factorisation theory in a non-commutative algebra

8/22/2019 Factorisation Theory in a Non-commutative Algebra

1/58

FACTORISATION THEORY IN A NON-COMMUTATIVEALGEBRA

Stephen Ozvatic

Supervised by Dr Daniel Chan

School of Mathematics,The University of New South Wales.

October 30, 2009

Submitted in partial fulfillment of the requirements of the degree of

Bachelor of Science with Honours


2/58

I hereby declare that this submission is my own work and to

the best of my knowledge it contains no materials previouslypublished or written by another person, nor material which toa substantial extent has been accepted for the award of anyother degree or diploma at UNSW or any other educationalinstitution, except where due acknowledgement is made inthe thesis. Any contribution made to the research by others,with whom I have worked at UNSW or elsewhere, is explicitlyacknowledged in the thesis.

I also declare that the intellectual content of this thesis is theproduct of my own work, except to the extent that assistance

from others in the projects design and conception or in style,presentation and linguistic expression is acknowledged.

Stephen Ozvatic

i


3/58

Acknowledgements

I dont really know what my mum wanted me to be when I grew up. She might saya pilot among other things since she has always liked traveling, planes and Europe,her birthplace. She was told by one of my primary school teachers many years agothat I was good at maths, so perhaps she had hope for a few short years that itwould eventuate, but I quickly threw that idea out. Now Im not saying Im goodat maths (I found out Im probably the opposite), but Ive always been interestedin it and it was probably a surprise when my mum found out I decided I was going

to study it at university, since my family migrated here to Australia from a farm inEurope. Now four years later Im here and have this thesis to show for it. It mayonly be an honours thesis to some, but whether this thesis is good or terrible, trivialor complicated, or if I get a mark of 50 (hopefully I wont get below this) or 95,it wont impact on how proud I am to have completed this; though 95 would be nice.

Id like to thank my supervisor Daniel along with Hendrik, Gary, Ian and Anthonywho helped me along the way during my studies. My thanks also go to my supportand distraction Sonia, my colleagues Hugh, Oliver and Roland among others, myfriends, and my family: Ana, Maria and Joe.

Stephen Ozvatic

ii


4/58

Prerequisites

Though most definitions and the proofs of theorems are included within, at leastan elementary knowledge of ring theory, factorisation theory and module theoryis required. If you have not heard the words homomorphism, integral domain ormodule, then refer to the references page for a list of suitable sources. For anyproofs that are not included here, there will be references pointing to proofs inwidely accepted sources; and for any material that is required conceptually, therewill be references given when required.

iii


5/58

Introduction

The theory of Dedekind domains comes from the studies of the factorisation prop-erties of the ring of algebraic integers in an algebraic number field . RichardDedekind (1831-1916) in the latter half of his life, knowing that was not alwaysa unique factorisation domain, looked at fractional ideals of rather than ele-ments and saw that there was always a unique factorisation of them. Through hiswork, he provided the concepts ofideals, modules (though he wrote modul), definedprime ideals (that is, generalised prime numbers), introduced the word field and

perhaps most importantly, the concept of a ring is due to him, though the termring first appears due to Hilbert and our axiomatic definitions were not introduceduntil the 1920s by Emmy Noether and Krull.

The work done on these rings by Dedekind were part of the third (1879) andfourth (1894) editions of Vorlesungen ber Zahlentheorie and the notions createdeventually lead to be fundamental to ring theory, allowing him to provide an alge-braic proof to the Riemann-Roch Theorem after only three years.

So what is a Dedekind domain? They are basically rings where each ideal isable to be factorised into prime ideals. If this sounds familiar it probably is sinceeach integer, an element of, has that same property. That is, each integer can befactorised into prime integers. It turns out that the integers are also a Dedekinddomain themselves, but this understates their usefulness. If we look at roots ofmonic polynomials

+ 11 + + 1 + 0

with coefficients in , then the set of these roots are called algebraic integers anddenoted . If we now look at a finite extension of the field , then the set is actually a Dedekind domain denoted by and called the ring of algebraicintegers of . This is the ring studied by Dedekind in which he found the uniquefactorisation property of ideals, even though the unique factorisation of elements in

this ring does not always hold.With a more modern theory available, this construction which is now often used

as an introduction to algebraic number theory, is mainly used as an example asDedekind domains are more powerful than this. For example, every principal idealdomain is a Dedekind domain. So in Chapter 1 we will look at Dedekind domainsand see their remarkable property of ideal factorisation and furthermore we willsee that in fact the set of all these ideals, to be called fractional ideals, forms anabelian group and is generated by the prime ideals of the Dedekind domain andtheir inverses, which will also be looked at.

This theory will then act as background knowledge as in the next two chapters

we generalise it to a non-commutative setting. We do this by definingorders

in

iv


6/58

semisimple algebras, which can be conceptually thought of as a subring that is alattice in space (the space here being a semisimple algebra). The theory of ordershere depends heavily on the requirement that they be maximal in the sense thatthere is no other order larger than it (perhaps other than the algebra it is in). Usingthis conceptual definition, a Dedekind domain also forms a lattice in space that is

a subring of its quotient field. For example the integers form a lattice in therationals and are a Dedekind domain, as has been mentioned.

However when moving to orders, there is actually two kinds of ideal factorisationto consider. The first is two-sided ideals, the second one-sided ideals. Obviouslythe one-sided case encompasses the two-sided case, but there are certain propertiesin the two-sided case that can be related back to Dedekind domains, where thesefail in the one-sided case. More specifically, the set of all two-sided ideals in amaximal order is an abelian group generated by the prime ideals of the maximalorder. However, in the one-sided case no such nicety occurs. The set of one-sidedideals of a single maximal order can not even be considered. It turns out that we

have to consider all ideals of all maximal orders and in that case it turns out to bea groupoid, a generalisation of a group, which will be called the Brandt Groupoidassociated to the semisimple algebra.

Finally, we will look briefly at some nice results due to the theory from Chapters1,2 and 3. This will be in regard to seeing when the factorisation of elements inDedekind domains and orders occurs (and also when unique factorisation of elementsin an arbitrary ring implies the same for ideals), as well as looking at the first caseof a Dedekind domain, the ring of algebraic integers. But nicest of all (for theauthor of this thesis anyway), we will look at the rational quaternions, an extensionof the complex numbers, and then finish with a proof due to the preceding theory

in Chapters 1,2 and 3 of Lagranges Four Squares Theorem: That every naturalnumber can be written in the form

2 + 2 + 2 + 2

for integers ,, and .

v


7/58

Contents

Chapter 1 Dedekind Domains and Fractional Ideals 11.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Dedekind Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The Factorisation of Ideals . . . . . . . . . . . . . . . . . . . . . . . 7

Chapter 2 Dedekind Domains and Orders 132.1 Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Maximal Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 The Factorisation of Two-Sided Ideals . . . . . . . . . . . . . . . . 20

Chapter 3 Maximal Orders and Groupoids 263.1 One-Sidedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.2 Generalising a Familiar Sight . . . . . . . . . . . . . . . . . . . . . 303.3 The Factorisation of One-Sided Ideals . . . . . . . . . . . . . . . . . 33

Chapter 4 Factorisation...of Elements 394.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.2 Factorisation, the Class Group and The Class Number . . . . . . . 404.3 Finiteness of The Class Number of the Ring of Integers . . . . . . . 424.4 Factorisation in an Order and the Jordan-Zassenhaus Theorem . . . 444.5 Quaternions and a Sum of Four Squares . . . . . . . . . . . . . . . 47

References 51

vi


8/58

Chapter 1

Dedekind Domains and Fractional Ideals

To start with, we will look at a standard view of the theory, with the end of therainbow not being home to an Irishman with a pot of gold, but rather a uniquenessresult about Dedekind domains. Chapter 1 is mainly to do with the commutativecase, however a more general case will be done after this. So to keep a standard,unless defined or stated otherwise commutativity is not assumed and every ring isunital. Firstly though, some preliminaries must be carried out, but most of these

should be familiar to the reader. If is a domain, it is meant that is a ring withno zero divisors. A ring is then an integral domain if it is a commutative domain.Assume hereon for the whole document that is a ring with quotient ring where = and note that if is an integral domain then is a field. For clarity,the notation will denote a strict subset whereas indicates the possibility ofequality.

1.1 Background

For a ring , not necessarily commutative, a left -module is an additive abeliangroup with a map

: (, )

such that for ,, 1

and ,

( + ) = + ( + ) = + () = () 1 = .

Similarly, a right and two-sided module can be defined. The left -module iscalled finitely generated if it can be written as = 1 + 2 + + for afinite number, , of elements . An -algebra is a ring such that thereexists a homomorphism : () where () = { = for every } is the centre of . In this document, an algebra is always assumed to beassociative. To see an as an -module, we define

= () for

and

. The dimension of a -algebra over the field is its dimension as avector space over .

Given an integral domain , let be a finite dimensional -algebra. An element is integral over if there is some monic polynomial () [] with () =0 and similarly a subring is integral over if every element of is integralover . Hence the set of all elements of integral over is called the integral closure

of in and is denoted

. is then called integrally closed if the integral closureof in is itself. For an integral domain and finite dimensional -algebra ,the following two results will be useful at times throughout this document. Theyare standard results so will not be proved here, but proofs can be found in Reiner

[1, p.3-6], Jacobson [6, p.408-409] and Janusz [7, p.5-7].

1


9/58

Proposition 1.1.1 The following are equivalent for an element .a) is integral over .b) [] is a finitely generated -module.c) There is finitely generated -module that is a subring of containing .

Proposition 1.1.2 If is integrally closed, then is integral over if andonly if the minimal polynomial of over , (), is an element of [].

Some concepts of factorisation theory are required in a later chapter, but arealso needed conceptually at times throughout the document, so a basic introductionwill be given here. For the reader not familiar with this material, it can be foundin Jacobson [5, p.140-149], Stewart & Tall [8, p.81-99] and Sivaramakrishnan [10,2,3]. For an integral domain , an element is called a unit if there is some such that = = 1. The set of units in , (), is then an abeliansubgroup of under multiplication. Then if two elements , satisfy =

for some () we call and associates

. If = for ,, , we say that (or equivalently ) divides and denote this by . An element is calledprime when for two elements , , if then either or . A non-zeronon-unit element is irreducible if for every factorisation = for , ,either () or (). See that both and can not be in () since thisforces (). Also note that irreducible elements are prime, but the converse isnot necessarily true.

Example 1.1.3 Every irreducible element in an integral domain is prime sincea factorisation = with , has, without loss of generalisation, ().So and are associates and .

Now we can define certain types of rings. Here they are commutative, but thedefinitions can be generalised to a non-commutative case. The integral domain is called an Unique Factorisation Domain (or UFD) when for every element ,there exists a factorisation of into irreducible elements of and if there are twofactorisations = 1 = 1 with , irreducible for all and ,then = and = () for some permutation perm{1, , } and unit () for every . That is, we can swap around the s and multiply them byunits to get 1 .

Certain rings have a nice property when looking at their ideals, so we will definethose rings here. Given a subset , the ideal generated by is andis denoted

. An ideal is then called a principal ideal if =

= for some . That is, it is generated by one element of . An integral domain is thuscalled a Principal Ideal Domain (or PID) if every ideal of is a principal ideal. Itturns out that the following results hold, but note importantly that the converse ofTheorem 1.1.5 does not hold.

Proposition 1.1.4 Every prime element in a UFD is irreducible.

Proof. Let be a UFD and be prime. If = for two elements , ,then and without loss of generality say that . Then = for some . But then = = and hence = 1 since is an integral domain,

implying () and is irreducible.

2


10/58

Theorem 1.1.5 Every PID is a UFD.

Proof. Let be a PID and first suppose that there is some that is notexpressible as a product of irreducibles and so can not be irreducible itself. Then = 11 for some non-units 1 and 1 where at least one is not expressible as aproduct of irreducibles as well. Suppose without loss of generalisation this is 1.Then we can write 1 = 22 for some non-units 2 and 2 where 2 is not expressibleas a product of irreducibles. Continuing this gets an infinite chain of proper ideals 1 2 . Using Proposition 1.1.7 from ahead gets that this chaineventually stabilises, so one of the s must be ireeducible, contradictiong that is not expressible as a product of irreducibles.

Secondly, write as = 12 = 12 for irreducible and. Since they are also prime we have 1 12 , so 1 for some andsince 1 and are irreducible, 1 = 1 for some unit 1 (). Now 2 =1

1+1

, so doing this repeatedly gets = and each an associate of

some with 1 = 1 . Thus is uniquely expressible as a product of irreducibleelements of up to rearrangement and associates, so is a UFD.

Example 1.1.6 The integers have ideals of the form , so they are a PID andthus a UFD. The units of are () = {1, 1} and the irreducible integers are theprime integers.

Finally, a ring is called left noetherian if its left ideals satisfy the ascendingchain condition (that is, a chain of left ideals 0 1 2 of eventually has = for every .) and left artinian if its left ideals satisfy the descendingchain condition (similarly to before but with a decreasing chain instead). This

similarly holds for right noetherian and artinian, and a ring is just called noetherianor artianian if the respective chain conditions hold for both right and left ideals.Similarly, an -module is called noetherian or artinian if its submodules satisfythose same conditions. A prime ideal of an integral domain is a non-zero properideal such that / has no zero divisors. This definition is equivalent to sayingthat given ideals , and , if implies or , then is prime(which is analogous to the usual definition of an element being prime, as givenabove). The following result about a PID will be useful. In fact, we have alreadyused it.

Proposition 1.1.7 Any PID is noetherian.

Proof. Let be a PID, 0 1 2 be an infinite chain of ideals in and = 0. Let , , so and for some and and without loss ofgenerality suppose that . Then since , + since is a groupand for , also, showing is in fact an ideal of. Since is a PID,every ideal is principal, so = for some irreducible . Now , so for some and 0 1 = +1 = , showing is noetherian.

Example 1.1.8 A maximal ideal of an integral domain is a prime ideal since/ is a field.

3


11/58

1.2 Dedekind Domains

Now using these definitions we can define a certain sort of ring, called a Dedekinddomain or Dedekind ring, which is the assumption taken upon many rings in thisdocument. As will be found by the end of the document, these types of rings all

have a property that can help when trying to establish when elements of can beexpressed uniquely in terms of its irreducible elements, that is, when is a uniquefactorisation domain. But their main property, as we will find, is their ability tofactorise of ideals.

The theory of Dedekind domains comes from the studies of the factorisationproperties of the ring of algebraic integers in an algebraic number field .Dedekind (1831-1916) in the latter half of his life, knowing that was not alwaysa UFD, looked at fractional ideals of rather than elements and saw that therewas always a unique factorisation of them. This lead to the use of the word idealin its common usage and the use of his name for the ring we are about to define.

References for the rest of this chapter are Reiner [1, p.44-50], Curtis & Reiner [4,

18], Jacobson [6, 10], Janusz [7, p.8-18], Samuel [9, p.47-52] and Sivaramakrishnan[10, p.394-410]. There are other definitions of a Dedekind ring involving projectivityand localizations and discrete valuation rings, but they will not be looked at. Ourdefinition will be as follows.

Definition 1.2.1 An integral domain is called a Dedekind domain if it is noethe-rian, integrally closed and such that every non-zero prime ideal is maximal.

This definition is equivalent to a remarkable fact about the factorisation ofideals in the integral domain into irreducibles. You might guess that theseirreducibles are the maximal ideals of and in that case, you would be right.

Except when is a Dedekind domain prime ideals coincide with the maximal ideals,or irreducible ideals, which is reminiscent of the property in unique factorisationdomains that prime elements are irreducible and every element can be written as aunique product of prime elements. The reader may also guess that principal idealdomains may be important here, and again they would be right as the followingexamples give an indication of.

Example 1.2.2 As with many things in number theory, some concepts are vastgeneralisations of properties held by the integers. The PID is a noetherian inte-grally closed integral domain with quotient field . Since its prime ideals are of theform for a non-zero prime integer, the prime ideals coincide with the maximal

ideals of and hence it is a Dedekind domain.Proposition 1.2.3 Every PID is a Dedekind domain.

Proof. Proposition 1.1.7 shows us every PID is noetherian, so we only need to showthe other two requirements. Let be a PID and be a prime ideal in . Thenfor two ideals and of that have = , it follows that .But then or since is prime, so we have either or ,showing is prime as an element of . Theorem 1.1.4 then implies is irreducible,so is actually a maximal ideal.

For the proof of being integrally closed, the usual proof will do. Let

and

suppose it is a root of a polynomial + 11 + + 1+ 0 []. Then

4


12/58

since is a UFD, =

where and are coprime. Then substituting

into thepolynomial and multiplying by gets

+ 11 + + 11 + 0 = 0.

Thus must follow and = 1, so and is integrally closed, provingthe proposition.

Example 1.2.4 The ring [5] = 5 is a Dedekind domain. Firstly,

it is noetherian since is noetherian. Secondly, it is integrally closed since if =

+

5 [5] with (, ) and (, ) pairwise coprime is a root of a monicpolynomial + 1

1 + + 1+ 0 [5][], then substituting =

and multiplying by () gets

() + 1()()1 + + 1()1() + 0() = 0,

a ploynomaial in [5][] But divides 0, so it must divide the left hand side,so must divide (). But = +

5, so must divide both and ,implying both and are either 1 or 1, so [5].

Thirdly, let be a prime ideal of [5]. If + 5 , then both of

5 + 525, 25 5 , so adding gets 2 + 52 also. But an integer

if and only if5 too. Thus is of the form +5 for some . If = for some , [5], then = (+ 5)(+ 5) = ,so (Note the multiplication of ideals is given by (1.2) ahead, if required).But and , so can not be prime. Thus must be an irreduciblein [

5]. Let be an ideal of[

5] such that

. Then there is some

such that . But by the Euclidean algorithm there exists , such that + = 1, so = and is maximal.

This example was given for a particular reason. Since 6 = (1+5)(15) =

2.3, it is not a UFD and thus not a PID. This shows the converse of Proposition1.2.3 does not necessarily hold.

Example 1.2.5 For another example, given a finite field extension / consider

the integral closure of in and let = . Then is a ring called thealgebraic integers of (or the ring of integers of ) and is in fact a Dedekinddomain. This example will be reintroduced later in Chapter 4 when talking aboutthe class number.

However from here the relationship between unique factorisation of elements andof ideals will be left until Chapter 4. What we need to focus on is what we are tryingto look at. If we want to factorise ideals lets make sure that we know exactly whatwe mean when we say an ideal. Before we do this, we again need to build up somestandard theory. Tensor products will be used here, but not in their full power, soonly a small notion of them is required. For readers unfamiliar with tensor productsee Curtis and Reiner [4, 12] and Jacobson [6, p.125-133]. For a basic view of aconstruction of a tensor product of two modules, let be an integral domain and

5


13/58

consider a right -module M, left -module N, and the module generated bytheir product

= {finite sums

(, ) , , }.

Now look at the subgroup of generated by the following subset of , for all, , , and .

( + , ) (, ) (, )(, + ) (, ) (, )(,) (, )(,) (, )

Then we define the tensor product of and over to be = /. Herethe tensor product is actually an -module. This is because for , and , ,

((, )) = (,) = ((), ) = ((), ) = ()(, ).

So since is an integral domain, = allows to be an -module. Ifthe tensor product is over the quotient field , then is more specificallya vector space. In the case we deal with, = , the quotient field of , and is commutative. So it is actually just the vector space and for ease ignorethe and just write

= = {finite sums

, }. (1.1)

That is all that is required of tensor products here (actually (1.1) is the themain point), so we will move onto some more definitons. For an integral domain and -module , the annihilator of is defined by = { = 0}and similarly the annihilator of an element is defined by = { = 0}. The element is called an -torsion element if = 0and is called -torsionfree if the only -torsion element in is 0.

The point here of torsion elements is that we want to visualize as 1 inside

. If there is a -homomorphism :

:

(1, ),

then {} is the submodule of these torsion elements in , and being -torsionfree sets {} = 0, giving us what we want. See Reiner [1, p.32-34, 44]for more details on this. Now we can define the following for an integral domain .An -lattice is a finitely generated -torsionfree -module. Each -lattice is an-submodule of the finite dimensional vector space = = . Hence is called a full -lattice in to emphasize that has a -basis of. This leads usto the following, as descibed in Reiner [1, p.47-48] and Swan & Evans [2, p.83-84],which turns out to be the same as the other references listed earlier.

Definition 1.2.6 For a Dedekind domain , a non-zero full -lattice in is calleda fractional -ideal.

6


14/58

Note here that since the vector space of a fractional -ideal is actuallythe field , it trivially follows that a finitely generated -submodule of is -torsionfree (since is an integral domain) and satisfies = = (since is a field). So a fractional -ideal can be defined more simply to be just afinitely generated -submodule in . Some important things to note are that

for a fractional -ideal there is an satisfying since is finitelygenerated in . Secondly, every ideal of is itself a fractional -ideal since it isa -submodule of which is noetherian. Thirdly, the multiplication of twofractional ideals and is defined by

= {finite sums

, }. (1.2)

Since and are finitely generated, must be finitely generated and is thus afractional ideal too.

These fractional ideals are the key part of the theory for factorisation of idealsinto prime ideals in a Dedekind domain. But to show this requires a bit more work.It turns out to be of immeasurable use to look at the set of elements of that senda fractional ideal inside . The reason being that the product of two fractionalideals is another fractional ideal and is istelf a fractional -ideal, so this set ofelements gives the possible notion of a group with identity . As we will soon seein the next section, this indeed gives us what we want. So to finish this section,given a Dedekind domain define for a fractional -ideal

1 = { }. (1.3)

1

is clearly an -module in and by the note above there is an such that , so 1 = 0. Now a non-zero satisfies 1 , so for every 1. So 1 and 1 is an -submodule of the principal -module 1and is thus finitely generated itself. So 1 is also a fractional -ideal.

Example 1.2.7 As before, take the Dedekind domain . Then (/) is a fractionalideal for non-zero , and ((/))1 is given by (/). Actually, all fractionalideals of are of that form. More generally, for any Dedekind domain , is afractional ideal for a non-zero and ()1 is given by 1. But all idealsof are of that form only if is a PID.

1.3 The Factorisation of IdealsFrom here we can start to think about if we can factorise ideals of a Dedekinddomain into maximal (or prime) ideals and even better if we can factorise allfractional ideals of into prime ideals, which will be the aim of the rest of thischapter. The order of the results here is sometimes in the opposite direction of theorder given in sources containing this material, but the content is often the same.References for this section were mentioned at the definition of a Dedekind domainand are Reiner [1, p.44-50], Curtis & Reiner [4, 18], Jacobson [6, 10], Janusz [7,p.8-18], Samuel [9, p.49-52] and Sivaramakrishnan [10, p.394-410]. From here, onlya few results are needed to get the group structure that was mentioned before, but

they also are our stepping stones towards our aim of factorising ideals.

7


15/58

Proposition 1.3.1 Every proper ideal of a Dedekind domain contains a productof prime ideals of and has 1.

Proof. For completeness, note the ideal = contains any product of prime ideals,but 1 = 1 = . Let be the set of proper ideals of that do not contain

a product of prime ideals. Since is noetherian there must exist an ideal in that is contained in no other ideal in , that is, a maximal element of . Now cannot be prime and any ideal properly containing must contain a product ofprime ideals. Since is not prime, there are elements , such that and hence + and + are ideals properly containing , so each contains aproduct of prime ideals. But

( + )( + ) + + +

and so must contain a product of prime ideals, contradicting the existence of a

maximal element of the set , and is empty.For the second part of the proposition let be a proper ideal of. Then clearly1 , so we must show 1 is not empty. Let be non-zero. Then by theprevious part there are prime ideals 1, . . . , such that 1 . . . , andselect 1, . . . , such that is minimal. As is noetherian and there is somemaximal ideal with 1 . . . . If = for every , then there are where / and 1 . . . , so and 1 . . . = 1 . . . ,contradicting being prime (since maximal ideals are prime ideals). Thus = for some and

1 . . . = 1 . . . 1+1 . . .

.

Now since is minimal, 1 . . . 1+1 . . . , so select some element 1 . . . 1+1 . . . . Thus 1 / and

1 1 11 . . . 1+1 . . . 1() =

and hence 1 1 .

Proposition 1.3.2 For a Dedekind domain and fractional -ideal , the set{ } = .

Proof. Let = { } and note that since is an -module, sowe only need to show the reverse inclusion. Let , then for

= 1() 1 . . . ,

so [] is an -submodule of and is thus finitely generated. If , then since is integrally closed it follow by Proposition 1.1.1 that . If then []is a finitely generated submodule of [] and similarly .

Corollary 1.3.3 For a fractional ideal of a Dedekind domain , 1 = 1 =

.

8


16/58

Proof. By definition 1 and since 1 is itself a fractional ideal, it is anideal in and so (1)1(1) is also an ideal in . By Proposition 1.3.1, if1 is proper in then (1)1 . Now since (1)1(1) wehave (1)11 1 and thus (1)1 by Proposition 1.3.2. So 1can not be proper and we must have 1 = . Also, 1 = since is

commutative.

Note also that for a fractional -ideal , this imples that (1)1 = . Nowdo we have a group structure? Firstly, the product of two fractional ideals is afractional ideal (and is commutative since is). Secondly the product of fractionalideals is clearly associative since is. Thirdly and finally; is the unity elementand there exist inverses for every fractional ideal that satisfy Corollary 1.3.3. Thuswe have just proven the following theorem.

Theorem 1.3.4 For a Dedekind domain , the set of fractional -ideals is amultiplicative abelian group with identity and the inverse of

given by

1.

But what about our aim? If we want to be able to factorise every ideal of aDedekind domain , or even better, every fractional ideal of in terms of primeideals of , then it makes sense that the set of fractional -ideals would begenerated by the prime ideals of. Indeed this is the case and the following resultslead to its proof. But first, say for two fractional ideals and that divides when there is an ideal such that = and denote this by . Thenthere is the following equivalence.

Proposition 1.3.5 Let and be fractional ideals of the Dedekind domain .

Then if and only if there is a proper ideal such that = .Proof. One way is trivial since if = then = = . So let .Then = 1 1 and 1 is a proper ideal in . Letting = 1 gets = 1 = as required.

Lemma 1.3.6 The ideals in a Dedekind domain can be written uniquely in termsof its prime ideals up to rearrangement.

Proof. First show an ideal in the Dedekind domain can be written as a productof prime ideals and afterwards show this expression is unique up to rearrangement.We define the ideal of to be an empty multiplication of primes, so excludethis case. Let be the set of proper ideals in that can not be expressed as aproduct of prime ideals and suppose it is not empty. Since is noetherian there is amaximal element that can not be a maximal ideal in (since maximal idealsare prime). Let be a maximal ideal of strictly containing . By Proposition1.3.5 there is a proper ideal in such that = . Now = = , sowe have , and by the maximality of both and are products ofprime ideals. But then so is = and thus must be empty.

Now suppose that two products of prime ideals are equal, say 12 . . . =12 . . . for prime ideals , and positive integers , . As in Proposition

1.3.1, let be a maximal ideal in such that 12 . . . . Then if = for

9


17/58

every there are where / such that 12 . . . . So forevery and 12 . . . = 12 . . . , contradicting being prime. Solet = for some . Similarly 12 . . . and thus = for some . So = and we get

12 . . . . . . = 12 . . . . . .

112 . . . 1+1 . . . = 112 . . . 1+1 . . .

12 . . . 1+1 . . . = 12 . . . 1+1 . . . .

By doing this procedure another 2 times it is clear that = and each coincides with some . Thus the factorisation is unique up to rearrangement.

We now know every proper ideal of a Dedekind domain can be written inthe form = 12 . . . for some positive integer and prime ideals , so wecan rewrite this as = 11

22 . . .

for some positive integer where all the

are distinct and > 0. This remarkable (and only slightly set up) property asmentioned earlier is reminiscent of the integers and principal ideal domains and iswhere the this theory, the theory of Dedekind domains, was born. Also, it gives usa nice generalisation of the greatest common divisor and lowest common multiple.

Corollary 1.3.7 The greatest common divisor of two ideals , of , denotedgcd(, ), and lowest common multiple, denoted lcm(, ), both exist and are a

product of prime ideals of .

Proof. Write = 11 . . . and =

11 . . .

where and are natrural

numbers as in the previous lemma. Then gcd(, ) = min{1,1}

1 . . . min{,} and

lcm(, ) = max{1,1}1 . . . max{,} .

This last Lemma also gives us a nice fact. If = 11 22 . . .

is a proper ideal

in , then can we use this to write 1 in a similar form? Consider 11 22 . . .

where = 1 1

times

. Then

11 22 . . .

=

11

22 . . .

11

22 . . .

=

and it follows that 1 = 11 22 . . .

. This then allows us to extend the

preceding Lemma to get one of the main theorems of this chapter.

Theorem 1.3.8 The abelian group of fractional ideals of a Dedekind domain is generated by the prime ideals of and their inverses.

Proof. This will be proved by showing every fractional ideal can be written as aproduct of primes and prime inverses. Let be any fractional ideal of . If ,then this is immediate by Lemma 1.3.6 so let . As mentioned earlier, since is finitely generated there is a non-zero such that , so is an ideal in. By Lemma 1.3.6 we can write this as = 11

22 . . .

for some non-negative

integer , positive integers and prime ideals . Similarly = 11

22 . . .

for some non-negative integer , positive integers and prime ideals . So =

10


18/58

() and thus = ()111 22 . . .

=

11

22 . . .

11

22 . . .

.

That is, can be written as the product of prime ideals and inverses of primeideals and this factorisation is unique (up to rearrangement) since the factoriza-tions of and are themselves. Thus the set of fractional ideals is generatedby the prime ideals of and their inverses.

Example 1.3.9 As in Example 1.2.7, fractional ideals of are of the form for

. So let and be coprime and write and as their prime factorisations = 11 and = 11 . Then

=

11 11

= (11 ) ( )(11 ) ( )

= (1)1 ()(1)1 ().

Also, the abelian group of fractional ideals is just the group generated by

{} {2, 3, 5, 7, 11, } {12,

1

3,

1

5,

1

7,

1

11 }.

So in a similar fashion to the previous Lemma, we can now write any fractionalideal uniquely in the form = 11 22 . . . for some non-negative integer with each distinct. But this time each is a non-zero integer, not necessarilypositive. This theorem effectively gives us a characterisation of Dedekind domainsand to conclude the chapter, it turns out that if we went backwards, then everythingworks out nicely. But as the next chapters will show us, the uniqueness result about

Dedekind domains that we have found can be generalised. So if we are at the endof the rainbow now, then maybe the pot of gold would of been better?

Theorem 1.3.10 The following are equivalent for an integral domain .

a) The ring is a Dedekind domain.b) The ring is noetherian, integrally closed and every non-zero prime ideal in

is maximal.c) Every proper ideal in is expressible as a product of prime ideals of

uniquely up to rearrangement as 11 . . . with each distinct and > 0.

d) The set of fractional ideals of is generated by the prime ideals of andtheir inverses.

Proof. By definition, a) and b) are equivalent and Theorem 1.3.8 c) and d) mustbe equivalent. Also, we have just shown that b) implies c), so we only need to showthat c) or d) implies b) to get the theorem. Assume d) and consider an infinitechain of ideals 1 2 3 . Then write = 1 for prime ideals not necessarily distinct. Then the chain can have at most distinct properideals, so eventually stabilizes and thus is noetherian. Secondly, let be a root of the polynomial + 1

1 + 1 + 0 []. Then =11 1 0, so letting = 1, , , 1 we see that andthus

. But = 1

is a product of prime ideals of, so multiplying

11


19/58

both sided by 11 1 gets , so . Finally, clearly the prime idealscoincide with the maximal ideals. So is a Dedekind domain, showing d) impliesb) and proving the theorem.

12


20/58

Chapter 2

Dedekind Domains and Orders

In Chapter 1, we went through what a Dedekind domain was and how ideals inDedekind domains could be written as a product of prime ideals. But what aboutdoing this more generally? What do we mean here by more generally? We meanfor much larger rings than Dedekind domains, say for rings containing Dedekinddomains. Dedekind domains are of course commutative, so to do this more generally,we generalise the ideas in Chapter 1 to a non-commutative setting. So as mentioned

at the beginning of Chapter 1, commutativity is not assumed. To start, we extendour reach and consider algebras over a Dedekind domains quotient field, then take asubring of this algebra and see if this subring (which is not necessarily commutativeor a domain) has a similar type of factorisation into prime ideals.

Some preliminaries are required to move forward, so we will say these here.Recalling a definition from Chapter 1, For a commutative ring , an -algebra is aring such that there exists a homomorphism : () where () = { = for every } is the centre of . So to see as an -module, wedefine = () for and . An -algebra is then called simple ifit has no non-trivial two-sided ideals and semisimple if is the direct product of

simple subalgebras.The use of fields in this chapter require some properties, so again let us define

them here. Let be a field and / be a field extension. Then is called separableover if the minimal polynomial of every over factors into distinct linearfactors over . The dimension of an -algebra over is its dimension as avector space over . For the finite dimensional -algebra , if is semisimple andthe centre of each simple simple summand of is a separable field extension of ,then is called separable over .

Also recall from Chapter 1 that for a Dedekind domain with quotient field and vector space over , a full -lattice in is a finitely generated -

torsionfree -module in that satisfies = = by the equality(1.1). This allows us to define a major part of the theory from hereon. So for therest of this chapter let be a Dedekind domain with quotient field , be afinite dimensional separable semisimple -algebra and be a full -lattice in .Note that is always associative, but not necessarily commutative and that isa subring of ().

2.1 Orders

In Chapter 1 we defined an integral domain to be a Dedekind domain if

I It was noetherian.

II It was integrally closed.

13


21/58

III Every prime ideal was maximal.

If we use these three properties as a hint to help find subrings of able to factoriseideals as we can in Dedekind domains, then these subrings might not be as compli-cated or hard to find as first thought. Actually, as references Reiner [1, p.108-110],Swan & Evans [2, p.83-84], Bass [3, p.152-156] and Curtis & Reiner [4, p.515-517]point out, these types of subrings are starring us in the face; as the next seeminglyabstract definition of an -order points out.

Definition 2.1.1 A unital subring of that is a full -lattice in is called an-order in .

As with Dedekind domains there are also other definitions of an -order thatinvolve Krull rings and primary ideals, but they will not be looked at. So for thisreason Reiner [1] and Swan & Evans [2] are the main references for this chapter,though Bass [3] and Curtis & Reiner [4] still provide some input, but not as much.(Note that conceptually orders can be thought of as follows: Starting with a finite

number of points in space, we can extend these by adding an subtracting pointsfrom each other to make a lattice in that space, and if that lattice turns out to bea subring of the space without needing infinitely many points to generate it, thenit is an order).

So does this definition satisfy the three properties above? Since is noetherianand the -order is finitely generated, must also be noetherian for both leftand right ideals. So we will just call noetherian to indicate both left and rightnoetherian. Thus property above is satisfied for and we are off to a goodstart. After a few examples of some orders we will focus our aims at the next twoproperties.

Example 2.1.2 If we let be the matrices over , (), then the homo-morphism : shows is a -algebra. Let be the subring and-module (). See that is generated by {(1) 0 , } where (1)is the matrix with a 1 in the position and zeros elsewhere and so is finitelygenerated by 2 elements. Since = and contains the identity with 1sin the main diagonal and zeros elsewhere, it is a unital subring and a full -latticein , so an -order in .

Example 2.1.3 Let = and = be the rationalquaternions, where = is an extension of called the quaternionswhere 2 = 2 = 2 = =

1. This is a -algebra since is embedded in .

If we let be the subring , then is also a finitely generated-module containing the identity 1 with = , so it is a -order in .Example 2.1.4 It is important to note that both the algebras in the previous twoexamples were separable and semisimple. In fact, both were simple algebras. But iforders are to generalise the notion of a Dedekind domain, then letting = and = should work and as expected, it does.

In Chapter 1 we defined the inverse of a fractional -ideal in (1.3) by elements that sent inside , and then showed in Proposition 1.3.2 that { } = . So what if we consider something similar for a full -lattice in? Take the set =

{

}. Then is a unital subring of and

clearly an -module. But what else can we say about it?

14


22/58

Since is a full -lattice in , is an -lattice in for every (notnecessarily a full -lattice). Now we can write = 11+ for some and since = . Since , there is an such that .Thus there is an such that = 11 + + and thus .So = 1

and we have

. Hence =

. If we now take

= 1, then there is an with . So and 1. Since1 is an -lattice by above thus so is . Hence is actually an -order in .Similarly, the set { } is an -order and we can define the following.Definition 2.1.5 The left and right orders of are defined respectively as

() = { },() = { }.

Now returning to before to see if the other two properties hold, we first notethat they do not quite do what we want them to at the moment. Looking at thesecond property, integrally closed is defined for a ring with respect to its quotientfield, but we only want to look at an -order of , not the whole of or of theintegral closure of. So we just want an -order to be integral over . This queryis then solved by the following theorem.

Theorem 2.1.6 The-order is integral over and thus the minimal polynomialof every over , (), is in [].

Proof. Let . Since is an -lattice and [] , then [] is a finitelygenerated -module. By Proposition 1.1.1 it follows that is integral over .Since was arbitrary, we have integral over . The second statement is then aresult of Proposition 1.1.2.

Example 2.1.7 As in Example 2.1.4, letting = and = gets beingintegral over as expected.

We now have properties and and look at the third. From here we let bean -order in and be a full -lattice in . We defined a prime ideal of anintegral domain in Chapter 1 and here we define one in an order, referencing Reiner[1, p.190] and Swan & Evans [2, p.89]. Since is commutative, ideals in are both

left and right ideals. In however, left ideals are not necessarily right ideals. Soat first we will stick only to two-sided ideals (unless stated otherwise) and look atone-sided ideals later in the chapter. When prime ideals are spoken about it will beclear what kind of prime is being referred to, to avoid confusion. We assume thatall ideals of are full -lattices in .

Definition 2.1.8 A prime ideal of is a proper two-sided ideal that is afull -lattice in such that for two-sided ideals , in , if then or .

15


23/58

The multiplication of ideals here is the same as in (1.2) and note that the lastpart of the definition is equivalent to saying for two-sided ideals , in /,

if = 0 then = 0 or = 0, (2.1)

reminiscent of an (possibly integral) domain.Example 2.1.9 Every maximal two-sided ideal of is a prime ideal of . Bymaximal, it is meant that is a proper ideal of that is not contained in anyother proper ideal of .

Proof. This example requires some proof, so we sill do it here. Let be a two-sidedmaximal ideal of . Two two-sided ideals , of that have , fall into thefollowing cases:

1. If one of the ideals contains , say , then either = or = . Thefirst shows and the second shows = = , so is prime.

2. Otherwise let and . Since , ( + ) . But + and so + falls into the first case. So if + = , then and if + = , then and again is prime.

This example can also be proven a different way by using (2.1) and the followingLemma, which will be proven for ideals in general, not just two-sided ideals.

Lemma 2.1.10

a) For an ideal of , / is an ideal of / if and only if is an ideal of containing .

b) / is simple if and only if is a maximal two-sided ideal in .

Proof. Consider left ideals and let be one such of . To Prove part a, first let/ be an ideal of /. Note that since otherwise / is not welldefined. Let and , so + / and + /. Then( + )( + ) = + + + /. But + + ,so must follow. Thus is an ideal of containing . Now to provethe converse, let be an ideal of containing and show / is an ideal in/. Again let and , so + / and + /. Then( + )( + ) = + + + / with + + and . Thus ( + )( + ) / and / is an ideal in /, provingpart a.

The proof of part b follows from part a since if / is simple then there areno proper two-sided ideals of strictly containing , so is a maximal two-sidedideal. Conversely, if is a maximal two-sided ideal, then the only two-sided idealsof / are / and /, that is, the trivial ones. So / has no non-trivialtwo-sided ideals and is thus simple, proving part b.

This proves the Example 2.1.9 trivially since simple algebras have no non-trivialideals, but it also gives us an important application of the next Theorem, which canbe found in Reiner [1, p.190-191] and also in Swan & Evans [2, p.95-96]. The latter,

however, uses a different proof to the one given here, instead using the Chinese

16


24/58

Remainder Theorem. But first we must state a standard theorem from the theoryof modules, which can be found in Curtis & Reiner [4, 25].Theorem 2.1.11 The ring is semisimple as a left -module if and only if everyleft -module is semisimple. Every semisimple ring can be decomposed into a

sum of minimal left ideals of where are orthogonal idempotents. That is2 = and for = , = 0.

Theorem 2.1.12 If is a prime ideal of then

a) = is a prime ideal of . Moreover gives a surjective relationbetween the prime ideals of and the prime ideals of .

b) / is a finite dimensional simple algebra over the field /.

Proof. Let be a prime ideal of , = and = /. We prove part afirst as we need / to be a field. Since is a full -lattice in , by the argument

before Definition 2.1.5 there is an such that , so is not empty. Also,since , 1 / and hence 1 / , implying . Now is an ideal of since = ( ) = () = = since is a full -lattice in .Now if , satisfy , then ()() . So or ,implying that or and is a prime ideal of .

Now let be a prime ideal of . Since is an -lattice and , isa proper two-sided ideal of and so is contained in some maximal ideal in .Now and since both and are prime in it follows that = proving the surjectivity of the relation = and thus parta.

For the proof of part b, / is a field so is trivially artinian as a ring. Now is a

finitely generated -module, so / must be a finitely generated /-module andis thus artinian, as well as a finite dimensional /-algebra since / (/).Suppose first that / is semisimple. Now two separate simple components and of have = 0, so by (2.1) one of them must be zero. Thusthere can only be one simple component and / is actually simple. So it sufficesto show that / is semisimple.

Suppose that there is a two-sided ideal such that / 0 is minimal.Such an exists since / is artinitan. Now (/)2 /, so since / isminimal either (/)2 = 0 or (/)2 = /. The first implies / = 0 via (2.1)so it must be the latter. But in that case, it must be also that 2 = in . Let

= 1 + for some , + and no equal to a sum = or product

=

= involving the other terms of since if any is,

removing that will not affect the ideal (so we have a minimal amount of ).Then 2 = 21 + 1 + 12 + 2 + + 1 + 2 and thussince each cannot be expressed in terms of the other elements of , they mustall be idempotent (that is 2 = ). But as

2 = then = 0 (and so they areorthogonal idempotents) must also be the case for all = . By Theorem 2.1.2.1this means is actually semisimple. Doing similarly for another minimal two-sidedideal it follows that is a direct product of ideals generated by idempotentsof that cancel each other so is semisimple as well. Thus every two-sided ideal in

/ is semisimple, so by Theorem 2.1.2.1/ is semisimple itself.

17


25/58

This result is quite remarkable and provides a correspondence between the orig-inal definition of a prime ideal in Chapter 1 and the abstracted definition here inChapter 2. Just as importantly, the succeeding corollary of Theorem 2.1.12 allowsus to get the last property we require of , property .

Corollary 2.1.13 The prime ideals of are maximal two-sided ideals in .

Proof. For a prime ideal of , Theorem 2.1.12 says that / is simple andLemma 2.1.10 then implies is maximal.

This gives us the equivalence of maximal ideals and prime ideals of , so theorder satisfies the three properties held by a Dedekind domain and we are off toa good start. Though this is very nice, it still does not mean too much to us aboutthe factorisation of ideals in this order without actually looking at the ideals of .So next we will discuss the ideals of , with references for the next section Reiner[1, p.108-111,192,204-205], Swan & Evans [2, p.83-84,88] and Bass [3, p.152-156].

2.2 Maximal Orders

In the last section, we went briefly into the ideals of to show property ofa Dedekind domain held for . Left ideals in all have = and areassumed to be full -lattices in , but what if we consider such not containedin . We did this in Chapter 1 with the definition of a fractional ideal and sawthat each fractional ideal could be factorised uniquely as a product of prime idealsand inverses of prime ideals and also that the set of these fractional ideals was anabelian group generated by the prime ideals and their inverses. So we search forthe Irishman the the end of the rainbow and again we do it here with a similar

definition with the aim of seeing if a similar factorisation occurs in an -order .Definition 2.2.1 For the -order , a non-zero full -lattice in that is a finitelygenerated -module is called a fractional -ideal of .

Example 2.2.2 As in Example 2.1.7, letting = and = gets the fractional-ideals here and the fractional -ideals in Chapter 1 coinciding.

We will denote fractional -ideals as we did in Chapter 1 by using , and soon and as usual the product of two fractional -ideals is another fractional -idealgiven by (1.2). We do however drop the word fractional for ease and just refer tothem as ideals of , assuming that they are in fact fractional -ideals of . If a full-lattice is a two-sided ideal of , it is then just an ideal of that is a two-sided-module. We will soon only look at two-sided ideals and then look at a moregeneral case later, but for now this two-sidedness is not assumed. First note theuseful fact via the argument given before Definition 2.1.5, that for every ideal of, the intersection = 0.

The ideals considered from Definition 2.1.8 until this point are all fractional -ideals since they are full -lattices in and are contained in , an -lattice itself,implying they are finitely generated -modules. To emphasize when an ideal of is contained in we call an integral ideal of . As usual, is a maximal integralideal of when it is contained properly in no other integral ideal of .

So now we have generalised some of the definitions of Chapter 1, in particular

Dedekind domains and the corresponding fractional ideals. But is this all that is

18


26/58

required to generalise the results for Dedekind domains from Chapter 1? At themoment there are no restrictions on the order we choose. In particular, as we willsoon find to be important, we have not established any sort of maximality for orders,which leads to the following.

Definition 2.2.3 An -order that is not properly contained in any other -orderis called a maximal -order.

Example 2.2.4 As in Example 2.2.2 (and as mentioned just before the previousdefinition), if = and = then is a maximal -order in since addingany element to gets an infinitely generated -module, so + cannotbe an -order.

Example 2.2.5 As in Example 2.1.2, letting = () and = (), then was an -order in . Actually, is a maximal -order in since if you addanother matrix to with, say, an element in the position. Since is generated by the matrices {(1) 0 , } where (1) isthe matrix with a 1 in the position and zeros elsewhere, the element can beplaced in any position. Thus any multiple of can be placed in any position andsimilarly to Example 2.2.4 + is not an -order.

Example 2.2.6 As in Example 2.1.3 with = and = , then was an order in . But is not a maximal order. Let = (1++ +)/2 and = . Then is a subring (since it can bewritten as = {(++ +)/2 mod 2) that contains (since = 2 1 ) and is generated by 1, , , , so is an -order of. Further, it ismaximal. This follows since is a non-commutative PID, which will be looked at

in Chapter 4. So suppose there is an order

, then is a -ideal also. Thereturns out to be an such that (which will be an element of 1). So = for some . But then = () = () = () = () = and is maximal.

The reason we need maximal orders will soon become apparent as we look intothe ideals of . But first we must make one more extrapolation from Chapter 1.As mentioned earlier, in (1.3) we defined for a fractional -ideal a set of elements that sent inside , which happened to be the inverse of in . Thisbecomes a slight problem here due to the issue of non-commutativity, since for afractional -ideal and element , is not necessarily the same as . Solet

=

{

}

and

={

}

. For arguments sake, let be a two-sided ideal, so for we have , so and similarly forany , also. So let = { }. We have just seen that , but are they equal? In general the answer is no.

The most we can say is that, since , we have () and (). But if is a maximal order the the answer is better than yes. In this case() , so () = and similarly () = . So we have = { ()} = { } = and also = { ()} = { } = , exactly what we want since it is the counterpart to (1.3) in amaximal order. That is, when is maximal we get = = . So we will use this

19


27/58

idea as motivation and define in general for an ideal (not necessarily two-sided)of

1 = { }, (2.2)

or equivalently

1 = { ()} = { ()}. (2.3)

So we then see that when is maximal and is strictly two-sided

1 = { } = { }. (2.4)

We see that 1 is clearly an -module, but is it an ideal in ? Since thereis an , then 1 (). Thus 1 1(), which shows 1 isfinitely generated and -torsionfree and thus an -lattice. If = (), then as

in the discussion before Defintion 2.1.5 there is an such that . So 1 and = (1) = { 1 } { } = 1.So 1 is full and indeed an ideal of itself. Similarly, 1 is an -ideal also when = ().

We know now that 1 is an ideal, but what else do we know? Using (2.3) wehave that 1 (), so (1()) () and thus 1() 1 whichshows 1 is a right ()-module. Similarly,

1 () gets (()1) () and ()

1 1 which shows that 1 is a left ()-module. Thesealso give us two nice containments,

(1)

() and

(1)

(). (2.5)

This allows us to think of the ideals of an order as a set that contains inverses,but we want to see if it is a group. Before we see if this is the case, we use thepreceding results to aquire the following useful facts. For the full -lattice , if() is maximal we have

1 () by definition. Firstly, since 1 is an idealand the product of two ideals is another ideal, it follows that 1 is an integralideal in (). Now by (2.5) (

1) = () and thus 1 is actually a two-

sided integral ideal of (). Secondly, if , then 1 1 () andso 1 1. Reiterating this instead with () maximal gets 1 a two-sidedintegral ideal of () and for

, yet again 1

1. Repeating the second

fact here, we get

if then 1 1. (2.6)

2.3 The Factorisation of Two-Sided Ideals

So now we have the necessary utensils to see if the set of these fractional ideals is agroup. We do however have to revert to the case of two-sided ideals (two-sidednesswill be stated explicitly), like in the previous section. But this turns out to benecessary here for our new aim, to see if the two-sided ideals of can be factorisedin a similar fashion to the ideals in a Dedekind domain (which may not be the

case for one-sided ideals). So we continue on with this, trying to establish a group

20


28/58

structure along the way. The method we will use to see if these cases hold may seemfamiliar to the reader and can be found in Reiner [1, p.204-207], Swan & Evans [2,p.88-92] and Bass [3, p.155-159]. In fact most of the results (and proofs) in thissection naturally follow from their counterparts in Chapter 1. For example, thenext three results resemble Proposition 1.3.1 and Corollary 1.3.3 and are again our

first stepping stones.

Lemma 2.3.1 Every two-sided ideal of contains a product of prime ideals of .

Proof. Note first that ideal here means two-sided ideal and any ideal containsany product of prime ideals of , so only consider integral ideals from hereon. Let be the set of proper ideals of that do not contain a product of prime ideals.Since is noetherian there must exist a maximal ideal in . Now cannot beprime and any ideal properly containing must contain a product of prime ideals.Since is not prime, there are two two-sided ideals , such that , and

. But then ( + )(+ )

also and both + , +

.

Both + and + are two-sided ideals and by the maximality of theycontain a product of prime ideals. But then so does since ( + )(+ ) ,contradicting the existence of a maximal element of the set , and is empty.

Lemma 2.3.2 Let be a two-sided ideal of the maximal order . If , then1 .

Proof. Let be a two-sided ideal of . Clearly 1 , so suppose 1 = .Then there is a maximal ideal such that and let . Then byLemma 2.3.1, contains a product of prime ideals

12 and choose these prime ideals such that is minimal. If = for every , thenthere are where / for every but 12 . So 12 but for every . So since is prime and 1 , then 2 must follow. But then 2 , so again 3 follows. Doing thisanother 3 times gets either 1 or , which is a contradiction.Thus = for some and

1

1+1

.

Let = 1 1 and = +1 and see that and are both two-sided ideals of . So gets 1 and 1 . Thus1 () = and so by (2.6) we get 1 1 1 = . But thisgets , so contains a product of 1 prime ideals of , contradictingthe minimality of . Thus 1 = .

Theorem 2.3.3 Let be a full -lattice of . If () is a maximal order then 1 = (). Similarly, if () is a maximal order then 1 = ().

21


29/58

Proof. Suppose () is maximal. We know that 1 is a two-sided ideal con-

tained in = (). Then we also get (1)(1)1 , implying 1(1)1

1 and thus (1)1 (1). But by (2.5) (1) , a maximal order, so(

1) = and we get (1)1 . By Lemma 2.3.2 it follows that 1 = .The second part of the theorem works similarly, but will be shown to indicate the

importance of this result. Again, 1 is a two-sided ideal contained in = (),so (1)1(1) gets (1)11 1 and hence (1)1 (1).By (2.5) (

1) , a maximal order, so (1) = and (1)1 ,where Lemma 2.3.2 again finishes off the proof.

The preceding theorem (which was proven for a full -lattice, not a two-sidedideal) gives us exactly what we want when is a maximal order since if ismaximal, for every two-sided ideal of we have () = () = . Also, seethat 1 = (1)11. But does (1)1 = ? In fact it does, not only fortwo-sided ideals, but also for one-sided ideals as the following result shows.

Proposition 2.3.4 Let be a full -lattice in and = () or (). If is a maximal order then (1)1 = . Also, any ideal of that contains is theinverse of an integral ideal of .

Proof. Let = () be maximal. Since is a full -lattice, then so is 1. So by

Theorem 2.3.3 and (2.5) we have (1)11 = (1) = . So (1)11 =

(1)1() = and thus (1)1 since () is a unital subring. To show

the reverse inclusion, by (2.5) we get

(1)1 = ((1)1)(1)1 (1)(1)1 = (1)1.

This then gets

(1)1 1(1)1 = (1) () = ,

and the reverse inclusion holds, implying (1)1 = . The proof is done similarlyfor = ().

For the second part, again let = () be maximal but also . Then1 1 = , so 1 () = and 1 is integral. Letting = 1, weget 1 = (1)1 = , so is the inverse of an integral ideal of . Again, theproof is done similarly for = ().

So now we only need one more lemma to get what we want, and then ourobjective of ideal factorisation into prime ideals is done. Unlike the Dedekinddomain case, we have the answer to our aim before before we certify that the setof fractional ideals is a group. By what we have so far, products of ideals of arealso ideals and this is associative since is, acts as an identity and there is aninverse for every ideal of . So we have a group structure, but we want to know ifit is abelian since at the moment there is no reason to see that it is. ReferencingReiner [1, p.206-207], Swan & Evans [2, p.91-92] and Bass [3, p.158], this query isalso answered by the following, which achieves our aim.

22


30/58

Lemma 2.3.5 Let be a maximal order. Then multiplication of prime ideals of is commutative and every integral two-sided ideal of can be written uniquely asa product of prime ideals of up to rearrangement.

Proof. Let an be distinct prime ideals of . Then see that 1

1 = by Theorem 2.3.3. Now (1) = , but since is prime either or 1 . The first implies = , which cannot be true, so wemust have 1 and thus . Doing exactly the same procedurewith and swapped gets the reverse inclusion and thus = , showingmultiplication of prime ideals is commutative.

Now for the second part of the proof, suppose that there are two-sided -idealsthat cannot be written as a product of prime ideals and let the set of these be .Then there is a maximal two-sided ideal which obviously is not prime. Sothere is a prime ideal such that and hence 1 1 andmultiplying by gets

1

. If = 1 then 1

() = ,

contradicting Lemma 2.3.2. So 1 and thus 1 is a two-sided idealstrictly containing , so by the maximality of 1 is expressible as a productof prime ideals, say 1 = 12 for some . But then = 1 , aproduct of prime ideals, contradicting the existence of a maximal element of andthus is empty.

Finally we must show that if two expressions of prime ideals are equal thenthey are the same prime ideals, but rearranged. So for prime ideals and , let12 = 12 . since both expressions are a product of prime ideals,they are contained in some prime ideal, say . So 12 and by theproof of Lemma 2.3.2 it follows that = for some . Similarly 12 and so = for some , so we get

1 1+1 = 1 1 + 1 11 1+1 = 11 1 + 1

1 1+1 = 1 1 + 1 .

Doing this another 2 times gets = and each coinciding with some .Thus the factorisation is unique up to rearrangement.

So as in Chapter 1 we can write two-sided ideals of in the form =11

22

for distinct prime ideals of and positive integers . But how

do we write 1? If we look only when is maximal so 1 = , then similarlyto Chapter 1 yet again, we see that due to the commutativity of the prime ideals,1 = 12 . This observation then leads to the major theorem ofthis section and our aim, the generalisation of Theorem 1.3.8. But first, we state acorollary, which yet again generalises the notion of a greatest common divisor andlowest commom multiple from Chapter 1.

Corollary 2.3.6 The greatest common divisor of two ideals , of , denotedgcd(, ), and lowest common multiple, denoted lcm(, ), both exist and are a

product of prime ideals of .

23


31/58

Proof. Write = 11 . . . and =

11 . . .

where and are natrural

numbers as in the previous lemma. Then gcd(, ) = min{1,1}1 . . .

min{,} and

lcm(, ) = max{1,1}1 . . .

max{,} .

Theorem 2.3.7 Let be a maximal order. Then the set of two-sided ideals of, , is an abelian group under multiplication generated by the prime ideals of with identity and inverse of two-sided ideal given by 1.

Proof. This will be proved by showing every two-sided ideal of can be writtenas a product of prime ideals of and prime inverses. Let be an ideal of .If , then this follows straight away from Lemma 2.3.5 so let . Let 1 so is integral and by Lemma 2.3.5 can be written as aproduct = 1 of prime ideals of . But = (), so since isan -module we can also write = 1 for prime ideals of . Thus = ()11

=

11

1 1

is written as a product of prime ideals

of and prime inverses. This proves the theorem.

So just like the fractional ideals of a Dedekind domain, Theorem 2.3.7 showsevery two-sided fractional ideal of a maximal order can be written as =11

22 for distinct prime ideals of and each integer = 0. That is,

there is unique factorisation of two-sided fractional -ideals up to rearrangement.

Example 2.3.8 We look at factorising some two-sided ideals in a maximal order inthe rational quaternions given in Example 2.2.6. So we let = and = where = (1 + + + )/2. Then we saw that was amaximal order (note the change in notation from that example) that could also be

written as = {( + + + )/2 mod 2}.So what are the maximal two-sided ideals in ? It turns out that is actually

a non-commutative version of a PID (see Chapter 4), so every left ideal in canbe written as for some . As it is a PID, there are irreducible elementsin also. It turns out (which will be proven again in Chapter 4) that if isirreducible, then writing = + + + gets the norm of to be

= = ( + + + )( ) = 2 + 2 + 2 + 2.

Furthermore, it is a positive prime integer. So from this we see that all maximalideals and maximal two-sided ideals are either of the form or

for someirreducible . Visually,

{Maximal Left -ideals} {Maximal Two-sided -ideals}

{-ideals of form or Irreducible }.

So when is an an ideal two-sided? Clearly the ones of the form are two-sided, but are there any irreducible such that is a two-sided ideal? Straightaway we can see that (1 + ) is actually two-sided since (1 + )( + + + ) =

( + + )(1+ ). So are there any more? If we let = + + +

24


32/58

be irreducible, if = 2, then two of ,, and are 1 and the other two 0, so itis identical to the case 1 + , so lets look when > 2.

If is two-sided, then for every there must be a such that = ,so =

. That is,

= . When substituting = 1 + and rearranging, we get

that 2+2+2+2

. When = we get that 2+2+2+2

+,

.

Thus it must be that 2 + 2 + 2 + 2 (since 2 + 2 + 2 + 2 > 2), implyingthat one of and is 0. But then 2 + 2 + 2 + 2 + = implying that oneof and is 0. So without loss of generality = + . By substituting = 1 +again yields 2 + 2 2 so one of and must be zero, implying a prime numberis an integer squared which is obviously nonsense.

So by this long-winded argument every ideal of the form for irreduciblewith > 2 a maximal left ideal. Thus are the maximal two-sided -ideals,that is, the prime ideals of in addition to (1 + ). But what about 2? Letsapply some of the theory. Since 2 = (1 + )(1 ) we have that

2 = (1 + )(1 ) = ((1 + ))2.Now for any positive rational , write = 2

11

11

where each and is a

positive odd prime and , > 0 (assuming no common factors on numerator anddenominator), but a non-zero integer. Then we have

= (2)(11 ) ( )(11 ) ( )= (2)(1)

1 ()(1)1 ()= ((1 + ))2(1)

1 ()(1)1 ().

Which seems almost identical to the factorisation of the ideals in the Dedekinddomain given in Example 1.3.9. Note if is negative, then there is no differencesince = (). Finally the set of two-sided ideals in , , is just the set

{, (1 + ), 1 + 2

} {3, 5, 7, 11, } {13

,1

5,

1

7,

1

11, }.

25


33/58

Chapter 3

Maximal Orders and Groupoids

The final result of the last chapter is indeed a nice consequence for two-sided idealsin a maximal -order over a finite dimensional separable semisimple -algebra,but if that is the only gold at the end of the rainbow, then the Irishman has runaway with our gold and left only a few pieces behind. So we chase after him bynow considering one-sided ideals to see yet again if they can be factorised uniquelyin terms of irreducible ideals and not only this, but we will again see if the set of

all these one-sided ideals forms a group. So now we focus at both of these aims.Strictly one-sided ideals, however, cannot be a product of prime ideals since primeideals here are by defintion two-sided. But there is still a relation between the primeideals and maximal one-sided ideals, and this lends a hand later in the piece. Sofrom hereon we let all ideals be one-sided unless stated otherwise.

3.1 One-Sidedness

To start ourselves off, we will actually give a conclusion to the previous chapterthat shall give us some preliminary argument towards our new aim. In the previouschapter we saw that the set of two-sided ideals,

, of the maximal -order in the

finite dimensional separable semisimple -algebra was actually an abelian groupgenerated by the prime ideals of . But if we look at two different maximal orders,do we get two different groups? This query will be answered shortly, but first weneed some requisites. The reference for this section is Reiner [1, p.197-199,204-205].So similarly to last section, let be a Dedekind domain with quotient field , a finite dimensional separable semisimple -algebra and an order of .

If we take a full -lattice in , then () is not necessarily the same as(). In fact, if

= () = () then is a two-sided fractional -ideal,

so if is strictly a one-sided -ideal for some order , then () = (). Thisthen introduces some problems. Firstly, what is an integral ideal now in the one-

sided case? Secondly, given two orders and

, is there an -ideal (also

-ideal) with () = and () =

and does the multiplication of ideals now changesince () = ()?

Upon considering the definition of an integral ideal: a -ideal is an integralideal of if ; It is seen that this is independent of left or right multiplication,so still holds in the one-sided case. Similarly, so does the definition for a maximalintegral ideal. However, it would be nice to know when integrality holds for anorder on each side. We get some insight into this if the left ideal has = (),allowing us the following result, taken for a full -lattice in .

26


34/58

Proposition 3.1.1 For any full -lattice in , () if and only if (). That is, is an integral ideal of () if and only if is anintegral ideal of ().

Proof. Let be a full -lattice in and let

(). Then

() = and it follows by definiton that (). Similarly, if (), then () = and (), completing the proof.

Now looking at the second problem, we see that the multiplication of ideals is asusual defined by (1.2) since we are in the algebra . But for the full -lattices and , = ( ()) = (()). So it would again seem optimal that() = (). Thus for full -lattices 1, 2, , the product 12 is called proper if() = (+1) for every {1, 2, , 1}. It is then seenthat

(

1

2

)

(

1) and

(

1

2

)

(

). (3.1)

Now for two orders and , is there a full -lattice with () = and() =

? The answer in general is not clear, but this is not needed. If and are maximal, then the answer is yes. Let = , then () , so() = and similarly () =

, so shows this (Note that the fact wasnot constructed via a proper product is not important, since all we wanted was for to be a left -ideal and right -ideal).

Using this, we create some notation by writing a full -lattice as wherewe denote = () and = (). Thus a proper product of full -latticescan be written as 1223

+1. So for a full lattice = with both

and maximal, we still have the inverse defined by (2.2) or (2.3). But canwe write 1 in the same way? By (2.5), we have (

1) () = , so(

1) = and similarly (1) = . Hence inverting swaps the orders

over and we write 1 = ()1 = 1 seeing in fact that (

1 ) = and

(1 ) = .

One final thing to note for a proper product is given in the following result.

Proposition 3.1.2 A proper product of integral ideals is an integral ideal.

Proof. Let 1223 +1 be a proper product of integral ideals. Then

1223 +1 223 +1 = 23 +1 334 +1 = 34 +1

...

+1 = +1 +1.

Thus 1223 +1 is an integral ideal, proving the proposition.Now returning to see if = for two maximal orders and , the

following result gives us the answer to our query.

27


35/58

Theorem 3.1.3 Let and be maximal orders in . For an ideal with() = and () = , the map

: : 1

is an isomorphism. Furthermore, is independent of the choice of .

Proof. Let = and the ideal = have () = and () =. Then for , by (3.1) we have (1) = (1) = and(

1) = () = , so 1 is a two-sided ideal and

1 .Also see that () =

1 = 1 = , so the identity in is sent to

the identity in . Now to show is a homomorphism, let , . Then() = 1 = 1

= 11 = ()() and is ahomomorphism.

Considering similarly the homomorphism = sending to 1 , then it is seen that

() = 1

1

= and

() = 1

1

=. So is the inverse of and they are both isomorphisms. Now for anotherideal = , see that

1 and 1 are two-sided -ideals since they bothhave left and right orders equal to , so they are elements of . Thus since is abelian, for we have

= = 1 =

1 = 11 = 11.

So using this with gets

1 = () = ( 11) = 1 11 = 1.

So is independent of the choice of = , proving the theorem.

So we see that = does not quite hold, but = always does, show-ing us that the group generated by the prime ideals of a maximal order essentiallydoes not depend on the maximal order. This is a wonderful result and it will guideus in the right direction via two of its corollaries. However, we cannot go into theseat the moment without some further theory, so they will be left to the next section.Instead, we start with some results about one-sided ideals, some of which generalisefurther the results of Chapter 2. As before, the reference for the remaineder of thissection is Reiner [1, p.195-196,204]. This will focus on left ideals, but the right case

follows equivalently.Getting back to one of our usual and aforementioned aims of seeing if the set

of the one-sided ideals is a group, we see that ideals are actually -ideals for someorder , so they are dependent on . Since a full -lattice in does notnecessarily have () = (), we see that for another full -lattice such that() = () and () = (), and are left ()-ideals, but one isa right ()-ideal, and the other a ()-ideal, so considering the set of idealsis not quite what we want. Now we want () or () to be maximal so thatTheorem 2.3.3 gets one of the products 1 = () or 1 = ()equal to a maximal order, allowing an order to be the identity of the potential

28


36/58

group. Thus we must consider a more general case than an ideal over a maximalorder and define the following.

Definition 3.1.4 A full -lattice in is called a normal ideal of if is amaximal order in .

Example 3.1.5 For a maximal order , left ideals of are normal idealssince () = . So this definition is extending the reach of all the ideals we areconsidering to all arbitrary maximal orders.

It will be seen more clearly why this definition is needed later. The definition,which says the left order is maximal, however says nothing about the maximalityof , so though we want the identity to be a maximal order, this is currentlyonly plausible for left multiplication. We soon see that for a normal ideal that must also be maximal, but we first need to state the following theorem,which actually needs another standard theorem, which can be found in Jacobson[6, p.200-201].

Theorem 3.1.6 is a simple artinian ring if and only if is artinian and has asimple module such that ann = 0.

Now the following theorem is a remarkably important result and will be usedthroughout the following sections. So for and as in the theorem below, we saythat belongs to .

Theorem 3.1.7 Let be a maximal left integral ideal of maximal order . Then

a) There is a unique prime ideal of such that and = /.Moreover gives a surjective relation between the maximal integralideals of and the prime ideals of .

b) The simple ring / has simple left module /.c) Each prime ideal determines a maximal left integral ideal that belongs to it.

Proof. Since is a maximal left integral -ideal we see that = ann/ = { } and it is a two-sided ideal. Choose an , then = .But is a two-sided ideal, so write it as 11 for disctinct prime ideals and non-zero . Then / / has only a finite number of two-sided idealsand is thus artinian, so / is artinian also. Now

ann// =

{

/

(/) = (/) = 0

}=

{

/

,

}= 0

(note this is 0 in /), so since / is a simple module (since is maximal) byTheorem 3.1.6 / is simple, proving part b and implying is actually a primeideal of . This is unique since if is another prime ideal contained in , then ann/, implying = .

Now given any prime ideal of there is a maximal left integral ideal such that . Since /(ann/) is simple, then ann/ and thus = ann/, proving part c and the surjectivity of the relation , provingpart a.

29


37/58

3.2 Generalising a Familiar Sight

So now we want to look at our aims of factorising ideals of maximal orders intoirreducible ideals and looking to see if the set of them is a group. But as we neededto do in Chapter 2 when generalising Chapter 1, we need to extend the results of

Chapter 2 to one-sided ideals here, where the main reference is Reiner [1, p.201,207-209]. Also, since we introduced normal ideals and intend on using them, we hadbetter look at them to see their properties. As we have suggested, this section ismainly for the creation of utensils to use for looking at our aims in the next section.Also, as suggested by the name of this section, Theorem 3.1.7 will have its first usein the following very familiar Lemma, which is courtesy of Reiner [1, p.207-208].

Lemma 3.2.1 Let be a proper integral ideal of . Then 1 .

Proof. If is not maximal then for some maximal integral ideal . But then1 1. So it suffices to prove the lemma for a maximal integral ideal. Let = ann(/) be the prime ideal to which belongs. Then / is a simple leftmodule over the simple ring = /. Hence is the inverse image, under the map , of some maximal left ideal of . Therefore we may write = (1 ) + ,where is such that its image is a primitive idempotent in . Set = + ,a right ideal in . Since (1 ) , it follows that . On the other hand, is a two-sided ideal of containing 2. It follows that is either or 2.If = 2, then 2; multiplying by 1, we deduce that , which isimpossible. This shows = .

Now choose a non-zero and write as a product of prime idealsof . Since , one of the factors must equal , and so we may write = = , where is some two-sided ideal of . If 1 = , then1 1 = implies , which in turn implies 1 = ,so . But = + , so the last inclusion is impossible. This shows that1 and completes the proof of the lemma.

This then allows the following fundamental result, which was alluded to earlierand allows the notion of a proper product to be more familiar and not a rare novelty.

Theorem 3.2.2 For a full -lattice in , is a maximal order if and only if is a maximal order.

Proof. Let = be a normal ideal. We first prove the theorem for integral

ideals, but see that if = , then = () = is trivially maximal. Soassume from hereon that is proper integral. For a contradiction, suppose that is a maximal counterexample to the theorem. That is, is not maximal and if is also a left -ideal, then is maximal. In fact, let = be one suchthat / is simple, that is, is a minimal left ideal in /. Let =

1, then 1 = since if = 1 = 1 = then = , a contradiction.Also, since / is simple it follows that is a maximal integral ideal in . Now() () = , so () = and () () = . But if (),then = ( ) () = , so () and () = () = and = .

30


38/58

Since is not maximal, let be maximal and consider the product

1. It is a full -lattice containing the unity since

1 1 = ,and is a unital full -lattice. It is also a subring of since

factorisation theory in a non-commutative algebra

Documents