1 An Introduction to Algebra - Auburn Universitywebhome.auburn.edu/~goetehp/algtex/intro.pdf1 An...

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1 An Introduction to Algebra by Pat Goeters Department of Mathematics Auburn University Auburn, AL 36849-5310 These notes cover a standard one-term course in Groups, Rings and Modules, as presented at Auburn University, Fall 2001. The notes were also presented to a large group of Central American Mathematics Professors who gathered at the University of Nicaragua at Managua, Nicaragua, during the Summer of 2002. Before presenting these notes at UNAM, copies were made by the department head and sold at a vending table to interested faculty for $ 2. Copies of my colleagues’ notes on Harmonic Analysis were also on sale at the vending table; the price, $ 3. His must have weighed more.

Transcript of 1 An Introduction to Algebra - Auburn Universitywebhome.auburn.edu/~goetehp/algtex/intro.pdf1 An...

Page 1: 1 An Introduction to Algebra - Auburn Universitywebhome.auburn.edu/~goetehp/algtex/intro.pdf1 An Introduction to Algebra by Pat Goeters Department of Mathematics Auburn University

1

An Introduction to Algebra

by

Pat Goeters

Department of MathematicsAuburn University

Auburn, AL 36849-5310

These notes cover a standard one-term course in Groups, Ringsand Modules, as presented at Auburn University, Fall 2001. The noteswere also presented to a large group of Central American MathematicsProfessors who gathered at the University of Nicaragua at Managua,Nicaragua, during the Summer of 2002. Before presenting these notesat UNAM, copies were made by the department head and sold at avending table to interested faculty for $ 2. Copies of my colleagues’notes on Harmonic Analysis were also on sale at the vending table; theprice, $ 3. His must have weighed more.

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Chapter 1

An Introduction to SomeImportant Topics in GroupTheory

1.1 Permutations on n Letters

By a permutation on n letters we merely mean a one-to-one, onto func-tion σ : {1, 2, . . . , n} → {1, 2, . . . , n}. The set or collection of all per-mutations on n letters is denoted by Sn (S refers to the term symmetric,which will be used later on).

One way to express a permutation σ is:

σ =

(1 2 3 . . . ni1 i2 i3 . . . in

),

where i1, i2, . . . , in is just a reordering of 1, 2, . . . , n. Using this nota-tion, we can count the number of permutations on n letters; there are nchoices for the number i1, n−1 choices remaining for i2, n−2 remainingfor i3, and so on. So, there are n(n− 1)(n− 2) · · · 2 · 1 permutations onn letters;

|Sn| = n!

By a cycle, we mean the expression

σ = (i1, i2, · · · , im),

3

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4 CHAPTER 1. AN INTRODUCTION

where i1, i2, . . . , im are distinct numbers in {1, 2, . . . , n}. This cyclerepresents the permutation that sends

i1 7→ i2, i2 7→ i3, . . . , im 7→ i1,

and all numbers other than i1, i2, . . . , im are left fixed (i.e., σ(i) = i ifi is not one of i1, i2, . . . , im).

The function composition of σ and δ, where σ, δ in Sn, producesthe element σδ in Sn by Exercise 1.2. By this we mean that (σδ)(j) =σ(δ(j)).

Observations:

(i) The function ι : {1, 2, . . . , n} → {1, 2, . . . , n} defined by ι(j) = jbelongs to Sn, and is called the identity map.

(ii) If σ ∈ Sn, then σ2, σ3, . . . ∈ Sn, as well (Exercise 1.2).

(iii) The functional inverse of σ ∈ Sn, σ−1, is also in Sn. By definitionσ−1(i) = j exactly when σ(j) = i. Another point of view yieldsthat σ−1 is the only permutation for which σσ−1 = ι = σ−1σ.

Example 1 (a) If

σ =

(1 2 3 . . . ni1 i2 i3 . . . in

),

then

σ−1 =

(i1 i2 i3 . . . in1 2 3 . . . n

)

(and then put the top row of the latter expression in order). Forexample, if

σ =

(1 2 3 4 52 4 1 5 3

),

then

σ−1 =

(1 2 3 4 53 1 5 2 4

).

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1.2. DECOMPOSITIONS OF PERMUTATIONS 5

(b) Given a transposition τ = (i1, i2, . . . , im), τ−1 = (im, im−1, . . . , i1).For example, (1, 4, 3, 5, 2)−1 = (2, 5, 3, 4, 1).

(iv) σk = ι for some positive integer k, and therefore σ−1 = σk−1. Tosee this, Sn is finite so there are distinct positive integers m, ` withσ` = σm. If m < ` multiply both sides by σ−m to get ι = σ`−m.

1.2 Decompositions of Permutations

By decomposition we mean an appropriate way to break down a per-mutation into more elementary parts. This does not directly relate tothe word composition from the phrase function composition.

Permutation Decomposition: Any permutation σ ∈ Sn can beexpressed as a product (or composition) of cycles. Moreover, the cyclesare disjoint, in that no number appears in two different cycles.

Here is how this works:Choose any i ∈ {1, 2, . . . , n}, and call i1 = i. Take i2 = σ(i1),

i3 = σ(i2), and so on; we quit this process as soon as σ(ik) appearsamong i1, i2, . . . , ik. We claim that σ(ik) = 11.

Note that, in this procedure, is = σs−1(i1) where σ0 is defined to beι. If σ(ik) = is with 1 < s ≤ k, then σs−1(i1) = σk−1(i1); and so,

i1 = σ1−sσk−1(i1) = σk−s(i1),

which contradicts our choice of k.We have just produced the cycle τ1 = (i1, i2, . . . , ik) which conforms

to σ at each of i1, i2, . . . , ik;

σ(i1) = i2 = τ1(i1)

σ(i2) = i3 = τ1(i2)

. . .

σ(ik) = i1 = τ1(ik).

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6 CHAPTER 1. AN INTRODUCTION

Choose j not among i1, i2, . . . , ik (if there is no such j, then wehave finished). Set j1 = j, and j2 = σ(j1), and proceed as before.We have obtained a second cycle τ2 = (j1, j2, . . . , j`) which agrees withσ on all of the numbers j1, j2, . . . , j`. We continue producing cyclesτ1, τ2, . . . , τs until all of the numbers 1, 2, . . . , n have been exhausted.Then

σ = τ1τ2 · · · τs. ¦A cycle

(i1, i2, · · · , ik),

is said to be of length k and we will refer to this cycle as a k-cycle.We can express the cycle above as a product of 2-cycles

(i1, i2, . . . , ik) = (i1, ik)(i1, ik−1) · · · (i1, i2).This expression is not unique since for example

(i1, i2, . . . , ik) = (i2, i3, . . . , ik, a1) = (i2, i1)(i2, ik) · · · (i2, i3).A 2-cycle is also called a transposition.

Even/Odd Permutation Theorem: Every permutation σ is a prod-uct of transpositions; the number of which is always even or always odd(depending upon σ and not the technique we use to decompose σ).

Proof: It remains to compare the parities (remainders when dividedby 2) of the number of factors in two decompositions

(a1, b1)(a2, b2) · · · (ak, bk) = (c1, d1)(c2, d2) · · · (cm, dm).

Getting all cycles to one side we find

(cm, dm) · · · (c1, d1)(a1, b1) · · · (ak, bk) = ι.

So, in order to argue that k and m have the same parity (both are evenor both are odd), we must show that m + k is even. That is, if ι canbe expressed as a product of say ` transpositions, then ` is even.

So we can forget about the notation used thus far and assume thatι can be expressed as

(†) (a`, b`)(a`−1, b`−1) · · · (a1, b1) = ι,

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1.2. DECOMPOSITIONS OF PERMUTATIONS 7

in order to show that ` is even. Let c be any number appearing in oneof the transpositions in (†), and let k be the smallest index for which cis one of ak, bk.

Write (ak, bk) = (c, d) and (ak+1, bk+1) = (a, b) and note that ` > ksince c cannot appear for the first time in the leftmost transposition.

One of the following applies:

(i) (a, b)(c, d) = ι when (a, b) = (c, d);

(ii) (a, b)(c, d) = (c, d)(a, b) when (a, b) and (c, d) are disjoint;

(iii) (c, b)(c, d) = (c, d, b) = (d, b, c) = (c, d)(d, b) when c = a; likewise(c, a)(c, d) = (c, d, a) = (c, d)(a, d), or the last possibilities,

(iv) (a, b)(c, a) = (a, c, b) = (b, a, c) = (b, c)(b, a) when a = d, and(a, b)(c, b) = (a, c)(a, b), when b = d.

When (i) holds we can reduce the number of transpositions in (†)by two and produce a representation of ι as a product of ` − 2 trans-positions. Otherwise, (ii) − (iv) apply and we can rewrite † to obtainan expression of ι as a product of ` transpositions where c appears forthe first time in the (k + 1)st transposition.

Focusing on the incident that one of (ii)− (iv) are applicable, oncewe have rewritten (†) so that the first occurrence of c is in the (k +1)st

transposition, we repeat the above steps to either cancel two transpo-sitions from the rewritten form of (†), or rewrite (†) yet again in orderto move the first occurrence of c into the (k + 2)nd transposition.

Again, since c cannot appear for the first time in the left-most trans-position, we are eventually led to canceling two transpositions.

Repeating the above steps, leads to reducing transpositions in pairsfrom (†); eventually leaving only ι on the left-hand side. So, ` must beeven as desired. ¦

We call a permutation odd or even depending upon whether it canbe expressed as either an odd-numbered product or even-numberedproduct of transpositions. The collection An consisting of the evenpermutations plays a prominent role in the history of Algebra:

An = {ρ ∈ Sn | ρ is even}

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8 CHAPTER 1. AN INTRODUCTION

By (1, 2)An we mean the set of permutations which are of the form(1, 2)ρ with ρ ∈ An. It is easy to see that (1, 2)An is the set of all oddpermutations from Sn, and that

|(1, 2)An| = |An|.So,

2|An| = |Sn| = n!

and therefore|An| = n!/2 when n ≥ 2.

1.3 The Integers

Given two integers n,m, the greatest common divisor of n and m isthe positive integer d that divides both n and m, with the additionalfeature that (∗) if k > 0 divides both n, m, then d ≥ k.

Quotient and Remainder: If n,m are positive integers, then thereare non-negative integers q, r such that

n = qm + r, and either r = 0 or 0 < r < m.

Euclid’s Division Algorithm: Given two positive integers n,m; Setr0 = m, and r1 equal to the remainder of n when divided by m. Ifr1 6= 0, set r2 equal to the remainder of r0 when divided by r1. If r2 6= 0,set r3 equal to the remainder of r1 when divided by r2. Note that aslong as zero is not encountered for any ri, we have r0 > r1 > r2 > · · · .So, continuing this precess must eventually lead to some subsequentrj = 0 (assume that no previous remainder was zero; i.e., j is the firstindex for which rj = 0).

A. rj−1 is the greatest common divisor of n,m.

B. The greatest common divisor of n,m can be expressed as kn+`mfor some integers k, `.

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1.3. THE INTEGERS 9

To see why Euclid’s Division Algorithm is true, we must first unravelthe sequence of quotient/remainder equations:

n = q1m + r1

m = q2r1 + r2

r1 = q3r2 + r3

· · ·rj−3 = qj−1rj−2 + rj−1

rj−2 = qjrj−1.

Disregarding the last equation for the moment, we will work back-wards from the second-to-the-last:

rj−1 = rj−3 − qj−1rj−2

rj−1 = rj−3 − qj−1(rj−4 − qj−2rj−3)

rj−1 = (1 + qj−1qj−2)rj−3 − qj−1rj−4.

We then replace rj−3 with rj−5 − qj−3rj−4 to express rj−1 in termsof rj−4 and rj−5, and likewise are able to express d = rj−1 in terms ofany pair rt and rt−1 where t < j − 1. In particular, we can express

d = ur0 + vr1

and sod = um + v(n− q1m) = vn + (u− q1)m.

From the equation d = vn+(u−q1)m, we see that any common divisorof n,m must also divide d.

We next observe that d is a common divisor of n and m; from thelast equation of our quotient/remainder equations, rj−2 = qj−1d, andfrom the second to the last d = rj−3− qj−1rj−2; from these we find thatd divides both rj−2 and rj−3. By d = (1 + qj−1qj−2)rj−3 − qj−1rj−4 wesee that d divides rj−3 and rj−4. Eventually we observe that d dividesall of the ri’s; in particular d divides r0 and r1.

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10 CHAPTER 1. AN INTRODUCTION

So, d divides r0 = m and r1 = n − q1m, and so d divides n,m.Since we have already noticed that any common divisor of n,m mustalso divide d, we now see that d is a common divisor of n,m that isgreater than or equal to any common divisor of n,m. So, d = rj−1 isthe greatest common divisor of n,m. ¦

Example 2 The greatest common divisor of 225 and 360 is found asfollows:

360 = (1)(225) + 135

225 = (1)(135) + 90

135 = (1)(90) + 45

90 = (2)(45).

So, 45 is the greatest common divisor, and

45 = 135− 90 = 135− (225− 135) = 2(135)− 225

= 2(360− 225)− 225 = 2(360)− 3(225).

So, the greatest common divisor 45 is obtained as a linear combination

45 = 2(360)− 3(225).

1.4 Complex Numbers

A complex number is an expression

a + bi

where i is a symbol whose square is −1. Multiplication and additionare said to extend linearly:

(a + bi) + (c + di) = (a + c) + (b + d)i,

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1.5. EXERCISES 11

and(a + bi) · (c + di) = (ac− bd) + (ad + bc)i.

The conjugate (or more precisely, the complex conjugate) of a + biis defined to be

a + bi = a− bi.

The length of a complex number a + bi is written as |a + bi|, and isdefined to be (a + bi)(a+bi) = (a+bi)(a + bi) = a2+b2. So, 1

a2+b2(a+bi)

has length 1.The multiplicative inverse of a non-zero complex number a + bi is

readily seen to be

(a + bi)−1 =1

|a + bi|(a + bi) =1

a2 + b2(a− bi).

Using the Maclaurin series for ex, cos x, and sin x, it is easy toestablish;

Demoivre’s Formula:

eθi = cos θ + isin θ.

From what we have stated, the numbers of the form eθi are the complexnumbers of length 1.

1.5 Exercises

1. Show that a function σ : {1, 2, . . . , n} → {1, 2, . . . , n} is one-to-one if and only if the function σ is onto.

2. Assume that σ, δ ∈ Sn.

(a) Show that σ ◦ δ ∈ Sn. Henceforth we write σδ for σ ◦ δ

(b) Show that σ−1 exists and belongs to Sn.

3. Express the following as products of disjoint cycles, and thenproducts of transpositions.

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12 CHAPTER 1. AN INTRODUCTION

(a)

σ =

(1 2 3 4 5 6 7 83 2 1 8 6 4 7 5

)

(b) δ = (5, 7, 2, 3)(3, 4, 1)(8, 6, 2)

(c) α = (1, 3, 5)(3, 4)(4, 5)(2, 6, 8)

(d) β = (1, 2, 3, 4, 5, 6, 7, 8)2.

4. Given σ = (1, 3, 4, 5)(2, 5, 7) and δ = (6, 5, 4)(1, 3, 2), compute:

(a) σδ.

(b) σ−1, δ−1.

(c) all distinct powers of σ and of δ.

(d) the powers of σ and δ which produce σ−1 and δ−1 respec-tively.

5. List all of the six elements of S3.

6. List all of the cycles in S4.

7. Write out A3 and A4.

8. How many cycles are there in A4?

9. Use Euclid’s Algorithm to find the greatest common divisors ofthe following pairs of numbers, and express the greatest commondivisors as linear combinations of the numbers involved.

(a) 1115, 25.

(b) 473, 86.

(c) 110, 283.

10. Show that if n,m are positive integers, and d is the greatest com-mon divisor of n,m, then the greatest common divisor of m/dand n/d is 1. Hint: Use B. of Euclid’s Division Algorithm.

11. Show if n and m have greatest common divisor equal to 1, and nand m divide an integer k, then nm divides k. Hint: Use B. fromEuclid’s Division Algorithm.

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1.5. EXERCISES 13

12. Assume that n and m have greatest common divisor 1. Show, forany integer k, if n divides mk, then n divides k. Hint: Use B.from Euclid’s Division Algorithm.

13. The least common multiple of positive integers n,m is the smallestpositive integer that is divisible by both n and m. Show that theleast common multiple k of n and m can be found as

k = nm/d,

where d is the greatest common divisor of n and m. Hint: UseProblem 11 and some elbow grease.

14. Find the least common multiples of the pairs of numbers fromProblem 9. Hint: Use Problem 12.

15. Important to the study of groups and rings, are the numbersρ = e(2π/n)i, where n is a positive integer. Show that each powerof ρ, ρj, with j an integer, satisfies (ρj)n = 1, and that ρj 6= 1 if0 < j < n.

16. How many permutations in S5 are not cycles?

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14 CHAPTER 1. AN INTRODUCTION

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Chapter 2

An Introduction to GroupTheory

2.1 Introduction

The study of groups arose from attempting to extend the the quadraticformula to polynomials of higher degree. Since that time, Group Theoryhas been employed in a variety of diverse fields in Mathematics andthe Sciences. We will offer an general presentation here, and providenumerous examples to promote the concepts.

A binary operation on a set G is any function ∗ : G× G → G. Wewill write a ∗ b for the image of (a, b) under ∗.

A group is a set G along with a binary operation ∗ on G such thatthe following hold:

(i) ∗ is associate; i.e., a ∗ (b ∗ c) = (a ∗ b) ∗ c for all a, b, c,∈ G.

(ii) There is an element e ∈ G such that e ∗ a = a ∗ e = a for alla ∈ G.

(iii) Given a ∈ G, there is an element a′ ∈ G such that a∗a′ = a′ ∗a =e, where e is described in (ii).

In honor of N. H. Abel for his triumphant work in showing thegeneral quintic is not solvable by radicals, a group (G, ∗) is called abelian

15

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16 CHAPTER 2. GROUPS

if a ∗ b = b ∗ a for all a, b ∈ G. Furthermore, it is conventional to use +for the binary operation when G is abelian, so a ∗ b is written a + b.

When a group is specified, the underlying set G and the binaryoperation ∗ must be described. In practice, we down-play references to∗, and write ab for a ∗ b, and write G instead of writing (G, ∗). Withthis notation, a−1 is written for the (unique) element a′ associate witha ∈ G described in (iii), except when the additive notation = is usedfor the abelian group G; then −a represents the element from (iii).Also, because of (i), parentheses are dropped when appropriate; i.e.,abc represents (a ∗ b) ∗ c.

2.2 Standard Examples

Example 3 Z is a group under the addition of integers binary opera-tion.

Example 4 Sn is a group under the composition of permutations bi-nary operation.

Once a group G has been found (selected), other groups may befound inside G.

Example 5 If G is a group, then a subset H of G is called a subgroup,provided H is again a group under the binary operation of G. It followsthat a non-empty subset H is a subgroup of G if and only if H is closedunder the binary operation of G and the unary operation a 7→ a−1 fora ∈ H.

A theorem called Cayley’s Theorem, asserts that for any group Gwith n elements, if we encode G as the numbers 1, 2, . . . , n, then G canbe regarded as a subgroup of Sn when G is identified with its encoding.

Example 6 Since permutation multiplication on An is closed and theinverse of an even permutation is even, An is a group under permuta-tion multiplication by the last two examples.

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2.2. STANDARD EXAMPLES 17

Example 7 The group G is called cyclic if G = 〈a〉 for some a ∈ G.In this case, the element a is called a generator for the cyclic group G.Prototypical examples of cyclic groups are the groups Z(n) = 〈1〉 andZ = 〈1〉.Example 8 Given a group G and an element a ∈ G, the subset 〈a〉 ={an | n ∈ Z} is a subgroup of G. Here, an is defined recursively; a0 = eand an = aan−1 for positive n; when n = −m for some positive m,an = (am)−1. The subgroup 〈a〉 is called the cyclic subgroup generatedby a.

The element σ = (1, 2, . . . , n) ∈ Sn has n distinct powers:

σ0 = ι, σ, σ2, · · · , σn−1.

Also, σn = ι. Consequently, the cyclic group 〈σ〉 has order n. Thus,for every n, there is a group of order n.

Example 9 Suppose G is a group and H is a subgroup of G. There isan equivalence relation ∼ defined on G, where a ∼ b if b−1a ∈ H. Theequivalence class of a ∈ G is called the congruence class of a moduloH. If a denotes this congruence class, then a = {b ∈ G | b−1a ∈ H} ={b ∈ G | a−1b ∈ H} = {b ∈ G | b ∈ aH} = aH.

Now assume that G is abelian (written additively). The set of con-gruence classes is denoted by G/H and because G is abelian, G/H isalso a group in a natural manner determined by G:

(a + H) + (b + H) = (a + b) + H.

The symbol + separating (a + H) and (b + H) refers to adding the setsa + H and b + H together, and so the equality is readily checked.

Some specific applications of the previous example yield some well-known groups.

Example 10 In the previous example, take G = Z under additionand H = nZ. Then Z/nZ is a group consisting of exactly n elements0, 1, . . . , n− 1, where i denotes the congruence class of i modulo nZ.

Addition is performed as i+ j = k with k taken to be the remainder(or residue) of the sum of i and j in Z when divided by n. We denotethe group Z/nZ by Z(n).

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18 CHAPTER 2. GROUPS

Example 11 Given a prime p, Zp = { nm| n,m ∈ Z, p 6| m } is a

subgroup of Q. Thus, Q/Zp is a group under addition of congruenceclasses. This group is infinite and is denoted by Z(p∞).

The general element of Z(p∞) can be written as npk where n, k are

integers. From this it follows that every subgroup of Z(p∞) is cyclic.Observe that Z(p∞) is not cyclic.

Example 12 One can form the external direct product of groupsG1, G2, . . . , Gn to obtain a group

G = G1 ×G2 × · · · ×Gn,

whose elements are the (formal) ordered n-tuples (g1, g2, . . . , gn), wheregj ∈ Gj for all j, and whose binary operation is given by

(g1, g2, . . . , gn) · (g′1, g′2, . . . , g′n) = (g1g′1, g2g

′2, . . . , gng′n).

The direct product G is abelian exactly when each Gj is abelian. Inthis case we will use the notation

G = G1 ⊕G2 ⊕ · · · ⊕Gn,

and refer to G as the direct sum of the Gi’s.

2.3 Subgroups

We find that a nonempty subset H of a group G is a subgroup if andonly if H is closed with respect to the binary operation on G and closedunder the formation of inverses.

Example 13 All subgroups of Z(12) can be found below;

Z(12), {0}, {2, 4, 6, 8, 10, 0}, {3, 6, 9, 0}, {4, 8, 0}, {6, 0}

Example 14 The subgroups of Z(4)⊕Z(6) can be computed as follows:Cyclic Subgroups:

C0 = { (0, 0) }

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2.3. SUBGROUPS 19

C1 = 〈(0, 1)〉 = {(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 0)} = 〈(0, 5)〉C2 = 〈(0, 3)〉 = {(0, 3), (0, 0)}

C3 = 〈(0, 2)〉 = {(0, 2)(0, 4), (0, 0)} = 〈(0, 4)〉C4〈(1, 3)〉 = {(1, 3), (2, 0), (3, 3), (0, 0)} = 〈(3, 3)〉

C5 = 〈(2, 3)〉 = {(2, 3), (0, 0)}C6 = 〈(1, 2)〉 = {(1, 2), (2, 4), (3, 0), (0, 2), (1, 4), (2, 0), (3, 2),

(0, 4), (1, 0), (2, 2), (3, 4), (0, 0)} = 〈(1, 4)〉C7 = 〈(1, 5)〉 = {(1, 5), (2, 4), (3, 3), (0, 2), (1, 1), (2, 0), (3, 5), (0, 4),

(1, 3), (2, 2), (3, 1), (0, 0)} = 〈(1, 1)〉 = 〈(3, 1)〉 = 〈(3, 5)〉C8 = 〈(1, 0)〉 = 〈(3, 0)〉C9 = 〈(2, 2)〉 = 〈(2, 4)〉C10 = 〈(2, 1)〉 = 〈(2, 5)〉

C11 = 〈(3, 2)〉 = 〈(3, 4)〉 = 〈(1, 4)〉 = 〈(1, 2)〉(Isn’t this fun!)Non-Cyclic Subgroups:

N0 = Z(4) ⊕ Z(6) = 〈(1, 0), (0, 1)〉 = 〈(3, 0), (0, 1)〉 = 〈(1, 5), (2, 1)〉(N0 can be obtained for example by taking any of the elements (a, b)which generate a subgroup of order 12 above, then including a singleelement (c, d) not in 〈(a, b)〉)

N1 = 〈(2, 0), (0, 3)〉N2 = 〈(1, 0), (0, 2)〉N3 = 〈(2, 0), (0, 1)〉

Example 15 The subgroups of Z(4) ⊕ Z(9) are as follows;

〈(0, 0)〉Z(4) ⊕ Z(9)

〈(a, b)〉where (a, b) belongs to Z(4)⊕Z(9) (Of course, as above, it may turn outthat 〈(a, b)〉 = 〈(c, d)〉 when (a, b) 6= (c, d)).

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20 CHAPTER 2. GROUPS

2.4 Cosets and Counting

From Example 9, when H is a subgroup of G we have an equivalencerelation

a ∼ b ↔ b ∈ aH.

From the nature of an equivalence relation, either

aH = bH or aH ∩ bH = φ.

Therefore, G is the disjoint union of distinct congruence classes, thenumber of which is equal to the number of cosets of H in G.

Furthermore, each congruence class has the same number of ele-ments (namely, |H|); to see this we observe the bijection between aHand H given by:

ah 7→ h = a−1ah

The notation for the number of cosets of H in G is [G : H]; and sowe have the following important result:

Lagrange’s Theorem: If H is a subgroup of G, then

|G| = |H| · [G : H].

The order of an element a 6= e in a finite group G is the least positiveinteger m such that

am = e.

Notice that for any integer k, write

k = qm + r,

with r either 0 or 0 < r < m. Then ak = aqm+r = aqmar = (am)qar = ar

(in particular, am = a · am−1 = am−1 · a = e so a−1 = am−1). Therefore,

〈a〉 = {ak | k ∈ Z} = {a, a2, . . . , am = e}.

Furthermore, {e, a, a2, . . . , am−1} has m distinct elements since if

ai = aj with i < j,

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2.5. CYCLIC GROUPS 21

then

aj−i = e,

contrary to the selection of m. By Lagrange’s Theorem, m divides theorder of G.

If a` = e for some integer `, write ` = qm + r as in the Quo-tient/Remainder Theorem. Then, as above,

a` = e = aqm+r = ar,

and since m is the smallest positive integer for which am = e, r mustbe zero. Thus, m divides `.

2.5 Cyclic Groups

The Fundamental Theorem of Arithmetic asserts that any positive in-teger n is uniquely expressible as a product of powers of distinct primes

n = pe11 pe2

2 · · · pekk

(primes are integers with no proper divisors).We will provide a classification of finite abelian groups; this can be

established from first principals. We only require the definition of agroup and a weak version of The Fundamental Theorem of Arithmetic;every positive integer n has a prime divisor. To see this, suppose n isnot prime. Then

n = n0m0

where n0 and m0 are both proper divisors of n (i.e., neither of them isn). If n0 is not prime, then

n0 = n1m1

where n1 and m1 are both proper divisors of n0. Continuing along thispath, we produce n > n0 > n1 > ... and since positive integers cannot

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22 CHAPTER 2. GROUPS

descend infinitely, at some point we obtain that ni is prime. I.e., n hasa prime divisor.

Subgroups of Cyclic Groups: Subgroups of cyclic groups are againcyclic.

Proof: Let G = 〈g〉 and suppose that H is a subgroup of G. Ifgk ∈ H, then g−k = (gk)−1 ∈ H, so either H = 〈e〉, or there is a leastpositive integer m such that gm ∈ H. In the former case, H is cyclic,so assume that latter. Now suppose gn ∈ H. Write

n = qm + r as in the Quotient/Remainder Theorem.

Then gn = gqm+r = (gm)qgr and so, gr = (gm)−qgn ∈ H due to thefact that gm hence gmq and (gm)−q ∈ H for every integer q. But m wasselected as the smallest positive integer for which gm ∈ H, so r cannotbe positive, and therefore must be 0. Thus, H = 〈gm〉. ¦

Theorem on Orders In Cyclic Groups: Let g ∈ G have ordern. Then h = gk, where k is a positive integer, has order n/(n, k) where(n, k) is the greatest common divisor of n and k.

Proof:Let d = (m, k) and set ` = |gk|. Then

(gk)(n/d) = g(n/d)k = gn(k/d) = (gn)k/d = ek/d = e.

By the remarks at the end of the last section, ` must divide n/d. But

(gk)` = e,

implying that n divides k` (by the previous section again), and so

k`

n=

(k/d)`

n/d,

and n/d divides (k/d)`.By Exercise 10 on page 12, k/d and n/d are relatively prime (i.e.,

their gcd is 1), and therefore by Exercise 12 on page 13, n/d divides `.Thus, ` = n/d as claimed. ¦

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2.6. ABELIAN GROUPS 23

Corollary 16 Let G = 〈g〉 be of order n. The only subgroups of G are〈h〉, where h = gd and d divides n.

Proof:Let H be a subgroup of G. From above, H is cyclic, and so H = 〈gk〉.

Let d = (n, k). From Euclid’s Division Algorithm,

d = uk + vn

for some integers u, v. We claim that H = 〈gd〉. First, gd = guk+vn =(gk)u(gn)v = (gk)u, so that gd ∈ H. Conversely, if we write k = ad forsome integer a, then gk = (gd)a ∈ 〈gd〉. Thus, H = 〈gd〉. ¦

Corollary 17 Let G be a finite group of order n. Then G is cyclic ifand only if for every m dividing n, G has an element of order m.

Proof: If G is cyclic of order n and m divides n, then set k = n/m,so that H = 〈gk〉 has order n/(n, k) = n

n/m= m by the Theorem on

Orders in Cyclic Groups. Conversely, if for every m dividing n, G hasan element of order m, then G must have an element of order n. I.e.,G is cyclic.

¦

Using the examples we can construct endless finite abelian groups

G = Z(n1) ⊕ Z(n2) ⊕ · · · ⊕ Z(nk).

It turns out that all abelian groups (up to renaming the elements) canbe obtained in this way.

2.6 Abelian Groups

Given groups G1, G2, . . . , Gn we have already encountered the (carte-sian) direct sum

G = G1 ⊕G2 ⊕ · · · ⊕Gn = {(g1, g2, . . . , gn) | gi ∈ Gi}.

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24 CHAPTER 2. GROUPS

On the other hand, given an abelian group G with subgroupsH1, H2, . . . , Hn, we say that G is the direct sum of the Hi’s and writeG = H1 ⊕H2 ⊕ · · · ⊕Hn provided

G = Σnj=1Hj = {g ∈ G | g = h1 +h2 + · · ·+hn where hi ∈ Hi for all i},

and for any i,Hi ∩ Σj 6=iHj = 0.

Stickleberger Theorem: Let G be a finite abelian group. If g ∈ Gis an element of maximal order, then G = 〈g〉 ⊕H for some subgroupH of G.

Proof: Let H be a subgroup of G maximal with respect to H∩〈g〉 = 0(as always, 0 serves many purposes; here 0 denotes the subgroup of Gcontaining only the additive identity). Since G is finite, H is easy tofind. In order to show that G is the direct sum of H and 〈g〉 we mustshow that any a ∈ G can be written as kg + h for some integer k andsome h ∈ H.

There is an integer m such that ma ∈ H ⊕ 〈g〉 (in fact, ma = 0for some m). From the opening remarks of this section, m has a primedivisor, p say. We wish to show that ma ∈ H implies a ∈ H. Factoringm as pm′, if we show that pm′a ∈ H implies m′a ∈ H, then we canrepeat this argument again with m′a instead of ma and eventuallydeduce that a ∈ H.

Thus, it suffices to argue the implication

pa ∈ H ⊕ 〈g〉 ⇒ a ∈ H ⊕ 〈g〉.We now forget about the above use of the letter m, and concern our-selves with p. Write pa = h + kg for some integer k.

First, we may as well assume that p divides m = |a|. For if p doesnot divide m, then up + vm = 1 for some integers u, v by Euclid’sAlgorithm. But then, a = upa+ vma = uh+ukg ∈ H⊕〈g〉 as desired.

Now we show that p divides n = |g|. If not, then b = (m/p)a is anelement of order p by the Theorem on Orders of Cyclic Groups, and so,

〈b〉 ∩ 〈g〉 = 0

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2.7. STACKED BASIS THEOREM 25

(by Lagrange’s Theorem). This implies, b+g ∈ 〈b〉⊕〈g〉 is an element oforder pn, the least common multiple of the orders of b and g (Exercise9); this contradicts g having maximal order.

We have reduced to the situation that p divides both |g| and |a|.To see that p divides k, write m = pim′ and n = pjn′ where p doesnot divide n′, or m′. Then m′a has order pi and pjg has order n′. ByLagrange’s Theorem, 〈m′a〉 ∩ 〈pjg〉 = 0 and therefore m′a + pjg hasorder equal to the least common multiple of |m′a|, and |pjg| which ispin′.

But pa ∈ H ⊕ 〈g〉 has order equal to the lcm{|h|, |kg|} and since pj

divides |kg|, pj+1 divides m. But then i ≥ j + 1 so m′a + pjg has orderat least pn, a contradiction.

Write k = pk′ and consider c = a−k′g and H ′ = H +〈c〉. If H ′ = Hthen we conclude a ∈ H ⊕ 〈g〉. Otherwise, H ′ ∩ 〈g〉 6= 0 by the choiceof H; say, k0g = h0 + `0c 6= 0. If p divides `0, then `0 = p`′0, and so0 6= k0g = h0 + `′0h ∈ H; which is contrary to 〈g〉 ∩ H = 0. So, pdoes not divide `0 and up + v`0 = 1 for some integers u, v. Therefore,a = upa+ v`0a = upa+ v`0c+ v`0k

′g = h+ kg + v(k0g−h0)+ v`0k′g ∈

H ⊕ 〈g〉 as needed. ¦

Corollary 18 Any finite abelian group G is a direct sum of cyclicgroups. Moreover, the cyclic groups can be taken to have primary order.

2.7 Stacked Basis Theorem

The last corollary is a version of the Fundamental Theorem for FiniteAbelian Groups; namely that every finite abelian group is a direct sumof cyclic groups. Although the proof relies solely on the SticklebergerTheorem which was proven in the late 1800′s, most textbooks prefer touse the following theorem to prove the Fundamental Theorem.

Stacked Basis Theorem: Let F = Z ⊕ Z ⊕ · · · ⊕ Z (n copies) andsuppose that K is a subgroup of F such that mF ⊆ K for some integerm. Then, there is a generating set z1, z2, . . . , zn of F , and positiveintegers d1, d2, . . . , dn such that d1z1, d2z2, . . . , dnzn generates K.

Proof:

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26 CHAPTER 2. GROUPS

¦

2.8 Normal Subgroups and Homomorphisms

When considering a direct product

G = G1 ×G2 × · · · ×Gn,

for any i, the subgroup N = {(e1, . . . , ei−1, gi, ei+1, . . . , en) | gi ∈ Gi}of G satisfies the condition

gNg−1 = N.

Subgroups enjoying this property are said to be normal and play anelemental role in the study of groups.

Proposition 19 The following conditions are equivalent for a subgroupH of a group G:

(1) gHg−1 = H for all g ∈ G.

(2) G/H is a group under the induced coset operation.

(3) Every left coset is a right coset.

Proof: (1) is equivalent to gH = Hg for every g ∈ G, so (3) followsfrom (1). Conversely, if aH = Hb, then Ha ∩ Hb 6= ∅ implying thataH = Ha for every a ∈ G, so (1) follows from (3). Condition (2)implies (gH)(g−1H = H and so gHg−1 ⊆ H for all g ∈ G. So, H =g−1gHg−1g ⊆ g−1Hg for all g ∈ H and (1) follows from (2). Condition(2) follows from (1) by the computation (aH)(bH) = abHb−1bH =abH. ¦

Given groups G and H, a function φ : G → H is called a homomor-phism provided

φ(ab) = φ(a)φ(b),

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2.9. EXERCISES 27

for all a, b ∈ G. The homomorphism φ is called a monomorphismwhen φ is One-to-one, and an epimorphism when φ is onto. A one-to-one, onto homomorphism is called an isomorphism. The kernel of φ isdefined as

Ker φ = {g ∈ G | φ(g) = eH},where eH is the identity element of H.

Correspondence Theorem: Given an epimorphism φ : G → H,there is a one-to-one correspondence between subgroups B of H andsubgroups A of G containing Ker φ given by

B 7→ φ−1(B)

andA 7→ φ(A).

Moreover, normal subgroups are sent to normal subgroups under thiscorrespondence. In particular, Ker φ is a normal subgroup of G.Proof: Exercise. ¦

When there exists an isomorphism φ : G → H we write G ∼= H.

Fundamental Theorem For Finite Abelian Groups: Any finiteabelian group is the finite direct sum of cyclic groups of primary or-ders. Furthermore, if G1 ⊕G2 ⊕ · · · ⊕Gn

∼= H1 ⊕H2 ⊕ · · · ⊕Hm withG1, G2, . . . , Gn, H1, H2, . . . , Hm cyclic of primary orders, then n = mand after reindexing Gj

∼= Hj for all j.

2.9 Exercises

1. It is possible to find all groups of a given order by examining thepossible binary operation tables. For example, there is only onepossibility for a group of order 2:

* e ae e aa a e

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28 CHAPTER 2. GROUPS

Using ab = ac → b = c and ca = ba → b = c for all a, b, cin the group, find all groups of orders 3, 4, and 5 (apart fromrelabeling).

2. Let H be a subgroup of a finite group G. If [G : H] = 2, showthat H is a normal subgroup of G.

3. What are the elements of maximal order in a direct product offinite cyclic groups C1 × C2 × · · · × Ck?

4. (a) Show that Z(2) ⊕ Z(2) is not cyclic, so it is has a differentstructure than Z(4).

(b) Show that Z(n)⊕Z(m) is cyclic if and only if m, n are relativelyprime (i.e., their gcd is 1).

5. Show that for every positive natural number n, there is a groupof order n.

6. Let g belong to a finite group G, and let m be the least positiveinteger such that gm = 1. We call m is the order of g in G. Showthat m = |〈g〉| and that m divides |G|.

7. List all abelian groups of order 24. Do not list isomorphic groups.Do the same for p3, p4, and p5.

8. If G is a finite abelian group, then G has an element of order pfor every p dividing |G|.

9. (a, b) ∈ A × B has order [n,m] (the lcm of m,n) where n is theorder of a ∈ A and m is the order of b ∈ B.

10. The largest order of an element in Z(n) ⊕ Z(m) is [n,m].

11. d = (n,m) if and only if nZ+ mZ = dZ.

12. ` = [n,m] if and only if nZ ∩mZ = `Z.

13. Can you describe all groups of order 3? Order 4? Order 5? Order6? Order 7?

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2.9. EXERCISES 29

14. Use Lagrange’s Theorem to argue that every group of order p iscyclic.

15. Let G be a group of even order. Show that G has an elementa 6= e such that a2 = e.

16. Find all generators for Z(12)? Do the same for Z(24).

17. Let ϕ be defined on the positive integers by setting ϕ(n) equal tothe number of positive integers less than or equal to n that arerelatively prime to n. (ϕ is called the Euler φ-function). Showthat ϕ(pe) = pe−1(p−1) for any prime p. Hint: Count the numberof positive integers less than or equal to pe that are not relativelyprime to pe.

18. How many generators does Z(n) have?

19. With ϕ equal to the Euler φ-function, show that ϕ(nm) = ϕ(n)ϕ(m)if m and n are relatively prime.

20. Suppose that n can be factored into n = pe11 pe2

2 · · · pekk where

p1, p2, . . . , pk are distinct primes and e1, e2, . . . , ek are positive in-tegers. Show that ϕ(n) = pe1−1

1 (p1−1)pe2−12 (p2−1) · · · pek

k (pk−1).

21. Consider the quaternion group H = {±1,±i,±j,±i} where i2 =j2 = i2 = −1, and ij = k = −ji, jk = i = −kj, and ki = j =−ik. Show that every subgroup of H is normal but that H is notabelian. Note also that every proper subgroup of H is abelian aswell.

22. Write down the cyclic group that represents the given factorgroup:

(a) Z(4)/〈2〉(b) Z(8)/〈3〉(c) Z(18)/〈3〉(d) Z(24)/〈6〉(e) Z(125)/〈10〉

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30 CHAPTER 2. GROUPS

(f) Z(72)/〈18〉

23. Classify the groups below according to the Fundamental Theoremfor Finite Abelian Groups:

(a) (Z(2) ⊕ Z(4))/〈(0, 1)〉(b) (Z(2) ⊕ Z(4))/〈(1, 0)〉(c) (Z(2) ⊕ Z(4))/〈(0, 2)〉(d) (Z(4) ⊕ Z(8))/〈(1, 2)〉(e) (Z(4) ⊕ Z(4) ⊕ Z(8))/〈(1, 2, 4)〉(f) (Z(4) ⊕ Z(4))/H, where H = {(0, 0), (2, 0), (0, 2), (2, 2)}

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2.9. EXERCISES 31

24. Give an example of a nonzero homomorphism, if one exists. If nosuch homomorphism exists, state this.

(a) φ : Z(12) → Z(5)

(b) φ : Z(12) → Z(15)

(c) φ : Z(12) → Z(4)

(d) φ : Z(2) ⊕ Z(4) → Z(2) ⊕ Z(5)

(e) φ : Z→ Z(6)

(f) φ : Z(4) → Z(12)

25. Give a proper normal subgroup of the given group, if one exists.If no such normal subgroup exists, state this.

(a) S3

(b) A3

(c) Z(13)

(d) Z(12)

(e) S4

(f) For G equal to the subgroup of S5 generated by the cycle(1, 2, 3, 4, 5) ∈ S5.

(g) Z(2) ⊕ Z(2)

26. State Lagrange’s Theorem, and use the theorem as an aid to findall of the cosets of the given H in the associated G:

(a) H = A3 in G = S3

(b) H = 〈3〉 in G = Z(12)

(c) H = 〈(1, 2)〉 in G = Z(2) ⊕ Z(4)

(d) H = 〈(1, 2)〉 in G = S3

(e) H = {(0, 0), (2, 0), (0, 2), (2, 2)} in G = Z(4) ⊕ Z(4)

(f) H = 〈6〉 in G = Z(15)

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32 CHAPTER 2. GROUPS

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Chapter 3

An Introduction toCommutative Rings

3.1 Introduction

A ring consists of a set R with two binary operations, + and ·, definedon it such that

(i) There is a element denoted by 1 ∈ R such that

1 · r = r · 1 = r for all r ∈ R.

(ii) R is an abelian group under the binary operation +.

(iii) The distributive laws hold: for every r, s, t ∈ R,

r · (s + t) = r · s + r · t and (r + s) · t = r · t + s · t.

(iv) The binary operation · is associative: for every r, s, t ∈ R,

(r · s) · t = r · (s · t).

If R is a ring with the additional property that r · s = s · r for everyr, s ∈ R, then R (together with the two binary operations) is called a

33

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34 CHAPTER 3. COMMUTATIVE RINGS

commutative ring.

Examples: Under the usual binary operations, Z, Q, R, and Z(n) arecommutative rings.

Example: The collection R of all 2× 2 matrices with entries in R;

(a bc d

)

with a, b, c, d ∈ R, forms a non-commutative ring under the standardmatrix multiplication and addition.

An element r 6= 0 in a ring R is called a zero divisor, if r · s = 0for some 0 6= s ∈ R. The previous example is a ring with many zerodivisors.

Example: Let H be the collection of elements of the form

a + bi + cj + dk

where a, b, c, d ∈ R and i, j, and k behave as described in Exercise2.1. Addition and multiplication extend linearly to make H a non-commutative ring without zero divisors. The ring H is called the ringof Hamiltonian Quaternions.

An element r of R is called a unit if there exists an s ∈ R such that

r · s = s · r = 1.

Commutative rings without zero-divisors are called integral domains.A ring for which every non-zero element is a unit is called a divisionring. A commutative division ring is called a field.

Example: If d is any square-free integer, then

R = Z[√

d] = {n + m√

d | n,m ∈ Z}

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3.2. THE FUNDAMENTAL THEOREM OF ARITHMETIC 35

is an integral domain under the standard operations. None of the ele-ments from Z (except ±1) are units in R.

Example: If d is any square-free integer, then

F = Q[√

d] = {n + m√

d | n,m ∈ Q}

is an field under the standard operations.

3.2 The Fundamental Theorem of Arith-

metic

In this section R is an integral domain. Given a, b ∈ R, we say that adivides b in R, and write a | b, if there exists a c ∈ R such that ac = b.

Definition: Let R be an integral domain.

- An element a ∈ R that is not a unit is called irreducible, if a = bcfor some b, c ∈ R can only occur when c or b is a unit in R.

- An element a ∈ R that is not a unit is called prime, if a | (bc) forsome c, b ∈ R, implies a | b or a | c.

Prime elements are irreducible; if p is prime and p = ab for somea, b ∈ R, then p | ab and so p | a or p | b. If p | a, then pc = a andso p = ab = pcb implying 1 = bc. Likewise, if p | b, then a is a unit.However, as well will see in our first theorem, irreducible elements arequite common-place while primes are not.

Example: The domain R = Z[√

10] affords that every non-zero, non-unit can be factored into a product of irreducible elements, and

6 = 2 · 3 = (4 +√

10)(4−√

10).

But none of 2, 3, 4 +√

10, 4−√10 divides the other elements so theseelements are not prime. It follows that 6 cannot be factored into a

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36 CHAPTER 3. COMMUTATIVE RINGS

product of primes in R.

We say that an ideal I of R is finitely generated if there exist asubset {a1, a2, . . . , an} of I such that

a ∈ I =⇒ ∃ r1, r2, . . . , rn ∈ R such that a = Σni=1riai.

In this case we writeI = (a1, a2, . . . , an).

Conversely, given a1, a2, . . . , an ∈ R we can define an ideal I by assert-ing

a ∈ I ⇔ ∃ r1, r2, . . . , rn ∈ R such that a = Σni=1riai,

in which case I = (a1, a2, . . . , an). Note that for nonzero elements a, bin a domain R, a | b if and only if (b) ⊆ (a).Definition: An integral domain R is called noetherian if every idealis finitely generated.

Proposition 20 The following are equivalent on an integral domainR:

(a) R is noetherian.

(b) If I1 ⊆ I2 ⊆ · · · ⊆ In ⊆ · · · are ideals of R, then there is an indexm such that n ≥ m implies Im = In.

(c) Any nonempty collection of ideals has a maximal member.

(d) A submodule of a finitely generated R-module is itself finitely gen-erated.

Proof: (a) → (b). Given such a chain of ideals, set I = ∪nIn. Sincethe ideals I1, I2, . . . form a chain, I is again an ideal (check this!). ButI is finitely generated, so there exists a finite generating set a1, . . . , an.For each i, there is an index mi such that ai ∈ Imi

. Let

m = max{m1, m2, . . . , mn}.

Then each aj ∈ Im and so I = Im.

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3.2. THE FUNDAMENTAL THEOREM OF ARITHMETIC 37

(b) → (c). Let I be any nonempty collection of ideals. ChooseI1 ∈ I. If I1 is not maximal among all ideals in I, choose I2 ∈ Ithat properly contains I1. Proceeding like this, in order that (b) is notviolated, this process must stop, and it terminates in selecting an idealthat is not smaller than any other ideal of I.

(c) → (a). Given an ideal I, let I be the set of all finitely generatedideals J such that J ⊆ I. I has a maximal member J , and if J 6= Ithere is an element a ∈ I \ J so that J ⊆ J + (a) ⊆ I, in contradictionto (c).

(d) → (a). R is generated by 1 and so is every submodule of R; i.e.,every ideal of R, must be finitely generated.

(a) → (d). We know that every ideal of R is finitely generated. Weinduct on n to show that every submodule of a direct sum of n copiesof R, which I’ll write as F = ⊕nR, is finitely generated. The inductionis easy since if K is a submodule of F and π : F → R is the projectionmap onto the first component (i.e., π(r1, r2, . . . , rn) = r1), then we havean exact sequence

0 → H → Kf→ I → 0,

where f is the restriction of π to K. Since H ≤ ⊕n−1R, and I ≤ R, byinduction H and I are finitely generated. Therefore, if x1, x2, . . . , xn

are such that f(x1), f(x2), . . . , f(xn) generate I, and y1, y2, . . . , ym gen-erate H = Kernel f , then check that x1, x2, . . . , xn, y1, y2, . . . , ym gen-erate K.

Theorem 21 Suppose every chain of principal ideals of R stabilizes.Then, any nonzero nonunit in R is a product of irreducible elements.

Proof: Suppose some nonzero, nonunit in R is not a product ofirreducible elements. Let I be the collection of all principal ideals{(a) | a is not a product of irreducible elements}. Any chain in I;

(a1) ⊆ (a2) ⊆ · · · ⊆ (an) ⊆ · · · ,

must stabilize (by hypothesis) meaning that for some index m, (aj) =(am) for all j ≥ m. Therefore, I contains a maximal element, call it(a).

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38 CHAPTER 3. COMMUTATIVE RINGS

We cannot have a irreducible by the description of I, so there existelement b, c ∈ R, both non-units, such that a = bc. But then (a) isproperly contained in both (b) and (c), and so by the choice of (a),both b and c are products of irreducible elements. But a = bc so a is aproduct of irreducible elements. This contradiction shows that I mustbe nonempty.Definition: A domain R is called a unique factorization domain, orUFD for short, if every nonzero, nonunit can be (uniquely) factoredinto a product of primes of R.

The uniqueness is provided for free (Mathematicians speak of ”pay-ing a price” for a particular hypothesis) for if

p1p2 · · · pn = q1q2 · · · qm,

with pi, qj primes, then p1 divides q1q2 · · · qm, and so by the obvious(correct) extrapolation, p1 divides some qj, which after re-indexing weassume to be q1. Since q1 is irreducible, p1 = u1q1 for some unit u1.After canceling we obtain

p2p3 · · · pn = u1q2q3 · · · qm,

and by induction we obtain n = m, and after re-indexing, pi = uiqi forsome unit ui ∈ R.

Uniqueness is not assured, in general, for factorizations into irre-ducibles.

Proposition 22 The following are equivalent for an integral domainR:

(a) R is a UFD.

(b) (i) Every non-zero, nonunit of R can be factored into a productof irreducible elements, and

(ii) If a1a2 · · · an = b1b2 · · · bm occurs with ai and bj irreducibleelements, then n = m and after re-indexing ai = uibi forsome units ui ∈ R.

Proof: By the remarks after the definition of UFD, it is sufficientto show, under either (a) or (b), that irreducible elements are prime.

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3.2. THE FUNDAMENTAL THEOREM OF ARITHMETIC 39

Assuming (a), any irreducible element r is a product of primes, and soby the definition of irreducible, r must be a single prime. Assuming(b), suppose that an irreducible element s divides a product rt. Sayst′ = rt.

Factoring t′, t and r into products of irreducibles, we find by theuniqueness in part (b), that s must be one of the irreducibles factors(up to unit multiple) of either r or t. That is, s | r or s | t.

A word of caution here: By Theorem 2 and its converse (Exercise 6.),R is a UFD if and only if every proper chain of principal ideals is finite.However, UFD’s need not be noetherian or even seem noetherian-like.

Example 23 It is an exercise at the end of the section that when Ris a UFD, then so is R[x]. Repeating, so is (R[x])[y] = R[x, y], and sois R[x11, x2, . . . , xn] for any number of indeterminates x1, x2, . . . , xn.By R[x1, x2, . . . ] we mean the polynomials with coefficients in R thatinvolve only finitely many of the xj’s, so it follows that R[x1, x2, . . . ] isa UFD as well. The ideal (x1, x2, . . . ) is obviously not finitely generated.

As a corollary to our theorem, we can deduce the FundamentalTheorem of Arithmetic, as it relates to domains.

The classical approach to studying rings is to view them throughtheir ideals (this is due to Dedekind).Definition: Let P and M be ideals of the integral domain R, withP, M 6= R.

- P is said to be a prime ideal, if for any r, s ∈ R, rs ∈ P impliesr ∈ P or s ∈ P .

- M is said to be a maximal ideal, if for any ideal J containing M ,either J = M or J = R.

There is an arithmetic that we can perform on ideals. Given idealsI and J , define the addition of the two ideals as

I + J = {a + b | a ∈ I, b ∈ J},

and the multiplication as

IJ = {x ∈ R | ∃ n ∈ Z+, ai ∈ I, bi ∈ J, for i = 1, 2, . . . , n

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40 CHAPTER 3. COMMUTATIVE RINGS

such that x = Σni=1aibi}.

Certainly IJ contains ab where a ∈ I and b ∈ J but the set {ab | a ∈I, b ∈ J} is not usually closed under + and will not be an ideal in thosecases. Later we will consider multiplicative ideal theory.

We will make several observations:

Observation: An ideal P 6= R is prime if and only if for any ideals I, Jof R with IJ ⊆ P , either I ⊆ P or J ⊆ P .Proof: If P is prime and the ideals I, J are given, with IJ ⊆ P whileJ and I are both not contained in P , then let a ∈ J \ P and b ∈ I \ P .Then ab ∈ P but a, b /∈ P , a contradiction. Conversely, if ab ∈ P ,then (a)(b) = (ab) ⊆ P implies a ∈ (a) ⊆ P or b ∈ (b) ⊆ P , so theideal-theoretic property leads to the elemental property.Observation: An ideal M 6= R is maximal if and only if for any r ∈R \M , M + (r) = R.Proof: If M is maximal and r ∈ R \ M , then M + (r) is an idealproperly containing M . Hence M + (r) = R. Conversely, if M satisfiesM + (r) = R for any r ∈ R \ M , and J is an ideal containing M ,then either J = M , or there exists r ∈ J \ M . In the latter case,R = M + (r) ⊆ J ⊆ R, and so J = R.Observation: Maximal ideals are prime.Proof: If rs ∈ M and r /∈ M , then M + (r) = R and so m + ar = 1for some m ∈ M and a ∈ R. Then, s = sm + sar ∈ M .Observations: Every ideal I unequal to R is contained in some max-imal ideal M .Proof: Consider the set M of all ideals J such that I ⊆ J andJ 6= R. Then I ∈ M so M is nonempty. If {Jα}α<λ is a chain ofelements from M (i.e., α < β ⇔ Jα ≤ Jβ), then set J = ∪α<λJα. Itis easy to see that J ∈M. Therefore, by Zorn’s Lemma, M contains amaximal element M . If J is an ideal that contains M and J 6= R, thenJ ∈ M and so J must be equal to M by the choice of M . Thus, M isa maximal ideal of R.

The following are easy exercises:Show: P is prime if and only if R/P is an integral domain.Show: M is maximal if and only R/M is a field.Show: Finite integral domains are fields, so if P is a prime ideal suchthat R/P is finite, then M is a maximal ideal.

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3.3. INTEGRAL CLOSURE 41

Definition: An integral domain R is called a principal ideal domain,or PID for short, if every ideal is principal (i.e., generated by a singleelement).

Theorem 24 Any PID is a UFD.

Proof: Let R be a PID. Since, obviously R is noetherian, any nonzero,nonunit can be factored into a product of irreducible elements by The-orem 2. It remains to show that any irreducible element in a PID isprime.

Let a be an irreducible element, and suppose a | bc for some b, c ∈ R.Since a is not a unit, (a) 6= R and so (a) is contained in a maximalideal M . Write M = (d). Then a ∈ (d) implies a = de for some e ∈ R,but because a is irreducible, e must be a unit. That is, (a) = (d) = M .Since maximal ideals are prime, bc ∈ (a) implies c ∈ (a) (i.e., a | c) orb ∈ (a) (i.e., a | b).

3.3 Integral Closure

When we study domains, a primary tool is the integral closure of thedomain, and to use it we need to develop it. If we just wish to charac-terize Dedekind domains, we can be satisfied with a simpler descriptionof the integrally closed property, one that is readily usable. So, we needto balance the two objectives. (That is, the definitions we consider nowwill be towards a more general perspective in order to accommodatethe homework, rather than the definition of integral closure given inclass for the purpose of proving our Dedekind domains theorem).Definition: An element t ∈ Q will be said to be integral over R ifthere is a finitely generated (as an R-module) over-ring S of R insideQ which contains t. Given a domain R, an over-ring S of R in thequotient field Q is said to be integral over R, if every t ∈ S is integralover R. If every element in Q that is integral over R belongs to R, thenwe say that R is integrally closed.

Proposition 25 The following are equivalent for an integral domainR and an element t in a field F containing R:

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42 CHAPTER 3. COMMUTATIVE RINGS

(1) t is integral over R.

(2) R[t] is a (module) finitely generated over-ring of R.

(3) There is a polynomial f(x) = xn +rn−1xn−1 + · · ·+r0 ∈ R[x] such

that f(t) = 0.

Proof: (2) → (1) is obvious. (3) → (2) is easy because tn =−Σn−1

j=1 rjtj implying that R[t] is a ring that is generated as an R-module

by 1, t, . . . , tn−1.(1) → (3). Let S be a subring of F containing R and t such that

S = Rt1 + Rt2 + · · · + Rtn. For each i there exists rij ∈ R with jrunning from 1 to n, such that

t · ti = Σnj=1rijtj.

Let A be the n × n matrix whose i, jth entry is rij. Then, we canrephrase the above equation as the matrix equation

AX = tX,

where X is the n-tuple (t1, t2, . . . , tn) (actually, stand X up to multi-ply). Or, in the other familiar form

(A− tI)X = 0,

where I is the n× n identity matrix.If we let x be an indeterminate, then the determinant of A−xI (com-

puted using cofactor expansion for example), is a monic polynomial ofdegree n, for which t is a root. This works the same way it always hasin undergraduate linear algebra. Therefore, t is a root to det(A− xI),which as you may recall, is equal to xn+an−1x

n−1+· · ·+a1x+a0 ∈ R[x].Here is the definition given in class describing integrally closed.

Corollary 26 The noetherian domain R is integrally closed if and onlyif whenever S is an over-ring of R, S−1 = ∅.Proof: This is due to the fact that S−1 6= 0 (i.e., S is a fractionalover-ring) if and only if S is finitely generated:

0 6= t ∈ S−1 ⇒ tS ⊆ R ⇒ S ∼= tS is a (finitely generated) ideal of R,

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3.3. INTEGRAL CLOSURE 43

andS finitely generated over R ⇒ S = Rt1 + . . . + Rtn

for some tj ∈ Q ⇒ rS ⊆ R for some 0 6= r ∈ R

(clear the denominators of the tj’s).Some well-known examples of integrally closed domains are the rings

we have just finished studying.

Example 27 If R is a UFD, then R is integrally closed.

Proof: Suppose t = r/s ∈ Q is integral over R. We can assume thatr/s is in lowest form so that the gcd(r, s) = 1 (r, s do not share a primefactor). By the proposition above, t is a root to some xn + rn−1x

n−1 +· · ·+ r0 ∈ R[x], and so plugging in t and clearing the powers of s fromthe denominators we have

rn + rn−1srn−1 + · · ·+ sn−1r1r + snr0 = 0.

But then

s(rn−1rn−1 + · · ·+ sn−2r1r + sn−1r0) = −rn,

and so any prime factor of s must divide r as well. By design, s doesnot share any primes with r, so s must be a unit, implying t ∈ R.

A way to create examples of integrally closed domains is to take thewidely-understood domain Z, and a field containing Z, and computethe ”integral closure” of Z inside the field.

Example 28 Let F be any field containing Z. The collection of allelements α ∈ F such that α is a root to some monic polynomial in Z[x]is integrally closed, and is called the integral closure of Z in F .

Proof: Let R be the collection of all the α’s just described. As in theproof of the proposition, α ∈ R if and only if the ring Z[α] is finitelygenerated as an abelian group. If α, β ∈ R, then αβ, and α + β bothbelong to S = Z[α, β] = (Z[α])[β].

Something in S looks like

f0(α) + f1(α)β + · · ·+ fk(α)βk

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44 CHAPTER 3. COMMUTATIVE RINGS

for some fj(x) ∈ Z[x]. But α, β ∈ R implies that there exists integersn and m such that any f(α) can be generated by 1, α, . . . , αn−1 overZ, and any g(β) by 1, β, . . . , βm−1 over Z. It follows that any elementin S can be generated by

1, αiβj, i < n, j < m.

For example, for j > m, one can write βj = Σmi=0`ijβ

i, so that

f0(α) + f1(α)β + · · ·+ fk(α)βk = Σmi1=0fi1(α)βi1+

Σkj=m+1[fj(α)Σm

i=0`ijβi].

Likewise we can express each fi(α) exclusively in terms of 1, α, . . . , αn

using integer coefficients. Thus, S is finitely generated and so R is asubring of F .

If γ ∈ F is integral over R, then γ is a root to some xn +rn−1xn−1 +

· · · + r0 ∈ R[x]. But Z[rj] is finitely generated for each j, from whichit follows as above that

Z[r1, r2, . . . , rn−1]

is finitely generated over Z as well. Hence

Z[r1, r2, . . . , rn−1, γ]

is finitely generated over Z too and so γ ∈ R to begin with. That is, Ris integrally closed.

Recall that an element α of a field F containing Z is called algebraicover Z if there exists

0 6= f(x) ∈ Z[x] such that f(α) = 0.

In this case, the minimal polynomial for α over Z is the monic poly-nomial of smallest positive degree in Q[x] having α as a root. Note,if

f(x) = anxn + an−1xn−1 + · · ·+ a0 ∈ Z[x]

has α as a root, with an 6= 0, then so does

g(x) = xn + bn−1xn−1 + · · ·+ b0,

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3.3. INTEGRAL CLOSURE 45

where bj = aj/an. While there are infinitely many choices for f(x),recall that the monic g(x) is unique provided we are considering thepolynomials of smallest degree. Furthermore, it is an easy exerciseusing long division to show that

h(x) ∈ Q[x] with h(α) = 0 ⇐⇒ g(x) | h(x) in Q[x].

Example 29 Let α be in a field F containing Z. The following areequivalent:

(1) α is integral over Z.

(2) α is algebraic over Z and the monic minimal polynomial for α isin Z[x].

(3) α is algebraic but is not the root of some polynomial in Z[x] withleading coefficient greater than 1 while the coefficients are rela-tively prime.

Proof: (1) → (2) is a proof given in class (essentially). Let g(x) ∈Z[x] be the monic polynomial of smallest degree having α as a root(α integral) and let f(x) ∈ Q[x] be the monic polynomial of smallestdegree having α as a root (clearly, α is algebraic over Z). Performinglong division, we can factor g(x) = f(x)h(x) for some monic polynomialh(x) ∈ Q[x].

There are positive integers m and n such that nf(x),mh(x) ∈ Z[x],and the gcd’s of the coefficients of nf(x) and also of mh(x) are both 1(for example, choose only the smallest n that will make nf(x) ∈ Z[x]).Then mng(x) = (nf(x))(mh(x)). Suppose there is a prime p dividingn. Considering the polynomials ”mod p”, we obtain

0 = mng(x) = nf(x) ·mh(x).

(What we mean is that the coefficients of the polynomials are reducedmodulo p, which is signified by the bar over the polynomials; for-mally, we have a ring epimorphism from Z[x] → Z[x]/pZ[x] given byy(x) 7→ y(x) + pZ[x], and the isomorphism Z[x]/pZ[x] ∼= Zp[x] affordsthe reduction of the coefficients).

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46 CHAPTER 3. COMMUTATIVE RINGS

But Zp[x] is an integral domain, and nf(x)·mh(x) = 0 implies eithernf(x) = 0 or mh(x) = 0. Neither is possible since the coefficientsof nf(x) and of mh(x) are relatively prime. Therefore, m = 1 andg(x) = f(x) ∈ Z[x].

(2) → (3). For the moment just assume that α is algebraic. Iff(x) ∈ Q[x] is the monic polynomial of least degree with α as a root,then for some integer n, nf(x) has the gcd of its coefficients equal to1. Conversely, if f0(x) is a polynomial of least degree in Z[x] with thegcd of its coefficients equal to 1 and α as a root, then f0(x)/a ∈ Q[x]is the monic polynomial for α of least degree where a is the coefficientof the highest power of x in f0(x). It follows that, under the full forceof (2), a has to be 1. Thus, every polynomial in Z[x] with the gcd ofits coefficients equal to 1 and having α as a root must be monic.

(3) → (1). Again, if f(x) ∈ Q[x] is the monic irreducible polynomialfor α, and f(x) /∈ Z[x], then we could multiply by an n > 1 to obtain apolynomial nf(x) ∈ Z[x] such that the gcd of the coefficients is 1. Weconclude that f(x) ∈ Z[x] and α is integral over Z.

For example, the following algebraic elements are roots to the givenpolynomials:

α =√

2 +√

3 ⇐⇒ f(x) = x4 − 4x2 + 1,

β =3√

2 + 3√

5 ⇐⇒ g(x) = x6 − 4x3 − 41,

γ =

√2 +

√3/5 ⇐⇒ h(x) = 5x4 − 20x2 + 17,

δ = (√

2+√

3)/(√

1 +√

2) ⇐⇒ k(x) = x8−60x6+134x4+60x2+1.

From this data and the previous example we find that α, β, δ areintegral over Z while γ is not. To see how to obtain the polynomials,let us consider obtaining k(x) for δ. Square both sides and eliminatethe denominator:

δ2(1 + 2 + 2√

2) = 2 + 3 + 2√

6;

(legally) remove the square-root on the√

6 term:

(δ2(3 + 2√

2)− 5)2 = 4(6) = 24;

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3.4. LOCALIZATIONS 47

or;(δ4(3 + 2

√2)2 − 2(5)δ2(3 + 2

√2) + 25 = 24;

or;(δ4(9 + 8 + 12

√2)− 20δ2

√2 = −1 + 30δ2;

isolate the remaining square-root;√

2(12δ4 − 20δ2) = −1 + 30δ2 − 17δ4;

square both sides;

2(144δ8 + 400δ4− 480δ6) = 1 + 900δ4 + (17)2δ8− 1020δ6− 60δ2 + 34x4;

and since 172 = 288 + 1 = 2(144) + 1, when we replace δ with x, weobtain

k(x) = x8 − 60x6 + 134x4 − 60x2 + 1.

3.4 Localizations and Over-rings of R

Definition: If P is a prime ideal in an integral domain R, then

RP = {r

s| r ∈ R, s ∈ R \ P},

is an over-ring of R (inside Q), called the localization of R at P.More generally, if S is a nonempty multiplicatively closed subset of

R that does not contain 0; that is, S 6= ∅, 0 /∈ S and for every s1, s2 ∈ S,s1s2 ∈ S, then one can form a localization, S−1(R), where

S−1(R) = {r

s| s ∈ S, r ∈ R}.

Using the principal that S is multiplicatively closed, we see immediatelythat S−1(R) is an over-ring of R (regard r ∈ R as sr

swhere s ∈ S).

In the case of the localization of R at a prime P , S is taken tobe R \ P . In one instance below we form a localization that is not ata prime ideal so we must consider the broader notion of localization,S−1(R) (though it is no harder).

Proposition 30 Let S be a multiplicatively closed subset of R.

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48 CHAPTER 3. COMMUTATIVE RINGS

(a) J is an ideal of S−1(R) if and only if J = S−1(I) = {as| s ∈

S, a ∈ I}, for some ideal I of R.

(b) P ′ is a prime ideal of S−1(R) if and only if P ′ = S−1(P ) ={a

s| s ∈ S, a ∈ P}, for some prime ideal P of R such that

P ∩ S = ∅.

Proof: (a). Given an ideal J of S−1(R), we will consider I = R∩J . Ifb ∈ J , then sb ∈ R∩J = I for some s ∈ S, so that b = sb

s∈ S−1(I). On

the other hand, I ⊆ J implies that S−1(R)I = S−1(I) ⊆ J . Therefore,J = S−1(I). Conversely, S−1(I) = S−1(R)I is an ideal of S−1(R) whenI is an ideal of R, so the proof of (a) is complete.

(b). From (a) we note that P ′ = S−1(P ) where P = P ′ ∩ R. Withthis notation, we will show that

P ′ is prime in S−1(R) ⇐⇒ P is prime in R with P ∩ S = ∅.

If P ′ is prime in S−1(R), then for any r1, r2 in R (in particular), r1r2 ∈P ′ implies r1r2 ∈ P ′ ∩ R = P . We are given that r1 ∈ P ′ (hencer1 ∈ P ′ ∩ R = P ) or r2 ∈ P ′ (hence r2 ∈ P ′ ∩ R = P ), and because1 /∈ P ′, 1 /∈ P so P is prime.

Conversely, if P = P ′∩R is a prime ideal of R with P ∩S = ∅, thenfirst of all, 1 /∈ P ′, so P ′ 6= S−1(R). If r1

s1

r2

s2∈ P ′ = S−1(P ) for some

r1, r2 ∈ R and s1, s2 ∈ S, then, multiplying by s1s2, r1r2 ∈ P ′∩R = P ,implies r1 ∈ P ⊆ P ′ or r2 ∈ P ⊆ P ′. Hence r1

s1∈ P ′ = S−1(P ) or

r1

s1∈ P ′.Do you wonder which domains have the property that every over-

ring S of R in Q must be a localization? This is what commutativering theorists do, and the next section provides a partial answer.

3.5 Dedekind Domains

Characterization of Dedekind Domains The following are equiv-alent for an integral domain R that is not a field.

(1) R is Dedekind (i.e., every proper ideal is a product of prime ide-als).

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3.5. DEDEKIND DOMAINS 49

(2) Every nonzero ideal is invertible.

(3) R is noetherian, every nonzero prime ideal is maximal, and R isintegrally closed.

(4) R is noetherian and for every maximal ideal M of R, RM is alocal PID.

Proof:(1) → (2). Observe that a product of ideals I1I2 · · · In is invertible

if and only if each Ij is invertible. Therefore, it is sufficient to showthat every nonzero prime ideal is invertible. Let P be a nonzero primeideal of R. Given 0 6= a ∈ P , by (1), we can write Ra = P1P2 · · ·Pn.By the preceding remark, each Pj is invertible.

Since P1P2 · · ·Pn = Ra ⊆ P and P is prime, P contains one of theinvertible primes Pj (for some j). In order to show that P is invertible,it is sufficient to show that Pj is maximal. Therefore, we have reducedthe problem to showing that every invertible prime ideal is maximal, sowithout loss of generality, let P be an invertible prime (in other words,forget all the previous uses of letters).

In order to show that P is maximal we must show that if r ∈R \ P , then P + Rr = R. Suppose that P + Rr 6= R. Then, by(1), P + Rr = P1P2 · · ·Pn for some prime ideals P1, P2, . . . , Pn. Sincer /∈ P , r2 /∈ P as well and we can write P + Rr2 = Q1Q2 · · ·Qm forsome prime ideals Q1, Q2, . . . , Qm. Each Pj contains P1P2 · · ·Pn andtherefore must contain P . Likewise, Qi contains P for all i. Hence, wecan factor modulo P :

(P + Rr)/P = P1P2 · · · Pn,

and(P + Rr2)/P = Q1Q2 · · · Qm,

where Pj = Pj/P , R = R/P , and Qi = Qi/P . Furthermore,

((P + Rr)/P )2 = (P + Rr2)/P,

so thatP 2

1 P 22 · · · P 2

n = Q1Q2 · · · Qm.

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50 CHAPTER 3. COMMUTATIVE RINGS

Each Pj and Qi is a prime ideal in the integral domain R/P ; forinstance, R/Pj = (R/P )/(Pj/P ) = R/Pj is an integral domain. Ad-ditionally, each Pj and Qi is invertible because (P + Rr)/P and (P +Rr2)/P are principal ideals of R (the general element in (P + Rr2)/Pfor example is the coset sr2 + P for some s ∈ R).

We can prove by induction on m that 2n = m and after re-indexing,Q2i−1 = Q2i = Pi for i = 1, 2, . . . n. We will show how reduction of theindex works. Among each Pi and Qj, choose one that is minimal withrespect to containment. If it is a Pi, then reorder so that i = 1. ThenP1 contains Q1Q2 · · · Qm and hence must contain some Qj since P1 is aprime ideal. Reorder so that j = 1, then cancel

P1P22 · · · P 2

n = Q2 · · · Qm.

If the original minimal prime was chosen as some Qj, reorder so thatj = 1, and as before Q1, must contain some Pi, which we take to beP1. Again, we cancel to obtain

P1P22 · · · P 2

n = Q2 · · · Qm.

In either case, induction applies (or, we can just repeat the argumentas many times as necessary).

From Q2i−1 = Q2i = Pi for i = 1, 2, . . . n, we see that Q2i−1 = Q2i =Pi for i = 1, 2, . . . n. Therefore,

(P + Rr)2 = P 21 P 2

2 · · ·P 2n = Q1Q2 · · ·Qm = P + Rr2.

So, P ⊆ P +Rr2 = P 2+Rr2+Pr. Given p ∈ P , write p = p2+sr2+p1r,so that sr2 = p− p2 − p1r ∈ P . Since r2 does not belong to the primeideal P , s ∈ P and so P ⊆ P + Pr2 + Pr = P 2 + Pr = P (P + Rr).But since P is invertible, R = P−1P ⊆ P−1P (P + Rr) = P + Rr ⊆ R.This contradiction, shows that indeed P +Rr = R (from the start) andtherefore P is maximal as claimed.

(2) → (3). For any proper ideal I,

I−1I = R ⇐⇒ Σni=1biai = 1 for some bi ∈ I−1, ai ∈ I.

Under (2), I is invertible, and so a = Σni=1(abi)ai for any a ∈ I, and since

abi ∈ I for all i, I is generated by a1, a2, . . . , an. I.e., R is noetherian.

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3.5. DEDEKIND DOMAINS 51

If P is a nonzero prime ideal, then P is contained in a maximal idealM . Since M is invertible, M−1P ⊆ R, and M(M−1P ) ⊆ P . Since Pis prime, either M−1P ⊆ P or M ⊆ P . If the former were to hold,M−1P ⊆ P , then P invertible implies M−1 = M−1PP−1 ⊆ PP−1 =R. But forming inverses is order reversing and M ⊆ R (so M−1 ⊇R). Taken together, we have M−1 = R, from which MM−1 = M ; acontradiction. Therefore, the latter case holds; P = M .

Suppose S is a fractional over-ring of R in the quotient field Q;R ⊆ S implies S−1 ⊆ R is an ideal and by (2), (S−1)−1S−1 = R. ButS is a ring, and (S−1)−1S−1 = R is an S-module, so S · R ⊆ R, andS = R.

(3) → (4). Recall that RM = { rs| r, s ∈ R, s /∈ M}. By Proposition

11, the ideals of S = RM are extended; i.e., the ideals of S must be ofthe form I · RM = { r

s| r ∈ I, s ∈ R \ M} for some ideal I of R.

Furthermore, the prime ideals of S = RM are of the form P ·RM whereP is a prime ideal contained in M . Therefore, from (3), every idealof S is finitely generated (since the ideals of R are finitely generated),with 0 and N = M ·RM as the only prime ideals of S.

We will first show that the maximal ideal of S is principal. Byhomework problem number 2 from the current set, it suffices to showthat N is invertible (relative to S). But, in a manner applied earlier,N−1 = {t ∈ Q | tN ⊆ S}, contains S, and so N ⊆ N−1N ⊆ S. IfN−1N = N , then N−1N−1N = N−1N = N , implying that N−1N−1 =N−1; i.e., T = N−1 is a fractional over-ring of S. But, we claim that Sis integrally closed; a property inherited from R.

If S[t] is an over-ring of S, then there exists an equation tn +sn−1t

n−1 + · · · + s0 = 0 where sj ∈ S and n ≥ 1. Let r ∈ R \ Mbe such that rsj ∈ R for all j. Then rn(tn + sn−1t

n−1 + · · · + s0) =(rt)n + rsn−1(rt)

n−1 + · · · + rns0 = 0, implying that rt is integral overR. Therefore, rt = a ∈ R and consequently t = ar−1 ∈ S. Therefore,S is integrally closed by Proposition 6, and so N−1 = S.

We will show in this paragraph that N−1 is in fact unequal to S.Fix 0 6= y ∈ N and consider the cyclic S-module

C = 〈1y

+ S〉

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52 CHAPTER 3. COMMUTATIVE RINGS

inside Q/S. Observe that C 6= 0, because N does not contain a unit ofS. The collection I of all ideals I of S such that I = {s ∈ S | s( t

y+S) =

0}, for some fixed t ∈ S such that ty

/∈ S. Since S is noetherian, unionsof chained ideals of S stabilize and so I contains maximal members.Let P be maximal in I, say P = annS x + S for some x = t

y∈ Q \ S

where t ∈ S, and suppose rs ∈ P , for some r, s ∈ S with r /∈ P . Then,r(x+S) 6= 0 but (P +(s))(rx+S) = 0. This implies P = P +(s) and sos ∈ P and P is prime. But P 6= 0 since 0 6= Sy ∈ I, therefore P = N .That is, there exists an element q = t

y∈ Q \ S such that Nq ∈ S and

so q ∈ N−1 \ S.

Therefore, N−1N = S is the only possibility, and since S is local,by homework problem 2, N must be principal. Write N = (b) for someb ∈ S. Now, S is a noetherian domain with principal maximal idealN = (b) and N and 0 are the only primes. We need to show that S isa PID.

Our first claim is that ∩∞k=1Nk = 0. If we set I = ∩∞k=1N

k thenclearly bI = I: check that if a = bc, then a ∈ (bn) if and only ifc ∈ (bn−1); so a ∈ (bn) for all n if and only if c ∈ (bn) for all n. Thus,1bI = I, and I is an ideal in S[1

b] = Q. To see that S[1

b] = Q observe

that S[1b] is a localization of S using the set {bj}∞j=1, and by Proposition

11, S[1b] has no prime ideals save 0; consequently, S[1

b] must be a field;

and consequently, the field Q. But having I a module over Q can onlyhappen when I = 0. Thus, ∩∞k=1N

k = 0.

Let J be a proper ideal of S. Then J ⊆ N . Since ∩k(bk) = 0, there

is a least positive integer such that J ⊆ (bk) but J is not contained in(bk+1). If a ∈ J \ (bk+1), write a = ubk where u ∈ R. Then, obviously,u /∈ (b), implying that u is a unit (since S is local with maximal ideal(b)). Therefore, J = (bk) and S is a PID.

(4) → (1). Let P be a prime ideal of R that is contained in amaximal ideal M , P 6= M . Then, from Proposition 11, P · RM is aprime ideal in the ring RM . But, RM is a local PID so the only primeideals are 0 and M ·RM . Also,

r ∈ M \ P =⇒ r ∈ (M ·RM) \ (P ·RM) (check!)

and so P must be zero. Hence any nonzero prime ideal of R is maximal.

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3.5. DEDEKIND DOMAINS 53

By the Primary Decomposition Theorem (to follow), every idealI 6= 0 has a (reduced) primary decomposition:

I = I1 ∩ I2 ∩ · · · ∩ In,

where the Ij’s are primary ideals associated with distinct primes (max-imal ideals) Mj’s. When the radicals, M1,M2, . . . , Mn are coprime inpairs (i.e., Mi + Mj = R for all i 6= j), the intersection can be replacedby the product:

I = I1 · I2 · · · In (see Corollary 11 below).

Once we show that the primary ideal Ij must be equal to Mnj

j for somepositive integer nj, the proof will be complete.

Let J be any nonzero primary ideal and let M = rad(J). Note thatM is a maximal ideal. Since RM is a local PID, J ·RM = Mn ·RM forsome positive integer n. We claim that J = Mn. Note, that Mn too isprimary by Proposition 14.

We claim that because J is primary,

J = R ∩ (J ·RM),

where M = rad(J). Once this has been established we can show toothat Mn = R∩ (Mn ·RM), since Mn is primary with radical M as well.This leads to

J = R ∩ (J ·RM) = R ∩ (Mn ·RM) = Mn,

as desired.Clearly, J ⊆ R∩ (J ·RM). Suppose a ∈ R∩ (J ·RM) \ J . Then, for

some r ∈ R \M , ra ∈ J . But J is primary and a /∈ J implies rm ∈ Jfor some m ≥ 1. However, r /∈ M implies rm /∈ M for any m since Mis prime; a contradiction. Therefore, a primary ideal must be Mn forsome maximal ideal M and integer n, completing the proof.

Dedekind domains have ample structure with which we can discoverthe wonders of ring theory. Many of the assertions that were madeduring the course of the proof no longer hold once we leave the structureof Dedekind domains.

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54 CHAPTER 3. COMMUTATIVE RINGS

Example 31 Let R = F [x, y] where F is a field. The quotient field Qof R is then F (x, y) = { f(x, y)/g(x, y) | f(x, y), g(x, y) ∈ R, g(x, y) 6=0}. The ideal generated by x and y, M = (x, y), is maximal. But

if q(x, y) ∈ M−1, then q(x, y) · x ∈ R implies q(x, y) = h(x,y)x

(recallthat R is a UFD). But then q(x, y) · y ∈ R, implies x|h(x, y) so thatq(x, y) ∈ R. That is,

M−1 = R,

unlike the case when the domain is noetherian such that every nonzeroprime ideal is maximal.

A class of domains receiving the most attention these days is theclass of Prufer domains. A domain is called Prufer if every finitelygenerated ideal is invertible. There are numerous examples of non-noetherian Prufer domains. Also, it can be shown that a maximal idealM of a Prufer domain is infinitely generated if and only if M2 = M ,and there are many example of such rings.

Example 32 The following conditions are equivalent for an integraldomain R:

(1) R is Prufer.

(2) For all ideals I, J,K of R, I(J ∩K) = (IJ) ∩ (IK).

(3) For all ideals I, J,K of R, I ∩ (J + K) = (I ∩ J) + (I ∩K).

(4) Every over-ring S of R in Q is Prufer.

(5) For every prime ideal P of R, R/P is Prufer.

It can be shown for a Dedekind domain R, that every over-ring Sof R in Q is of the form

S = ∩M∈MRM ,

where M is a collection of maximal ideals of R, and S is like-wiseDedekind.EXERCISES of SECTION 3:

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3.6. PRIMARY DECOMPOSITIONS OF IDEALS 55

1. A domain R is called a divisorial domain if every proper ideal Isatisfies (I−1)−1 = I.

(i) Show that Dedekind domains are divisorial.

(ii) If R is divisorial and M is a maximal ideal of R, show thatM−1/R ∼= R/M .

2. (i) Show that a PID is Dedekind.

(ii) Show that if every maximal ideal in a Dedekind domain Ris principal, then R is a PID.

(iii) Find Dedekind domain that is not a PID. Hint: considerZ[√−5].

3.6 Primary Decompositions of Ideals

We have seen that any ideal in a PID is a product of primary ideals(i.e., powers of a prime ideal). In order to establish this for more generaldomains, we need to weaken the definition of primary.Definition: An ideal I of an integral domain R is called primary,if I 6= R and whenever rs ∈ I with r /∈ I, there is an n ≥ 1 such thatsn ∈ I.Definition: The radical of an ideal I, rad(I), is the ideal consistingof the elements r ∈ R such that rn ∈ I for some n ≥ 1.

Sometimes rad(I) is called the nilradical of I, and sometimes it iswritten as

√I.

Proposition 33 Given an ideal I 6= R of the noetherian domain R,one can say the following:

(i) The radical of I, rad(I), is the intersection of all prime idealscontaining I.

(ii) There is an n ≥ 1 such that rad(I)n ⊆ I.

(iii) If I is primary, then rad(I) is prime.

(iv) If rad(I) is a maximal ideal, I is primary.

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56 CHAPTER 3. COMMUTATIVE RINGS

Item (i) prompts the phrase, prime radical, for rad(I). Proof: (i).Let J = ∩{P | I ⊆ P} (P will always represent a prime unless other-wise indicated). If r ∈ rad(I), then rn ∈ I for some n ≥ 1. But then,for any P ⊇ I, rn ∈ P and since P is prime, r ∈ P . I.e., rad(I) ⊆ J .Conversely, if r ∈ J , but rn /∈ I for any n, then consider the situationwith the multiplicatively closed set S = {rn}∞n=1. We have S ∩ I = 0and so S−1(I) is a proper ideal of S−1(R). However, there are no primesin S−1(R) containing S−1(I) since r belongs to every P containing I;a contradiction. Thus, rn ∈ I for some n and J ⊆ rad(I).

(ii). Since R is noetherian, rad(I) is finitely generated; say rad(I)is generated by r1, r2, . . . , rk. For each j there exists an nj such thatr

nj

j ∈ I. Set n = Σjnj. Given r = Σjajrj ∈ rad(I), we will convinceourselves that rn ∈ R by considering the case of two rj’s:

rn = (a1r1+a2r2)n = Σn

i=0cn,i(a1r1)i(a2r2)

n−i = Σni=0cn,i(a

i1a

n−i2 )[ri

1rn−i2 ],

where cn,i is the (integral) number of combinations of n things taken iat a time without regard to order. Observe that when i < n1, n−i > n2

and so rn−i2 ∈ I and when i ≥ n1, ri

1 ∈ I. Thus

rn ∈ I.

That is, rad(I)n ⊆ I. The case when k > 2 is analogous, one must usethe multinomial expansion formula rather than the binomial expansionformula.

(iii). Suppose rs ∈ rad(I) for some r, s ∈ R with r /∈ rad(I). Bythe definition of rad(I), there is an n ≥ 1 such that rnsn = (rs)n ∈ I.But I is primary, and rn /∈ I (since r /∈ rad(I)), implies that somepower of sn, smn ∈ I. That is, s ∈ rad(I). Therefore, rad(I) is prime.

(iv). Let I be an ideal such that rad(I) = M is maximal andsuppose that ab ∈ I, with a /∈ I. By (ii), there is an n ≥ 1 such thatMn ⊆ I, so if we show that b ∈ M , then we will have bn ∈ I. Supposeb /∈ M . Then M + (b) = R. Consider

(M + (b))n = Mn + bMn−1 + · · · bn−1M + (bn) = R.

If we multiply this by a we obtain

aMn + (ab)Mn−1 + · · ·+ (abn−1)M + (abn) = aR.

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3.6. PRIMARY DECOMPOSITIONS OF IDEALS 57

But the left-hand-side is contained in I while the right-hand-side is not;a contradiction. Thus, b ∈ M and so bn ∈ I as required for a primaryideal.

Corollary 34 If J is a primary ideal in a noetherian domain and P =rad(J), then for some positive integer n, P n ⊆ J .

Example 35 Let R = F [x, y] where F is a field, and observe thatM = (x, y) is a maximal ideal of R. Then each of the ideals givenbelow is primary with radical M :

(x2, y), (x2, y2), (x2, y3), (x3, y3), . . .

Observe that a power of M is primary but does not appear in this list.Check for M or M2 = (x2, xy, y2) for example.

Given any module B, we can consider the annihilator ideal associ-ated with an element b ∈ B

annR b = {r ∈ R | rb = 0}.

Of course annR b may be 0 or may be R, but for certain modules B,annR b will always be a proper ideal when b 6= 0. For example, whenB = R/I for a proper ideal I of R. Then, for any 0 6= b ∈ B = R/I,annR b is a proper ideal, properly containing I.

Let us restrict ourselves to torsion (or bounded modules B), andassume that for any b ∈ B there exists an 0 6= r ∈ R such that rb = 0.The associated primes of B are the prime ideals P of R such that

P = annR b,

for some b ∈ B. With A = {J | J = annR b for some 0 6= b ∈ B}, themaximal elements in A are associated primes of B.

To see this, let P be a maximal element in A and write P = annR bfor some b ∈ B. If rs ∈ P but r /∈ P , then rb 6= 0, but

(P + (s)) · rb ⊆ Pb = 0.

Thus, (s) ⊆ P , and P is prime.

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58 CHAPTER 3. COMMUTATIVE RINGS

Proposition 36 Let I be a proper ideal of the noetherian domain R.Then R/I has a single associated prime P if and only if I is primarywith radical P .

Proof: Assume that I is primary with radical P , and let P ′ be anassociated prime of the (bounded) module R/I (a · R/I = 0 for some(any) nonzero a ∈ I). Say P ′ = annR r + I for some r ∈ R \ I, andobserve that I ⊆ P ′. Suppose there exists an a ∈ P ′ \ P . Then ar ∈ Iand r /∈ I implies an ∈ I ⊆ P for some n, and because P is prime a ∈ P ;i.e., P ′ ⊆ P . But I contains a power of its radical P , so Pm ⊆ I ⊆ P ′

implies P = P ′.Conversely, suppose R/I has the lone associate prime P . Note that

P ⊇ I. Suppose rs ∈ I with r /∈ I. Then b = r + I 6= 0 andthe annihilator of b is contained in a maximal element in the set of allannihilators of R/I. I.e., P (r+I) = 0 and also (P +(s))(r+I) = 0. Tocomplete the proof it is enough to show that every prime P ′ containingI, contains P (for then P = rad(I) by Proposition 14, and by Corollary15, sn ∈ P n ⊆ I as needed).

Suppose there exists an a ∈ P \ P ′. Since I ⊆ P ′, a /∈ I and soa + I 6= 0. But then P is the unique maximal annihilator, and soPa ⊆ I ⊆ P ′. In particular, a2 ∈ P ′; a contradiction. Thus P ⊆ P ′

and the proof is complete.There are a lot of different proofs of the result that every ideal has

a primary decomposition, that is, for any proper ideal I, there existprimary ideals I1, I2, . . . , In such that

I = I1 ∩ I2 ∩ · · · ∩ In.

We opt for the direct approach, though this is not the traditional ap-proach.

An ideal J that is different from R is called irreducible, if J =J ′ ∩ J ′ for some ideals J ′, J”, implies one of J ′ of J” must be J (soundfamiliar?). An intersection I = J1∩J2∩ · · ·∩Jn is called irredundant iffor no proper subset U of {1, 2, . . . , n} is I = ∩j∈UJj. So J is irreducibleif J is not the irredundant intersection of two ideals J ′ and J ′′.

Theorem 37 Let I be a proper ideal in a noetherian domain R. Then,there exists primary ideals I1, I2, . . . , Ik such that

I = I1 ∩ I2 ∩ · · · ∩ Ik.

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3.6. PRIMARY DECOMPOSITIONS OF IDEALS 59

Proof: We claim that any proper ideal I can be expressed as

I = I1 ∩ I2 ∩ · · · ∩ Ik,

where the ideals I1, I2, . . . , In are irreducible. Perhaps you can do thisyourselves? While this is straightforward, we can shorten the processby using a slight technique.

If some proper ideal is not a finite intersection of irreducible ide-als, then let I be the nonempty collection of all such ideals. Since Ris noetherian, I contains a maximal member, call it J , by Proposi-tion 1. Every ideal properly containing J must be an intersection ofirreducibles.

If J is irreducible, then J is an intersection of irreducibles. Other-wise, J = J1∩J2 with J1, J2 both properly containing J . But then J1, J2

are intersections of irreducibles, and so is J = J1 ∩ J2; a contradiction.Therefore I is empty.

It remains to show that irreducible ideals are primary. Let J be anirreducible ideal and consider R/J . If R/J has two associated prime, P1

and P2, then there exist elements r1, r2 ∈ R such that Pi = annR ri+J .Note that this last condition implies that for Ji = Rri+J , Ji/J ∼= R/Pi

for i = 1, 2. Since J1, J2 properly contain J , J1 ∩ J2 = J is impossible.Let x ∈ J1∩J2 but x /∈ J . The element x+J1 corresponds to a nonzeroelement r + P1 in R/P1. But P2(x + J) = 0 implies that P2 annihilatesr + P1 and so P2r ⊆ P1. This is impossible because r /∈ P1 and P2 andP1 are distinct primes. Thus, R/J has only 1 associated prime and byProposition 17, J is primary.

Corollary 38 If I1, I2, . . . , In are primary ideals whose radicals are thedistinct maximal ideals M1,M2, . . . , Mn, then

I1 ∩ I2 ∩ · · · ∩ In = I1 · I2 · · · In.

Proof: Each of the primary ideals Ij contains a power of their radicalMj by Corollary 8, so there exists a k ≥ 1 such that Mk

j ⊆ Ij for all j.Note that

Mki + Πj 6=iM

kj = R

for all i. To see this observe that if M is a maximal ideal containingMk

i +Πj 6=iMkj then M contains Mi and M contains some Mj with j 6= i

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60 CHAPTER 3. COMMUTATIVE RINGS

(since M is prime), which is impossible. It follows that for any i,

Ii + Πj 6=iIj = R.

We have I1I2 · · · In ⊆ I1∩ I2∩ · · · ∩ In, and we can go the other wayby induction on n. The inductive step is handled by multiplying bothsides of the above equation by I1 ∩ I2 ∩ · · · ∩ In:

I1 ∩ I2 ∩ · · · ∩ In = [Ii ∩ ∩j 6=iIj](Ii + Πj 6=iIj) =

[Ii ∩ (Πj 6=iIj)](Ii + Πj 6=iIj) = [Ii ∩ (Πj 6=iIj)]Ii + [Ii ∩ (Πj 6=iIj)]Πj 6=iIj

⊆ I1I2 · · · In.

3.7 Exercises

1. Show that any non-zero element a + bi + cj + dk in the ring ofHamiltonian Quaternions H has a multiplicative inverse in H.

2. Determine all zero-divisors in the ring R of all 2 × 2 matriceswhose entries lie in R.

3. (a) Determine all zero-divisors in Z(n).

(b) Determine all units in Z(n).

(c) Conclude that any non-zero element of Z(n) is either a zero-divisor or is a unit.

4. Give an example of a ring R and a non-zero element r ∈ R forwhich r is neither a zero-divisor nor a unit.

5. Show that if F is a field, then R = F [x] is a PID. Hint: use thedegree function deg : F [x] → N and the division algorithm inF [x] to show that if f(x) 6= 0 is an ideal of smallest degree in anideal I 6= 0, that I = (f(x)).

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3.7. EXERCISES 61

6. Suppose α ∈ C is the root to a monic polynomial f(x) ∈ Z[x],and put R = Z[α]; the collection of all polynomials in Z evaluatedat α.

(a) Show that R is an integral domain that is finitely generatedover Z.

(b) Argue that the quotient field of R is Q[α], and that for anyβ ∈ R, there exists an integer k 6= 0 such that kβ−1 ∈ R.

(c) Using (b), argue that any nonzero ideal of R contains aninteger, and so R/I is finite.

(d) Using (c) conclude that R is a noetherian domain such thatevery non-zero prime ideal is maximal.

7. Let R be a UFD with quotient field Q, and let f(x) ∈ R[x].

(a) Show that there is an element r ∈ R and a polynomial g(x) ∈R[x], such that f(x) = rg(x) and the gcd of the coefficients ofg(x) is 1. (r is called the content of f and we write c(f) = r).

(b) Show that if f(x) = g(x)h(x) with g(x), h(x) ∈ Q[x], thenthere exists polynomials g1(x), h1(x) ∈ R[x] each having con-tent 1 such that ag(x) = g1(x) and bh(x) = h1(x), andabf(x) = g1(x)h1(x), for some nonzero a, b ∈ R.

(c) With the notation of part (b), show that if c(f) = 1, thatab = 1.

(d) Conclude that f(x) ∈ R[x] is irreducible in R[x] if and onlyif f(x) is irreducible in Q[x].

8. Let R be a UFD. Using Problem 3. show that R[x] is a UFD.

9. The intersection of all maximal ideals of R is called the Jacobsonradical of R. Suppose J = ∩M maxM .

(a) Show that a ∈ J if and only if for every r ∈ R, 1 + ra is aunit. Hint: ⇒ ra ∈ J for every r ∈ R, and so 1 + ra mustbe a unit. ⇐ a /∈ M for some M implies M + (a) = R and1− ra ∈ M for some r ∈ R.

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62 CHAPTER 3. COMMUTATIVE RINGS

(b) Deduce a version of Nakayama’s Lemma: If K is a finitelygenerated module, that is, K = Rx1 + Rx2 + · · · + Rxn,such that JK = K, then K = 0. Hint: Assume n is theminimal number of generators required to generate K; writex1 = a1x1 + · · · anxn ∈ JK = K with aj ∈ J for all j. Invert1− a1.

(c) Prove Nakayama’s Lemma: Let B be a finitely generatedmodule, and let A be a submodule such that B = A + JB.Then B = A. Hint: K = B/A is finitely generated andsatisfies K = JK.

10. The Hilbert Basis Theorem asserts that if R is noetherian, thenso is R[x]. Assuming R is a noetherian domain, show that R[x]is noetherian. Hint: If J is an ideal in R[x], let In be the idealof elements that occur as coefficients of xn in some polynomial inJ . Observe that I1 ⊆ I2 ⊆ I3 ⊆ · · · . Use this to pick generatorsfor J .

11. Show that any non-zero element of F = Q[√

d] is a unit.

12. Let R = Z[√

d] = {a + b√

d | a, b ∈ Z} with d a square-freeinteger.

(a) Show that the function

N : R → Z

defined by N(a + b√

d) = a2 − db2, is multiplicative in thatN(αβ) = N(α)N(β) for any α, β ∈ R.

(b) Show that 0 6= a + b√

d ∈ R is a unit of R if and only ifN(a + b

√d) = ±1.

13. This exercise examines the ring R = Z[√

10].

(a) Show that 2, 3, 4 +√

10, 4−√10 are all non-units.

(b) Show that 2, 3, 4 +√

10, 4 − √10 are all irreducible. Hint:Use N as defined in Exercise 3.12.

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3.7. EXERCISES 63

(c) From

6 = 2 · 3 = (4 +√

10)(4−√

10)

deduce that 2 is not a unit multiple of either 4 +√

10, or4−√10.

14. Show that R = Z[√

10] is not a UFD. Hint: use Exercise 13.

15. Regarding R = Z[√−5], show that the ideal I = (2, 1 +

√−5) isnot principal.

16. Determine the number of units in the respective rings:

(a) R = Z[√

10]

(b) R = Z[√

5]

(c) R = Z[√

d] with d < 0, d 6= −1.

(d) R = Z[√−1].

17. Show that if α ∈ R = Z[√

d] with N(α) = ±p, an integral prime,then α is irreducible. Must α be prime in R?

18. Show that if F is a field, then every irreducible element of R =F [x] is prime.

19. Let F be a field and f(x) ∈ F [x] have degree 2 or 3. Show thatf(x) is irreducible if and only if f(x) has no roots in F .

20. For the following fields, find at least 3 irreducibles in F [x].

(a) F = Z(p) for your choice of p.

(b) F = Q.

(c) F = R.

21. Given n and m, express

n = pe11 pe2

2 · · · pekk and m = pf1

1 pf22 · · · pfk

k

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64 CHAPTER 3. COMMUTATIVE RINGS

where ei, fi are non negative integers (but may be zero). Showthat the greatest common divisor (n,m) and least common mul-tiple [n,m] of m,n are given by

(n,m) = ps11 ps2

2 · · · pskk and [n,m] = pt1

1 pt22 · · · ptk

k

where si is the minimum of ei, fi and tj is the maximum of ej, fj.

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Chapter 4

An Introduction to ModuleTheory

4.1 Introduction

Given an integral domain R, an abelian group M , written additively,constitutes a left R-module or a left module over R if there exists afunction on sets

R×M → M ;

with the image of the tuple (r,m) written (generically) as r ·m - and∀ r, s ∈ R, ∀ m,n ∈ M the following properties hold:

r · (s ·m) = (rs) ·m,

r · (n + m) = r · n + r ·m,

(r + s) ·m = r ·m + s ·m,

and1 ·m = m.

In this case the function R×M → M is called the module action.If M is an R-module, a nonempty subset N of M is called a sub-

module of M if the restriction of the module action R × M → M toR × N has image in N (i.e., N is closed with respect to the moduleaction), and N is closed under +.

65

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66 CHAPTER 4. MODULES

The ideals of R are naturally R-modules and are submodules of R.Given a natural number n we can form the direct sum of R, taken ntimes, as

⊕nR = {(r1, r2, . . . , rn) | ri ∈ R, i = 1, 2, . . . , n}.

More generally, given any collection of modules Mi where i belongsto the index set I, we define the direct sum of the Mi’s to be

⊕i∈IMi = {(mi | i ∈ I) | mi ∈ Mi ∀ i ∈ I, and, almost all m′is = 0}.

The notation (mi | i ∈ I) denotes a sequence whose ith entry for i ∈ Iis mi.

The direct sum ⊕i∈IMi of modules is again a module under themodule and group actions

r(mi | i ∈ I) = (rmi | i ∈ I),

and(mi | i ∈ I) + (ni | i ∈ I) = (mi + ni | i ∈ I).

A module M over an integral domain R is called torsion-free if

rm = 0 ⇒ r = 0 or m = 0 ∀ r ∈ R, m ∈ M.

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Chapter 5

An Introduction to GaloisTheory

5.1 Introduction

Perhaps one of the mostly intensely studied problems in mathematicswas the issue of trying to obtain formulas, analogous to the quadraticformula, where one can express the roots of a polynomial in terms ofradicals of the coefficients of the polynomial. Almost from the begin-ning of recorded time, there was some knowledge of obtaining solutionsof quadratic equations using completion of the square and the quadraticformula. The ancient Greeks were interested in

67

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68 CHAPTER 5. FIELDS

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Bibliography

[1] Arnold, David M.; Finite Rank Torsion-Free Abelian Groupsand Rings, LNM 931, Springer-Verlag, 1982.

[2] Reinhold Baer, Abelian groups without elements of finite order,Duke Math. J. 3 (1934), 68-122.

[3] M. C. R. Butler, A class of torsion-free abelain groups of finiterank, Proc. London Math. Soc. (3) 15 (1965), 680-698.

[4] M. C. R. Butler, Torsion-free modules and diagrams of vectorspaces, Proc. London Math. Soc. (3) 18 (1968), 635-652.

[5] Laszlo Fuchs; Infinite Abelian Groups, Volume I, AcademicPress 1970.

[6] Laszlo Fuchs; Infinite Abelian Groups, Volume II, AcademicPress 1973.

[7] H. Pat Goeters; On number rings and a problem of P. Hill,Comm. in Alg. 23 (10), (1995), 3677-3683.

[8] Daniel Marcus; Number Fields, Springer-Verlag 1977.

[9] Eben Matlis; Torsion-Free Modules, University of ChicagoPress, 1972.

[10] Eben Matlis; 1-Dimensional Cohen-Macaulay Rings, LNM 327,Springer-Verlag, 1973.

[11] Matsumura, H.; Commutative Ring Theory, Cambridge Univer-sity Press, 1980.

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70 BIBLIOGRAPHY

[12] Rotman, Joseph J.; An Introduction to Homological Algebra,Academic Press, 1979.