University of South CarolinaMath 701Fall 2010

Course Notes

Rings and Modules

Instructor:Dr. George F. McNulty

Last compiled: January 21, 2011

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Contents

Syllabus for Mathematics 701, Fall 2010 2Lecture 0. The Basics of Algebraic Systems 3

0.1. Algebraic Systems 30.2. Homomorphisms and their relatives 40.3. Subuniverses and sublagebras 50.4. Congruence relations 50.5. A comment of mathematical notation 6

Lecture 1. Algebras with simplest signature and direct products 71.1. The Homomorphism Theorem for the empty signature 71.2. Direct Products 8

Lecture 2. The Isomorphism Theorems 10Lecture 3. Comprehending plus and times 16Lecture 4. Zorn’s Lemma 17Lecture 5. Submodules of Free Modules over a PID 19Lecture 6. The Direct Decomposition of Finitely Generated Modules over a PID: The First

Step 22Lecture 7. Direct Decomposition of Finitely Generated Torsion Modules over a PID: The

Second Step 25Lecture 8. The Structure Theorem for Finitely Generated Modules over a PID 29Lecture 9. Decomposition of Vector Space with a Designated Linear Operator 34Solutions to Problem Sets 39Solutions for Midterm Examination I 60Solutions for Midterm Examination II 63Solutions to Midterm Examination III 67Solutions to Final Examination 71

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Syllabus for Mathematics 701, Fall 2010

Text: Basic Algebra I and II by Nathan JacobsonClass Meetings: TTh 2:00pm–3:15 pm in LeConte 310Instructor: George McNultyOffice: 302 LeConteOffice Hours: Monday and Thursday from 3:30pm to 5:00pm.

Actually, anytime I am in my office(The best times are on Monday and Wednesday)

Contact Info: 781-9505 (Home) email: mcnulty@math.sc.edu

Examination ScheduleMidterm 0 Midterm 1 Midterm 2 FinalThursday Thursday Thursday Wednesday16 September 21 October 18 November 8 December at 2 pmIf you cannot take an exam at the scheduled time, see me to arrange an alternate time.

What you should work to acheive in this course• To acquire a reliable algebraic intuition,• To come to friendly terms with the basic concepts and examples in algebra,• To master a number of the techniques and methods of algebra in general and the theories

of groups and rings in particular.• To acquire a fluency with algebraic ideas, results, methods, and examples that will lead to

success on the Ph.D. qualifying examination in algebra.

Weekly Problem SetsThere will be roughly one problem set each week, generally due at the Thursday meetings of theclass. The problems, for the most part, will be like those on the Ph.D. Qualifying exam. Studentsare encouraged to collaborate on their solution.

ExaminationsThe examinations will be cumulative with 4 problems on Midterm 0, 8 problems on Midterm 1,and 12 problems on Midterm 2 and on the final. The problems accumulate in the sense that, forexample Problem 4 will always test roughly the same material on each examination. Once youmaster a particular problem, it is not necessary to attempt it on later exams. So the idea is toattempt 3 or 4 problems on each of the Midterms (even though more problems are offered) and usethe final to attempt those problems that you have yet to master.

Course GradesI keep track of how you do on each examination problem. At the end of the semester, if you havemastered at least 8 of the problems I will assign an A, while mastery of between 5 and 7 problemswill merit some kind of B. Weaker performances (none I hope) will earn lower grades. An A in thiscourse means I think that you have a good shot at passing the Algebra portion of the QualifyingExam, provided you work during the summer to keep the ideas and theorems fresh in your mind.A B in the course means I think you have a shot at passing the Algebra portion of the QualifyingExam, but you will need to work pretty hard over the summer to gain a better command of theideas and methods needed.

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Lecture 0. The Basics of Algebraic Systems

0.1. Algebraic Systems. In your undergraduate coursework you have already encounteredmany algebraic systems. These probably include some specific cases, like 〈Z,+, ·,−, 0, 1〉 which isthe system of integers equipped with the usual two-place operations of addition, multiplication,the one-place operation of forming negatives, and two distinguished integers 0 and 1, which wewill construe as zero-place operations (all output and no input). You have also encountered wholeclasses of algebraic systems such as the class of vector spaces over the real numbers, the class ofrings, and the class of groups. You might even have encountered other classes of algebraic systemssuch are Boolean algebras and lattices.

The algebraic systems at the center of this two-semester course are rings, modules, groups, andfields. Vector spaces are special cases of modules. These kinds of algebraic systems arose in thenineteenth century and the most of the mathematics we will cover was well-known by the 1930’s.This material forms the basis for a very rich and varied branch of mathematics that has flourishedvigorously over the ensuing decades.

Before turning to rings, modules, groups, and fields, it pays to look at algebraic systems from afairly general perspective. Each algebraic system consist of a nonempty set of elements, like theset Z of integers, equipped with a system of operations. The nonempty set of elements is calledthe universe of the algebraic system. (This is a shortening of “universe of discourse”.) Each ofthe operations is a function that takes as inputs arbitrary r-tuples of elements of the universe andreturns an output again in the universe—here, for each operation, r is some fixed natural number.In the familiar algebraic system 〈Z,+, ·,−, 0, 1〉, the operations of addition and multiplication areof rank 2 (they are two-place operations), the operation of forming negatives is of rank 1, and thetwo distinguished elements 0 and 1 are each regarded as operations of rank 0.

Aside. Let A be a set and r be a natural number. We use Ar to denote the set of all r-tuples ofelements of A. An operation F of rank r on A is just a function from Ar into A. There is a curiouscase. Suppose A is the empty set and r > 0. Then Ar is also empty. A little reflection shows thatthe empty set is also a function from Ar into A, that is the empty set is an operation of rank r.The curiosity is that this is so for any positive natural number r. This means that the rank ofthis operation is not uniquely determined. We also note that A0 actually has one element, namelythe empty tuple. This means that when A is empty there can be no operations on A of rank 0.On the other hand, if A is nonempty, then the rank of every operation of finite rank is uniquelydetermined and A has operations of every finite rank. These peculiarities explain, to some extent,why we adopt the convention that the universe of an algebraic system should be nonempty.

The notion of the signature of an algebraic system is a useful way to organize the basic notionsof our subject. Consider these three familiar algebraic systems:

〈Z,+, ·,−, 0, 1〉〈C,+, ·,−, 0, 1〉〈R2×2,+, ·,−, 0, 1〉

The second system consists of the set of complex numbers equipped with the familiar operations,while the third system consists of the set of 2× 2 matrices with real entries equipped with matrixmultiplication, matrix addition, matrix negation, and distinguished elements 0 for the zero matrix,and 1 for the identity matrix. Notice that + has a different meaning on each line displayedabove. This is a customary, even well-worn, ambiguity. To resolve this ambiguity let us regard +not as a two-place operation but as a symbol for a two-place operation. Then each of the threealgebraic systems gives a different meaning to this symbol—a meaning that would ordinarily beunderstood from the context, but could be completely specified as needed. A signature is just a

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set of operation symbols, each with a uniquely determined natural number called its rank. Moreformally, a signature is a function with domain some set of operation symbols that assigns to eachoperation symbol its rank. The three algebraic systems above have the same signature.

“Algebraic system” is a mouthful. So we shorten it to “algebra”. This convenient choice is inconflict another use of this word to refer to a particular kind of algebraic system obtained byadjoining a two-place operation of a certain kind to a module.

As a matter of notation, we tend to use boldface A to denote an algebra and the correspondingnormalface A to denote its universe. For an operation symbol Q we use, when needed, QA todenote the operation of A symbolized by Q. We follow the custom of writing operations like +between its inputs (like 5 + 2) but this device does not work very well if the rank of the operationis not two. So in general we write things like QA(a0, a1, . . . , ar−1) where the operation symbol Qhas rank r and a0, a1, . . . , ar−1 ∈ A.

Each algebra has a signature. It is reasonable to think of each algebra as one particular way togive meaning to the symbols of the signature.

0.2. Homomorphisms and their relatives. Let A and B be algebras of the same signature.We say that a function h : A → B is a homomorphism provided for every operation symbol Qand all a0, a1, . . . , ar−1 ∈ A, where r is the rank of Q, we have

h(QA(a0, a1, . . . , ar−1)) = QB(h(a0), h(a1), . . . , h(ar−1)).That is, h preserves the basic operations. We use h : A→ B to denote that h is a homomorphismfrom the algebra A into the algebra B. For example, we learned in linear algebra that the deter-minant det is a homomorphism from 〈R2×2, ·, 0, 1〉 into 〈R, ·, 0, 1〉. The key fact from linear algebrais

det(AB) = detA detB.We note in passing that the multiplication on the left (that is AB) is the multiplication of matrices,while the multiplication on the right is multiplication of real numbers.

In the event that h is a homomorphism from A into B that happens to be one-to-one we call itan embedding and express this in symbols as

h : A � B.In the event that the homomorphism h is onto B we say that B is a homomorphic image of Aand write

h : A � B.In the event that the homomorphism h is both one-to-one and onto B we say that h is an isomor-phism and we express this in symbols as

h : A �� Bor as

Ah∼= B.

It is an easy exercise, done by all hard-working graduate students, that if h is an isomorphism fromA to B then the inverse function h−1 is an isomorphism from B to A. We say that A and B areisomorphic and write

A ∼= Bprovided there is an isomorphism from A to B.

The algebra 〈R,+,−, 0〉 is isomorphic to 〈R+, ·,−1, 1〉, where R+ is the set of positive real numbers.There are isomorphisms either way that are familiar to freshman in calculus. Find them.

A homomorphism from A into A is called an endomorphism of A. An isomorphism from A toA is called an automorphism of A.

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0.3. Subuniverses and sublagebras. Let A be an algebra. A subset B ⊆ A is called asubuniverse of A provided it is closed with respect to the basic operations of A. This means thatfor every operation symbol Q of the signature of A and for all b0, b1, . . . , br−1 ∈ B, where r is therank of Q we have QA(b0, b1, . . . , br−1) ∈ B. Notice that if the signature of A has an operationsymbol c of rank 0, then cA is an element of A and this element must belong to every subuniverseof A. On the other hand, if the signature of A has no operation symbols of rank 0, then the emptyset ∅ is a subuniverse of A.

The restriction of any operation of A to a subuniverse B of A results in an operation on B. Inthe event that B is a nonempty subuniverse of A, we arrive at the subalgebra B of A. This isthe algebra of the same signature as A with universe B such that QB is the restriction to B of QA,for each operation symbol Q of the signature. B ≤ A symbolizes that B is a subalgebra of A.

Here is a straightforward but informative exercise for hard-working graduate students. Let N ={0, 1, 2, . . . } be the set of natural numbers. Discover all the subuniverses of the algebra 〈N,+〉.

0.4. Congruence relations. Let A be an algebra and h be a homomorphism from A to somealgebra. We associate with h the following set, which called here the functional kernel of h,

θ = {(a, a′) | a, a′ ∈ A and h(a) = h(a′)}.This set of ordered pairs of elements of A is evidently an equivalence relation on A. That is, θ hasthe following properties.(a) It is reflexive: (a, a) ∈ θ for all a ∈ A.(b) It is symmetric: for all a, a′ ∈ A, if (a, a′) ∈ θ,then (a′, a) ∈ θ.(c) It is transitive: for all a, a′, a′′ ∈ A, if (a, a′) ∈ θ and (a′, a′′) ∈ θ, then (a, a′′) ∈ θ.This much would be true were h any function with domain A. Because θ is a binary (or two-place)relation on A it is useful to use the following notations interchangeably.

(a, a′) ∈ θa θ a′

a ≡ a′ mod θ

Here is another piece of notation which we will use often. For any set A, any a ∈ A and anyequivalence relation θ on A we put

a/θ := {a′ | a′ ∈ A and a ≡ a′ mod θ}.We also put

A/θ := {a/θ | a ∈ A}.Because h is a homomorphism θ has one more important property, sometimes called the substi-

tution property:For every operation symbol Q of the signature of A and for all a0, a

′0, a1, a

′1, . . . , ar−1, a

′r−1 ∈ A,

where r is the rank of Q,if

a0 ≡ a′0 mod θ

a1 ≡ a′1 mod θ

...ar−1 ≡ a′r−1 mod θ

thenQA(a0, a1, . . . , ar−1) ≡ QA(a′0, a′1, . . . , a′r−1) mod θ.

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An equivalence relation on A with the substitution property above is called a congruence relationof the algebra A. The functional kernel of a homomorphism h from A into some other algebra isalways a congruence of A. We will see below that this congruence retains almost all the informationabout the homomorphism h.

As an exercise to secure the comprehension of this notion, the hard-working graduate studentsshould try to discover all the congruence relations of the familiar algebra 〈Z,+, ·〉.

0.5. A comment of mathematical notation. The union A∪B and the intersection A∩B ofsets A and B are familiar. These are special cases of more general notions. Let K be any collectionof sets. The union

⋃K is defined via

a ∈⋃

K⇔ a ∈ C for some C ∈ K.

Here is the special case A ∪B rendered in this waya ∈ A ∪B ⇔ a ∈

⋃{A,B} ⇔ a ∈ C for some C ∈ {A,B} ⇔ a ∈ A or a ∈ B.

Similarly, the intersection⋂

K is defined viaa ∈

⋂K⇔ a ∈ C for all C ∈ K.

Notice that in the definition of the intersection, each set belonging to the collection K imposes aconstraint on what elements are admitted to membership in

⋂K. When the collection K is empty

there are no constraints at all on membership in⋂

K. This means⋂

∅ is the collection of allsets. However, the having the collection of all sets in hand leads to a contradiction, as discoveredindependently by Ernst Zermelo and Bertrand Russell in 1899. To avoid this, we must avoid formingthe intersection of empty collections. This situation is analogous to division by zero. Just as whendivision of numbers comes up, the careful mathematician considers the possibility that the divisoris zero before proceeding, so must the careful mathematician consider the possibility that K mightbe empty before proceeding to form

⋂K.

We also use the notation ⋃i∈I

Ai and⋂i∈I

Ai

to denote the union and intersection of K = {Ai | i ∈ I}. The set I here is used as a set of indices.In using this notation, we impose no restrictions on I (save that in forming intersections the set Imust not be empty). In particular, we make no assumption that the set I is ordered in any way.

The familiar set builder notation, for example {n | n is a prime number}, has a companion in thefunction builder notation. Here is an example

f = 〈ex | x ∈ R〉.The function f is just the exponential function on the real numbers. We take the words “func-tion”, “sequence”, and “system” to have the same meaning. We also use the notation f(c) and fcinterchangeably when f is a function and c is a member of its domain.

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Lecture 1. Algebras with simplest signature and direct products

The simplest signature is, of course, empty. It provides no operation symbols. In this setting,algebras have nothing real to distinguish them from nonempty sets. Every function between twononempty sets will be a homomorphism. Every subset will be a subuniverse. Every equivalencerelation will be a congruence. Isomorphisms are just one-to-correspondences between nonemptysets and two nonempty sets will be isomorphic just in case they have the same cardinality. So doingalgebra in the empty signature if a branch of combinatorics.

Nevertheless, there is an important lesson for us to learn here.

1.1. The Homomorphism Theorem for the empty signature. Suppose that A is anonempty set. By a partition of A we mean a collection P of subsets of A with the followingproperties:(a) Each member of P is a nonempty subset of A.(b) If X,Y ∈ P and X 6= Y , then X and Y are disjoint.(c) Every element of A belongs to some set in the collection P.

There is a close connection between the notion of a function with domain A, the notion of anequivalence relation on A, and the notion of a partition of A. You may already be familiar withthis connection. We present it in the following theorem:

The Homomorphism Theorem, empty version. Let A be a nonempty set, let f : A � B bea function from A onto B, let θ be an equivalence relation on A, and let P be a partition of A. Allof the following hold.(a) The functional kernel of f is an equivalence relation on A.(b) The collection A/θ = {a/θ | a ∈ A} is a partition of A.(c) The map η that assigns to each a ∈ A the set in P to which it belongs is a function from A

onto P; moreover P is the collection of equivalence classes of the functional kernel of η.(d) If θ is the functional kernel of f , then there is a one-to-one correspondence g from A/θ to B

such that f = g ◦ η.

Here is a diagram that displays something of what this theorem asserts.

The empty version of the Homomorphism Theorem is almost too easy to prove. One merely hasto check what the definitions of the various notions require. The map η is called the quotient

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map. That it is a function, i.e. that

{(a,X) | a ∈ A and a ∈ X ∈ P}

is a function, follows from the dijointness of distinct members of the partition. That its domainin A follows from condition (c) in the definiton of partition. The one-to-one correspondence gmentioned in assertion (d) of the Homomorphism Theorem is the following set of ordered pairs:

{(a/θ, f(a)) | a ∈ A}.

The proof that this set is a one-to-one function from A onto B is straightforward, the most amusingpart being the demonstration that it is actually a function.

1.2. Direct Products. Just as you are familiar with A∪B and A∩B, you probably alreadyknow that A × B denotes the set of all ordered pairs whose first entries are chosen from A whilethe second entries are chosen from B. Just as we did for unions and intersections we will extendthis notion.

Let 〈Ai | i ∈ I〉 be any system of sets. We call a function a : I →⋃i∈I Ai a choice function for

the system 〈Ai | i ∈ I〉 provided ai ∈ Ai for all i ∈ I. It is perhaps most suggestive to think of aas an I-tuple (recalling that we are using function, tuple, system, and sequence interchangeably).The direct product of the system 〈Ai | i ∈ I〉 is just the set of all these choice functions. Here isthe notation we use:∏

〈Ai | i ∈ I〉 :=∏i∈I

Ai := {a | a is a choice function for the system 〈Ai | i ∈ I〉}.

This sets Ai are called the direct factors of this product. If any of the sets in the system 〈Ai | i ∈ I〉is empty, then the direct product is also empty. On the other hand, if I is empty then the directproduct is {∅}, since the empty set will turn out to be a choice function for the system. Noticethat {∅} is itself nonempty and, indeed, has exactly one element.

Observe that∏〈A,B〉 = {〈a, b〉 | a ∈ A and b ∈ B}. This last set is, for all practical purposes,

A×B.Projection functions are associated with direct products. For any j ∈ I, the jth projection function

pj is defined, for all a ∈∏〈Ai | i ∈ I〉, via

pj(a) := aj .

The systems of projection functions has the following useful property: it separate points. Thismeans that if a, a′ ∈

∏〈Ai | i ∈ T 〉 and a 6= a′, then pj(a) 6= pj(a′) for some j ∈ I. Suppose that

〈Ai | i ∈ I〉 is a system of sets, that B is some set that 〈fi | i ∈ I〉 is a system of functions suchthat fi : B → Ai for each i ∈ I. Define the map h : B →

∏〈Ai | iI〉 via

h(b) := 〈fi(b) | i ∈ I〉.

Then it is easy that fi = pi ◦ h for all i ∈ I. If the system 〈fi | i ∈ I〉 separates points, then thefunction h defined just above will be one-to-one, as all hard-working graduate students will surelycheck.

We form direct products of systems of algebras in the following way. Let 〈Ai | i ∈ I〉 be a systemof algebras, all with the same signature. We take

∏〈Ai | i ∈ I〉 to be the algebra P with universe

P :=∏〈Ai | i ∈ I〉 and where the operations on P are defined coordinatewise. This means that for

each operation symbol Q and all a0, a1, . . . , ar−1 ∈ P , where r is the rank of Q we have

QP(a0, a1, . . . , ar−1) = 〈QAi(a0,i, a1,i, . . . , ar−1,i) | i ∈ I〉.

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To see more clearly what is intended here, suppose that Q has rank 3, that I = {0, 1, 2, 3}, andthat a, b, c ∈ P . Then

abc

QP(a, b, c)

====

〈a0, a1, a2, a3〉〈b0, b1, b2, b3〉〈c0, c1, c2, c3〉

〈QA0(a0, b0, c0), QA1(a1, b1, c1), QA2(a2, b2, c2), QA1(a3, b3, c3)〉In this way, the direct product of a system of algebras, all of the same signature, will be again an

algebra of the common signature and it is evident that each projection map is a homomorphismfrom the direct product onto the corresponding direct factor. Even the following fact is easy toprove.

Fact. Let 〈Ai | i ∈ I〉 be a system of algebras, all of the same signature. Let B be an algebra ofthe same signature as A and let 〈fi | i ∈ I〉 be a system of homomorphisms so that fi : B → Ai

for all i ∈ I. Then there is a homomorphism h : B →∏i∈I Ai so that fi = pi ◦ h for all i ∈ I.

Moreover, if 〈fi | i ∈ I〉 separates points, then h is one-to-one.

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Lecture 2. The Isomorphism Theorems

The four theorems presented today arose over a period of perhaps forty years from the mid 1890’sto the mid 1930’s. They emerged from group theory and the theory of rings and modules chieflyin the work of Richard Dedekind and Emmy Noether and it was Noether who gave their first clearformulation in the context of module theory in 1927. You have probably already seen versions ofthese theorems for groups or rings is an undergraduate abstract algebra course.

We will frame them is the broader context of algebras in general. That way it will not be necessaryto do more than add a comment or two when applying them in the context of groups, rings, andmodules (these being our principal focus). In addition, you will be able to apply them in the contextof lattices, Boolean algebras, or other algebraic systems.

At the center of this business is the notion of a quotient algebra. Let A be an algebra and letθ be a congruence of A. Recall that for each a ∈ A we use a/θ to denote the congruence class{a′ | a′ ∈ A and a ≡ a′ mod θ}. Moreover, we use A/θ to denote the partition {a/θ | a ∈ A} of Ainto congruence classes. We make the quotient algebra A/θ by letting its universe be A/θ and, foreach operation symbol Q of the signature of A, and all a0, a1, . . . , ar−1 ∈ A, where r is the rank ofQ, we define

QA/θ(a0/θ, a1/θ, . . . , ar−1/θ) := QA(a0, a1, . . . , ar−1)/θ.Because the elements of A/θ are congruence classes, we see that the r inputs to QA/θ must becongruence classes. On the left side of the equation above the particular elements ai have nospecial standing—they could be replaced by any a′i provided only that ai ≡ a′i mod θ. Looselyspeaking, what this definition says is that to evaluate QA/θ on an r-tuple of θ-classes, reach into eachclass, grab an element to represent the class, evaluate QA at the r-tuple of selected representative toobtain say b ∈ A, and then output the class b/θ. A potential trouble is that each time such a processis executed on the same r-tuple of congruence classes, different representatives might be selectedresulting in, say b′ instead of b. But the substitution property, the property that distinguishescongruences from other equivalence relations, is just what is needed to see that there is really notrouble. To avoid a forest of subscripts, here is how the argument would go were Q to have rank 3.Suppose a, a′, b, b′, c, c′ ∈ A with

a/θ = a′/θ

b/θ = b′/θ

c/θ = c′/θ

So a and a′ can both represent the same congruence class—the same for b and b′ and for c and c′.Another way to write this is

a ≡ a′ mod θ

b ≡ b′ mod θ

c ≡ c′ mod θ

What we need is QA(a, b, c)/θ = QA(a′, b′, c′)/θ. Another way to write that is QA(a, b, c) ≡QA(a′, b′, c′) mod θ. But this is exactly what the substitution property provides. Hard-workinggraduate students will do the work to see that what works for rank 3 works for any rank.

The theorem below, sometimes called the First Isomorphism Theorem, is obtained from its versionfor the empty signature replacing arbitrary functions by homomorphism and arbitrary equivalenacerelations by congruence relations.

The Homomorphism Theorem. Let A be an algebra, let f : A � B be a homomorphism fromA onto B, and let θ be a congruence relation of A. All of the following hold.

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(a) The functional kernel of f is an congruence relation of A.(b) A/θ is an algebra of the same signature as A.(c) The map η that assigns to each a ∈ A the congruence class a/θ is a homomorphism from A

onto A/θ and its functional kernel is θ.(d) If θ is the functional kernel of f , then there is an isomorphism g from A/θ to B such that

f = g ◦ η.The proof of this theorem has been, for the most part, completed already. We just saw how to

prove part (b) and part (a) was done when the notions of congruence relation and functional kernelwere introduced. Even parts (c) and (d) were most established in the version of the theorem foralgebras with empty signature. It only remains to prove that the quotient map η in part (c) andthe map g in part (d) are actually homomorphisms. With the definition of how the operations inthe quotient algebra work, this only requires checking that the basic oeprations are preserved by ηand by g. This work is left to the diligent graduate students.

From parts (a) and (c) of the Homomorphism Theorem we draw the following corollary.

Corollary 0. Let A be an algebra. The congruence relations of A are exactly the functional kernelsof homomorphisms from A into algebras of the same signature as A.

It will be necessary, as we develop the theory of rings, modules, and groups, to determine whethercertain equivalence relations at hand are in fact congruence relations. Of course, we can alwayscheck the conditions defining the concept of congruence relation. But sometimes it is simpler toshow that the relation is actually the functional kernel of some homomorphism.

Now let us suppose that θ is a congruence of A and that B is a subalgebra of A. By θ � B wemean the restriction of θ to B. That is

θ � B := θ ∩ (B ×B).Now θ partitions A into congruence classes. Some of these congruence class may include elementsof B while others may not. We can inflate B using θ to obtain the set θB of all elements of Arelated by θ to some element of B. That is

θB := {a | a ≡ b mod θ for some b ∈ B}.The diagram below illustrates in inflation of B by θ, where we have drawn lines to indicate the

partition of A into θ-classes.

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The Second Isomorphism Theorem. Let A be an algebra, let θ be a congruence of A, and letB be a subalgebra of A. Then each of the following hold.(a) θ � B is a congruence relation of B.(b) θB is a subuniverse of A.(c) θB/(θ � θB) ∼= B/θ � B.

Proof. For part (a) we have to see that θ � B is an equivalence relation on B and that it has thesubstitution property. Hard-working graduate students will check that it is indeed an equivalencerelation. To see that the substitution property holds, let Q be an operation symbol. Just forsimplicity, let us suppose the rank of Q is 3. Pick a, a′, b, b′, c, c′ ∈ B so that

a ≡ a′ mod θ � B

b ≡ b′ mod θ � B

c ≡ c′ mod θ � B.

We must show that QB(a, b, c) ≡ QB(a′, b′, c′) mod θ � B. Because all those elements come fromB we see that

a ≡ a′ mod θ

b ≡ b′ mod θ

c ≡ c′ mod θ,

and that both QB(a, b, c) = QA(a, b, c) and QB(a′, b′, c′) = QA(a′, b′, c′). It follows from the sub-stitution property for θ that QA(a, b, c) ≡ QA(a′, b′, c′) mod θ. But since both QA(a, b, c) =QB(a, b, c) ∈ B and QA(a′, b′, c′) = QB(a′, b′, c′) ∈ B, we draw the desired conclusion thatQB(a, b, c) ≡ QB(a′, b′, c′) mod θ � B.

For part (b) we have to show that θB is closed under all the basic operations of A. So let Q bean operation symbol, which without loss of generality we assume to have rank 3. Let a, b, c ∈ θB.Our goal is to show that QA(a, b, c) ∈ θB. Using the definition of θB pick a′, b′, c′ ∈ B so that

a ≡ a′ mod θ

b ≡ b′ mod θ

c ≡ c′ mod θ.

Because B is a subuniverse, we see that QA(a′, b′, c′) ∈ B. Because θ is a congruence, we see thatQA(a, b, b) ≡ QA(a′, b′, c′). Putting these together, we find that QA(a, b, c) ∈ θB, as desired.

For part (c) we will invoke the Homomorphism Theorem. Define the map h from B to θB/(θ � θB)via

h(b) := b/(θ � θB).We have three contentions, namely that h is a homomorphism, that h is onto B/(θ � θB), and thatthe functional kernel of h is θ � B. Given these, the Homomorphism Theorem provides that desiredisomorphism.

To see that h is a homomorphism we have to show it respects the operations. So again take Q beto an operation symbol, of rank 3 for simplicity. Let a, b, c ∈ B. Now observe

h(QB(a, b, c)) = QB(a, b, c)/(θ � θB)= QθB(a, b, c)/(θ � θB)

= QθB/(θ�θB)(a/(θ � θB), b/(θ � θB), c/(θ � θB))

= QθB/(θ�θB)(h(a), h(b), h(c)).

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In this way we see that h respects Q. So h is a homomorphism.To see that h is onto, let b′ ∈ θB. Pick b ∈ B so that b′ ≡ b mod θ. We assert that h(b) = b′/(θ �

θB). So what we have to demonstrate is that

b/(θ � θB) = b′/(θ � θB)

or what is the sameb ≡ b′ mod θ � θB.

Now both b and b′ belong to θB, so all that remains is to see that b ≡ b′ mod θ. But we alreadyknow this.

Finally, we have to understand the functional kernel of h. Let a, b ∈ B and observe

h(a) = h(b)⇔ a/(θ � θB) = b/(θ � θB)⇔ a ≡ b mod θ � θB

⇔ a ≡ b mod θ � B.

The last equivalence follows since a and b both belong to B. So we see that θ � B is the functionalkernel of h, completing the proof. �

Let A be an algebra and let θ and ϕ be congruences of A with θ ⊆ ϕ. Let

ϕ/θ := {(a/θ, a′/θ) | a, a′ ∈ A with a ≡ a′ mod ϕ}.

So ϕ/θ is a two place relation on A/θ.

The Third Isomorphism Theorem. Let A be an algebra and let θ and ϕ be congruences of Awith θ ⊆ ϕ. Then(a) ϕ/θ is a congruence of A/θ, and(b) (A/θ)/(ϕ/θ) ∼= A/ϕ.

Proof. Define the function h from A/θ to A/ϕ so that for all a ∈ A we have

h(a/θ) := a/ϕ.

Here we have to worry again whether h is really a function—the definition above uses a represen-tative element a of the congruence class a/θ to say how to get from the input to the output. Whatif a/θ = a′/θ? Then (a, a′) ∈ θ. Since θ ⊆ ϕ, we get (a, a′) ∈ ϕ. This means, of course, thata/ϕ = a′/ϕ. So we arrive at the same output, even using different representatives. This means ourdefinition is sound.

Let us check that h is a homomorphism. So let Q be an operation symbol, which we suppose hasrank 3 just in order to avoid a lot of indices. Pick a, b, c ∈ A. Now observe

h(QA/θ(a/θ), b/θ, c/θ) = h(QA(a, b, c)/θ)= QA(a, b, c)/ϕ

= QA/ϕ(a/ϕ, b/ϕ, c/ϕ)

= QA/ϕ(h(a/θ), h(b/theta), h(c/θ))

In this way we see that h respects the operation symbol Q. We conclude that h is a homomorphism.Notice that h is onto A/ϕ since any member of that set has the form a/ϕ for some a ∈ A. This

means that h(a/θ) = a/ϕ.Now lets tackle the functional kernel of h. Let a, b ∈ A. Then observe

h(a/θ) = h(b/θ)⇔ a/ϕ = b/ϕ⇔ a ≡ b mod ϕ.

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So (a/θ, b/θ) belongs to the functional kernel of h if and only if a ≡ b mod ϕ. That is, thefunctional kernel of h is ϕ/θ. From the Homomorphism Theorem we see that ϕ/θ is a congruenceof A/θ. Also from the Homomorphism Theorem we conclude that (A/θ)/(ϕ/θ) ∼= A/ϕ. �

The set inclusion relation ⊆ is a partial ordering of the congruence relations of an algebra A.Some of the secrets of about A can be discovered by understanding how the congruence relationsare ordered. The next theorem, sometimes called the Fourth Isomorphism Theorem, is a firstand useful step along this road. To understand it we need the notion of isomorphism of relationalstructures (as opposed to algebras). Let A and B be nonempty sets and let v be a two-place relationon A and � be a two-place relation on B. A function h from A to B is called an isomorphismbetween 〈A,v〉 and 〈B,≺〉 provided h is one-to-one, h is onto B, and for all a, a′ ∈ A we have

a v a′ if and only if h(a) � h(a′).

As a matter of notation, let Con A be the set of congruence relations of A.

The Correspondence Theorem. Let A be an algebra and let θ be a congruence of A. LetP = {ϕ | ϕ ∈ Con A and θ ⊆ ϕ}. Then the map from P to Con A/θ that sends each ϕ ∈ P to ϕ/θis an isomorphism between the ordered set 〈P,⊆〉 and the ordered set 〈Con A/θ,⊆〉.

Proof. Let Ψ denote the map mentioned in the theorem. So

Ψ(ϕ) = ϕ/θ

for all ϕ ∈ Con A with θ ⊆ ϕ.To see that Ψ is one-to-one, let ϕ, ρ ∈ Con A with θ ⊆ ϕ and θ ⊆ ρ. Suppose that Ψ(ϕ) = Ψ(ρ).

This means ϕ/θ = ρ/θ. Now consider for all a, a′ ∈ A

(a, a′) ∈ ϕ⇔ (a/θ, a′/θ) ∈ ϕ/θ⇔ (a/θ, a′/θ) ∈ ρ/θ⇔ (a, a′) ∈ ρ

So ϕ = ρ. Notice that the first equivalence depends of θ ⊆ ϕ while the last depends of θ ⊆ ρ. Wesee that Ψ is one-to-one.

To see that Ψ is onto Con A/θ, let µ be a congruence of A/θ. Define

ϕ := {(a, a′) | a, a′ ∈ A and (a/θ, a′/θ) ∈ µ}.

This two-place relation is our candidate for a preimage of µ. First we need to see that ϕ is indeed acongruence of A. The checking of reflexivity, symmetry, and transitivity is routine. To confirm thesubstituion property, let Q be an operation symbol (with the harmless assumption that its rank is3). Pick a, a′, b, b′, c, c′ ∈ A so that

a ≡ a′ mod ϕ

b ≡ b′ mod ϕ

c ≡ c′ mod ϕ.

We must see that QA(a, b, c) ≡ QA(a′, b′, c′) mod ϕ. From the three displayed conditions wededuce

a/θ ≡ a′/θ mod µ

b/θ ≡ b′/θ mod µ

c/θ ≡ c′/θ mod µ.

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Because µ is a congruence of A/θ, we obtainQA/θ(a/θ, b/θ, c/θ) ≡ QA/θ(a′/θ, b′/θ, c′/θ) mod µ.

But given how the operations work in quotient algebras, this givesQA(a, b, c)/θ ≡ QA(a′, b′, c′)/θ mod µ.

Then the definition of ϕ supports the desired conclusion that QA(a, b, c) ≡ QA(a′, b′, c′) mod ϕ.So ϕ is a congruence of A. But we also need to see that θ ⊆ ϕ to get that ϕ ∈ P . So suppose thata, a′ ∈ A with (a, a′) ∈ θ. Then a/θ = a′/θ. This entails that (a/θ, a′/θ) ∈ µ since µ is reflexive. Inthis way, we see that (a, a′) ∈ ϕ. So θ ⊆ ϕ and ϕ ∈ P . Now consider

Ψ(ϕ) = ϕ/θ

= {(a/θ, a′/θ) | a, a′ ∈ A and (a, a′) ∈ ϕ}= {(a/θ.a′/θ) | a, a′ ∈ A and (a/θ, a′/θ) ∈ µ}= µ.

In this way, we see that Ψ is onto Con A/θ.Last, we need to show that Ψ respects the ordering by set inclusion. So let ϕ, ρ ∈ Con A with

θ ⊆ ϕ and θ ⊆ ρ. Let us first suppose that ϕ ⊆ ρ. To see that Ψ(ϕ) ⊆ Ψ(ρ), let a, a′ ∈ A andnotice

(a/θ, a′/θ) ∈ Ψ(ϕ) =⇒ (a/θ, a′/θ) ∈ ϕ/θ=⇒ (a, a′) ∈ ϕ=⇒ (a, a′) ∈ ρ=⇒ (a/θ, a′/θ) ∈ ρ/θ=⇒ (a/θ, a′/θ) ∈ Ψ(ρ)

So we find if ϕ ⊆ ρ, then Ψ(ϕ) ⊆ Ψ(ρ). For the converse, suppose Ψ(ϕ) ⊆ Ψ(ρ). Let a, a′ ∈ A andnotice

(a, a′) ∈ ϕ =⇒ (a/θ, a′/θ) ∈ ϕ/θ=⇒ (a/θ, a′/θ) ∈ Ψ(ϕ)=⇒ (a/θ, a′/θ) ∈ Ψ(ρ)=⇒ (a/θ, a′/θ) ∈ ρ/θ=⇒ (a, a′) ∈ ρ.

So we find that if Ψ(ϕ) ⊆ Ψ(ρ), then ϕ ⊆ ρ. So we have for all ϕ, ρ ∈ Pϕ ⊆ ρ if and only if Ψ(ϕ) ⊆ Ψ(ρ).

Finally, we can conclude that Ψ is an isomorphism between our two ordered sets of congruences. �

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Lecture 3. Comprehending plus and times

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Lecture 4. Zorn’s Lemma

Zorn’s Lemma is a transfinite existence principle which has found a number of useful and infor-mative applications in algebra and analysis. While the Lemma bears the name of Max Zorn, equalcredit should be extended to Felix Hausdorff and Kazimierz Kuratowski who found closely relatedresults decades before Zorn.

A chain or linearly ordered set is just a partially ordered set in which any two elements arecomparable. We also refer to any subset of a partially ordered set as a chain when it is linearlyordered by the ordering inhereted from the larger ordered set. This means that, where a and b areelements of the chain and ≤ denotes the order relation, we have either a ≤ b or b ≤ a. Let C be asubset of a partially ordered set P and b ∈ P . We say that b is an upper bound of C provideda ≤ b for all a ∈ C. We say b is a strict upper bound provided a < b for all a ∈ C. An elementd is maximal in C if d ∈ C and whenever d ≤ a ∈ C it follows that d = a.

Zorn’s Lemma. Let P be a partially ordered set and suppose that every chain in P has an upperbound in P . Then P has a maximal member.

Proof. Let g be a function which chooses an element from each nonempty subset of P . That is thedomain of g is the collection of nonempty subsets of P and g(D) ∈ D for each nonempty subsetD ⊆ P . The function g, which is called a choice function, exists according to the Axiom of Choice.

Denote the ordering on P by ≤. For each set C ⊆ P let C denote the set of all strict upperbounds of C. Notice that the empty set ∅ is a chain in P . According to our hypothesis it musthave an upper bound in P . Since ∅ is empty this upper bound must be a proper upper bound.This means ∅ is nonempty. (Hence, P is nonempty.)

We will say that K ⊆ P is a g-chain provided• K is not empty.• K is a chain.• if C ⊆ K and C has a strict upper bound in K, then g(C) is a minimal member of C ∩K.

Here is a useful fact about how elements in g-chains compare.

Fact. Let K and J be g-chains so that a ∈ K − J and b ∈ J . Then b < a.

Proof. Let C = {d | d ∈ K ∩ J and d < a}. So C has a strict upper bound in K. Since K is ag-chain, we have g(C) is a minimal member of C ∩K. Also, g(C) ≤ a. Now C ∩ J must be empty,since otherwise g(C) ∈ K ∩ J , putting g(C) ∈ C, which is impossible. So if b ∈ J then there isd ∈ C with b ≤ d < a. Hence, b < a. �

Claim. The union of any nonempty collection of g-chains is a g-chain.

Proof. Let L be a union of some nonempty family F of g-chains. We first have to check that L islinearly ordered. So suppose a, b ∈ L. Pick K,J ∈ F so that a ∈ K and b ∈ J . We need to showthat a and b are comparable. We might as well consider that a /∈ J , since if a ∈ J we see, J beinga chain, that a and b are comparable. But the Fact above then tells us that b < a, so a and b arecomparable. This means that L is a chain.

Of course, L is not empty since it is union of a nonempty collection of nonempty sets. So itremains to verify the last condition in the definition of g-chain. To this end, let C ⊂ L such thatC has a strict upper bound b ∈ L. Pick J ∈ F so that b ∈ J . To see that C ⊆ J , pick a ∈ Cand, for contradiction, suppose a /∈ J . Pick K ∈ F so that a ∈ K. Now the Fact yields b < a.But here we also have a < b. So we find C ⊆ J . Since J is a g-chain, we have g(C) is a minimalmember of C ∩ J . But we need to see that g(C) is a minimal member of C ∩L. Suppose not. Picka′ ∈ C ∩ L so that a′ < g(C). To simplify notation, let g(C) = b′. So a′ < b′. Now pick K ′ ∈ F so

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that a′ ∈ K ′. Now the Fact above again yields b′ < a′, which is contrary to a′ < b′. This verifiesfor L the last condition in the definition of g-chain. So L is a g-chain, as claimed. �

Now let M be the union of the collection of all g-chains. Were M nonempty we could formM ∪ {g(M)}, which would be a chain properly extending M . A routine check of the definitionshows that M ∪ {g(M)} would again be a g-chain. This produces g(M) ∈ M ∪ {g(M)} ⊆ M .But we know g(M) /∈ M . So M must be empty. So M has no strict upper bounds. But by thehypothesis, every chain has an upper bound. So M must have a largest element m. That is a ≤ mfor all a ∈ M . As there is no strict upper bound of M , there can be no element which is strictlyabove m. That is m is the maximal element we seek.

Zorn’s Lemma is proven.�

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Lecture 5. Submodules of Free Modules over a PID

The objective here is to prove that, over a principal ideal domain, every submodule of a free isalso a free module and that the rank of a free submodule is always at least as large of the ranks ofits submodules.

So let R be a (nontrivial) principal ideal domain. We know that R is a free R-module of rank 1.What about the submodules of R? Suppose E is such a submodule. It is clear that E is an idealand, in fact, that the ideals of R coincide with the submodules of R. In case E is trivial (that is thesole element of E is 0) we see that E is the free R-module of rank 0. So consider the case that E isnontrivial. Since R is a principal ideal domain we pick w 6= 0 so that E is generated by w. That isE = {rw | r ∈ R}. Since we know that R has {1} as a basis, we see that the map that sends 1 to wextends to a unique module homomorphism from R onto E. Indeed, notice h(r · 1) = r · h(1) = rwfor all r ∈ R. But the homomorphism h is also one-to-one since

h(r) = h(s)rh(1) = sh(1)rw = sw

r = s

where the last step follows because integral domains satisfy the cancellation law and w 6= 0. In thisway we see that E is isomorphic to the free R-module of rank 1. We also see that {w} is a basisfor E.

So we find that at least all the submodules of the free R-module of rank 1 are themselves freeand have either rank 0 or rank 1. We can also see where the fact that R is a principal ideal domaincame into play.

The Freedom Theorem for Modules over a PID.Let R be a principal ideal domain, let F be a free R-module and let E be a submodule of F. ThenE is a free R-module and the rank of E is no greater than the rank of F.

Proof. Since trivial modules (those whose only element is 0) are free modules of rank 0, we supposebelow that E is a nontrivial module. This entails that F is also nontrivial.

Let B be a basis for F and C ⊆ B. Because F is not the trivial module, we see that B is notempty. Let FC be the submodule of F generated by C. Let EC = E ∩ FC . Evidently, C is a basisfor FC . To see that EC is free we will have to find a basis for it.

Suppose, for a moment, that C has been chosen so that EC is known to be free and that w ∈ Bwith w /∈ C. Put D := C ∪{w}. Consider the map defined on D into R that sends all the elementsof C to 0 and that sends w to 1. This map extends uniquely to a homomorphism ϕ from FD ontoR and it is easy to check (as hardworking graduate student will) that the kernel of ϕ is just FC .By the Homomorphism Theorem, we draw the conclusion that FD/FC is isomorphic to R and thatit is free of rank 1. What about ED/EC? Observe that EC = E ∩ FC = E ∩ FD ∩ FC = ED ∩ FC .So we can apply the Second Isomorphism Theorem:

ED/EC = ED/ED ∩ FC ∼= ED + FC/FC .

But ED + FC/FC is a submodule of FD/FC . This last is a free R-module of rank 1. We saw abovethat every submodule of a free R-module of rank 1 must be itself a free R-module and have rankeither 0 or 1. In this way, we find that either ED = EC (in the rank 0 case) or else ED/EC is a freeR-module of rank 1. Let us take up this latter case. Let X be a basis for EC , which we assumed,for the moment, was free. Pick u ∈ ED so that {u/EC} is a basis for ED/EC .

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We contend that X∪{u} is a basis for ED. To establish linear independence, suppose x0, . . . , xn−1are distinct element of X, that r0, . . . , rn ∈ R and that

0 = r0x0 + · · ·+ rn−1xn−1 + rnu.

First notice thatrn(u/EC) = rnu/EC = (r0x0 + · · ·+ rn−1xn−1 + rnu)/EC = 0/EC .

Since {u/EC} is a basis for ED/EC , we must have rn = 0. This leads to0 = r0x0 + · · ·+ rn−1xn−1.

But now since X is a basis for EC we see that 0 = r0 = · · · = rn−1. So we find that X ∪ {u} islinearly independent.

To see that X ∪ {u} generates ED, pick z ∈ ED. Since {u/EC} is a basis for ED/EC , pick r ∈ Rso that

z/EC = ru/EC .

This means that z−ru ∈ EC . ButX is a basis of EC . So pick x0, . . . , xn−1 ∈ X and r0, . . . , rn−1 ∈ Rso that

z − ru = r0x0 + · · ·+ rn−1xn−1.

Surely this is enough to see that z is in the submodule generated by X ∪{u}. So this set generatesED and we conclude that it must be a basis of ED.

In this way we see that for C ⊆ D ⊆ B where D arises from adding an element to C, if EC isfree, then so is ED and that either ED = EC or a basis for ED can be produced by adding just oneelement to a basis for EC .

With this in mind, we can envision a procedure for showing that E is free and its rank cannotbe larger than that of F. Notice that E = E ∩ F = E ∩ FB. So E = EB. The idea is simple. Wewill start with ∅ ⊆ B. We observe that F∅ = E∅ is the module whose sole element is 0. It isfree of rank 0. Next we select an element w ∈ B and form ∅ ∪ {w} = {w}. We find that E{w} isfree of rank 0 or rank 1. We select another element and another and another. . . until finally all theelements of B have been selected. At this point we would have EB is free and its rank can be nomore than the total number of elements we selected, namely |B| which is the rank of F.

To carry out this program, in case B were finite or even countable, we could mount a proof byinduction. You can probably see how it might be done. But we want to prove this for arbitary setsB. We could still pursue this inductive strategy openly by well-ordering B and using transfiniteinduction. By using the well-ordering we would always know what was meant by “pick the nextelement of B.”

Instead, we will invoke Zorn’s Lemma to short-circuit this rather long induction.Let F = {f | f is a function with dom f ⊆ B and range f a basis for Edom f}. Recalling that

functions are certain kinds of sets of order pairs, we see that F is paritally ordered by set inclusion.Maybe it helps to realize that to assert f ⊆ g is the same as asserting that g extends f . We note thatF is not empty since the empty function (the function with empty domain) is a member of F. Toinvoke Zorn’s Lemma, let C be any chain included in F. Let h =

⋃C. Evidently f ⊆ h for all f ∈ C.

So h is an upper bound of C. We contend that h ∈ F. We ask the hard-working graduate studentsto check that the union of any chain of functions is itself a function. Once you do that bit of work,it should be evident that dom h =

⋃{dom f | f ∈ C} and that rangeh =

⋃{range f | f ∈ C}. So

it remains to show that rangeh is a basis for Edomh. To see that rangeh is a generating set, let zbe an arbitrary element of Edomh = E ∩Fdomh. Hence z must be generated by some finitely manyelements belong in dom h. This means there are finitely many functions f0, . . . , fn−1 ∈ C so thatz is generated by finitely many elements of dom f0 ∪ · · · ∪ dom fn−1. But dom f0, . . . ,dom fn−1,rearranged in some order, forms a chain under inclusion. So z ∈ Fdom f` for some ` < n. Hence

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z ∈ Edom f` . But range f` is a basis for Edom f` . Because range f` ⊆ rangeh we find that rangehhas enough elements to generate z. Since z was an arbitrary element of Edomh we conclude thatrangeh generates Edomh. It remains to show that rangeh is linearly independent. But rangeh isthe union of the chain {range f | f ∈ C}. I ask the hard-working graduate students to prove thatthe union of any chain of linearly independent sets must also be linearly independent. Once youhave done this you will be certain that h belongs to F. By Zorn, let g be a maximal element of F.

We would be done if dom g = B, since then E = E ∩ F = E ∩ FB = EB = Edom g. In which case,range g would be a basis for E and rank E = | range g| ≤ |dom g| = |B| = rank F.

Consider the possibility that dom g is a proper subset of B. Put C = dom g and put X = range g.Let w ∈ B with w /∈ dom g. Put D = C ∪ {w}. As we have seen above, either ED = EC orX ∪ {u} is a basis for ED, for some appropriately chosen u. We can now extend g to a functiong′ by letting g′(w) be any element of range g in the case when ED = EC and by letting g′(w) = uis the alternative case. In this way, g′ ∈ F, contradicting the maximality of g. So we reject thispossibility.

This completes the proof. �

Corollary 1. Let R be a principal ideal domain. Every submodule of a finitely generated R-modulemust itself be finitely generated.

Proof. Suppose M is an R-module generated by n elements. Let N by a submodule of M.Now let F be the free R-module with a basis of n elements. There is a function that matches this

basis with the generating set of M. So, appealing to freeness, there is a homomorphism h from Fonto M. Let E = {v | v ∈ F and h(v) ∈ N}. It is straightforward to check (will you do it?) thatE is closed under the module operations. So we get a submodule E of F. Moreover, the restrictionof h to E is a homomorphism from E onto N. But by our theorem E is generated by a set with nomore than n elements. Since the image, under a homomorphism, of any generating set for E mustbe a generating set of N (can you prove this?), we find that N is finitely generated. �

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Lecture 6. The Direct Decomposition of Finitely Generated Modules over a PID:The First Step

The objective here is to show how to obtain a direct decomposition of a finitely generated moduleover a principal ideal domain. We would like that the direct factors admit no further nontrivialdirect decomposition.

The operations of modules work in a direct product of modules in a coordinatewise manner. Soknowing how to perform the operations in each direct factor leads immediately to a knowledge ofhow the operations work in the direct product. One point of inconvenience with direct productsis that very few modules actually arise as direct products—simply because the elements of yourfavorite module are not tuples of any kind. So our direct decompositions make use of isomorphisms.

For the task at hand, our direct decompositions turn out to have only finitely many direct factors.In this situation, it is easy to replace the direct product with the notion of a direct sum. Supposethat we have the following direct decomposition of the R-module M:

M ∼= N× L.Then composing the isomorphism with the projection functions on the direct product, we findtwo homomorphisms f : M � N and g : M � L and these homomorphisms have the followingproperties:(a) For every v ∈ N and w ∈ L there is u ∈M so that f(u) = v and g(u) = w.(b) For every u ∈M , if f(u) = 0 and g(u) = 0, then u = 0.Another way to frame these two properties is in terms of the kernels of these homomorphism. LetN′ be the submodule that is the kernel of g and let L′ be the submodule that is the kernel of f .(a’) For every u ∈M there are v ∈ N ′ and w ∈ L′ so that u = v + w.(b’) The intersection of N ′ and L′ is trivial.Here is how to derive (a’) from (a) and (b). Use (a) to get w ∈ M such that f(w) = 0 andg(w) = g(u). The g(u − w) = g(u) − g(w) = 0. Now observe that u = (u − w) + w andu−w ∈ ker g = N ′ and w ∈ ker f = L′ as desired. I leave it to the hard-working graduate students toshow that these two views (one from homomorphisms and one from kernels) are logically equivalent.The Homomorphism Theorem, after just a bit of work, yields that N ∼= N′ and L ∼= L′. This leadsto the following definition. Let N′ and L′ be submodules of M that satisfy (a’) and (b’). We saythat M is a direct sum of N′ and L′ and we write M = N′ ⊕ L′. Evidently, N′ ⊕ L′ ∼= N′ × L′.

We can extend this notion to three submodules N0,N1, and N2. Here is what works.(a’) For every u ∈M there are v0 ∈ N0, v1 ∈ N1, and v2 ∈ N2 so that u = v0 + v1 + v2.(b’) The intersection N0 ∩ (N1 +N − 2), N1 ∩ (N0 +N2), and N2 ∩ (N0 +N1) are all trivial.The hard-working graduate students should verify that this works and also that the obvious exten-sion to any finite number of direct summands also succeeds.

Now let us turn to our task of decomposing modules. Here is a first step.Fact. Let R be a nontrivial integral domain. As an R-module R is directly indecomposable.Proof. We know that R can be itself construed as an R-module and as such it is a free R-module ofrank 1. To see that this module is directly indecomposable, suppose that M and N are R-modulesand that ϕ is an isomorphism from R onto M×N. Let M ′ = {r | ϕ(r) = (u, 0) for some u ∈M}.Likewise, let N ′ = {r | ϕ(r) = (0, v) for some v ∈ N}. Plainly, M ′ and N ′ are ideals in R andM ′ ∩N ′ = {0} since ϕ is one-to-one. Since R is nontrivial, we see that M and N cannot both betrivial. Suppose, without loss of generality, that M is nontrivial. So M ′ is nontrivial. Pick r ∈M ′with r 6= 0. We want to see that N must be a trivial module, or, what is the same, the N ′ is atrivial ideal. Let s be an arbitrary element of N ′. Then rs ∈ M ′ ∩ N ′ = {0}. That is, rs = 0.Since r 6= 0 and R is an integral domain, we conclude that s = 0. Since s was an arbitrary element

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of N ′, we have that N ′, and hence N, is trivial. This means that R is a directly indecomposableR-module, since R is itself nontrivial but in any direct decomposition we find that one of the directfactors must be trivial. �

This means that over any nontrivial integral domain any free module of finite rank directlydecomposes into the direct product of finitely many copies of the ring and this direct decompositionis into directly indecomposable modules.

Here is another step we can take in directly decomposing a module.

Fact. Let R be a commutative ring. Suppose M is an R-module, that F is a free R-module, andthe f is a homomorphism from M onto F with kernel N. Then there is a free R-module E so thatM ∼= N×E.

Proof. Let B be a basis for F. For each u ∈ B pick vu ∈ M so that f(vu) = u. The setC = {vu | u ∈ B} is a linearly independent subset of M . Here is how to see it:

Let w0, . . . , wn−1 be finitely many distinct elements of C and let r0, . . . , rn−1 ∈ Rwith

r0w0 + · · ·+ rn−1wn−1 = 0.Applying f to both sides we obtain

r0f(w0) + · · ·+ rn−1f(wn−1) = 0.But f(w0), . . . , f(wn−1) are distinct elements of B, which is linearly independent.So r0 = · · · = rn−1 = 0, as desired.

Now let E be the submodule of M generated by C. So E is free since C is a basis. I contend thatM ∼= N×E. Here is how to define the isomorphism ϕ:

Let w be an arbitrary element of M . Let f(w) = r0u0 + · · · + rn−1un−1 whereu0, . . . , un−1 are distinct elements of B and all the ri’s are nonzero. Let v0, . . . , vn−1be elements of C so that f(vi) = ui for all i < n. Put x = r0v0 + · · · + rn−1vn−1.Then x ∈ E and f(x) = f(w). This means w − x ∈ N since N is the kernel of f .So define ϕ(w) = (w − x, x).

It is a straightforward piece of work (done by all hard working graduate students) to see that ϕ isan isomorphism. �

To invoke this last fact for a particular module M we essentially have to find a free submodule ofM. Such a submodule would have a basis C. For w ∈ C we would have to have the implication

rw = 0 =⇒ r = 0, for all r ∈ R,since this is just part of the definition of linear independence. Indeed, when R is an integral domain,all the elements of E, not just those in C would have to have this property. This suggests that Nshould consist of those elements that fail this property. That is elements x ∈ M such that rx = 0for some r 6= 0. Such elements are called torsion elements. The 0 of a module is always a torsionelement, provided the ring in nontrivial. The module M is said to be torsion free provided 0 isits only torsion element. The step along the way is the following fact.

Fact. Let R be a nontrivial integral domain and let M be an R-module. Then the set T of torsionelements is a submodule of M and M/T is torsion free.

Proof. We have already noted that 0 ∈ T . To see that T is closed under addition, let u, v ∈ T .Pick nonzero elements r, s ∈ R so that ru = 0 = sv. Then rs 6= 0 since R is an integral domain.Now observe (rs)(u + v) = (rs)u + (rs)v = (sr)u + (rs)v = s(ru) + r(sv) = 0 + 0 = 0 is R iscommutative. So u+ v ∈ T . Finally, suppose that t ∈ R. Then r(tu) = (rt)u = (tr)U = t(ru) = 0,

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so tu ∈ T . In this way, we see that T is closed under the module operations and we can form thesubmodule T.

To see that M/T is torsion free, pick a nonzero element u/T of M/T and a scalar r ∈ R so thatr(u/T ) = 0/T . Since u/T is nonzero we know that u /∈ T . On the other hand r(u/T ) = (ru)/T =0/T means that ru ∈ T . So pick a nonzero s ∈ R so that s(ru) = 0. This means that (sr)u = 0.But, since u /∈ T we know that u is not a torsion element. So sr = 0. Since s 6= 0 and R is anintegral domain, we see that r = 0. This means that u/T is not a torsion element. So M/T is atorsion free module. �

So when is a torsion free module actually free?Fact. Let R be a nontrivial integral domain. Every finitely generated torsion free R-module isfree of finite rank.Proof. Let M be a torsion free R-module generated by the finite set X. Let Y be a maximallinearly independent subset of X. Let F be the submodule of M generated by Y . Of course, F isfree of finite rank. For each x ∈ X pick sx ∈ R so that sxx ∈ F . This is possible since if x ∈ Ywe can let sx = 1, while if x /∈ Y , then Y ∪ {x} is linearly dependent. This means that for somedistinct y0, . . . , yn−1 ∈ Y there are sx, r0, . . . , rn−1 ∈ R \ {0} so that

sxx+ r0y0 + · · ·+ rn−1yn−1 = 0.In this way, sxx ∈ F . Now let s be the product of all the sx’s as x runs through X. Then sx ∈ Ffor all x ∈ X. Since X generates M, we see that sv ∈ F for all v ∈ M . Now let ϕ get the mapfrom M into F defined via

ϕ(v) := sv for all v ∈M.

It is routine to check that ϕ is a homomorphism. The kernel of ϕ must be trivial since M is torsionfree. So ϕ is one-to-one. This means that M is isomorphic with a submodule of the free moduleF. Since R is a principal ideal domain, by the Freedom Theorem we conclude that M is free.Moreover, since F has finite rank, so must M. �

The First Decomposition Theorem for Modules over an Integral Domain.Let R be a nontrivial integral domain, let M be a finitely generated R-module, and let T be thetorsion submodule of M. There is a free module F of finite rank such that

M ∼= T× F.Moreover, the rank of F is determined by M.

Proof. According to the Facts established above, we can take F to be M/T . So only the “moreover”part of the theorem remains to be established. To this end, suppose that F′ is some free moduleso that

M ∼= T× F′.The conclusion we want is F ∼= F′.

What are the torsion elements of T× F′? Suppose (u, v) is torsion. Pick r 6= 0 so that r(u, v) =(0, 0). So rv = 0. But v ∈ F ′, which being free is also torsion free. So v = 0. This means that thetorsion elements of T× F′ are exactly the elements of T ′ := {(u, 0) | u ∈ T}. In this way we see

F = M/T ∼= (T× F′)/T ′ ∼= F′.The rightmost isomorphism above comes from the Homomorphism Theorem since T ′ is the kernelof the project of the direct product onto its rightmost direct factor. �

Both the torsion module T and the free module F may admit further direct decomposition. Asregards the free module, we know it can be decomposed as the direct product of n copies of theR-module R, which we have seen is directly indecomposable.

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Lecture 7. Direct Decomposition of Finitely Generated Torsion Modules over aPID: The Second Step

Let M be any R-module and X be any subset of M . DefineannX := {r | r ∈ R and rx = 0 for all x ∈ X}.

This set is called the annihilator of X. It is routine to check that annX is always an ideal of R,provided R is commutative. Running this game in the other direction, let S ⊆ R and define

M [S] := {u | u ∈M and ru = 0 for all r ∈ S}.Again, it is routine to check that M [S] is closed under the module operations, provided that R iscommutative. So we obtain a submodule M[S] of the module M. For a single element r ∈ R wewrite M[r] for M[{r}].

Let T be a torsion module of a principal ideal domain R. Suppose that the finite set X generatesT. As we did in one of the proofs in the preceding lecture, for each x ∈ X we can pick a nonzerosx ∈ R so that sxx = 0. Let s be the product of these finitely many sx’s as x runs through X.Since R is an integral domain we see that s 6= 0 and since R is commutative and X generates Twe see that su = 0 for all u ∈ T . This means that annT is a nontrivial ideal. Because R is aprincipal ideal we can pick r ∈ R so that (r) = annT . This nonzero element r, which is unique upto associates, is called the exponent of T.

More generally, if u is a torsion element of an R-module, where R is a principal ideal domain thenthere will be a nonzero element r so that (r) = ann{u}. We call r the order of u and sometimesrefer to ann{u} as the order ideal of u. If v is also a torsion element and s is the order of v, wherer and s are relatively prime, then rs will be the order of u + v. (This could be proven by a hardworking graduate student.)

We are ready to begin decomposing our torsion module.

Fact. Let T be a torsion R-module with exponent r, where R is a principal ideal domain. Supposethat r = sq where s and q are relatively prime. Then T ∼= T[s]×T[q].

Proof. Using the relative primeness of s and q select elements a, b ∈ R so that 1 = as+ bq. So forany u ∈ T we have

u = 1 · u = (as+ bq)u = q(bu) + s(au).Observe that q(bu) ∈ T [s] and s(au) ∈ T [q]. So every element of T can be expressed as a sumof an element of T [s] and an element of T [q]. The expression is unique since if u = v + w wherev ∈ T [s] and w ∈ T [q], then v − qbu = sau− w ∈ T [s] ∩ T [q]. But the order of any element of thisintersection must divide both s and q, which are relatively prime. So the intersection is just {0}and it follows that v = qbu and w = sav. The map that sends u to (v, w) where v ∈ T [s], w ∈ T [q],and u = v + w is easily seen to be an isomorphism. �

Suppose that r is the exponent of our torsion R-module T, where R is a principal ideal domain.Let p0, . . . , pn−1 ∈ R be distinct primes and let e0, . . . , en−1 be positive integers so that

r = pe00 . . . p

en−1n−1 .

Then applying the Fact above over and over again we findT ∼= T[pe0

0 ]× · · · ×T[pen−1n−1 ].

Modules of the form T[pe] where p ∈ R is prime and e is a positive integer are said to be primaryor sometimes more specifically p-primary. These are just the torsion modules of exponent a powerof p. So now we know that every finitely generated torsion module over a principal ideal domaincan be decomposed as a product of primary modules. We state this as a theorem.

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The Primary Decomposition Theorem.Every finitely generated module over a principal ideal domain is isomorphic to a direct products offinitely many primary modules.

Still, these primary modules may admit further direct decomposition.Which p-primary modules are directly indecomposable?

Fact. Let R be a principal ideal domain and p ∈ R be prime. Every cyclic p-primary R-module isdirectly indecomposable.

Proof. Let T be an R-module of exponent pe that is generated by w. Suppose that T ∼= M ×N.We need to argue that one of M and N is trivial. We know that pew = 0 but pe−1w 6= 0. Picku ∈ M and v ∈ N so that (u, v) generates M×N. Then we have peu = 0 and pev = 0 and eitherpe−1u 6= 0 or pe−1v 6= 0. Without loss of generality, suppose pe−1u 6= 0. This makes M nontrivial,so our ambition is to show that N is trivial. Now since the element (u, v) generates all elements ofM×N, we see that every element of M×N can be obtained by multiplying (u, v) by an appropriatescalar. So pick r ∈ R so that r(u, v) = (0, v). It follows that ru = 0 and rv = v. Since the order ofu is pe, we have that pe | r. So r = spe for some s. But then v = rv = spev = 0. But this entailsthat N is trivial. �

Of course, we can get a cyclic submodule easily enough just by selecting a single element andusing it to generate a submodule. Something more clever is possible.

Fact. Let R be a principal ideal domain and let T be a torsion R-module of exponent r. Then Thas an element of order r.

Proof. Pick distinct primes p0, . . . , pn−1 ∈ R and positive integers e0, . . . , en−1 so that

r = pe00 . . . p

en−1n−1 .

For each j < n, let rj = rpj

= pe0 . . . pej−1j−1 p

ej−1j p

ej+1j+1 . . . p

en−1n−1 . Notice that r - rj for all j < n. This

allows us, for each j < n, to pick uj ∈ T so that rjuj 6= 0. Now, for each j < n, put

vj = r

pejj

uj .

Then, for each j < n, we have pejj vj = 0 but pej−1j vj 6= 0. So vj is an element of order pejj . It now

follows that v0 + · · ·+ vn−1 is an element of order r, as desired. �

There is one additional fact that proves useful.

Fact. Let T be a torsion R-module of exponent r, where R is a principal ideal domain. Let M andN be submodules of T, where M is generated by an element of order r. Let f be a homomorphismfrom N into M. Finally, let v ∈ T with v /∈ N . Then f can be extended to a homomorphism fromthe submodule generated by N ∪ {v} into M.

Proof. Let u be the element of order r that generates M. Let N′ be the submodule generated byN ∪ {v}. Evidently, rv = 0 ∈ N , so v/N has some nonzero order s in N′/N and s | r. So sv ∈ N .Pick p ∈ R so that

r = ps.

Now f(sv) ∈M and pf(sv) = f(psv) = f(rv) = f(0) = 0. So the order of f(sv) divides p.Since f(sv) ∈ M and M is generated by u pick q ∈ R so that f(sv) = qu. Then we see

0 = pf(sv) = pqu. This meansr | pq.

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Hence, ps | pq. Since R is an integral domain, we haves | q.

So pick t ∈ R such thatq = st.

This entails f(sv) = qu = stu. Let w = tu ∈M . So f(sv) = sw.Now every element of N ′ can be written in the form y+ av for some choices of y ∈ N and a ∈ R.

There may be several ways to make these choices.Here is how we define our extension f ′ of f :

f ′(y + av) = f(y) + aw.

It is not clear that this definition is sound. Let us verify that. Suppose that y+av = y′+a′v wherey, y′ ∈ N and a, a′ ∈ R. We need to see that

f(y) + aw = f(y′) + a′w or written another way f(y)− f(y′) = (a′ − a)w.But notice that y − y′ = (a′ − a)v. So (a′ − a)v ∈ N . This means s | a′ − a. Pick m ∈ R so thata′ − a = ms. But this leads to

f(y)− f(y′) = f(y − y′) = f((a′ − a)v) = f(msv) = mf(sv) = msw = (a′ − a)w,just as we desire. So our definition is sound. Since f ′(y) = f ′(y + 0v) = f(y) + 0w = f(y), for ally ∈ N , we see that f ′ extends f . We must also check that f ′ is a homomorphism, a task we leaveto hard working graduate students. �

So our scheme is to grab an element whose order is the exponent of T, let M be the submodulegenerated by that element, and hope to find another submodule N so that T ∼= M×N. If we arelucky maybe the exponent of N will be smaller. Here is what we need.

Fact. Let R be a principal ideal domain and let T be a torsion R-module of exponent r. ThenT has a cyclic submodule M of exponent r and a submodule N of exponent s so that s | r andT ∼= M×N.

Proof. Let u ∈ T have order r and let M be the submodule of T generated by u. Let f be theidentity map on M. LetF = {g | g : N→M is a homomorphism extending f for some submodule N with M ⊆ N ⊆ T}.

We will apply Zorn’s Lemma to F, which is partially ordered by set-incluion. Notice that f ∈ F, soF is not empty. Let C be a nonementy chain in F. Certainly

⋃C is an upper bound on C. We must

show it is in F. We have noted before that the union of a chain of functions is again a function.The hard-working graduate students will see that the union of a chain of homomorphism is itselfa homomorphism. It is routine to see that the union of a chain of submodules (the domains ofthose homomorphisms) is again a submodule. So we see indeed that

⋃C belongs to F. Let g be a

maximal element of F. Since the fact just above would otherwise allow the extension of g to largermember of F, a thing impossible by the maximality of g, we see that g is a homomorphism from Tinto M which extends the identity map on M. Let N be the kernel of g. Let s be the exponent ofN. Evidently, s | r.

For any w ∈ T we have g(w) ∈ M . But then g(g(w)) = f(g(w)) since g extends f . But f is theidentity map. So we see that g(g(w)) = g(w) for all w ∈ T . But notice

g(w − g(w)) = g(w)− g(g(w)) = g(w)− g(w) = 0.This means that w − g(w) ∈ N for all w ∈ T , since N is the kernel of g. So we find that

w = g(w) + (w − g(w)) ∈M +N, for all w ∈ T.

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Another way to write this isT = M +N.

Now suppose w ∈ M ∩N . Then, on the one hand, g(w) = f(w) = w since w ∈ M and g extendsf (which is the identity function), while on the other hand g(w) = 0 since w ∈ N = ker g. Takentogether we find that w = 0. This means

M ∩N = {0}.So we see that T = M⊕N. This yields our desired conclusion. �

The Invariant Factor Theorem.Let T be a nontrivial finitely generated torsion R-module, where R is a principal ideal domain.Then for some natural number n there are r0, r1, . . . , rn ∈ R with

rn | rn−1 | · · · | r1 | r0

and cyclic submodules M0 of exponent r0,. . . , Mn of exponent rn so thatT ∼= M0 × · · · ×Mn.

Proof. Let the order of T be r0. (Recall that we have already proven that a nontrivial finitelygenerated torsion module has an exponent.) Let u0 ∈ T have order r0 and let M0 be the submodulegenerated by u0. By the preceding Fact, there is a submodule N0 of order r1 with r1 | r0 such thatT ∼= M0×N0. If N0 is trivial, we can stop since T ∼= M0 in that case. Otherwise, pick u1 ∈ N0 withorder r1. Take M1 to be the submodule of generated by u1 and invoke the immediately precedingfact to get a proper submodule N1 of N0 of exponent r2 so that r2 | r1 and N0 ∼= M1 ×N1. Atthis stage we have

T ∼= M0 ×M1 ×N1 and r2 | r1 | r0 and N1 $ N0,

where the exponent of M0 is r0, the exponent of M1 is r1, and the exponent of N1 is r2. Againour process terminates in the event N1 is trivial, but otherwise the process can be continued. Inthis process to chains of submodules of T are constructed. One is the descending chain consistingof the submodules Nj . The other is the ascending chain

M0 $ M0 ⊕M1 $ . . . .

Now we know, as a corollary of the Freedom Theorem that every submodule of T is finitely gen-erated. We saw for rings that if every ideal was finitely generated then there could be no infiniteascending chains of ideals. The same reasoning applies here (as the hard working graduate studentwill establish) to see that there can be no infinite ascending chair of submodules of T. This mustmean the process described above terminates at some finite stage. This completes our proof. �

The r0, r1, . . . , rn mentioned in the theorem are called invariant factors.Now we are ready for the chief existence theorem for direct decomposition of finitely generated

modules.The Elementary Divisor Theorem.Let T be a nontrivial finitely generated torsion R-module, where R is a principal ideal domain.Then for some natural number n, there are cyclic primary submodules M0 . . .Mn of T so that

T ∼= M0 × · · · ×Mn.

The exponents of the various cyclic primary submodules are referred to as the elementary di-visors of T. The proof of the Elementary Divisor Theorem is obtained by applying the InvariantFactor Theorem to each of the direct factors arising from an application of the Primary Decom-postion Theorem.

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Lecture 8. The Structure Theorem for Finitely Generated Modules over a PID

Here is one of the key results in our course.

The Structure Theorem for Finitely Generated Modules over a Principal Ideal Domain.Let M be a finitely generated R-module, where R is a principal ideal domain. There is a naturalnumber n such that:(a) there are n finitely generated directly indecomposable submodules M0, . . . ,Mn−1 of M such that

M ∼= M0 × · · · ×Mn−1, and

(b) for any natural number m, if N0, . . . ,Nm−1 are directly indecomposable R-modules such thatM ∼= N0 × · · · × Nm−1, then n = m and there is permutation σ of {0, . . . , n − 1} so thatMi∼= Nσ(i) for all i < n.

Moreover, the finitely generated directly indecomposable R-modules are, up to isomorphism, theR-module R (that is the free R-module of rank 1), and the R-modules of the form R/(r) where ris a positive power of some prime element of the ring R (these are the cyclic primary R-modules).Finally, the free R-module of rank 1 is not primary and if r, s ∈ R are prime powers and R/(r) ∼=R/(s), then r and s are associates in the ring R.

Before turning to the proof a few remarks are in order.First, we have allowed n = 0. This results in the direct product of an empty system of R-modules.

A careful, but easy, examination of the definition of direct products reveals that such a directproduct produces the trivial R-module—that is the module whose only element is 0. Evidently,the trivial R-module is the direct product of exactly one system of directly indecomposable R-modules, namely of the empty system.

This theorem has three parts:• the assertion of the existence of a decomposition into indecomposables,• the assertion that such a decomposition is unique, and• a description of the indecomposables.

These are the hallmarks of a good structure theorem. There are other theorems of this kind inmathematics. Perhaps the most familiar is the Fundamental Theorem of Arithmetic. To makethe connection plain, consider the least complicated algebraic systems, namely just nonempty setsequipped with no operations. Then the finitely generated algebraic systems are just the finitenonempty sets and isomorphisms are just one-to-one correspondences. So two of these algebraicsystems will be isomorphic if and only if they have the same number of elements. The FundamentalTheorem of Arithmetic says that every finite set is isomorphic to a direct product of directly inde-composable finite sets in a way that is unique up to isomorphism and rearranging the factorizationand that a set is directly indecomposable if and only if it has a prime number of elements.

This theorem also resonates with the notion of a unique factorization domain. We could refor-mulate our structure theorem to make this more apparent. Each finitely generated R-module isisomorphic to a lot of other R-modules (in fact, to a proper class of R-modules). Pick a represen-tative from each of these isomorphism classes, taking care to include among these representatvesthe R-modules R and R/(r) where r ∈ R is a positive power of some prime element of R. LetM be the set of all these representative and let 1 be the representative trivial R-module. Then〈M,×,1〉 is an algebraic system (actually a monoid) with the unique factorization property.

Finally, the structure theorem above has a number of far-reaching consequences. Taking R to beZ we obtain a structure theorem for finitely generated Abelian group. Taking R to be F[x], whereF is field leads to the canonical form theorems of linear algebra.

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Proof. Let us first dispose of the descriptions of the directly indecomposable R-modules that couldarise in any factorization of M. These must be finitely generated because they will be isomorphicto submodules of M and, according to the Corollary of the Freedom Theorem every submodule ofM must be finitely generated. We have already seen that the free R-module of rank 1 (namelythe module R) is directly indecomposable and every other free R-module of finite n > 1 is adirect product of n-copies of R. We have also seen that the cyclic primary R-modules are theonly finitely generated directly indecomposable torsion modules. Can there be any others finitelygenerated directly indecomposable R-module? By the First Decomposition Theorem every finitelygenerated R-module is isomorphic to a direct product of the form T × F, where T is the torsionsubmodule and F is a submodule that is free. For a directly indecomposable module we must haveeither F trivial (and the our module would be torsion) or else T trivial (and the our module wouldbe free). So the only finitely generated directly indecomposable R-modules are the ones already inhand, the R-module R and the cyclic primary R-modules.

We can say more about the cyclic primary R-modules. Let r ∈ R be a positive power of a primeelement of R. Then the ideal (r) is a submodule of the R-module R. The element 1/(r) of thequotient module R/(r) generates the quotient module and has order r. So the quotient module iscyclic and of exponent r. In this way we know cyclic R-modules of exponent r exist. Suppose thatN is an R-module of exponent r which is generated by the single element u. Since {1} is a basisfor R, we know there is a homomorphism h from R onto N that takes 1 to u. Now for all s ∈ Rwe have h(s) = h(s · 1) = sh(1) = su. From this we see that

s ∈ kerh⇔ h(s) = 0⇔ su = 0⇔ r | s⇔ s ∈ (r).

That is, kerh = (r). So by the Homomorphism Theorem N ∼= R/(r). So, up to isomorphism, theonly cyclic R-module of exponent r (where r is a positive power of some prime) is R/(r).

Now observe that the free R-module R of rank 1 is not a torsion module since s·1 = 0 =⇒ s = 0.So the R-module R cannot be isomorphic with any of the modules R/(r) where r is a positivepower of some prime. (One needs to observe here, as the hard-working graduate students will verify,that r cannot be a unit.) Now suppose that r and s are both positive powers of primes (we don’tassume the primes are tne same) and that R/(r) ∼= R/(s). Then r is the order of a generator ofthis cyclic module and so is s. This means that r | s and s | r. Consequently, (r) = (s) and r ands are associates.

Now consider part (a) of the theorem. This is an immediate consequence of the First Decompo-sition Theorem and the Elementary Divisor Theorem.

Finally, consider part (b). Some of the Ni’s can be cyclic primary modules and others can be freeof rank 1, according to our description of the directly indecomposable finitely generated modules.Without loss of generality, we assume that the primary modules come first. So pick k ≤ m so thatN0, . . . ,Nk−1 are cyclic primary modules and Nk, . . . ,Nm−1 are free of rank 1. Let T be the directproduct of the first group and F be the direct product of the second. So we find M ∼= T×F. It isroutine (according to hard-working graduate students) that T is a torsion module and also that Fis free of m−k. Let (v, u) ∈ T ×F be a torsion element of order r. Then (0, 0) = r(v, u) = (rv, ru).In particular, ru = 0. The element u can be written as a linear combination of the basis elementsof F. By distributing r through the linear combination and invoking both linear independence andthe fact the r is a nonzero element of an integral domain, we see that u = 0. What we concludeis that the torsion elements of T × F are exactly those of the form (v, 0) where v ∈ T is nonzero.Thus under the isomorphism M ∼= T× F, the module T corresponds to the torsion submodule ofM. Then according to the First Decomposition Theorem the rank of F is determined by M.

It remains to show that if T ∼= M0× · · · ×M`−1 ∼= N0× · · · ×Nk−1, where all the Mi’s and Nj ’sare cyclic primary R-modules, then ` = k and, after a suitable reindexing Mi

∼= Ni for all i < `.

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Let p ∈ R be prime. For any R-module Q letQ(p) = {v | v ∈ Q and pev = 0 for some positive integer e}.

It is routine to check that this set is closed under the module operations, show we have the sub-module Q(p). It is also not hard to see (as hard-working graduate students will check) that

T(p) ∼= M0(p)× · · · ×M`−1(p) ∼= N0(p)× · · · ×Nk−1(p).In this decomposition, if Mi (or Nj) were primary with respect to a prime not associate to p, thenthe module Mi(p) (respectively Nj(p)) would be trivial. On the other hand, if they were primarywith respect to an associate of p, then Mi(p) = Mi and Nj(p) = Nj . Since this holds for arbitraryprimes p, we do not lose any generality by assuming the the primes underlying all the Mi’s andNj ’s are the same prime p.

Now suppose Q is a cyclic primary R-module, where pe is the exponent and u is a generator.Then Q[p] is generated by pe−1u. So Q[p] is cyclic of exponent p. In this case, we know thatQ[p] ∼= R/(p). Now (p) is a prime ideal of the ring R. In a principal ideal domain, the maximalideals and the prime ideals coincide. So the ring R/(p) is a field. This allows us to construethe R-module Q[p] as a one-dimensional vector space over the field R/(p). In doing this, we arechanging the scalar multiplication, but leaving the addition and the zero the same. Now we have

T[p] ∼= M0[p]× · · · ×M`−1[p] ∼= N0[p]× · · · ×Nk−1[p]construed as vector spaces over the field R/(p), with each of the direct factors being a copy of theone-dimensional vector space. This means

` = dim T[p] = k.

So we have discovered that ` = k, one of our desired conclusions.So we are reduced to considering the following situation:

T ∼= M0 × · · · ×M`−1∼= N0 × · · · ×N`−1

where Mi is a cyclic module of exponent pei and Ni is a cyclic module of exponent pfi for all i < `and

e0 ≥ e1 ≥ · · · ≥ e`−1

f0 ≥ f1 ≥ · · · ≥ f`−1.

It remains only to show that ei = fi for all i < `. Suppose, for the sake of contradiction, thatthis were not so. Let i be as small as possible so that ei 6= fi. It is harmless to also suppose thatei > fi. Let r = pfi . Now multiplication by r is homomorphism and it is easy to also see that

rT ∼= rM0 × · · · × rMi−1 × rMi × · · · × rM`−1∼= rN0 × · · · × rNi−1 × rNi × · · · × rN`−1.

Being homomorphic images of cyclic modules, each of the direct factors above is also cyclic. Becauser is a positive power of the prime p, we see that the factor modules above are either primary (withprime p) or trivial. But exponents of all the Nj ’s where i ≤ j are factors of r, we see that thesemodules are all trivial. On the other hand, the exponent of Mi is pei whereas r = pfi with ei > fi.So rMi is not trivial. This would mean

rT ∼= rM0 × · · · × rMi−1 × rMi × · · · × rM`−1∼= rN0 × · · · × rNi−1,

where the top direct factorization has at least i+1 nontrivial cyclic primary factors but the bottomhas only i. But we have just proven that the number of such factors must be the same no matter

32

how the direct factorization is accomplished. This contradiction means our supposition must berejected. So ei = fi for all i < `. This establishes the uniqueness of our direct factorization intodirectly indecomposable modules. The proof of the last remaining part of our theorem, namelypart (b), is complete. �

The Structure Theorem above is an extension of the Elementary Divisor Theorem formulated inthe previous lecture. We can also extend the Invariant Factor Theorem.

The Extended Invariant Factor Theorem.Let T be a nontrivial finitely generated torsion R-module, where R is a principal ideal domain.Then for some natural number n there are r0, r1, . . . , rn ∈ R with

rn | rn−1 | · · · | r1 | r0

and cyclic submodules M0 of exponent r0,. . . , Mn of exponent rn so that

T ∼= M0 × · · · ×Mn.

Moreover, the natural number n is uniquely determined by T, the sequence rn | rn−1 | · · · | r1 | r0is uniquely determined up to associates, and cyclic submodules M0, . . . ,Mn are determined but toisomorphism.

Only the various aspects of uniqueness require proof at this point. However, these proofs followthe lines of the uniqueness portion of the proof and the Structure Theorem. We leave the detailsin the hands of the hard working graduate students. It is useful to note that the cyclic moduleswhich are the factors in this direct decomposition may not themselves be directly indecomposable.

Using the Structure Theorem, for each principal ideal domain R we can define a function d suchthat d(pe,M) to be the number of direct factors isomorphic to the module R/(pe) in any directfactorization of M into directly indecomposable modules, where p ∈ R is prime, e is a positivenatural number, and M is a finitely generated R-module. In addition, we take d(0,M) to be thenumber of direct factors isomorphic to the R-module R (that is, the directly indecomposable freemodule).

Then we have the useful

Corollary. Let R be and principal ideal domain and M and N be finitely generated R-modules.Then M ∼= N if and only if d(q,M) = d(q,N) for all q such that either q = 0 or q is a positivepower of a prime in R.

What this corollary asserts is that the system natural numbers

〈d(q,M) | q = 0 or q is the positive power of a prime of R〉

is a complete system of invariants of M—that is this system of natural numbers determines M upto isomorphism.

As noted earlier, modules over the ring Z of integers are essentially the same as Abelian groupssince, for instance, 3u = (1+1+1)u = u+u+u and −7v = −(v+v+v+v+v+v+v). In this way, wesee that for a Z-module M = 〈M,+,−, 0, a·〉a∈Z the scalar multiplication is expressible by means ofthe additive structure +,−, and 0. In particular, any map between Z-modules that respects +,−,and 0 must also respect the scalar multiplication, any subset of a Z-module that is closed under+, 1, and contains 0 will also be closed under all the scalar multiplications, and a similar remarkholds for direct products—in any of these constructions one may ignore the scalar multiplicationsalong the way, but impose then on the result (the homomorphic image, the subalgebra, or the directproduct) by means of repeated addition.

With this in mind, noting that Z is a principal ideal domain, we obtain

33

The Fundamental Theorem for Finitely Generated Abelian Groups.Let A be a finitely generated Abelian groups.0 There is a natural number n such that:(a) there are n finitely generated directly indecomposable subgroups A0, . . . ,An−1 of A such that

A ∼= A0 × · · · ×An−1, and(b) for any natural number m, if B0, . . . ,Bm−1 are directly indecomposable Abelian groups such

that A ∼= B0 × · · · × Bm−1, then n = m and there is permutation σ of {0, . . . , n − 1} so thatAi∼= Bσ(i) for all i < n.

Moreover, the finitely generated directly indecomposable Abelian groups are, up to isomorphism, thegroup 〈Z,+,−, 0〉 of integers with respect to addition (that is the free Abelian group of rank 1),and the cyclic groups of prime power order (these are the groups Zq where q is a positive power ofa prime number, the set of elements is {0, 1, . . . , q − 1}, and addition works modulo q). Finally,the free Abelian group of rank 1 is not of prime power order and if r, s ∈ R are prime powers andZr ∼= Zs, then r = s.

This could be regarded as the elementary divisor version of the structure theorem for finitelygenerated Abelian groups. One could as easily formulate a structure theorem from the invariantfactor point of view. To see how these two points of view compare consider a description, up toisomorphism, of all the Abelian groups of order 100. The prime factorization gives 100 = 2252.Using the elementary divisor perspective we see that the list, representative up to isomorphism andalso pairwise nonisomorphic, of Abelian groups of order 100 is

Z4 × Z25 Z2 × Z2 × Z25 Z4 × Z5 × Z5 Z2 × Z2 × Z5 × Z5

while from the invariant factor perspective the list isZ100 Z2 × Z50 Z5 × Z20 Z10 × Z10.

At work here are the following direct decompositions:Z100 ∼= Z4 × Z25 Z50 ∼= Z2 × Z25 Z20 ∼= Z4 × Z5 Z10 ∼= Z2 × Z5.

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Lecture 9. Decomposition of Vector Space with a Designated Linear Operator

Let V be a finite dimensional vector space over a field F and let T be a linear operator on V—that is, T is an endomorphism of the vector space V. Our objective here is to decompose V as adirect product of subspaces that are invariant with respect to T . The most straightforward wayto proceed with such a project is to adjoin T to the vector space as a new one place operation.This new algebraic system would have a binary operation + (the old vector addition), a designatedelement 0 (the zero vector), a one-place operation − of forming negations, a one-place operation aIfor each a ∈ F (the scalar muliplications), and the new one-place operation T . The program wouldthen become the direct decomposition of this new algebraic system into directly indecomposablefactors. It is possible to carry out this program, to prove the corresponding structure theorem(which would prove the existence and uniqueness of such decompositions and describe the directlyindecomposable algebras, much as in the last section).

However, there is an alternate route to the same result that allows us to take advantage of thework we have done with modules. The idea is to regard V as a module over the principal idealdomain F[x] instead of over the field F. This means we have to define what f(x) ·v means for everyvector v ∈ V and every polynomial f(x) ∈ F[x]. Here is the definition:

f(x) · v := f(T )(v).Here f(T ) = a0I+a1T +a2T

2 + · · ·+anTn where f(x) = a0 +a1x+ · · ·+anxn and T 2 = T ◦T, T 3 =T ◦T ◦T , and so on. It is easy to see that each f(T ) is a linear operator (that is, an endomorphismof the vector space V). The polynomials of degree 0 provide the ordinary scalar multiplications ofthe vector space. So construing V as a module over F[x] in effect adjoins many more one-placeoperations than our first approach, but they are all built up from the a · I and T by addition andcomposition. This is why the two approaches are equivalent.

Recall from linear algebra that the linear operators on a finite-dimensional vector space Vconstitute a finite dimensional vector space themselves. So for any linear operator T the set{I, T, T 2, T 3, . . . } is linearly dependent. This means that for some natural number m there area0, a1, . . . , am ∈ F with am 6= 0 so that a0I + a1T + · · · + amT

m = 0. In other words, there is anonzero polynomial f(x) ∈ F[x] so that f(T ) is the zero linear operator (the map taking all vectorsto the zero vector). Evidently, {f(x) | f(T ) is the zero operator} is an ideal of F[x]. Since F[x] isa principal ideal domain first ideal is generated by a single polynomial. In fact we can take thispolynomial to be the monic polynomial mT (x) of least degree in this ideal. This polynomial iscalled the minimal polynomial of T .

Now fix a linear operator T on the finite dimensional vector space V. We use VT to denote themodule over F[x] described above. We know that this module can be decomposed into a directproduct of cyclic submodules. What do these cyclic submodules look like? Well, suppose thatv ∈ V is a generator. Then the submodule consists of all the vectors of the form f(T )(v) as f(x)runs through the ring of polynomials. Hence the linear span of the set {v, Tv, T 2v, . . . } is thewhole submodule. Since this submodule is, among other things, a subspace of V (with additionaloperations), we know that some finite subset must span the submodule. Let m be as small aspossible so that {v, Tv, . . . , Tmv} spans the submodule. Then there are a0, . . . , am ∈ F so that

Tm+1v = a0v + · · ·+ amTmv.

This leads toTm+2v = a0Tv + · · ·+ amT

m+1v = a0Tv + · · ·+ (a0v + · · ·+ amTmv).

In this way we see that m is also the smallest natural number so that Tm+1v is a linear combinationof {v, Tv, . . . , Tmv}. I contend that this set is linearly independent. Suppose

b0v + b1Tv + · · ·+ bmTmv = 0.

35

Now bm must be 0, otherwise Tmv = − b0bmv − · · · − bm−1

bmTm−1v. Once the term bmT

mv has beeneliminated (because it is 0), we can apply the same reasoning the see that bm−1 = 0, and then thatbm−2 = 0, and so on. In this way we establish the linear independence of {v, Tv, . . . , Tmv}. Thuswe see that our cyclic submodule, construed as an ordinary vector space over the field F has a verynice basis. We call this kind of basis a T -cyclic basis.

Here is what happens if we represent T with respect to this basis. Putv0 = v, v1 = Tv, . . . , vm = Tmv.

ThenTv0 = v1 = 0v0 + 1v1 + 0v2 + · · ·+ 0vmTv1 = v2 = 0v0 + 0v1 + 1v2 + · · ·+ 0vm

...Tvm−1 = vm = 0v0 + 0v1 + 0v2 + · · ·+ 1vmTvm = a0v0 + a1v1 + · · ·+ amvm

This produces the matrix

0 0 0 . . . 0 a01 0 0 . . . 0 a10 1 0 . . . 0 a2...

...... . . . ...

...0 0 0 . . . 0 am−10 0 0 . . . 1 am

This is a very pleasant matrix with lots of entries 0 and the nonzero entries located in ratherrestrained positions. Let us rewrite that last equation:

Tvm = a0v0 + a1v1 + · · ·+ amvm

Tm+1v = a0v + a1Tv + · · ·+ amtmv

0 = −Tm+1v + amTmv + · · ·+ a1Tv + a0v

0 = Tm+1v − amTmv − · · · − a1Tv − a0v

0 = (Tm+1 − amTm − · · · − a1T − a0)v0 = mT (T )v

where mT (x) = xm+1 − amxm − · · · − a0 ∈ F[x]. Notice that this is a monic polynomial of leastdegree which belongs to the annihilator of VT . So it is an exponent of the module VT .

We could start with any monic polynomial f(x) = b0 +b1x+ · · ·+bmxm+xm+1 of positive degree.

If this polynomial happened to be the minimal polynomial of some linear operator T so that VT

was a cyclic module, then the associated matrix, as above, would be

Cf =

0 0 0 . . . 0 −b01 0 0 . . . 0 −b10 1 0 . . . 0 −b2...

...... . . . ...

...0 0 0 . . . 0 −bm−10 0 0 . . . 1 −bm

.

This matrix is called the companion matrix of the polynomial f(x). Observe that this is am + 1 ×m + 1 matrix, where m + 1 is the degree of f(x). It easy to write down the companion

36

matrix given the monic polynomial and, vice versa, given the companion matrix to write down themonic polynomial. A routine calculation also reveals that for any monic polynomial f(x) we have

f(x) = det(xI − Cf ).This means that f(x) is the characteristic polynomial of its companion matrix.

We summarize these findings in the following Fact.

Fact. Let V be a finite dimensional vector space over a field F and T be a linear operator on Vsuch that VT is a cyclic module over F[x] with generator v and with minimal polynomial mT (x)of degree n + 1. Then {v, Tv, T 2v, . . . , Tnv} is a basis for VT and, with respect to this basis, thematrix of T is the companion matrix of mT (x) and the characteristic polynomial of T is the sameas the minimal polynomial of T .

Now let’s apply the Invariant Factor Theorem:The Rational Canonical Form Theorem.Let V be a finite dimensional vector space over the field F. Let T be a linear operator of V .Then for some natural number n there are monic polynomials f0(x), f1(x), . . . , fn(x) ∈ F[x] withf0(x) = mT (x), the minimal polynomial of T , such that

fn(x) | fn−1(x) | · · · | f1(x) | f0(x)and cyclic submodules of VT (sometimes called T -cyclic subspaces of V) V0 of exponent f0(x),. . . ,Vn of exponent fn(x) so that

VT∼= V0 × · · · ×Vn.

Moreover, the natural number n is uniquely determined by T , the sequence fn(x) | fn−1(x) | · · · |f1(x) | f0(x) of monic polynomials is uniquely determined, and cyclic submodules V0, . . . ,Vn aredetermined up to isomorphism. Furthermore, each of the submodules Vk for k ≤ n has a T -cyclicbasis Bk, and Bn ∪ · · · ∪ B0 is a basis for V. The linear operator T is represented with respect tothis basis by

Cfn 0 0 . . . 00 Cfn−1 0 . . . 00 0 Cfn−2 . . . 0...

...... . . . ...

0 0 0 . . . Cf0

where the companion matrices of the fk(x)’s are placed in order as square blocks along the diagonal,with all remaining entries of the matrix 0.

The matrix representing T in this theorem is said to be in rational canonical form. Theuniqueness assertions of the Extended Invariant Factor Theorem ensure that a linear operator hasexactly one rational canonical form. The following important theorem is now a corollary.The Cayley-Hamilton Theorem.Let T be a linear operator on a finite dimensional vector space over a field. The minimal polynomialof T divides the characteristic polynomial of T . Hence, f(T ) is the constantly zero linear operator,when f(x) is the characteristic polynomial of T .

Actually, it is easy to see that the characteristic polynomial is just the product of the invariantfactors.

Recall from linear algebra that if A and B are m×m matrices with entries in the field F, then wesay that A and B are similar provided there is a linear operator T on the m-dimensional vectorspace over F such that T can be represented by both A and B (using appropriated bases). So wefind

37

Rational Canonical Form Theorem: Matrix Version.Let F be a field and m be a positive natural number. Every m × m matrix with entries is F issimilar to exactly one matrix in rational canonical form.

Now let’s turn to the elementary divisor perspective. Consider the case when VT is a cyclicprimary F[x] module. In this case, there is a vector v ∈ V , an irreducible monic polynomialf(x) ∈ F[x], and a positive natural number e so that (f(x))e is the order of v. As long as thefield F is arbitrary, the polynomial f(x) could be quite complicated—for instance it might havearbitrarily large degree. In such a situation, it would be difficult to improve on the process weused above the obtain the rational canonical form. However, two fields immediately come to mindwherre the situation is much more restrained. The field C of complex numbers has the propertythat all irreducible polynomials in C[x] have degree 1, while over the field R of real numbers therecan also be irreducible polynomials of degree 2 but of no higher degrees. Both of these facts willlbe proved in the next semester. So let us consider that f(x) = x− a for some a ∈ F . Then put

v0 = v

v1 = (T − aI)v0 = Tv0 − av0

v2 = (T − aI)v1 = Tv1 − av1

...ve1 = (T − aI)ve−2 = Tve−2 − ave−2

0 = (T − aI)ve−1 = Tve−1 − ave−1

Rearranging this just a bit we getTv0 = av0 + v1

Tv1 = av1 + v2

...Tve2 = ave−2 + ve−1

Tve−1 = ave−1

Now by an argument similar (hard working graduate students will provide the variations needed)to the ones used above, we can see that the e distinct vectors v0, v1, . . . , ve−1 form a basis for thevector space V. With respect to this basis, the linear operator T has the following matrix

a 0 0 0 . . . 0 01 a 0 0 . . . 0 00 1 a 0 . . . 0 00 0 1 a . . . 0 0...

......

... . . . ......

0 0 0 0 . . . a 00 0 0 0 . . . 1 a

.

A matrix of this form is called a Jordan block and the basis underlying is called a Jordan basis.Observe that it is easy given the polynomial (x − a)e to write down its Jordan block and, viceversa, given the Jordan block, the poynomial can be recovered at once. Moreover, (x − a)e is thecharacteristic polynomial det(xI − J) of the Jordan block J .

This time, appealing the the Structure Theorem we getThe Jordan Canonical Form Theorem.Let V be a finite dimensional vector space over the field F. Let T be a linear operator of V such

38

that the irreducible factors of the minimal polynomial of T are all of degree 1. Then for somenatural number n there are polynomials (x−a0)e0 , (x−a1)e1 , . . . , (x−an−1)en−1 ∈ F[x], namely theelementary divisors of VT , and and cyclic primary submodules of VT V0 of exponent (x−a0)e0,. . . ,Vn of exponent (x− an−1)en−1 so that

VT∼= V0 × · · · ×Vn−1.

Moreover, the natural number n is uniquely determined by T , the polynomials (x− ak)ek for k < nare uniquely determined, and cyclic submodules V0, . . . ,Vn are determined up to isomorphism.Furthermore, each of the submodules Vk for k ≤ n has a Jordan basis Bk, and Bn ∪ · · · ∪ B0 is abasis for V. The linear operator T is represented with respect to this basis by

J0 0 0 . . . 00 J1 0 . . . 00 0 J2 . . . 0...

...... . . . ...

0 0 0 . . . Jn−1

where the Jordan blocks of the (x − ak)ek ’s are placed in some order as square blocks along thediagonal, with all remaining entries of the matrix 0.

The matrix mentioned at the conclusion of this theorem is said to be in Jordan canonical form.It should be noted here, that there may be several matrices in Jordan form associated with thelinear operator T according to this theorem. This happens because the order in which the Jordanblocks appear along the diagonal is arbitrary. The uniqueness part of the Structure Theorem Thepermutation mentioned in the statement of the Structure Theorem reflects the same point. Itis clear that if A and B are two matrices in Jordan form that can be obtained from each otherby rearranging the Jordan blocks along the diagonal, the A and B are similar. So the StructureTheorem gives usJordan Canonical Form Theorem: Matrix Version.Let F be a field and m be a positive natural number. Let A m ×m matrix with entries is F withthe additional property that the irreducible factors of the minimal polynomial of the matrix are allof degree 1. Then A is similar to a matrix in Jordan canonical form and any matrices in Jordancanonical form that are similar to A can be obtained from each other by rearranging the Jordanblocks.

39

Solutions to Problem Sets

Algebra Homework Solutions, Edition 0Due 26 August 2010

Problem 0.Classify up to similarity all the square matrices over the complex numbers with minimal polynomialm(x) = (x− 1)2(x− 2)2 and characteristic polynomial c(x) = (x− 1)6(x− 2)5.

SolutionObserve that the degree of the characteristic polynomial is 11 so what is needed is to list the JordanNormal Forms of 11× 11 matrices with eignevalues 1 (of multiplicity 6) and 2 (of mulitiplicity 5) whichhave the correct minimal polynomial. I counted six of these. In view of the minimal polynomial thelargest Jordan block has to be 2× 2 and for each of the two eignenvalues there must be a Jordan blockof that size. For the eigenvalue 1 there are three possibilities: three 2× 2 blocks, two 2× 2 blocks, orone 2 × 2 block. For the eigenvalue 2 there are only two possibilities: two 2 × 2 blocks or one 2 × 2block. The Jordan Normal Form Theorem tells us that no two of these six Jordan matrices are similarand on the other hand that any matrix with the given characteristic and minimal polynomials must besimilar to one of these.

For completeness, these six Jordan matrices are displayed below.

1 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 1 1 0 0 0 0 00 0 0 0 0 0 2 0 0 0 00 0 0 0 0 0 1 2 0 0 00 0 0 0 0 0 0 0 2 0 00 0 0 0 0 0 0 0 1 2 00 0 0 0 0 0 0 0 0 0 2

1 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 1 1 0 0 0 0 00 0 0 0 0 0 2 0 0 0 00 0 0 0 0 0 1 2 0 0 00 0 0 0 0 0 0 0 2 0 00 0 0 0 0 0 0 0 0 2 00 0 0 0 0 0 0 0 0 0 2

1 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 2 0 0 0 00 0 0 0 0 0 1 2 0 0 00 0 0 0 0 0 0 0 2 0 00 0 0 0 0 0 0 0 1 2 00 0 0 0 0 0 0 0 0 0 2

1 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 2 0 0 0 00 0 0 0 0 0 1 2 0 0 00 0 0 0 0 0 0 0 2 0 00 0 0 0 0 0 0 0 0 2 00 0 0 0 0 0 0 0 0 0 2

40

1 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 2 0 0 0 00 0 0 0 0 0 1 2 0 0 00 0 0 0 0 0 0 0 2 0 00 0 0 0 0 0 0 0 1 2 00 0 0 0 0 0 0 0 0 0 2

1 0 0 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 2 0 0 0 00 0 0 0 0 0 0 2 0 0 00 0 0 0 0 0 0 1 2 0 00 0 0 0 0 0 0 0 0 2 00 0 0 0 0 0 0 0 0 0 2

Problem 1.Let T : V → V be a linear transformation of rank 1 on a finite dimensional vector space V overany field. Prove that either T is nilpotent or V has a basis of eigenvectors of T .

SolutionWe give two solutions.

For the first, let M be an n × n rank 1 matrix and let T be linear transformation represented byM . Since rank T + nullityT = n we see that the nullspace of T must have dimension n − 1. Let{v1, v2, . . . , vn−1} be a basis for the nullspace of T and let u be any vector not in the nullspace. Itfollows that {u, v0, . . . , vn−1} is linearly independent. Since it has n elements it must be a basis for thewhole space. So there are scalars a0, a1, . . . , an−1 so that

Tu = a0u+ a1v1 + a2v2 + · · ·+ an−1vn−1 = w.

Moreover, since u is not in the nullspace of T then Tu = w 6= 0. So at least one of the ai’s are not0. Now observe that Tw = a0Tu = a0w. This means that w is an eigenvector of T belonging to theeigenvalue a0. Now if a0 6= 0 we put v0 = w and otherwise we put v0 = u. In either case v0 is not inthe nullspace of T and so {v0, v1, . . . , vn−1} is a basis for the whole space. With respect to this basisthe matrix for T is either

a0 0 0 . . . 00 0 0 . . . 0...

... . . . · · ·...

0 0 0 . . . 0

or

0 0 0 . . . 0a1 0 0 . . . 0...

... . . . · · ·...

an−1 0 0 . . . 0

depending on whether a0 6= 0 (the left matrix) or a0 = 0 (the right matrix). Thus M is similar to eithera diagonal matrix or to a nilpotent matrix.

Our second proof uses the rational canonical form. There are two kinds of rank one companionmatrices: (

0 01 0

)and (λ)

where λ 6= 0. The left matrix is the companion matrix of x2 while the one on the right is the companionmatrix of x − λ. Now every rank one matrix is similar to a rank one matrix in rational form and thisrational form is comprised of campanion matrices along the diagonal with 0 elsewhere. Because ourrational form has rank one, these companion matrices must all be the trivial (0) companion matrix

41

except for one, which must be one of the rank one companion matrices. If it is the kind on the left,then the matrix is nilpotent, while if it is the kind on the right, then the matrix is diagonal.

Problem 2.Let V be a vector space over a field K.(a) Prove that if U0 and U1 are subspaces of V such that U0 * U1 and U1 * U0, then V 6= U0 ∪U1.(b) Prove that if U0, U1, and U2 are subspaces of V such that Ui * Uj when i 6= j and K has at

least 3 elements, then V 6= U0 ∪ U1 ∪ U2.(c) State and prove a generalization of (b) for n subspaces.

SolutionThe general version is

If U0, U1, . . . , Un−1 are pairwise incomparable subspaces of V and K has at least nelements, then V 6= U0 ∪ U1 ∪ · · · ∪ Un−1.

We prove this by induction on n.Base Step: n = 2. Pick u0 ∈ U0 − U1 and u1 ∈ U1 − U0. Then u0 + u1 /∈ U0 since otherwiseu1 = u0 + u1 − u0 ∈ U0. Likewise, u0 + u1 /∈ U1. So u0 + u1 /∈ U0 ∪ U1, which entails U0 ∪ U1 6= V .Inductive Step Here n > 2. We can assume that Ui is not included in the union of the remainingsubspaces, since otherwise Ui can be omitted and an immediate appeal to the inductive hyupothesiswill complete the proof. So for each i < n+ 1 pick an element ui ∈ Ui so that ui is not in any of theremaining subspaces. Now for each k < n we pick a scalar ak so that

u0 + a1u1 + · · ·+ am−1um−1 /∈ Umfor all m < n+1. At the start we put a0 = 1. Suppose that a0, a1, . . . , ak−1 have already been selected.Since n is at least 3 we know that there are at least two nonzero scalars b and c. Were it the case that

u0 + a1u1 + · · ·+ ak−1uk−1 + buk ∈ Uk+1 andu0 + a1u1 + · · ·+ ak−1uk−1 + cuk ∈ Uk+1 then

(b− c)uk ∈ Uk+1

But this entails that uk ∈ Uk+1 which is false.Let w = u0 + a1u1 + · · · + an−1un−1. So w /∈ Un. For the sake of contradiction, suppose that

V = U0 ∪ · · · ∪ Un. Now there are at least n + 1 scalars. So there are at least n + 1 vectors of theform w + bun where b is a scalar. Since w /∈ Un none of these n + 1 vectors belongs to Un. So theyall belong to U0 ∪ U1 ∪ · · · ∪ Un−1. So by the Pigeonhole Principle there are two distinct scalars b andc and a k < n so that w + bun, w + cun ∈ Uk. It follows that (b − c)un ∈ Uk. But this entails thatun ∈ Uk for some k < n, contrary to the selection of un.

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Algebra Homework, Edition 1Due 2 September 2010

Problem 3.Prove that the congruence relations of A are exactly those subuniverses of A ×A which happento be equivalence reations on A.

SolutionSuppose first that θ is a congruence of A. Then θ is surely an equivalence relation on A. So we haveto argue that θ is closed under all the operations of A ×A. Take Q to be a basic operation symbol,say of rank 3 to make notation easy, and suppose

(a0, a′0) ∈ θ

(a1, a′1) ∈ θ

(a2, a′2) ∈ θ.

Then (QA(a0, a1, a2), QA(a′0, a′1, a′2)) ∈ θ by the substitution property. Now according to the coordi-natewise definition of operations in the direct product, we know

QA×A((a0, a′0), (a1, a

′1), (a2, a

′2)) = (QA(a0, a1, a2), QA(a′0, a′1, a′2)).

As a result QA×A((a0, a′0), (a1, a

′1), (a2, a

′2)) ∈ θ and we see that θ is closed with respect to the

operations. So θ is a subuniverse of A×A.For the converse, now suppose that θ is a subuniverse of A×A which is also an equivalence relation

on A. To conclude that θ is a congruence of A we only need to demonstrate the substitution propertyfor θ. So suppose

(a0, a′0) ∈ θ

(a1, a′1) ∈ θ

(a2, a′2) ∈ θ.

Since θ is a subuniverse of A × A we know it is closed with respect to all the operations. SoQA×A((a0, a

′0), (a1, a

′1), (a2, a

′2)) ∈ θ. Now according to the coordinatewise definition of operations

in the direct product, we knowQA×A((a0, a

′0), (a1, a

′1), (a2, a

′2)) = (QA(a0, a1, a2), QA(a′0, a′1, a′2)).

So we have (QA(a0, a1, a2), QA(a′0, a′1, a′2)) ∈ θ and we conclude that θ has the substitution propertyas desired.

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Problem 4.Prove that the homomorphisms from A to B are exactly those subuniverses of A × B which arefunctions from A to B.

SolutionFirst suppose that Φ is a homomorphism from A to B. Evidently Φ is a function from A to B, so itremains to show that Φ is a subalgebra of A × B. Let Q be an operation symbol, which to simplifynotation we suppose has rank 3. So take three elements of Φ:

(a0, b0) ∈ Φ(a1, b1) ∈ Φ(a2, b2) ∈ Φ

We wish to show that QA×B((a0, b0), (a1, b1), (a2, b2)) ∈ Φ, that is by the coordinatewise definition ofoperations in the product, what we need is

(QA(a0, a1, a2), QB(b0, b1, b2)) ∈ Φ.Recalling that (x, y) ∈ Φ is just another way to write Φ(x) = y we see our task is the following. From

Φ(a0) = b0

Φ(a1) = b1

Φ(a2) = b2

we want to deduceΦ(QA(a0, a1, a2)) = QB(b0, b1, b2).

But of course this is obvious since Φ is a homomorphism.Second, for the reverse direction, we suppose that Φ is a subuniverse of A×B which happens to be

a function from A to B. We want the conclusion that Φ is a homomorphism from A to B. That is wehave to argue that Φ preserves the operations. So again let Q be an operation and for simplicity let 3be the rank of Q. Let a0, a1, and a2 be elements of A. What we need is

Φ(QA(a0, a1, a2)) = QB(Φ(a0),Φ(a1),Φ(a2)).Another way to write this is

(QA(a0, a1, a2), QB(Φ(a0),Φ(a1),Φ(a2))) ∈ Φ.Now we know that Φ is a function, so we have

(a0,Φ(a0)), (a1,Φ(a1)), (a2,Φ(a2)) ∈ Φ.Since Φ is also a subuniverse of A×B we can draw the conclusion that

QA×B((a0,Φ(a0)), (a1,Φ(a1)), (a2,Φ(a2))) ∈ Φ.But recalling the coordinatewise manner in which the operations in the direct product work, we get

(QA(a0, a1, a2), QB(Φ(a0),Φ(a1),Φ(a2))) ∈ Φ,just as desired.

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Problem 5.Prove that the projection functions associated with A×B are homomorphisms.

SolutionLet π : A × B → A be defined so that π(a, b) = a for all a ∈ A and b ∈ B. So π is the projection ofA × B onto A. To check that π is a homomorphism, let Q be an operation symbol, and to simplifynotation, suppose the rank of Q is 3. We pick arbitrary elements (a0, b0), (a1, b1), (a2, b2) ∈ A × B.Then just note

π(QA×B((a0, b0), (a1, b1), (a2, b2))) = π(QA(a0, a1, a2), QB(b0, b1, b2))= QA(a0, a1, a2)= QA(π(a0, b0), π(a1, b1), π(a2, b2)).

Consequently, π is a homomorphism. In a like manner, the projection onto the other factor is also seento be a homomorphism.

Algebra Homework, Edition 29 September 2010

Problem 6.

(1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I ∩ J .(2) Let I, J, and K be ideals of a principal ideal domain. Prove that I∩(J+K) = I∩J+I∩K.

SolutionFor (1) pick a ∈ I and b ∈ J so that a + b = 1. To see that I ∩ J ⊆ IJ let d be any element ofI ∩ J . Then d = d · 1 = d(a+ b) = da+ db = ad+ db where the last equality holds by commutativity.Since a ∈ I and d ∈ J we see that ad ∈ IJ , while since d ∈ I and b ∈ J we see that db ∈ IJ . Sod = ad + db ∈ IJ . Thus, I ∩ J ⊆ IJ . But IJ ⊆ I since I is closed with resect the multiplication onthe right by any element of R (much less those in J). Likewise IJ ⊆ J . So IJ ⊆ I ∩ J , giving theopposite inclusion.

For Part (2), It pays to think about this equality in lattices more generally. Lattices are just partiallyordered sets in which any elements x and y have both a least upper bound (denoted by x ∨ y) andgreatest lower bound (denoted by x ∧ y). Rendered in this notation the equality we are interested inbecomes

x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).Consider the case of a linearly ordered set, like the real numbers. Here x ∧ y = min(x, y) and x ∨ y =max(x, y). The equality above holds in linearly ordered sets for the following reason. The elements yand z are comparable, and in view of the symmetry of the roles of y and z in the equality, it does noharm to suppose that y ≤ z. Then y∨z = z and the equality becomes obvious. Turning a lattice upsidedown interchanges the notion of least upper bound with the notion of greatest lower bound. Since theupside down version of a linearly ordered set is also lineaerly ordered we see that a linearly ordered setalso makes the following equality true.

x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z).

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While it might not be so clear, this holds more generally: a lattice that satisfies one distributive lawmust also satisfy the other.

Returning to our ideals in the principal ideal domain, pick elements a, b, and c so that I = (a), J = (b),and K = (c). Since our domain is a unique factorization domain, we pick primes p0, p1, . . . , pn−1 sothat each of a, b, and c can be written, to within unit multipliers as a product of these primes. Thismeans we can pick natural numbers k0`0,m0, k1, `1,m1, . . . , kn−1, `n−1,mn−1 and units u, v and w sothat

a = upk00 p

k11 . . . p

kn−1n−1

b = vp`00 p`11 . . . p

`n−1n−1

c = wpm00 pm1

1 . . . pmn−1n−1

Now J +K = (d) where d is any greatest common divisor of b and c. This means we can take

d = pmin(`0,m0)0 p

min(`1,m1)1 . . . p

min(`n−1,mn−1)n−1

So the left side of the equality becomes (a)∩ (d) = (e) where e is any least common multiple of a andd. We can take

e = pmax(k0,min(`0,m0))0 p

max(k1,min(`1,m1))1 . . . p

max(kn−1,min(`n−1,mn−1))n−1

Now let us consider the right side of the equality. There (a)∩ (b) = (q) where q is any least commonmultiple of a and b. We can take

q = pmax(k0,`0)0 p

max(k1,`1)1 . . . p

max(kn−1,`n−1)n−1 .

Likewise (a) ∩ (c) = (r) where r is any least common multiple of a and c. We put

r = pmax(k0,m0)0 p

max(k1,m1)1 . . . p

max(kn−1,mn−1)n−1 .

So the whole right side becomes (q) + (r) = (s) where s is any greatest common divisor of q and r.We put

s = pmin(max(k0,`0),max(k0,m0))0 p

min(max(k1,`1),max(k1,m1))1 . . . p

min(max(kn−1,`n−1),max(kn−1,mn−1))n−1

Our problem will be finished if we can show that e and s are associates. This amounts to showing foreach i < n that

max(ki,min(`i,mi)) = min(max(ki, `i),max(ki,mi))This equality concerns natural numbers, which are linearly ordered. So rendering this equation in latticenotation, we see

ki ∨ (`i ∧mi) = (ki ∨ `i) ∧ (ki ∨mi).But we saw at the start that this distributive law holds in linearly ordered sets, like the nautral numbers.So we are done.

Problem 7.Let R be a commutative ring and I be a proper prime ideal of R such that R/I satisfies thedescending chain condition on ideals. Prove that R/I is a field.

SolutionLet S = R/I. Then S is a nontrivial integral domain with the descending chain condition. Our problemis to show that it is a field. So let a ∈ S be any nonzero element. Consider the elements a, a2, a3, . . . . Ifwe look at the ideals generated by these elements we see the descending chair (a) ⊇ (a2) ⊇ (a3) ⊇ · · ·

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and conclude that it must be finite. So pick n large enough so that (an) = (an+1). This entails thatan = an+1b for some b. But since a 6= 0 and S is an integral domain, then an 6= 0 and we can cancelit, leaving 1 = ab. This means that a is invertible, and so S is a field.

For a counterexample to the problem without the added stipulation of primeness, take R to be thering of integers and let I = (6). Then R/I = Z6 which is a finite ring (and hence has the descendingchain condition) which is not even an integral domain (much less a field).

Problem 8.Let R be a commutative ring and I be an ideal which is contained in a prime ideal P . Prove thatthe collection of prime ideals contained in P and containing I has a minimal member.

SolutionLet us try using the upside down version of Zorn’s Lemma. So let

F = {J | I ⊆ J ⊆ P and J is a prime ideal}.Notice that P ∈ F. Let C be a chain in F. We need to argue that C has a lower bound in F. First let’stake care of the case when C is empty. In this case P will be a lower bound. Now suppose C is not empty.We argue that

⋂C is the desired lower bound. The definition of intersection gives us I ⊆

⋂C ⊆ J ⊆ P

for all J ∈ C. So the only thing that remains is to show that⋂

C ∈ F—that is that⋂

C is a primeideal containing I and contained in P . Everything is clear except that

⋂C is prime. (Recall that the

intersection of any nonempty collection of ideals is always an ideal.) To see the primeness, let ab ∈⋂

C.For contradiction, suppose that a /∈

⋂C and also that b /∈

⋂C. So pick Ja, Jb ∈ C so that a /∈ Ja and

b /∈ Jb. Since C is a chain, we can suppose without loss of generality, that Ja ⊆ Jb. Therefore we alsoget that b /∈ Ja. On the other hand, ab ∈ Ja since it belongs to every member of C. This means wehave contradicted the primeness of Ja. So we conclude that the hypotheses for the minimal-elementversion of Zorn’s Lemma hold. Thus F must have a minimal element, as desired.

Problem 9.Let X be a finite set and let R be the ring of functions from X into the field R of real numbers.Prove that an ideal M of R is maximal if and only if there is an element a ∈ X such that

M = {f | f ∈ R and f(a) = 0}.

SolutionFirst, suppose that a ∈ X and that M = {f | f ∈ R and f(a) = 0}. Upon immediate reflection, wesee that M is an proper ideal of R. Let J be an ideal of R so that M ⊆ J with M 6= J . Pick g ∈ Jso that g /∈M . Let X = {x0, . . . , xn−1} and suppose a = x0. Observe that g(x0) 6= 0. For each i < nwith 0 < i we define

fi(xj) ={

1 if i = j

0 otherwise.So f1, f2, . . . , fn−1 ∈M ⊆ J . We also define

h(x) ={

0 if x 6= x01

g(x0) otherwise.

Now notice thath(x)g(x) + f1(x) + f2(x) + · · ·+ fn−1(x) ∈ J.

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But this function is constantly 1. This means that J = R and so M is a maximal proper ideal.For the converse, suppose that I is an ideal of R but that for each i < n there is fi(x) ∈ M with

fi(xi) 6= 0. Define gi(x) via

gi(xj) ={ 1fi(xi) if i = j

0 otherwise.So gi(x)fi(x) ∈ I for all i < n. But then

g0(x)f0(x) + · · ·+ gn−1(x)fn−1(x) ∈ I.But this function is constantly 1. So the ideal I is not proper. This means that for an ideal M which ismaximal among proper ideals there must be a ∈ {x0, x1, . . . , xn−1} so that f(a) = 0 for all f(x) ∈M .Hence M ⊆ {f(x) | f(a) = 0 and f(x) ∈ R}. As this last set was already shown to be a maximalideal, it must be that M = {f(x) | f(a) = 0 and f(x) ∈ R}.

Algebra Homework, Edition 3Due 16 Septemer 2010

Problem 10.Let R be a commutative ring and let n be a positive integer. Let J, I0, I1, . . . , In−1 be ideals of Rso that Ik is a prime ideal for every k < n and so that J ⊆ I0 ∪ · · · ∪ In−1. Prove that J ⊆ Ik forsome k < n.

SolutionWe proceed by induction on n. The base step, when n = 1 is evident. For the induction step, we cansuppose that J is not included in the union of fewer than all n of the Ik’s, for otherwise we can appealto the induction hypothesis. We must reject this supposition. So for each i < n we pick ai ∈ J ∩ Iiso that ai /∈ Ij for any j 6= i. The element a0 + a1a2 . . . an−1 certainly belongs to J since J is anideal. Therefore pick k < n so that a0 + a1 . . . an−1 ∈ Ik. There are two cases, both of which we canreject. The first case is that k = 0. Then we have both a0 and a0 + a1 . . . an−1 in I0. This entails thata1 . . . an−1 ∈ I0. Now I0 is a prime ideal so there must be j with 1 ≤ j < n and aj ∈ I0. But thiscontradicts the choice of aj . The second case is that k > 0. Then both a0 +a1 . . . an−1 and a1 . . . an−1belong the Ik, the latter membership follows since Ik is an ideal and a1 . . . an−1 is a multiple of ak. Inconsequence a0 ∈ Ik, contradicting the choice of a0.

Problem 11.Let R be a nontrivial commutative ring and let J be the intersection of all the maximal properideals of R. Prove that 1 + a is a unit of R for all a ∈ J .

SolutionSuppose that a belongs to all the proper maximal ideals. Were 1 + a not a unit, then (1 + a) wouldbe a proper ideal. By the Maximal Ideal Theorem there would be a maximal proper ideal M so that1 + a ∈M . But also a ∈M since it is in every maximal proper ideal. This would force 1 ∈M and wewould have the contradiction that M is not a proper ideal.

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Algebra Homework, Edition 4Due 23 September 2010

Problem 12.Let F be a field and let p(x) ∈ F[x] be a polynomial of degree n. Prove that p(x) has at most ndistinct roots in F.

SolutionSuppose r is a root of p(x). We know that (x − r) is a factor of p(x). So pick q(x) ∈ F[x] suchthat p(x) = (x − r)q(x). Now let s be a root of p(x) distinct from r. Now both x − r and x − sare prime and x − s | (x − r)q(x). We see x − s cannot divide x − r since r 6= s. So x − s dividesq(x). So we can pick t(x) ∈ F[x] with p(x) = (x − r))(x − s)t(x). Continuing in this way, we seethat if r0, r1, . . . rm−1 are distinct roots of p(x) then there is a polynomial u(x) ∈ F[x] such thatp(x) = (x− r0)(x− r1) . . . (x− rm−1)u(x). So m can be no bigger than the degree of p(x).

Problem 13.Let F be a field and let F∗ be its (multiplicative) group of nonzero elements. Let G be any finitesubgroup of F∗. Prove that G must be cyclic.

SolutionOur gorup G is a finite Abelian group, so it factors into a direct product of cyclic groups of prime powerorder. In the case that each of these direct factors is associated with a different prime, we see that Gis cyclic (by taking the product of the generators of each factor. So let us suppose this is not the case.Then there is a prime p and two subgroups H and K of F∗ such that |H| = pn, |K| = pm, where nand m are positive integers, and H ∩K is trivial. We invoke Cauchy to have n = m = 1. This meansthat there are at least 2p− 1 elements of order p in F ∗. These elements would all have to be roots ofxp − 1. On the other hand, this polynomial can have at most p roots in F .

Problem 14.Suppose that D is a commutative ring such that D[x] is a principal ideal domain. Prove that D isa field.

SolutionThe polynomial x ∈ D[x] is irreducible and hence prime. This means that the ideal (x) generated byx is a prime ideal. In a principal ideal domain, every prime ideal in maximal. Hence D[x]/(x) is afield. So we need to see that D ∼= D[x]/(x). Let h be the (unique) homomporphism from D[x] intoD that fixes each element of D (that is h(d) = d for all d ∈ D and sends x to 0. So h maps D[x]onto D. Since h(d0 + d1x+ · · ·+ dnx

n) = d0 for any polynomial d0 + d1x+ · · ·+ dnxn, we see that

d0 + d1x + · · · + dnxn ∈ kerh iff d0 = 0 iff d0 + d1x + · · · + dnx

n ∈ (x). So (x) = kerh. By theHomomorphism Theorem, we see that D[x]/(x) ∼= D. So D is a field.Here is a second solution:We only nee to see that every nonzero element of D has a multiplicative inverse. So let d ∈ D withd 6= 0. Now the ideal (d, x) of D[x] generated by {d, x} is principal. Say it is generated by the

49

polynomial f(x). Then d = f(x)g(x) for some polynomial g(x). This entails that the degree of f(x)is 0. Say f(x) = a. Now there must also be a polynomial h(x) so that x = f(x)h(x) = ah(x). Soevidently h(x) = cx and ac = 1. This means that a is a unit. Because D[x] is a principal ideal domain,there must be polynomials u(x) and w(x) such that a = du(x) + xw(x). Let b be the constant termof u(x). It must turn out that a = db. So we find that a and d are associates. So d must also be aunit. This means d is invertible, as desired.

Algebra Homework, Edition 5Due 30 September 2010

Problem 15.Is the polynomial y3 − x2y2 + x3y + x+ x4 irreducible in Z[x, y]?

SolutionConstrue y3 − x2y2 + x3y + x + x4 as a polynomial (in y) with coefficients from Z[x]. Thus thecoefficients of this polynomial are 1,−x2, x3, and x4 + x. Since these coefficients have no commonirreducilbe factor (in Z[x]) our polynomial is primitive. Now x ∈ Z[x] is irreducible and

x - 1, x | −x2, x | x3, x | x4 + x, and x2 - x4 + x = (x4 + 1)x.This means that Eisenstein Criterion applies, and our polynomial is irreducible.

Problem 16.Let R be a principal ideal domain, and let I and J be ideals of R. IJ denotes the ideal of Rgenerated by the set of all elements of the form ab where a ∈ I and b ∈ J . Prove that if I +J = R,then I ∩ J = IJ .

SolutionIn any case, IJ ⊆ I and IJ ⊆ J since I and J are ideals. This means that IJ ⊆ I ∩ J . We need towork for the reverse inclusion.

Let I = (a) and J = (b). In a principal ideal domain we see that (d) = (a) ∩ (b) means the samething as d is a least common multiple of a and b.

On the other hand, ra and sb are arbitrary elements of I and J and (ra)(sb) = (rs)(ab). So theproduct of an element of I with an element of J turns a multiple of ab. From this it is easy to see thatIJ = (ab). So we would be done, if we could prove that ab is a least common multiple of a and b. Thisis where the hypothesis I + J = R comes in. We could restate this as (a) + (b) = (1). Another way tosay it is that a and b are relatively prime. For a and b have no common prime factors. For any prime pthe number of times it occurs as a factor in d must be the sum of the number of times it occurs in aand the number of times it occurs in b. So d and ab must be associates.

Problem 17.Let D be a unique factorization domain and let I be a nonzero prime ideal of D[x] which is minimalamong all the nonzero prime ideals of D[x]. Prove that I is a principal ideal.

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SolutionWe know that D[x] is also a unique factorizaton domain. First pick a polynomial p(x) ∈ I with p(x) 6= 0.Next factor p(x) into primes. Because I is a prime ideal, at least one of the prime factors of p(x) mustbe in I. So it does no harm to suppose p(x) is prime to begin with. Let J be the ideal generated byp(x). Because p(x) is prime J is a prime ideal. But J ⊆ I and I is minimal among nonzero primeideals. So J = I. Since J is principal, so is I.

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Algebra Homework, Edition 6Due 7 October 2010

Problem 18.a. Prove that (2, x) is not a principal ideal of Z[x].b. Prove that (3) is a prime ideal of Z[x] that is not a maximal ideal of Z[x].

SolutionFor part (a) suppose otherwise. Pick f(x) ∈ (2, x) so that f(x) generates (2, x). It is routine to check(and hard working grad students do it) that

(2, x) = {g(x) | g(x) ∈ Z[x] and the constant term of g(x) is even}.So we know that f(x) has an even constant term and that f(x) | 2 and f(x) | x since both 2 and xmust be multiples of f(x). Because f(x) | 2 we see that f(x) must be one of 1,−1, 2, or −2. Sinceneither 1 nor −1 is even we see that f(x) must be either 2 or −2. But neither of these divides x. Sowe have the desired contradiction.

For part (b) we first argue that (3) is a prime ideal. Plainly(3) = {f(x) | f(x) ∈ Z[x] and all the coefficients of f(x) are multiples of 3}.

Suppose g(x)h(x) ∈ (3) and g(x) /∈ (3). Let g(x) = b0 + b1x+ · · ·+ bnxn and h(x) = c0 + c1x+ · · ·+

cmxm. We have to show that 3 | cj for all j. Let k be as small as possible so that 3 - bk. Observe that

the kth coefficient of g(x)h(x) is∑i+j=k

bicj = b0cj + b1cj−1 + · · ·+ bk−1c1 + bkc0.

Now 3 this sum and it also divides the first k terms of the sum. Therefore it divides the last term aswell. Since 3 is prime and 3 - bk, we conclude that 3 | c0. We can continue in this way to see that 3 | cjfor all j. I’ll do one more step and leave it to you to reorganize the proof into a real proof by inductionon j. The k + 1st coefficient of g(x)h(x) is∑

i+j=k+1bicj = b0ck+1 + · · ·+ bkc1 + bk+1c0.

This time it is clear that 3 divides all the terms (the problem term is bkc1 this time). Since 3 | bkc1 andsince 3 - bk we find that 3 | c1. So after some more moves like this we find that 3 | cj for all j. Soh(x) ∈ (3) as desired. We conclude that (3) is a prime ideal of Z[x].

Finally, we have to argue that (3) is not a maximal ideal. Consider the ideal (3, x). It is clearly anideal which is properly larger than (3) and on the other hand it consists of all those polynomials withconstant terms divisible by 3. Since this is not the whole ring Z[x] we find that (3) is a prime idealwhich fails to be maximal.

Problem 19.Show that any integral domain satisfying the descending chain condition on ideals is a field.

SolutionLet D be an integral domain satisfying the descending chain condition. Let 0 6= a ∈ D. For each

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natural number n let In = (an). Then D = I0 ⊇ (a) = I1 ⊇ (a2) ⊇ . . . is a descending chain of ideals.By the descending chain condition, there is m such that

Im = Im+1 = Im+2 = . . . .

This means, in particular that am ∈ (am+1. So there is r ∈ D so that am = ram+1. But thenam · 1 = am(ra). Since we are in an integral domain and a 6= 0, we see that am 6= 0. But in integraldomains the cancellation law holds. So we get that 1 = ra. This means that r is a multiplicative inverseof a. Therefore, every nonzero element of D has a multiplicative inverse. Hence D is a field.

Problem 20.Prove the following form of the Chinese Remainder Theorem: Let R be a commutative ring withunit 1 and suppose that I and J are ideals of R such that I + J = R. Then

R

I ∩ J∼=R

I× R

J.

SolutionDefine the map Φ : R → R/I × R/J via Φ(r) = (r/I, r/J) for all r ∈ R. The hardworking graduatestudent will carry out the details to prove that Φ is a homomorphism.

I contend that the kernel of Φ is I ∪ J . Here is why:r ∈ ker Φ⇔ Φ(r) = (0/I, 0/J)

⇔ (r/I, r/J) = (0/I, 0/J)⇔ r/I = 0/I and r/J = 0/J⇔ r ∈ I and r ∈ J⇔ r ∈ I ∩ J

Now the idea is to apply the Homomorphism Theorem to obtain the desired isomorphism. But itremains to argue that Φ maps R onto R/I × R/J . To this end, pick a, b ∈ R obtaining (a/I, b/J),which is an arbitrary element of R/I × R/J . To complete the solution to this problem, we must findr ∈ R so that r/I = a/I and r/J = b/J . Another way to phrase this is that we must find r so thatr − a ∈ I and r − b ∈ J . Yet a third way is that we must find r so that

r ≡ a (mod I)r ≡ b (mod J)

Because R = I + J we can pick a0, b0 ∈ I and a1, b1 ∈ J so that a = a0 + a1 and b = b0 + b1. Here isour value: r = a1 + b0. Then

r − a = a1 + b0 − a0 − a1 = b0 − a0 ∈ Ir − b = a1 + b0 − b0 − b1 = a1 − b1 ∈ J

In this way we have found the desired r and established that Φ maps R onto R/I ×R/J with ker Φ =I ∩ J . An appeal to the Homomorphism Theorem completes this solution.

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Algebra Homework, Edition 7Due 21 October 2010

Problem 21.Let D be an integral domain and let c0, . . . , cn−1 be n distinct elements ofD. Further let d0, . . . , dn−1be arbitrary elements of D. Prove there is at most one polynomial f(x) ∈ D[x] of degree n − 1such that f(ci) = di for all i < n.

SolutionSuppose

f(x) = a0 + a1x+ · · ·+ anxn

g(x) = b0 + b1x+ · · ·+ bnxn

where an 6= 0 6= bn and f(ci) = di = g(ci) for all i < n. Now consider the polynomial f(x)−g(x). Thispolynomial has n distinct roots in D. Let F be the field of fractions of D. We can regard f(x)− g(x)as a polynomial with coefficients in F. This polynomial is either the zero polynomial (our desire) orit has degree no more than n − 1. But F[x] is a unique factorization domain, and polynomials of anydegree k can only have k roots. Since f(x)− g(x) has at least n roots, too many for a polynomial ofdegree no more than n − 1, we see that f(x) − g(x) is the zero polynomial. Hence, f(x) = g(x) asdesired.

Problem 22.Let F be a field and let c0, . . . , cn−1 be n distinct elements of F . Further let d0, . . . , dn−1 be arbitraryelements of F . Prove there is at least one polynomial f(x) ∈ F [x] of degree n such that f(ci) = difor all i < n.

SolutionTry this:

f(x) =∑i<n

di(x− c0)(x− c1) . . . (x− ci−1)(x− ci+1) . . . (x− cn−1)

(ci − c0)(ci − c1) . . . (ci − ci−1)(ci − ci+1) . . . (ci − cn−1)

Problem 23.Let R be the following subring of the field of rational functions in 3 variables with complex coeffi-cients:

R ={f

g: f, g ∈ C[x, y, z] and g(1, 2, 3) 6= 0

}Find 3 prime ideals P1, P2, and P3 in R with

0 $ P1 $ P2 $ P3 $ R.

SolutionConsider the three very nice polynomial x− 1, y− 2, and z − 3. Let P3 be the ideal of R generated bythese three. A typical element of P3 looks like

f0g0

(x− 1) + f1g1

(z − 2) + f2g2

(z − 3).

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If there were some way to chose the fi’s and gi’s so this turned out to be 1, then with a bit ofmultiplication we would have

g0g1g2 = f0g1g2(x− 1) + f1g0g2(y − 2) + f2g0g1(z − 3).

Now substitute x = 1, y = 2, and z = 3 to get

g0(1, 2, 3)g1(1, 2, 3)g2(1, 2, 3) = 0.

But this is impossible since none of the gi evaluate to 0 at (1, 2, 3). This means that 1 cannot be in P3so P3 $ R.

Let P2 be the ideal generated by x− 1 and y − 2. Suppose

z − 3 = f0g0

(x− 1) + f1g1

(y − 2).

Evaluating both sides at (1, 2, z), after a bit of multiplication, gives

z − 3 = 0.

Since z − 3 is not the zero polynomial, we conclude P2 $ P3.Take P1 to be se ideal generated by x− 1. It is straightforward to see

〈0〉 $ P1 $ P2 $ P3 $ R.

All the remains is the argue that our three ideals are prime.Define h1 : R → R by

h1

(f(x, y, z)g(x, y, z)

):= f(1, y, z)

g(1, y, z) .

This map is easily seen to be a homomorphism (as the hardworking graduate students know). Theimage of R with respect to this map is an integral domain because it is a nontrivial subring of R andwe easily see that R is an integral domain. So the kerh1 must be a prime ideal. But now

f

g∈ kerh1 ⇔ h1

(f

g

)0⇔ f(1, y, z) = 0.

But this last condition, viewed in C[y, z][x], is the same as saying x − 1 | f(x, y, z). That’s the sameas saying f

g ∈ P1. So kerh1 = P1 and we see that P1 is a prime ideal.Now let R1 = h(R) and P ′2 = h1(P2) Observe

R1 ={f(1, y, z)g(1, y, z) | f(x, , y, z), g(x, y, z) ∈ C[x, y, z] and g(1, 2, 3) 6= 0

}={h(y, z)`(y, z) | h(y, z), `(y, z) ∈ C[y, z] and `(2, 3) 6= 0

}

P ′2 ={f(1, y, z)(y − 2)

g(1, y, z) | f(x, , y, z), g(x, y, z) ∈ C[x, y, z] and g(1, 2, 3) 6= 0}

={h(y, z)(y − 2)

`(y, z) | h(y, z), `(y, z) ∈ C[y, z] and `(2, 3) 6= 0}

We know that P ′2 is an ideal of R1, because kerh1 = P1 ⊆ P2, and that h1(fg

)∈ P ′2 if and only if

fg ∈ P2.

55

Now just has we proved that P1 is a prime ideal of R, we can prove that P ′2 is a prime ideal of R1.Here is how to use this to prove that P2 is a prime ideal of R.f0g0

f1g1∈ P2 =⇒ h1

(f0g0

)h1

(f1g1

)∈ P ′2 =⇒ h1

(f0g0

)∈ P ′2 or h1

(f1g1

)∈ P ′2 =⇒ f0

g0∈ P2 or f1

g1∈ P2

By using the same procedure we can prove that P3 is a prime ideal.

Algebra Homework, Edition 8Due 28 October 2010

Problem 24.Prove that the polynomial x3y + x2y − xy2 + x3 + y is irreducible in Z[x, y].

Problem 25.Let D be a subring of the field F. An element r ∈ F is said to be integral over D provided thereis a monic polynomial f(x) ∈ D[x] such that r is a root of f(x). For example, the real number

√2

is integral over the ring of integers since it is a root of x2 − 2.Now suppose D is a unique factorization domain and F is its field of fractions. Prove that the set

of elements of F that are integral over D coincides with D itself.

Problem 26.Prove that there is a polynomial f(x) ∈ R[x] such that(a) f(x)− 1 belongs to the ideal (x2 − 2x+ 1);(b) f(x)− 2 belongs to the ideal (x+ 1), and(c) f(x)− 3 belongs to the ideal (x2 − 9).

For the solutions of Edition 8 see the solutions to Midterm Exam II below.

Algebra Homework, Edition 9Due 4 November 2010

Problem 27.Let A be the 4× 4 real matrix

A =

1 1 0 0−1 −1 0 0−2 −2 2 11 1 −1 0

(a) Determine the rational canonical form of A.(b) Determine the Jordan canonical form of A.

56

SolutionSome computation leads to the conclusion that the characteristic polynomial isx4−2x3+x2 = x2(x−1)2.By the Cayley-Hamilton Theorem, the minimal polynomial is a factor of the characteristic polynomial.There are only a few factors to consider. By plugging A into each one we discover that the only oneto annihilate A is the characteristic polynomial itself—so the minimal polynomial and the characteristicpolynomial are the same. This means that in our decomposition of the module R4

A there must be anelement that is of order x2(x − 1)2. Such an element must be an A-cyclic vector which gives of abasis for the whole four-dimensional space. With respect to this basis the matrix will be the companionmatrix of x4 − 2x3 + x2. Namely, the following matrix

0 0 0 01 0 0 00 1 0 −10 0 1 2

.That is the rational canonical form.

Now the irreducible factors of the characteristic polynomial are all of degree 1 so we know the matrix hasa Jordan canonical form with real entries. The eigenvalues are just 0 and 1, both with multiplicity 2. Soalong the diagonal we must have two consecutive 0’s and two consecutive 1’s to make the characteristicpolynomial come out right. So the only issue is where to locate the 1’s along the subdiagonal. Thereare two such spots (between the 0’s and between the 1’s). If we omit the one between the 0’s and keepthe other 1, then x2 annihilates the matrix. If we omit the 1 between the 1’s and keep the 1 betweenthe 0’s, then x2(x − 1) annihilates the matrix. If we have no 1’s on the subdiagonal then x(x − 1)annihilates the matrix. So the Jordan Form must be what remains, namely

0 0 0 01 0 0 00 0 1 00 0 1 1

.

Problem 28.Suppose that N is a 4× 4 nilpotent matrix over a field F with minimal polynomial x2. What arethe possible rational canonical forms for N?

SolutionThe characteristic polynomial has the same irreducible factors has the minimal polynomial and it musthave degree 4. So it must be x4. Now the companion matrices appearing along the diagonal in therational canonical form have corresponding polynomials and the product of those polynomials is thecharacteristic polynomial, namely x4. So the polynomials associated with the companion matrices mustbe powers of x no larger than the fourth power. Such companion matrices have 1’s along the subdiagonaland 0’s everywhere else. Because the minimal polynomial is x2 the largest of the companion matricescan only be a 2× 2 matrix. So there are just two possibilities:

0 0 0 00 0 0 00 0 0 00 0 1 0

and

0 0 0 01 0 0 00 0 0 00 0 1 0

.

57

Algebra Homework, Edition 10Due 16 November 2010

Problem 29.Let R be a nontrivial integral domain and M be an R-module. Prove the set T of torsion elementsis a submodule of M and M/T is torsion free.

SolutionWe must show that T is closed under the module operations. Suppose u, v ∈ T . Pick r, s ∈ R withr 6= 0 6= s and ru = 0 = sv. Now rs 6= 0 since R is an integral domain. But notice

(rs)(u+ v) = (rs)u+ (rs)v = (sr)u+ (rs)v = s(ru) + r(sv) = s · 0 + r · 0 = 0.So u + v ∈ T and T is closed under module addition. Notice also that 0 ∈ T since 1 6= 0 and1 · 0 = 0. Finally suppose u ∈ T and s ∈ R. Pick r ∈ R with r 6= 0 so that ru = 0. Thenr(su) = (rs)u = (sr)u = s(ru) = s · 0 = 0. So su ∈ T . Thus we can form the submodule T.

To see that M/T is torsion free, suppose that r ∈ R with r 6= 0 and u ∈M so that r (u/T ) = 0/T .We need to show that u/T = 0/T . That is, we must show u ∈ T . Now

0/T = r (u/T ) = ru/T.

So we find ru ∈ T . That is, we see that ru is a torsion element. Pick s ∈ R with s 6= 0 so thats(ru) = 0. But this means (sr)u = 0 and we know that sr 6= 0 since R is an integral domain. So wesee that u ∈ T , as desired.

Problem 30.Let R be a principal ideal domain and let T be a torsion R-module of exponent r. Prove that Thas an element of order r.

SolutionPick distinct primes p0, . . . , pn−1 ∈ R and positive integers e0, . . . , en−1 so that

r = pe00 . . . p

en−1n−1 .

For each j < n, let rj = rpj

= pe0 . . . pej−1j−1 p

ej−1j p

ej+1j+1 . . . p

en−1n−1 . Notice that r - rj for all j < n. This

allows us, for each j < n, to pick uj ∈ T so that rjuj 6= 0. Now, for each j < n, put

vj = r

pejj

uj .

Then, for each j < n, we have pejj vj = 0 but pej−1j vj 6= 0. So vj is an element of order pejj . It now

follows that v0 + · · ·+ vn−1 is an element of order r, as desired.

58

Problem 31.Prove that the sequence of invariant factors (i.e. the sequence r0, r1, . . . , rn) mentioned in theInvariant Factor Theorem is uniquely determined by the module.

SolutionOnly minor adjustments in the proof of uniqueness that we did for the Structure Theorem are needed.

Problem 32.Let M be a finitely generated R-module, where R is a principal ideal domain. Prove each of thefollowing.(a) The direct decomposition using the Invariant Factor Theorem is the one using the smallest

number of factors that are all cyclic.(b) The direct decomposition using the Elementary Divisor Theorem is the one using the largest

number of factors that are all cyclic.

SolutionFor Part (a) let us first see a simple case. Suppose M = M0⊕M1 where both M0 and M1 are cyclic.Suppose r0 in the exponent of M0 and r1 is the exponent of M1. Now factor these exponents intoprimes:

r0 = pe00 . . . p

em−1m−1

r1 = pf00 . . . p

fm−1m−1

where we further suppose that the indexing is arranged so that for some j < m we have0 ≤ fi ≤ ei for all i < j and 0 ≤ ei < fi for all i with j < i < m.

Now we knowM0 = M0[r0] = M0[pe0

0 ]⊕ · · · ⊕M0[pem−1m−1 ]

M1 = M1[r1] = M1[pf00 ]⊕ · · · ⊕M1[pfm−1

m−1 ]Let

N0 = M1[pf00 ]⊕ · · · ⊕M1[pfjj ]⊕M0[pej+1

j+1 ]⊕ · · · ⊕M0[pem−1m−1 ]

N1 = M0[pe00 ]⊕ · · · ⊕M0[pejj ]⊕M1[pfj+1

j+1 ]⊕ · · · ⊕M1[pfm−1m−1 ]

What we need to observe is that M0 ⊕M1 = N0 ⊕N1 where both N0 and N1 are cyclic and theexponent of N0 is the greatest common divisor of r0 and r1 while the exponent of N1 is the leastcommon multiple of r0 and r1. Recall that we have proven in class that M[pq] = M[p]⊕M[q] providedp and q are relatively prime. We have also noted (but left it in the hands of the students) that if u hasorder p and v has order q, where p and q are relatively prime, then u+ v has order pq. It also if w hasorder pq then qw has order p and pw has order q. This is enough to put together an inductive proofsupporting our observations.

In the end, we see that we can replace a direct sum of two nontrivial cyclic modules by a direct sum oftwo cyclic modules of orders ` and d so that d | `. Now d might be a unit, in which case we have thatthe original direct sum is itself a cyclic module. Consequently, our direct sum of two cyclic modules canbe replaced a direct sum with one or two summands that fulfills the desired divisibility conditions.

Now we just apply this a lot of times to obtain the general case. Here is how. Suppose M is the directsum of m cyclic submodules of exponents s0, s1, . . . , sm−1. Let r0 be the least common multiple of allthe si’s. By using our method above no more that m− 1 times we can replace the direct sum by one

59

with no more than m summands so that one summand, call it M0 has exponent r0 and the exponentsof of the other summands divide r0. Now consider the remaining summands, numbering fewer than m.Appeal to induction to replace their direct sum by one with no more that m − 1 summands but thatobeys the divisibility stipulations of the Invariant Factor Theorem. Put M0 together with the last directsum to obtain the desired result.

For Part (b) consider any direct decomposition of M into a direct product of nontrivial modules. Ifeach of the factor modules is indecomposable, then by the uniqueness part of the Structure Theoremthe number of factors is same as the number in the Elementary Divisor Theorem. If, on the otherhand, some of the factors are directly decomposable, then invoke the Elementary Divisor Theorem todecompose them. This increases the number of direct factors and leads, by the uniqueness part of theStructure Theorem, to the same number as in the Elementary Divisor Theorem.

Algebra Homework, Edition 11Due 2 December 2010

Problem 33.Let R be a commutative ring and let S be a subring of R so that S is Noetherian. Let a ∈ R andlet S′ be the subring of R generated by S ∪ {a}. Prove that S′ is Noetherian.

Problem 34.Let R be a commutative ring. An R-module P is said to be projective provided for all R-modulesM and N and all homomorphisms f from M onto N, if g is a homomorphism P into N, then thereis a homomorphism h from P into M so that f ◦ h = g.

Prove that every free R-module is projective.

Problem 35.Let R by a principal ideal domain and let M and N be finitely generated R-modules such thatM×M ∼= N×N. Prove M ∼= N.

Problem 36.Give an example of two 4 × 4 matrices with real entries that have the same minimal polynomialand the same characteristic polynomial but are not similar.

For the solutions of Edition 11 see the solutions to Midterm Exam III below.

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Solutions for Midterm Examination I

Problem 0. Let R be a commutative ring. Define

N := {a | an = 0 for some positive integer n}.

a. Prove that N is an ideal of R.b. Prove that N ⊆ P for every prime ideal P of R.

SolutionPart a.There was a flaw in the statement of the problem. The set N should have had the following definition:

N := {a | a ∈ R and an = 0 for some positive integer n}For this part we have only to check that N satisfies all the conditions in the definition of ideal.

Condition: 0 ∈ NSince 01 = 0 we see that 0 ∈ N .Condition: If a, b ∈ N , then a+ b ∈ N .Suppose a, b ∈ N . Pick positive numbers n and m so that an = 0 and bm = 0. We need to find apositive number k so that (a+ b)k = 0. Let us try k = m+ n. Then

(a+ b)m+n =∑

i≤m+n

(m+ n

i

)am+n−ibi

=∑i<n

(m+ n

i

)(am)an−ibi +

∑n≤i≤m+n

(m+ n

i

)am+n−ibi

=∑i<n

0an−ibi +∑

n≤i≤n+mam+n−i0

= 0

The first equality above is just the Binomial Theorem, which holds in every commutative ring.Condition: If a ∈ N and r ∈ R, then ar ∈ N .So pick a natural number n so that an = 0. Just observe that (ra)n = rnan = rn · 0 = 0, whichwitnesses that ra ∈ N .

Hence, N is an ideal. Moreover, this ideal is so important that it has its own name. It is called thenilradical of R.Part b.Suppose P is a prime ideal of R. To establish N ⊆ P , pick an arbitrary a ∈ N and prove that itbelongs to P . We know there is a positive integer n so that an = 0 and we know that 0 ∈ P sinceP is an ideal. So we know that there is a positive integer n so that an ∈ P . Let m be the least suchpositive integer. In the event that m = 1 we immediately obtain a = a1 = am ∈ P , as desired. Soconsider the possibility that m > 1. In this case, a · am−1 = am ∈ P . But we know that P is a primeideal. So either a ∈ P or am−1 ∈ P . Each of these alternatives violates the minimality of m. So wereject the possibility that m > 1. Hence a ∈ P and N ⊆ P , for every prime ideal P of R.

Problem 1. Let R be a commutative ring and let a ∈ R with an 6= 0 for every natural number n.Prove that R has an ideal P such that each of the following properties holds:

i. an /∈ P for every natural number n, andii. for all ideals I of R, if P ⊆ I and P 6= I, then an ∈ I for some natural number n.

61

SolutionThe commutative ring R and one its elements a is fixed at the outset. Moreover, we know that an 6= 0for every natural number n. We are trying to prove the existence of an ideal that has the two conditionsstipulated in the problem. An examination of the two conditions reveals that we are after an ideal thatis maximal among all the ideals satisfying condition (i). So let

F = {I | I in an ideal of R and an /∈ I for every natural number n}.What we need is to see that F has a maximal member. To invoke Zorn’s Lemma, we need to verify

that every chain in F has an upper bound in F. So let C ⊆ F be a chain. If the chain C is empty, thenthe trivial ideal {0} can serve as an upper bound. It is in F since an 6= 0 for every natural number n.So suppose that C is not empty. It is evident that I ⊆

⋃C for every I ∈ C. That is,

⋃C is an upper

bound of C. So we need to prove that⋃

C belongs to F. We proved in class that the union of anynonempty chain of ideals is always an ideal. So all that remains is to see that an /∈

⋃C for every natural

number n. For contradiction, suppose otherwise. Pick k so that ak ∈⋃C. Next pick I ∈ C so that

ak ∈ I. But C ⊆ F. So I ∈ F. This contradicts the definition of F. So we have that an /∈⋃

C. Thus⋃C ∈ F, and we have verified the hypothesis of Zorn’s Lemma. We conclude that F has a maximal

element. Such a maximal element is the P we needed.

Problem 2. Let R be a commutative ring and I be an ideal of R. Prove each of the following:a. Suppose P0 and P1 are ideals of R. If I ⊆ P0 ∪ P1, then either I ⊆ P0 or I ⊆ P1.b. Suppose P0, P1, and P2 are prime ideals of R. if I ⊆ P0 ∪P1 ∪P2, then either I ⊆ P0 or I ⊆ P1

or I ⊆ P2.

SolutionPart a.Let us assume that I ⊆ P0 ∪ P1 and that I * P0. We need only prove that I ⊆ P1. To this end,let a ∈ I. We only need to prove that a ∈ P1. Since I * P0, pick a1 ∈ I \ P0. We see thata1 ∈ I ⊆ P0 ∪ P1. So we have a1 ∈ P1. Now a+ a1 ∈ I ⊆ P0 ∪ P1. So there are two cases:Case: a+ a1 ∈ P0In this case we have, as desired, that a ∈ P1, since otherwise this puts both −a and a + a1 into P0.Since P0 is an ideal and ideals are closed under addition, we would have a1 = −a+a+a1 ∈ P0, contraryto the way a1 was chosen.Case: a+ a1 ∈ P1In this case, we have both −a1 and a + a1 in P1. This puts the sum a = −a1 + a + a1 into P1, asdesired.Part b.

This time let us assume I ⊆ P0 ∪ P1 ∪ P2. According to part (a) above, it will suffice to show thatone of the following three alternatives must hold:

I ⊆ P0 ∪ P1 or I ⊆ P0 ∪ P2 or I ⊆ P1 ∪ P2.

To do this we will also assume that I * P0 ∪P1 and that I * P0 ∪P2. So our ambition is to show thatI ⊆ P1 ∪ P2.

To this end, let a be an arbitrary element of I. We need to prove that a ∈ P1∪P2. Since I * P0∪P1,pick a2 ∈ I \ (P0 ∪ P1). Likewise, pick a1 ∈ I \ (P0 ∪ P2). Surely a1 ∈ I ∩ P1 and a2 ∈ I ∩ P2 (do yousee why?).

62

Now consider the element a+a1a2. We see that this element is in I and therefore also in P0∪P1∪P2.So we have three cases.Case: a+ a1a2 ∈ P0Were a in P0 then −a would be as well. But then we would get a1a2 = −a + a + a1a2 ∈ P0. SinceP0 is a prime ideal, this would lead to at least one of the two alternatives: a1 ∈ P0 or a2 ∈ P0, both ofwhich are impossible. So a /∈ P0. But since a ∈ I ⊆ P0 ∪ P1 ∪ P2, we conclude that a ∈ P1 ∪ P2, asdesired.Case: a+ a1a2 ∈ P1We know that a1 ∈ P1 so a1a2 ∈ P1. Therefore a ∈ P1. Hence a ∈ P1 ∪ P2, as desired.Case: a+ a1a2 ∈ P2After seeing the case above, I bet you can do this one too.

63

Solutions for Midterm Examination II

Problem 0. Let R be a commutative ring and I be an ideal of R. DefineN := {a | an ∈ I for some positive integer n}.

a. Prove that N is an ideal of R.b. Prove that N ⊆ P for every prime ideal P of R such that I ⊆ P .

SolutionFor part (a) just check the conditions that define the notion of ideal.

We see 0 ∈ N because 01 = 0 ∈ I since I is an ideal. So we can let n = 1.Suppose that a, b ∈ N . We need to prove that a + b ∈ N . Pick positive integers ` and m so that

a`, bm ∈ I. Observer that

(a+ b)`+m =∑

k≤`+m

(`+m

k

)akb`+m−k

=∑k≤`

(`+m

k

)akbm+(`−k) +

∑`<k≤`+m

(`+m

k

)a`+(k−`)b`+m−k

This works because the Binomial Theorem holds in commutative rings. In the last line the leftmost sumbelongs to I since bm ∈ I while the rightmost sum belong to I since a` ∈ I. So (a+ b)`+m ∈ I. Thismeans we can take n = `+m to conclude that a+ b ∈ N .

Finally suppose a ∈ N and r ∈ R. Pick a positive integer n so that an ∈ I. The rnan ∈ I since I isan ideal. Hence ra ∈ N .

So N is an ideal of R.For part (b), let P be a prime ideal so that I ⊆ P . To see that N ⊆ P , let a be an arbitrary element

of N . We need to show that a ∈ P . Because a ∈ N we pick a positive integer n so that an ∈ I ⊆ P .The following contention finishes this proof.Contention: For each positive integer n that following holds: if an ∈ P , then a ∈ P .

We argue this by induction on n.Base Step: Take n = 1. Nothing more needs saying.Inductive Step: Assume the implication at n and prove it as n + 1. We have an+1 = a · an ∈ P .Since P is a prime ideal either a ∈ P (as desired) or an ∈ P (in which case we appeal to the inductivehypothesis to conclude a ∈ P ). So in either case a ∈ P .

64

Problem 1. Let R be a commutative ring and let I be a finitely generated nontrivial ideal of R.Prove that R has an ideal M such that each of the following properties holds:

i. I is not a subset of M , andii. for all ideals J of R, if M ⊆ J and M 6= J , then I ⊆ J .

SolutionLet F = {J | J is an ideal of R and I is not a subset of J}. This family is not empty since the trivialideal is a member. We invoke Zorn’s Lemma to obtain a maximal member M ∈ F. To see that Zorn’sLemma applies, let C be any nonempty chain in F. We know

⋃C is an ideal. To see that it is in F we

need to prove that this union is not included in I. Were it otherwise, consider a finite generating set Xof I. For each x ∈ X pick Jx ∈ C so that x ∈ Jx. There are finitely many of those J ′s and they form achain under inclusion. This chain has a largest element—an that member is at once an ideal belongingto C ⊆ F and it must contain all of X and thus I itself. This would be a contradiction.

So F has a largest element M that satisfies (i) because it belongs to F and satifies (ii) because it isa maximal member of F.

Problem 2. Let R be a commutative ring and (p) be a principal prime ideal of R. Let P be aprime ideal of R that is properly included in (p). Prove that P ⊆ (pn) for all natural numbers n.

SolutionWe do this by induction on the natural number n.Base Step: n = 0Notice that p0 = 1. Evidently P ⊆ R = (1) = (p0).Inductive Step: Assume P ⊆ (pn)We must prove P ⊆ (pn+1). Let a be an arbitrary element of P . So a ∈ (pn) by the inductivehypothesis. So pick r ∈ R such that a = rpn. This means that rpn = a ∈ P . But P is a prime ideal soeither r ∈ P or pn ∈ P . Just as in Problem 0, the last alternative entails that p ∈ P . But P is properlyincluded in (p), so that alternative is impossible. We conclude that r ∈ P ⊆ (p). Pick s ∈ Rso thatr = sp. Then we find that a = rpn = sppn = spn+1. This means that a ∈ (pn+1). So P ⊆ (pn+1), asdesired. The induction is complete.

Problem 3. Provide an example of a unique factorization domain that is not a principal idealdomain. Prove that your example has the desired properties.

SolutionSince the ring Z integers is a principal ideal domain. So we know Z is a uniques factorization domain.Thus Z[x] is also a unique factorization domain. We will prove that (2, x) is not a principal ideal. Thiswill show that Z[x] is not a principal ideal domain.

Suppose, to the contrary, that (2, x) = (f(x)). Then we have the following constraints on thepolynomial f(x):(a) f(x) ∈ (2, x),(b) there is a polynomial, call it r(x), so that r(x)f(x) = 2, and(c) there is a polynomial, call it s(x), so that s(x)f(x) = x.

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Now it is easy (the diligent grad student does this part) to show that (2, x) consists of those polynomialsin Z[x] with even constant term. So constraint (a) tells us that the constant term of f(x) is even. NowZ and hence also Z is an integral domain. So constraint (b) says, in part, that f(x) cannot be the 0polynomial. But also, over an integral domain the degree of the product of two nonzero polynomials isjust the sum of the degrees of the factors. So constraint (b) also ensures that the degree of f(x) is 0.So combining our conclusions from constraints (a) and (b) we find that f(x) must be an even factorof 2. That is f(x) = 2 or f(x) = −2. But now constraint (c) cannot hold, since neither 2 nor −2 is afactor of x.

So in this way we see that (2, x) is not a principal ideal.

Problem 4. Prove that the polynomial x3y + x2y − xy2 + x3 + y is irreducible in Z[x, y].

SolutionWe construe Z[x, y] as Z[y][x] and write

x3y + x2y − xy2 + x3 + y = (y + 1)x3 + yx2 − y2x+ y.

Now y is an irreducible in the ring Z[y]. (The diligent grad student would explain why.) Observe thaty - (y + 1), y | y, y | −y2, and y2 - y. Moreover, our polynomial is primitive since (y + 1), y, and−y2 have no common factors apart from units. By Eisenstein’s Criteria, our polynomial is irreducible inZ[x, y]. Notice that the primitivity of the polynomial is crucial to this conclusion.

Problem 5. Let D be a subring of the field F. An element r ∈ F is said to be integral over Dprovided there is a monic polynomial f(x) ∈ D[x] such that r is a root of f(x). For example, thereal number

√2 is integral over the ring of integers since it is a root of x2 − 2.

Now suppose D is a unique factorization domain and F is its field of fractions. Prove that the setof elements of F that are integral over D coincides with D itself.

SolutionFirst observe that every element a ∈ D is integral since it is the root of the monic polynomial x − a.For the converse, suppose r ∈ F is integral over D. Let f(x) ∈ D[x] be a monic polynomial with ras a root. Then f(x) will factor in D[x] into a product of polynomials that are irreducible in D[x],since D[x] is a unique factorization domain. The element r must be a root of one of these monicirreducible factors (since D[x] is an integral domain). Thus, we do not lose generality by assuming f(x)is itself irreducible in D[x]. But we proved in class that every irreducible of positve degree in D[x] isalso irreducible in F[x] where F the field of fractions of D. But in F[x] we know that (x − r) | f(x)by an elementary result about roots. Since f(x) is monic, this must mean that f(x) = x − r. Sincef(x) ∈ D[x], we find that r ∈ D, as desired.

Problem 6. Prove that there is a polynomial f(x) ∈ R[x] such that(a) f(x)− 1 belongs to the ideal (x2 − 2x+ 1);(b) f(x)− 2 belongs to the ideal (x+ 1), and(c) f(x)− 3 belongs to the ideal (x2 − 9).

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SolutionWe see that x2 − 2x+ 1 = (x− 1)2 and x2 − 9 = (x− 3)(x+ 3). Since R[x] is a unique factorizationdomain, we see that the polynomials x2 − 2x + 1, x + 1, and x2 − 9 are relatively prime. This meansthat the following equations involving sums of ideals holds:

(x2 − 2x+ 1) + (x+ 1) = (x2 − 2x+ 1) + (x2 − 9) = (x+ 1) + (x2 − 9) = R[x].Now a direct appeal to the Chinese Remainder Theorem tells us there is f(x) ∈ R[x] such that(a) f(x) ≡ 1 mod (x2 − 2x+ 1)(b) f(x) ≡ 2 mod (x+ 1)(c) f(x) ≡ 3 mod (x2 − 9)Simply rewriting this gives our conclusion.

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Solutions to Midterm Examination III

Problem 0. You have already mastered this problem!

Problem 1. Let M be a module over a commutative ring and let X be a linearly independentsubset of M . Prove that there is a linearly independent subset Y of M so that both(a) X ⊆ Y , and(b) if Y ⊆ Z ⊆M and Y 6= Z, then Z is not linearly independent.

SolutionLet F = {Y | X ⊆ Y ⊆ M and Y is linearly independent}. What the problem asks for is a maximalmember of F. To show one exists, we invoke Zorn’s Lemma.

First we note that F is not empty since X ∈ F. Now let C be any nonempty chain of sets, eachbelonging to F. Then certainly

⋃C includes every set along the chain C. That is,

⋃C is an upper

bound for C. We contend that⋃

C ∈ F. It is clear thatX ⊆

⋃C ⊆M

sinceX ⊆ Y ⊆M

for all Y ∈ C. So it only remains to show that⋃

C is linearly independent. So pick any finitely manydistinct elements u0, u1, . . . , un−1 ∈

⋃C and any a0, . . . , an−1 for the ring of scalars so that

0 = a0u0 + a1u1 + · · ·+ an−1un−1.

Now pick Yi ∈ C, for each i < n, so that ui ∈ Yi. Since C is a chain, one of these finitely many Yi’smust include all the others. Call it Y . Now Y is linearly independent since it belongs to C ⊆ F. So wefind that

0 = a0 = a1 = · · · = an−1.

In this way, we see tha that⋃

C is linearly independent. So⋃

C ∈ F. That is, every nonempty chain inF has an upper bound in F. According to Zorn, F must have a maximal member M , as we desired.

Problem 2. You have already mastered this problem!

Problem 3. Give an example of a ring R and a prime ideal P over R so that P is not a maximalideal of R. Of course, prove your example has the required properties.

SolutionLet R = Z[x]. Let h be the map from Z[x] onto Z that sends each polynomial to its constant term. Itis straightforward to check (and hard working graduate students will) that h is a homomorphism. LetP be the kernel of h. Since Z is an integral domain, we see that P must be a prime ideal of Z[x]. SinceZ is not a field, we see that P is not a maximal ideal of Z[x].

One could also argue directly as follows. Let P is the ideal generated by x. (This is the same Pactually.) Then it is easy to see that P is comprised of those polynomial with constant term 0.

Suppose f(x)g(x) ∈ P . Then the constant term of f(x)g(x) must be 0. But the constant term off(x)g(x) is just the product of the constant term of f(x) with the constant term of g(x). Since Z isan integral domain, one of these constant terms must be 0. So one of f(x) and g(x) already belongsto P .

Now P is properly contained in the ideal of all polynomials with even constant term—this is the idealgenerated by {2, x} which is itself a proper ideal. So we see that P is not maximal.

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Problem 4. Prove that x4 − x2y2 + y is an irreducible element of the ring Z[x, y].

SolutionConsider x4 − x2y2 + y = x4 + (−y2)x2 + y has a polynomial in Z[y][x], that is as a polynomial inx with coefficients in the ring Z[y]. We know that Z[y] is a unique factorization domain, since Z isa unique factorization domain. We also see that x4 − (y2)x2 + y is a primitive polynomial in Z[y][x],and that y is irreducible in Z[y]. We can invoke Eisentstein to obtain our desired conclusion becausey - 1, y | −y2, y | y, and y2 - y.

Problem 5. Let R be the subring of the ring of rational numbers consisting of all those ratios ab

of integers with b odd.(a) Describe all the units of R.(b) Prove that the set of nonunits is the unique maximal ideal of R.(c) Prove that R is a principal ideal domain.

SolutionSince R is a subring of the field of rational numbers, the ring operations work in R in the same familiarway they work in the rationals. Now every nonzero rational number a

b is a unit. That is it has a (uniquelydetermined) multiplicative inverse b

a . So the units of R must be those nonzero rationals whose inversesbelong to R. So the units of R are just those rationals a

b where both a and b happen to be odd. Thatdoes Part (a).

Let M = {ab | a ∈ Z and a is even while b is odd}. Then M is the set of nonunits. It is verystraightforward to check (as all be hard working graduate students did) that M is an ideal of R. Ofcourse, this ideal is maximal, since any element not already in it must be a unit—and once a unit isadded the resulting ideal generated must be the whole of R. On the other hand, every proper ideal I ofR cannot have any units as elements. So I ⊆M . This means that M is the only maximal ideal. Thatdoes Part (b).

Now for Part (c). Let I be a proper nontrivial ideal (of course the ideal R is generated by the singleelement 1 and the ideal {0} is generated by 0). Observe that for any even integer a there are (uniquelydetermined) n and c so that n is a positive natural number and c is odd and a = 2nc. This means thatab = 2n

1cb . That is, every nonzero nonunit has the form 2nu where u is a unit and n is a positive natural

number. So let n be the smallest positive natural number so that 2n ∈ I. Now other than 0 everyelement of I has the form 2mu where u is a unit and m is a natural number with n ≤ m. But then justnotice that 2mu = (2m−nu)2n. In this way, we see that every nonzero element of I is a multiple of 2n.Since 0 = 0 · 2n, we see that I is generated by 2n. So every ideal is principal.

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Problem 6. Prove that there is a polynomial f(x) ∈ R[x] such that(a) f(x)− x belongs to the ideal (x2 + 2x+ 1);(b) f(x)− x2 belongs to the ideal (x− 1), and(c) f(x)− x3 belongs to the ideal (x2 − 4).

SolutionThe three polynomials x2 − 2x+ 1, x− 1, and x2 − 4 are relatively prime in R[x], has can be seen byfactoring them and appealing to unique factorization. So the Chinese Remainder Theorem says that thedesired f(x) ∈ R[x] must exist.

Problem 7. Let R be a commutative ring and let S be a subring of R so that S is Noetherian.Let a ∈ R and let S′ be the subring of R generated by S ∪ {a}. Prove that S′ is Noetherian.

SolutionSince S is Noetherian, Hilbert’s Basis Theorem tells us that S[x] is Noetherian. Let h : S[x] → R bethe (unique) homomorphism such that h(s) = s for all s ∈ S and h(x) = a. The image of S[x] underh is, evidently, the subring S′ of R generated by S ∪ {a}. In this way we see that S′ is a homomorphicimage of a Noetherian ring. Now the Correspondence Theorem tells of that the lattice (i.e. the orderedsystem) of ideals of S′ is isomorphic to the lattice of ideals of S[x] that include the kernel of h. Sincethis latter lattice cannot have a strictly ascending infinite chain of ideals, neither can the lattice of idealsof S′. So S′ is Noetherian.

Problem 8. Let R be a commutative ring. An R-module P is said to be projective provided forall R-modules M and N and all homomorphisms f from M onto N, if g is a homomorphism fromP into N, then there is a homomorphism h from P into M so that f ◦ h = g.

Prove that every free R-module is projective.

SolutionLet F be a free R-module. Let B be a basis for F. Let M and N be R-modules and let f be ahomomorphism from M onto N. Finally, let g be any homomorphism from F into N. For each b ∈ Bpick mb ∈ M so that f(mb) = g(b). This is possible since f is onto N. Let h be the (unique)homomorphism from F into M so that h(b) = mb for all b ∈ B. To check that f ◦ h = g let u be anarbitrary element of F and check each side of the equation. Since B is a basis for F we can have somefinitely many elements b0, b1, . . . , bn ∈ B and (nonzero) elements a0, a1, . . . , an ∈ R so that

u = a0b0 + a1b1 + · · ·+ anbn.

Then to see that (f ◦ h)(u) = g(u) just look at(f ◦ h)(u) = f(h(u))

= f(h(a0b0 + · · ·+ anbn))= f(a0h(b0) + · · ·+ anh(bn))= f(a0mb0 + · · ·+ anmbn)= a0f(mb0) + · · ·+ anf(mbn))= a0g(b0) + · · ·+ ang(bn)= g(a0b0 · · ·+ anbn)= g(u).

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Problem 9. Let R by a principal ideal domain and let M and N be finitely generated R-modulessuch that M×M ∼= N×N. Prove M ∼= N.

SolutionFor any finitely generated R-module Q and and q ∈ R so that q is a positive power of some primein R let d(q,Q) denote the number of cyclic factors of exponent q in a direct decomposition of Qinto directly indecomposable factors. Let d(0,Q) be the number of directly indecomposable facotorsisomorphic to the module R. The Structure Theorem asserts that this function d is well-defined andthat for any finitely generated R-modules Q and Q′ we have Q ∼= Q′ if and only if d(q,Q′) = d(q,Q′)for all q either 0 or a prime power. Evidently, for any finitely generated R-modules Q and S and anyappropriate q we have

d(q,Q× S) = d(q,Q) + d(q,S).So now just consider for any appropriate q

M×M ∼= N×Nd(q,M×M) = d(q,N×N)

2d(q,M) = 2d(q,N)d(q,M) = d(q,N)

Since this works for any q among 0 and the positive powers of primes in R, we conclude that M ∼= N.

Problem 10. Give an example of two 4× 4 matrices with real entries that have the same minimalpolynomial and the same characteristic polynomial but are not similar.

SolutionThere are, of course many examples. Here is one of the simplest. Let us take the minimal polynomialto be x2 and the characteristic polynomial to be x4. The companion matrix of x2 (= 0 + 0x+ x2) is(

0 01 0

)Considering the invariant factor formulation of the Structure Theorem and knowing that characteristicpolynomial is the product of the invariant factors, we see that there are two possibilities: x | x | x2 andx2 | x2. Now notice that the companion matrix of the monic polynomial x is the 1× 1 matrix (0). Inthis way we come to two distinct matrices in rational canonical form:

0 0 0 00 0 0 00 0 0 00 0 1 0

and

0 0 0 01 0 0 00 0 0 00 0 1 0

.By the uniquenes part of the Structure Theorem, these matrices cannot be similar but each has minimalpolynomial x2 and characteristic polynomial x4. It is interesting to notice that we could have usedelementary divisors and the Jordan canonical form and we would be obtained the same two matrices.

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Solutions to Final Examination

Problem 0. You have already mastered this problem!

Problem 1. You have already mastered this problem!

Problem 2. You have already mastered this problem!

Problem 3. You have already mastered this problem!

Problem 4. You have already mastered this problem!

Problem 5. Let F be the subring of the field of complex numbers consisting of those numbers ofthe form a + ib where a and b are rational. Let G be the subring of the field of complex numbersconsisting of those numbers of the form m+ ni where m and n are integers.(a) Describe all the units of G.(b) Prove that F is (isomorphic to) the field of fractions of G.(c) Prove that G is a principal ideal domain.[Hint: In this problem it is helpful to consider the function that sends each complex number z tozz = |z|2.]

SolutionLets define N(z) = zz = |z|2. This makes sense for any complex number. Notice N(a+ bi) = a2 + b2

when a and b are real numbers. In particular, if z ∈ G, then N(z) is a natural number. It is easy tocheck that N(zw) = N(z)N(w), which is why the function N is useful to us here.

For part (a), suppose z, w ∈ G and zw = 1. Then N(z)N(w) = N(1) = 1 and both N(z) and N(w)must be natural numbers. The only natural number that is a factor of 1 is 1 itself. So we see thatN(z) = 1 = N(w). This means that z and w must lie on the unit circle in the complex plane. Theonly member of G with this property are 1,−1, i, and −i. These are the only possibilities for the unitsof G. Evidently, each of these four is a unit, so we have found them all.

For part (b) we first establish that F is a field. It is straightforward to check that F is closed withrespect to addition and multiplication and that 0, 1 ∈ F . So F is certainly a subring. We have to showthat every nonzero element of F has a mutliplicative inverse in F . Let a+ bi ∈ F where a, b ∈ Q anda+ bi 6= 0. Notice this means a− bi 6= 0 as well. But now observe

1a+ bi

= 1a+ bi

a− bia− bi

= a− bia2 + b2 = a

a2 + b2 + −ba2 + b2 i.

So the multiplicative inverse in C of a + bi already belongs to F . So F is a field. It is evidently thesmallest subfield of C that contains G. We know from class that such a smallest subfield is isomorphicto the field of fractions of G.

For part (c) it helps to see that on the plane G consists of all the points “with integer coordinates”.For any complex number z there is usually a member of G closest to z. We let bze denote a memberof G closest to z, favoring the points of G to the left and above z to break any ties. For exampleb1

2 + 12 ie = 1 + i. For any complex number z = x + yi we see that bze must be one of the following

four members of G:bxc+ byci bxc+ dyei dxe+ dyei dxe+ byci.

But notice that |x− bxc| ≤ 12 for any real number. Similar inequalities hold for dxe. This means that

N(z − bze) ≤(1

2

)2+(1

2

)2= 1

2 .

Let I be a nontrivial ideal of G. The set {N(z) | z ∈ I and z 6= 0} is a nonempty set of positiveintegers. So it has a least element. Pick z ∈ I with z 6= 0 and N(z) as small as possible. We need to

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see that if u ∈ I then u = zw for some w ∈ G. Let us try w = buz e. At least this belongs to G. Soboth u ∈ I and buz ez ∈ I. Consequently, u− buz ez ∈ I. I contend that

N

(u−

⌊u

z

⌉z

)< N(z).

Once this is shown, given the minimality of z, we see that N(u− buz ez) = 0. This in turn implies thatu = buz ez, as desired. Here is how to establish the contention

N

(u− bu

zez)

= N

((u

z−⌊u

z

⌉)z

)= N

(u

z−⌊u

z

⌉)N(z)

≤ 12N(z)

< N(z)The ring G is called the ring of Gaussian integers.

Problem 6. You have already mastered this problem!

Problem 7. Let R and S be commutative Noetherian rings. Prove that R×S is also Noetherian.

SolutionLet I be an ideal of R × S. We will show that I is finitely generated. Suppose that (a, b) ∈ I. Since(1, 0) ∈ R× S and I is an ideal, we see that (a, 0) = (1 · a, 0 · 0) = (1, 0) · (a, b) ∈ I. This means thatfor any a ∈ R we get (a, b) ∈ I for some b ∈ S if and only if (a, 0) ∈ I. Likewise, for any b ∈ S we get(a, b) ∈ I for some a ∈ R if and only if (0, b) ∈ I.

Now let I∗ = {a | (a, 0) ∈ I} and let I∗ = {b | (0, b) ∈ I}. It is straightforward to prove that I∗ isan ideal of R and that I∗ is an ideal of S. Since these two rings are Noetherian we know that boththese ideals are finitely generated. Let the elememts a0, a1, . . . , an generate I∗ and let the elementsb0, b1, . . . , bm generate I∗. I contend that the finitely many elements

(a0, 0), (a1, 0), . . . , (an, 0), (0, b0), (0, b1), . . . , (0, bm)generate the ideal I. It is evident that each one of them belongs to I. Now let (u, v) be an arbitraryelement of I. As pointed out above, this means (u, 0) and (0, v) both belong to I. Hence u ∈ I∗ andv ∈ I∗. So pick r0, . . . , rn ∈ R and s0, . . . , sm ∈ S so that

u = r0a0 + · · ·+ rnan v = s0b0 + · · ·+ smbm.

From this we get(u, 0) = (r0, 0)(a0, 0) + · · ·+ (rn, 0)(an, 0) (0, v) = (0, s0)(0, b0) + · · ·+ (0, sm)(0, bm).

But then (u, v) = (u, 0) + (0v) is generated by the elements(a0, 0), (a1, 0), . . . , (an, 0), (0, b0), (0, b1), . . . , (0, bm).

So we conclude that these finitely many elements generate I. Consequently, R × S is Noetherian.

Problem 8. Let F and M be modules over the same ring and let F be a free module. Leth : M � F be a homomorphism from M onto F. Prove each of the following.

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(a) There is an embedding g : F � M of F into M such that h ◦ g = idF . (Here idF denotes theidentity map of the set F .)

(b) M = kerh⊕ F′, where F′ is the image of F with respect to g.

SolutionLet B be a basis for F. For each b ∈ B pick mb ∈M so that h(mb) = b. This is possible since h mapsM onto F . Since B is a basis for the free module F there is a unique homomorphism g : F→M suchthat g(b) = mb for each b ∈ B. Observe that the map h ◦ g is a homomorphism from F to F suchthat each b ∈ B is mapped to itself. By the freeness of F there is only one such homomorphism. Ofcourse, the identity map of F has this property. So it must be that h ◦ g is the indentity map on F .This entails that g must be one-to-one. So g embeds F into M. This establishes part (a).

For part (b) we need to show two things: that kerh ∩ F ′ is trivial and that every element of M canbe written as the sum of an element of kerh and an element of F ′. For the first, suppose v kerh ∩ F ′.Since v ∈ F ′ pick u ∈ F so that v = g(u). Since v ∈ kerh, we find

0 = h(v) = h(g(u)) = u

by part (a). Since u = 0 and v = g(u), we see that v = 0. In this way, we find that ker∩F ′ is trivial.For the second condition in the definition of direct sum, let w ∈M . Since w = (w−g(h(w)))+g(h(w))and g(h(w)) ∈ F ′, it will be enough for us to show that w − g(h(w)) ∈ kerh. So look at

h(w − g(h(w))) = h(w)− h(g(h(w))) = h(w)− h(w) = 0and w − g(h(w)) ∈ kerh, has desired.

Problem 9. Let M and N by finitely generated modules over the same principal ideal domain.Prove that if M×N×M ∼= N×M×N, then M ∼= N.

SolutionLet R denote the principal ideal domain. Let d(q,P) be the function defined, whenever q is a positivepower of some prime element of r or when q = 0 and whenever P is a finitely generated R-module,so that d(q,P) counts the number of directly indecomposable direct factors in any complete directdecomposition of P which are of order q (if q is the power of a prime) or which are free (if q = 0).The Structure Theorem for Finitely Generated Modules over of PID ensures that this function d existsand that finitely generated R-modules P and P′ are isomorphic if and only if d(q,P) = d(q,P) for allvalues of q.

So be know for all appropriate qd(q,M×N×M) = d(q,N×M×N).

But now considerd(q,M) = d(q,M) + d(q,N) + d(q,M)− d(q,M)− d(q,N)

= d(q,M×N×M)− d(q,M)− d(q,N)= d(q,N×M×N)− d(q,M)− d(q,N)= d(q,N) + d(q,M) + d(q,N)− d(q,M)− d(q,N)= d(qN).

So we conclude that M ∼= N.

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Problem 10. Give an example of two dissimilar 4 × 4 matrices A and B with real entries thathave all the following properties:(a) A and B have the same minimal polynomial,(b) A and B have the same characteristic polynomial, and(c) The common minimal polynomial has no real roots.

SolutionWe know that every square matrix is similar to exactly one matrix in rational canonical form. So we canrestrict our attention to matrices in ratinal canonical form. Every such matrix is comprised of companionmatrices along the diagonal whose associated polynomials are invariant factors and satisfy the familiardivisibility condition. The dimensions of the companion matrices are the same as the degrees of theassociated polynomials:

fn−1(x) | · · · | f1(x) | f0(x).Now it is easy to see that the graph of any polynomial in R[x] of odd degree must cross the X-axis.

This means that any polynomial of odd degree must have a (real) root. Constraint (c) tells us thatnone of the companion matrices involved here can have odd size.

Now the minimal polynomial is f0(x) and the characteristic polynomial is fn−1(x) . . . f1(x)f0(x).Since we are looking for two dissimilar matrices, we must have a second sequence of polynomialsgm−1(x) | · · · | g1(x) | g0(x) as well.

The constraint (a) tells us that f0(x) = g0(x). The constraint (b) tells us that fn−1(x) . . . f1(x)f0(x) =gm−1(x) . . . g1(x)g0(x). Recall that constraint (c) entails that all these polynomials have even degree.We also know that they are all monic.

Now we are in trouble. The degree of the minimal polynomial must be either 4 or 2. It cannot be4 since then it is also the characteristic polynomial and we see the m = n = 0. In other words, thosedivisibility chains only have one polynomial each. So there is only one rational canonical matrix ratherthan the two we need. On the other hand, if the degree is 2, then we see that f1(x) and g1(x) must alsobe of degree 2. This time the chains have just two polynomials apiece. Since we know f0(x) = g0(x)and f1(x)f0(x) = g1(x)g0(x), we also get f1(x) = g1(x). So again there is only one rational canonicalmatrix.

The professor must have screwed up the statement of the problem. Would it help to look at largermatrices? Surely, it cannot be done with 5 × 5 matrices, since constraint (c) would be violated. Thiswould apply to any odd-sized matrices.

How about 6 × 6? Just considering the sizes of the companion matrices, we see three schemes: thefirst giving one single 6× 6 companion matrix, the second giving one of size 2× 2 and a second of size4× 4, and the third scheme giving three companion matrices each of dimension 2× 2. Under the firstscheme the degree of the minimal polynomial will be 6, under the second it will be 4, and under thethird it will be 2. So our counterexamples, having the same minimal polynomial, must both come fromthe same scheme. So we reject the first scheme outright. In the third scheme, since all the associatedpolynomials are monic, of the same degree, and satisfy the divisibility conditions, we see that they are allthe same as the minimal polynomial. For this reason we reject the third scheme. Unfortunately, we haveto reject the remaining possibility has well, Since we know f0(x) = g0(x) and f1(x)f0(x) = g1(x)g0(x),we also get f1(x) = g1(x). So again there is only one rational canonical matrix.

How about 8× 8? Here we can succeed! Look at

x2 + 1 | x2 + 1 | (x2 + 1)2 and (x2 + 1)2 | (x2 + 1)2.

Here the minimal polynomial isx4 + 2x2 + 1 = (x2 + 1)2

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and the characteristic polynomial isx8 + 4x6 + 6x4 + 4x2 + 1 = (x2 + 1)4.

The two dissimilar 8× 8 matrices are displayed below.

0 −1 0 0 0 0 0 01 0 0 0 0 0 0 00 0 0 −1 0 0 0 00 0 1 0 0 0 0 00 0 0 0 0 0 0 −10 0 0 0 1 0 0 00 0 0 0 0 1 0 −20 0 0 0 0 0 1 0

0 0 0 −1 0 0 0 01 0 0 0 0 0 0 00 1 0 −2 0 0 0 00 0 1 0 0 0 0 00 0 0 0 0 0 0 −10 0 0 0 1 0 0 00 0 0 0 0 1 0 −20 0 0 0 0 0 1 0

.

Perhaps this demonstrates that the professor believes 4 = 8.