Algebra and Rings Introduction to...

Introduction to AbstractAlgebra and Rings

W. Ethan Duckworth

Loyola University Maryland

c©2014

Contents

1 The integers 41.1 What are the integers? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3 Unique factorization and the integers . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 Congruence in Z and Modular Arithmetic 212.1 Congruence and Congruence Classes . . . . . . . . . . . . . . . . . . . . . . . 212.2 Modular Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3 The structure of Zp and Zn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3 Rings 393.1 Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.2 Basic Algebraic Properties of Rings . . . . . . . . . . . . . . . . . . . . . . . . 463.3 Isomorphisms and Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . 54

4 Polynomial rings 674.1 Definition of a Polynomial Ring . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.1.1 Formally Defining the Polynomial Ring . . . . . . . . . . . . . . . . . . 72

6 Ideals and Quotient Rings 746.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.2 Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

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Preface

This text started as a set of lecture notes based on courses I taught at Loyola College inMaryland from the years 2004 to 2008. The students who take this course at Loyola have awide range of mathematical readiness. Some are eager to start working with abstract math-ematics and proofs, and some are still learning to write proofs, some prefer computation totheory. These notes are distinguished from similar texts by the inclusion of more discussionof abstraction and proofs, include more computations, and more applications.

The text presents a “rings first” approach, but starts with the integers and modular num-bers before doing rings from an abstract approach. An application of some kind is givenevery few sections. Examples include: solving a linear Diophantine equation in two vari-ables, solving a linear modular equation, check-digit schemes, RSA encryption and AESencryption.

A quick instructor might be able to cover all the topics in this text, but I prefer to covera little less and have a better time doing it. One “less is more” approach is to try to get toRSA and AES as quickly as possible.

One goal is to get to AES. Here’s how to find the geodesic path to AES, by going back-wards. AES depends on finite fields, which in turn depend on polynomials over a field,which depend on polynomials, and the division algorithm. (Technically, it doesn’t requirethe division algorithm, but this is the constructive approach, and gives unique results inthe calculations. Also, I prefer to assume results about which polynomials are irreducible.)Constructing quotient rings of the polynomial ring should be patterned after the similarconstruction of the modular integers from the integers.

Along this path, I prefer to add a few applications to keep the student’s interest. Also,I include material that will help them understand difficult points on the geodesic path. Forinstance, I slow down on modular arithmetic and do it twice. I do RSA as a warm up toAES. I discuss the role of axiomitization and rigorous proofs. Finally, I add a small amountof material about rings (basic definitions, homomorphisms and ideals), since this materialis assumed (by other teachers, other schools, etc.) to be covered in such a course.

For readers who are interested in further information about these topics, seeMen of Mathematics by E.T. Bell. This book is famous for having entertaining and short

biographies of a large number of prominent mathematicians (up to 1900).Unknown Quantity by John Derbyshire. This is a history of algebra, written for a popular

audience (and based on secondary sources).Integers, polynomials, rings by Irving Reiner. This is a textbook similar in aim and scope

to the present set of lecture notes.History of Mathematics by John Stillwell. This is not a history book per se, but rather

a math book that works its way through the history of mathematics. In other words, itconcentrates more on the mathematics than on the history. Written as a linear combinationof a math and history text, it’s about 1

3 history and 23 math text.

Chapter 1

The integers

1.1 What are the integers?

What are the integers? How should we even try to answer this question? Let’s cop-out andstart with the dictionary. The American Heritage Dictionary says an integer is

A member of the set of positive whole numbers (1, 2, 3, . . . ), negative wholenumbers (−1, −2, −3, . . . ), and zero (0).

Of course a dictionary definition is not meant to be complete in a technical sense, rather,it is meant to indicate how the word is used, assuming that the meaning behind the wordis already understood. In any case, ignoring whatever information is given in the phrase“whole numbers” (which American Heritage defines poorly), one might be led to the fol-lowing statement: “An integer is a member of the set . . . , −3, −2, −1, 0, 1, 2, 3, . . . ”.

Again, this statement is useful if one already knows the set being described, and justneeds to connect this to the word “integer”. But, this definition is the sort that one mightgive to a small child. It is similar to defining “cat” by saying “A cat is the animal pictured tothe right” (and including a picture).

In contrast, American Heritage gives a grown up definition of cat

A small carnivorous mammal (Felis catus or F. domesticus) domesticated sinceearly times as a catcher of rats and mice and as a pet and existing in severaldistinctive breeds and varieties.

By the way, this definition is not accompanied by a picture! It is interesting that the dictio-nary gives a sophisticated, grown up definition of cat, but an immature definition of integer.

I would like to give a definition of integers, that is modelled very closely on the dictionarydefinition of cat. It starts by defining the thing in question, cat, as a single instance of a widercategory of things, small carnivorous mammal. It then describes a property, domesticated.Then what a cat does, catches rats and mice. Finally, something about the appearance.

We’ll follow a similar pattern for the integers, in a slightly different order. In order todescribe the integers as an example of something from a wider category, we must definethe category. It is the category of rings, and we will define what a ring is. We will define

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what the integers, and other rings, do: they have operations that given them a certain life,that make them more about process than about static existence. Mathematically, this meansthat rings are more than just sets, they are sets with additional structure. Finally, we willdescribe the properties that integers have, both in common with other rings, and uniquelyto themselves.

Definition 1.1.1. The integers are a set Z of elements

. . . ,−4,−3,−2,−1, 0, 1, 2, 3, 4, . . .

that have the usual operations1 of + and ×, the usual relation2 <, and that satisfy thefollowing properties, for all a, b, c ∈ Z:

1. a+ (b+ c) = (a+ b) + c,

2. a+ b = b+ a,

3. 0 + a = a+ 0 = a,

4. −a ∈ Z and a+ (−a) = 0,

5. a× (b× c) = (a× b)× c,

6. a× b = b× a

7. a× 1 = a

8. a× (b+ c) = (a× b) + (a× c)

9. if a < b then a+ c < b+ c,

10. if 0 < a and 0 < b then 0 < ab.

11. Well-Ordering Axiom If S is a nonempty subset of positive integers, and every ele-ment of S is ≥ 0, then S contains a smallest element.

We can derive all other familiar facts about the integers from the above properties. Wewill not do that here, but will list a few, and prove one.

Theorem 1.1.2. The following properties hold for any a, b, c ∈ Z:

(a) If a+ b = a+ c then b = c.

(b) 0× a = a× 0 = 0.1If we wanted to define the integers more formally, we would use a more precise statement of what “+” and

“×” are.2If we were being more formal, we’d list the hidden assumptions that we’re making about “<”: that it is total,

meaning given any two integers we can say that one is smaller than the other, and that it is transitive, meaningthat if a < b and b < c then a < c.

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(c) −(−a) = a.

(d) a× (−b) = (−a)× b = −(a× b).

(e) (−a)× (−b) = a× b.

(f) a× (b− c) = a× b− a× c.

(g) If a× b = a× c, and a 6= 0, then b = c.

(h) If a < b and 0 < c then ac < bc.

(i) If a 6= 0 and a× b = a× c then b = c.

(j) −a = −1× a.

(k) 1 > 0.

(l) If a < b, and c < 0 then a× c > b× c.

(m) If a < 0 and b > 0 then a× b < 0. If a < 0 and b < 0 then a× b > 0.

(n) There does not exist d ∈ Z satisfying a < d < a+ 1.

(o) There exists some n ∈ Z such that n× b > a.

Proof. We will prove just a couple of these properties. Allthough we are not going to proveall of them here, we will assume them going forward. In particular, we will assume whenwe prove part (l), that the previous parts have already been proven.

Part (a). Suppose a+ b = a+ c. Then −a ∈ Z by property Z4. Adding −a to both sideswe get −a+(a+ b) = −a+(a+ c). Using property Z1 we have (−a+ a)+ b = (−a+ a)+ c.Using property Z6 we have (a + (−a)) + b = (a + (−a)) + c. Using property Z4 we have0 + b = 0 + c. Using property Z3 we have b = c.

Part (l). Suppose that a < b and c > 0. Add −a to both sides to get 0 < b − a. Multiplyboth sides by c to get 0 < c(b− a). Finally, add ac to both sides got get ac < bc.

There are a couple of stylistic points to make about the previous proofs. (1) It’s very hardto prove things that are so “obvious” and “elementary” without skipping steps. When youare proving a statement that is supposed to follow only from the most basic definitions, agood rule of thumb is to make sure that you use exactly one basic definition at each step. (2)Writing the proof of part (a) was a bit tedious, with all the phrases like “using property. . . ”,and with no simplifications taking place until the next step. Not skipping the simplificationswas part of the point. Although the phrases seemed repetitious, the goal in every proofshould be to write complete sentences, with references to which property is being used. Asan alternative, one could write a two column proof:

proof of part (a)Assertion Justification

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Assertion Justificationa+ b = a+ c Given−a ∈ Z Z4−a+ (a+ b) = −a+ (a+ c) adding −a to both sides(−a+ a) + b = (−a+ a) + c Z1(a+ (−a)) + b = (a+ (−a)) + c. Z60 + b = 0 + c. Z4b = c Z3

The advantages of a two-column proof are that we don’t have to write as many words,it’s more obvious what the logical steps are, and the justifications are clearly separated.However, two-column proofs are annoying to read for anything that is longer than 5–10steps, or that would be clearer with some explanations. They are useful for the simplestproofs, with the smallest steps.

My final comment is about the style of the proof given for part (l). It’s closer to the spiritof how a student in this class should write proofs. It’s similar to the proof of part (a), but asmall steps were skipped, and instead of saying things like “by Property Z23” we just said“Add . . . to both sides to get”. The student should aim their style somewhere between whatwe’ve done for part (a) and for part (l), perhaps being closer to part (a) at the beginning ofthe semester, and being closer to part (l) by the end of the semester.

Now we turn to division. But our goal is to work entirely within the integers, andso when we divide 5 by 2 we won’t, ever, write something like 5

2 or 12 . We won’t write

fractions, ever, as part of our formal work. We will of course allow them informally whenwe’re trying to understand what’s going on. The next example illustrates how to work withdivision without using fractions.

Example 1.1.1. Perform the division with remainder of 11) 1700 . Double check your an-swer (using only integers!).

Solution:

11)154 R 6

1700116005550446

One way to write this answer (the “forbidden” way) is: 170011 = 154 + 6

11 . To double checkthis over the integers, we can “clear denominators”, i.e. multiply by 11 to get

1700 = 154× 11 + 6.

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This equation can be verified (the right hand side is 1694 + 6 = 1700), and the calculationsare just over the integers.

The following result states division with remainder over the integers. The actual processof doing the division is not mentioned, just the result. And, the result is stated without usingfractions. In essence, the way we double checked our work in the last example is the result.

Theorem 1.1.3 (Division algorithm). Let a, b ∈ Z with b > 0. There exist unique elementsq, r ∈ Z such that a = qb+ r with 0 ≤ r < b.

This is where we ended on Wednesday, September 3

The numbers q and r are the familiar quotient and remainder that appear in long divi-sion. Perhaps they will be more familiar written like this

q R r

b) a . ora

b= q +

r

b.

Here’s a geometric way to picture this result. We partition the number line into a se-quence of intervals

. . . , [−4b,−3b), [−3b,−2b), [−2b,−b), [−b, 0), [0, b), [b, 2b), [2b, 3b), . . . .

The statement of the theorem is equivalent to the fact that a is contained in a unique in-terval, [qb, qb + b), at a unique distance r from the left edge of the interval. The reader isinvited to see how the geometric description just given gets translated into the steps in theproof below.

Proof. The proof has two main parts: first we show that q and r exist, then we show thatthey are unique.

Step 0: Define the set S:

S = {a− bx | q ∈ Z, a− qp ≥ 0}

= {all numbers of the form a − xb as x ranges over the integers andwith a− bx ≥ 0. }

Our basic approach will be to apply the W.O.P. to the set S.Step 1: Show that S is nonempty. To do this we need to show that some value of x will

make a − bx ≥ 0. If a ≥ 0 then let x = 0. Note that a − bx = a ≥ 0 and so a − bx ∈ S. Ifa < 0 then let x = a. Note that a − bx = a − b(a) = a(1 − b). Note that a is negative, and1− b is ≤ 0 (since b > 0). Therefore, a(1− b) ≥ 0, and so a− bx ∈ S.

Step 2: Identify q and r. Since S is nonempty, and since it contains only elements thatare ≥ 0, the W.O.P. shows that S contains a smallest element: call it r. By definition of Swe have that r = a− bx for some x ∈ Z: let q = x.

Step 3: We show that 0 ≤ r < b. By construction we know that r ≥ 0. Suppose, forcontradiction, that r ≥ b. Then r − b ≥ 0 and so r − b = a − qb − b = a − (q + 1)b ≥ 0.

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Therefore r− b ∈ S. But we also have r− b < r and this contradicts the construction of r asthe smallest element of S.

Step 4: We show that r and q are unique. We do this by proving the following statement:if q and r also satisfy the conclusion of the theorem, then q = q and r = r.

Suppose a = qb + r, with 0 ≤ r < b. Since 0 ≤ r. One of the following must be true:r ≤ r or r ≤ r. Without loss of generality, we may assume that r ≤ r. Then

r − r ≥ 0

qb− qb ≥ 0

(q − q)b ≥ 0

q − q ≥ 0

On the other hand, we have

r ≥ 0

−r ≤ 0

r < b

r − r < b

(q − q)b < b

q − q < 1

So now we have0 ≤ q − q < 1.

Since q− q is an integer, and is between 0 and 1, it must equal 0. Therefore q = q and r = r.

Example 1.1.2. Use your calculator to apply the division algorithm and divide a = −5432by b = 234. Double check your answer.

Solution: Here’s what we enter, and what we get, using the calcultor:

-5432/234

-23.21367521

It is a mistake to use q = −23, because a is negative. We want the integer to the left of−23.213 . . . , so we take q = −24. Now we calculate r:

-5432-(-24*234)

184

Thus r = 184. Now we double check our answer:

234*(-24)+184

-5432

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1.2 Divisibility

In this section we give the first result about the integers that is not at the most basic level:what we state here is not quite obvious, and can be proved using the definitions given inthe previous sections.

The main idea is to talk about division in the integers. As we do this, the reader shouldtake note that we always avoid working with the rational numbers, i.e. fractions: we keepall statements strictly within Z using addition and multiplication. Partly this is stylistic,partly with a view towards working in other rings that are not contained in something likeQ, partly as a matter of practice, partly as a way to avoid making mistakes.

We start with the crucial definition (later stated for any ring).

Definition 1.2.1. Let a, b ∈ Z. We say that a divides b if b = ac for some c ∈ Z. We writethis as a|b. Synonyms are “b is divisible by a”, “b is a multiple of a”, “a is a factor of b”, etc.

Note that “a|b” is a statement about relationship between a and b. It does not say “dividea by b” or “divide b by a” or “find a ÷ b” or anything similar. You are not meant to do anycalculation for “a|b” and in fact “a|b” is not equal to a number.

Example 1.2.1. Determine which of the following divide which others. Identify all possiblecombinations.

−2,−3, 2, 3, 4, 6, 10, 12, 15, 6 + 15

Solution: We could write down a huge list of things like 2 - 3, 4÷ 12, etc. But since we haveto figure out all possible combinations, we might as well make a table:

−2 −3 2 3 4 6 10 12 15 6 + 15−232346101215

6 + 15

We’ll write Y or N in each spot, so that 3|6 is recorded with a Y in the spot with 3 to the left,and 6 on top.

This is where we ended on Friday, September 5

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−2 −3 2 3 4 6 10 12 15 6 + 15−2 Y N Y N Y Y Y Y N N3 N Y N Y N Y N Y Y Y2 Y N Y N Y Y Y Y N N3 N Y N Y N Y N Y Y Y4 N N N N Y N N Y N N6 N N N N N Y Y Y N N10 N N N N N N Y N N N12 N N N N N N N Y N N15 N N N N N N N N Y N

6 + 15 N N N N N N N N N Y

Hopefully you looked for patterns in the previous example. Maybe you wondered aboutwhy I wrote 6 + 15 instead of 21. The reason was to help you think about the pieces in thisnumber. So, if you have 3|6 and 3|15 what does that say about 3|(6 + 15)? If you have 2|6and 2 - 15 what does that say about 2|(6 + 15)? Some of the answers are recorded in thenext lemma.

Lemma 1.2.2. For all a, b, c ∈ Z the following properties hold.

(a) If a|b and b|c, then a|c (1.2#3).

(b) If a|b then a|bc (i.e. a divides the product bc).

(c) If a|b and a|c then a|(b+ c) (1.2#4a).

(d) If a|b and a - c, then a - (b+ c).

(e) If a, b ≥ 1 and a|b then a ≤ b.

This is where we ended on Monday, September 8

Proof. Part (b). Suppose a|b. Then b = ax for some x ∈ Z. Then bc = axc. Let z = xc. Thenz ∈ Z and bc = az. Therefore a|bc.

Definition 1.2.3. Let a, b ∈ Z, not both 0. A common divisor of a and b is a number thatdivides a and divides b. We say that d ∈ Z is a greatest common divisor of a and b if d is acommon divisor of a and b and d is ≥ any other common divisor of a and b. If gcd(a, b) = 1then we say that a and b are relatively prime.

The next result is quite surprising in many ways. It has all sorts of implications forcommon divisors, and work in modular numbers (Chapter 2). And, it provides the mainreason that unique factorization holds in Z. Note: the statement is made completely insidethe integers, and without the use of fractions.

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Theorem 1.2.4 (GCD equals Z-linear combination). Let a, b ∈ Z, not both 0. Then

gcd(a, b) = na+mb

where (1) n,m ∈ Z, (2) na+mb ≥ 0, and (3) na+mb is the smallest number satisfying (1)and (2). In other words, gcd(a, b) is the smallest positive Z-linear combination of a and b.

Example 1.2.2. Through trial and error, show how to write gcd(15, 55) as a linear combi-nation of 15 and 55.

Solution: We can easily see that gcd(15, 55) = 5. It’s “clear” that we should subtract 15 from55.

55− 15 = 40

55− 2× 15 = 25

55− 3× 15 = 10

This got us close to 5 but we can’t keep going because the next step would have a negativenumber. So, we try using a multple of 55:

2× 55− 15 = 95

2× 55− 2× 15 = 80

2× 55− 3× 15 = 65

2× 55− 4× 15 = 50

2× 55− 5× 15 = 35

2× 55− 6× 15 = 20

2× 55− 7× 15 = 5

Thus, our answer is5 = 2× 55− 7× 15.

The previous example gives a hint as to the technique of proof we’ll use in the theorem.We looked at subtracting one number from another, and we looked at how small we couldmake the result, while still keeping it positive. This should suggest the W.O.P.

Proof. Step 0: Define the set S:

S = {xa+ yb | m,n ∈ Z, xa+ ny > 0}

={

all numbers of the form xa+ yb where this is positive and x and yare integers.

}Our basic approach will be to apply the W.O.P. to the set S.

Step 1: Show that S is nonempty. We may assume that a 6= 0. If a > 0 then a =1 · a+ 0 · b ∈ S. If a < 0 then −a = −1 · a+ 0 · b ∈ S. This shows that S 6= ∅.

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Step 2: Identify d = gcd(a, b). Since S is nonempty, and since all the elements of S arepositive, by definition, the W.O.P. shows that S contains a smallest element, d. By definitionof S, we can write d = ma+ nb for some m,n ∈ Z.

Step 3: Show that d is a common divisor of a and b. Suppose for contradiction that ddoes not divide a. Apply the division algorithm (Theorem 1.1.3) and let a = qd + r with0 ≤ r < d. Since d does not divide a, we see that r > 0. Then

0 < r = a− qd = a− q(ma+ nb) = (1− qm)a+ (−qn)b < d

This gives a Z-linear combination of a and b which is positive and < d, a contradiction.Step 4: Show that d ≥ any other common divisor of a and b. Let d be a common divisor

of a and b. Let a = d · r and b = d · s. Then d = ma+ nb = mdr + nds = (mr + ns)d. Sinced is a multiple of d, we see that d ≥ d.


Theorem 1.2.5. If a|bc and gcd(a, b) = 1, then a|c.

Proof. Suppose a|bc and gcd(a, b) = 1. Apply Theorem 1.2.4 to write the gcd as a linearcombination and get

1 = ma+ nb

for some m,n ∈ Z. Multiply both sides of this equation by c to get

c = mac+ nbc.

Since a|bc we can write bc = ax for some x ∈ Z. Therefore, the previous equation becomes

c = mac+ nax.

Factoring, we getc = (mc+ nx)a.

Let z = mc+ nx, then z ∈ Z andc = za

which means a|c.

1.3 Unique factorization and the integers

In this section we study the final, important property of the integers: that they have uniquefactorization. You probably take for granted that each integer can be uniquely factored intoprimes, but this property is both subtle and important. Subtle, as witnessed by the factthat for 2000 years the result existed in a limbo state: pretty well assumed, and pretty wellproven, but not 100% explicit. It was Gauss who finally gave a statement that we wouldaccept as completely precise (see [?collison] for an article about exactly what Euclid andGauss stated). Important for some applications that we’ll mention later, as well as for ad-vanced research in topics such as Fermat’s Last Theorem. In fact, it was exactly the failureof unique factorization to hold in a different kind of number system that doomed some

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mathematician’s efforts to prove Fermat’s Last Theorem, and through efforts of others ledto the invention of parts of modern algebra. (See [?edwards] for more of this history andmathematics.)

Definition 1.3.1. Let p ∈ Z with p 6= 0,±1. We say that p is prime if the only integers thatdivide p are ±1 and ±p.

Definition 1.3.2. Let n ∈ Z. We say that n is a product of primes if (1) n is prime or (2)n = p1 . . . pr for some primes p1, . . . , pr.

Example 1.3.1. Write 24255 as a product of primes. How do you know you’ll ever finish?

Solution: It’s easy to see that 24255 is divisible by 5 so we start by factoring out 5:

24255 = 5× 4851.

Now 4851 is divisible by 3:

24255 = 5× 4851 = 5× 3× 1617.

Now 1617 is divisible by 3 again:

24255 = 5× 3× 1617 = 5× 3× 3× 539.

Now 539 is divisible by 7:

24255 = 5× 3× 3× 539 = 5× 3× 3× 7× 77.

Finally, 77 is divisible by 7:

24255 = 5× 3× 3× 7× 77 = 5× 3× 3× 7× 7× 11.

We knew that this process would eventually finish because the part we hadn’t factored yetkept getting smaller. It can’t get smaller forever because the unfactored part is always apositive integer.

The point of the previous example is to think about how we would prove the followingtheorem.

Theorem 1.3.3 (Fundamental Theorem of Arithmetic I: Existence of factorization). Let n ∈Z with n 6= 0,±1. Then we can write n as a product of primes.

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Proof. Step 0: suppose we have proven it for n ≥ 2. Then we are done: if n ≤ −2 then wecan apply the theorem to −n to get −n = p1 · · · , pr, and then conclude n = (−p1)p2 · · · pr.

Step 1: Assume n ≥ 2 and assume for contradiction that n cannot be written as a productof primes.

Step 2: Define the set S:

S = {x ∈ Z | x ≥ 2, x 6= a product of primes}.

Step 3: Show that S is nonempty. We have assumed for contradiction that n ≥ 2 existssuch that n is not equal to a product a primes. Thus n ∈ S.

Step 4: Identify minimal element of S. Since S is nonempty, and since all the elementsof S are positive by definition, the W.O.P. shows that S contains a smallest element m.

Step 5: Get a contradiction. By definition, m ≥ 2 and m cannot be written as a productof primes. Therefore m is not prime. Therefore m = ab for some factors a, b with a, b 6=±1,±m. Since m is positive we may choose a and b to be positive, so 1 < a, b < m.Therefore both a and b are smaller than m. Since m is the smallest element of S, this meansthat a and b are not in S. By definition of S this means that a and b can be written as aproduct of primes:

a = p1 · · · pr, b = q1 · · · qs,where all the pi and qj are prime. Therefore

m = ab = p1 · · · pr · q1 · · · qs.

But this means that m is a product of primes, a contradiction.

Note that the previous proof was spread out, broken into multiple steps, and explainedat some length. This is done on purpose with the hope that every reader can follow everystep. However, once the reader has understood the proof, or if the reader is already moreexperienced in these sorts of proofs, the argument can be shortened, and this may make iteasier to remember or follow. Here’s a shortened version:

Proof. It suffices to prove the result for positive numbers, since otherwise we can apply thetheorem to −n. Suppose for contradiction that there exists n ≥ 2 such that n cannot bewritten as a product of primes. Then the set

S = {x ∈ Z | x ≥ 2, x 6= a product of primes}

is nonempty. Apply the W.O.P. and let m be the smallest element of S. Then m is not primeand so we can factor m = ab, where 1 < a, b < m. Then a, b 6∈ S and so we can write a andb as products of primes

a = p1 · · · pr, b = q1 · · · qs.Therefore

m = ab = p1 · · · pr · q1 · · · qs,which is a contradiction.

Perhaps, for amusement, or maybe as a useful exercise, it helps to see an even shorterversion (I would point out that such a version is not very nice to give to someone else toread).

(version 2014/12/03) 16

Proof. If n is negative, multiply by −1. Suppose that n ≥ 2 is not a product of primes, andthat it is the smallest such. Since n is not prime we can write n = ab where 1 < a, b < n.Writing a and b as a product of primes gives that n is as well, a contradiction.

To illustrate an extremism, we give a hiku version (it has 5, 7, and 5 syllabules, whenread correctly):

Proof. Make n positive.Factors are smaller, productof primes. So is n.

We turn now to uniqueness of factorization. Since we are working over the integersinstead of just N, uniqueness will always be up to multples of +1 or −1. This even showsup in the next lemma.

Lemma 1.3.4. If a and b are any two integers with a|b and b|a then a = ±b.

Proof. Homework.

The following result is the main “engine” in the uniqueness of factorization theorem thatis coming up.

Theorem 1.3.5 (Euclid’s Lemma). Let p ∈ Z with p 6= 0,±1. Then p is prime if and only if itsatisfies this property:

∀a, b ∈ Z, if p|ab then p|a or p|b (∗)

Proof. “=⇒”. Suppose p is prime. We need to prove the p satisfies (∗). So, let p|ab. We needto prove that p|a or p|b. If p|a, then we are done. Otherwise, suppose p - a.

Since p is prime, we have gcd(a, p) equal 1 or p. Since p - a we have gcd(a, p) = 1.Apply Theorem 1.2.4 and write 1 = ma + np. Multiply both sides of this equation by b toget b = mab+ npb. Then p|npb (since p|p) and p|mab (since p|ab). Therefore p|(mab+ npb),whence p|b.


“⇐=”. Suppose p satisfies (∗). Let d|p. Then p = dx for some x ∈ Z. Since p|p we havep|dx. By property (∗) we have that p|d or p|x. If p|d then d = ±p by Lemma 1.3.4. If p|xthen x = pz for some z ∈ Z. Then p = dx = dpz. Cancelling p from both side we see1 = dz. Thus d = ±1. Therefore, any integer which divides p must equal ±1 or ±p, and sop is prime.

Proof. The first part of the previous theorem can be proven directly, without using Theo-rem 1.2.5. Suppose p is prime. We need to prove the following: if p|ab then p|a or p|b.Suppose p|ab. We need to prove that p|a or p|b. Suppose p - a. Let d = gcd(p, a). Since p isprime we have that d equals 1 or p. Since d|a, and p - a, we see that d 6= p, and so d = 1.Now Theorem 1.2.5 shows that p|b.

(version 2014/12/03) 17

Corollary 1.3.6. Let p be prime and suppose that p|a1a2 · · · ar. Then p|ai for some i.

Proof. Induct on r.

Example 1.3.2. How many ways can you write −30 as a product of primes?

Solution: The prime factors are easily found

−300 = −2× 3× 5.

But there are lots of ways we can write this:

−30 = (−2)× 3× 5

= (−2)× 5× 3

= (−3)× 2× 5

...

= 2× (−3)× 5

= 2××(−5)...

= (−5)× (−2)× (−3)= (−5)× (−3)× (−2)

(Why not try to calculate how many ways we rearrange the factors and the negative signs?)But of course, most of the time we don’t view the different ways of writing these prime

factors as making any real difference: we ignore the differences shown here.

Nowe we are ready to state and prove the uniqueness part of prime factorization.

Theorem 1.3.7 (Fundamental Theorem of Arithmetic II: Uniqueness of factorization). Letn ∈ Z, n 6= 0,±1. Then the prime factorization given by Theorem 1.3.3 is unique. In otherwords, if

n = p1 · · · pr = q1 · · · qswhere each pi and each qi is a prime, then r = s and, if necessary, we may rearrange the factorson one side, so that pi = ±qi for each i.

Note that if we require that n, as well as each pi and qi are required to be in N, asopposed to Z, then we can remove the “±” in the last line of this theorem.

Proof. Step 0: Set up an index for induction. By Theorem 1.3.3, we know that n has a primefactorization. Let r be the number of factors which appear in this factorization. In other

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words, r is the number of primes which it is possible to write n as a product of. We inducton r.

Step 1: Base case, r = 1. It is possible to write n as a product of 1 prime, so we haven = p1. Suppose we can also write n = q1 · · · qs. Then p = q1 · · · qs. Since p is prime thisimplies that s = 1 and q1 = p1.

Step 2: Induction hypothesis. Suppose now that the theorem has been proven for all rin a range {1, . . . , k}. In other words, the uniqueness statement holds for any number thanequals a single prime, or a product of two primes, or a product of three primes, or . . . , aproduct of k primes.

Step 3: Induction step. Let r = k + 1. Then it is possible to write n as a product of k + 1primes, so we have

n = p1 · · · pk+1.

Suppose we can also write n = q1 · · · qs. Then p1|n and so p1|q1 · · · qs, and so by the corollaryto Lemma 1.3.5 we have that p1 divides some qi. Rerrange the factors if necessary sothat p1|q1. Then p1 = ±q1. Then we can cancel and get p2 · · · pk+1 = ±q2 · · · qs. Letn = p2 · · · pk+1. Then n has k prime factors. Therefore the uniqueness statement holdsfor n. Therefore, after rearranging the factors we can conclude p2 = ±q2, p3 = ±q3, . . . ,pk+1 = ±qs. Also, we have the number of factors in p2 · · · pk+1 equals the number of factorsin q2 · · · qs, i.e. k = s− 1, and so r − 1 = s− 1 and r = s.

1.3.1 Applications

In this subsection we allude to some important applications that come later, and explicitlygive a somewhat whimsical application in the meantime. Perhaps though we should startby clarifying the word “application”. In most people’s minds, if we say that X has an appli-cation, it means that X can be used in practical sense, or to accomplish something concrete.In mathematics, it simply means that X implies Y where Y has some sort of value, such asbeing well known, or interesting, or important, etc. So, the applications that come later areof this type: mathematical facts that have some value. The application that comes presentlyis concrete, but not exactly realistic on its own terms.

Unique factorization of the integers is what makes the following mathematical factswork:

• The fact that every rational number ab can be written, uniquely, in reduced form, i.e.

wwith a and b having no common factors,

• The fact that√2 is irrational,

• The fact that we can use an alternative notation for integers, writing them as productsof primes,

• The fact that we can define and/or calculate various functions such as gcd, lcm, andothers, on the integers in terms of their product of prime notation.

Now we give our whimsical, or not exactly realistic, application.

Definition 1.3.8. The prime cipher encrypts text as follows:

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• Encode letters as prime numbers as follows:

a=2b=3c=5d=7

e=11f=13g=17h=19

i=23j=29k=31l=37

m=41n=43o=47p=53

q=59r=61s=67t=71

u=73v=79w=83x=89

y=97z=101

• Take a message in plain text, remove the spaces, break the letters up into blocks of 3,or as long as possible without including a repeated letter.

• For each block of text replace each letter with it’s prime, and raise the prime to the npower where n is the position of the letter within the block.

• For each block, multiply the primes, including their powers, together. The resultingnumber is the encrypted version of the block, and the list of all blocks is the encryptedversion of the message.

• To decrypt you factor each block, put each prime in the correct position using thepower, and then turn the primes back into letters.

Example 1.3.3. Encrypt the message: “attack at dawn”.

Solution:

attackatdawn remove spacesat tac kat daw n break into blocks

2× 712 71× 22 × 53 31× 22 × 713 7× 22 × 833 43 encode in prime powers10082 35500 44380864 16010036 43 multply out

So, the final encrypted message is

10082 35500 44380864 16010036 43


Figure 1.1 shows a TI program to find the prime factors of numbers (the program is fromthe extra resources material for the Hungerford textbook, available at cengagebrain.com).

Example 1.3.4. Decrypt the following message.

13451794 22471119 772650766 1445166215

Solution: The hard part is finding the factors. Start with easy primes such as 2, 3, 5, etc.

13451794 = 2× 6725897.

(version 2014/12/03) 20

Figure 1.1: Hungerford’s Factor program

:ClrHome

:Prompt N

:1ÔI

:2ÔD

:Lbl 1

:√(N)ÔC

:If D>C:Goto 2

:If I>6

:Goto 3

:If fPart(N/D)=0

:Goto 4

:If D=2

:Goto 5

:D+2ÔD

:Goto 1

:Stop

:Lbl 4

:Disp D

:I+1ÔI

:N/DÔN

:If N=1

:Stop

:Goto 1

:Lbl 2

:Disp N

:Stop

:Lbl 3

:Disp "CONTINUES"

:Pause

:1ÔI

:ClrHome

:Goto 1

:Lbl 5

:3ÔD

:Goto 1

Now we know the other factors are not even. So there is only one factor of 2, and the otherfactors must be of the form p2 or p3. We divide 6725897 by 3, by 5, by 11, by 13 and then by17. Since 17 works, we divide by 172, and again by 173

6725897 = 172 × 1369.

Now we divide 23273 by 19, 23, 29, 31 and 37. It turns out that 372 = 1369. So we have ourfirst decryption:

13451794 = 2× 173 × 372 −→ alg.

Now we repeat this work for the other blocks:

22471119 772650766 1445166215 what we start with32 × 11× 613 2× 473 × 612 5× 312 × 673 a lot of work!

ebr aro cks

So our plain text message is

alg ebr aro cks

algebra rocks

Chapter 2

Congruence in Z and ModularArithmetic

In this chapter we learn about one of the most important types of rings, the modular num-bers. These numbers are centrally important in ring theory, group theory and number theory.They also have a huge number of important applications from check digits, to bar codes, andpublic key encryption. Given all this, their most suprising property might be that they arequite simple.

People think about modular numbers in two ways: one that is conceptually simple, butsomewhat difficult to prove things about; and a second way that is conceptually harder, butthat is easier to prove things about. The second approach is what is almost always used intextbooks, because it yields the best proofs, and generalizes to other settings. We will followthis approach.

2.1 Congruence and Congruence Classes

Definition 2.1.1. Let n ∈ Z, n ≥ 2. For all a, b ∈ Z, if n|(a− b) then we write a ≡ b (mod n)and we say that a is congruent to b modulo n. If the value of n is clear in context, then wewill write just a ≡ b, dropping “(mod n)”.

Example 2.1.1. Identify for each combinition of a, b and n whether it’s true or false thata ≡ b (mod n):

(a, b) ∈{(−4, 6), (4, 4), (4, 5), (4, 17), (4, 2),(5, 17), (5, 25), (6, 21), (17, 5)

}n ∈ {5, 6, 7}

Solution: We will organize our answers in the table shown in Figure 2.1The resulting work is shown in Figure 2.2

21

CHAPTER 2. CONGRUENCE IN Z AND MODULAR ARITHMETIC 22

Figure 2.1: Table for answers to example

a and b n = 5 n = 6 n = 10

−4, 6

4, 4

4, 5

4, 17

4, 25

5, 17

5, 25

6, 21

17, 5


Figure 2.2: Answers to example

a and b n = 5 n = 6 n = 7

−4, 65

?

| (−4− 6)

5?

| −10T

6?

| (−4− 6)

6?

| −10F

10?

| (−4− 6)

10?

| −10T

4, 45

?

| (4− 4)

5?

| 0T

6?

| (4− 4)

6?

| 0T

10?

| (4− 4)

10?

| 0T

4, 55

?

| (4− 5)

5?

| −1F

6?

| (4− 5)

6?

| −1F

10?

| (4− 5)

10?

| −1F

4, 175

?

| (4− 17)

5?

| −13F

6?

| (4− 17)

6?

| −13F

10?

| (4− 17)

10?

| −13F

4, 255

?

| (4− 25)

5?

| −21F

6?

| (4− 25)

6?

| −21F

10?

| (4− 25)

10?

| −21T

5, 175

?

| (5− 17)

5?

| −12F

6?

| (5− 17)

6?

| −12T

10?

| (5− 17)

10?

| −12F

5, 255

?

| (5− 25)

5?

| −20T

6?

| (5− 25)

6?

| −20F

10?

| (5− 25)

10?

| −20T

6, 215

?

| (6− 21)

5?

| −15T

6?

| (6− 21)

6?

| −15F

10?

| (6− 21)

10?

| −15F

17, 55

?

| (17− 5)

5?

| 12F

6?

| (17− 5)

6?

| 12T

10?

| (17− 5)

10?

| 12F

(version 2014/12/03) 24

Warning: there is a similar, but slightly different meaning for the notation for mod n. Wehave used it here as it appears in almost all math textbooks; it goes along with ≡ and twonumbers a and b. However, a very similar, but slightly different meaning is also common:mod is used as a function so that a mod n means “give me the smallest number r ≥ 0 suchthat a ≡ r (mod n).” It turns out that r in this case is the remainder of a divided by n.

The notation is not ambiguous if only one number is given, as in “the answer is 235 mod11”. This always means take the remainder of 235 after division by 11. The notation is notambiguous if two numbers are given, both of which are known, as in “we see that 290 ≡ 235mod 11”. The one case where ambiguity can occur is where one number is known, but notthe other, for instance “let x ≡ 15 (mod 6)”. Here, one could mean x = 3, but it would alsobe correct to allow x = 45.

In any case, the notation ≡ was introduced by Gauss. It was probably chosen as some-thing that looks like “=”, and indeed ≡ works similarly to =, as the next few results show.

Theorem 2.1.2 (Congruence modulo n is an Equivalence Relation). Let n ∈ Z, n ≥ 1. Therelation a ≡ b (mod n) is an equivalence relation. In other words, ∀a, b, c ∈ Z, we have thefollowing:

1. a ≡ a (mod n) (reflexive property),

2. if a ≡ b (mod n), then b ≡ a (mod n) (symmetric property),

3. if a ≡ b (mod n), and b ≡ c (mod n), then a ≡ c (mod n) (transitive property).

Proof. I will write these proofs as calculations, althought it pains me to do so, for in almostall cases calculations should be mixed with words that clarify, explain and motivate. But inthis case, the calculations are so simple that I think the student may understand them betterwithout too much extra verbiage.

1.

a?≡ a (mod n)

⇐⇒ n?

| (a− a)

⇐⇒ n?

| 0 X

In other words, a ≡ a (mod n) is equivalent to n|0, and n|0 is true.

2.

a ≡ b (mod n)

⇒ n|(a− b)⇒ a− b = nx, x ∈ Z⇒ b− a = n(−x), −x ∈ Z⇒ n|(b− a)⇒ b ≡ a (mod n)

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3.a ≡ b (mod n) ⇒ n |(a− b)b ≡ c (mod n) ⇒ n |(b− c)

}⇒ n

∣∣∣((a− b) + (b− c))

⇒ n|(a− c)⇒ a ≡ c (mod n)

Theorem 2.1.3. Let n ∈ Z with n ≥ 1. Let a ≡ b (mod n) and c ≡ d (mod n). Then thefollowing hold:

1. a+ c ≡ b+ d (mod n),

2. ac ≡ bd (mod n).

Proof. The proofs are again mainly calculations. In both parts we assume that a ≡ b(mod n) and c ≡ d (mod n).

1. A sequence of calculations shows that

a ≡ b (mod n) ⇒ n |(a− b)c ≡ d (mod n) ⇒ n |(c− d)

}⇒ n

∣∣∣((a− b) + (c− d))

⇒ n∣∣∣((a+ c)− (b+ d)

)⇒ a+ c ≡ b+ d (mod n).


2.n |(a− b) ⇒ n |c(a− b)n |(c− d) ⇒ n |b(c− d)

}⇒ n

∣∣∣(c(a− b) + b(c− d))

⇒ n|(ca− cb+ bc− bd)⇒ n|(ca− bd).⇒ ac ≡ bd (mod n).

Definition 2.1.4. Let n ∈ Z, n ≥ 1. Given any b ∈ Z we define

[b] = {a ∈ Z | a ≡ b (mod n)}.

For example, we have

[−2] = {a ∈ Z | a ≡ −2 (mod n)}[0] = {a ∈ Z | a ≡ 0 (mod n)}[3] = {a ∈ Z | a ≡ 3 (mod n)}

In other words, [b] is the set of all integers that are congruent to b modulo n. We call this setthe congruence class of b modulo n.

(version 2014/12/03) 26

There is another useful way to describe the set [b]:

[b] = {b+ kn | k ∈ Z}.

This formula describes the same set since a ≡ b ⇐⇒ n|(a− b) ⇐⇒ a− b = kn ⇐⇒ a =b+ kn.

Example 2.1.2. Describe all the congruence classes of Z modulo 7.Solution:

[0] = {. . . ,−21,−14,−7, 0, 7, 14, 21, . . .}[1] = {. . . ,−20,−13,−6, 1, 8, 15, 22, . . .}[2] = {. . . ,−19,−12,−5, 2, 9, 16, 23, . . .}[3] = {. . . ,−18,−11,−4, 3, 10, 17, 24, . . .}[4] = {. . . ,−17,−10,−3, 4, 11, 18, 25, . . .}[5] = {. . . ,−16,−9,−2, 5, 12, 19, 26, . . .}[6] = {. . . ,−15,−8,−1, 6, 13, 20, 27, . . .}

Note that [7] = {. . . ,−14,−7, 0, 7, 14, . . . } = [0], so we don’t bother putting [7] on the list.Similarly, [−1] = l . . . ,−15,−8,−1, 6, 13, . . . } = [6], so we don’t bother putting [−1] on thelist. Similar comments apply to all the other congruence classes.

Theorem 2.1.5. Let n ∈ Z, n ≥ 1. For all a, b ∈ Z we have

[a] = [b] if and only if a ≡ b (mod n).

Proof. “=⇒”. Let [a] = [b]. These are two sets that are equal, so that every element of thefirst set is also an element of the second set, and vice versa. By the reflexive property wehave a ≡ a and so a ∈ [a]. Since [a] = [b] we have a ∈ [b]. Therefore, by definition of [b], wehave a ≡ b.

“⇐=”. Let a ≡ b. To prove that [a] = [b] we need to prove that the sets are equal. Letc ∈ [a]. Then c ≡ a. Since a ≡ b the transitive property shows that c ≡ b. Therefore c ∈ [b].This shows that [a] ⊆ [b]. We can give the same argument for c ∈ [b], which proves [b] ⊆ [a].Thus, [a] = [b].

In general, when we intersect two sets, we expect to see some nontrivial intersection

A B

A ∩B

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But in extreme cases the picture might look different: the overlap might be everything, ornothing:

A = B A B

A = B = A ∩B A ∩B = ∅

However, when the sets in question are congruence classes, only the last two pictures arepossible.

Example 2.1.3. Let n = 13. Explain as explicitly as possible how you know the followingare the same sets

[7] = {. . . ,−19,−6, 7, 20, 33, . . . }[59] = {. . . , 33, 46, 59, 72, 85, . . . }

Solution: We can see explicitly that the sets have an element in common: 33. Now we use33 to connect elements from [7] to elements of [59], and vice versa.

If we look at an element like−19 ∈ [7], we see that−19 ≡ 33 because 33 ∈ [7]. Now since33 ∈ [59] we see that −19 ≡ 33 ≡ 59, and so −19 ∈ [59]. Similarly, 85 ∈ [59], 85 ≡ 33 ≡ 7and so 85 ∈ [7].


Corollary 2.1.6. Let a, b ∈ Z. Then

[a] = [b] or [a] ∩ [b] = ∅.

In other words: two congruence classes are either the same class or disjoint.

Proof. If [a] ∩ [b] = ∅ then we are done.Otherwise, suppose [a] ∩ [b] 6= ∅. Let c ∈ [a] ∩ [b]. Then a ≡ c and b ≡ c. Let d ∈ [a].

Then d ≡ a ≡ c ≡ b so d ≡ b and d ∈ [b]. This shows that [a] ⊆ [b]. The same argumentshows that [b] ⊆ [a] and so the sets are equal.

Corollary 2.1.7. Let n ∈ Z, n ≥ 2. The following hold:

1. If a ∈ Z and r is the remainder of a divided by n, then [a] = [r],

2. There are exactly n distinct congruence classes:

[0], [1], . . . , [n− 1].

We call 0, 1, . . . , n− 1 the canonical representatives.

(version 2014/12/03) 28

Definition 2.1.8. The set of all equivalence classes modulo n is denoted by Zn.

Warning 1: the elements of Zn are sets. Thus, it is correct to write [5] ∈ Z7 but not5 ∈ Z7.

Warning 2: Sometimes other notations are used: Z/(n) instead of Zn.

2.2 Modular Arithmetic

One of the topics of this section is how we can do arithmetic with infinite sets. The next twoexamples should warm us up.

Example 2.2.1. Fill in the following tables:

+ even oddevenodd

× even oddevenodd

How do you prove these? What does this have to do with arithmetic of infinite sets?

Solution:+ even odd

even even oddodd odd even

× even oddeven even evenodd even odd

These rules are (hopefully) really familiar and obvious. But wait! Is everything so obvious?What about the proofs?

To prove one of these rules we go back to definitons and equations. For instance, weneed to know that an even number can be written as 2x and an odd number as 2y + 1. Soadding two even numbers looks like 2x+2y = 2z where z = x+y. Adding two odd numberslooks like (2x+ 1) + (2y + 1) = 2(x+ y) + 2 = 2z where z = x+ y + 1. To prove that thesetables are correct one would check equations like this for each entry in the tables.

What about infinite sets? There’s an infinite number of even integers, and an infinitenumber of odd integers, so these rules describe how to add and multiply infinite sets!

Example 2.2.2. Greatly generalize the previous example as follows. “Even” and “Odd” arebased on properties of the number 2. Generalize these properties to be based on the number3. Start by making up names and meanings analagous to “even” and “odd”, then make tableslike in the previous example. Explore the possibilities, and don’t be afraid to try the “wrong”thing.

Solution: Let’s start by naming things: as soon as we have a name, then we can objectifythe name and manipulate it (in a good way).

Just to rhyme with the previous example, let’s make up the following names:

(version 2014/12/03) 29

pretty = all numbers that are multiples of 3,normal = all numbers that are not multiples of 3 (warn-

ing: this definition will be changed in a couple ofparagraphs).

Now, once we’ve made these names up, we would hope that they would satisfy rules likethe ones for even and odd:

+ pretty normalpretty pretty normalnormal normal pretty

× pretty normalpretty pretty prettynormal pretty normal

?So let’s check some of the above rules out.

pretty + pretty ?= pretty

3x+ 3y?= 3z

z = x+ y Xcorrect

pretty + normal ?= normal

3x+ (3y + 1) or 3x+ (3y + 2)eq3z + 1 or 3z + 2

z = x+ y Xcorrect

normal + normal ?= pretty

(3x+ 1) + (3y + 1) or (3x+ 1) + (3y + 2) or (3x+ 2) + (3y + 2)?= 3z + 1 or 3z + 2

3x+ 3y + 2 or 3x+ 3y + 3 or 3x+ 3y + 4?= 3z + 1 or 3z + 2

3x+ 3y + 3 = 3z + 1 or 3z + 2 IMPOSSIBLE

What this means is that we have chosen the wrong sets of numbers to add and multiply.The failed proof has hints of what we should have done. The proof failed because of thecombinatios of the +1 and the +2 in our normal numbers. In fact, there are two kinds ofnormal numbers, and they should each get a name and they should each define a set:

pretty = all numbers that are multiples of 3,normal = all numbers that are 1 bigger than a multiple of 3,ugly = all numbers that are 2 bigger than a multiple of 3.

Now we would try to fill in the following:

+ pretty normal uglyprettynormal

ugly

and× pretty normal ugly

prettynormal

ugly

(version 2014/12/03) 30

A little experimentation should provide the results we need. For instance 2 + 5 = 7 meansugly +ugly =normal and 2× 5 = 10 means ugly ×ugly =normal. The full results are shownbelow:

+ pretty normal uglypretty pretty normal uglynormal normal ugly pretty

ugly ugly pretty normal

and× pretty normal ugly

pretty pretty pretty prettynormal pretty normal ugly

ugly pretty ugly normal

To prove these in general, we would work with the definitions. For example, if a isnormal and b is ugly, then a = 3n+1 and b = 3m+2 for some n,m ∈ Z. Then a+ b = 3(n+m+1), and is pretty, and a×b = (3n+1)(3m+2) = 9nm+6n+3m+2 = 3(3nm+2n+m)+2,which is ugly.


The previous two examples show how easily we can add certain infinite sets of numberstogether. However, it is not always so easy, as the next example shows.

Example 2.2.3 (Hungerford). Define the following sets:

A = {. . . ,−14,−8,−2, 0, 6, 12, 18, . . . }= {−2− 6n | n ∈ N} ∪ {0 + 6n | n ∈ N}

B = {. . . ,−11,−7,−3, 1, 6, 9, 13, . . . }= {4n+ 1 | n ∈ Z}

C = {. . . ,−9,−5,−1, 3, 7, 11, 15, . . . }= {4n+ 3 | n ∈ Z}

D = {. . . ,−16,−10,−4, 2, 8, 14, 20, . . . }= {6n+ 2 | n ∈ Z}

E = {. . . ,−18,−12,−6, 4, 10, 16, 22, . . . }= {−6− 6n | n ∈ N} ∪ {4 + 6n | n ∈ N}

(a) Is every integer in one of these sets?

(b) Do these sets have the property described in Corollary 2.1.6: either they are equal orthey are disjoint?

(c) Can you define an addition table for these sets based on adding elements in the setstogether?

Solution:

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(a) Yes, every integer is in one of A, B, C, D, and E.

(b) Yes, these sets are all disjoint from each other.

(c) Let’s try. What about A + B? If we pick 0 ∈ A and 1 ∈ B, we should have 0 + 1 ∈ Band so A + B = B. On the other hand, if we pick −14 ∈ A and 13 ∈ B then−14 + 13 = −1 ∈ C. So which is it? Should we define A+B = B or A+B = C. Wecan’t answer “it depends on which of A and B you use” and we are only interested indefining addition based on using elements. So no, it’s impossible. (Of course we canjust make up what A+ B should be without looking at all at the elements, and that’sfine, but it’s not what the point of this example is about).

The next theorem shows that the problems we had in the previous example do not arisewhen we are dealing with congruence classes.

Theorem 2.2.1. Let n ∈ Z, n ≥ 1. Let A and C be equivalence classes in Zn. Let a, b ∈ A andc, d ∈ C. Then

[a+ c] = [b+ d] and [ac] = [bd].

In other words, it doesn’t matter which elements we use in A and C to define addition andmultiplicaiton.

Proof. Since a, b ∈ A we have a ≡ b (mod n) and similarly c ≡ d (mod n). Thereforea+ b ≡ c+ d (mod n) and ab ≡ cd (mod n) by Theorem 2.1.3.

Definition 2.2.2. Let n ∈ Z, n ≥ 1. Let Zn be the collection of all equivalence classes ofintegers modulo n. The canonical representatives of Zn are as follows:

Zn ={[0], [1], . . . , [n−1]

}.

We define addition and multiplication in Zn as follows:

[a]⊕ [c] = [a+ c] and [a]� [c] = [ac].

Example 2.2.4. Fill in the following addition and multiplation tables for Z2 and Z3.

⊕ [0] [1][0][1]

� [0] [1][0][1]

⊕ [0] [1] [2][0][1][2]

� [0] [1] [2][0][1][2]

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Solution:⊕ [0] [1][0] [0] [1][1] [1] [0]

� [0] [1][0] [0] [0][1] [0] [1]

⊕ [0] [1] [2][0] [0] [1] [2][1] [1] [2] [0][2] [2] [0] [1]

� [0] [1] [2][0] [0] [0] [0][1] [0] [1] [2][2] [0] [2] [1]

Now we compare the algebraic properties in Z, as summarized in Definition 1.1.1, withthe algebraic properties in Zn.

Theorem 2.2.3. For all [a], [b], [c] in Zn the following hold:

1. [a]⊕ (b⊕ c) = ([a]⊕ b)⊕ c,

2. [a]⊕ b = b⊕ [a],

3. [0]⊕ [a] = [a]⊕ [0] = [a],

4. [a]⊕ [−a] = 0,

5. [a]� ([b]� [c]) = ([a]� [b])� [c],

6. [a]� [b] = [b]� [a]

7. [a]� [1] = [a]

8. [a]� ([b]⊕ [c]) = ([a]� [b])⊕ ([a]� [c])

The properties Z9–Z11 from Definition 1.1.1 do not apply to Zn, because we have notdefined what [a] < [b] would mean. In fact, there is no good way to define what < shouldmean for Zn so that it satisfies property Z9.

Definition 2.2.4. Let [a] ∈ Zn and let k ∈ N. We define

k[a] = [a]⊕ · · · ⊕ [a] (repeated k times)

[a]k = [a]� · · · � [a] (repeated k times)

The previous definition carries some of the notation that we are used to for Z, R, etc.over to Zn. In fact, we will in general apply notation and conventions to Zn, when theymake sense. For example, we will assume that the order of operations is applie in Zn as inR so that [a]⊕ [b]� [c] means multiplication first, then addition.

Example 2.2.5. Find all the solutions of x2⊕ [4]� x = [0] in Z7. (Try first just guessing andchecking, but then try and be more clever.)Solution:

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x?= [0]

[0]2 ⊕ [4]� [0] = [0]⊕ [0]

= [0]X

x?= [1]

[1]2 ⊕ [4]� [1] = [1]⊕ [4]

= [5] no

x?= [2]

[2]2 ⊕ [4]� [2] = [4]⊕ [8]

= [13]

= [6] no

x?= [3]

[3]2 ⊕ [4]� [3] = [9]⊕ [12]

= [21]

= [0]X

x?= [4]

[4]2 ⊕ [4]� [4] = [16]⊕ [16]

= [32]

= [4] no

x?= [5]

[5]2 ⊕ [4]� [5] = [25]⊕ [20]

= [45]

= [3] no

x?= [6]

[6]2 ⊕ [4]� [6] = [36]⊕ [24]

= [60]

= [4] no

We see that the only solutions are x = [0], [3]. Now we find them in a more clever way.How would we solve this equation over R, or over Z? We would factor. What makes

factoring work? The first step is to apply the distributive law, backwards. Since this law stillholds for Zn, we can apply it here too:

x2 ⊕ [4]� x = [0]

x(x⊕ [4]) = [0] (2.1)

x = [0] (2.2)

x⊕ [4] = [0] (2.3)

x = [−4]= [3]

WARNING: in this problem factoring gave the correct solutions, but this will not always bethe case. The problem isn’t the step where we factored, i.e. Equation (2.1). The problem isthat Equation (2.1) will not always imply Equations (2.2) or (2.3). In any case, it’s a goodidea to double check your answers, and we’ve done that above.


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2.3 The structure of Zp and ZnDefinition 2.3.1. We define new notation for the elements of Zn:

old notation: [0], [1], [a], etc.

new notation: 0, 1, a

and addition and multiplcation:

old notation: 1⊕ 2, a⊕ b, 1� 2, a� b, etc.

new notation: 1 + 2, a+ b, 1 · 2 or 1× 2, ab, etc.

and even for equivalence:

old notation: 2 + 5 ≡ 3 (mod 4)

new notation: 2 + 5 = 3.

The new notation has the advantage that there’s a lot less to write, and things look moresimilar to what we are used to with algebra in Z and R. It has the disadvantage that wehave to remember that we’re not working over Z anymore, even though it looks like we are.

Example 2.3.1. On the next page we have shown the addition and multiplication tablesfor Z11 and Z12. Identify as many patterns and facts as you can within each table, and alsoidentify differences between the Z11 and Z12 tables.

Solution: For the addition tables we noticed these patterns:

• Each row equals the previous row shifted to the left one place.

• The entries down the cross-diagonals (the diagonals with slope = 1) are constant.

For the addition and multiplication tables we noticed these patterns

• The entries in row x are equal to the entries in column x.

For multiplication we noticed these:

• Conjecture: The number of distinct elements in row x equals xgcd(n,x) .

• In Z11, the entries in each row/column have no repeats. As a consquence, each entryappears exactly once in each row and column.

• In Z12, if x is a nontrivial divisor of 12, or a multiple of one of these divisors, then rowx has repeats.

• In Z12, if x is not one of the above, then row x has no repeats.


Z11

+ 0 1 2 3 4 5 6 7 8 9 100 0 1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 10 02 2 3 4 5 6 7 8 9 10 0 13 3 4 5 6 7 8 9 10 0 1 24 4 5 6 7 8 9 10 0 1 2 35 5 6 7 8 9 10 0 1 2 3 46 6 7 8 9 10 0 1 2 3 4 57 7 8 9 10 0 1 2 3 4 5 68 8 9 10 0 1 2 3 4 5 6 79 9 10 0 1 2 3 4 5 6 7 810 10 0 1 2 3 4 5 6 7 8 9

· 0 1 2 3 4 5 6 7 8 9 100 0 0 0 0 0 0 0 0 0 0 01 0 1 2 3 4 5 6 7 8 9 102 0 2 4 6 8 10 1 3 5 7 93 0 3 6 9 1 4 7 10 2 5 84 0 4 8 1 5 9 2 6 10 3 75 0 5 10 4 9 3 8 2 7 1 66 0 6 1 7 2 8 3 9 4 10 57 0 7 3 10 6 2 9 5 1 8 48 0 8 5 2 10 7 4 1 9 6 39 0 9 7 5 3 1 10 8 6 4 210 0 10 9 8 7 6 5 4 3 2 1

Z12

+ 0 1 2 3 4 5 6 7 8 9 10 110 0 1 2 3 4 5 6 7 8 9 10 111 1 2 3 4 5 6 7 8 9 10 11 02 2 3 4 5 6 7 8 9 10 11 0 13 3 4 5 6 7 8 9 10 11 0 1 24 4 5 6 7 8 9 10 11 0 1 2 35 5 6 7 8 9 10 11 0 1 2 3 46 6 7 8 9 10 11 0 1 2 3 4 57 7 8 9 10 11 0 1 2 3 4 5 68 8 9 10 11 0 1 2 3 4 5 6 79 9 10 11 0 1 2 3 4 5 6 7 810 10 11 0 1 2 3 4 5 6 7 8 911 11 0 1 2 3 4 5 6 7 8 9 10

· 0 1 2 3 4 5 6 7 8 9 10 110 0 0 0 0 0 0 0 0 0 0 0 01 0 1 2 3 4 5 6 7 8 9 10 112 0 2 4 6 8 10 0 2 4 6 8 103 0 3 6 9 0 3 6 9 0 3 6 94 0 4 8 0 4 8 0 4 8 0 4 85 0 5 10 3 8 1 6 11 4 9 2 76 0 6 0 6 0 6 0 6 0 6 0 67 0 7 2 9 4 11 6 1 8 3 10 58 0 8 4 0 8 4 0 8 4 0 8 49 0 9 6 3 0 9 6 3 0 9 6 310 0 10 8 6 4 2 0 10 8 6 4 211 0 11 10 9 8 7 6 5 4 3 2 1

(version 2014/12/03) 36

After being prompted to think about when we have a row that equals contains a 1 or anontrivial 0, we came up with the following:

• In Z12, if x shares a factor with 12 then row x contains a 0.

• In Z12, if x is not one of the above, the row x contains a 1.

• In Z11, every row contains a 1.

Some of the patterns and facts we observed in the previous example are recorded in thetwo theorems of this section.

Definition 2.3.2. Let a ∈ Zn. If ax = 1 has a solution we call a a unit. If ax = 0 has asolution with a 6= 0 and x 6= 0 we call a a zero divisor

Theorem 2.3.3 (Zp has only units, and no zero divisors). Let p ∈ Z, p > 1 (we do not assumethat p is prime). The following statements are equivalent:

1. p is prime,

2. for any a ∈ Zp , if a 6= 0 then the equation ax = 1 has a solution x ∈ Zp ,

3. for all a, b ∈ Zp , if ab = 0 then a = 0 or b = 0.

Example 2.3.2. For each a ∈ Z11, identify a solution of ax = 1.

Solution:a x1 12 63 44 35 96 27 88 79 510 10

Proof. “(1)⇒(2)” Suppose p is prime. Let a ∈ Zp such that a 6= 0.

[a] 6= [0] in Zp given

a 6≡ 0 (mod p) in Z restatement

p - a in Z restatement

gcd(a, p) = 1 in Z p is prime and previous line

(version 2014/12/03) 37

am+ pn = 1 in Z Previous line and Theorem 1.2.4

am = 1− pn in Z previous line

am ≡ 1 (mod p) in Z[am] = [1] in Zp

[a][m] = = [1] in Zp

x = [m] is a solution


“(2)⇒(3)” Suppose that for any a ∈ Zp , a 6= 0, the equation ax = 1 has a solutionx ∈ Zp. Let ab = 0 in Zp. If a = 0 we are done. Otherwise, a 6= 0 and let x = u be a solutionof ax = 1.

ab = 0 given

au = 1 for a 6= 0

abu = 0u mult 1st line by u

(au)b = 0 assoc. prop., × by 0 prop.

1b = 0 2nd line

b = 0 mult by 1 prop.

“(3)⇒(1)” Suppose that for all a, b ∈ Zp , if ab = 0 then a = 0 or b = 0.

∀[a], [b] ∈ Zp, if [a][b] = [0] then [a] = [0] or [a] = [0]

∀a, b ∈ Z, if ab ≡ 0 (mod p) then a ≡ 0 (mod p) or b ≡ 0 (mod p)

∀a, b ∈ Z, if p|ab then p|a or p|b

Then p is prime by Theorem 1.3.5.

Theorem 2.3.4 (Classifying units and zero divisors in Zn). Let a ∈ Zn , with a 6= 0.

1. ax = 1 has a solution x ∈ Zn if and only if gcd(a, n) = 1.

2. ax = 0 has a nonzero solution x ∈ Zn if and only if gcd(a, n) 6= 1.

Proof. (1) “=⇒” direction. Suppose ax = 1 has a solution in Zn.

au = 1 in Zn given

[a][u] = [1] in Zn restatement

au ≡ 1 (mod n) in Z restatement

au = 1 + nk in Z definition of ≡au− nk = 1 restatement

gcd(a, n) = 1 Theorem 1.2.4

(1)“⇐=” direction. Suppose gcd(a, n) = 1. Repeat the second half of the proof ofTheorem 2.3.3 where we showed “(1)⇒(2)”.

ar + ns = 1 in Z Theorem 1.2.4

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ar = 1− ns in Z previous line

ar ≡ 1 (mod n) in Z[ar] = [1] in Zn

[a][r] = = [1] in Zn

x = [r] is a solution

Scholium. If gcd(a, n) = 1, then the solution of ax = 1 in Zn is given by x = r wherear + ns = 1 in Z.

Example 2.3.3. Identify the units and zero divisors in Z12. In each case identify both a andx.

Solution:Units Zero Divisors ax = 1 ax = 0

1, 5, 7, 11 2, 3, 4, 6, 8, 9, 10 a x1 15 57 711 11

a x2 63 44 36 28 39 410 6

Chapter 3

Rings

3.1 Definitions and Examples

Definition 3.1.1. A ring is a set R together two operations + and · that satisfy propertiesR1–R8 below:

1. ∀a, b ∈ R we have a+ b ∈ R (closure of addition)

2. ∀a, b, c ∈ R we have a+ (b+ c) = (a+ b) + c. (associative addition)

3. ∀a, b ∈ R we have a+ b = b+ a (addition is commutative)

4. ∃e ∈ R such that ∀a ∈ R we have e+ a = a. The usual notation for e is 0.(additive identity)

5. ∀a ∈ R, ∃b ∈ R such that a+ b = 0. The usual notation for b is −a. (additive inverses)

6. ∀a, b ∈ R we have a · b ∈ R. The usual notation for a · b is ab.(closure of multiplication)

7. ∀a, b, c ∈ R we have a(bc) = (ab)c (multiplication is associative)

8. ∀a, b, c ∈ R we have a(b+ c) = ab+ bc and (a+ b)c = ac+ bc. (distributive law)

This is where we ended on Wednesday, October 1

Definition 3.1.2. We say that R is a commutative ring if ∀a, b ∈ R we have ab = ba.

Definition 3.1.3. We say that R is a ring with identity if ∃e ∈ R such that ∀a ∈ R we haveea = ae = a. The usual notation for e is 1.

39

ring? commutative? with identity? integral domain? field?

N

Z

R

Q

C

R = 2Z

R = 2Z+ 1

R = {2Z, 2Z+ 1}

Zn

Zp

R = {0, a, b}(defined below)

R =M(R)(defined below)

R = RR

(defined below)

R = Z5 × Z3

(defined below)

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Example 3.1.1. Let 2Z denote the set of even integers, and let 2Z+1 denote the set of oddintegers. Fill in those parts of the chart on the following page that we know so far, justifyanswers where possible.

Solution: Z, R, Q and C are all commutative rings with identities. The usual rules ofalgebra are stated for R and apply equally to all these sets.

The set R = 2Z is a ring. The main properties to check are closure of addition, closureof multiplication and additive inverses. The other properties follow for 2Z since they holdfor Z (e.g. since a+ b = b+ a for all integers, it still holds if a and b are even integers). Theset R is commutative, but it does not have identity, since 1 6∈ R.

The set R = 2Z+ 1 is not a ring. It is not closed under addition.The set R = {2Z, 2Z + 1} is a ring. The operations on R are more famliar as the parity

rules for even and odd numbers. It is commutative and it is a ring with identity.The set R = N is not a ring. It does not contain additive inverses.The set Zn is a ring. It is commutative and it has identity.

Example 3.1.2. Let R = {0, a, b} with all these elements distinct.

(a) Fill in the rest of the following tables below so that R forms a ring.

(b) Is R commutative?

(c) Does R have identity?

+ 0 a b

0 0 a b

a a

b b

· 0 a b

0 0 0 0

a 0

b 0

Solution: We start by filling in the addition table, mostly by contradiction. If a+ b = b, thena = 0⇒⇐. If a+ b = a, then b = 0⇒⇐. Therefore, a+ b = 0.

If a+ a = 0, then a+ b = a+ a and a = b⇒⇐. Therefore a+ a = b.Commutativity implies b+ a = 0.If b+ b = b then b = 0⇒⇐. If b+ b = 0 then b+ b = a+ b and a = b⇒⇐.Therefore b+ b = a.

+ 0 a b0 0 a ba a b 0b b 0 a

Now we turn to multiplication.Since a + b = 0 we have 0 = a(a + b) = a2 + ab and similarly a2 + ba = 0. Thus

ab = ba = −a2 and similarly ab = ba = −b2. One way to make this multiplication work is toset all these products equal to 0. This is one valid way to make a ring, but we will constructsomething more interesting.

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Evidently, to make at least some of the products nonzero we must have ab = a or ab = b.It turns out to make little difference, so we arbitrarily choose ba = ab = a. Then a2 = −a = band b2 = −a = b.

· 0 a b0 0 0 0a 0 b ab 0 a b

This is where we ended on Friday, October 3

Example 3.1.3. Let M(R) be the set of all 2× 2 matrices components in R. Is R a ring? Ifyes, then is it commutative, and does it have identity?

Solution: Yes, R is a ring. If A and B are 2×2 matrices then it’s standard to note that A+Bis also a 2× 2 matrix, that A+B = B +A, etc.

The distributive law is not as obvious, but still standard.The ring R is not commutative, which is to say that in general AB 6= BA. To prove this

note that (0 10 0

)(0 01 0

)=

(1 00 0

)but

(0 01 0

)(0 10 0

)=

(0 00 1

).

The ring R has identity, mainly the identity matrix(1 00 1

).

Definition 3.1.4. A ring R is an integral domain if it is commutive, with identity, andsatisfies the following property: ∀a, b ∈ R, if ab = 0 then a = 0 or b = 0.

Example 3.1.4. Let R = RR denote the set of all functions from R to R. We can imaginethat R is just what you studied in Calculus I, ordinary functions, although it contains a fairnumber of functions that you haven’t ever seen.

For f, g ∈ R define f + g and fg in the usual ways

(f + g)(x) = f(x) + g(x) and (fg)(x) = f(x)g(x)

Is R a ring? Is it an integral domain?

Solution: Yes R is a ring. For instance, the additive identity is the function f that is alwaysequal to 0, i.e. f(x) = 0∀x. It is commutative, it does have identity, namely the functionthat is always equal to 1.

This is where we ended on Monday, October 6

The ring R is not an integral domain. Define two piecewise functions:

f(x) =

{0 if x ≤ 0

x if x > 0

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g(x) =

{x2 if x ≤ 0

0 = if x > 0

Then f 6= 0 and g 6= 0 (meaning that neither f nor g is the 0 function, i.e. the function thatis always 0). But fg = 0, meaning that f(x)g(x) = 0∀x.

Example 3.1.5. Out of all the examples of rings given above, which are integral domains?

Solution: We have already seen that RR is not an integral domain, no we look at the others.If R = Z then R is an integral domain. It should be very familiar that if ab = 0 then

a = 0 or b = 0. In fact, the name “integral domain” is supposed to suggest something like“similar to integer ring”. Similarly, if R is any one of R, Q or C then R is an integral domain.

If R = 2Z then R has no identity, and so it is not an integral domain.If R = {2Z, 2Z+1} then R is an integral domain. Here, 2Z plays the role of 0, and 2Z+1

plays the role of the nonzero element 1. We have 1 · 1 = 1.If R = Zn, and if n is not prime, then R is not an integral domain. Let n = ab where we

view n, a and b as integers. Then a 6= 0 and b 6= 0 in Zn, and yet ab = 0 in Zn.If R = Zp, then R is an integral domain, because there are no zero divisors.If R = M(R) then R is not an integral domain because R is not commutative. In

addition, R also fails the third condition: it is possible to have A 6= 0 and B 6= 0, but stillhave AB = 0, for instance

A =

(1 00 0

)and B =

(0 00 1

)then

AB = 0.

Definition 3.1.5. A ring R is a field if it commutative, with identity, and satisfies the fol-lowing property: ∀a ∈ R, if a 6= 0 then ax = 1 has a solution x ∈ R. If x is the solution ofthis equation we call x the inverse of a. The usual notation for the inverse is a−1.

Example 3.1.6. Out of all the examples of rings given above, which are fields?

Solution: If R = Z then R is not a field. For instance, the equation 2x = 1 has no solutionx ∈ Z.

The rings R, Q and C are all fields. In each case ax = 1 has a solution x, usually writtenas x = a−1. If a ∈ Q then we usually write a = b

c and then x = cb . If a ∈ R we don’t have

a special formula for a−1. If a ∈ C then we usually write a = b + ci and then the inverseformula is given by

(b+ ci)−1 =b√

b2 + c2− c√

b2 + c2i

The ring 2Z does not have identity, and so it cannot be a field.

(version 2014/12/03) 44

The ring R = {2Z, 2Z+1} is a field. The only nonzero element is 2Z+1 and it is its owninverse.

The ring Zn where n is composite is not a field. As we know, if n is not prime, thereexists an element a ∈ Zn such that ax = 1 has no solution in Zn.

The ring Zp where p is prime is a field.The ring R = {0, a, b} is a field.The ring M(R) is not a field: it is not commutative, and many matrices do not have

inverses.The ring RR is not a field: many functions do not have multiplicative inverses that are

defined on all of R. For instance, let f(x) = x2. Then g(x) = 1/x2 is only defined on R∗, theset of nonzero elements of R. If this isn’t extreme enough, consider

f(x) =

{0 if x ≤ 0

x if x > 0

Then a multiplicative inverse of f(x) cannot be defined for any x ≤ 0.

Theorem 3.1.6 (Product of rings). Let R and S be rings and R × S the Cartesian product oftheir sets. Define addition and multiplication on R× S as

(a, b) + (c, d) = (a+ c, b+ d) and (a, b)(c, d) = (ac, bd)

for all (a, b), (c, d) ∈ R× S. Then R× S is a ring.

Example 3.1.7. Consider the ring Z5 × Z3.

(a) Calculate (1, 2) + (4, 1) and (2, 2)(4, 2).

(b) Is Z5 × Z3 commutative?

(c) Does Z5 × Z3 have identity?

(d) Is Z5 × Z3 an integral domain?

Solution:

(a)

(1, 2) + (4, 1) = (1 + 4, 2 + 1)

= (5, 3)

= (0, 0) (reducing mod5 and mod3)

(2, 2)(4, 2) = (2 · 4, 2 · 2)= (8, 4)

= (3, 1) (reducing mod5 and mod3)

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(b) Yes, Z5 × Z3 commutative because it is commutative in each component. More pre-cisely:

(a, b)(c, d) = (ac, bd) = (ca, db) = (c, d)(a, b).

(c) Yes, Z5 × Z3 has identity. It is (1, 1).

(d) No, Z5 × Z3 is not an integral domain. For instance (1, 0) and (0, 1) are both nonzeroelements of Z5 × Z3, but their product is (0, 0), which is the 0 element.

Theorem 3.1.7. Let R and S be rings. If both R and S are both commutative, then so is R×S.If R and S both have identity then so does R× S.

Definition 3.1.8. Let R be a ring and let S be a subset of R. If S satisfies all the propertiesR1–R8 in Definiton 3.1.1, using the addition and multiplication defined in R, then we callS a subring. If R is a field and S is a subring that also satisfies Definition 3.1.5, then wecall S a subfield.


Theorem 3.1.9. Let R be a ring and S be a subset of R. If properties (1)–(4) listed below aretrue, then S is subring:

1. ∀a, b ∈ S, we have a+ b ∈ S,

2. ∀a, b ∈ S, we have ab ∈ S,

3. 0 ∈ S,

4. ∀a ∈ S, we have −a ∈ S.

Proof. The properties listed here are directly from the definition of ring. The remainingproperties from the definition are: the associative and commutative properties of addition,the associative property for multiplication, and the distributive law. These follow for S sincethey hold in R. For example, if a, b ∈ S and want to verify that a + b = b + a, simply notethat a and b are also in R. Since commutative holds in R we conclude that a+ b = b+a.

Corollary 3.1.10. Let R be a field and S a subring of R. Suppose that ∀a ∈ S, if a 6= 0 thenax = 1 has a solution x ∈ S. Then S is a subfield.

Example 3.1.8. Out of all the rings and fields we have studied above, which are subringsand and which are subfields.

Solution: We indicate below the relations:

2Zsubring⊆ Z

subring⊆ Q

subfield⊆ R

subfield⊆ C

Some of the other subsets are not subrings: for intstance, N ⊆ Z, but N is not a subring.Simlarly, R = 2Z+ 1 is a subset of Z but not a subring.

(version 2014/12/03) 46

Some of the others rings are close to being subrings in the following sense: technically,one set might not be a subset of another, but there might be a copy of the first inside thesecond. For instance, there is a “copy” of R inside of M(R): consider the subset of M(R)that has 0’s except in the first entry:{

all matrices of form(a 00 0

)}= copy of R.

In a similar sense, one could say that there is a copy of Z5 inside of Z5 × Z3, namely all theentries in the first coordinate.

3.2 Basic Algebraic Properties of Rings

Theorem 3.2.1. Let R be a ring. The followming hold:

1. ∀a ∈ R, there is a unique element b ∈ R such that a+ b = 0.

2. ∀a, b, c ∈ R, if a+ b = a+ c then b = c (additive cancelation law)

3. ∀a ∈ R, we have a0 = 0a = 0.

4. ∀a, b ∈ R, we have a(−b) = (−a)b = −ab.

5. ∀a ∈ R, we have −(−a) = a.

6. ∀a, b ∈ R, we have −(a+ b) = (−a) + (−b).

7. ∀a, b ∈ R, we have (−a)(−b) = ab.

Proof. In the proofs below I will show more steps than may be necessary. I don’t mind ifyou skip some steps like this on your homework. But, be very careful: the reason I showedthese steps is that each one is very explicitly true by one of the ring properties. If you skip astep like this, you better check, at least in your head, that it follows from the ring propertiesand that you’re not assuming a property that’s not always true.

Part 1. Suppose a+ b = 0 and a+ c = 0. Add c to the first equation to get

a+ b = 0

c+ (a+ b) = c+ 0

c+ (a+ b) = c+ 0

(a+ c) + b = c

0 + b = c

b = c

Part 2. Suppose a+ b = a+ c. Add −a to both sides to get

a+ b = a+ c

(version 2014/12/03) 47

−a+ (a+ b) = −a+ (a+ c)

(−a+ a) + b = (−a+ a) + c

0 + b = 0 + c

Part 3. After thinking about it, it should be clear that the only way to prove this is touse the distributive law. The reason is that propertise R1–R7 each only describe somethingabout addition, or something about multiplication, but not both. In other words, if we useonly R1–R7 we will never be able to make any connection between addition and multiplica-tion. But, part 3 requires a connection between addition and multiplication: the statementinvolves multiplication, and 0 involves, by definition, addition.

0 + 0 = 0

a(0 + 0) = a · 0a · 0 + a · 0 = a · 0

a · 0 + a · 0− (a · 0) = a · 0− (a · 0)a · 0 + 0 = 0

a · 0 = 0

The remaining parts are left as exercises.

Definition 3.2.2. Let R be a ring. Let a ∈ R and let k ∈ N. We define

ka = a+ . . .+ a (repeated k times)

ak = a · . . . · a (repeated k times)

Furthermore, define

−ka = (−a) + . . .+ (−a) (repeated k times)

0a = 0

a0 = 1 if (and only if) R has identity

Example 3.2.1. Let R be any ring and a, b ∈ R. Can you rewrite (a + b)2? Under whatconditions can rewrite it?Solution: The point of the question is: to what extend can you foil in a general ring?

(a+ b)2 = c2 let c = a+ b

= c · c definition of c2

= c(a+ b) definition of c

= ca+ cb distributive law

= (a+ b)a+ (a+ b)b definition of c

= a · a+ ba+ ab+ b · b distributive law

= a2 + ba+ ab+ b2 definition of a2 and b2

(version 2014/12/03) 48

This is as far as we can go in general. But we can extend this in two ways if R has extraproperties:

If R is commutative: (a+ b)2 = a2 + ba+ ab+ b2 = a2 + 2ab+ b2

If R has also identity: (a+ b)2 =

2∑i=0

(2

i

)a2−ibi

Of course, we don’t need the second statement just for squaring things, but if you have(a+ b)3 or (a+ b)n, then the Binomial Theorem applies and you can write:

(a+ b)n =

n∑i=0

(n

i

)an−ibi

for any n ∈ N and any commutative ring with identity.

This is where we ended on Friday, October 10

Definition 3.2.3. Let R be a ring and let a ∈ R. If ax = 0 or xa = 0 has a solution witha 6= 0 and x 6= 0 we call a a zero divisor. If R has an identity, and ax = 1 and xa = 1 has asolution x ∈ R then we call a a unit.

Fact 3.2.4. Using the language above, we can restate two of our earlier definitions:

1. A ring is an integral domain if and only if it is commutive, with identity, and has nozero divisors.

2. A ring is a field if and only if it is commutative, with identity, and every nonzeroelement is a unit.

Example 3.2.2. Let R = M(R), the 2 × 2 matrices over R. Describe the units and zerodivisors of R.Solution: Let A ∈ R. Then A is a unit if and only if A−1 exists. As we recall from linear

algebra, this is true if and only if det(A) 6= 0 (in fact, in this case, if A =

(a bc d

)then

A−1 = 1ad−bc

(d −b−c a

)).

Now suppose det(A) = 0. Then one row of A must equal a multiple of the other row,say Row 2 equals a multiple of Row 1:

A =

(a bra rb

).

Then it’s easy to find another matrix to multiply by A to get 0:

B =

(b b−a −a

)

(version 2014/12/03) 49

Note that AB = 0. It also not hard to find a matrix to multiply by on the other side:

C =

(r −1r −1

)Note that CA = 0.

We summarize our results in the theorem below.

Theorem 3.2.5 (Classification of units and zero divisors in matrix ring). Let R =M(R) andlet A ∈ R.

1. A is a unit if and only if detA 6= 0.

2. A is a zero divisor if and only if detA = 0.

Theorem 3.2.6. Let R be any ring with identity and let a ∈ R. Then a cannot be both a unitand a zero divisor.

Proof. Note that if a = 0 then a cannot be a unit (0 times anything equals 0) and it alsocannot be a zero divisor (by definition zero divisors do not include 0). So, let a 6= 0 andsuppose for contradiction that a is both a unit and a zero divisor. Then we have ab = ba = 1and ac = 0 for some b, c ∈ R with c 6= 0. Multiply the second equation by b to get

ac = 0

b(ac) = b0

(ba)c = 0

1c = 0

c = 0

which contradicts the fact that c 6= 0.

Question. Here’s what we know about units and zero divisors so far:

1. In all rings, you can never have an element that is both a unit and a zero divisor.

2. In Zn, every element is either a unit or a zero divisor (but not both).

3. In M(R), every element is either a unit or a zero divisor (but not both).

4. In a field, every nonzero element is a unit, no elements are zero divisors.

This leads to some questions:

1. Is there a ring with nonzero elements that are neither units or zero divisors?

2. Is there a nontrivial ring where all nonzero elements are zero divisors?

Theorem 3.2.7 (Multiplicative cancelation law). Let R be an integral domain. The followingholds:

∀a, b, c ∈ R, if a 6= 0 and ab = ac then b = c (multiplicative cancellation law).

(version 2014/12/03) 50


Proof. Assume that a 6= 0 and ab = ac. We calculate

ab = ac

ab− ac = 0

a(b− c) = 0

a = 0 or b− c = 0 (because we are in an integral domain)

b− c = 0 (because a 6= 0)

b = c

Example 3.2.3. Let R = Z3[i]. In other words, we have

R = {a+ bi | a, b ∈ Z3, i2 = −1 = 2}

where we define addition and multiplication as follows

∀a+ bi, c+ di ∈ R : (a+ bi) + (c+ di) = a+ c+ (b+ d)i

(a+ bi)(c+ di) = ac+ 2bd+ (bc+ ad)i

Assume that R is a ring.

(a) Using the formulas for multiplication, show that R has no zero divisors.

(b) Fill in the multiplication table of R.

(c) Is R a field?

Solution:

(a) Suppose that (a + bi)(c + di) = 0, but that a + bi 6= 0. We will prove that c + di = 0.By the formula for multiplication above, we have

(a+ bi)(c+ di) = ac+ 2bd+ (bc+ ad)i = 0

which we turn into two equations:

ac+ 2bd = 0 (3.1)

bc+ ad = 0 (3.2)

Now we do some calculations that may seem to come from nowhere, but which wewill see below are somewhat predictable:

a2, b2 = 0 or 1 (02 = 0, 12 = 1, 22 = 1 in Z3)

a+ bi 6= 0

a 6= 0 or b 6= 0

(version 2014/12/03) 51

(a2, b2) = (0, 1), (1, 0), or (1, 1)

a2 + b2 = 1 or 2

a2 + b2 6= 0

Now we will combine Equations (3.1) and (3.2) in a rather complicated fashion:

a ∗ (Eq.3.1) + b ∗ (Eq.3.2) : a(ac+ 2bd) + b(bc+ ad) = 0

2b ∗ (Eq.3.1) + a ∗ (Eq.3.2) : 2b(ac+ 2bd) + a(bc+ ad) = 0

Now we simplify:{a(ac+ 2bd) + b(bc+ ad) = 0

2b(ac+ 2bd) + a(bc+ ad) = 0

}{

a2c+ 2abd+ b2c+ abd = 0

2abc+ 2b2d+ abc+ a2d = 0

}{

a2c+ 3abd+ b2c = 0

3abd+ 2b2d+ a2d = 0

}{

a2c+ b2c = 0

2b2d+ a2d = 0

}since we’re in Z3{

(a2 + b2)c = 0

(a2 + b2)d = 0

}{c = 0

d = 0

}since Z3 is an integral domain and a2 + b2 6= 0

So, why are the above calculations predictable? We have basically just done linear al-gebra without matrices (we skipped the matrices only because using them may requirethat the student believe too many new things at once!):(

a 2bb a

)(cd

)=

(00

)(a2 + b2)−1

(a b2b a

)(a 2bb a

)(cd

)=

(00

)(a2 + b2)−1

(a2 00 b2

)(cd

)=

(00

)(1 00 1

)(cd

)=

(00

)(cd

)=

(00

)Or, even more briefly: (

a 2bb a

)(cd

)=

(00

)

(version 2014/12/03) 52

det

(a 2bb a

)6= 0(

cd

)=

(00

)

(b) Here’s the multiplication table for R.

· 0 1 2 i 1 + i 2 + i 2i 1 + 2i 2 + 2i0 0 0 0 0 0 0 0 0 01 0 1 2 i 1 + i 2 + i 2i 1 + 2i 2 + 2i2 0 2 1 2i 2 + 2i 1 + 2i i 2 + i 1 + ii 0 i 2i 2 2 + i 2 + 2i 1 1 + i 1 + 2i

1 + i 0 1 + i 2 + 2i 2 + i 2i 1 1 + 2i 2 i2 + i 0 2 + i 1 + 2i 2 + 2i 1 i 1 + i 2i 22i 0 2i i 1 1 + 2i 1 + i 2 2 + 2i 2 + i

1 + 2i 0 1 + 2i 2 + i 1 + i 2 2i 2 + 2i i 12 + 2i 0 2 + 2i 1 + i 1 + 2i i 2 2 + i 1 2i

(c) Yes, R is a field. Every nonzero element of R has an inverse, because we can find theelement 1 in each row of the above table.


Theorem 3.2.8. Every field is an integral domain.

Proof. Let F be a field. Then every nonzero element of F is a unit. Therefore, by Theo-rem 3.2.6, every nonzero element of F is not a zero divisor. Therefore, F is an integraldomain.

Theorem 3.2.9. Every finite integral domain is a field.

Proof. Let F be a finite integral domain. Fix a ∈ F with a 6= 0. Define a function f(x) = ax.Then f : F → F . If f(x1) = f(x2) then ax1 = ax2, and the multiplicative cancellation lawshows that x1 = x2. This shows that f is injective. Since f is injective, as we apply it to allthe elements of F , none of the elements in F show up twice in the range of f . Since F isfinite, this means that every element shows up in the range f exactly once. Thus, f(x) = 1for some x ∈ F . This means ax = 1, as desired.

Theorem 3.2.10 (Subring Test). Let S be a subset of a ring R. If S satisfies the propertiesbelow, then S is a subring:

1. ∀a, b ∈ S, we have a− b ∈ S (closed under subtraction),

2. ∀a, b ∈ S, we have ab ∈ S (closed under multiplication).

(version 2014/12/03) 53

Proof. It suffices to show that this theorem implies the conditions of Theorem 3.1.9. Inother words, we assume that (i) and (ii) from this theorem hold, and then we ill prove thatall four conditions in Theorem 3.1.9 hold. Label the conditions in Theorem 3.1.9 as (1), (2)(3) and (4).

Note that condition (ii) is identical to condition (2), so (ii)⇒ (2).Now we prove (3). Let a ∈ S. Then (i) implies that a − a ∈ S. Therefore 0 ∈ S, so (3)

holds.Now we prove (4). Let a ∈ S. We already know that 0 ∈ S so (i) implies that 0− a ∈ S,

which means −a ∈ S and so (3) holds.Now we prove (1). Let a, b ∈ S. We know that −b ∈ S and so (i) implies that a− (−b) ∈

S. But a− (−b) = a+ b so a+ b ∈ S and (1) holds.

Example 3.2.4. Let R = Q[√2]. In other words

R = {a+ b√2 | a, b ∈ Q}

with addition and multplication inherited from R.

(a) What is the easiest way to show that R is a subring?

(b) What is the easiest way to show that R is an inegral domain?

(c) What is the easiest way to show that R is a field?

Solution:

(a) To see that it is a subring we note that subtracting and multiplying elements of R areclosed:

(a+ b√2)− (c+ d

√2) = (a+ c)− (b+ d)

√2

(a+ b√2)(c+ d

√2) = (ac+ 2bd) + (ad+ bc)

√2

(b) To see that it is an integral domain, not that R ⊆ R, and that 1 ∈ R. Therefore, R iscommutative, has identity, and has no zero divisors.

(c) To show that it is a field, we need to show that every nonzero element has an inversein R. We know that this inverse exists in R, but it might not be in R. For instance, 1√

2

exists in R, but does not appear to exist in R. However, once we think about it, wesee that 1√

2= 1

2

√2 ∈ R. A similar trick would work for all elements of R of the form

b√2. In fact, we claim that

(a+ b√2)−1 =

a

a2 − 2b2− b

a2 − 2b2

√2.

Note that the formula on the right gives an element of R, so once we verify thisformula we are done. (Note also that this formula is always defined: a2 − 2b2 6= 0

because a2

b2 6= 2 for any rational numbers because ab 6=√2 because

√2 is irrational!)

(a+ b√2)

(a

a2 − 2b2− b

a2 − 2b2

√2

)= (a+ b

√2)

(a− b√2)

a2 − 2b2

(version 2014/12/03) 54

=a2 − 2b2

a2 − 2b2

= 1

Example 3.2.5. Can you fill in the following chart with T or F, and justify your answerswith a theorem, example, or quick proof? Put a T in a spot if a given property is preservedor inherited by a given ring construction. So, for instance, you would put a T in the spot forfield and subring if you thought that given a field R, every subring of R was also a subfield.

Here’s an explanation for each property along the top. Subring: true if the given propertyis inherited by subrings, i.e. for every ringR, ifR has propertyX then every subring ofR hasproperty X. Cartesian product: true if the given property is inherited by cartesian products,i.e. for every pair of rings R1 and R2, if R1 and R2 have property X then the cartesianproduct R1 × R2 has property X. Extension: true if thea given property is inherited bylarger rings, i.e. for every ring R, if R has property X then any ring that contains R as asubring also has property X.

subringscartesianproducts extensions

commutative

identity

has some zero divisors

has some units

all nonzero elements areunits or zero divisors

field

unit/zero divisor dichotomy

3.3 Isomorphisms and Homomorphisms

Example 3.3.1. Recall Example , where we constructed the ring R = {0, a, b} with additionand multiplication defined by

+ 0 a b0 0 a ba a b 0b b 0 a

and

· 0 a b0 0 0 0a 0 b ab 0 a b

(version 2014/12/03) 55

This ring is the same (in some sense) as another ring we’ve studied. What other ring? Howprecise can you make the statement that they are the same?Solution: R is the same as the ring Z3.


If we simply replace b with 1, and a with 2 we obtain Z3:

+ 0 �a 2 �b 10 0 �a 2 �b 1

�a 2 �a 2 �b 1 0

�b 1 �b 1 0 �a 2

and

· 0 �a 2 �b 10 0 0 0

�a 2 0 �b 1 �a 2

�b 1 0 �a 2 �b 1

Here’s another way to say this. If we define f : R→ Z3 as follows

f : 0 7→ 0a 7→ 2b 7→ 1

then f turns the addition and multiplication table for R into the addition and multiplicationtable for Z3.

Example 3.3.2. Recall from Example 3.1.7 the ring Z5×Z3. This ring is the same as anotherring we’ve seen (although we haven’t written out the second one explicitly). What do youthink the second ring is? Can you make a conjecture as to how to show that they are thesame?Solution: The ring Z5 × Z3 is the same as the ring Z15. The easiest way to make this guessis that Z5 × Z3 has exactly 15 elements, as does Z15. To prove this is true we would definef : Z5 × Z3 → Z15 and then show that f turns the addition and multiplication tables forZ5 × Z3 into the corresponding tables for Z15.

We don’t want to write both tables for both rings, and part of what we learn below willbe how to show that two rings are the same without writing out the tables, which would beespecially hard for an infinite ring! But for now, we can probably guess what f should looklike.

We should guess that f(0, 0) = 0 because f should take the additive identity for one ringto the additive identity for the other ring. The same reasoning, to the multiplicative identity,shows that should guess f(1, 1) = 1. In fact, we can make further guesses, but we will waituntil we give our definition.

Example 3.3.3. While reading an ancient Chinese text in the library one day, you comeacross the following

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

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!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

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!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

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!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

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!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

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!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

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!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

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<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

(version 2014/12/03) 56

As you gaze at them, you begin to recognize a pattern. Where have you seen that before?Can you make your observation more precise?Solution: Aha! It’s Z4. But is it really the same? How can you tell, it looks so different.

Define a map f : {

!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

,

!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

,

!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

,

!"#$#%&'#%("#%#)('*#$%*)%("#%&++*(*,)%&)+%-./(*0/*1&(*,)%(&2/#$%3,'%456%*)%("#%('&+*(*,)&/%1"*)#$#%1"&'&1(#'$7%%8%"&9#%),(%/*$(#+%("#%!'$(%',:%,'%1,/.-)%"#&+*);$7%%

!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

} → {0, 1, 2, 3} via

f :

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

7→ 0

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

7→ 1

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!"#%#)('*#$%<=>?%1,''#$0,)+%(,%@A%BA%CA%DA%'#$0#1(*9#/E7%%

<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

7→ 2

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<=>?=>?<>?<=?<=>

<<<<<=>?<><><?>=

7→ 3

Note that if you apply f to the Chinese tables above, you get the following:

0 1 2 31 2 3 02 3 0 13 0 1 2

0 0 0 00 1 2 30 2 0 20 3 2 1

These are the usual tables for Z4 (minus the row and column that we usually use for labellingthe going on inside.

Definition 3.3.1. Let R and S be rings, with addition and multiplication written as +R

,

·R

, +S

and ·S

. Let f : R → S a map. We say that f is an isomorphism if it satisfies the

following.

1. f is a bijection from R onto S, (i.e. f is one-to-one and onto),

2. ∀a, b ∈ R we have f(a +Rb) = f(a) +

Sf(b), (i.e. f preserves addition)

3. ∀a, b ∈ R we have f(a ·Rb) = f(a) ·

Sf(b) (i.e. f preserves multiplication).

If such an f exists we say thatR and S are isomorphic and we writeR ∼= S. We usually dropthe extra notation of +

R, +

S, etc. and just write +, relying on the reader to remember

which ring the operation is working in.

The previous three examples show the right way to think about isomorphisms: theymerely translate the way we write a ring into different language/notation. Similarly, if tworings are isomorphic, then for all intents and purposes they are the same. Every equationthat is solvable in one is solvable in the other, every property involving units, zero divisors,number of elements, etc., is the same between the two rings.

The previous definition tells us how to be more precise in what it means to translate onering into another: we have to translate what we call each element (this is what f does), andthe translation also has to apply to the operations of + and ·.

(version 2014/12/03) 57

Example 3.3.4. Return to Example 3.3.2 and see if you can finish proving that Z5×Z3 andZ15 are isomorphic.

Solution: We saw above that we should define

f : Z5 × Z3 → Z15

and that we should have f(0, 0) = 0 and f(1, 1) = 1. We now have some idea of additionalproperties that f should have: it should be bijective, it should preserve addition and mul-tiplication. It turns out that these properties basically determine f in the sense that there’sonly one way to define f on all the elements of Z5 × Z3.

If we define f(1, 1) = 1, and we require that f preserve addition, then we must define

f

((1, 1) +

Z5×Z3

(1, 1)

)= f(1, 1) +

Z15

f(1, 1).

This means that me must define f(2, 2) = 1 +Z15

1, which means f(2, 2) = 2. If we define

f(2, 2) and we requre that f preserve addition, then we must define

f

((2, 2) +

Z5×Z3

(1, 1)

)= f(2, 2) +

Z15

f(1, 1).

This implies that f(3, 0) = 2+1 = 3. Continuing in this way we have an inductive argumentthat requires we define f as follows:

If f(1, 1) = 1 and f preserves addition, then:

f(2, 2) = 2

f(3, 0) = 3

f(4, 1) = 4

f(0, 2) = 5

f(1, 0) = 6

f(2, 1) = 7

f(3, 2) = 8

f(4, 0) = 9

f(0, 1) = 10

f(1, 2) = 11

f(2, 0) = 12

f(3, 1) = 13

f(4, 2) = 14

f(0, 0) = 0

(By the way, can you figure out a formula for f?) We should pause to verify, by looking atthe table, that we have not defined f in a contradictory fashion. In other words, each valueof (a, b) only shows up once on the left hand side as f(a, b). We should pause to verify that

(version 2014/12/03) 58

we have a bijection. This is easy to check since in this case we have listed 15 distinct thingson the left, and 15 distinct things on the right, and the two sets each had 15 elements. Weshould pause to verify that f preserves multiplication.

Well, this isn’t so obvious. We can verify this for a couple of examples. The main ideais that we have to evaluate f in two ways: once doing all the calculations in side theparanenthes, and then applying f , and then the opposite, applying f as soon as possibleand then doing the calculations.

calculate inside of f first, then apply f :

f

((2, 2) ·

Z5×Z3

(3, 2)

)= f(6, 4)

= f(1, 1)

= 1

apply f to parts first, then calculate:

f

((2, 2) ·

Z5×Z3

(3, 2)

)= f(2, 2) ·

Z15

f(3, 2)

= 2 ·Z15

8

= 16

= 1

So these two calculations came out the same, meaning that f preserved the multiplicationof (2, 2) and (3, 2). But in theory we need to double check all the calculations or do a generalproof. We cannot do a general proof until we have a formula for f . It’s not so easy to find aformula for f , so we postpone finishing this proof.

Example 3.3.5. Let R be the set of 2× 2 matrices of the following form

R =

{(a b−b a

)| a, b ∈ R

}.

Assume that R is a ring under the usual matrix addition and multiplication. Show that R isisomorphic to C.Solution: We define f as follows:

f

(a b−b a

)= a+ bi.

We will verify below that

• f is surjective,

• f is injective,

• f preserves addition,

(version 2014/12/03) 59

• f preserves multiplication.


• Surjective: Let a+ bi ∈ C. Then f(a b−b a

)= a+ bi.

• Injective: Suppose f(a b−b a

)= f

(c d−d c

). Then a+ bi = c+di which means a = c

and b = d. This shows(a b−b a

)=

(c d−d c

).

• Preserves addition:

f

((a b−b a

)+

(c d−d c

))= f

(a+ c b+ d−b− d a+ c

)= (a+ c) + (b+ d)i (3.3)

f

(a b−b a

)+ f

(c d−d c

)= (a+ bi) + (c+ di)

= (a+ c) + (b+ d)i (3.4)

Since Equations (3.3) and (3.4) are equal to each other, we see that f preserves addi-tion.

• Preserves multiplication:

f

((a b−b a

)(c d−d c

))= f

(ac− bd ad+ bc−bc− ad −bd+ ac

)= (ac− bd) + (ad+ bc)i (3.5)

f

(a b−b a

)f

(c d−d c

)= (a+ bi)(c+ di)

= (ac− bd) + (ad+ bc)i (3.6)

Since Equations (3.5) and (3.6) are equal to each other, we see that f preserves mul-tiplication

Theorem 3.3.2 (Inverses of isomorphisms are isomorphisms). Let f : R → S be an isomor-phism. Then f has an inverse function f−1, and f−1 : S → R is an isomorphism.

Proof. Since f is a bijection the inverse function exists: Given y ∈ S we let x ∈ R satisfyf(x) = y and then define f−1(y) to be equal to x.

Now we need to prove that f−1 preserves addition and multiplication.This is where we ended on Friday, October 24

Fix a, b ∈ S. We calculate and justify:

f−1(a+ b) = f−1(f(x) + f(y)) f is onto: let x, y ∈ R such that f(x) = a and f(y) = b

(version 2014/12/03) 60

= f−1(f(x+ y)) f preserves addition

= x+ y inverse functions

= f−1(a) + f−1(b) solve f(x) = a and f(y) = b for x and y

The proof that f−1 preserves multiplication is identical with the change of + to ·.

Example 3.3.6. Finish Example 3.3.4 by proving that Z5 × Z3 and Z15 are isomorphic.Solution: The trick is to work with f−1 instead of f . Here is the table that defines f−1

f(1) = (1, 1)

f(2) = (2, 2)

f(3) = (3, 0)

f(4) = (4, 1)

f(5) = (0, 2)

f(6) = (1, 0)

f(7) = (2, 1)

f(8) = (3, 2)

f(9) = (4, 0)

f(10) = (0, 1)

f(11) = (1, 2)

f(12) = (2, 0)

f(13) = (3, 1)

f(14) = (4, 2)

f(0) = (0, 0)

Note that f([a]15) = ([a]5, [a]3) where [a]15 means that we take the class in Z15 representedby the number a, and similarly for [a]5 and [a]3. Now that we have a general formula forf−1, we can show that it presevers addition in general:

calculate inside of f−1 first, then apply f−1:

f−1([a]15 + [b]15

)= f−1([a+ b]15) adding in Z15

=([a+ b]5, [a+ b]3

)definition of f−1

=([a]5 + [b]5, [a]3 + [b]3

)adding in Z5 and Z3

apply f−1 to parts first, then calculate:

f−1([a]15 + [b]15

)= f−1([a]15) + f−1([b]15) applying f−1 to parts

= ([a]5, [a]3) + ([b]5, [b]3) definition of f

= ([a]5 + [b]5, [a]3 + [b]3) addition in Z5 × Z3

= ([a+ b]5, [a+ b]3) addition in Z5 and Z3

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There are functions from one ring to another that are not isomorphism, but are still prettyuseful to study. Which properties do you think we can weaken and still have a function thatmeans somethin? Bijection? Preserving addition? Preserving multiplication?

Example 3.3.7. Define the function Z→ Z7 via x 7→ [x]7.

(a) Which properties in the definition of isomorphism does f satisfy?

(b) Although f fails some properties, is it still useful for studying Z or Z7 or anything?

Solution: It’s easy to see that f is surjective since f(0) = [0], f(1) = [1], . . . , f(6) = []6. But,f is far from injective: f(0) = f(7) = f(14) = . . . , and so on.

On the other hand, f , does preserve addition (it’s essentially the same calculation thatwe gave above for the function Z15 → Z5 × Z3):

calculate inside of f first, then apply f :

f(a+ b) = [a+ b]7 definition of f

apply f to parts first, then calculate:

f(a) + f(b) = [a]7 + [b]7

= [a+ b]7 addition in Z7

A similar proof shows that f preserves multiplication.Here’s one application of this particular function. It’s not the kind of application you

might think of ahead of time, the point is just that it turns a question that is hard to answerin Z into one that is easy to answer in Z7. Suppose I ask the following question: “Can I findtwo integers a and b such that a2 + b2 = 7777?”

The answer is no. To prove this, suppose for contradiction that such integers exist:

a2 + b2 = 7777, a, b ∈ Z, a, b 6= 0.

Then we apply f and calculate:

f(a2 + b2) = f(7777), in Z7

f(a2) + f(b)2 = [0], in Z7

[a]2 + [b]2 = [0]

We calculate directly what the possibilities are in Z7:

[a]2, [b]2 ∈ {[0], [1], [4], [2]}.

No two of these add up to [0], unless we use [0] itself. So we must have [a] = [b] = [0]. Thismeans that 7|a and 7|b (in Z). Thus, a = 7x and b = 7x and so

a2 + b2 = 7777 in Z(7x)2 + (7y)2 = 7777

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72x2 + 72y2 = 7777

72(x2 + y2) = 7777

which implies 72 divides 7777. But this is false, and so our original assumption is wrong.

Example 3.3.8. Defined the function f : R→M(R) via f(r) =(r 00 0

).


(b) Although f fails some properties, is it still useful for studying R or M(R) or anything?


Solution: It’s easy to see that f is injective, for if f(r) = f(s) then r = s. But, f is far from

surjective, we cannot find f(r) =

(a bc d

)unless b = c = d = 0. It’s easy to check that f

preserves addition and multiplication:

f(r + s) =

(r + s 00 0

)f(r) + f(s) =

(r 00 0

)+

(s 00 0

)=

(r + s 00 0

)In this sense, f gives us an isomorphism between R and a subring of M(R).

It is useful in the following sense: it allows us to make precise the statement that M(R)contains a copy of R, or that R can be identified with a subring of M(R). Can we use this toanswer questions about R? Not so likely. What about questions about M(R)? Yes, certainly.

For instance: “Is there a 2× 2 matrix A such that 2A2 − 3A+

(1 00 0

)=

(0 00 0

)?” Yes, let

r ∈ R be a root of the equation 2x2 − 3x + 1 = 0. Then A =

(r 00 0

)satisfies the matrix

equation since

2r2 − 3r + 1 = 0

f(2r2 − 3r + 1) = f(0)

f(r2 + r2 − r − r − r + 1) =

(0 00 0

)f(r)2 + f(r)2 − f(r)− f(r)− f(r) + f(1) =

(0 00 0

)A2 +A2 −A−A−A+

(1 00 0

)=

(0 00 0

)2A2 − 3A+

(1 00 0

)=

(0 00 0

)

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Example 3.3.9. Define the function f : R→ R via f(x) = 0.3x+ 1.


(b) Although f fails some properties, is it still useful for studying R or anything?

Solution: It is easy to see that f is a bijection, after all the graph of y = 0.3x+ 1 passes thehorizontal line test, and has range equal to all of R.

On the other hand, f does not preserve addition or multiplication. For instance

f(2 + 5) = f(7) = 0.3(7) + 1 = 3.1

f(2) + f(5) =(0.3(2) + 1

)+(0.3(5) + 1

)= 1.6 + 2.5 = 4.1

f(2 · 5) = f(10) = 0.3(10) + 1 = 4

f(2) · f(5) =(0.3(2) + 1

)·(0.3(5) + 1

)= (1.6)(2.5)

= 4

Wait, does f preserve multiplication?!! No, it was just bad luck:

f(3 · 5) = f(15) = 0.3(15) + 1 = 5.5

f(3) · f(5) =(0.3(3) + 1

)·(0.3(5) + 1

)= (1.9)(2.5)

= 4.75

The function f does not appear to reveal anything useful about R, after all we are notcomparing R with something simpler (like we compared Z with Z7 in an earlier example),or something more complicated (like we compared R with M(R) in another example). Toexagerate a bit, you could say that there is nothing useful about studying functions of theform y = 0.3x+ 1 or in general y = mx+ b.

Definition 3.3.3. Let R and S be rings. Let f : R → S a function. We say that f is ahomomorphism if it satisfies the following.

1. ∀a, b ∈ R we have f(a+ b) = f(a) + f(b), (i.e. f preserves addition)

2. ∀a, b ∈ R we have f(ab) = f(a)f(b) (i.e. f preserves multiplication).

How can one interpret homomorphisms? Ones like in Example 3.3.7 are simplifications,and these can be useful. Metaphorically, this reminds me of discusssing the world with a

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child. When I did this with my son, when he was 6, the result was often simplified. Forinstance, although there are an infinite number of gradiations of motivations, justificationsand consequences for the conflicts in the world, at the time my son would simply ask, “Arethey good, or are they bad?” This is like a homomorphism from an infinite, complicatedring, to Z2, the simplest (nontrivial) ring.

Example 3.3.10. Which of the previous three examples were homomorphisms? What otherproperties did they have?Solution: The function Z → Z7 given above is a homomorphism that is surjective but notinjective. The function R→M(R) given above is a homomorphism that is injective but notsurjective. The function R→ R give above is not a homomorphism, but it bijective.

Example 3.3.11.

(a) Does every ring have at least one homomorphism?

(b) How far from injective or surjective can a homomorphism be?

(c) Suppose one ring is a subring of a another. Is there a homomorphism involved?

Solution: Let R be any ring. The identity function

Id : R−→Rr 7→ r

is a homomorphism. It is actually an isomorphism.Let R and S be any rings. The zero map

z : R−→Sr 7→ 0

is a homomorphism. It is as far from injective as you can get, and as far from surjective asyou can get (unless S is very small, e.g. unless S = {0}).

Let R be any ring and S a subring. The inclusion map

IncS : S−→Rs 7→ s

is a homomorphism. It is injective, but not surjective (unless S = R).


Theorem 3.3.4. Let f : R→ S be a ring homomorphism. Then

1. f(0R) = 0S

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2. ∀a ∈ R, we have f(−a) = −f(a)

3. ∀a, b ∈ R, we have f(a− b) = f(a)− f(b)

Suppose that R has identity and f is surjective.

4. S has identity and f(1R) = 1S .

5. If u is a unit in R, then f(u) is a unit in S and f(u−1) = (f(u))−1.

Proof. Here is a sketch of a proof. (Do not write your proofs this way! It is more like a Haikuversion of a proof.)

1. f(0) = f(0 + 0) = f(0) + f(0), now subtract.

2. 0 = f(0) = f(a− a) = f(a+ (−a)) = f(a) + f(−a), now subtract.

3. f(a− b) = f(a+ (−b)) = f(a) + f(−b) = f(a)− f(b).

4. f(1)s = f(1)f(r) = f(1r) = f(r) = s.

5. 1 = f(1) = f(uv) = f(u)f(v).

Definition 3.3.5. Let f : R→ S be a ring hommorphism. We define two sets ker f ⊆ R andIm f ⊆ S

ker f = {r ∈ R | f(r) = 0},Im f = {s ∈ S | ∃r ∈ R with s = f(r)}.

We call ker f the kernel of f , and call Im f the image of f .

The sets ker f and Im f are in some sense dual to each other: they are sort of upsidedown mirror images. In some sense, the larger ker f is, the smaller Im f is, and vice versa.In linear algebra we have something similar where f is replaced by a linear transformation.The kernel is then the null space and the image is the column space. The larger the nullspace (in terms of dimension), the smaller the column space, and vice versa.

Corollary 3.3.6 (Kernel and Image are subrings). Let f : R → S be a ring homomorphsim.Then the kernel, ker f and the image, Im f , are subrings, of R and S respectively.

Proof. Let a, b ∈ ker f . Then f(a − b) = f(a) − f(b) = 0 − 0 = 0. Thus, a − b ∈ ker f . Alsof(ab) = f(a)f(b) = 0 · 0 = 0, so ab ∈ ker f .

Let u, v ∈ Im f . Then u = f(a) and v = f(b) for some a, b ∈ R. Then u−v = f(a)−f(b) =f(a − b). Since a − b ∈ R, this shows u − v = f(r) for some r ∈ R and u − v ∈ Im f . Alsouv = f(a)f(b) = f(ab). Since ab ∈ R, this shows that uv = f(r) for some r ∈ R, and souv ∈ Imf.

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Example 3.3.12. What are the kernels and images of the homomorphisms we saw in Ex-amples 3.3.7 and 3.3.8?

Solution: The function f : Z → Z7 has kernel given by all multiples of 7, and image givenby all of Z7.

The other function, f : R→M(R) has kernel given by 0 and image given by the subringS,

S =

{(r 00 0

) ∣∣∣∣ r ∈ R}.

Example 3.3.13. Let R and S be any rings and let R × S be the resulting product ring.Define functions π1 and π2 as shown

π1 : R× S−→R(r, s) 7→ r

π2 : R× S−→R(r, s) 7→ s

(i.e. π1(r, s) = r and π2(r, s) = s)

Assume that π1 and π2 are surjective homomorphisms (see Exercise 13 in 3.1).

(a) Describe their kernel and image.

(b) Apply this examle to the special case of R2.

Solution:

(a) Since we are assuming that π1 and π2 are surjective, we have Imπ1 = R and Imπ2 =S.

We claim that kerπ1 equals the natural “copy” of S that is inside of R× S:

kerπ1 = {(r, s) | r = 0}.

To see that this is true, we calculate:

(r, s) ∈ kerπ1

⇐⇒ π1(r, s) = 0

⇐⇒ r = 0

This is where we ended on Monday, November 3

(b) In R2 we have R = S = R and R × S = R2. Then we usually write (x, y) ∈ R2. Thefunction π1 simply gives the x-coordinate of a point. The kernel, kerπ1, equals they-axis and the image Imπ1 equals the x-axis.

Chapter 4

Polynomial rings

4.1 Definition of a Polynomial Ring

Throughout this section R is a commutative ring with identity.

So far we know one way to make rings bigger: the Cartesian product, R × S is in somesense bigger than R and S. Now we learn another way to make bigger rings, a far moreimportant way: polynomial rings.

Definition 4.1.1. Let R be a commutative ring with identity, and let the symbol x be distinctfrom all the element of R. A polynomial over R with indeterminate x equals a formal sumof the form

anxn + an−1x

n−1 + · · ·+ a2x2 + a1x+ a0

where each ai is an element in R.

“Formal sum” means that we do not claim that this sum equals an element of R or thatx2 is calculated in R; we do not assume anything about this expression except for it’s formalproperties which we will give below; otherwise it is taken just as a symbol. (This means thatwe have not given a mathematically complete definition: for readers who are interested weoffer one as an appendix to this section.)

Definition 4.1.2. Let R[x] denote the set of all polynomials over R. We define what itmeans for two polynomials to be equal to each other:

1. For any polynomial anxn + · · ·+ a0, we declare the following are equal

anxn + · · ·+ a0 = 0xn+1 + anx

n + · · ·+ a0.

We can apply this declaration recursively so that there can be more than one 0 termwith higher powers of x.

67

(version 2014/12/03) 68

2. For any two polynomials anxn + · · ·+ a0 and bmxm + · · ·+ b0 we declare

anxn + · · ·+ a0 = bmx

m + · · ·+ b0 ⇐⇒ ai = bi, ∀i ≥ 0.

Definition 4.1.3. Let anxn + · · · + a0 ∈ R[x]. If an 6= 0, then we call an the leadingcoefficient of this polynomial, and we call n the degree of this polynomial. Note that ifa0 6= 0 then deg a0 = 0. As a matter of convenience, we set deg 0 = −∞. If an = 1 we saythat this polynomial is monic.

Definition 4.1.4. The usual short hand notation for a polynomial in R[x] is f(x). Thus wecould write f(x) = anx

n + · · ·+ a0 and deg f(x) = n, etc.

Definition 4.1.5. Let R be a commutative ring with identity and let R[x] be the polynomialsas defined above. Let f(x), g(x) ∈ R[x], write

f(x) = anxn + · · ·+ a0

g(x) = bmxm + · · ·+ b0

We define f(x) + g(x) and f(x)g(x) as follows

f(x) + g(x) = ctxt + ct−1x

t−1 + · · ·+ c1x+ c0

where ci = ai + bi for all i ≥ 0

f(x)g(x) = ctxt + ct−1x

t−1 + · · ·+ c1x+ c0

where c0 = a0b0c1 = a1b0 + a0b1c2 = a2b0 + a1b1 + a0b2. . .

ci =∑

j+k=i

ajbk

. . .ct = anbm

Note that we can apply the formula given above for ci even when i is relativelely large.For instance, suppose deg f(x) = n and deg g(x) = m. Then

cnm = anam

= a0bn+m + a1bn+m−1 + a2bn+m−2 + · · ·+ anbm + · · ·+ an+m−1b1 + an+m+1b0

= 0 + 0 + 0 + · · ·+ anbm + 0 + · · ·+ 0 + 0

Using the formula in this way is not practical, but makes it easier to write general results,proofs, etc. For instance, we could just give one formula for all the ci in the previousdefinition of multiplication.

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Example 4.1.1. Let f(x), g(x) ∈ Z11[x] be defined as

f(x) = 4x8 + 2x7 + x6 + 6x5 + 9x4 + 10x3 + 5x2 + 4x+ 7

g(x) = 5x9 + 3x8 + 10x7 + 4x6 + 5x5 + 2x4 + 3x3 + 4x2 + 5x+ 6

Find f(x) + g(x) and the x9-coefficient of f(x)g(x).Solution:

f(x) + g(x) =(4x8 + 2x7 + x6 + 6x5 + 9x4 + 10x3 + 5x2 + 4x+ 7

)+(5x9 + 3x8 + 10x7 + 4x6 + 5x5 + 2x4 + 3x3 + 4x2 + 5x+ 6

)= 5x9 + 7x8 + 12x7 + 5x6 + 11x5 + 11x4 + 13x3 + 9x2 + 9x+ 13

= 5x9 + 7x8 + x7 + 5x6 + 2x3 + 9x2 + 9x+ 2

c9 = 4 · 5 + 2 · 4 + 1 · 3 + 6 · 2 + 9 · 5 + 10 · 4 + 5 · 10 + 4 · 3 + 7 · 5= 20 + 8 + 3 + 12 + 45 + 40 + 50 + 12 + 35

≡ 9 + 8 + 3 + 1 + 1 + 7 + 6 + 1 + 2

= 38

≡ 5 mod 11

This is where we ended on Wednesday, November 5

Theorem 4.1.6 (Polynomial rings inherit commutativity and identity). Let R be a com-mutative ring with identity, and let R[x] be the set of all polynomials. Then R[x] is also acommutative ring with identity.

Proof. We will only sketch most of these properties. For instance, for the commutative prop-erty of addition, note that for ci = ai + bi = bi + ai, since adding ai and bi is commutative.Most of the other properities are similar. Here are the full details for the associative propertyof multiplication.

Let f(x) = anxn + · · ·+ a0, g(x) = bmx

m + · · ·+ b0, h(x) = ctxt + · · ·+ c0. Let αi be the

coefficient of xi in f(x)g(x), so we have

αi =∑

j+k=i

ajbk

and let β` be the coefficient of x` in(f(x)g(x)

)h(x).

We will calculate β`. It suffices to show that the final formula we obtain for β` does notchange under the operation of switching f , g and h. We calculate as follows

β` =∑

i+z=`

αicz =∑

i+z=`

∑j+k=i

ajbk

cz =∑

i+z=`

∑j+k=i

ajbkcz

=∑

{i,z,j,k | i+z=` and j+k=i}

ajbkcz =∑

{z,j,k | j+k+z=`}

ajbkcz

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The last expression does not depend on which order one chooses i, or j or k. Thus, it doesnot depend on combining f(x) and g(x) first, and then combining this with h(x). The sameresult will be obtained if we combine g(x) and h(x) first, and then with f(x).

One of the big ideas in this section is that for a given a ring R, we can make a biggerring R[x]. The previous theorem justifies calling R[x] a ring, now we explain how R[x] isbigger. Since R[x] equals all polynomials anxn + . . . a1x+ a0, it contains all polynomials ofthe form a0. But a0 varies as any element of R. Thus, there is a “copy” of R inside of R[x],namely all polynomials of degree ≤ 0. This copy even has the same addition: a0 + b0 = c0and multiplication c0 = a0b0.

Now we illustrate how polynomials in algebra are not the same as polynomials in pre-calculus. When we first learn algebra, in middle school, we are told that x is an unknownreal number. In that case, anxn + · · · + a0 would also be an unkown real number. That isnot the view we take in Abstract Algebra: for us, anxn+ · · ·+ a0 is not an element of R, it isoutside of R. In calculus we might view f(x) = anx

n + · · · + a0 as a function. But this notthe view we take in abstract alegbra, as the next example illustrates.

Example 4.1.2. Let f(x), g(x) ∈ Z5[x] be define by f(x) = x6 + 4x2 + x and g(x) = x.Define the functions

f : Z5−→Z5

a 7→ f(a)and g : Z5−→Z5

a 7→ g(a).

(a) Do we have f(x) = g(x)?

(b) Do we have f = g?

(c) Do the previous two parts appear to contradict each other, and if so, how should weresolve the conflict?

Solution:

(a) f(x) 6= g(x) since their coefficients are different.

(b) To find out if f = g we evalute these functions:

f(0) = 0 g(0) = 0

f(1) = 6 = 1 g(1) = 1

f(2) = 64 + 16 + 2 = 82 = 2 g(2) = 2

f(3) = f(−2) = 64 + 16− 2 = −78 = −3 = 3 g(3) = 3

f(4) = f(−1) = 1 + 4− 1 = 4 g(4) = 4

Since f(a) = g(a) for all a ∈ Z5 we see that f = g.

(c) The two previous parts sort of seem to contradict each other: the polynomials aren’tequal but the functions are. We don’t really have any choice on the functions: ifyou get the same outputs from them, the functions must be equal. We could think

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about defining what equality of polynomials means differently. But it would be morecomplicated than the definition we’ve given above. Besides, we will see later thatpolynomials are more general than functions in the following sense: we can start withthe polynomial ring and turn it into the function ring (by something called “taking thequotient”). It’s not possible to go the other way, and so it’s more useful to start withthe polynomial ring as we’ve defined it here.

What do you need to remember? Polynomials are not the same thing as functions.Polynomials are formal combinations of powers of x and coefficients, subject to certainrules. They are not graphs, or functions, or unknowns, etc.

Theorem 4.1.7 (Degree is Ultrametric and Logarithmic). Let R be an integral domain andlet f(x), g(x) ∈ R[x]. Then we have

1. deg(f(x) + g(x)) ≤ max{deg f(x),deg g(x)} (degree is ultrametric1).

2. deg(f(x)g(x)) = deg f(x) + deg g(x) (degree is logarithmic2).

Proof. Write f(x) and g(x) as

f(x) = anxn + · · ·+ a0

g(x) = bmxm + · · ·+ b0

and assume that an and bm are nonzero. Then

the leading coefficient of (f(x)+g(x)) =

an if n > m

bm if m > n

an + bn if m = n and an + bn 6= 0

ai + bi, for some i < n if m = n and an + bn = 0

In the first two cases, deg(f(x) + g(x) = max{n,m}. In the third case the degree equalsn = m = max{n,m}. In the last case the degree is smaller than n.

When we calculate the product f(x)g(x) we get

f(x)g(x) = ctxt + ct−1x

t−1 + · · ·+ c1x+ c0

where t = n +m and ct = ambm. Since R has no zero divisors, we see that ct 6= 0 and sodeg(f(x)g(x)) = n+m.

Corollary 4.1.8 (Polynomials inherit integral domainity3). If R is an integral domain, thenso is R[x].

1As far as I know this property is not usually given a name. However, the ultrametric property of valuationsis |x + y| ≤ max{|x|, |y|}. This is called “ultrametric” because the usual metric space inequality is the triangleinequality: |x+ y| ≤ |x|+ |y|.

2Because logarithms turn products into sums.3Yes, I made up the word “domainity”, as an extension of the theme “commutativity” and “identity”. Yes, it’s a

horrible word.

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Proof. If f(x), g(x) are nonzero, then deg f(x) ≥ 0 and deg g(x) ≥ 0. Therefore deg(f(x)g(x)) =deg f(x) + deg g(x) ≥ 0. Since deg 0 = −∞, this shows that f(x)g(x) 6= 0, and so R[x] hasno zero divisors.

This is where we ended on Friday, November 11

4.1.1 Formally Defining the Polynomial Ring

A careful look at polynomial rings shows that the role of x is simply as a place holder:the different powers of x indicate which coefficients we add together and which ones wemultiply together. So, to define a polynomial in a rigorous fashion, we need to give arigorous definition of something which allows us to keep track of different coefficients indifferent places. How many places? Only a finite number at a time, but it neeeds to bea finite number of an indefinite size, a size that is allowed to grow as we multiply twopolynomials together. One standard definition is given below.

Definition 4.1.9. Let R be a commutative ring with identity. A finite sequence in R is anordered n-tuple, (a0, a1, . . . , an) where each ai is an element of R. A sequence in R is aninfinite ordered list (a0, a1, . . . ) where each ai is an element of R. Denote such a sequenceby (ai)i≥0, or, more briefly, as (ai). A sequence (ai)i≥0 is of finite type if there exists ann ∈ N such that ai = 0 for all i ≥ n. Let x be the notation for the sequence (0, 1, 0, 0, . . . ),i.e. x = (ai)i≥0 where a1 = 1 and ai = 0 for all i 6= 1. Let R[x] be the collection of allsequences in R of finite type. Define addition on R[x] as follows:

(ai) + (bi) = (ci)

where for all i ≥ 0 we set ci = ai + bi. Define multiplication on R[x] as follows:

(ai)(bi) = (ci)

where for all i ≥ 0 we set ci =∑

j+k=i

ajbk.

The whole point of this definition is just to make our definition of polynomial rigorous.We should not start writing our polynomials using the notation of sequences. But, just tomake it clear how sequences and the operations we’ve defined on them here are equiva-lent to polynomials with which we are familiar, we will translate a previous example intosequence notation.

Example 4.1.3. Translate Example 4.1.1 (involving two polynomials in Z11[x]) into se-quence notation.Solution: We had

f(x) = 4x8 + 2x7 + x6 + 6x5 + 9x4 + 10x3 + 5x2 + 4x+ 7

g(x) = 5x9 + 3x8 + 10x7 + 4x6 + 5x5 + 2x4 + 3x3 + 4x2 + 5x+ 6

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This becomes

f = (7, 4, 5, 10, 9, 6, 1, 2, 4, 0, 0, . . . )

g = (6, 5, 4, 3, 2, 5, 4, 10, 3, 5, 0, 0, . . . )

Then

f + g = (7, 4, 5, 10, 9, 6, 1, 2, 4, 0, 0, . . . ) + (6, 5, 4, 3, 2, 5, 4, 10, 3, 5, 0, 0, . . . )

= (13, 9, 9, 13, 11, 11, 5, 12, 7, 5, 0, 0 . . . )

= (2, 9, 9, 2, 0, 0, 5, 1, 7, 5, 0, 0 . . . )

and

c9 = 4 · 5 + 2 · 4 + 1 · 3 + 6 · 2 + 9 · 5 + 10 · 4 + 5 · 10 + 4 · 3 + 7 · 5= 20 + 8 + 3 + 12 + 45 + 40 + 50 + 12 + 35

≡ 9 + 8 + 3 + 1 + 1 + 7 + 6 + 1 + 2

= 38

≡ 5 mod 11

Chapter 6

Ideals and Quotient Rings

Throughout this chapter, we consider only rings with identity.Although rings weren’t studied as a stand alone topic until the 20th century, various

results that came before this can now be seen as laying the foundation for ring theory. Someof these results were created by Ernst Kummer in the 1840’s in his work on higher reciprocitylaws in number theory, and Fermat’s Last Theorem.

In particular, Kummer introduced something that he called ideal numbers, which evolvedinto what we call ideals (defined below). He introduced ideal numbers to get a set of thingswhich had unique factorization, similar to the unique factorization property that the integershave. In the 1870’s Dedekind reworked Kummer’s ideas and gave the axioms we now use forideals. Finally, in the 1920’s Emmy Noether made the final steps in a full, abstract treatmentof rings, ideals, quotients, homomorphisms and isomorphisms.

These days, ideals tend to be used in an introductory algebra class more for definingquotient rings (as in Dedekind and Noether) than for their factorization properties (as inKummer). In this section we will see quotient rings, and in the next section we will learn alittle bit about factorization of ideals.

6.1 Ideals

Example 6.1.1. Theorem 2.2.1 was the result where we defined addition and multiplicationon Zn, as [a] + [b] = [a+ b] and [a][b] = [ab]. But this theorem depended almost entirely onTheorem 2.1.3, where we proved that if a ≡ a′ and b ≡ b′, then a+ b ≡ a′+ b′ and ab ≡ a′b′.Our goal below is to learn how to extend these kinds of operations to an arbitrary ring R.

More specifically, we want to replace conditions of the form “n|a” with a ∈ S for a set Scontained in a ring R. To do this, we need to learn what conditions we need to impose onS.

Return to the proof of Theorem 2.1.3, identify the crucial properties that “n| ” satisfies,and translate those into properties about a subset S.

Solution: We start with the proof that a ≡ a′ and b ≡ b′ implies a+ b ≡ a′ + b′. The crucial

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property that “n| ” satisfies for this proof is this:

n|(a− a′) and n|(b− b′)⇒ n∣∣∣((a− a′) + (b− b′)

).

Our goal is to get rid of the parts with “n|”. We can replace this with a set. Let nZ be the setof all multiples of n. Then the above statement becomes

a− a′ ∈ nZ and b− b′ ∈ nZ⇒ (a− a′) + (b− b′) ∈ nZ.

Now our goal is to replace nZ with some other kind of set, S, with S a subset of a ring, butkeep the property just defined:

a− a′ ∈ S and b− b′ ∈ S ⇒ (a− a′) + (b− b′) ∈ S.

But now that we look at it this way, we don’t really need a − a′ and b − b′, we could statethe property this way:

r ∈ S and s ∈ S ⇒ r + s ∈ S.

In other words, we need S to be closed under addition.Now let’s look the proof that a ≡ a′ and b ≡ b′ implies ab ≡ a′b′. The crucial property

that “n| ” satisfies for this proof is this:

n |(a− a′) ⇒ n |b(a− a′)n |(b− b′) ⇒ n |a′(b− b′)

}⇒ n

∣∣∣(b(a− a′) + a′(b− b′))

But actually, the last property shown here is already the property that we identified upabove. So, let’s rewrite the first two parts:

a− a′ ∈ nZ ⇒ b(a− a′) ∈ nZb− b′ ∈ nZ ⇒ a′(b− b′) ∈ nZ

Now our goal is to replace nZ with some other kind of set, S, with S a subset of a ring, butkeep the property just defined:

a− a′ ∈ S ⇒ b(a− a′) ∈ Sb− b′ ∈ S ⇒ a′(b− b′) ∈ S.

But now that we look at it this way, we don’t really need a − a′ and b − b′, we could statethe property this way:

r ∈ S ⇒ xr ∈ Ss ∈ S ⇒ ys ∈ S.

In other words, we need S to be closed under multiplication by R:

∀s ∈ S,∀x ∈ R, we need rs ∈ S.

Definition 6.1.1. Let R be a ring with identity and S a subset of R. We say that S is anideal if it satisfies the following properties.

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1. if a, b ∈ S, then a+ b ∈ S and ab ∈ S (closed under addition)

2. if a ∈ S and r ∈ R then ar ∈ S and ra ∈ S (closed under super-multiplication1).

Corollary 6.1.2. If R is a ring with identity and I an ideal then I is a subring of R.

Proof. Let a, b ∈ I. Since b ∈ R we have ab ∈ I by I2. Since R has identity 1, we also have−1 ∈ R and so I2 implies that (−1)b = −b ∈ I. Then property I1 shows a− b ∈ I.

Example 6.1.2. Let R = Z and S = 6Z, i.e. S equals all the multiples of 6. Show that S isan ideal.Solution: Let a, b ∈ S. Then we can write a = 6n and b = 6m for some n,m ∈ Z. Soa+ b = 6(n+m) ∈ 6Z. Also, if x ∈ Z then xa = x(6n) = 6(xn) ∈ 6Z.

Theorem 6.1.3 (Multiples of an element form an ideal). Let R be any commutative ring withidentity, fix an element a ∈ R and let I = Ra, i.e. I equals all the multiples of a. Then I is anideal in R.

Definition 6.1.4. An ideal of the form Ra is called a principal ideal. We say that a gener-ates Ra. Other common notation for the ideal generated by a is 〈a〉 and (a).


Proof. Let c, d ∈ I. Then we can write c = ra and d = sa for some r, s ∈ R. So c + d =ra+ sa = (r + s)a ∈ Ra. Let x ∈ R. Then xc = x(ra) = (xr)a ∈ Ra.

Example 6.1.3. Consider the ring Z[x] and consider the subset I defined as follows

I = {f(x) ∈ Z[x] | f(x) has constant term that’s even}

Show that I is an ideal, but not a principal ideal.Solution: Let f(x), g(x) ∈ I. Write f(x) = anx

n + · · · + a0 and g(x) = bmxm + · · · + b0

where a0 and b0 are even. Then f(x) + g(x) has constant term a0 + b0, which is even. Leth(x) = ctx

t + · · · + c0 ∈ Z[x]. Then f(x)g(x) has constant term a0c0. Even though c0 maynot be even, the product a0c0 is even, since c0 is an integer. This shows that I is closedunder super-multiplication, and so I is an ideal.

Now we show that I is not a principal ideal. Suppose to the contrary that I is generatedby a polynomial, f(x). Note that 2 ∈ I. So if 2 = f(x)h(x) for some h(x) ∈ Z[x], thendeg 2 = 0 and 0 = deg f(x)h(x) = deg f(x) + deg h(x). This shows that f(x) is a constant.But the only constants that divide 2 are±1 and±2. If f(x) = ±1 then, since I is closed undersuper-multiplication, we would have I = Z[x] which is false. So suppose f(x) = ±2. Thenthere’s no multiple of f(x) which will equal x. However, x ∈ I, so this is a contradictiontoo.

1Hungerford calls this the “absorption” property because the product is “absorbed” into S. I like “super-multiplication” for two reasons: (1) we are multiplying a by an element in a super-set, R, (2) because a setthat is closed under super-multiplication is automatically closed under multiplication.

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Example 6.1.4. Consider the ring Z5[x] (see Example 4.1.2). Define a subset I ⊆ Z5[x] asfollows:

I = {f(x) ∈ Z5[x] | f(a) = 0, ∀a ∈ Z5}

= {f(x) ∈ Z5[x] | f equals the 0 function from Z5 to Z5}

Show that I is an ideal in Z5[x].Solution: Let f(x), g(x) ∈ I. Then for all a ∈ Z5 we have (f + g)(a) = f(a) + g(a) = 0 + 0.Thus f + g is the 0 function on Z5 and so f + g ∈ I. Let h(x) ∈ Z5[x]. Then for all a ∈ Z5

we have (fh)(a) = f(a)h(a) = 0h(a) = 0. Thus, fh is the 0 function on Z5 and so fh ∈ I.

Example 6.1.5. Show that Z is not an ideal in Z[i].Solution: Let a, b ∈ Z. Then a+ b ∈ Z. (Actually, this is clear since Z is a subring of Z[i].)

Let c + di ∈ Z[i]. Then a(c + di) = ac + adi 6∈ Z. Thus, Z is not closed under super-multiplication.

We return now to our original goal: to perform over an arbitrary ring R the same kindsof construction we did with Z and Zn.

Definition 6.1.5. Let I be an ideal in the ring R. For all a, b ∈ R, if a− b ∈ I then we writea ≡ b (mod I) and we say that a is congruent to b modulo I. If the set I is clear in context,then we will write just a ≡ b, dropping “(mod I)”.

Theorem 6.1.6 (Congruence modulo I is an Equivalence Relation). Let I be an ideal in thering R. The relation a ≡ b (mod I) is an equivalence relation. In other words, ∀a, b, c ∈ R, wehave the following:

1. a ≡ a (mod I) (reflexive property),

2. if a ≡ b (mod I), then b ≡ a (mod I) (symmetric property),

3. if a ≡ b (mod I), and b ≡ c (mod I), then a ≡ c (mod I) (transitive property).

Proof. 1. We have a− a = 0, and 0 ∈ I since I is a subring. Therefore a ≡ a.

2. Suppose a ≡ b. Then a− b ∈ I and so b− a ∈ I, which means b ≡ a.

3. Suppose a ≡ b and b ≡ c. Then a−b ∈ I and b−c ∈ I. Therefore a−b+b−c = a−c ∈ I,which means a ≡ c.

Theorem 6.1.7 (Adding and Multiplying equivalence classes is well defined). Let I be anideal in R. Let a ≡ a′ (mod I) and b ≡ b′ (mod I). Then the following hold:

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1. a+ b ≡ a′ + b′ (mod I),

2. ab ≡ a′b′ (mod n).


Proof. Suppose a ≡ a′ and b ≡ b′.Then a−a′, b−b′ ∈ I. Closure under addition implies that a−a′+b−b′ = (a+b)−(a′+b′) ∈

I and so a+ b ≡ a′ + b′.Closure under super-multiplication implies that b(a−a′) ∈ I and a′(b− b′) ∈ I. Then we

have b(a− a′) + a′(b− b′) = ab− a′b′ ∈ I which means ab ≡ a′b′.

Theorem 6.1.8 (Equivalence classes are cosets). Let I be an ideal in R. For any a ∈ Z wedefine

[a] = {b ∈ R | b ≡ a (mod I)}.

In other words [a] is the set of all elements in R that are equivalent to a. Define the set a+ I asfollows

a+ I = {b | b = a+ x for some x ∈ I}.

Then[a] = a+ I

Proof. Let b ∈ R. Then we have

b ∈ [a]

⇐⇒ b ≡ a (mod I)

⇐⇒ b− a ∈ I⇐⇒ b− a = x for some x ∈ I⇐⇒ b = a+ x for some x ∈ I⇐⇒ b ∈ a+ I

Example 6.1.6. Consider the ring and ideal shown below:

R = Z[x]I = {f(x) ∈ Z[x] | f(x) has constant term that’s even}

(See Example 6.1.3). Can you list all the equivalence classes of Z[x] modulo I?

Solution: We start with two classes

0 + I = I, and 1 + I

We should justify that these are not equal:

1 + I = 0 + I

⇐⇒ 1 ≡ 0 (mod I)

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⇐⇒ 1− 0 ∈ I

Since 1 6∈ I we see that 1 + I 6= 0 + I.Now we claim that these are all the equivalence classes. Let f(x) ∈ Z[x]. Our goal is to

show that f(x) + I = I or f(x) + I = 1 + I. Suppose f(x) has even constant term. Then

f(x) ∈ If(x)− 0 ∈ I

f(x) ≡ 0 (mod I)

f(x) + I = 0 + I

Now suppose that f(x) has odd constant term. Then f(x) − 1 has an even constant term.Thus,

f(x)− 1 ∈ If(x) ≡ 1 (mod I)

f(x) + I = 1 + I

Example 6.1.7. Consider the ring and ideal shown below:

R = Z5[x],

I = {f(x) ∈ Z5[x] | f(a) = 0, ∀a ∈ Z5}.

(See Examples 6.1.4 and 4.1.2).

(a) Can you list all the equivalence classes of Z5[x] modulo I?

(b) How can you tell when two classes are equal?

(c) Is there another way to describe a set of canonical represenatitves besides listing them?


Solution: (a) We can start with the constant classes:

0 + I = I, 1 + I, 2 + I, 3 + I, 4 + I.

At this point, one thing is clear and one thing is not: It’s not clear if some of these classesare equal; it’s clear that there are probably other classes. Before we see how to resolve thefirst issue, let’s illustrate the second. Here are some more classes:

x+ I, (x+ 1) + I, x2 + I, (x6 + 4x2 + x) + I, (3x9 + 2x7 + x6) + I, . . .

There might be an infinite number of classes, after all there are an infinite number of poly-nomials. But there might not be an infinitee number of classes because perhaps many ofthem are equal, despite appearances. Listing them at this point appears to be hopeless.

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(b) Here’s how to tell that some of the above are equal, and that some are not.

f(x) ≡ g(x) (mod I)

⇐⇒ f(x)− g(x) ∈ I⇐⇒ f(a)− g(a) = 0, ∀a ∈ Z5

⇐⇒ f = g as functions on Z5

So, we can say for instance2 + I 6= 4 + I

because the functionsf : Z5−→Z5

a 7→ 2and g : Z5−→Z5

a 7→ 4

are not equal. Similarly, we can say that

4 + I 6= (x+ 1) + I

because the functions

f : Z5−→Z5

a 7→ 4and g : Z5−→Z5

a 7→ a+ 1

are not equal.Finally, we can say that

(x6 + 4x2 + x) + I = x+ I

because the functions

f : Z5−→Z5

a 7→ f(a)and g : Z5−→Z5

a 7→ g(a).

are equal (as we showed earlier).(c) So, finally, can we describe canonical representatives in some way besides listing

them? Let PolyFun(Z5) denote the set of all polynomial functions defined Z5 → Z5. ThenPolyFun(Z5) is finite: it is at most 55 in size because the set of all functions, polynomial andotherwise, is at most 55. Then

classes in Z5[x] = {f(x) + I | f ∈ PolyFun(Z5)}

6.2 Quotient Rings

Theorem 6.2.1. LetR be a commutative ring with identity and I an ideal inR. Let a, a′, b, b′ ∈R such that a+ I = a′ + I and b+ I = b′ + I. Then

(a+ b) + I = (a′ + b′) + I

(ab) + I = (a′b′) + I

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Proof. Essentially this proof is just to restate Theorem 6.1.8 in different notation, in thesame way that Theorem 2.2.1 was simply a restatement of Theorem 2.1.3. The comparisonsbetween these theorems are shown below:

Theorem 2.1.3 Theorem 2.2.1notation: a ≡ b (mod n) [a] = [b]

Theorem 6.1.8 Theorem 6.2.1notation: a ≡ a′ (mod I) a+ I = a′ + I

Definition 6.2.2. Let R be a ring with identity and let I be an ideal in R. Let R/I be thecollection of all equivalence classes of elements of R modulo I:

R/I = {a+ I | a ∈ R}.

We call a set of the form a+ I a coset.We define addition, ⊕, and multiplication, �, in R/I as follows:

(a+ I)⊕ (b+ I) = (a+ b) + I, and (a+ I)� (b+ I) = (ab) + I

Example 6.2.1. Let R = Z and fix n ≥ 2. Let I = nZ. Describe R/I as a set and as a ring.

Solution:

R/I = Z/nZ= {a+ nZ | a ∈ Z}= {[a] | a ∈ Z}= {[a] | 0 ≤ a ≤ n− 1}= Zn

Finally, we note that the addition and multiplication of R/I as defined here for ideals isthe same addition and multiplication as defined for Zn:

(a+ I)⊕ (b+ I) = (a+ b) + I

(a+ nZ)⊕ (b+ nZ) = (a+ b) + nZ[a]⊕ [b] = [a+ b]

(a+ I)� (b+ I) = (ab) + I

(a+ nZ)� (b+ nZ) = (ab) + nZ[a]� [b] = [ab]


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Theorem 6.2.3. LetR be a ring with identity, and I an ideal. LetR/I be as in Definition 6.2.2.The following hold:

1. R/I is a ring with identity (we call R/I a the quotient ring formed by R and I).

2. If R is commutative, then R/I is commutative.

We would, at this point, ideally look at 3 or 4 or 5 examples of constructing quotientrings, and develop some intuition about what they look like, how they work, how theycompare to the original ring, etc. But there are two problems with this approach. First, weare too short on time and space to look at that many examples. Second, all of our examplesfall into one of four categories: (1) Z, (2) polynomial rings, (3) matrix rings, (4) subringsof C, usually of the form R[α] where R could be Z, or Q or R and α is some element not inR such as

√2, or i. There is perhaps not much point in returning to Z, after all quotients

of the form Z/I are all equal to Zn for some value of n. I think the reader/student maybe tired of polynomial rings at this point. I think the reader/student may not be excitedby matrix rings, although they are our best sources of noncommutative examples. So thatleaves subrings of C. There are many interesting examples of subrings in C, but we haveonly looked at relatively small ones. In particular, the ones we have looked at don’t havethat exciting of quotient rings. None the less, that’s what we return to now.

Example 6.2.2. Let R and I be as shown

R = Z[x]I = {f(x) | f(x) has even constant term}

(See Examples 6.1.3 and 6.1.6.) Describe R/I including the addition and multiplication.

Solution: Although we didn’t use the notattion R/I in Example 6.1.6, we did find theclasses:

R/I = {0 + I, 1 + I}.

Now we add and multpiply:

⊕ 0 + I 1 + I0 + I 0 + I 1 + I1 + I 1 + I 0 + I

� 0 + I 1 + I0 + I 0 + I 0 + I1 + I 0 + I 1 + I

To verify one or two of these we calculate

(1 + I)⊕ (1 + I) = (1 + 1) + I definition of ⊕= 2 + I addition in Z (or in Z[x])= 0 + I because 2 ∈ I

(1 + I)� (1 + I) = (1 · 0) + I definition of �= 0 + I multiplication in Z (or in Z[x])

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Example 6.2.3. In each case below, describe I and R/I.

(a) R = Q[√5] = {a+ b

√5 | a, b ∈ Q} and I = R

√5.

(b) R = Z[√5] = {a+ b

√5 | a, b ∈ Z} and I = R

√5.

Solution:

(a) I is the set of all multiples of√5 where the multiplier comes from Q[

√5]. Some typical

elements of I are

7√5 ∈ I (since 7 ∈ R)(

−113

+7

17

√5

)√5 ∈ I

(since

−113

+7

17

√5 ∈ R

)√5 ∈ I (this equals 1

√5 where 1 ∈ R)

5 ∈ I (this equals√5√5 where

√5 ∈ R)

Note also the following element of I:

1 ∈ I(

since 1 =1

5

√5 ·√5 where

1

5

√5 ∈ R

)Now that 1 ∈ I, it’s all over.

Let r ∈ R. Then r1 ∈ I, so r ∈ I for all r ∈ R. In other words, I = R. Then for alla, b ∈ R we have a− b ∈ R. So, all elements in R are equivalent modulo I. ThereforeR/I = R/R = 0 + I. In other words, R/I has only one element, and so R/I is the 0ring.


(b) I is the set of all multiples of√5 where the multiplier comes from Z[

√5]. We claim

that a canonical set of representatives is given as follows:

R/I = {0 + I, 1 + I, 2 + I, 3 + I, 4 + I}.

To prove this we have to show (i) that none of the classes we have given are equal toeach other, and (ii) that every coset in R/I equals one of the ones we have given.

For (i) let’s start by proving one pair of the given cosets are distinct:

1 + I = 4 + I

⇐⇒ 4− 1 ∈ I⇐⇒ 3 ∈ I

⇐⇒ 3 = (a+ b√5)√5 for some a, b ∈ Z

⇐⇒ 3 = 5b+ a√5

If 3 = 5b+a√5 for some a, b ∈ Z, then we must have a = 0 (otherwise

√5 is a rational

number). Then 3 = 5b which is impossible.

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Now we can see the general argument: let n ∈ {0, . . . , 4}. Then we cannot haven = 5b+a

√5 for integers a and b. Then n 6∈ I. Then c−d 6∈ I for 0 ≤ d ≤ c ≤ 4. Then

c+ I 6= d+ I.

For (ii) let (a+ b√5)+ I be an arbitrary element of R/I. Let r be the remainder of

a divided by 5. Then a = q5 + r with 0 ≤ r ≤ 4. We claim that (a+ b√5) + I = r + I.

The proof is essentially two calculations. First we show that (a+ b√5) + I = a+ I

b√5 ∈ I

⇒ a+ b√5− a ∈ I

⇒ (a+ b√5) + I = a+ I

Now we show that a+ I = r + I

a = q5 + r

⇒ a− r = 5q

⇒ a− r ∈ I, because 5 ∈ I⇒ a+ I = r + I

There’s a way of looking at this example, and similar examples, that makes it alittle bit easier to figure out the answer. We start with Z[

√5] and then we’re going to

make it smaller: you can imagine “erasing” part of it, or setting part of it equal to 0.Since we’re “dividing” by something with

√5 in it, we’re going to “erase”

√5. This

would turn Z[√5] into Z. But, in fact, we’re divding by something with all multiples

of 5 in it. So, all multiples of 5 will be set equal to 0. Well, we know a ring wheremultiples of 5 equal 0: Z5. So, that’s what we’ve got here, an isomorphic copy of Z5.

The previous example suggested something: that R/I ∼= Z5 where R = Z[√5] and

I = R√5. This is indeed the case, and we will now develop the results that will allow us to

prove this, and to prove a great many similar statements.Recall (Definition 3.3.5) that the kernel of a map is the set of points that get mapped to

0.

Theorem 6.2.4 (Kernels are Ideals and Ideals are Kernels).

1. Let f : R→ S be a ring homomorphism. Then ker f is an ideal.

2. Let I be an ideal in R. Then π : R−→R/Ir 7→ r + I

is a ring homomorphism, and kerπ = I.

(We call π the canonical homomorphism.)

Proof. (1) We know already that ker f is closed under addition (Corollary 3.3.6). Let a ∈ker f and r ∈ R. Then f(ar) = f(a)f(r) = 0f(r) = 0. Therefore, ar ∈ ker f . The same sortof proof shows that ra ∈ ker f and so ker f is closed under super-multiplication.

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(2) First we show that π is a homomorphism:

π(a+ b) = (a+ b) + I definition of π

π(a) + π(b) = (a+ I)⊕ (b+ I) definition of π

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

π(ab) = (ab) + I definition of π

π(a)π(b) = (a+ I)� (b+ I) definition of π

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

Now we show that kerπ = I.

r ∈ kerπ

⇐⇒ π(r) = 0

⇐⇒ r + I = 0 + I

⇐⇒ r − 0 ∈ I⇐⇒ r ∈ I

Example 6.2.4. Let R[x] be the polynomials over the real numbers. Let ϕ : R[x] → R bedefined by ϕ(f(x)) = a0 where a0 is the constant term of f(x).

(a) Show that ϕ is a homomorphism,

(b) describe kerϕ,

(c) describe a canonical set of representatives for R[x]/ kerϕ, and the addition and multi-plication for R[x]/ kerϕ


Solution: Let f(x) = anxn + · · · + a0 and g(x) = bmx

m + · · · + b0. Then f(x) + g(x) hasconstant term a0 + b0. Therefore, ϕ(f(x) + g(x)) = a0 + b0 = ϕ(f(x)) + ϕ(g(x)). Similarly,f(x)g(x) has constant term a0b0 and ϕ(f(x)g(x)) = a0b0 = ϕ(f(x))ϕ(g(x)).

The kernel of ϕ is

K = kerϕ

= {f(x) ∈ R[x] | ϕ(f(x) = 0}= {f(x) ∈ R[x] | f(x) has constant term 0}= {f(x) ∈ R[x] | every term in f(x) has an x}

Now we describe R[x]/K:

R[x]/K = {f(x) +K | f(x) ∈ R[x]}= {a+K | a ∈ R}

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This requires two justifications: (i) that all the cosets we have defined are all distinct and(ii) that we have listed all possible cosets.

For (i) note that

a+K = b+K

⇐⇒ a− b ∈ K

Note that a− b does not have an x, and so a− b 6∈ K.For (ii) note that

f(x)− a0 ∈ Kf(x) +K = a0 +K

and so f(x) +K is equal to one of the cosets we have defined.Now we claim that addition and multipliction in R[x]/K is essentially the same as addi-

tion and multiplication in R:

(a+K)⊕ (b+K)︸︷︷︸addition in R[x]/K

= (a+ b)︸︷︷︸addition in R

+K

(a+K)� (b+K)︸︷︷︸multiplication in R[x]/K

= (ab)︸︷︷︸multiplication in R

+K

Basically, when we add or multiply in R[x]/K we are just ignoring the K and looking at aand b.

Lemma 6.2.5. Let f : R → S be a ring homomorphism. Then f is injective if and only ifker f = {0}.


Proof. Suppose f is injective. Let r ∈ ker f . Then f(r) = 0. Since f(0) = 0 this meansf(r) = f(0). Since f is injective, this means r = 0.

Suppose ker f = {0}. Let f(r) = f(s) for r, s ∈ R. Then 0 = f(r) − f(t) = f(r − t) sor − t ∈ ker f . This means r − t ∈ {0} and so r − t = 0 and r = t.

Theorem 6.2.6 (First Isomorphism Theorem). Let f : R → S be a surjective ring homo-morphsim with kernel K. Then S is isomorphic to R/K, with isomorphism given by

ψ : R/K−→Sr +K 7→ f(r).

Proof. We ignore one small issue for now (skip to the end of the proof to see what it is) andproceed with the main part of the proof.

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We prove that ψ preserves addition. Let a+K, b+K ∈ R/K.

ψ((a+K)⊕ (b+K)

)= ψ

((a+ b) +K

)addition in R/K

= f(a+ b) definition of ψ

= f(a) + f(b) f preserves addition

ψ(a+K) + ψ(b+K) = f(a) + f(b) defiinition of ψ

We prove that ψ preserves multiplication. Let a+K, b+K ∈ R/K.

ψ((a+K)� (b+K)

)= ψ

((ab) +K

)multiplication in R/K

= f(ab) definition of ψ

= f(a)f(b) f preserves multiplication

ψ(a+K)ψ(b+K) = f(a)f(b) defiinition of ψ

We prove that ψ is surjective: Let s ∈ S. Since f is surjective we have f(r) = s for somer ∈ R. Then ψ(r +K) = f(r) = s, so ψ is surjective.

We prove that ψ is injective. We apply the previous lemma and calculate kerψ. Supposeψ(r+K) = 0. Then f(r) = 0. Then r ∈ ker f . Then r ∈ K. Then r+K = 0+K. Therefore,kerψ = {0 +K}.

Now we finish with one last small point: we need to know that ψ is well-defined. Hereis what this means: ψ is a function where each input is a set, a + K. We defined ψ usingjust one element of that set, namely a. We need to make sure that the definition of ψ doesnot depend on which element in the set a+K that we pick. In other words, if we wrote thesame set in a different way, such as b+K, and we defined ψ using b we should get the sameresult. Here’s the proof.

Suppose a+K = b+K. Then we have

a+K = b+K,

a− b ∈ K,f(a− b) = 0,

f(a)− f(b) = 0,

f(a) = f(b),

ψ(a+K) = ψ(b+K).

Example 6.2.5. Let R = R[x], let K = Rx, the principal ideal generated by x. Prove thatR/K = R[x]/Rx ∼= R.Solution: Let ϕ : R[x] → R be defined by ϕ(f(x)) = a0 where a0 is the constant term (seeExample 6.2.4).

This is where we ended on Monday, December 1

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Then ϕ is (i) a homomorphism, (ii) surjective and (iii) kerϕ = K = Rx. ThereforeR/K ∼= R with isomorphism give by ψ(f(x) +K) = ϕ(f(x)) = a0 where a0 is the constantterm of f(x).

Example 6.2.6. Let R be the ring of all matrices of the form(a b0 c

)where a, b, c ∈ R:

R =

{(a b0 c

) ∣∣∣∣ a, b, c ∈ R}.

Let K be the set of all matrices of the form(0 b0 c

). Show that K is an ideal and that

R/K ∼= R.

Solution: Define a function ϕ : R→ R via ϕ(a b0 c

)= a.

We show that ϕ preserves addition:

ϕ

((a b0 c

)+

(d e0 f

))= ϕ

(a+ d b+ e0 c+ f

)= a+ c

ϕ

(a b0 c

)+ ϕ

(d e0 f

)= a+ c.

We show that ϕ preserves multiplication:

ϕ

((a b0 c

)(d e0 f

))= ϕ

(ad ae+ bf0 cf

)= ad

ϕ

(a b0 c

)+ ϕ

(d e0 f

)= ad

We show that kerϕ = K. (a b0 c

)∈ kerϕ

⇐⇒ ϕ

(a b0 c

)= 0

⇐⇒ a = 0

⇐⇒(a b0 c

)=

(0 b0 c

)

If r ∈ R then ϕ(r b0 c

)= r, so ϕ is surjective.

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Therefore, by the 1st isomorphism theorem, we have

R/K ∼= R

and the isomorphsim is given by

ψ

((a b0 c

)+K

)= a.

Definition 6.2.7. Let R be ring with ideals I and J . Let I + J be the subset of R consistingof all sums of an element from I with an element from J :

I + J = {c | c = a+ b for some a ∈ I, b ∈ J}.

Lemma 6.2.8. Let R be ring with ideals I and J . Then I ∩ J and I + J are both ideals in R.In particular, I ∩ J is also an ideal in I.

Theorem 6.2.9 (Second Isomorphism Theorem). Let R be ring with ideals I and J . ThenI

I ∩ Jis isomorphic to

I + J

Jwith isomorphism given by

ψ :I

I ∩ J−→I + J

J

a+ I ∩ J 7→ a+ J

Proof. Define the function

f : I−→I + J

Ja 7→ a+ J

.

It is easy to show that (i) f is a homomorphism (if a, b ∈ I then f(a + b) = (a + b) + J =(a + J) + (b + J) etc.), (ii) f is surjective (given (a + b) + J with a ∈ I and b ∈ J we havef(a) = a + J = a + b + J), and (iii) that ker f = I ∩ J (if a ∈ I, then f(a) = 0 ⇐⇒ a ∈J ⇐⇒ a ∈ I ∩ J). Now apply the First Isomorphism Theorem.

The way to remember this result is to look at a diagram showing these ideals in R

I + J

I J

I ∩ J

where each line indicates that one ideal is contained in another, with the larger ideal higherup in the diagram. Then the second isomorphism theorem states that we have isomorphicquotients formed by taking pairs on opposite sides of this diagram.

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Example 6.2.7. Use the Second Isomorphim Theorem to prove that

7Z63Z

∼=Z9Z

Solution: The Second Isomorphism Theorem shows that

7Z7Z ∩ 9Z

∼=7Z+ 9Z

9Z

7Z+ 9Z

7Z 9Z

7Z ∩ 9Z

Now we simplify parts of this, starting with the intersection:

x ∈ 7Z ∩ 9Z⇐⇒ x is a multiple of 7 and a multiple of 9

⇐⇒ x is a multiple of 63

⇐⇒ x ∈ 63Z

This is where we ended on Wednesday, December 3

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