abstract algebra i & ii - gonzaga university chapter 0. welcome to abstract algebra stu↵”,...

Abstract Algebra I & II

Katharine Shultis

February 14, 2018

Contents

0 Welcome to Abstract Algebra 50.1 Course Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

0.1.1 “Daily” Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.1.2 Benefits of this order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.1.3 More details . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.1.4 Graded components of course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60.1.5 Grade Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.2 What is “abstract algebra”? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60.2.1 A first attempt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60.2.2 Abstract? Modern? Both? Neither? . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70.2.3 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70.2.4 Triangle Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1 Operations 9

2 What is a field? 13

3 What is a ring? 17

4 “Special” ring elements 21

5 Subrings 25

6 Ideals 29

7 Quotient Rings 33

8 The Integers 37

9 Ring Homomorphisms 41

10 Ring Isomorphism Theorems 45

11 What is a group? 47

12 Subgroups 51

13 Normal Subgroups 53

14 Quotient Groups 57

15 Group Homomorphisms 61

16 Group Isomorphism Theorems 63

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Chapter 0

Welcome to Abstract Algebra

The learning outcomes for this chapter are to...

• ...learn about the structure of the course.

• ...understand what we mean when we say “abstract algebra”.

0.1 Course Structure

0.1.1 “Daily” Structure

Approximately every pair of class days will follow the same plan: Starting at the end of class one day, I’ll givea very short lecture involving the main definitions of the next section. I will then post notes for that lecturealong with a list of problems for you to work on. Before the next class period, you’re expected to attempt allof the problems and bring your scratch work to class. Make sure that your scratch work includes commentsabout what you’re stuck on. That next class day, you’ll work in small groups to solve the problems. Wemay also have whole class discussions and I will certainly provide assistance to your groups on the problems.Hopefully by the next class day, you’re feeling good about the problems you’ve solved and can present someof them to the class. Most of that day will be spent on whole class discussions and we’ll finish up on a shortlecture for the next section. Unless specified otherwise, it is considered ACADEMIC DISHONESTY to useoutside resources on these problems. Academic dishonesty WILL NOT be tolerated.

0.1.2 Benefits of this order

This will allow you time between classes to work on those problems on your own, figure out where you’restuck, and spend your problem solving time productively. I hope that at the end of each chapter’s classroomtime, you’ll have a (mostly completed) solution for all of the required portfolio problems, leaving you withjust the task of writing them up (and possibly filling in a few minor details).

0.1.3 More details

I’m writing my own notes for the class with the assistance of several standard textbooks (and my own baseknowledge of the subject - this is, after all, a first course in my research area) and the plan is to get througha chapter every two days. In each chapter, the problems are labeled with letters, A stands for “absent bits”,meaning the stu↵ that I left out of the lecture for you to fill in on your own, B stands for “basics”, meaningthe stu↵ every student leaving the class should know, and problems labeled C and beyond are for “cool

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6 CHAPTER 0. WELCOME TO ABSTRACT ALGEBRA

stu↵”, meaning the more interesting, and possibly more challenging problems. In each chapter you shoulddo all of A and B and then choose one more letter of problems to do. It will become clear soon what I mean.

0.1.4 Graded components of course

Exam There will be one exam given during the regular semester. It is scheduled for Friday, October 20th.You should not expect this date to change. Nominally, it will have an in-class and a take-home portion,but in practice there may be much more of one than the other. The exam will be worth 25% of yourgrade.

Final Exam There will also be a cumulative final exam. As with the other exam, there will be an in-classand a take-home portion of the final exam. The in-class portion of the final exam is scheduled forThursday, December 14th from 8-10pm. The final exam will be worth 25% of your grade.

Reflection Assignments and Productive Failure There will be approximately seven written assign-ments asking you to reflect on your mathematical experiences. Each will have its own instructions andmay require you to read or watch information first. You will have about a week to complete mostassignments. One of the most important steps along the path to success is failure. As such, you willbe graded on having and describing productive failures in this class. Combined, productive failure andthe reflection assignments are worth 10% of your grade.

Portfolio The portfolio will be a compilation of your work done this semester. The problems you do in eachchapter (A, B, and one more) will compose your portfolio which will be graded for correctness (andcompleteness). You’ll turn in drafts of any part on which you’d like feedback on Wednesdays, and I’lldo my best to return your drafts with comments on Fridays. This will be done electronically, so youmay turn in drafts even on Wednesdays that we don’t have class (yes, the day before Thanksgivingtoo, although I probably won’t be as quick about returning work to you that week). The portfolio willbe worth 40% of your grade.

0.1.5 Grade Scale

I do not intend on sticking strictly to a 90-80-70 type scale for final grades. As assignments are returned, Iwill make it clear where the grade cuto↵s are. It is my intention that these cuto↵s will represent learning,not competition among your peers. I will do my best to keep you appraised of your standing in the class.Please let me know (sooner is better) if you have questions or concerns about your grade. My main goal isthat you’re learning, and earning a good grade should be a side e↵ect of doing that.

Question 0.1. What further information do you need from me?

0.2 What is “abstract algebra”?

0.2.1 A first attempt

My research is in “commutative algebra.” Here’s what I tell the Dean/AVP/random stranger on a planewhen they ask what I do/research: “You took algebra in high school, right? In that course you learnedthat addition and multiplication are associative (meaning the parentheses can be placed anywhere) andcommutative (meaning that the order doesn’t matter), and about other such properties of the real numbers,like the distributive property. Well, I look at those properties independent of the real numbers and ask whatcan be said about sets with those properties? What happens as you add in additional properties? Whathappens when you take properties away?” At this point in the conversation, most random strangers onplanes think I’m very smart (ha!) and their eyes have glazed over and I can go back to reading my book,thank you very much! For the Dean/AVP, who are trying hard to follow and understand, I might add inthat “My research specifically is about sets with both addition and multiplication, where I have most ofthe properties that you might want, except that you cannot necessarily divide; however, I do insist that

0.2. WHAT IS “ABSTRACT ALGEBRA”? 7

multiplication is commutative. I study objects (and related objects) with these properties.” Part of my goalin this course, is to make that description make sense to you.

0.2.2 Abstract? Modern? Both? Neither?

This course is titled “Abstract Algebra” and other institutions might call it “Modern Algebra.” Goodquestions here include:

1. What’s the di↵erence between abstract and modern algebra?

2. Why do we use the words abstract and modern to describe the subject?

3. Why can’t mathematicians agree on the name of the subject?

As for some short answers: 1. there is no di↵erence, 2. both adjectives (abstract and modern) might bemisnomers, and 3. I have no clue! More on the second one: the foundations of abstract algebra rely on“abstracting” the notions learned in high school/ancient algebra, where the goal is solving equations. This isa much more “modern” viewpoint than what was discovered/created before and by the ancient Greeks. Onthe flip side, what is considered abstract today, may be considered concrete in the future (ideas like negativeand complex numbers were once considered absurd and abstract, despite the fact that they’re abundant andimportant in modern STEM work). Similarly, much of modern algebra was actually discovered before theCivil War, and the goals of modern algebra were stated in the 17th century.

0.2.3 History

Etymology

The word “algebra” comes from the Arabic term al jebr which translates roughly as “reunion” and refers tothe method of combining/collecting like terms to solve algebraic equations. It was first used by MohammedKharizm who taught mathematics in Baghdad in the 9th century.

Solving polynomials

Most (possibly all) of the algebra you learned though high school was discovered no later than, and notationwas standardized in, the 16th-18th centuries. Certainly, the ancient Greeks knew how to solve quadraticequations (and that wasn’t even their discovery), but much more work was done in these centuries thanthose before or after on the subject. In this time, mathematical “duels” were common - the idea being tochallenge an opponent to solving lists of equations created by the other person. So mathematicians workedhard to be able to solve harder equations than their rivals. This is when general solutions to cubic andquartic equations were discovered. However, general polynomial equations of degree 5 and higher remaineda mystery.

The Beginnings of Modern Algebra

Fun fact: There is no formula for the solution to a general polynomial equation of degree 5 or higher! Thisdiscovery was made by Niels Abel in 1824. Around the same time, studies in other areas of mathematics ledto the study of unconventional “algebras” and were answering questions that had nothing to do with solvingequations. This leads to (among other things) matrix algebra and Boolean algebra. The new questionsbecome more about what algebras have in common rather than about solving equations. As with any areaof mathematics, setting down axioms are important. We’ll be clearly and precisely defining our algebraicobjects and then studying what objects with a given set of rules (axioms) have in common.

0.2.4 Triangle Activity

The goal of abstraction (particularly in algebra) is to understand the key characteristics without worryingabout the details. Everyone draw a triangle. They’re all di↵erent; however, they’re all still triangles. Why?

Chapter 1

Operations


• ...understand the definition of an operation.

• ...be able to give several examples of operations.

• ...understand some basic properties of operations and the consequences of those operations.

Let’s start with a definition of an operation.

Definition 1.1. A (binary) operation on a set A is a function � : A⇥A ! A.

Note that we usually use infix notation for operations. As an example, we write the sum and product of 28and 17 as 28 + 17 and 28 · 17 respectively, instead of as +((28, 17)) or ·((28, 17)). If nothing else, we do thisbecause that looks awkward, and who can remember to use four parentheses!? Moreover, for a multiplicativeoperation, we sometimes even remove the · and simply write ab for the product of a and b.

Example 1.2. You’re (presumably/hopefully) already comfortable with the operations of addition andmultiplication on C (and note that this includes R, N, Z, and Q as subsets).

Question 1.3. Actually...how do YOU KNOW that addition and multiplication are operations on C? R? Q?Z? N? Is it possible that the sum or product of two real numbers isn’t real? What about complex?

Definition 1.4. Consider a set A with an operation, denoted ?. We say the operation is associative if forevery a, b, c 2 A, then (a ? b) ? c = a ? (b ? c). We say the operation is commutative if a ? b = b ? a for everya, b 2 A.

Example 1.5. You (presumably/hopefully) already know that addition and multiplication on C are bothassociative and commutative. We should think about why this is though...at least for N.

Definition 1.6. Consider a set A with two operations, which we’ll denote here as + and ?. We say that ?distributes over + if for every a, b, c 2 A, we have a? (b+c) = a?b+a?c and (a+b)?c = a?c+b?c. Somefolks will separate these two and call these properties left distributivity and right distributivity (respectively);however, we won’t make that distinction in this course.

Example 1.7. Hopefully you’re already comfortable with the fact that multiplication distributes over ad-dition in C. Again, we should think about why this is though...at least for N.

Definition 1.8. Consider a set A with an operation we’ll denote ?. We say the operation has an identityif there is an element e 2 A such that e ? a = a ? e = a for every a 2 A. If the operation has an identity, itmight also be the case that an element, a 2 A, has an inverse. By this we mean that there is an elementa0 2 A such that a ? a0 = a0 ? a = e.

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10 CHAPTER 1. OPERATIONS

Question 1.9. Why did I say “if the operation has an identity” before defining an inverse? Must an operationhave an identity for inverses to make sense?

We sometimes write 0 for an additive identity and 1 for a multiplicative identity. When an operation isadditive, we sometimes write �a for the inverse of a (instead of a0). When the operation is multiplicative,we sometimes write a�1 or 1

a for the inverse. In C, this leads to the familiar notation b� a as shorthand forb+ (�a) and b

a for b · a�1. If a multiplication operation is associative, and n 2 N, we typically let an denotethe result of multiplying a by itself n times.

Example 1.10. We know that 1 · a = a · 1 = a and 0 + a = a + 0 = a for any a 2 C, so 1 and 0 are themultiplicative and additive identities, respectively. What elements in C have additive inverses? multiplicativeinverses?

Fact 1.11. Let X be a set with an operation, ?. If x, y, z 2 X and x = y, then x ? z = y ? z.

Exercises

(1.A) Complete the missing parts above. They’re listed below to help you out!

(i) The question posed in Question 1.3 is actually quite complex (no pun intended). Explain in wordswhy the sum and product of two natural numbers is a natural number.1 Extend your reasoningto all of the integers,2 and then use the definition of Q to show that addition and multiplicationare actually operations on Q.3 For a formal definition of R and C you’ll need to take an analysiscourse (Math 413/414/417 at GU). I suspect you believe me though when I say that the sum andproduct of two real/complex numbers is again real/complex.

(ii) Answer both questions asked in Question 1.9: Why did I say “if the operation has an identity”before defining an inverse? Must an operation have an identity for inverses to make sense?

(iii) Answer the questions from Example 1.10: Which elements in C have additive inverses, and whichhave multiplicative inverses?

(1.B) Here are some (non-)examples of operations. For each, explain briefly why it is or is not an operation.If it is an operation, determine whether or not if it satisfies the properties listed. When “inverses” islisted, you should identify which elements have inverses. Make sure to prove your answers.

(i) Addition. Identity, inverses.

(ii) Division on the set Q. Associativity, identity, inverses.

(iii) Subtraction on the set Z. Associativity, identity.

(iv) The operation a ⇤ b = aba+b+1 on the set Q. The expression on the right side of the equation is

using standard notation in Q. Commutative, associative.

(v) The operation a ⇤ b = max{a, b} on the set R. Associativity, commutativity, identity, inverses.

(vi) The set of two by two matrices with entries in R together with two operations: matrix additionand multiplication. Distributive.4

(vii) The set {a, b, c, d, e, f} together with the possible operation ? given in the table below. The entryin the row labeled a and column labeled c indicates that a ? c = f , whereas the entry in the rowlabeled c and column labeled a indicates that c ? a = d. Commutative, identity, inverses.5

? a b c d e f

a e d f b a cb f e d c b ac d f e a c bd c a b f d ee a b c d e ff b c a e f d

1I expect you’ll want to use your elementary school definitions of addition and multiplication for this.2What needs to change to include zero and the negative integers?3Here’s a definition of Q for you. It is the set

�ab | a, b 2 Z, b 6= 0, and a, b have no common factors

.

4Ask if you haven’t seen these operations before!5Notice that you probably don’t want to check this is associative - how many things to check would there be?

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(viii) Computer code is written in strings of 0’s and 1’s, for instance 0011010 and 1011 are both stringsof 0’s and 1’s. Concatenation, which we’ll temporarily denote ⇤ is the “operation” of putting twostrings adjacent to each other to form a new string. For instance, the concatenation of 0011010and 1011 is 0011010⇤1011 = 00110101011. The “operation” to check here is concatenation on theset of all strings of 0’s and 1’s. Associative, commutative, identity, inverses.

(1.C) In Definition 1.8 we insisted that an identity element commute with every other element, and that anelement commutes with its inverse. Is that part of the definition for the identity and inverse necessary?Also, are these elements unique? The more precise phrasing in problems C, D, and E should help yourexploration of these questions.

(i) Given a set A and operation ⇤, if we knew that there was an element e 2 A such that e ⇤ a = afor every a 2 A, would we necessarily know that a ⇤ e = a for every a 2 A as well?

(ii) What about the similar question for inverses?6

(1.D) In Definition 1.8 we insisted that an identity element commute with every other element, and that anelement commutes with its inverse. Is that part of the definition for the identity and inverse necessary?Also, are these elements unique? The more precise phrasing in problems C, D, and E should help yourexploration of these questions.

(i) Given a set A and operation ⇤, if we knew that there were elements e, e0 2 A such that e ⇤ a =a ⇤ e = a and e0 ⇤ a = a ⇤ e0 = a for every a 2 A, would we necessarily know that e = e0 as well?


(1.E) In Definition 1.8 we insisted that an identity element commute with every other element, and that anelement commutes with its inverse. Is that part of the definition for the identity and inverse necessary?Also, are these elements unique? The more precise phrasing in problems C, D, and E should help yourexploration of these questions.

(i) Now try taking away some of the commutativity. That is, what if we know there are elements eand e0 such that e ⇤ a = a and a ⇤ e0 = a for every a 2 A. Does this imply that e = e0?


(1.F) The table in exercise (vii) is called a Cayley table. It is a concise way of listing what the operationdoes on all ordered pairs. While such a table is impractical for operations on large sets, they’re veryuseful tools for operations on smaller sets. For the finite sets A = {a, e} and B = {a, b, e} create allpossible Cayley tables for an operation that is associative, commutative, has an identity, and for whichevery element has an inverse. How do you know when you’ve got an exhaustive list?7

(1.G) For the finite sets C = {a, b, c, e} create all possible Cayley tables for an operation that is associative,commutative, has an identity, and for which every element has an inverse. How do you know whenyou’ve got an exhaustive list?7

6This is intentionally a bit vague. Your first task here is to write the question in a concrete way so that you can answer it!7Hint: It may help to put your identity element in the first row and column. It may also help to think about what the

properties tell you about how the table MUST look. When checking for associativity (which is probably your last property toensure your tables have), you’ll need to determine how many things you need to check - try to make the list shorter by usingthe other properties.

Chapter 2

What is a field?


• ...understand the definition of a field.

• ...be able to give several examples of fields.

Let’s jump right into the definition of a field, since that’s our main goal of the chapter!

Definition 2.1. A field is a set, F , together with two operations, +, and ·, typically called addition andmultiplication respectively, such that:

(A) Both operations are associative.

(C) Both operations are commutative.

(D) Multiplication distributes over addition.

(Id) Both operations have identity elements, denoted 0 and 1 for addition and multiplication respectively.

(Inv) Every element has an inverse with respect to addition, and every element other than 0 has an inversewith respect to multiplication.

Such a field is denoted (F,+, ·), although F su�ces when the operations are clear.

Remark 2.2. Basically, we can do anything we want! So it’s easier to remember that you can do what youwant rather than trying to memorize the full list. The only thing missing is the possibility of having amultiplicative inverse for 0. Why is that?

Example 2.3. Have we already shown that C is a field? Which of R, Q, Z, and N are fields?

Example 2.4. The set {0} together with the operations defined the only way they can be, is a field. Fill inthe details for yourself!

Example 2.5. Let E denote the set of even integers and O denote the set of odd integers. We know thatthe sum of two even integers is even, the sum of two odd integers is even, and the sum of an even integerand an odd integer is odd. Letting F = {E ,O}, then we could define an operation + on F via: E + E = E ,E + O = O, O + E = O, and O + O = E . We also know that that the product of two integers is only oddwhen both integers are odd. We could hence define a second operation · on F via: E · E = E , E · O = E ,O · E = E , and O · O = O. With these operations, we can check that F is a field! Indeed,

(C) both operations are commutative. This is clear since for a, b 2 F to be distinct, it must be the casethat a = E and b = O (or visa versa). Then a+ b = b+ a and a · b = b · a by definition.

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14 CHAPTER 2. WHAT IS A FIELD?

(A) both operations are associative. In order to show this, we should go through the choices. Let a, b, c 2 F .If at least one of a, b, c is E , then a · (b · c) and (a · b) · c are both E . And in the case that each of a, b, c isO, we know that both products are also O. For addition, we consider the various cases, noting that weonly care how many of each of E and O we have since we already know the operation is commutative.

• E + (E + E) = E + E = (E + E) + E• E + (E +O) = E +O = (E + E) +O• E + (O +O) = E + E = E = O +O = (E +O) +O• O + (O +O) = O + E = O = E +O = (O +O) +O

(D) multiplication distributes over addition. Again, we know that multiplication and addition are commuta-tive, so there are fewer things to check. For any a, b 2 F we have E·(a+b) = E and E·a+E·b = E+E = E ,so we only need to check the cases where we have O · (a+ b):

• O · (E + E) = O · E = E = E + E = O · E +O · E• O · (E +O) = O · O = O = E +O = O · E +O · O• O · (O +O) = O · E = E = O +O = O · O +O · O

(Id) both operations have identity elements since E + a = a for every a 2 F and O · a = a for every a 2 F .Note here that the choices for a 2 F are O and E and that we no longer need to show commutativityas we’ve shown both operations are commutative.

(Inv) every element has an additive inverse, namely itself since E + E = E = O +O. Also, the only nonzeroelement is O and as it’s the multiplicative identity, it is its own multiplicative inverse.

There are many standard notations for this field. In this course, we’ll typically use Z/2Z. We’ll come backto both the notation and similar fields soon!

Exercises


(i) What would happen if you had a field F and 0, i.e. the additive identity, had a multiplicativeinverse? You may use the first part of problem (2.D) without proof as long as you do that problemlater; otherwise, you should include a proof here.

(ii) Determine which of C, R, Q, Z, and N are fields. Note that some properties1 are inherited from Cand others you’ll need to check are satisfied. You may assume (without proof) that the sum andproduct of two real/complex numbers is again real/complex, i.e. that addition and multiplicationare operations on R and C. You showed that addition and multiplication are operations on Q, Z,and N in Chapter 1.

(iii) Show that the “field” given in Example 2.4 is actually a field.

(2.B) Here are some (non-)examples of fields. Show that each is a field or is not a field.2

(i) The set Q together with typical addition and multiplication defined by a ⇤ b = (a+ b)(a+ b).

(ii) The set of two by two matrices with entries in Q together with typical matrix addition andmultiplication.

(iii) The set of two by two matrices with entries in Q with nonzero determinant, together with typicalmatrix addition and multiplication.

(iv) The set of two by two matrices with entries in Q with determinant 1, together with typical matrixaddition and multiplication.

(2.C) In Example 2.5, Dr. Shultis promised you more about the field Z/2Z soon. Here’s a start. You’llhave to wait longer for the explanation of the notation. This exercise simply generalizes the field andpossibly leads you to pose new and interesting questions about the field, including why we use thisnotation instead of something simpler (the Z appears twice, for Pete’s sake!)

1which ones?2Hint: To show a set with two operations is not a field, you only need to show ONE of the properties fails.

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(i) Let p � 2 be a prime integer and define a relation ⌘p on Z by a ⌘p b if and only if p | a� b. Thisis an equivalence relation, and hence partitions the set Z. Remind yourself of the meanings ofthese words and for each integer p � 2, identify how many equivalence classes there will be for therelation ⌘p. When p is clear, you’ll presumably want to just use the notation ⌘ for this relation.

(ii) You should have found that there will be p equivalence classes for the relation ⌘p. In fact, youshould recognize that you can describe the elements in a given equivalence class as all having thesame remainder when dividing by p. Make sure you understand this step before moving on aswe’ll use the classes [0], [1], . . . , [p� 1] as our field elements in the next step.3

(iii) Let Ap temporarily denote the set of equivalence classes under the equivalence relation ⌘p. Thatis, let Ap = {[0], [1], . . . , [p� 1]} and define the operations + and · on Ap via [a]+ [b] = [a+ b] and[a] · [b] = [ab] respectively. Show that Z/pZ, i.e. (Ap,+, ·) is a field. To show this is a field, wemust show several things: (a) the operations are well-defined,4 (b) the operations are associative,(c) the operations are commutative, (d) the distributive law holds, (e) both operations have anidentity, (f) every element has an additive inverse. We also need to show that every nonzeroelement has a multiplicative identity; however, that’s outside the scope of what we can do today.Let’s plan to come back to that thought later! For now, please simply find the inverses of thenonzero elements for Z/pZ for p = 2, 3, 5, 7. It may help throughout to notice that you need notuse the same representative of the class throughout. Indeed, [0] = [np] for all n 2 Z and 0 is notalways the convenient choice of representative.

(iv) Why was there the extra step of showing the operation was well-defined in the previous part?

(v) For those of you who took Fundamentals of Mathematics with Dr. Shultis last semester, hidingbehind the Z/3Z door are the “round,” “weird,” and “strange” integers we played with in chapter4. Which of [0], [1] and [2] is which set?5

(2.D) We all “know” that “zero times anything is zero” and that factoring to find the roots of a polynomialworks because the product of elements is only zero if one of the factors is zero.

(i) Let (F,+, ·) be a field whose additive identity is denoted 0. Prove that a · 0 = 0 · a = 0 for everya 2 A.

(ii) Which properties of a field did you use in your proof? Another way to phrase this (which willresult in a di↵erent answer) is which properties of a field can you remove and still have thata · 0 = 0 · a = 0 for every a 2 A.

(iii) Find a set A together with two operations + and · for which + has an identity denoted 0, but forwhich 0 · a 6= 0 for some a 2 A.

(2.E) We all “know” that factoring to find the roots of a polynomial works because the product of elementsis only zero if one of the factors is zero.

(i) Let (F,+, ·) be a field whose additive identity is denoted 0. Prove that if a, b 2 A, and ab = 0,then a or b must be 0.6

(ii) Which properties did you use in your proof?

(iii) Find a set A together with two operations + and · for which + has an identity denoted 0, but forwhich there are elements a, b 2 A \ {0} such that ab = 0.

3The notation [a] is easier to type for an equivalence class than a. However, both are commonly used, and a may be easierto handwrite. Which you use is your call!

4meaning that if [a] = [a0] and [b] = [b0] for some a, a0, b, b0 2 Z, then [a] + [b] = [a0] + [b0] and [a][b] = [a0][b0]. As a hint here,think about how you would show that a + b ⌘p a0 + b0 and ab ⌘p a0b0 and note that you’ll need to use the fact that a ⌘p a0

and b ⌘p b0.5If you’re doing this problem and didn’t take Fundamentals last semester, you should ask for those definitions from another

student or from Dr. Shultis!6Remember that in mathematics the word “or” is always meant inclusively, so it is entirely possible here that a = b = 0.

Chapter 3

What is a ring?


• ...understand the definition of a ring.

• ...be able to give several examples of rings.

• ...know some basic properties of rings.

Let’s get going right away with the definition of a ring.

Definition 3.1. A ring is a set, R, together with two operations, + and · such that:


(C) Addition is commutative.


(Id) Addition has an identity, denoted 0.

(Inv) Every element has an additive inverse.

Remark 3.2. Some people would argue with this definition of a ring. Most notably, some mathematiciansinsist that rings have a multiplicative identity, and will call a ring without such an element a rng (that’s nota typo!). My research is in commutative algebra, and so when I’m working with an arbitrary ring, it alwayshas a (multiplicative) identity and multiplication is commutative. I’ve given you this definition as it is themost general one possible; however, you should be aware that some books will use di↵erent definitions!

Remark 3.3. Since this list is a “sublist” of the one for fields, it might be easiest to remember what youaren’t allowed to do, rather than the full list: You’re not allowed to: assume multiplication is commutative,or that multiplication has an identity (and hence you are DEFINITELY not allowed to assume elementshave multiplicative inverses).

This next definition indicates some special kinds of rings. Namely, those that have the missing properties.These terms exist to make adding on a few basic properties easier. There are other properties we mightwant in rings, and that’s a large part of what ring theorists study - what does having a given property tellyou about your ring?

Definition 3.4. A ring with identity is a ring R with the additional property that:

(Id) Multiplication has an identity, denoted 1.

A commutative ring is a ring R with the additional property that:

(C) Multiplication is commutative.

17

18 CHAPTER 3. WHAT IS A RING?

A division ring is a ring R with the additional properties that:

(Id) Multiplication has an identity, denoted 1.

(Inv) Every element has a multiplicative inverse.

Question 3.5. In the same vein as Remark 3.3, what is the di↵erence between a division ring and a field?What is the di↵erence between a commutative ring with identity and a field?

Example 3.6. Since all the properties on this list are also on the list for fields, every field is automaticallya ring! In fact, one could define a field as a commutative division ring.

Example 3.7. Which of N, Z, Q, R, and C are rings? We’ve done a bunch of the legwork for this already.

Example 3.8. Since R = {0} with the operations + and · defined the only way they can be is a field, it’salso a ring. However, this is not necessarily an interesting ring.

Example 3.9. Let S be a ring and let R = S[X] be the set of all polynomials with coe�cients in S.1 Are wecomfortable adding and multiplying polynomials? Good! Then with standard addition and multiplication,R is a ring. Moreover, if S is a commutative ring or has an identity, then so does R. What actually needsto be shown here?

Example 3.10. Let S be a ring and let R = SJXK be the set of all (formal) power series with coe�cientsin S. Addition and multiplication is done “as if they were polynomials” with not even a passing thoughtabout convergence. That is, if

a =1X

n=0

anXn and b =

1X

n=0

bnXn

then

a+ b =1X

n=0

(an + bn)Xn and ab =

1X

n=0

nX

i=0

aibn�i

!Xn.

With these operations, R is a ring. What actually needs to be shown here?

Exercises

(3.A) Complete the missing parts above. They’re listed below to help you out! Both of the parts (iii) and(iv) are optional.

(i) While Q, R, and C were fields, and Z was not a field, we do have that Z is a ring (with identity).However, N is not a ring. What goes wrong with N; which properties does it fail and which doesit satisfy?

(ii) Write an arbitrary polynomial as p(X) = a0+a1X+a2X2+ · · ·+anX

n for some n 2 N. With thisnotation (or possibly concise summation notation), write down formulas for the sum and productof two polynomials.2

(iii) Show that with the standard polynomial addition and multiplication, S[X] is a ring and thatS[X] is commutative or has an identity when S has those properties.

(iv) Show that with the standard power series addition and multiplication, SJXK is a ring and thatSJXK is commutative or has an identity when S has those properties.

(3.B) Here are some (non-)examples of rings. Show that each is a ring or is not a ring. For those that arerings, identify whether or not they are (a) rings with identity, (b) commutative rings, and (c) divisionrings. Part (ii) is optional.

1It may be helpful here to pretend that S is a ring you’re already comfortable with, like Z, Q, R, or C. However, thestatement and proof will be identical for any ring S you’re imagining.

2Hint: It may help to first assume both polynomials are of the same degree, and then deal with the possibility that they havedi↵erent degrees.

19

(i) The set Q together with typical addition and multiplication defined by a ⇤ b = (a+ b) · (a+ b).

(ii) The set of two by two matrices with entries in R together with typical matrix addition andmultiplication.3

(iii) The set of two by two matrices with entries in R with nonzero determinant, together with typicalmatrix addition and multiplication.

(iv) The set of two by two matrices with entries in R with determinant 1, together with typical matrixaddition and multiplication.

(v) The set of upper triangular two by two matrices with entries in R, together with typical matrixaddition and multiplication.4

The remaining problems in this chapter are optional.

(3.C) Let’s generalize problem (2.C).5

(i) Let n � 2 be an integer and define a relation ⌘n on Z by a ⌘n b if and only if n | a� b. This isan equivalence relation, and hence partitions the set Z. Remind yourself of the meanings of thesewords and for each integer n � 2, identify how many equivalence classes there will be for therelation ⌘n. When n is clear, you’ll presumably want to just use the notation ⌘ for this relation.

(ii) You should have found that there will be n equivalence classes for the relation ⌘n. In fact, youshould recognize that you can describe the elements in a given equivalence class as all having thesame remainder when dividing by n. Make sure you understand this step before moving on aswe’ll use the classes [0], [1], . . . , [n� 1] as our ring elements in the next step.6

(iii) Temporarily let An denote the set {[0], [1], . . . , [n� 1]} and define the operations + and · on An

via [a] + [b] = [a + b] and [a] · [b] = [ab] respectively. Show that Z/nZ is a ring. To show thisis a ring, we must show several things: (a) the operations are well-defined,7 (b) the operationsare associative, (c) addition is commutative, (d) the distributive law holds, (e) addition has anidentity, (f) every element has an additive inverse.

(iv) Why was there the extra step of showing the operation was well-defined in the previous part?

(v) We know from problem (2.C) that when n is prime, this forms a field. What goes wrong when nis not prime? It may help to look at some small non-prime examples, like n = 4, 6, 8, 10.

(3.D) In this exercise we’ll explore a feature of the two by two matrices examples a bit more. You shouldhave determined that the set of two by two matrices with entries in R form a ring under standardmatrix addition and multiplication. The typical notation for this ring is M2(R). In fact, for any ringR, the ring of n⇥ n matrices with entries in R is denoted Mn(R).

(i) In problem (2.D) we showed that in a field, a · 0 = 0 · a = 0. Is that true in a ring as well?

(ii) Find two matrices, A and B, none of whose entries are zero, for which AB = 0.

(iii) Find two upper triangular matrices, A and B, neither of which is the zero matrix, for whichAB = 0.

(iv) Can you find two upper triangular matrices, none of whose entries are zero, for which AB = 0?Make sure to prove your answer.

3We saw these operations in Chapter 1

4Here upper triangular is meant to mean simply that the lower left entry is zero, so these matrices have the form

a b0 c

�

with a, b, c 2 R.5If you did problem (2.C) you should only read through this; the details haven’t really changed. If you did not do problem

(2.C) please do this one!6The notation [a] is easier to type for an equivalence class than a. However, both are commonly used, and a may be easier

to handwrite. Which you use is your call!7meaning that if [a] = [a0] and [b] = [b0] for some a, a0, b, b0 2 Z, then [a+ b] = [a0 + b0]. As a hint here, think about how you

would show that a+ b ⌘p a0 + b0 and note that you’ll need to use the fact that a ⌘p a0 and b ⌘p b0.

20 CHAPTER 3. WHAT IS A RING?

(v) In problem (2.E) we showed that in a field, if ab = 0, then a = 0 or b = 0. What property offields allowed us to do that?8

(3.E) We’re going to explore rings of functions in this exercise.

(i) Show that function composition in general is associative. That is, for sets A,B,C,D and functionsf : A ! B, g : B ! C, and h : C ! D show that h � (g � f) = (h � g) � f .9

(ii) Next, let QQ denote the set of all functions Q ! Q. Show that QQ is a ring under the operationsof function addition and composition. To show this is a ring, we must show several things: (a) ad-dition is associative10, (b) addition is commutative, (c) the distributive law holds, (d) additionhas an identity 11, and (e) every element has an additive inverse.

(iii) Is the ring QQ commutative? Does it have an identity element? Make sure to prove your results.

(iv) What characteristics about Q did you use in your proof? A similar, but di↵erently phrased,version of this question is: what can you ignore about Q and still have your proof work? Youranswer should be phrased in terms of operations and their properties as follows: “Let X be a setwith such and such operations and properties.12 Then the set of functions X ! X, denoted XX ,is a ring under the operations of function addition and composition.” Make sure to prove yourresult.

(v) Let C denote the set of continuous functions on R. Show that C is a noncommutative ring withidentity under the operations of function addition and composition.13 Can you think of othertypes of functions (probably on the set R) for which the sum of two such functions and thecomposition of two such functions is also that type of function? That is, what other rings can youfind in this way? Find at least two such rings.

Remark 3.11. A note on operations. Recall that an operation on a set A is defined as a function with domainA ⇥ A and codomain A. This means that for addition and composition on C to be operations on the set,we must have that the sum and composition of two continuous functions is continuous. So that’s the partwe’re simplifying! We’ll need to come back to this idea soon!

8Hint: It’s a property we no longer have in these rings.9Hint: What does it mean when we say two functions are equal?

10Make sure to note in your proof that the work done in the first part shows that your multiplication operation of compositionis associative.

11I haven’t told you what the identity is. Hopefully you have a first guess since you’ll need to know what it is to show itworks!

12Of course, this sentence needs to be “fleshed out.”13You may assume that the sum and composition of two continuous functions is continuous. A proof of this fact would be

content for Math 413/414.

Chapter 4

“Special” ring elements


• ...understand the definition of various special elements in rings.

• ...be able to give several examples of special ring elements.

• ...be able to describe the connections between di↵erent types of special ring elements.

Here’s a bunch of definitions involving “special” elements of rings. Examples of all of these will follow.

Definition 4.1. Let R be a ring.

• We say that a 2 R \ {0} is a zerodivisor if there is some b 2 R \ {0} such that ab = 0 or ba = 0.

• We say that a 2 R is nilpotent if there exists some n 2 N such that an = 0.

• We say that e 2 R is an idempotent if e2 = e.

• If R has an identity 1 6= 0, we say an element u 2 R is a unit if there is an element u0 2 R such thatuu0 = u0u = 1. It should come as no surprise that we say u0 is the inverse of the unit u.

• If R is a commutative ring with identity 1 6= 0 and no zerodivisors, we say R is an integral domain.

• If R is an integral domain, we say p 2 R is prime if p | ab implies1 p | a or p | b.• If R is an integral domain, we say p 2 R is irreducible if p is not a unit and p = ab implies a or b isa unit.

Remark 4.2. Note several things:

• It would be silly to include zero as a zerodivisor. Recall that 0a = a0 = 0 for every a 2 R.

• Every nilpotent element is a zerodivisor.

• In a ring with identity 1 6= 0, an element cannot simultaneously be a unit and a zerodivisor. Why?

• In any ring with identity, 1 is both an idempotent and a unit. In any ring, 0 is an idempotent.

• Every nonzero element in a field is a unit.

Example 4.3. The integers are an integral domain, as are Z[X] and ZJXK.1Note that our definition of divides extends to a more general ring, R: For a, b 2 R with a 6= 0, we say a divides b if there

is some c 2 R such that b = ac.

21

22 CHAPTER 4. “SPECIAL” RING ELEMENTS

Example 4.4. Consider the ring M2(R). In Exercise (3.D), you found some zerodivisors in this ring. If you

didn’t do that problem, here’s some:

1 00 0

�and

0 01 0

�. And here’s a more interesting pair:

1 23 6

�

and

3 �115 �5

�. Is

1 23 4

�a zerodivisor?

Example 4.5. Consider the ring Z/8Z. Here, [2] 6= [0]. However,

[2]3 = [2] · [2] · [2] = [2 · 2 · 2] = [8] = [0]

so [2] is nilpotent. Also, [3] 6= [1] is a unit. Indeed, [3] · [3] = [9] = [1]. So more is true: the element [3]is its own inverse in this ring. What other elements of Z/8Z are units? What other elements of Z/8Z arenilpotent?

Example 4.6. Let X be a set and let R = P(X) be the power set of X. Define addition and multiplicationon R via A+B = A4B = (A \B)[ (B \A) and A ·B = A\B for all A,B 2 R. It is a fact that with theseoperations, R is a ring.2 In this ring, every element is an idempotent since A ·A = A \A = A for every setA. Thus, we know that there exist nontrivial idempotents.

Exercises


(i) Let R be a ring with identity 1 6= 0 and a 2 R. Assume that a is both a unit and a zerodivisor,that is, there are elements a0, b 2 R such that aa0 = a0a = 1 and ab = 0 or ba = 0. Show thateither possibility (ab = 0 or ba = 0) results in a contradiction. Your final product here should bea proof (by contradiction) that no element of a ring with identity 1 6= 0 can be simultaneously aunit and a zerodivisor.

(ii) Is A =

1 23 4

�a zerodivisor?3

(iii) Which elements of Z/8Z are units? Which are nilpotent? Which are zerodivisors? Which areidempotents? Are there non-nilpotent zerodivisors in Z/8Z? Are there any interesting idempo-tents?

(4.B) Here are some fairly straight-forward things to prove about elements in rings. Throughout R is a ring.

(i) For any a, b 2 R, we have (�a)b = a(�b) = �(ab).

(ii) For any a, b 2 R, we have (�a)(�b) = ab.

(iii) In a ring with identity, �a = (�1)a for every a 2 R.

(iv) In a ring with identity, (�1)2 = 1.

(4.C) Let R be an integral domain. Prove that if p 2 R is prime, then p is irreducible. As a hint, this shouldfollow from the definitions: prime, divides, and integral domain together (and probably in that order)give that p is irreducible.

(4.D) Let R be a ring such that every element of R is an idempotent. Prove that a+ a = 0 for every a 2 R.As a hint, try fixing some a 2 R and looking at the square of a+a. Use the definition of an idempotentand some algebraic manipulations to show that a+ a = 0.

2The details require tedious set proofs, which I’d like to spare you from doing (unless you really want to, in which case,assuming you did (2.C) and therefore don’t need to do (3.C), you may choose to prove this is a ring for a third problem in therings chapter in addition to (3.A) and (3.B) - let’s call it (3.F)).

3It is possible that the easiest strategy is to assume there is some matrix B =

a bc d

�such that AB = 0 and work to see

what this implies about a, b, c, d and then repeat that with the similar assumption about BA = 0.

23

(4.E) Let R be a commutative ring and a, b 2 R be nilpotent elements. Show that a+ b is also nilpotent. Asa hint, try looking at smaller examples. For instance, if you know a2 = 0 and b3 = 0, can you find apower n such that (a+b)n = 0. Also, you may use (although it may or may not help you) the binomialformula for expressing powers of a sum:

(a+ b)n =nX

k=0

✓nk

◆akbn�k

where

✓nk

◆=

n!

k!(n� k)!.

Chapter 5

Subrings


• ...understand the definition of a subring.

• ...be able to give several examples of subrings.

• ...understand and know how to use the First and Second Subring Tests.

As is typical in algebra (and I’ll try to ensure you get this “joke” by the end of the semester), one definitionof a subring is that it is a SUB-RING. That is, it is a subset of a ring that is itself a ring. Here’s a moreformal definition.

Definition 5.1. Let (R,+, ·) be a ring. A set S is a subring of R provided that

(S) The set S is a (nonempty) subset of R.

(Cl) The sum and product of two elements in S is again an element of S. That is, if a, b 2 S, then so isa+ b and a · b.

(R) The set S together with the operations + and · is a ring. That is, it satisfies the axioms of a ring:


(C) Addition is commutative.


(Id) Addition has an identity, denoted 0.

(Inv) Every element has an additive inverse.

Question 5.2. Why is the word “nonempty” in parentheses? That is, is it a necessary part of the definition,or is it a consequence of everything else in the definition?

Example 5.3. Let R be a ring. Then, {0} and R are subrings of R. Is this clear?

Remark 5.4. The second item here is typically referred to as closure. We say that a set X is closed underan operation ? if x ? y 2 X for every x, y 2 X. We haven’t needed to remark on this yet since our operationswere originally defined as functions � : A⇥ A ! A. However, I did make awkward remarks in the exercisesfor Chapter 3 about how the sum and composition of two continuous functions is again continuous. Anotherway to say some of the stu↵ in Exercise (3.E) is that C is a subring of the set of all functions from R to R.Remark 5.5. Whew - that’s a long list! Good news: we can actually shorten the list.

Theorem 5.6 (First Subring Test). A nonempty subset S of a ring R is a subring provided that for all

a, b 2 S, the elements ab, a+ b and �a are also in S.

25

26 CHAPTER 5. SUBRINGS

Proof. To prove this, we must show that S satisfies ALL of the axioms of a subring:

(S) The statement includes that S is a nonempty subset of R.

(Cl) The statement includes that S is closed under addition and multiplication.

(R) Let’s go through the list:

(A) Associativity doesn’t change as we look at a smaller set. More formally, if a, b, c 2 S, thena, b, c 2 R as S ✓ R and so a+ (b+ c) = (a+ b) + c and a(bc) = (ab)c.

(C) Neither does commutativity.

(D) Nor distributivity.

(Id) As S is nonempty, we can say that there is some element, a 2 S. By the assumptions, �a 2 S,and then 0 = a+(�a) 2 S as well as S is closed under addition. So the additive identity elementis in S, and its properties don’t change as we look at a smaller set.

(Inv) The statement includes that S is closed under additive inverses.

Since we’ve shown all of the properties of a subring hold, we know that S ✓ R is a subring.

Theorem 5.7 (Second Subring Test). A nonempty subset S of a ring R is a subring provided that for all

a, b 2 S, a� b, ab 2 S.

Example 5.8. We have that C ◆ R ◆ Q ◆ Z are all rings, so Z,Q,R are all subrings of C, etc. Also, wehave that C (the ring of all continuous functions from R to R) is a subring of RR (the ring of all functions Rto R). For more on this, see Remark 5.4.

Example 5.9. Consider the set of even integers, which we’ll denote as 2Z here. This notation means thatthe set contains 2 times every integer, which is precisely the set of even integers. This is a commutativering (although it doesn’t have an identity). Why? Well, by the Second Subring Test, we simply need toshow that the product, and di↵erence of any even integers is again even. I bet you ALL did that in yourFundamentals class, so I’m not going to have you do it again. If you haven’t, you should though.

Question 5.10. Which rings from Chapter 3 could you show are rings with less work using either Theorem 5.6or Theorem 5.7?

Exercises


(i) Answer Question 5.2 as succinctly as you can.

(ii) Explain (briefly) the answer to the question in Example 5.3. That is, explain why {0} and R arealways subrings of a given ring R.

(iii) Prove Theorem 5.7. As a hint, use the First Subring Test to make your work easier.

(iv) If you aren’t comfortable with the fact that the product and di↵erence of two even integers isagain even, prove those two facts.1

(v) Go back to your Chapter 3 exercises and simplify any proofs that you can using either Theorem 5.6or Theorem 5.7.2

(5.B) Here are some (non-)examples of subrings, along with questions to answer. You may wish to appealto work done earlier. As a side note about LATEX, if you’d like to learn to make more complex LATEXdocuments with chapters in separate files and internal references created automatically, I’d be happyto share the TEX code for the first few chapters with you, and can help explain it as well. Just holler!

1I suspect most of you will skip this.2This exercise should not appear explicitly in your portfolio.

27

(i) Generalizing Example 5.9 is fairly easy. Let nZ denote the set of all integers which are multiplesof n. Equivalently (and you should think about why this is equivalent), nZ can be defined as theset that results from multiplying all integers by n. Is nZ a subring of Z for all n 2 Z? For whichvalues of m and n is mZ ✓ nZ? When containment holds, is mZ a subring of nZ?

(ii) We know that the set of all two-by-two matrices with entries in R is a non-commutative ring withidentity, and that it’s denoted M2(R). Find at least three subrings of this ring and three subsetswhich are not subrings. Please ensure that your examples are su�ciently di↵erent to be of someinterest.

(iii) Let K be a field3. Is K[X] a subring of KJXK?

(5.C) Actually, let’s skip this part in this chapter. It was originally combined with the next chapter, andwas split into two parts to avoid having a longer lecture.

3It may help to have a few fields in mind when this is said. Good go-to examples include Q, R, and C, also Z/pZ with pprime.

Chapter 6

Ideals


• ...understand the definition of an ideal.

• ...be able to give several examples of ideals.

I hope everyone knows which is left and which is right. If you’re struggling to remember, your left handmakes an “L” when you hold out your pointer finger and thumb. If you’re weird like me, you may use thistrick in reverse to remember that the bottom of an “L” heads right and that “ L” is not correct.

Definition 6.1. Let R be a ring and I a subset of R. We say that I is a left ideal of R if I is a subring of Rand I is closed under left multiplication by elements of R. That is, if ra 2 I for every r 2 R and a 2 I. Wesay that I is a right ideal of R if I is a subring of R and I is closed under right multiplication by elementsof R. That is, if ar 2 I for every r 2 R and a 2 I. We say that I is an ideal of R (or two-sided ideal foremphasis) if I is both a left and right ideal of R.

Remark 6.2. Something that helped me when I was first learning about ideals was to think of the elementsin an ideal as having “cooties”. When an element in the ring is touched (multiplied) by an element of theideal, the result is an element with “cooties.” For a left ideal, the elements have cooties on their left side,and for a right ideal, they have cooties on their right side. Elements in a two-sided ideal have cooties onboth sides.

Remark 6.3. Note that if R is a commutative ring, these three notions are all the same. Also, most of oure↵orts (past a definition and a few examples) will be on two-sided ideals.

Example 6.4. For any ring, R, both R and {0} are ideals of R. Why? These ideals are considered “trivial”in the sense that they are ideals of EVERY ring, so they’re always there. When we’re looking for “non-trivial”ideals, we’re hoping to find more than just these two.

Theorem 6.5 (Corn, J.). Let I be a subring of R. Then I is an ideal of R if and only if I is both a left and

right ideal of R.

Theorem 6.6 (Ideal Test). Let I be a nonempty subset of a ring R. The set I is an ideal if for any a, b 2 Iand r 2 R we have a� b 2 I, ar 2 I and ra 2 I.

Proof. Let I be a nonempty subset of a ring R and assume that for any a, b 2 I and r 2 R, we havea � b, ar, ra 2 I. To show this is an ideal, we need that I is a subring of R and that ar, ra 2 I for everya 2 I and r 2 R. So the only missing piece is showing it is a subring. We’ll do that with Theorem 5.7, i.e.the Second Subring Test. For all a, b 2 I, we were given that a� b 2 I. Also, as b 2 I ✓ R, we know ab 2 I.Hence, I is a subring of R by the Second Subring Test and so I is an ideal of R.1

1This is WAY too many words for such a straightforward proof.

29

30 CHAPTER 6. IDEALS

Example 6.7. Note that nZ is an ideal of the commutative ring Z. We already showed that nZ was asubring of Z, so we just need to show the multiplicative “cooties” property. Note that for any a 2 nZ, wemay write a = bn for some b 2 Z. Then, for any c 2 Z, we have ac = (bn)c = (bc)n and bc 2 Z, so ac 2 nZ.

Definition 6.8. Let R be a ring and a1, . . . , an 2 R. The ideal generated by a1, . . . , an is denoted(a1, . . . , an) and is the smallest ideal of R containing the set {a1, . . . , an}.Remark 6.9. Similar definitions can be made for the left and right ideals generated by a set of elements. Letme know if I should be more explicit here.

Question 6.10. Let R be a ring and a1, . . . , an 2 R. Can you write down a formula for an arbitrary elementof the ideal generated by {a1, . . . , an}?

Example 6.11. The left ideal ofM2(R) generated by

1 00 0

�is the set of all matrices of the form

a 0c 0

�

with a, c 2 R. What details need to be checked to show this?

Example 6.12. The ideal (x2) of Z[x] consists of all polynomials with integer coe�cients whose constantand linear terms are zero. Note that in all such polynomials, we can factor out an x2.

Example 6.13. The polynomials in the ideal (2, x2) of Z[x] are all those whose constant and linear termshave even coe�cients. Why? Note that this means that (x2) ( (2, x2). Is it always the case that

(a1, . . . , an) ( (a1, . . . , an, an+1)?

Theorem 6.14. Let I be an ideal of a ring R. For a, b 2 R, define a relation ⇠ on R by a ⇠ b if and only

if a� b 2 I. This is an equivalence relation.

Remark 6.15. Note that as an immediate consequence of Theorem 6.14, the equivalence classes

[a] = a = {b 2 R : a� b 2 I}

form a partition of the ring R.

Exercises


(i) Explain briefly (without a formal proof) why R and {0} are ideals of any ring R. As a notationalremark, as a subring of any ring R, the set containing only 0 is actually the ideal generated by 0,so we can denote this either by (0) or by {0}. The first is typically considered easier to write andtype. Some people get even lazier and write this ring as simply 0.

(ii) Let R be a ring. Write down a formula for an arbitrary element of the ideal generated by{a1, . . . , an} with a1, . . . , an 2 R. It may help to first use small examples (say n = 2 or 3) andthen generalize your work. No formal proof is required here.

(iii) Prove that the left ideal of M2(R) generated by

1 00 0

�is the set of matrices in M2(R) of the

form

a 0c 0

�with a, c 2 R.

(iv) Prove that the ideal (2, x2) of Z[x] consists of all polynomials with integer coe�cients whoseconstant and linear terms have even coe�cients. It may help to write a generic polynomial asa0 + a1x+ a2x

2 + · · ·+ anxn and show that a0, a1 are even.

(v) Let a1, . . . , an, an+1 be elements of a ring. Explain why the containment (a1, . . . , an) ( (a1, . . . , an, an+1)does not always hold and (a1, . . . , an) ✓ (a1, . . . , an, an+1) does hold.

(vi) Prove Theorem 6.14.

(6.B) Here are some (non-)examples and basic facts about ideals. Each problem has its own directions.

31

(i) Let R be a ring with identity. Show that (1) = R.

(ii) Let F be a field (and hence a ring). Show that the only ideals of F are (0) and F .

(iii) We know that Z ( Q ( R ( C are all rings. Are any of these ideals of eachother?

(iv) Go back to the list of subrings of M2(R) that you created for Exercise (5.B)(ii). Identify whichof those subrings are left, right, or two-sided ideals and which are not.

(6.C) Actually, let’s skip this part in this chapter. It was originally combined with the previous chapter, andwas split into two parts to avoid having a longer lecture.

Chapter 7

Quotient Rings


• ...understand the definition of a quotient ring.

• ...be able to give several examples of quotient rings.

• ...be able to explain why quotient rings are rings.

• ...understand why ideals are the necessary type of subring to “mod out” by.

In the last chapter, we learned about ideals of a ring. These are special subrings which allow us to “modout” or “factor out” by them to obtain new rings. Obtaining a longer list of rings and ways to create newrings is helpful if your goal is to understand rings in general. Let’s get going with the definitions.

Definition 7.1. LetR be a ring and I an ideal ofR. The cosets of I are the equivalence classes defined by theequivalence relation from Theorem 6.14. That is, the cosets are the sets r+ I = [r] = r = {s 2 R | r� s 2 I}for r 2 R. For a coset [r] of I, we say that r is a coset representative of [r].

Remark 7.2. Remember that since I is an ideal of R, we know that the cosets partition the ring R. So ifr1 + I = r2 + I, then r1 � r2 2 I.

Remark 7.3. Note that the notation r + I for the coset of r in R/I is actually sort of brilliant as the set ofelements r+ I can be described as those that can be written as r+ a with a 2 I. Here’s justification of that,which I’ve written down in class a couple of times now:

r + I = {s 2 R | s ⇠ r}= {s 2 R | s� r 2 I}= {s 2 R | s� r = a for some a 2 I}= {s 2 R | s = r + a for some a 2 I}= {r + a | a 2 I}.

Theorem 7.4. Let R be a ring, I an ideal of R, and define addition and multiplication on the cosets of Ivia (r + I) · (s+ I) = rs+ I and (r + I) + (s+ I) = (r + s) + I.1 With these operations, the set of cosets is

a ring, denoted R/I and pronounced “R mod I.”

Proof. We need to show first that these operations are well-defined, and then go through the list of conditionsfor a set with two operations to be a ring. We’ll do some of this together and leave parts for the portfolio.

1If you prefer the equivalence class notation, this is [r] · [s] = [rs] and [r] + [s] = [r + s] or r · s = rs and r + s = r + s.

33

34 CHAPTER 7. QUOTIENT RINGS

First, suppose [r] = [s] and [a] = [b]. In order to show that the operations are well-defined, we must showthat [r + a] = [s+ b] and [ra] = [sb]. As [r] = [s] and [a] = [b], we know that r � s, a� b 2 I. Then,

(r + a)� (s+ b) = (r + a)� s� b distributive

= (r � s) + (a� b) associative and commutative

which is an element of I as I is closed under addition and r� s, a� b 2 I. Hence, [r+a] = [s+ b] so additionis well-defined. Similarly,

(ra)� (sb) = ra� sa+ sa� sb adding a “fancy” 0

= (r � s)a+ s(a� b) distributive

which is an element of I as r � s, a � b 2 I and I is closed under addition and under multiplication byelements of R. Thus, both operations are well-defined, i.e. are actually operations and we can start the restof our proof. Here, we need to run through the axioms:

(A) I’ll leave this one for you.

(C) This one too.

(D) Let’s do this one together. Let [a], [b], [c] be elements of R/I, i.e. cosets of I. Then,

[a] · ([b] + [c]) = [a] · [b+ c] by definition of addition in R/I

= [a(b+ c)] by definition of multiplication in R/I

= [ab+ ac] because the distributive law holds in R

= [ab] + [ac] by definition of addition in R/I

= [a] · [b] + [a] · [c] by definition of multiplication in R/I.

A similar argument holds to show that ([a] + [b]) · [c] = [a] · [c] + [b] · [c].(Id) Consider the element [0] 2 R/I where 0 is the additive identity of R. For any [a] 2 R/I, we have

[a] + [0] = [a+ 0] = [a] as a+ 0 = a in R. Also, addition is commutative, so [a] + [0] = [0] + [a] for all[a] 2 R/I. Thus, [0] is the additive identity in R/I.

(Inv) I’ll leave this for you to do too.

Example 7.5. We can do this with any ring and any ideal of that ring. For instance, in Q[X,Y ]/(X2, Y 2),the coset representatives all look like a = a0 + a1X + a2Y + a3XY + (X2, Y 2) with ai 2 Q. To add, we seethat

a+ b = [a0 + a1X + a2Y + a3XY + (X2, Y 2)] + [b0 + b1X + b2Y + b3XY + (X2, Y 2)]

= (a0 + b0) + (a1 + b1)X + (a2 + b2)Y + (a3 + b3)XY + (X2, Y 2).

For multiplication,

a · b = [a0 + a1X + a2Y + a3XY + (X2, Y 2)] · [b0 + b1X + b2Y + b3XY + (X2, Y 2)]

= (a0b0) + (a0b1 + a1b0)X + (a0b2 + a2b0)Y + (a0b3 + a1b2 + a2b1 + a3b0)XY + (X2, Y 2).

Example 7.6. Certainly R[X]/(X2 + 1) is a quotient ring. What do its coset representatives look like?What does addition and multiplication look like?

Exercises


35

(i) Show the remaining parts of Theorem 7.4: Both operations are associative, addition is commuta-tive, and every element has an additive inverse.

(ii) State what a generic coset representative of R[x]/(x2 + 1) looks like. Then, use that notation togive the sum and product of two generic coset representatives in that form.

(7.B) Here are some basics of quotient rings. Each problem has its own directions.

(i) What if I is simply a subring of R and not an ideal? What could go wrong? Is R/I a ring inthat case? Here’s a strategy for answering these questions: Find where in the proof that R/I is aring we used the fact that I is an ideal, and then look at that part for an example of a ring anda subring which is not an ideal.

(ii) Use the work we’ve done in Chapter 6 and Chapter 7 to explain as succinctly as you can why Z/nZis a ring. While this is a MUCH simpler argument than the one you gave in Chapter 2 or Chapter 3you MAY NOT go back and simplify your work there using this. However, understanding thework done there may now be easier and you may be able to improve your proof with this newfoundunderstanding.

(iii) What do the coset representatives for Z[X]/I with I = (X2) and I = (2, X2) look like?

(7.C) Let M2(Z) denote the ring of two by two matrices with integer entries and let 2M2(Z) denote the setof two by two matrices with even integer entries. It is a fact (that you may use without proof) that

2M2(Z) is an ideal of M2(Z). In fact, it is the ideal generated by

2 00 0

�,

0 20 0

�,

0 02 0

�, and

0 00 2

�. What do the elements of the quotient ring M2(Z)/2M2(Z) look like? Are they the same or

di↵erent from the elements of the matrix ring M2(Z/2Z)?

(7.D) In Exercise (6.A)(i) you explained why R and (0) were both ideals of any ring R. Convince yourselfthat R/R looks an awful lot like the ring {0} and that R/(0) looks an awful lot like the ring R. Describethe di↵erence between R/R and {0} and the di↵erence between R/(0) and R. This is not expected tobe a formal proof. We’ll come back to this idea soon.

Chapter 8

The Integers


• ...understand the statements and proofs of the Division Theorem and Baby Bezout’s Identity.

• ...be able to use the ideas in the proofs of the Division Theorem and Baby Bezout’s Identity to computenot only inverses in Z/pZ, but also quotients and remainders for various division problems in N.

The main goal of this chapter is to understand why Z/pZ is a field when p is prime. Recall that p 2 Z isprime if the only positive factors of p are 1 and p. Along the way, we’ll use that any nonempty finite setof integers has both a maximum and a minimum. This is a consequence of the integers being well-ordered.I don’t want to get bogged down in those details, and since we’ve already explained why the operationmax{a, b} on R is associative and commutative, we’ll stop dwelling on this detail now. We’ll need a coupleof named theorems. But first, a “motivating” example!

Example 8.1. Consider a = 35 and b = 3. To divide 35 by 3, we ask ourselves how many groups of 3paperclips we can make with 35 paperclips and how many extra paperclips we’ll have. Since 11 groups of 3paperclips is 33 paperclips and 12 groups would need 36 paperclips (which we don’t have), we say that that35 divided by 3 is 11 and that there are 2 paperclips leftover (35�33 = 2). Notice that the number of leftoverpaperclips is non-negative and less than the number of paperclips we want in each group of paperclips.

Theorem 8.2 (Division Theorem). Let a, b 2 Z with b > 0 and a � 0. There exist integers q, r � 0 with

r < b such that a = bq + r.

Proof. Consider the set S = {x 2 Z | 0 bx a}. We claim that S is a nonempty finite set. Indeed,T = {y 2 Z | 0 y a} is a finite set (it has a + 1 elements), and S ✓ T . Additionally, 0 2 S, so S isnonempty. As S is a nonempty finite set of integers, maxS exists. Set q = maxS and r = a� bq. As q 2 S,we know bq a and so r � 0. It thus su�ces to show r < b, which we’ll do by contradiction.

So suppose for a contradiction that r � b. Then,

a = bq + r by definition of q, r

= bq + b� b+ r by adding 0

= b(q + 1) + (r � b) by distributivity and additive associativity and commutativity in Z.

Since r � b, we also know r� b � 0, giving that q+1 2 S. As q was defined to be the largest element of theset S, this is a contradiction. Hence, r < b, which completes the proof.

Definition 8.3. In the above theorem, q is called the quotient and r is the remainder.

Example 8.4. We’ll do Example 8.1 again using this new notation. We have S = {x 2 Z | 0 3x 35} ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} since 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 are all the integer multiples of threethat are non-negative and at most 35. So maxS = 11 = q and r = 35� 3 · 11 = 2. Note that 0 r < 3.

37

38 CHAPTER 8. THE INTEGERS

Example 8.5. Consider a = 35 and b = 5. Then S = {x 2 Z | 0 5x 35} = {0, 1, 2, 3, 4, 5, 6, 7}since 0, 5, 10, 15, 20, 25, 30, 35 are all the integer multiples of five that are non-negative and at most 35. SomaxS = 7 = q and r = 35� 5 · 7 = 0. Note that 0 r < 5.

Theorem 8.6 (Baby Bezout’s Identity). Let a, p 2 N with p prime and a < p. There exist integers x, y 2 Zsuch that ax+ py = 1.

Proof. Consider the set S = {ax+py | x, y 2 Z, 0 < ax+py p}. The set S is nonempty as p = a·0+p·1 2 S.Additionally, S is a finite set as it is a subset of the set T = {z 2 Z | 0 < z p} which has p+ 1 elements.As S is a nonempty finite set of integers, it has a smallest element. Set d = minS. We will show that d = 1.The first step in doing so is to show that d | s for all s 2 S, which we’ll do by contradiction.

So, assume s 2 S and d - s. As s, d 2 S, there exist integers x, y, x0, y0 such that

s = ax+ py and d = ax0 + py0. (8.1)

Note that s, d > 0. By Theorem 8.2 there exist integers q, r � 0 with r < d such that s = dq+ r. Since d - s,we must also have that r > 0. Rewriting the expression s = dq + r and using Equation 8.1, we then havethat

r = s� dq = (ax+ py)� (ax0 + py0)q = a(x� x0q) + p(y � y0q).

Note that x� x0q and y� y0q are both integers and that 0 < r < d < p. Thus, r 2 S, a contradiction as d isthe smallest element in S. This shows that d | s for all s 2 S.

Next, note that a = a · 1+ p · 0 2 S as well. So we know that d | a and d | p. As a < p we must have thatd < p and so by definition of prime, we know that d = 1.

We can generalize this quite a bit. With a careful definition of the greatest common divisor of two positiveintegers a and b, there exist integers x and y such that ax + by is the greatest common divisor of a and b.This more general version is called Bezout’s Identity and can be proven in a similar manner. The bad newsis that the proof of Bezout’s Identity does not enlighten us on how to actually find the integers x and y suchthat ax+ by = gcd(a, b). Finding the greatest common divisor of two positive integers can be done using theDivision Theorem repeatedly, in a process called the Division Algorithm. As our main goal in this chapteris to prove that Z/pZ is a field when p is prime, we’ll avoid getting excessively interested in the integers andinstead focus on said goal.

Theorem 8.7. When p 2 Z is prime, the ring Z/pZ is a field.

Proof. Note that as we know Z/pZ is a ring, it only remains to be shown that Z/pZ is commutative, has amultiplicative identity, and every nonzero element has a multiplicative inverse. We’ll start by showing R/Iis a commutative ring or has identity for any commutative ring or ring with identity R and ideal I of R. Leta+ I, b+ I 2 R/I. Then,

(a+ I)(b+ I) = ab+ I by definition of multiplication in R/I

= ba+ I as R is commutative

= (b+ I)(a+ I) by definition of multiplication in R/I

so that R/I is indeed a commutative ring. Let 1 denote the multiplicative identity in a ring R and a+I 2 R/I.Then,

(1 + I)(a+ I) = 1 · a+ I by definition of multiplication in R/I

= a+ I as 1 is the multiplicative identity of R

= a · 1 + I as 1 is the multiplicative identity of R

= (a+ I)(1 + I) by definition of multiplication in R/I

so that 1 + I is the multiplicative identity of R/I.

39

What remains is to show that every nonzero element of Z/pZ has a multiplicative inverse, so let [a] 2 Z/pZbe nonzero and note that we may assume 0 < a < p as every nonzero coset has a representative in thatrange. By Theorem 8.6 there exist integers x, y such that ax+ py = 1. Note that

[a] · [x] = [a] · [x] + [0] by definition of additive identity

= [a] · [x] + [py] as [py] = [0]

= [ax] + [py] by definition of multiplication in Z/pZ= [ax+ py] by definition of addition in Z/pZ= [1] as ax+ py = 1 in Z

so that [x] is the multiplicative inverse of [a].We have shown that Z/pZ is a commutative ring with identity in which every nonzero element has a

multiplicative inverse; hence, Z/pZ is a field.

Example 8.8. We can use repeated applications of the Division Theorem and the proof that Z/pZ is a fieldto find the inverse of any nonzero element in Z/pZ with p prime. For instance, to find the inverse of [10] inZ/17Z, we must first find integers x, y such that 10x+ 17y = 1. To do this, we’ll use the division algorithmseveral times:

17 = 10 · 1 + 7

10 = 7 · 1 + 3

7 = 3 · 2 + 1.

In each step, our new divisor b is the previous remainder and our new dividend a is the previous divisor.We stop when we obtain a remainder of 1 and start working backwards. The first backwards step is to solveeach of the existing equations for the remainders:

7 = 17� 10

3 = 10� 7

1 = 7� 3 · 2.

Then we substitute the remainder expressions back in one at a time, making sure to keep track of thingscarefully:

1 = 7� 3 · 2= 7� (10� 7) · 2 substitute back in using 3 = 10� 7

= 7 · 3 + 10 · (�2) simplify while keeping track of how many 7’s and 10’s we have

= (17� 10) · 3 + 10 · (�2) substitute back in using 7 = 17� 10

= 17 · 3 + 10 · (�5) simplify while keeping track of how many 10’s and 17’s we have.

We thus know that 1 = 17 · 3 + 10 · (�5) and so using the proof that Z/pZ is a field, we see that [�5] = [12]is the inverse of [10] in Z/pZ. To check this we can see that [10] · [�5] = [�50] = [17 · (�3) + 1] = [1].Alternately [10] · [12] = [120] = [17 · 7 + 1] = [1].

Exercises

(8.A) We didn’t skip anything in this chapter that I want you to spend any time on, so this problem doesn’texist!

(8.B) Here are some computational examples to help you understand these ideas better. Each has its owndirections.

(i) Following Example 8.4 and Example 8.5, find the quotient and remainder for the pairs of integers87 and 8, 138 and 17, and 192 and 12.

40 CHAPTER 8. THE INTEGERS

(ii) Following Example 8.8, find the inverse of [3] in Z/7Z, the inverse of [5] in Z/19Z, and the inverseof [17] in Z/37Z.

(8.C) As our goal was to prove Z/pZ is a field, let’s skip this too. That said, there’s interesting stu↵ to behad regarding the integers and we could come back to this next semester.

Chapter 9

Ring Homomorphisms


• ...understand the definition of ring homomorphisms and isomorphisms.

• ...be able to give several (non)-examples of homomorphisms and isomorphisms.

In this chapter we’re going to learn about “useful” maps between rings. Certainly, I can define a functionf : Z ! Z such that f(3) = 5 and f(n) = 0 for n 6= 3; however, this isn’t a particularly interesting map fromthe point of view of the ring structure on Z. Here’s the main definition of the chapter:

Definition 9.1. Let R and S be rings. A function ' : R ! S is called a ring homomorphism (or simplyhomomorphism when it’s clear we’re discussing rings) if for every r1, r2 2 R,

'(r1 + r2) = '(r1) + '(r2) and '(r1 · r2) = '(r1) · '(r2).

Remark 9.2. More colloquial ways to think about the definition of a ring homomorphism include (a) thathomomorphisms preserve the algebraic structure of the object, and (b) that you can do the operation andthen apply the function, or apply the function and then do the operation.

Example 9.3. Let R = Z and S = Z/2Z. The function ' : Z ! Z/2Z given by n 7! [n] is a ringhomomorphism. Indeed,

'(n+m) = [n+m] = [n] + [m] = '(n) + '(m)

and'(nm) = [nm] = [n][m] = '(n)'(m).

Example 9.4. More generally, there is a ring homomorphism R ! R/I for any ring R and ideal I of Rgiven by '(r) = r + I. (Possibly) you should prove this!?

Example 9.5. Also, for any ring R and subring S ✓ R, there is the “natural” inclusion map:

◆ : S ! R given by s 7! s.

Example 9.6. However, the map ' : Z ! Z given by n 7! 3n is not a ring homomorphism. Why?

Definition 9.7. Recall that a function f : X ! Y is called

• injective if for x1, x2 2 X, if f(x1) = f(x2) then x1 = x2,

• surjective if for every y 2 Y there exists an x 2 X such that f(x) = y, and

• bijective if it is both injective and surjective.

41

42 CHAPTER 9. RING HOMOMORPHISMS

Remark 9.8. In high school, we referred to injective functions as “one-to-one” and surjective functions as“onto”. While I won’t prevent you from continuing to use those terms, I also encourage embracing the “bigkid” terminology for these notions as we’re using more precise definitions that we did in high school.

Definition 9.9. Let R and S be rings. A function ' : R ! S is called a ring isomorphism (or simplyan isomorphism when it is clear we’re discussing rings) if ' is a bijective homomorphism. When such anisomorphism exists, we say that R and S are isomorphic.

Example 9.10. The homomorphisms in Example 9.3 and Example 9.4 are surjective homomorphisms asfor any r+ I in R/I, we know that r 7! r+ I. However, these are typically not injective as there are usuallymany representatives for each coset.

Example 9.11. The homomorphism in Example 9.5 is an injective homomorphism as ◆(s1) = ◆(s2) meanss1 = s2 by definition of the map ◆. However, when S is a proper subset of R, ◆ is not a surjective map.

Example 9.12. Recall the ring R[X]/(X2+1) from Example (7.A)(ii). We were awfully suspicious in classthat this looked a lot like C. In fact, these are isomorphic rings! However, the “obvious” isomorphisms totry: ' : C ! R[X]/(X2 + 1) given by a + bi 7! [a + bX] or its inverse : R[X]/(X2 + 1) ! C given by[a+ bX] ! a+ bi are surprisingly tricky to show are well defined bijective homomorphisms. We’ll come backto this in the next chapter once we have a few more tools and prove that these rings are isomorphic usingsaid tools.

Remark 9.13. Suppose R and S are isomorphic rings. As the isomorphism preserves the additive andmultiplicative structures in a bijective way, these two rings are e↵ectively the same. One common strategyin abstract algebra is to replace one structure with another one that is isomorphic.

Exercises


(i) Let R be a ring and I an ideal of R. From Example 9.4, show that the function ' : R ! R/Igiven by r 7! r + I is a surjective ring homomorphism.1

(ii) Let S be a subring of a ring R and ◆ the function from Example 9.5. Show that this is indeed aring homomorphism.

(iii) Show the function ' : Z ! Z given by n 7! 3n is not a ring homomorphism by providing anexplicit counterexample to one of the properties required.

(9.B) Here are some (non-)examples of homomorphisms and isomorphisms. Show that each is a homo-morphism or is not a homomorphism. For those that are homomorphisms, determine whether or notthey’re injective or surjective (or bijective). For those that are not homomorphisms, provide an explicitcounterexample to one of the properties required of homomorphisms.

(i) The function ' : 2Z ! 3Z given by 2n 7! 3n.

(ii) The function ' : 2Z ! Z given by 2n 7! n.

(iii) The function ' : M2(Z) ! Z given by A 7! detA. That is,

a bc d

�7! ad� bc.

(iv) The function 'a : R[X] ! R given by p(X) 7! p(a) for some a 2 R. Does this work for all, some,or no values of a?2

(v) It is a fact that for any rings R, T , the set R⇥T with operations (r1, t1)+(r2, t2) = (r1+r2, t1+t2)and (r1, t1) · (r2, t2) = (r1 · r2, t1 · t2) is a ring.3 The function in question is ' : R⇥ T ! R givenby (r, t) 7! r.

1Note that this means you must both show it is a ring homomorphism and a surjective function.2If the notation in this problem is daunting, start by working with examples, and let me know if it needs to be modified or

if I should ensure everyone is working with the same notation.3It is probably worth identifying for yourself the additive identity and inverses to ensure you understand the ring; however,

you need not prove this is a ring. The “interesting bits” all fall out easily from the relevant properties on R and T .

43

(vi) The function ' : R⇥ T ! T ⇥R given by (r, t) 7! (t, r).

(vii) The function ' : Q⇥Q ! M2(Q) given by (a, b) 7!

a 00 b

�.

(9.C) In this problem, we’ll explore a few additional properties of homomorphisms. Let R and T be ringsand ' : R ! T a ring homomorphism.

(i) Let 0R denote the additive identity in R and 0T denote the additive identity in T . Then '(0R) =0T . As a hint, it may help to write 0R in a “clever” way and then apply '.

(ii) For any r 2 R, we have '(�r) = �'(r). Colloquially, this says that“The homomorphic image of the additive inverse of an element is the additive inverse of its

homomorphic image.”

As a hint, it may help to use the first part of this problem and the definition and uniqueness ofan additive inverse.

(9.D) In this problem, we’ll explore the homomorphisms between rings of the form Z/nZ.

(i) List all the homomorphisms Z/2Z ! Z/4Z. As these are both very small rings, one strategy4 isto list all of the possible functions and identify which are homomorphisms.

(ii) List all the homomorphisms Z/4Z ! Z/2Z. As these are both very small rings, one strategy4 isto list all of the possible functions and identify which are homomorphisms.

(9.E) You may do any, all, or none of these problems in addition to either problem (9.D) or (9.C). They’reall bonus in your portfolio and related to problem (9.D)

(i) (Bonus!) If m | n, show that there is a surjective homomorphism Z/nZ ! Z/mZ.(ii) (Bonus!) Explain why any homomorphism Z/nZ ! Z/mZ with n < m is not surjective and any

homomorphism Z/nZ ! Z/mZ with n > m is not injective.

(iii) (Bonus!) List all the homomorphisms Z/2Z ! Z/3Z, Z/2Z ! Z/5Z and Z/2Z ! Z/7Z. Unlikethose in problem (9.D), these are large enough rings that it is a bad idea here to try and list allthe functions and identify the homomorphisms from that list. What patterns do you notice here?You need not prove the patterns hold true, just find at least one.

(iv) (Bonus!) List all the homomorphisms Z/6Z ! Z/4Z, Z/6Z ! Z/5Z, and Z/5Z ! Z/3Z. Unlikethose in problem (9.D), these are large enough rings that it is a bad idea here to try and list allthe functions and identify the homomorphisms from that list. What patterns do you notice here?You need not prove the patterns hold true, just find at least one.

4...although perhaps not the best strategy...

Chapter 10

Ring Isomorphism Theorems


• ...be able to state all three isomorphism theorems for rings.

• ...be able to use the isomorphism theorems in examples.

• ...understand the need in many proofs for showing that the maps are well-defined.

There are three isomorphism theorems (telling us that creating certain algebraic structures always results inisomorphic structures) and they hold for many di↵erent kinds of algebraic structures. In this chapter, we’llfocus on the ring versions of the theorems. First, we’ll need a definition:

Definition 10.1. Let R and S be rings and ' : R ! S a ring homomorphism. The kernel of ' is thesubset of R defined as

ker(') = {r 2 R | '(r) = 0S}where 0S represents the additive identity in S. Also, recall that the image of ' is the subset of S defined as

im(') = '(R) = {'(r) | r 2 R} = {s 2 S | s = '(r) for some r 2 R}.Theorem 10.2 (First Isomorphism Theorem for Rings). Let R and S be rings and ' : R ! S a ring

homomorphism. Then ker(') is an ideal of R, im(') is a subring of S, and

R/ ker(') ⇠= im(').

Theorem 10.3 (Second Isomorphism Theorem for Rings). Let R be a ring, A,B ✓ R with A a subring of

R and B an ideal of R. Then A+B is a subring of R, B is an ideal of A+B, A \B is an ideal of A, and

A/(A \B) ⇠= (A+B)/B.

Theorem 10.4 (Third Isomorphism Theorem for Rings). Let R be a ring, I ✓ J ✓ R with I and J ideals

of R. Then J/I is an ideal of R/I and

(R/I)/(J/I) ⇠= R/J.

Remark 10.5. My plan is to walk you through the proofs in the exercises.

Example 10.6. We can use the First Isomorphism Theorem to show that R[X]/(X2 + 1) ⇠= C. To that

end, let R = R[X] and S = C. Define a function ' : R ! S viaNX

n=0

anXn 7!

NX

n=0

anin, i.e. p(X) 7! p(i).

First, note that ' is a ring homomorphism since

'(p(X) + q(X)) = (p(X) + q(X))��X=i

= p(i) + q(i) = '(p(X)) + '(q(X))

45

46 CHAPTER 10. RING ISOMORPHISM THEOREMS

and'(p(X)q(X)) = (p(X)q(X))

��X=i

= p(i)q(i) = '(p(X))'(q(X)).

Next, we see that ' is a surjective function since for any a + bi 2 C, we have a + bX 7! a + bi. Also,X2 + 1 7! i2 + 1 = �1 + 1 = 0 and so X2 + 1 is in the kernel. As the kernel is an ideal of R[X], we know

ker(') ◆ (X2 + 1).

To see that we have equality, we need to show that for any p(X) = a0 + a1X + · · · + aNXN 2 ker', thenp(X) = (X2 +1)q(X) for some polynomial q(X) 2 R[X]. This is easy enough to show for N small; however,I find the full argument showing this to be unenlightening and notationally messy. It boils down to the factthat the coe�cient of Xj becomes the coe�cient of ij in the image and we can look at the coe�cients basedupon the equivalence class of j modulo 4 (remember i0 = 1, i1 = i, i2 = �1, i3 = �i, i4 = 1, i5 = i, etc.).

Exercises

(10.A) Prove the First Isomorphism Theorem for Rings. The steps below should guide your proof.

(i) Show that ker(') is an ideal of R (probably using the ideal test). This tells us that R/ ker(') isactually a ring (which we need in order to proceed).

(ii) Show that im(') is a subring of S (probably using one of the subring tests).

(iii) Define the “obvious” function to use here f : R/ ker(') ! im('). As you may not yet haveenough experience to have an instinct as to what the “obvious” function is, I’d be happy to havea whole class discussion about ideas/suggestions that you do have.

(iv) Since elements of R/ ker(') are cosets (and hence typically have multiple coset representatives),we’ll need to show the function we wrote down is “well-defined.” That is, for r + ker(') =r0 + ker('), we need to show that f(r + ker(')) = f(r0 + ker(')). If this step fails, that’s a goodhint that our function in the previous step is wrong.

(v) Now that we know f is a well-defined function between the two rings we want to show areisomorphic, we need to show that f is a ring homomorphism. This step should be straight-forward. Use the definition of a homomorphism and the way that f is defined.

(vi) We also need that f is injective. As a reminder, to show a function ↵ : X ! Y is injective, we letx1, x2 2 X with the assumption that ↵(x1) = ↵(x2) and use the properties of X and Y and thedefinition of the function ↵ to show that this means x1 = x2.

(vii) We also need that f is surjective. As a reminder, to show a function ↵ : X ! Y is surjective, welet y 2 Y and use the properties of X and Y and the definition of the function ↵ to demonstratethe existence of an element x 2 X such that ↵(x) = y.

(10.B) These two exercises are some basic examples of using the isomorphism theorems.

(i) Show M2(Z)/2M2(Z) ⇠= M2(Z/2Z). Which isomorphism theorem should you use to do this?

(ii) For notational ease, let Zn denote the ring Z/nZ. Show Z6/[2]Z6⇠= Z2. Note that [2]Z6 here

denotes the ideal generated by the coset [2] in Z6 = Z/6Z. Which isomorphism theorem shouldyou use to do this?

(iii) Use Theorem 10.3 to show 6Z/30Z ⇠= 2Z/10Z. What should you use for R, A, B?

(10.C) Prove the Second Isomorphism Theorem for Rings. To prove it, we’ll need to show that A + B is asubring of R, B is an ideal of A+B, and A\B is an ideal of A. We’ll then need to go about showingthe claimed isomorphism holds. To show that isomorphism, we can either appeal to Theorem 10.2, orwrite down a function and show it is a (well-defined) bijective homomorphism.

(10.D) Prove the Third Isomorphism Theorem for Rings. To prove it, we’ll need to show that J/I is an ideal ofR/I. We’ll then need to go about showing the claimed isomorphism holds. To show that isomorphism,we can either appeal to Theorem 10.2, or write down a function and show it is a (well-defined) bijectivehomomorphism.

Chapter 11

What is a group?


• ...understand the definition of a group.

• ...understand some basic properties of groups.

• ...be able to give several examples of groups.

Let’s jump in headfirst with our definitions:

Definition 11.1. A group (G, ?) is a set G together with an operation ? such that:

(A) The operation is associative.

(Id) The operation has an identity element, denoted e.

(Inv) Every element in G has an inverse with respect to ?.

Definition 11.2. We say that a group (G, ?) is abelian if it has the additional property that:

(C) The operation is commutative.

Definition 11.3. Let G be a group with finitely many elements, say |G| = n. Then G is a finite groupand the order of G is n.

Remark 11.4. We commonly use multiplication or addition notation for groups. Additive notation is mostcommon when a group is abelian and multiplicative notation is standard for a group when it is either knownto be non-abelian or when it is unknown whether or not the group is abelian.

Example 11.5. Every ring is an abelian group under addition.

Example 11.6. Let n 2 N be fixed and let Xn denote the set {1, 2, . . . , n}. The set of bijections Xn ! Xn

form a group under the operation of composition. This group is called the symmetric group on n lettersand denoted Sn. What do we need to do in order to show this is a group? Presumably this is a finite group.What is its order?

Remark 11.7. Let n = 5, so that Xn = {1, 2, 3, 4, 5}. As a set, we have that Sn is all the bijections from{1, 2, 3, 4, 5} to {1, 2, 3, 4, 5}. We can think of these as the ways to rearrange five things. For instance, thefunction defined by �(1) = 5, �(2) = 3, �(3) = 4, �(4) = 2, and �(5) = 1 is such a bijection. Here’s twodi↵erent ways to write this function down in compact notation:

two-line notation:

✓1 2 3 4 55 3 4 2 1

◆

cycle notation: (15)(234)

47

48 CHAPTER 11. WHAT IS A GROUP?

In two-line notation, we list the elements of Xn in order along the top line, and their images on the secondline. In cycle notation, we list the cycles in parentheses in the order that the elements cycle through. The(15) portion tells us about the cycle where 1 gets mapped to 5 and 5 gets mapped back to 1. The (234)portion tells us about the other cycle: 2 gets mapped to 3, 3 gets mapped to 4 and 4 gets mapped back to 2.

Example 11.8. The Quaternion group, denoted Q8 is a group consisting of the 8 elements ±1,±i,±j,±kwhere 1 is the identity element, along with the rules (�1)2 = 1, (�1)a = �a for any a 2 Q8, i2 = j2 = k2 =�1, ij = k, jk = i, ki = j, and ji = �k, kj = �i, ik = �j. Here’s a picture that may help. Multiplyingin the direction of the arrows yields the next element in the circle, multiplying against the direction of thearrows yields the negative of the next element in the circle.

i

jk

Example 11.9. The dihedral group of a regular n-gon is the group of symmetries of that n-gon anddenoted Dn (although some texts denote this by D2n as it is a finite group of order 2n).1

Exercises


(i) In Example 11.6, we claimed that Sn is a finite group for all n 2 N. To complete the proof of this,we need to show that the claimed operation is actually an operation (i.e. the composition of twobijections is again a bijection), that the operation is associative, that the group has an identityelement, that every element has an inverse, and we need to determine the order of Sn. For thefirst two parts here, do as much as you need to do, and no more. For the third part, you shouldfind the identity element in the group, and explain why it is an identity. For the fourth part, youshould find the inverse of an arbitrary element and show it is indeed the correct inverse. For thefifth part, you should identify the order of Sn and give an explanation for your answer; however,a formal proof is not required.

(ii) In Example 11.8, we claimed that Q8 is a group. Write out the Cayley table for this group.Explain how the Cayley table shows that the claimed operation is an operation, that the groupindeed has an identity and every element has an inverse. Also, recognize that you don’t want todo the work for the associativity and hope that there’s a shortcut involving changing notation tosomething like functions, or matrices (where we already know associativity holds). When you’reready, talk to Dr. Shultis about a strategy for proving associativity.

(iii) In Example 11.9, we only sort of described the dihedral group. In this exercise, you’ll workexplicitly with D4. Draw and cut out a square of paper. Label each corner (on one side of thepaper) with a unique symbol. On the other side of the paper, label the corners with the samesymbols (peeking over the corner should show the same symbol on both sides in any given corner).Your goal is to find all of the ways to pick up your square and put it back down in the same place(with the symbols now in di↵erent places). For instance, you could reflect the square about avertical line (perhaps denoted f) or rotate 90� counter-clockwise (perhaps denoted r). Can youdescribe all of the ways to do this as repeated iterations of r and f?

(11.B) Here are some (non-)examples of groups. Show that each is a group or is not a group. Remember thatto show a set is not a group, we need a single (explicit) counterexample of a property that fails. If itis a group, is it abelian? Is it a finite group? If it is finite, what is the order of the group?

(i) The set of n ⇥ n matrices with entries in R with determinant 1 under the operation of matrixmultiplication.

1Note that this group is a subset of the symmetric group, and so we want to call it a subgroup. More on that soon!

49

(ii) The operation from Exercise (1.B)(vii). When you’re stuck on the associativity as it still soundslike a pain, talk to Dr. Shultis about a strategy (it will be similar in some ways to showing thequaternion group is associative).

(iii) The operation in the Cayley table below:

? e a b c d

e e a b c da a e c d ab b c e a bc c d a e cd d a b c e

(11.C) This problem has two (unrelated, except that they’re both about groups) parts:

(i) Let G be a group written multiplicatively and g, h 2 G. Show that (gh)�1 = h�1g�1, i.e. “the

inverse of a product is the product of the inverses (in the other order)”.

(ii) Let G be a finite group. Show that in every row and column of the Cayley table for G, eachelement of G must appear exactly once.

(11.D) Let G = {e, a, b, c}. Find both operations on G that make G a group. It may help to start with ablank Cayley table, fill in what you must have, and figure out which of the remaining choices you canand cannot make.

(11.E) So far, we know of two groups of order 6, namely Z/6Z under addition, and S3. Prove that these arethe only two groups of order 6.

Chapter 12

Subgroups


• ...understand the definition of a subgroup.

• ...be able to give several examples of subgroups.

As is typical in algebra, one definition of a subgroup is that it is a SUB-GROUP. That is, it is a subset of agroup that is itself a group. Here’s a more formal definition.

Definition 12.1. Let (G, ?) be a group. A set H is a subgroup of G if:

(S) The set H is a (nonempty) subset of G.

(Cl) If h, k 2 H, then so is h ? k.

(G) The set H together with the operation ? is a group. That is, it satisfies the axioms of a group:

(A) The operation is associative.

(Id) The operation has an identity element, denoted e.

(Inv) Every element has an inverse with respect to ?.

Remark 12.2. As with subrings, the word nonempty is in parentheses because the emptyset is not a group,and so cannot be a subgroup of any group. The issue is that a group must contain at least one element: theidentity element.

Remark 12.3. Some (possibly most or all) texts use H G to denote that H is a subgroup of G. Thenotation H � G then means that H is a proper subgroup of G, meaning that H 6= G and H is a subgroupof G.

Example 12.4. Let G be a group with identity element e. Then {e} and G are subgroups of G. Is thisclear?

Remark 12.5. Note that as with subrings, another way to rephrase item (Cl) above is that we say the set His closed under the operation ?.

Theorem 12.6 (Subgroup Test). Let (G, ?) be a group and H a nonempty subset of G. For any g 2 G,

we let g�1denote the inverse of g in G. The set H is a subgroup of G if and only if g ? h�1 2 H for all

g, h 2 H.

Proof. (=)) First, we assume H is a subgroup of G and we must show that g ? h�1 2 H for all g, h 2 H.So let g, h 2 H. As H is a subgroup of G, it is a group, and so the inverse of h is an element of H, i.e.h�1 2 H. As H is a subgroup of G, it is closed under multiplication and so we have g ? h�1 2 H asdesired.

51

52 CHAPTER 12. SUBGROUPS

((=) Next, we assume g ? h�1 2 H for all g, h 2 H and we must show that H is a subgroup of G. To dothis, we must show all of the axioms of a subgroup hold:

(S) In the hypotheses of the theorem, we were told that H is a nonempty subset of G, so thisrequirement is satisfied.

(Cl) Let h, k 2 H. Then by assumption we have k ? k�1 2 H. As k ? k�1 = e, we know e 2 H.Now, we can see that our hypothesis gives k�1 = e ? k�1 2 H, and using it one more time givesh ? k = h ? (k�1)�1 2 H as desired.

(G) We must now show that the set H satisfies the axioms of a group:

(A) I’ll leave this one for you.

(Id) Looking carefully at the proof of closure, did we already do this?

(Inv) Did we do this one too?

Example 12.7. Recall that every ring is an abelian group under addition. Which of Z,Q,R are subgroupsof C? of R of Q? Name as many subgroups of Z as you can.

Example 12.8. Let n,m 2 N and assume that n m. Then Sn Sm. Is this clear?

Exercises


(i) Explain as briefly as possible why {e} and G are subgroups of a group G.

(ii) Complete the proof of Theorem 12.6. In particular, you must show that in the backwards direction,we have that associativity holds, that the set H has an identity element, and that every elementhas an inverse in H.

(iii) Succinctly answer the questions in Example 12.7.

(iv) Prove the claim in Example 12.8. That is, prove that Sn is a subgroup of Sm when n,m 2 N withn m.

(12.B) Here are some basics of subgroups. Each problem has its own directions.

(i) Let IM2(R) denote the set of invertible 2⇥2 matrices with entries in R. Recall from linear algebrathat this is the set {A 2 M2(R) | detA 6= 0}. Recall from Exercise (11.B)(i) that IM2(R) is agroup under matrix multiplication. Find at least three subgroups of this group, and three subsetswhich are not subgroups.

(ii) Let G be a group and g1, . . . , gn 2 G. The subgroup of G generated by g1, . . . , gn is definedas the smallest subgroup of G which contains {g1, . . . , gn}. What is the subgroup of S4 generatedby (12)(34)? What about the subgroup generated by (12) and (34)? What about the subgroupgenerated by (123) and (12)(34)?

(12.C) As with the subrings and ideals chapters, we’ll skip this as it was split from the normal subgroupschapter to ensure a shorter lecture.

Chapter 13

Normal Subgroups


• ...understand the definition of a normal subgroup.

• ...be able to give several examples of normal subgroups.

• ...understand some important properties of normal subgroups.

Recall that with rings, the definition of an ideal was ideally suited to define quotient rings and have theresulting operations be well-defined. The notion of a normal subgroup is analogous in that our ultimate goalin the sense that we want something ideally suited to define a quotient group and have the operation bewell-defined.

Let’s think backwards a bit, and work with a group written additively. If G is a group, and N is a“normal” subgroup of G, we might want to define the “cosets” of N , i.e. the elements of the quotient groupG/N , to be of the form [g] = g + N = {g + n | n 2 N} with g 2 G. If we want the operation on these tobe given by [g] + [h] = [g + h] for [g], [h] 2 G/N . For this to be well-defined, we’d need that for [g] = [g0]and [h] = [h0], then [g + h] = [g0 + h0], i.e., we’d need that �(g0 + h0) + (g + h) 2 N whenever �g0 + g 2 Nand �h0 + h 2 N . Because we might be more prone to assuming commutativity in additive notation, andwe should be careful about not assuming things we’re not given, let’s switch now to multiplicative notation.

If G is a group written multiplicatively, with identity element e and N a “normal” subgroup of G, wehave “cosets” of N being of the form g = [g] = gN = {gn | n 2 N}. For the operation (gN)(hN) = (gh)Nto be well-defined, we will need to take a, a0, b, b0 2 G such that a0N = aN and b0N = bN and need to showthat (a0b0)N = (ab)N , i.e. that (ab)�1(a0b0) 2 N . For now, note that nN = eN for any n 2 N , and fix someg 2 G and n 2 N . Then, for the operation to be well-defined, we’ll need that (nN)(gN) = (eN)(gN), i.e.we need that (eg)�1(ng) 2 N . Recall from Exercise (11.C)(i) that (ab)�1 = b�1a�1 for any a, b 2 G. Also,note that the identity element always has an inverse as it is always its own inverse. Now, note that for theoperation to be well-defined, we need that

(eg)�1(ng) = (g�1e�1)(ng) = g�1eng = g�1ng

is an element of N . We’re ready for the definition of a normal subgroup now, as we’ll create that definitionto make this work.

Definition 13.1. Let (G, ?) be a group and N a subset of G. We say that N is a normal subgroupof G, and denote this as N C G, if N is a subgroup of G and for every g 2 G and n 2 N , we have thatg�1 ? n ? g 2 N as well.

Remark 13.2. The LATEX code to obtain the symbol C is \lhd.

Example 13.3. Our favorite abelian group might be Z, and our favorite subgroup of this is probably nZwhere n is our favorite nonzero integer. As Z uses additive notation, for nZ to be a normal subgroup ofZ, we’d need that �g + nh + g 2 nZ whenever nh 2 nZ and g 2 Z. This is “easily” seen to work, as thecommutativity of the operation allows for us to rewrite this as (�g + g) + nh = nh 2 NZ.

53

54 CHAPTER 13. NORMAL SUBGROUPS

Theorem 13.4. The example above seems to imply that every subgroup of an abelian group is normal. We

should probably prove that!

Since every subgroup of an abelian group is normal, we should really look at a non-abelian example too:

Example 13.5. Our smallest non-abelian group is S3. Consider the set A3 = {(1), (123), (132)} ⇢ S3. Wecan see1 from this smaller part of the group table for S3 that A3 is a subgroup of S3:

? (1) (123) (132)

(1) (1) (123) (132)(123) (123) (132) (1)(132) (132) (1) (123)

To check that it is normal, we need that g�1ng 2 A3 for every n 2 A3 and every g 2 S3. Note that it isclear that g�1ng 2 A3 if g, n 2 A3, so we really only need to test this for g 2 S3 \ A3. You should do thesecomputations to get more comfortable with cycle notation for elements of a permutation group.

Recall 13.6. Let G be a group written multiplicatively and H a subgroup of G. Recall that the left cosetsof H in G are the sets of the form gH = {gh | h 2 H}. We can similarly define the right cosets of H in Gto be the sets of the form Hg = {hg | h 2 H}.Theorem 13.7. Let G be a group and H a subgroup of G. The following are equivalent:

(a) The subgroup H is a normal subgroup of G.

(b) For every g 2 G, we have gH = Hg.2

(c) For every g 2 G, we have g�1Hg = H where g�1Hg denotes the set of all elements of the form g�1hgwith h 2 H.

3

Proof. I’m going to have you prove this one in the exercises.

Remark 13.8. The equivalence of the first two items above is typically phrased to say that “A subgroup His normal if and only if its left and right cosets are the same.”

Recall 13.9. Recall that when G is a finite group, |G| is the order of the group G and defined to be thenumber of elements of G. Also, recall that if G is a finite group and H is a subgroup of G, then the orderof H divides the order of G.

Definition 13.10. Let H be a subgroup of a finite group G. We call the quotient|G||H| the index of H in

G and denote it by [G : H].

Theorem 13.11. Let G be a finite group. If H is a subgroup of G of index 2, then H is a normal subgroup

of G.

Proof. I’d like for “us” to prove this. However, maybe we’ll delay on that for now. Here’s a sketch to get youstarted. Of course, we need to assume that H is a subgroup of G and that H has index 2. Our goal/task isto show that H is normal, which, by Theorem 13.7, we can do by proving gH = Hg for every g 2 G. So letg 2 G.

• If g 2 H, then gH = H = Hg as H is a subgroup of G.

• if g /2 H, then gH \H = Hg \H = ;. As H has index 2 in G, then H is half of the elements of G andthe other half must be both Hg and gH so that gH = Hg as desired.

1How!?2This is a claim about set equality.3The claim g�Hg = H is also a claim about set equality.

55

Exercises


(i) Let G be an abelian group and H any subgroup of G. Prove that H is a normal subgroup of G.

(ii) Do the 3 · 2 = 6 computations necessary to show that A3 C S3. Note that there are 3 choices forg 2 S3 \A3 and only two interesting choices for h 2 A3.4

(iii) Prove Theorem 13.7. Note that to do this, it su�ces to show one of the following chains ofimplications:

(a) =) (b) =) (c) =) (a)

or(a) =) (c) =) (b) =) (a).

It’s unclear to me which of these you’ll find easier.

(13.B) Here are some basics of normal subgroups. Each exercise has its own instructions.

(i) On the final exam, you identified all six subgroups of Q8. They were: {1}, {1,�1}, {1, i,�1,�i},{1, j,�1,�j}, {1, k,�1,�k}, and Q8. Which of these are normal and which are not normal?

(ii) In Exercise (12.B)(ii) you identified some subgroups of S4:

h(12)(34)i = {(1), (12)(34)},h(12), (34)i = {(1), (12), (34), (12)(34)}, and

h(123), (12)(34)i = {(1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)}.

Which of these are normal and which are not?

(13.C) As with the subrings and ideals chapters, we’ll skip this as it was split from the subgroups chapter toensure a shorter lecture.

4For your own benefit of learning, it would be best if you did this without the Cayley table for the group and only usedcycle notation for the computations!

Chapter 14

Quotient Groups


• ...understand the definition of a quotient group.

• ...be able to give several examples of quotient groups.

• ...be able to explain why quotient groups are groups.

• ...understand that quotient groups and quotient rings are analogous structures.

Let’s start o↵ with working on drawing parallels between quotient rings and quotient groups. In particular,it would be good to have a better understanding of the cosets of a normal subgroup of a group and theirrelationship with cosets of an ideal of a ring.

Recall (probably from Math 301) that equivalence relations define a partition on a set, and a partition ofa set defines an equivalence relation. In ring land, we first defined an equivalence relation, and defined ourcosets to be the equivalence classes that form the associated partition. Since we already have a definitionof cosets for groups, perhaps we should think about things in reverse here by showing our cosets form apartition and “reverse engineering” the equivalence relation out of that.

Proposition 14.1. Let G be a group and N a subgroup of G. Then, the left cosets of N form a partition

of G, i.e. [

g2G

gN = G

and for g, h 2 G, either gN = hN or gN \ hN = ;.Proof. Let G be a group and N a subgroup of G. As gN ✓ G for every g 2 G, then the containment[

g2G

gN ✓ G holds. To see that we have equality, let x 2 G and note that since e 2 N , we have x = xe 2 xN

so that x 2[

g2G

gN as well. What remains then is to let g, h 2 G, assume that gN \ hN 6= ; and show that

gN = hN . So let x 2 gN and write x = gn for some n 2 N . Also, since gN \ hN 6= ;, we know there issome y 2 gN \ hN . Write y = ga = hb for some a, b 2 N . Multiplying by a�1 yields g = hba�1. Thus,x = gn = hba�1n. Since a, b 2 N , we then have ba�1n 2 N and so x = h(ba�1n) 2 hN , which shows thatgN ✓ hN . A similar argument shows that hN ✓ gN so that gN = hN as desired.

Now that we know the left cosets of a normal subgroup form a partition of the group, we know there isan associated equivalence relation. Let’s work out the details of that. Let [x] denote the set of a partitionwhich contains the element x. The equivalence relation defined from a partition is

x ⇠ y () [x] = [y].

57

58 CHAPTER 14. QUOTIENT GROUPS

Using the notation we have for the cosets, this becomes

g ⇠ h () gN = hN

() g 2 hN by Proposition 14.1

() there exists an n 2 N such that g = hn

() there exists an n 2 N such that h�1g = n

() h�1g 2 N

and so our equivalence relation boils down to being

g ⇠ h () h�1g 2 N

which, we might have guessed based on the equivalence relation for ideals.

Theorem 14.2. Let G be a group, N a normal subgroup of G, and define an operation ? on the left cosets

of N via (gN) ? (hN) = (gh)N . With this operation, the set of left cosets is a group, denoted G/N and

pronounced “G mod N .”

Proof. I’m going to leave this whole proof to you. Make sure you show all of the necessary parts: theclaimed operation is well-defined, the operation is associative, the operation has an identity element, andevery element has an inverse. Don’t forget that the set on which this operation is defined is the set of leftcosets: G/N = {gN | g 2 G}.

Definition 14.3. Let G be a group and N a normal subgroup of G. Then the group G/N is called aquotient group of G.

Example 14.4. Consider the group Q8 and the normal subgroup {1,�1} of Q8, which we’ll denote Z. Weshould figure out if Q8/Z is a group we’ve seen before by computing its elements and a Cayley table for thegroup. We’ll fill in this table during class and create the Cayley table. Afterwards, I’ll update these noteswith the answer(s). We’ll know we’ve got all the cosets when every element of the group is in a coset. Wewill ensure we only have one representative of each coset by only listing each element once.

coset name coset elements

Z {1,�1}iZ {i,�i}jZ {j,�j}kZ {k,�k}

Instead of this notation, let’s use [1], [i], [j], and [k] for the cosets as we write our Cayley table:

⇤ [1] [i] [j] [k]

[1] [1] [i] [j] [k][i] [i] [1] [k] [j][j] [j] [k] [1] [i][k] [k] [j] [i] [1]

Exercises


(i) Prove Theorem 14.2. It may help to look back on our proof of Theorem 7.4.

(ii) What group does Q8/Z look an awful lot like?

(14.B) Here are some basics of quotient groups. Each problem has its own directions.

59

(i) (Bonus!) As with rings, if H is a subgroup of G and is not a normal subgroup, things can goterribly wrong. Identify a non-normal subgroup of S4 which I’ll call H. Then identify an element� 2 S4 such that �H 6= H�. Conclude that the operation on the quotient of S4 by the non-normalsubgroup you’ve identified is not well-defined by demonstrating an explicit example. For whichelements � of S4 is it the case that �H = H�?

(ii) The center of a group G is the subgroup Z(G) = {g 2 G | gh = hg for all h 2 G}. This notationcomes from the word “zentrum” which is German for center.1 Prove that for any group G, Z(G)is a normal subgroup of G, so that G/Z(G) is always a group.

(iii) Let G be a group and H a subgroup of G. The normalizer of H in G is the set

NG(H) = {g 2 G | gHg�1 = H},

i.e. the largest subgroup of G containing H in which H is normal. Compute NG(H) for

H = h(12)i = {(1), (12)}

and G = S4.

(14.C) Let G be a finite group and N a normal subgroup of G. Prove that |G/N | = [G : N ], that is, the orderof G/N is the index of N in G.

(14.D) Let G be an abelian group and N a subgroup of G. Prove that G/N is abelian. Also, give an exampleof a non-abelian group G and normal subgroup N such that G/N is abelian.

(14.E) Prove that Z(Sn) = {(1)} for all n 2 N.

1Where’s the other Michaela when we need her for her German!?

Chapter 15

Group Homomorphisms


• ...understand the definition of group homomorphisms and isomorphisms.

• ...be able to give several (non)-examples of homomorphisms and isomorphisms.

Just like ring homomorphisms were the “useful” maps between rings, group homomorphisms can be viewedas the “useful” maps between groups. Recall also that the “useful” ring maps were ones that preserved theoperations. Here’s the main definition of this chapter:

Definition 15.1. Let G and H be groups, with operations $ and ? respectively1. A function ' : G ! H iscalled a group homomorphism (or simply homomorphism when it’s clear we’re discussing groups) if forevery g1, g2 2 G,

'(g1 $ g2) = '(g1) ? '(g2).

Example 15.2. Let G = Z and H = Z/2Z. The function ' : Z ! Z/2Z given by n 7! [n] is a grouphomomorphism. Indeed,

'(n+m) = [n+m] = [n] + [m] = '(n) + '(m).

Example 15.3. More generally, there is a group homomorphism G ! G/N for any group G and normalsubgroup N of G given by '(g) = gN . (Possibly) you should prove this!?

Example 15.4. Also, for any group G and subgroup H ✓ G, the “natural” inclusion map:

◆ : H ! G given by h 7! h

is a group homomorphism.

Example 15.5. However, the map ' : S3 ! S3 given by � 7! (12)� is not a group homomorphism. Why?

Take a look back at Definition 9.7 if you need reminders about the words injective, surjective, and/orbijective.

Definition 15.6. Let G and H be groups. A function ' : G ! H is called a group isomorphism (orsimply isomorphism when it is clear we’re discussing groups) if ' is a bijective homomorphism. When suchan isomorphism exists, we say that G and H are isomorphic and denote this by G ⇠= H.

Example 15.7. The homomorphisms in Example 15.2 and Example 15.3 are surjective homomorphisms asfor any gN 2 G/N , we know that g 2 G and '(g) = gN 2 G/N .

Example 15.8. The homomorphism in Example 15.4 is an injective homomorphism as ◆(h1) = ◆(h2) meansh1 = h2 by definition of the map ◆. However, when H is a proper subset of G, ◆ is not a surjective map.

1Hannah suggested the symbol $ for an operation on the plane back from Nebraska.

61

62 CHAPTER 15. GROUP HOMOMORPHISMS

Example 15.9. Recall that the group Q8/Z from Example 14.4 looked an awful lot like Z/2Z⇥ Z/2Z. infact, these are isomorphic groups! We can demonstrate this by comparing Cayley tables and explicitly givingthe bijection; however, showing the map is a homomorphism without demonstrating that the Cayley tablesare just relabeled is harder to do. We’ll come back to this in the next chapter once we have a few more toolsto prove these are isomorphic in a nicer way.

Exercises


(i) Let G be a group and H a subgroup of G. From Example 15.3, show that the function ' : G !G/H given by g 7! gH is a surjective homomorphism.2

(ii) Let H be a subgroup of a group G and ◆ the function from Example 15.4. Show that this is indeeda group homomorphism.

(iii) Recall that every ring is a group under addition, so when we discuss the group Z, we’re using onlythe addition operation. Either prove the function ' : Z ! Z given by n 7! 3n is a group homo-morphism, or demonstrate an explicit counterexample showing it is not a group homomorphism.

(15.B) Here are some (non-)examples of group homomorphisms and isomorphisms. Show that each is a grouphomomorphism or is not a homomorphism. For those that are group homomorphisms, determinewhether or not they’re injective or surjective (or bijective). For those that are not group homomor-phisms, provide an explicit counterexample to the definition of a group homomorphism.

(i) The function ' : 2Z ! 3Z given by 2n 7! 3n.

(ii) The function ' : 2Z ! Z given by 2n 7! n.

(iii) The function Q⇥Q ! M2(Q) given by (a, b) 7!

a 00 b

�.

(iv) The function ' : S4 ! Z/4Z given by � 7! [�(1)]. As the notation here may be hard to follow,here’s an example with � = (142) 2 S4:

'(�) = '((142)) = [�(1)] = [4] = [0].

(15.C) Recall from Example 11.9 that the group Dn is the group of symmetries of a regular n-gon and has 2nelements. Prove that D3

⇠= S3.

(15.D) Recall from Example 11.9 that the group Dn is the group of symmetries of a regular n-gon and has 2nelements. Prove that there is an injective homomorphism 'n : Dn ! Sn for all n 2 N.

(15.E) You may do any, all, or none of these problems in addition to either problem (15.C) or (15.D). They’reall bonus in your portfolio and related to problem (9.D).

(i) (Bonus!) If m | n, show that there is a surjective group homomorphism Z/nZ ! Z/mZ.(ii) (Bonus!) Explain why any group homomorphism Z/nZ ! Z/mZ with n < m is not surjective

and any homomorphism Z/nZ ! Z/mZ with n > m is not injective.

(iii) (Bonus!) List all the group homomorphisms Z/2Z ! Z/3Z, Z/2Z ! Z/5Z and Z/2Z ! Z/7Z.Unlike those in problem (9.D), these are large enough rings that it is a bad ida to try and list allthe functions and identify the homomorphisms from that list. What patterns do you notice here?You need not prove the patterns hold true, just find at least one.

(iv) (Bonus!) List all the group homomorphisms Z/6Z ! Z/4Z, Z/6Z ! Z/5Z and Z/5Z ! Z/3Z.Unlike those in problem (9.D), these are large enough rings that it is a bad ida to try and list allthe functions and identify the homomorphisms from that list. What patterns do you notice here?You need not prove the patterns hold true, just find at least one.

2Note that this means you must both show it is a ring homomorphism and a surjective function.

Chapter 16

Group Isomorphism Theorems


• ...be able to state all three isomorphism theorems for groups.

• ...be able to use the isomorphism theorems in examples.

• ...understand the need in many proofs for showing that the maps are well-defined.

There are three isomorphism theorems (telling us that creating certain algebraic structures always results inisomorphic structures) and they hold for many di↵erent kinds of algebraic structures. In this chapter, we’llfocus on the group versions of the theorems. First, we’ll need a definition:

Definition 16.1. Let G and H be groups and ' : G ! H be a group homomorphism. The kernel of ' isthe subset of G defined as

ker(') = {g 2 G | '(g) = eH}where eH denotes the identity element in H. Also, recall that the image of ' is the subset of H defined as

im(') = '(G) = {'(g) | g 2 G} = {h 2 H | h = '(g) for some g 2 G}.

We’ve done less with groups thnn we have with rings so far, and the following facts could be useful inyour proofs. The proofs of these facts are similar to the proofs of the similar facts for rings; however, Ineither want to include them in lecture, nor ask you to put them in your portfolio. Make Dr. Shultis provethem in class some time soon!

Fact 16.2. Let G and H be groups written multiplicatively1 and ' : G ! H a group homomorphism.

1. If eG and eH denote the identity elements of G and H respectively. Then '(eG) = eH .

2. For any g 2 G, we have '(g�1) = '(g)�1.

Theorem 16.3 (First Isomorphism Theorem for Groups). Let G and H be groups and ' : G ! H a group

homomorphism. Then ker(') is a normal subgroup of G, im(') is a subgroup of H and

G/ ker(') ⇠= im(').

Theorem 16.4 (Second Isomorphism Theorem for Groups). Let G be a group written multiplicatively

2and

A, B subgroups of G with ACG. Also, let AB denote the subset of {ab | a 2 A, b 2 B} of G. Then, AB is

a subgroup of G, A is a normal subgroup of AB, A \B is a normal subgroup of B, and

AB/A ⇠= B/(A \B).1for notational convenience2for notational convenience

63

64 CHAPTER 16. GROUP ISOMORPHISM THEOREMS

Theorem 16.5 (Third Isomorphism Theorem for Groups). Let G be a group, K ✓ H ✓ G with K,Hnormal subgroups of G. Then, H/K is a normal subgroup of G/K and

(G/K)/(N/K) ⇠= G/N.

Definition 16.6. Let X be a (probably finite for now) set. Let F (X) denote the set of all “words” in theelements of X and their “inverses”. So some examples of elements of F ({a, b, c, d, e}) include abbca�1d�1 andada�1d�1db. Let ? be an operation on F (X) defined by concatenation so that for our example, abbca�1d�1 ?ada�1d�1db = abbca�1d�1ada�1d�1db. Note also that we typically don’t write x and x�1 immediately nextto one another since they’re inverse of each other and so cancel out. Also, when we have a string of severalx’s in a row, we typically use exponential notation, so aaa = a3 and b�1b�1 = b�2, and our product aboveshould be simplified to ab2ca�1d�1ada�1b. Then, F (X) is a group under this operation and is called thefree group on X.

Example 16.7. Let G = {r, f} and let F be the free group on G. Then, let N be the normal subgroup ofF generated3 by R = {f2, r3, fr(r2f)�1}. Then F/N ⇠= D3. Indeed, define a homomorphism ' : F ! D3

via r 7! r and f 7! f . Then, r3, f2, fr(r2f)�1 2 ker(') which means that N ✓ ker(') as N is the normalsubgroup of F generated by r3, f2, and fr(r2f)�1. It can also be shown that ker(') ✓ N . Moreover, asim' = D3, then T/N ⇠= D3. We call G the set of generators of D3 and R the set of relations of D3

4. Infact, every group can be constructed in this way with a set of generators and relations! However, we’ll leavethat as an unproven fact for now.

Exercises

(16.A) Prove the First Isomorphism Theorem for Groups. The steps below should guide your proof.

(i) Show that ker(') is a normal subgroup of G. Note that this requires proving that ker(') is asubgroup and that it’s normal - we, unfortunately, don’t have a “normal subgroup test.”

(ii) Show that im(') is a subgroup of G (presumably using the subgroup test).

(iii) Define the “obvious” function to use here f : G/ ker(') ! im('). As you may not yet haveenough experience to have an instinct as to what the “obvious” function is, I’d be happy to havea whole class discussion about ideas/suggestions that you do have.

(iv) Since elements of G/ ker(') are cosets (and hence typically have multiple coset representatives),we’ll need to show the function we wrote down is “well-defined.” That is, using N to denoteker('), we need that for any gN = g0N , we have f(gN) = f(g0N). If this step fails, that’s a goodhint that our function in the previous step is wrong.

(v) Now that we know f is a well-defined function between the two groups we want to show areisomorphic, we need to show that f is a group homomorphism. This step should be straight-forward. Use the definition of a homomorphism and the way that f is defined.

(vi) We also need to show that f is a bijection.

(16.B) These are some basic exercises on using the isomorphism theorems.

(i) Prove that Q8/Z ⇠= Z/2Z⇥ Z/2Z using the First Isomorphism Theorem.

(ii) Let F be a field and let 1 denote the multiplicative identity in F . The notation GL2(F ) denotesthe group of 2 ⇥ 2 matrices with nonzero determinant and entries in F . The notation SL2(F )denotes the subgroup of GL2(F ) where the matrices in GL2(F ) have determinant 1. The notationF ⇤ denotes the nonzero elements of F , i.e. the elements of F which have multiplicative inverses.

Lastly, the notation F ⇤ · I2 denotes the 2⇥ 2 matrices of the form

a 00 a

�with a 2 F ⇤. Use the

Second Isomorphism Theorem for Groups to show that GL2(C)/(C⇤ · I) ⇠= SL2(C)/{±I2} where

{±I2} =

⇢1 00 1

�,

�1 00 �1

��.

3The G is for “generators” and the R is for “relations.” Moreover, when we say N is the normal subgroup generated by R,we mean that N is the smallest set in F which contains R and is a normal subgroup of F .

4Hence, the notational choices!

65

(iii) Note that H = {(1), (12)(34), (13)(24), (14)(23)} is a normal subgroup of S4 which is contained in

A4 = {(1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)}.

In the next chapter, we’ll learn more about Sn and the notation A3 and A4 will be explained.For now, use the Third Isomorphism Theorem for Groups with these groups to prove a newisomorphism.5

(16.C) Prove the Second Isomorphism Theorem for Groups. To prove it, we’ll need to show that AB is asubgroup of G, A is a normal subgroup of AB, and A\B is a normal subgroup of A+B. We’ll then needto go about showing the claimed isomorphism holds. To show that isomorphism, we can either appealto Theorem 16.3, or write down a function and show it is a (well-defined) bijective homomorphism.

(16.D) Prove the Third Isomorphism Theorem for Groups. To prove it, we’ll need to show that H/K is anormal subgroup of G/K. We’ll then need to go about showing the claimed isomorphism holds. Toshow that isomorphism, we can either appeal to Theorem 16.3, or write down a function and show itis a (well-defined) bijective homomorphism.

5Is this too vague of a problem? Do you need more guidance?

abstract algebra i & ii - gonzaga university chapter 0. welcome to abstract algebra stu↵”,...

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