group homomorphism:

−1 ψff (g + h) = f(g + h)f

= fgf −1 + fhf −1

= ψff (g) + ψff (h).

ψff is ψf -semilinear:

−1 ψff (αgβ) = fαgβf

= fαf −1fgf −1fβf −1

= ψf (α)ψff (g)ψf (β).

ψff is bijective since it is invertable via ψgf −1 .

Theorem 4.6. Let τ ∈ I(P ) and deﬁne Mτ : FI(P ) → FI(P ) by Mτ (α) = τ ? α.

Then deﬁne Mfτ : I(P ) → I(P ) by Mfτ (g) = τ ? g. Then Mfτ |FI(P ) = Mτ and is both a group homomorphism and Mτ -semilinear.

Proof. As the Hadamard product is a component wise multiplication, Mfτ is clearly a group homomorphism:

Mfτ (g + h)(x, y) = (τ ? (g + h))(x, y)

= τ(x, y)(g + h)(x, y)

= τ(x, y)(g(x, y) + h(x, y))

30 = τ(x, y)g(x, y) + τ(x, y)h(x, y)

= (τ ? g)(x, y) + (τ ? h)(x, y)

= Mfτ (g)(x, y) + Mfτ (h)(x, y).

Mfτ is Mτ -semilinear: For the left hand side:

Mfτ (αgβ)(x, y) = (τ ? αgβ)(x, y)

= τ(x, y)(αgβ)(x, y) X = τ(x, y) (αg)(x, z)β(z, y) x≤z≤y " # X X = τ(x, y) α(x, v)g(v, z) β(z, y). x≤z≤y x≤v≤z For the right hand side:

X ((τ ? α)(τ ? g)(τ ? β)) (x, y) = ((τ ? α)(τ ? g)) (x, z)(τ ? β)(z, y) x≤z≤y " # X X = (τ ? α)(x, v)(τ ? g)(v, z) (τ ? β)(z, y) x≤z≤y x≤v≤z " # X X = τ(x, v)α(x, v)τ(v, z)g(v, z) τ(z, y)β(z, y) x≤z≤y x≤v≤z " # X X = τ(x, z)α(x, v)g(v, z) τ(z, y)β(z, y) x≤z≤y x≤v≤z " # X X = τ(x, z) α(x, v)g(v, z) τ(z, y)β(z, y) x≤z≤y x≤v≤z " # X X = τ(x, y) α(x, v)g(v, z) β(z, y) x≤z≤y x≤v≤z " # X X = τ(x, y) α(x, v)g(v, z) β(x, y), x≤z≤y x≤v≤z since τ is multiplicative.

Note, since τ is multiplicative τ(x, y) 6= 0 for all x ≤ y. Hence Mfτ is bijective

∗ −1 since in is invertable via Mτ ? where τ (x, y) = τ(x, y) . 31 Theorem 4.7. Let σ be an automorphism of P . Then σ induces σˆ ∈ Aut FI(P ) deﬁned by (ˆσ(α))(x, y) = α(σ(x), σ(y)). Deﬁne σeˆ : I(P ) → I(P ) by σeˆ(g) (x, y) = g(σ(x), σ(y)). Then σeˆ|FI(P ) =σ ˆ and is both a group homomorphism and is σˆ- semilinear.

Proof. σeˆ is clearly a group homomorphism:

σeˆ(g + h)(x, y) = (g + h)(σ(x), σ(y))

= g(σ(x), σ(y)) + h(σ(x), σ(y))

= σeˆ(g)(x, y) + σeˆ(h)(x, y).

σeˆ isσ ˆ-semilinear: For the left hand side:

σeˆ(αgβ) (x, y) = (αgβ)(σ(x), σ(y)) X = (αg)(σ(x), z)β(z, σ(y)) σ(x)≤z≤σ(y)   X X =  α(σ(x), v)g(v, z) β(z, σ(y)). σ(x)≤z≤σ(y) σ(x)≤v≤z

For the right hand side:

σˆ(α)σeˆ(g)ˆσ(β) (x, y) = σˆ(α)σeˆ(g) σˆ(β) (x, y) X = σˆ(α)σeˆ(g) (x, z0)ˆσ(β)(z0, y) x≤z0≤y X = σˆ(α)σeˆ(g) (x, z0)β(σ(z0), σ(y)) x≤z0≤y " # X X = σˆ(α)(x, v0)σeˆ(g)(v0, z0) β(σ(z0), σ(y)) x≤z0≤y x≤v0≤z " # X X = α(σ(x), σ(v0))g(σ(v0), σ(z0)) β(σ(z0), σ(y)) x≤z0≤y x≤v0≤z0   X X 0 0 0 0 =  α(σ(x), σ(v ))g(σ(v ), σ(z )) β(σ(z ), σ(y)), x≤z0≤y σ(x)≤σ(v0)≤σ(z0)

32 since σ is a bijection call σ(v0) = v and σ(z0) = z; then   X X =  α(σ(x), v)g(v, z) β(z, σ(y)). σ(x)≤z≤σ(y) σ(x)≤v≤z

σeˆ is a bijection since it is invertible via σgˆ−1, where σ−1 the inverse of the poset isomorphism σ.

Theorem 4.8. Given ρ ∈ Aut(FI(P )), there exists an extension ρe : I(P ) → I(P ) such that ρe is ρ-semilinear and ρe|FI(P ) = ρ and ρe is a group automorphism of I(P ).

Furthermore, ρe is unique.

Proof. Since ρ = ψβ ◦ Mσ ◦ αˆ, we may deﬁne ρe : I(P ) → I(P ) by deﬁning ρe =

ψff ◦ Mfτ ◦ σeˆ.

Since ρe = ψff ◦ Mfτ ◦ σeˆ is the composition of bijections, it is a bijection and clearly ρe|FI(P ) = ρ. Also we now have that ρe is ρ-semilinear.

Furthermore, since each of ψff , Mfτ and σeˆ are group automorphisms, so is ρe. Now suppose λ ∈ Aut(FI(P )) is also ρ-semilinear. Then, let g ∈ I(P ) and recall that ex, ey, exgey ∈ FI(P ). Then

−1 −1 ρe (λ(g))(x, y) = (exρe (λ(g))ey)(x, y)

−1 −1 −1 = (ρe (ρ(ex))ρe (λ(g))ρe (ρ(ey)))(x, y)

−1 = (ρe (ρ(ex)λ(g)ρ(ey)))(x, y)

−1 = (ρe (λ(exgey))(x, y)

−1 = (ρe (ρ(exgey)))(x, y)

= (exgey)(x, y)

= g(x, y),

−1 for all x ≤ y ∈ P . Thus ρe ◦ λ = idI(P ) so ρe = λ.

33 CHAPTER FIVE

The Idealization of FI(P )

5.1 Deﬁnition

Deﬁnition 5.1. Let R be a commutative ring and M an R-module. Nagata introduced the idealization R ⊕ M of M. Here R ⊕ M is a commutative ring with product

(r1, m1)(r2, m2) = (r1r2, r1m2 + m1r2).

The name comes from the fact that if N is a submodule of M, then 0 ⊕ N is an ideal of R ⊕ M. In particular 0 ⊕ M is an ideal. This sometimes also known as a Dorroh extension.

Although not commutative, FI(P ) is a ring and I(P ) is a module over it.

This is suﬃcient for our needs to form the idealization D(P ) = FI(P ) ⊕ I(P ) with multiplication as deﬁned above, with 0 ⊕ I(P ) being and ideal of D(P ).

Deﬁnition 5.2. Let D(P ) be the Nagata idealization of I(P ) as an FI(P )-bimodule with the standard multiplication.

Since FI(P ) is an associative algebra and I(P ) is a module over it [5], we can form the Dorroh extension D(P ) = FI(P ) ⊕ I(P ), a ring with unity. Speciﬁcally multiplication is deﬁned by

(α ⊕ β)(γ ⊕ δ) = αγ ⊕ (αγ + βγ), from which it is easy to see that the identity is e ⊕ 0.

5.2 Basics

Lemma 5.1. α ⊕ β is invertible if and only if α is invertible. Furthermore

(α ⊕ β)−1 = α−1 ⊕ −α−1βα−1.

34 Proof. If α is invertible then

(α ⊕ β)(α−1 ⊕ −α−1βα−1) = (α−1 ⊕ −α−1βα−1)(α ⊕ β) = e ⊕ 0.

If α ⊕ β is invertible, then for some γ ⊕ δ

(α ⊕ β)(γ ⊕ δ) = (γ ⊕ δ)(α ⊕ β) = e ⊕ 0, which implies that αγ = γα = 1 in FI(P ). Hence γ = α−1.

Lemma 5.2. If α ⊕ β ∈ D(P ) is an idempotent then α ∈ FI(P ) is an idempotent and β = αβ + βα.

Proof. Simply multiplying α ⊕ β by itself and setting it equal to itself immediately yields the result:

(α ⊕ β)(α ⊕ β) = α2 ⊕ αβ + βα = α ⊕ β.

Let J(D) be the Jacobson Radical of D.

Recall that J(D) can be characterized as the maximal ideal of D(P ) with respect to the property that every element is left quasi-regular.

Theorem 5.1.

J(D(P )) = J(FI(P )) ⊕ I(P ).

Proof. Let α ∈ J(FI(P )) and β ∈ I(P ). We will ﬁrst show the element α⊕β in D(P ) is left quasi-regular. Hence we must ﬁnd a left inverse m ⊕ n of (e ⊕ 0) − (α ⊕ β) =

(e − α) ⊕ −β.

(m ⊕ n)((e − α) ⊕ −β) = m(e − α) ⊕ (−mβ + n(e − α)) = e ⊕ 0.

We can deﬁne m := (e − α)−1 since α is left quasi-regular, being in J(FI(P )).

Setting the second term equal to 0 and multiplying through by m yields

−mβm + n(e − α)m = −mβm + n = 0, 35 since left and right inverses coincide in FI(P ) by [5]. Thus setting n := mβm yields the necessary inverse and we have J(FI(P )) ⊕ I(P ) ⊆ J(D(P )).

Now let α ⊕ β be an element of J(D(P )). Thus 1 − α ⊕ −β is invertible which means there exists some m ∈ FI(P ) such that m(1 − α) = 1 and hence α is left quasi-regular in FI(P ). Let t ∈ FI(P ). Then (t ⊕ 0)(α ⊕ β) = tα ⊕ tβ which is still in the ideal J(D(P )). As above we again have that tα is left quasi-regular in FI(P ).

Since t was arbitrary we have that FI(P )α is an ideal consisting entirely of left quasi-regular elements and thus FI(P )α ⊆ J(FI(P )). Hence α ∈ J(FI(P )) which implies α ⊕ β ∈ J(FI(P )) ⊕ I(P ). Therefore we have J(D(P )) ⊆ J(FI(P )) ⊕ I(P ), completing the proof.

Recall Deﬁnition 2.4.

Theorem 5.2. Given α ⊕ β ∈ D(P ) an idempotent, α ⊕ β is primitive if and only if

α is a primitive idempotent in FI(P ).

Proof. Suppose α ⊕ β is a primitive idempotent and αγ = γα = γ for some idempotent γ. Since γ is idempotent, so is γ ⊕ 0 and since α ⊕ β is primitive we have

(α ⊕ β)(γ ⊕ 0) = (γ ⊕ 0)(α ⊕ β) = γ ⊕ 0, implies γ ⊕ 0 = 0 ⊕ 0 or γ ⊕ 0 = α ⊕ β. Hence γ = α or γ = 0.

Now suppose α is primitive and α ⊕ β is an idempotent and (α ⊕ β)(γ ⊕ δ) =

(γ ⊕ δ)(α ⊕ β) = γ ⊕ δ. Multiplying the previous equation out we get

αγ = γα = γ (5.1) and

αδ + βγ = γβ + δα = δ. (5.2)

Since α is primitive and γ is an idempotent we have either γ = α or γ = 0.

If γ = 0 then by 5.2 we have δ = γδ + δγ = 0, and hence γ ⊕ δ = 0 ⊕ 0. 36 If γ = α, substituting into equation 5.2 we get γδ + βγ = γβ + δγ = δ. By

Lemma 5.2 we have γδ + βγ = γβ + δγ = γδ + δγ. Canceling we get βγ = δγ and

γβ = γδ. Adding yields γβ + βγ = γδ + δγ. Substituting γ = α on the left and we see the left hand side is exactly β while the right hand side is δ. Hence β = δ and thus γ ⊕ δ = α ⊕ β.

Corollary 5.1. An idempotent α ⊕ β is primitive if and only if α is conjugate to ex for some x ∈ P

Proof. Suppose α ⊕ β is primitive. By Theorem 5.2 α is primitive which implies α is conjugate to ex by Lemma 1 of [5].

Now suppose α is conjugate to ex for some x ∈ P . Then by Lemma 1 of [5] α is primitive and by Theorem 5.2 α ⊕ β is primitive.

Since ex is clearly conjugate to itself we immediately get that ex ⊕ 0, an idempotent, is primitive.

Theorem 5.3. Let P and Q be arbitrary partially ordered sets. Then

D(P ) ∼= D(Q) implies P ∼= Q.

Proof. Let Φ : D(P ) ∼= D(Q) be an isomorphism. For each x ∈ P the image

P P Φ(ex ⊕ 0) = α ⊕ β of ex ⊕ 0 is primitive and by Corollary 5.1 α is conjugate to

Q ey , for some y ∈ Q. By Remark 2.4 y is unique. Thus Φ generates the bijection Q ϕ : P → Q such that the ﬁnitary part of Φ(ex ⊕ 0) is conjugate to eϕ(x). Let us prove that ϕ preserves the order. Recall by Corollary 2.1, α ∈ J(FI(P )) if and only if a(x, x) = 0 for all x ∈ P . Also we know that J(D(P )) = J(FI(P )) ⊕

I(P ) and ex(x, x) = 1. Suppose then, that x ≤ y. Then

P P P P P P P P (ex ⊕ 0)(e ⊕ 0)(ex ⊕ 0) = ex eex ⊕ ex 0ex = ex ⊕ 0.

37 P P Since ex ∈/ J(FI(P )) then ex ⊕ 0 ∈/ J(D(P )). Also, note that if x > y, or x and y are incomparable, then

P P (ex ⊕ 0)(α ⊕ β)(ey ⊕ 0) = α(x, y)δxy ⊕ β(x, y)δxy = 0 ⊕ 0 ∈ J(D(P )).

Hence we get the result that for all x, y ∈ P

P P x ≤ y ⇔ (ex ⊕ 0)D(P )(ey ⊕ 0) * J(D(P )).

Similarly, for all u, v ∈ Q

P P u ≤ v ⇔ (eu ⊕ 0)D(Q)(ev ⊕ 0) * J(D(P )).

Q −1 P Let x ≤ y and for Φ(ex ⊕ 0) = α1 ⊕ β1 let α1 = γ1eϕ(x)γ1 and for Φ(ey ⊕ 0) = Q −1 α2 ⊕ β2 let α2 = γ2eϕ(y)γ2 for some invertable elements γ1 and γ2 in FI(P ). From the bijectivity of Φ we obtain

P P P P Φ(ex ⊕ 0)D(Q)Φ(ey ⊕ 0) = Φ((ex ⊕ 0)D(P )(ey ⊕ 0)) * J(D(Q)).

Then

P P Φ(ex ⊕ 0)D(Q)Φ(ey ⊕ 0) = (α1 ⊕ β1)D(Q)(α2 ⊕ β2)

Q −1 Q −1 = γ1eϕ(x)γ1 FI(Q)γ2eϕ(y)γ2 ⊕ ζ, ζ ∈ I(P )

Q Q −1 = γ1eϕ(x)FI(Q)eϕ(y)γ2 ⊕ ζ,

since γ1 and γ2 are invertible. Hence if ϕ(x) > ϕ(y) then

Q Q −1 γ1eϕ(x)FI(Q)eϕ(y)γ2 ⊕ ζ = 0 ⊕ ζ ∈ J(D(Q)).

This is a contradiction and thus ϕ preserves order.

38 CHAPTER SIX

Automorphisms of D(P )

We will deﬁne three types of automorphisms of D(P ) and their associated subgroups of Aut(D(P )). We will then show that and map from Aut(D(P ) can be decomposed as a composition of automorphisms we have deﬁned.

Theorem 6.1. Let ρ ∈ Aut(FI(P )) and let ρ˜ be its ρ-semilinear extension as in

Theorem 4.8. Now deﬁne gρ : D(P ) → D(P ) by   ρ 0   gρ :=   . 0 ρe

Then gρ is an automorphism of D(P ).

Proof. We can show directly that gρ is a ring homomorphism:           ρ 0 α α ρ 0 α α    1   2     1 2        =     0 ρe β1 β2 0 ρe α1β2 + β1α2

= ρ(α1α2) ⊕ ρe(α1β2 + β1α2)

= ρ(α1)ρ(α2) ⊕ ρe(α1β2) + ρe(β1α2)

= ρ(α1)ρ(α2) ⊕ ρ(α1)ρe(β2) + ρe(β1)ρ(α2)     ρ(α ) ρ(α )  1   2  =     ρe(β1) ρe(β2)         ρ 0 α ρ 0 α    1     2  =         , 0 ρe β1 0 ρe β2 since ρe is ρ-semilinear, and clearly           ρ 0 α α ρ 0 α + α    1   2     1 2      +   =     0 ρe β1 β2 0 ρe β1 + β2 39   ρ(α + α )  1 2  =   ρe(β1 + β2)     ρ(α ) ρ(α )  1   2  =   +   ρe(β1) ρe(β2)         ρ 0 α ρ 0 α    1     2  =     +     . 0 ρe β1 0 ρe β2 −1 Furthermore, since both ρ and ρe are bijective, then so is gρ since clearly gρ = gρ−1 , and then           ρ 0 ρ−1 0 α ρ 0 ρ−1(α)       =        −1       −1  0 ρe 0 ρe β 0 ρe ρe (β)   ρ (ρ−1(α)) =    −1  ρe(ρe (β))   α   =   . β

Deﬁnition 6.1. Deﬁne G(P ) as the set of all gρ for ρ ∈ Aut(FI(P )).

Theorem 6.2. G(P ) is a subgroup of Aut(D(P )). Furthermore G(P ) ∼= Aut(FI(P )).

Proof. To see that G(P ) is a subgroup simply let ρ1, ρ2 ∈ Aut(FI(P )) and then note       −1 −1 ρ1 0 ρ2 0 ρ1 ◦ ρ2 0 −1       gρ1 ◦ gρ = = , 2    −1   −1  0 ρe1 0 ρe2 0 ρe1 ◦ ρe2 −1 −1 where ρ1 ◦ ρ2 ∈ Aut(FI(P )) and its semilinear extension to I(P ) is ρe1 ◦ ρe2 . Then gρ ◦ g −1 1 ρ2 ∼ To see that G(P ) = Aut(FI(P )) simply deﬁne the mapping ρ 7→ gρ which is clearly bijective. The isomorphism follows trivially. 40 Theorem 6.3. Let D be a derivation of FI(P ). Deﬁne ∆D : D(P ) → D(P ) by   e 0   ∆D :=   . D e

Then ∆D is an automorphism of D(P ).

−1 Proof. ∆D is bijective since ∆D = ∆−D and           e 0 e 0 α e 0 α                 =     D e −D e β D e −D(α) + β   α   =   D(α) − D(α) + β   α   =   . β

Also ∆D is a ring homomorphism since       e 0 α + α α + α    1 2   1 2      =   D e β1 + β2 D(α1 + α2) + β1 + β2     α α  1   2  =   +   D(α1) + β1 D(α2) + β2         e 0 α e 0 α    1     2  =     +     . D e β1 D e β2 since D is a derivation and we also have           e 0 α α e 0 α α    1   2     1 2        =     D e β1 β2 D e α1β2 + β1α2   α α  1 2  =   D(α1α2) + α1β2 + β1α2

41   α α  1 2  =   α1D(α2) + D(α1)α2 + α1β2 + β1α2   α α  1 2  =   α1D(α2) + α1β2 + D(α1)α2 + β1α2     α α  1   2  =     D(α1) + β1 D(α2) + β2         e 0 α e 0 α    1     2  =         . D e β1 D e β2

Deﬁnition 6.2. Deﬁne ∆(P ) as the set of all ∆D for D a derivation of FI(P ).

Theorem 6.4. ∆(P ) is a subgroup of Aut(D(P )).

Proof. To see that ∆(P ) is a subgroup of Aut(D(P )) let D1 and D2 be derivations of FI(P ). Then       e 0 e 0 e 0 −1 ∆D ◦ ∆ =     =   , 1 D2       D1 e −D2 1 D1 − D2 e which is in ∆(P ) since the diﬀerence of two derivations is a derivation.

Theorem 6.5. Let f ∈ F , the ﬁeld over which we are deﬁning FI(P ) and I(P ). We

] deﬁne Ff : D(P ) → D(P ) by   e 0 F ] :=   . f   0 f

] Then Ff is an automorphism of D(P ). −1 ] ] ] Proof. Clearly Ff = Ff −1 , and then Ff is bijective since               e 0 e 0 α e 0 α α α                     =     =   =   . 0 f 0 f −1 β 0 f f −1β ff −1β β

42 ] Furthermore Ff is a ring homomorphism since       e 0 α + α α + α    1 2   1 2      =   0 f β1 + β2 fβ1 + fβ2     α α  1   2  =   +   fβ1 fβ2         e 0 α e 0 α    1     2  =     +     , 0 f β1 0 f β2 and           e 0 α α e 0 α α    1   2     1 2        =     0 f β1 β2 0 f α1β2 + β1α2   α α  1 2  =   fα1β2 + fβ1α2   α α  1 2  =   α1fβ2 + fβ1α2     α α  1   2  =     fβ1 fβ2         e 0 α e 0 α    1     2  =         . 0 f β1 0 f β2

] ] Deﬁnition 6.3. We will call the collection of Ff , for all f ∈ F , F (P ).

Theorem 6.6. F ](P ) is a subgroup of Aut(D(P )).

Proof. This follows immediately letting f1, f2 ∈ F . Then

43       e 0 e 0 e 0     =   ,    −1   −1  0 f1 0 f2 0 f1f2 which is clearly in F ](P ).

We may now present the following result:

] Theorem 6.7. Let P be a partially ordered set, ϕ ∈ Aut(D(P )) and ∆D, gρ and Ff as deﬁned above. Then

] ϕ = ∆D ◦ gρ ◦ Ff .

Proof. Let   ϕ ϕ  11 12  ϕ =   ∈ Aut(D(P )), ϕ21 ϕ22 where

ϕ11 : FI(P ) → FI(P )

ϕ12 : I(P ) → FI(P )

ϕ21 : FI(P ) → I(P )

ϕ22 : I(P ) → I(P ).

Also, let r ⊕ i and s ⊕ j be elements of D(P ). Then         r s rs ϕ (rs) + ϕ (is + rj)        11 12  ϕ     = ϕ   =   . i j is + rj ϕ21(rs) + ϕ22(is + rj)

On the other hand,         r s ϕ (r) + ϕ (i) ϕ (s) + ϕ (j)      11 12   11 12  ϕ   ϕ   =     i j ϕ21(r) + ϕ22(i) ϕ21(s) + ϕ22(j)   (ϕ (r) + ϕ (i))(ϕ (s) + ϕ (j))  11 12 11 12  =   . (ϕ11(r) + ϕ12(i))(ϕ21(s) + ϕ22(j)) + (ϕ21(r) + ϕ22(i))(ϕ11(s) + ϕ12(j))

44 Comparing ﬁrst coordinates we get

ϕ11(rs) + ϕ12(is) + ϕ12(rj) = ϕ11(r)ϕ11(s) + ϕ11(r)ϕ12(j)

+ ϕ12(i)ϕ11(s) + ϕ12(i)ϕ12(j).

By setting i = j = 0 we get

ϕ11(rs) = ϕ11(r)ϕ11(s). (6.1)

Moreover by setting r = s = 0 we get

0 = ϕ12(i)ϕ12(j). (6.2)

By setting r = j = 0 we get

ϕ12(is) = ϕ12(i)ϕ11(s). (6.3)

Finally by setting i = s = 0 we get

ϕ12(rj) = ϕ11(r)ϕ12(j). (6.4)

These equations hold for all r, s ∈ FI(P ) and i, j ∈ I(P ).

From the second coordinate we get

ϕ21(rs) + ϕ22(is) + ϕ22(rj) = ϕ11(r)ϕ21(s) + ϕ11(r)ϕ22(j) + ϕ12(i)ϕ21(s)

+ ϕ12(i)ϕ22(j) + ϕ21(r)ϕ11(s) + ϕ21(r)ϕ12(j)

+ ϕ22(i)ϕ11(s) + ϕ22(i)ϕ12(j).

It follows

ϕ21(rs) = ϕ11(r)ϕ21(s) + ϕ21(r)ϕ11(s) (6.5) by setting i = j = 0. Moreover,

0 = ϕ12(i)ϕ22(j) + ϕ22(i)ϕ12(j) (6.6)

45 by setting r = s = 0. Moreover,

ϕ22(is) = ϕ12(i)ϕ21(s) + ϕ22(i)ϕ11(s) (6.7) by setting j = r = 0. Finally,

ϕ22(rj) = ϕ11(r)ϕ22(j) + ϕ21(r)ϕ12(j) (6.8) by setting j = s = 0. Again, these equations hold for all r, s ∈ FI(P ) and i, j ∈ I(P ).

Since ϕ is an automorphism. Then     e e     ϕ   =   , 0 0 and thus       e ϕ (e) + ϕ (0) ϕ (e)    11 12   11  ϕ   =   =   . 0 ϕ21(e) + ϕ22(0) ϕ21(e)

So ϕ11(e) = e and ϕ21(e) = 0. Moreover, since FI(P ) = ϕ11(FI(P )) + ϕ12(I(P )),

∗ 0 letting ϕ12(i ) ∈ ϕ12(I(P )) and r ∈ FI(P ) then by 6.1, 6.2, and 6.3

0 ∗ ∗ r ϕ12(i ) = (ϕ11(r) + ϕ12(i))(ϕ12(i ))

∗ ∗ = ϕ11(r)ϕ12(i ) + ϕ12(i)ϕ12(i )

∗ = ϕ12(ri ),

2 is in ϕ12(I(P )) giving us that ϕ12(I(P ))¢FI(P ). Also, by 6.2 we have (ϕ12(I(P ))) = {0}. Using 6.3 and 6.4 we get

0 ϕ12(e)r = ϕ12(e)(ϕ11(r) + ϕ12(i))

= ϕ12(er)

= ϕ12(re)

= ϕ11(r)ϕ12(e) + ϕ12(i)ϕ12(e)

46 = (ϕ11(r) + ϕ12(r))ϕ12(e)

0 = r ϕ12(e).

Thus ϕ12(e) is in the center of FI(P ) and in J(FI(P )) since its also nilpotent. But ζ(FI(P )) ∩ J(FI(P )) = {0} and thus

ϕ12(e) = 0, (6.9) which implies

ϕ12(re) = ϕ12(er) = 0 (6.10) by 6.3 and 6.4 again.

Looking to 6.6, letting j = e and i = I(P ) we get

{0} = ϕ12(I(P ))ϕ22(e) + ϕ22(I(P ))ϕ12(e), which implies

ϕ12(I(P ))ϕ22(e) = −ϕ22(I(P ))ϕ12(e) = {0}, by 6.9. Similarly, letting i = e and j = I(P ) we get

ϕ22(e)ϕ12(I(P )) = {0}. (6.11)

Letting i = j = e and r = s in 6.7, along with 6.9 we get

ϕ22(er) = ϕ12(e)ϕ21(r) + ϕ22(e)ϕ11(r) = ϕ22(e)ϕ11(r), in 6.8

ϕ22(re) = ϕ11(r)ϕ22(e) + ϕ21(r)ϕ12(e) = ϕ11(r)ϕ22(e).

So together

ϕ22(e)ϕ11(r) = ϕ22(er) = ϕ22(re) = ϕ11(r)ϕ22(e), ∀r ∈ FI(P ). (6.12)

0 0 Letting r = ϕ11(r ) + ϕ12(i ) then

0 0 rϕ22(e) = ϕ11(r )ϕ22(e) + ϕ12(i )ϕ22(e) 47 0 = ϕ11(r )ϕ22(e)

0 = ϕ22(r e), by 6.11 and 6.12,

0 = ϕ22(er )

0 = ϕ22(e)ϕ12(r )

0 0 = ϕ22(e)ϕ11(r ) + ϕ22(e)ϕ12(i ), again by 6.11,

0 0 = ϕ22(e)(ϕ11(r ) + ϕ12(i ))

= ϕ22(e)r.

In I(P ) if rϕ22(e) = ϕ22(e)r for all r ∈ FI(P ) then

ϕ22(e) = fe (6.13)

for some f ∈ F . Since ϕ22 is injective, we have f 6= 0. By 6.11

{0} = ϕ22(e)ϕ12(I(P )) = feϕ12(I(P )) = efϕ12(I(P ))

= eϕ12(fI(P )) = eϕ12(I(P )) = ϕ12(I(P )).

Hence ϕ12 = 0. Hence ϕ has the form   ϕ 0  11  ϕ =   , ϕ21 ϕ22 where ϕ11 is an isomorphism of FI(P ), call it ρ. Now       ρ 0 ρ 0 e 0         =     , ϕ21 ϕ22 ϕ21 ϕe22 0 f

48 −1 where ϕe22 = ϕ22f . Then by 6.12 we have that

ϕe22|FI(P ) = ρ, and that it is ρ-semilinear. Since by Theorem 4.8 there is a unique semilinear extension of ρ we have that ϕ22 = ρe. Decomposing again we can then write           ρ 0 e 0 e 0 ρ 0 e 0     =       .      −1      ϕ21 ρe 0 f ϕ21 ◦ ρ e 0 ρe 0 f

Since ϕ11 = ρ, by 6.5 we get

−1 −1 −1 (ϕ21 ◦ ρ )(rs) = ϕ21(ρ (r)ρ (s))

−1 −1 −1 −1 = ρ(ρ (r))(ϕ21 ◦ ρ )(s) + (ϕ21 ◦ ρ )(r)ρ(ρ (s))

−1 −1 = r(ϕ21 ◦ ρ )(s) + (ϕ21 ◦ ρ )(r)s,

−1 so that ϕ21 ◦ ρ = D is a derivation. Thus we have any automorphism of D(P ) will take the form       e 0 ρ 0 e 0             . D e 0 ρe 0 f

Remark 6.1. Given this decomposition we may now write the automorphism group for D(P ) as a product of its subgroups:

Aut(D(P )) = ∆ · G(P ) · F ].

Theorem 6.8.

] Aut(D(P )) = (∆ o G(P )) o F is a semidirect product and this decomposition is unique.

Proof. By         ρ−1 0 e 0 ρ 0 e 0       =   ∈ ∆,  −1       −1  0 ρe D e 0 ρe ρe Dρ e 49 and         e 0 e 0 e 0 e 0               =   ∈ ∆, 0 f −1 D e 0 f f −1D e we see that ∆ is a normal subgroup of Aut(D(P )). Hence we can form the semidirect product

∆ o G(P ) with the subgroup G(P ). Then         e 0 ρ 0 e 0 ρ 0       =   ∈ ∆ o G(P ),  −1       −1  0 f Dρ ρe 0 f f Dρ ρe shows that ∆ o G(P ) is normal in Aut(D(P )) and thus we can form the semidirect product

] (∆ o G(P )) o F .

This decomposition provides a unique decomposition of Aut(D(P )).

50 CHAPTER SEVEN

Zero Product Determined Algebras

There is much ongoing work in the study of zero product determined algebras.

We will reproduce here the most signiﬁcant results relevant to our discussion so that the reader need not search out the relevant papers. Throughout, let F be a ﬁeld, A an F -algebra, and X an F -module.

7.1 Prior Results

Deﬁnition 7.1. An F -bilinear map B : A × A → X is zero preserving if, for all a, b ∈ A, the equation ab = 0 implies B(a, b) = 0.

Deﬁnition 7.2. We call A, zero product determined if for all zero preserving maps B there exists ϕ ∈ HomF (A, X) with B(a, b) = ϕ(ab) for all a, b ∈ A.

Theorem 7.1. [2] A is ZPD if and only if given B is a zero preserving map and

k X aibi = 0, i=1 then k X B(ai, bi) = 0. i=1 Furthermore, if A is unital then we may ﬁx k = 2.

Theorem 7.2. [1] If a unital algebra A is generated as an F -algebra by its idempotents, then A is ZPD.

The proof is reproduced with some notation changes from [1]

Proof. Let B : A×A → X be such that xy = 0 implies that B(x, y) = 0. If e = e2 ∈

A, then xe·(1−e)y = x(1−e)·ey = 0. Therefore B(xe, y −ey) = B(x−xe, ey) = 0.

That is, B(xe, y) = B(xe, ey) and B(x, ey) = B(xe, ey). Comparing, we get 51 B(xe, y) = B(x, ey). This means that the set R = {z ∈ A : B(xz, y) = B(x, zy) for all x, y ∈ A} contains all idempotents. As R is readily a subalgebra of A, it follows that R = A.

Therefore B(x, z) = B(xz, 1) = ϕ(xz), where ϕ(w) = B(w, 1).

Deﬁnition 7.3. Deﬁne the linear map µ : A ⊗ A → A by µ(a ⊗ b) = ab.

Theorem 7.3. [3] A is zero product determined if and only if

Ker µ = spanF {a ⊗ b : ab = 0} .

7.2 Cartesian Products of Zero Product Determined Algebras

Theorem 7.4. F as an F -algebra is ZPD.

Proof. Let B : F × F → X be F -bilinear and zero preserving and let

2 X aibi = 0, i=1 for ai, bi ∈ F . Then

2 2 X X B(ai, bi) = B(1, aibi) i=1 i=1 2 ! X = B 1, aibi i=1 = B(1, 0)

= 0.

It is a result of [3] that the class of ZPD algebras is closed with respect to direct sums. The following example demonstrates that this is not the case with Cartesian products.

Q Theorem 7.5. Let F be an infinite field and A := i∈I F ei with I infinite such that P |I| ≤ |F |, or equivalently the set of all formal sums i∈I fiei with fi ∈ F and the ei 52 are the standard basis vectors. Thus A is a commutative F -algebra where addition and multiplication of elements is component wise. Then A is not a zero product determined F -algebra.

Proof. We will be using Theorem 7.3, but we would like an alternate representation P P of A ⊗F A. Hence we construct the set i,j∈I (ei ⊗ ej)F := i,j∈I eijF . We can see P that A ⊗F A is a subset of i,j∈I eijF :

X X a ⊗ b = aiei ⊗ bjej i∈I j∈I ! X X = aiei ⊗ bjej i∈I j∈I X X = (aiei ⊗ bjej) i∈I j∈I X = aibj(ei ⊗ ej) i,j∈I X = aibjeij. i,j∈I

Now deﬁne the map X a ⊗ b 7→ aibjeij, i,j∈I where a = (ai)i∈I and b = (bi)i∈I . We start by showing that this map is an inclusion.

l l Deﬁne πi : A ⊗ A → A ⊗ A by πi(a ⊗ b) = (aiei) ⊗ b, i.e. aiei consists of the vector all of whose entries are 0 except the ith entry which is ai. We can similarly

r deﬁne πj . Then the map above is explicitly

X l r X X X πi ◦ πj (a ⊗ b) = (aiei ⊗ bjej) = aibj(ei ⊗ ej) = aibjeij. i,j∈I i,j∈I i,j∈I i,j∈I

Pn k k Note, that for an arbitrary element of y = k=1 a ⊗ b ∈ A ⊗F A, if the representation chosen is such that n is minimal, then the set bk : 1 ≤ k ≤ n is linearly independent over F .

53 n Pn−1 k To see this, without loss of generality we may assume that b = k=1 fkb some fk ∈ F . Then n n−1 ! n−1 X k k n X k X k k a ⊗ b = a ⊗ fkb + a ⊗ b k=1 k=1 k=1 n−1 n−1 X n k X k k = a ⊗ fkb + a ⊗ b k=1 k=1 n−1 n−1 X n k X k k = fka ⊗ b + a ⊗ b k=1 k=1 n−1 X n k k = fka + a ⊗ b , k=1 which is contradiction to the minimality of n.

Pn k k l Suppose k=1 a ⊗ b is in the kernel of πi and that the representation chosen minimizes n. Then n ! l X k k 0 = πi a ⊗ b k=1 n X k k = ai ei ⊗ b k=1 n X k k = ei ⊗ ai b k=1 n ! X k k = ei ⊗ ai b . k=1 k k However, since the b are linearly independent, this must mean that ai = 0 for all k.

Pn k k P j r Thus, if k=1 a ⊗ b is in the kernel of i,j∈I πi ◦ πj we must have that k k Pn k k ai = 0 = bj for all i, j ∈ I and 1 ≤ k ≤ n. Hence k=1 a ⊗ b = 0. Thus the map is an inclusion map. P We will call (γij)i,j∈I ∈ i,j∈I F eij separable if for each i, j ∈ I then γij = aibj for some a = (ai)i∈I and b = (bi)i∈I in A. Then it should be clear from the inclusion map above that ∼ A ⊗F A = spanF {γ : γ is separable} . 54 It should be noted that the kernel of µ is not trivial. For example, for all a, b ∈ A we have

µ(a ⊗ b − b ⊗ a) = µ(a ⊗ b) − µ(b ⊗ a) = ab − ba = 0, since A is commutative. However, letting θ : I → F be 1-1 (since |I| ≤ |F |) if we take a = (ai)i∈I with ai = θ(i) and b = (bj)j∈I with bj = θ(j) + 1 then

X X a ⊗ b − b ⊗ a = eijaibi − eijbiaj i,j∈I i,j∈I X = eij(aibj − biaj) i,j∈I X = eij(θ(i)(θ(j) + 1) − (θ(i) + 1)θ(j)) i,j∈I X = eij(θ(i) − θ(j)), i,j∈I which is not zero for all i 6= j. Furthermore, note that ab = ba 6= 0.

If A is ZPD, then by Theorem 7.3 we should have that y := a ⊗ b − b ⊗ a can be written: n X y = a(k) ⊗ b(k), with a(k)b(k) = 0 for all k. k=1

Suppose A is ZPD. Given a ∈ A deﬁne Supp(a) := {i ∈ I : ai 6= 0}. Note that ab = 0 if and only if Supp(a) ∩ Supp(b) = ∅ since multiplication is component wise. Also deﬁne ! X Supp (a ⊗ b) = Supp aibjeij i,j∈I

:= Supp (a) × Supp (b) .

Since y = a ⊗ b − b ⊗ a then Supp(y) = I × I − {(i, i): i ∈ I} := (I × I)∗ since the only zero terms were those along the diagonal where i = j. We should note that by

(k) (k) (k) (k) hypothesis a b = 0 and thus ai bi = 0 for all i and k. Hence ! (k) (k) X (k) (k) ∗ Supp a ⊗ b = Supp ai bj eij ⊆ (I × I) , i,j∈I 55 i.e. all diagonal terms are zero. We can then say the following

n ! X Supp(y) = Supp a(k) ⊗ b(k) k=1 n ! X X (k) (k) = Supp ai bj eij k=1 i,j∈I n ! [ X (k) (k) ⊆ Supp ai bj eij (7.1) k=1 i,j∈I n [ = Supp a(k) × Supp b(k) k=1 n [ = (Uk × Vk) , k=1

(k) (k) where Uk = Supp a and Vk = Supp b . We must have then

n ∗ [ (I × I) ⊆ (Uk × Vk) . k=1 It should be clear then that

n n [ [ I = Uk = Vk. k=1 k=1

Since I is inﬁnite, without loss of generality, we may assume U1 is inﬁnite. Now

(k) (k) we know that a b = 0. Thus by 7.1 we have (i, i) ∈/ Uk × Vk for all i ∈ I and

1 ≤ k ≤ n. Thus, for any k we must have that Uk ∩ Vk = ∅. We can then say that

∗ (U1 × U1) ∩ (U1 × V1) = ∅.

It is also clearly true that

n ∗ ∗ [ (U1 × U1) ⊆ (I × I) ⊆ (Uk × Vk) . k=1 Thus we must have that

n ∗ [ (U1 × U1) ⊆ (Uk × Vk). k=2

56 n n This must mean that U1 ⊆ ∪k=2Vk, for if not then there is some i ∈ U1 not in ∪k=2Vk,

∗ n but then for j ∈ U1, j 6= i we have (j, i) ∈ (U1 × U1) and (j, i) ∈/ ∪k=2(Uk × Vk) which is a contradiction.

Again without loss of generality we may assume that U1 ∩ V2 is inﬁnite. Fur- thermore since U2 ∩ V2 = ∅ we then have (U1 ∩ V2) ∩ U2 = ∅ and so

∗ ((U1 ∩ V2) × (U1 ∩ V2)) ∩ (U2 × V2) = ∅.

Thus we must have that

n ∗ [ ((U1 ∩ V2) × (U1 ∩ V2)) ⊆ Uk × Vk, k=3 since we still have that

n ∗ ∗ [ ((U1 ∩ V2) × (U1 ∩ V2)) ⊆ (U1 × U1) ⊆ Uk × Vk. k=2

n We thus now have that (U1 ∩V2) ⊆ ∪k=3Vk so as before we may assume (U1 ∩V2)∩V3 is inﬁnite.

Continuing in this manner we have

∗ ((U1 ∩ V2 ∩ · · · ∩ Vn−1) × (U1 ∩ V2 ∩ · · · ∩ Vn−1)) ⊆ Un × Vn,

with (U1 ∩V2 ∩· · ·∩Vn−1), an inﬁnite set. But this is a contradiction. To see this note that if there exists some i ∈ (U1 ∩ V2 ∩ · · · ∩ Vn−1) with i∈ / Vn then for j ∈ (U1 ∩ V2 ∩

∗ · · · ∩ Vn−1) with j 6= i then (j, i) ∈ ((U1 ∩ V2 ∩ · · · ∩ Vn−1) × (U1 ∩ V2 ∩ · · · ∩ Vn−1)) but (j, i) ∈/ Un × Vn. Thus

(U1 ∩ V2 ∩ · · · ∩ Vn−1) ⊆ Vn, and

(U1 ∩ V2 ∩ · · · ∩ Vn−1) ⊆ Un, which contradicts Un ∩Vn = ∅. Hence we have that y∈ / spanF {a ⊗ b : ab = 0}. Thus A is not ZPD. 57 Theorem 7.6. The epimorphic image of a ZPD algebra is ZPD.

Proof. Let A be an F -algebra and L the epimorphic image of A and K its kernel.

Then L ∼= A/K. Deﬁne ¯ : A → L byx ¯ = x + K, ie the coset containing x. Let ψ : L × L → X be a zero-preserving bilinear map. Now deﬁne h : i : A × A → X by hx : yi = ψ(¯x, y¯). Then h : i is bilinear:

hx + x0 : yi = ψ(x + x0, y¯)

= ψ(¯x + x0, y¯)

= ψ(¯x, y¯) + ψ(x0, y¯)

= hx : yi + hx0 : yi, and

hkx : yi = ψ(kx, y¯)

= ψ(kx,¯ y¯)

= kψ(¯x, y¯)

= khx : yi.

If xy = 0 this impliesx ¯y¯ = 0 so hx : yi = ψ(¯x, y¯) = 0 since ψ is zero preserving.

Since A is ZPD we have that there exists some T : A → X with hx : yi = T (xy).

Now let y ∈ K which implies that

T (y) = T (1y) = h1 : yi = ψ(1¯, y¯) = ψ(1¯, 0) = 0.

Thus K ⊆ Ker(T ). Deﬁne T : L → X by T (x + K) = T (¯x) = T (x). Then T is well deﬁned since K ⊆ Ker(T ). We then have

ψ(¯x, y¯) = hx : yi = T (xy) = T (xy) = T (¯xy¯).

Therefore L is ZPD.

58 Q Corollary 7.1. Let F be an infinite field and A := i∈I F ei with I infinite, or equiv- P alently the set of all i∈I fiei with fi ∈ F and the ei are the standard basis vectors. Thus A is a commutative F -algebra where addition and multiplication of elements is component wise. Then A is not zero product determined.

Proof. If |I| ≤ |F | then this is Theorem 7.5.

Suppose |I| > |F |. Now project I onto any subset I0 such that |I0| ≤ |F |.

Then A0 generated by this index is a projection of A and thus an epimorphic image of A. Now by Theorem 7.5 A0 is not ZPD and so by the contrapositive of Theorem

7.6 A is not ZPD.

Q Theorem 7.7. Let F be a field and I an infinite index set. Then A = i∈I F ei is generated as an F -algebra by idempotents if and only if F is finite.

∼ Proof. First, let us assume that I is countable (I = N). An element δ ∈ A is an idempotent if and only if all its entries are either 0 or 1. Since A is commutative, any ﬁnite set ∆0 of idempotents is contained in a ﬁnite set ∆ of idempotents which is closed under multiplication and generated by ∆0. Thus, the subalgebra generated

0 by ∆ is the F -algebra S∆ = spanF {∆}. Hence α = (αn)n∈I ∈ S∆ means that there P are qδ ∈ F for all δ ∈ ∆ such that α = δ∈∆ qδδ. It follows that for all n ∈ N we have αn ∈ spanZ {qδ : δ ∈ ∆} a ﬁnitely generated abelian group. Note that this implies that spanZ {an : n ∈ N} ⊆ spanZ {qδ : δ ∈ ∆}. Suppose that char(F ) = 0. If A is generated by idempotents then for α ∈ A there is some ﬁnite set of idempotents ∆0 contained in ∆ the multiplicatively closed

0 set of idempotents generated by ∆ such that α ∈ S∆. Since char(F ) = 0 then there is an embedding of Q in F and hence F may be viewed as a Q-vector space with

F = Q ⊕ V where V is the vector space complement. Now let π : F → Q denote 1 the natural projection with kernel V . Also, let α = (αn)n∈N with αn = 2n . Then 1 span {α : n ∈ } = span π(α ) = : n ∈ ⊆ span {π(q ): δ ∈ ∆} . Z n N Z n 2n N δ 59 But the last set is ﬁnitely generated over Z and the second set cannot be, thus a contradiction to α being generated by a ﬁnite number of idempotents.

Now suppose char(F ) = p with F still inﬁnite. Then F is a Zp-vector space.

F must have an inﬁnite dimension over Zp and hence there exists an inﬁnite subset

{βn : n ∈ N}, of F that is linearly independent in the Zp-module F . If A is generated by its idempotents then for α = (αn)n∈N with αn = βn there must exist some ﬁnite P set of idempotents ∆ such that αn = δ∈∆ zδqδ ∈ spanZ {qδ : δ ∈ ∆} for zδ ∈ Z. Similar to the previous case we then have

span {β : n ∈ } = span {β : n ∈ } ⊆ span {q : δ ∈ ∆} . Zp n N Z n N Zp δ

Again, the last is ﬁnitely generated while the ﬁrst is not. Hence again a contradiction.

If I is inﬁnite but not countable there exists a surjective ring homomorphism

ϕ : Q F e → Q F e . If Q F e were generated by idempotents then so would i∈I i n∈N n i∈I i Q F e . n∈N n Thus we can see for I and F infinite A is not generated by its idempotents. In the case where I is infinite but F is finite, then for α ∈ A we define P Uf = {i ∈ I : αi = f}. Then we define eUf = i∈I χei where χ = 1 for i ∈ Uf P and 0 otherwise. We can then write α = f∈F feUf an F -linear combination of idempotents.

7.3 FI(P ) as a ZPD Algebra

Theorem 7.8. Suppose F and P are inﬁnite. Then FI(P ) is not ZPD.

Proof. Assume FI(P ) is ZPD. Then by Theorem 7.6 FI(P ) modulo is Jacobson radical should also be. However we have

FI(P ) Y ∼= F e , J(FI(P )) x x∈P since all that is left are the diagonal terms. By Corollary 7.1 we know the latter cannot be ZPD. Thus a contradiction. 60 Theorem 7.9. Suppose P is ﬁnite. Then FI(P ) is ZPD.

Proof. Let α ∈ FI(P ). We will show that α can be generated as an F -linear combination of idempotents. We begin by deﬁning α∗:   α(x, y) x 6= y α∗(x, y) = . (7.2)  1 x = y

∗ ∗ Thus α is identically equal to α except along the diagonal. Now note that exα is idempotents since

∗ ∗ ∗ ∗ (exα )(exα ) = (exα ex)α

∗ ∗ = α (x, x)exα

∗ = exα .

∗ Similarly α ex is also idempotent. Then we simply note that

∗ X α = α − e + α(x, x)ex x∈P

X ∗ ∗ X = (exα + α ex) − e + α(x, x)ex x∈P x∈P

X ∗ X ∗ X = −e + exα + α ex + α(x, x)ex. x∈P x∈P x∈P Since P is finite, the final equality is a finite F -linear combination of idempotents.

Hence FI(P ) is generated by its idempotents and is thus ZPD.

Theorem 7.10. Suppose F and P are ﬁnite. Then A := Q FI(P ) is ZPD. N

Proof. Let α ∈ A. Since multiplication is componentwise, α = (α ) is idempotent n n∈N ∗ if αn is idempotent for each n ∈ N. Again using α as deﬁned in 7.2, for each n ∈ N we may write

X ∗ X ∗ X αn = −e + exαn + αnex + αn(x, x)ex. x∈P x∈P x∈P

61 Thus

α = (α ) n n∈N ! X ∗ X ∗ X = −e + exαn + αnex + αn(x, x)ex x∈P x∈P x∈P n∈N ! ! ! X ∗ X ∗ X = −(e)n∈N + exαn + αnex + αn(x, x)ex x∈P x∈P x∈P n∈N n∈N n∈N X X X = −(e) + (e α∗ ) + (α∗ e ) + (α (x, x)e ) . n∈N x n n∈N n x n∈N n x n∈N x∈P x∈P x∈P Unfortunately this decomposition is not enough. Speciﬁcally, the last term in this sum, although ﬁnite, is not represented as a linear combination of idempotents since

α(x, x)ex is not necessarily idempotent. It is here that we make use of the fact that

F is also finite. We define Uf := {x ∈ P : αn(x, x) = f} and define   1 x = y ∈ Uf eUf (x, y) = .  0 otherwise

Then we may rewrite the last term: ! ! X X X α (x, x)e = fen = f en . n x Uf Uf n∈N x∈P f∈F f∈F n∈N n∈N Note that this sum is ﬁnite since F is ﬁnite. Since en is clearly idempotent for each Uf n ∈ N we now have

X X X α = −(e) + (e α∗ ) + (α∗ e ) + f en , n∈N x n n∈N n x n∈N Uf n∈N x∈P x∈P f∈F a ﬁnite F -linear combination of idempotents. Hence A is generated by idempotents and thus is ZPD

Remark 7.1. Note that the algebra A in Theorem 7.10 is isomorphic to FI(Q) where

Q is the countable poset whose connected components are all isomorphic to P . Thus, for a finite field F , there are nontrivial infinite posets Q such that FI(Q) is ZPD.

62 CHAPTER EIGHT

Future Research

In attempting to determine the relationship between FI(P ), its idempotents, and for what conditions on F and P , the algebra FI(P ) is zero product determined, certain results have been more general than for the class of incidence algebras. In the future I would like to continue down this path and study the properties of algebras generated by their idempotents. In particular there are questions as to the closure properties under specific operations for such algebras. Moreover, given a finite field

F , for which inﬁnite posets P is FI(P ) generated by idempotents and/or ZPD?

I would also like to continue studying properties of FI(P ) and D(P ). In particular I have done some preliminary research into the ring theoretic properties of FI(P ) in how regular elements and units relate. In [7] Spiegel and O’Donnell proved a concise result as to this relationship. Although I have able to determine that the same result does not carry over into the case of FI(P ), I have yet to provide any satisfactory generalization.

63 BIBLIOGRAPHY

[1] Matej Breˇsar, Multiplication algebra and maps determined by zero products, Linear Multilinear Algebra 60 (2012), no. 7, 763–768.

[2] Matej Breˇsar,Mateja Graˇsiˇc,and Juana S´anchez Ortega, Zero product determined matrix algebras, Linear Algebra Appl. 430 (2009), no. 5-6, 1486–1498.

[3] Huajun Huang Daniel Brice, Direct sums of zero product determined algebras, arXiv (2011).

[4] Robert E. Hartwig and Mohan S. Putcha, When is a matrix a diﬀerence of two idempotents?, Linear and Multilinear Algebra 26 (1990), no. 4, 267–277.

[5] N. S. Khripchenko and B. V. Novikov, Finitary incidence algebras, Comm. Algebra 37 (2009), no. 5, 1670–1676.

[6] Nikolay Khripchenko, Automorphisms of ﬁnitary incidence rings, Algebra Dis- crete Math. 9 (2010), no. 2, 78–97.

[7] Eugene Spiegel and Christopher J. O’Donnell, Incidence algebras, Monographs and Textbooks in Pure and Applied Mathematics, vol. 206, Marcel Dekker Inc., New York, 1997.