7.38. Find the number of elements in the semigroup of the machine, given by

Figure 7.3, that controls the elevator.

7.39. Find the monoid of the machine in Figure 7.10.

a, b b

s2

s1 a

g

g

g

s3

a, b

Figure 7.10

7.40. A serial adder, illustrated in Figure 7.11, is a machine that adds two

numbers in binary form. The two numbers are fed in together, one digit

at a time, starting from the right end. Their sum appears as the output.

The machine has input symbols 00, 01, 10, and 11, corresponding to the

rightmost digits of the numbers. Figure 7.12 gives the state diagram of

such a machine, where the symbol “sij /j ” indicates that the machine is in

state sij and emits an output j . The carry digit is the number i of the state

sij . Find the monoid of this machine.

EXERCISES 153

Serial

adder

Figure 7.11

10, 01

00

00

s00 0 s 01 1

10, 01

11

11

00

11

10, 01 00

10, 01 s

s10 0 11 1

11

Figure 7.12. State diagram of the serial adder.

The circuits in Exercises 7.41 to 7.44 represent the internal structures of some

¬nite-state machines constructed from transistor circuits. These circuits are con-

trolled by a clock, and the rectangular boxes denote delays of one time unit. The

input symbols are 0 and 1 and are fed in at unit time intervals. The internal

states of the machines are described by the contents of the delays. Draw the state

diagram and ¬nd the elements in the semigroup of each machine.

7.41.

Delay

AND y

7.42.

Delay

NAND

y1

Delay

OR y2

7.43.

Delay Delay

AND

y1 y2

7.44.

Delay Delay

NOR y1 y2

154 7 MONOIDS AND MACHINES

7.45. In the spring, a plant bud has to have the right conditions in order to

develop. One particular bud has to have a rainy day followed by two

warm days, without being interrupted by cool or freezing days, in order to

develop. Furthermore, if a freezing day occurs after the bud has developed,

the bud dies. Draw a state diagram for such a bud using the input symbols

R, W , C, F to stand for rainy, warm, cool, and freezing days, respectively.

What is the number of elements in the resulting monoid of this bud?

7.46. A dog can either be passive, angry, frightened, or angry and frightened,

in which case he bites. If you give him a bone, he becomes passive. If

you remove one of his bones, he becomes angry, and, if he is already

frightened, he will bite you. If you threaten him, he becomes frightened,

but, if he is already angry, he will bite. Write out the table of the monoid

of the dog.

8

RINGS AND FIELDS

The familiar number systems of the real or complex numbers contain two basic

binary operations, addition and multiplication. Group theory is not suf¬cient to

capture all of the algebraic structure of these number systems, because a group

deals with only one binary operation. It is possible to consider the integers as a

group (Z, +) and the nonzero integers as a monoid (Z— , ·), but this still neglects

the relation between addition and multiplication, namely, the fact that multiplica-

tion is distributive over addition. We therefore consider algebraic structures with

two binary operations modeled after these number systems. A ring is a structure

that has the minimal properties we would expect of addition and multiplication.

A ¬eld is a more specialized ring in which division by nonzero elements is

always possible.

In this chapter we look at the basic properties of rings and ¬elds and con-

sider many examples. In later chapters we construct new number systems with

properties similar to the familiar systems.

RINGS

A ring (R, +, ·) is a set R, together with two binary operations + and · on R

satisfying the following axioms. For any elements a, b, c ∈ R,

(i) (a + b) + c = a + (b + c). (associativity of addition)

(ii) a + b = b + a. (commutativity of addition)

(iii) there exists 0 ∈ R, called the (existence of an additive identity)

zero, such that a + 0 = a.

(iv) there exists (’a) ∈ R such that (existence of an additive inverse)

a + (’a) = 0.

(v) (a · b) · c = a · (b · c). (associativity of multiplication)

(vi) there exists 1 ∈ R such that (existence of multiplicative identity)

1 · a = a · 1 = a.

Modern Algebra with Applications, Second Edition, by William J. Gilbert and W. Keith Nicholson

ISBN 0-471-41451-4 Copyright ™ 2004 John Wiley & Sons, Inc.

155

156 8 RINGS AND FIELDS

(vii) a · (b + c) = a · b + a · c and (distributivity)

(b + c) · a = b · a + c · a.

Axioms (i)“(iv) are equivalent to saying that (R, +) is an abelian group, and

axioms (v) and (vi) are equivalent to saying that (R, ·) is a monoid.

The ring (R, +, ·) is called a commutative ring if, in addition,

(viii) a · b = b · a for all a, b ∈ R. (commutativity of multiplication)

The integers under addition and multiplication satisfy all of the axioms above,

so that (Z, +, ·) is a commutative ring. Also, (Q, +, ·), (R, +, ·), and (C, +, ·)

are all commutative rings. If there is no confusion about the operations, we write

only R for the ring (R, +, ·). Therefore, the rings above would be referred to as

Z, Q, R, or C. Moreover, if we refer to a ring R without explicitly de¬ning its

operations, it can be assumed that they are addition and multiplication.

Many authors do not require a ring to have a multiplicative identity, and

most of the results we prove can be veri¬ed to hold for these objects as well.

Exercise 8.49 shows that such an object can always be embedded in a ring that

does have a multiplicative identity.

The set of all n — n square matrices with real coef¬cients forms a ring

(Mn (R), +, ·), which is not commutative if n > 1, because matrix multiplication

is not commutative.

The elements “even” and “odd” form a commutative ring ({even, odd}, +, ·)

where the operations are given by Table 8.1. “Even” is the zero of this ring, and

“odd” is the multiplicative identity. This is really a special case of the following

example when n = 2.

Example 8.1. Show that (Zn , +, ·) is a commutative ring, where addition and

multiplication on congruence classes, modulo n, are de¬ned by the equations

[x] + [y] = [x + y] and [x] · [y] = [xy].

Solution. It follows from Example 4.19, that (Zn , +) is an abelian group.

Since multiplication on congruence classes is de¬ned in terms of represen-

tatives, it must be veri¬ed that it is well de¬ned. Suppose that [x] = [x ] and

[y] = [y ], so that x ≡ x and y ≡ y mod n. This implies that x = x + kn

and y = y + ln for some k, l ∈ Z. Now x · y = (x + kn) · (y + ln) = x · y +

(ky + lx + kln)n, so x · y ≡ x · y mod n and hence [x · y] = [x · y ]. This

shows that multiplication is well de¬ned.

TABLE 8.1. Ring of Odd and Even Integers

+ ·

Even Odd Even Odd

Even even odd Even even even

Odd odd even Odd even odd

RINGS 157

The remaining axioms now follow from the de¬nitions of addition and mul-

tiplication and from the properties of the integers. The zero is [0], and the unit

is [1]. The left distributive law is true, for example, because

[x] · ([y] + [z]) = [x] · [y + z] = [x · (y + z)]

= [x · y + x · z] by distributivity in Z

= [x · y] + [x · z] = [x] · [y] + [x] · [z].

Example 8.2. Construct the addition and multiplication tables for the ring

(Z5 , +, ·).

Solution. We denote the congruence class [x] just by x. The tables are given

in Table 8.2.

√ √

Example 8.3. Show that (Q( 2), +, ·) is a commutative ring where Q( 2) =

√

{a + b 2 ∈ R|a, b ∈ Q}.

√

Solution. The set Q( 2) is a subset of R, and the addition and multiplication

is the same as that of real numbers. First, we check that + and · are binary

√

operations on Q( 2). If a, b, c, d ∈ Q, we have

√ √ √ √

(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 ∈ Q( 2)

since (a + c) and (b + d) ∈ Q. Also,

√ √ √ √

(a + b 2) · (c + d 2) = (ac + 2bd) + (ad + bc) 2 ∈ Q( 2)

since (ac + 2bd) and (ad + bc) ∈ Q. √

We now check that axioms (i)“(viii) of a commutative ring are valid in Q( 2).

(i) Addition of real numbers is associative.

(ii) Addition of real numbers is commutative.

√ √

The zero is 0 = 0 + 0 2 ∈ Q( 2).

(iii)

√ √ √

The additive inverse of a + b 2 is (’a) + (’b) 2 ∈ Q( 2), since

(iv)

(’a) and (’b) ∈ Q.

TABLE 8.2. Ring (Z5 , +, ·)

+ ·

0 1 2 3 4 0 1 2 3 4

0 0 1 2 3 4 0 0 0 0 0 0

1 1 2 3 4 0 1 0 1 2 3 4

2 2 3 4 0 1 2 0 2 4 1 3

3 3 4 0 1 2 3 0 3 1 4 2

4 4 0 1 2 3 4 0 4 3 2 1

158 8 RINGS AND FIELDS

(v) Multiplication of real numbers is associative.

√ √

(vi) The multiplicative identity is 1 = 1 + 0 2 ∈ Q( 2).

(vii) The distributive axioms hold for real numbers and hence hold for ele-

√

ments of Q( 2).

(viii) Multiplication of real numbers is commutative.

We have already investigated one algebraic system with two binary operations:

a boolean algebra. The boolean algebra of subsets of a set is not a ring under

the operations of union and intersection, because neither of these operations has

inverses. However, the symmetric difference does have an inverse, and we can

make a boolean algebra into a ring using this operation and the operation of

intersection.

, ©) is a commutative ring for any set X.

Example 8.4. (P (X),

Solution. The axioms (i)“(viii) of a commutative ring follow from Proposi-

tions 2.1 and 2.3. The zero is ˜, and the identity is X.

In the ring above, A © A = A for every element A in the ring. Such rings

are called boolean rings, since they are all derivable from boolean algebras (see

Exercise 8.13).

, ©), where X =

Example 8.5. Construct the tables for the ring (P (X),

{a, b, c}.

Solution. Let A = {a}, B = {b}, and C = {c}, so that A = {b, c}, B = {a, c},

and C = {a, b}. Therefore, P (X) = {˜, A, B, C, A, B, C, X}. The tables for the

symmetric difference and intersection are given in Table 8.3.

The following properties are useful in manipulating elements of any ring.

Proposition 8.6. If (R, +, ·) is a ring, then for all a, b ∈ R:

a · 0 = 0 · a = 0.

(i)

a · (’b) = (’a) · b = ’(a · b).

(ii)

(’a) · (’b) = a · b.

(iii)

(’1) · a = ’a.

(iv)

(’1) · (’1) = 1.

(v)

Proof. (i) By distributivity, a · 0 = a · (0 + 0) = a · 0 + a · 0. Adding

’(a · 0) to each side, we obtain 0 = a · 0. Similarly, 0 · a = 0.

(ii) Compute a · (’b) + a · b = a · (’b + b) = a · 0 = 0, using (i). Therefore,

a · (’b) = ’(a · b). Similarly, (’a) · b = ’(a · b).

(iii) We have (’a) · (’b) = ’(a · (’b)) = ’(’(a · b)) = a · b by (ii) and

Proposition 3.7.

INTEGRAL DOMAINS AND FIELDS 159

TABLE 8.3. Ring P ({a, b, c})

˜ A B C A B C X

˜ ˜ A B C A B C X

˜

A A C B X C B A

˜

B B C A C X A B

˜

C C B A B A X C

˜

A A X C B C B A

˜

B B C X A C A B

˜

C C B A X B A C

˜

X X A B C A B C

© ˜ A B C A B C X

˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜

˜ ˜ ˜ ˜

A A A A A

˜ ˜ ˜ ˜

B B B B B

˜ ˜ ˜ ˜

C C C C C

˜ ˜

A B C A C B A

˜ ˜

B A C C B A B

˜ ˜

C A B B A C C

˜

X A B C A B C X

(iv) By (ii), (’1) · a = ’(1 · a) = ’a.

(v) By (iii), (’1) · (’1) = 1 · 1 = 1.

Proposition 8.7. If 0 = 1, the ring contains only one element and is called the

trivial ring. All other rings are called nontrivial.

Proof. For any element, a, in a ring in which 0 = 1, we have a = a · 1 =

a · 0 = 0. Therefore, the ring contains only the element 0. It can be veri¬ed that

this forms a ring with the operations de¬ned by 0 + 0 = 0 and 0 · 0 = 0.

INTEGRAL DOMAINS AND FIELDS

One very useful property of the familiar number systems is the fact that if ab = 0,

then either a = 0 or b = 0. This property allows us to cancel nonzero elements

because if ab = ac and a = 0, then a(b ’ c) = 0, so b = c. However, this prop-

erty does not hold for all rings. For example, in Z4 , we have [2] · [2] = [0], and

we cannot always cancel since [2] · [1] = [2] · [3], but [1] = [3].

If (R, +, ·) is a commutative ring, a nonzero element a ∈ R is called a zero

divisor if there exists a nonzero element b ∈ R such that a · b = 0. A nontrivial

commutative ring is called an integral domain if it has no zero divisors. Hence

160 8 RINGS AND FIELDS

a nontrivial commutative ring is an integral domain if a · b = 0 always implies

that a = 0 or b = 0.

As the name implies, the integers form an integral domain. Also, Q, R, and C

are integral domains. However, Z4 is not, because [2] is a zero divisor. Neither

is (P (X), , ©), because every nonempty proper subset of X is a zero divisor.

2

01

= 0).

Mn (R) is not an integral domain (for example,

00

Proposition 8.8. If a is a nonzero element of an integral domain R and

a · b = a · c, then b = c.

Proof. If a · b = a · c, then a · (b ’ c) = a · b ’ a · c = 0. Since R is an inte-

gral domain, it has no zero divisors. Since a = 0, it follows that (b ’ c) = 0.

Hence b = c.

Generally speaking, it is possible to add, subtract, and multiply elements in a

ring, but it is not always possible to divide. Even in an integral domain, where

elements can be canceled, it is not always possible to divide by nonzero elements.

For example, if x, y ∈ Z, then 2x = 2y implies that x = y, but not all elements

in Z can be divided by 2.

The most useful number systems are those in which we can divide by nonzero

elements. A ¬eld is a ring in which the nonzero elements form an abelian group

under multiplication. In other words, a ¬eld is a nontrivial commutative ring R

satisfying the following extra axiom.

(ix) For each nonzero element a ∈ R there exists a ’1 ∈ R such that a · a ’1 = 1.

The rings Q, R, and C are all ¬elds, but the integers do not form a ¬eld.

Proposition 8.9. Every ¬eld is an integral domain; that is, it has no zero divisors.

Proof. Let a · b = 0 in a ¬eld F . If a = 0, there exists an inverse a ’1 ∈ F

and b = (a ’1 · a) · b = a ’1 (a · b) = a ’1 · 0 = 0. Hence either a = 0 or b = 0,

and F is an integral domain.

Theorem 8.10. A ¬nite integral domain is a ¬eld.

Proof. Let D = {x0 , x1 , x2 , . . . , xn } be a ¬nite integral domain with x0 as 0

and x1 as 1. We have to show that every nonzero element of D has a multiplica-

tive inverse.

If xi is nonzero, we show that the set xi D = {xi x0 , xi x1 , xi x2 , . . . , xi xn } is the

same as the set D. If xi xj = xi xk , then, by the cancellation property, xj = xk . Hence

all the elements xi x0 , xi x1 , xi x2 , . . . , xi xn are distinct, and xi D is a subset of D with

the same number of elements. Therefore, xi D = D. But then there is some element,

xj , such that xi xj = x1 = 1. Hence xj = xi’1 , and D is a ¬eld.

SUBRINGS AND MORPHISMS OF RINGS 161

Note that Z is an in¬nite integral domain that is not a ¬eld.

Theorem 8.11. Zn is a ¬eld if and only if n is prime.

Proof. Suppose that n is prime and that [a] · [b] = [0] in Zn . Then n|ab. So

n|a or n|b by Euclid™s Lemma (Theorem 12, Appendix 2). Hence [a] = [0] or

[b] = [0], and Zn is an integral domain. Since Zn is also ¬nite, it follows from

Theorem 8.10 that Zn is a ¬eld.

Suppose that n is not prime. Then we can write n = rs, where r and s are

integers such that 1 < r < n and 1 < s < n. Now [r] = [0] and [s] = [0] but

[r] · [s] = [rs] = [0]. Therefore, Zn has zero divisors and hence is not a ¬eld.

√

Example 8.12. Is (Q( 2), +, ·) an integral domain or a ¬eld?

√

Solution. From Example 8.3 we know that Q( 2) is a commutative ring. Let

√

a + b√2 be a nonzero element, so that at least one of a and b is not zero. Hence

√

a ’ b 2 = 0 (because 2 is not in Q), so we have

√

√

a’b 2

1 a b

√= √ =2 ’

√ 2.

a ’ 2b2 a 2 ’ 2b2

a+b 2 (a + b 2)(a ’ b 2)

√ √ √

This is an element of Q( 2), and so is the inverse of a + b 2. Hence Q( 2)

is a ¬eld (and an integral domain).

SUBRINGS AND MORPHISMS OF RINGS

If (R, +, ·) is a ring, a nonempty subset S of R is called a subring of R if for

all a, b ∈ S:

a + b ∈ S.

(i)

’a ∈ S.

(ii)

a · b ∈ S.

(iii)

1 ∈ S.

(iv)

Conditions (i) and (ii) imply that (S, +) is a subgroup of (R, +) and can be

replaced by the condition a ’ b ∈ S.

Proposition 8.13. If S is a subring of (R, +, ·), then (S, +, ·) is a ring.

Proof. Conditions (i) and (iii) of the de¬nition above guarantee that S is closed

under addition and multiplication. Condition (iv) shows that 1 ∈ S. It follows

from Proposition 3.8 that (S, +) is a group. (S, +, ·) satis¬es the remaining

axioms for a ring because they hold in (R, +, ·).

162 8 RINGS AND FIELDS

For example, Z, Q, and R are all subrings of C. Let D be the set of n — n

real diagonal matrices. Then D is a subring of the ring of all n — n real matri-

ces, Mn (R), because the sum, difference, and product of two diagonal matrices

is another diagonal matrix. Note that D is commutative even though Mn (R)

is not.

√ √

Example 8.14. Show that Q( 2) = {a + b 2|a, b ∈ Q} is a subring of R.

√ √ √

Solution. Let a + b 2, c + d 2 ∈ Q( 2). Then

√ √ √ √

(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 ∈ Q( 2).

(i)

√ √ √

’(a + b 2) = (’a) + (’b) 2 ∈ Q( 2).

(ii)

√ √ √ √

(a + b 2) · (c + d 2) = (ac + 2bd) + (ad + bc) 2 ∈ Q( 2).

(iii)

√ √

1 = 1 + 0 2 ∈ Q( 2).

(iv)

A morphism between two rings is a function between their underlying sets

that preserves the two operations of addition and multiplication and also the

element 1. Many authors use the term homomorphism instead of morphism.

More precisely, let (R, +, ·) and (S, +, ·) be two rings. The function f : R ’ S

is called a ring morphism if for all a, b ∈ R:

(i) f (a + b) = f (a) + f (b).

(ii) f (a · b) = f (a) · f (b).

(iii) f (1) = 1.

If the operations in the two rings are denoted by different symbols, for

example, if the rings are (R, +, ·) and (S, •, Ž), then the conditions for

f : R ’ S to be a ring morphism are:

(i) f (a + b) = f (a) • f (b).

(ii) f (a · b) = f (a)Žf (b).

(iii) f (1R ) = 1S where 1R and 1S are the respective identities.

A ring isomorphism is a bijective ring morphism. If there is an isomorphism

between the rings R and S, we say R and S are isomorphic rings and write

R ∼ S.

=

A ring morphism, f , from (R, +, ·) to (S, +, ·) is, in particular, a group

morphism from (R, +) to (S, +). Therefore, by Proposition 3.19, f (0) = 0 and

f (’a) = ’f (a) for all a ∈ R.

The inclusion function, i: S ’ R, of any subring S into a ring R is always

a ring morphism. The function f : Z ’ Zn , de¬ned by f (x) = [x], which maps

an integer to its equivalence class modulo n, is a ring morphism from (Z, +, ·)

to (Zn , +, ·).

SUBRINGS AND MORPHISMS OF RINGS 163

Example 8.15. If X is a one element set, show that f : P (X) ’ Z2 is a

ring isomorphism between (P (X), , ©) and (Z2 , +, ·), where f (˜) = [0] and

f (X) = [1].

Solution. We can check that f is a morphism by testing all the possibilities

for f (A B) and f (A © B). Since the rings are commutative, they are

f (˜ ˜) = f (˜) = [0] = f (˜) + f (˜)

f (˜ X) = f (X) = [1] = f (˜) + f (X)

f (X X) = f (˜) = [0] = f (X) + f (X)

f (˜ © ˜) = f (˜) = [0] = f (˜) · f (˜)

f (˜ © X) = f (˜) = [0] = f (˜) · f (X)

f (X © X) = f (X) = [1] = f (X) · f (X).

Both rings contain only two elements, and f is a bijection; therefore, f is an

isomorphism.

If f : R ’ S is an isomorphism between two ¬nite rings, the addition and

multiplication tables of S will be the same as those of R if we replace each

a ∈ R by f (a) ∈ S. For example, Tables 8.4 and 8.5 illustrate the isomorphism

of Example 8.15.

The following ring isomorphism between linear transformations and matrices

is the crux of much of linear algebra.

Example 8.16. The linear transformations from Rn to itself form a ring,

(L(Rn , Rn ), +, Ž ), under addition and composition. Show that the function

f : L(Rn , Rn ) ’ Mn (R)

TABLE 8.4. Ring P (X )

When X Is a Point

©

˜ ˜

X X

˜ ˜ ˜ ˜ ˜

X

˜ ˜

X X X X

TABLE 8.5. Ring Z2

+ ·

[0] [1] [0] [1]

[0] [0] [1] [0] [0] [0]

[1] [1] [0] [1] [0] [1]

164 8 RINGS AND FIELDS

is a ring morphism, where f assigns to each linear transformation its standard

matrix, that is, its n — n coef¬cient matrix with respect to the standard basis

of Rn .

Solution. If ± is a linear transformation from Rn to itself, then

®® ®

a11 x1 + · · · + a1n xn

x1 a11 . . . a1n

. . .

±° . » = °. .

» and f (±) = ° . .»

. . .

. . . . .

an1 x1 + · · · + ann xn

xn an1 . . . ann

Matrix addition and multiplication is de¬ned so that f (± + β) = f (±) + f (β)

and f (± Ž β) = f (±) · f (β). Also, if ι is the identity linear transformation then

f (ι) is the identity matrix.

Any matrix de¬nes a linear transformation, so that f is surjective. Fur-

thermore, f is injective, because any matrix can arise from only one linear

transformation. In fact, the j th column of the matrix must be the image of the

j th basis vector. Hence f is an isomorphism.

Example 8.17. Show that f : Z24 ’ Z4 , de¬ned by f ([x]24 ) = [x]4 is a ring

morphism.

Proof. Since the function is de¬ned in terms of representatives of equiva-

lence classes, we ¬rst check that it is well de¬ned. If [x]24 = [y]24 ,

then x ≡ y mod 24 and 24|(x ’ y). Hence 4|(x ’ y) and [x]4 = [y]4 , which

shows that f is well de¬ned.

We now check the conditions for f to be a ring morphism.

(i) f ([x]24 + [y]24 ) = f ([x + y]24 ) = [x + y]4 = [x]4 + [y]4 .

(ii) f ([x]24 · [y]24 ) = f ([xy]24 ) = [xy]4 = [x]4 · [y]4 .

(iii) f ([1]24 ) = [1]4 .

NEW RINGS FROM OLD

This section introduces various methods for constructing new rings from given

rings. These include the direct product of rings, matrix rings, polynomial rings,

rings of sequences, and rings of formal power series. Perhaps the most impor-

tant class of rings constructible from given rings is the class of quotient rings.

Their construction is analogous to that of quotient groups and is discussed in

Chapter 10.

If (R, +, ·) and (S, +, ·) are two rings, their product is the ring (R — S, +, ·)

whose underlying set is the cartesian product of R and S and whose operations

are de¬ned component-wise by

(r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 ) and (r1 , s1 ) · (r2 , s2 ) = (r1 · r2 , s1 · s2 ).

NEW RINGS FROM OLD 165

It is readily veri¬ed that these operations do indeed de¬ne a ring structure on

R — S whose zero is (0R , 0S ), where 0R and 0S are the zeros of R and S, and

whose multiplicative identity is (1R , 1S ), where 1R and 1S are the identities in R

and S.

The product construction can be iterated any number of times. For example,

(R , +, ·) is a commutative ring, where Rn is the n-fold cartesian product of R

n

with itself.

Example 8.18. Write down the addition and multiplication tables for Z2 — Z3 .

Solution. Let Z2 = {0, 1} and Z3 = {0, 1, 2}. Then Z2 — Z3 = {(0, 0), (0, 1),

(0, 2), (1, 0), (1, 1), (1, 2)}. The addition and multiplication tables are given in

Table 8.6. In calculating these, it must be remembered that addition and mul-

tiplication are performed modulo 2 in the ¬rst coordinate and modulo 3 in the

second coordinate.

We know that Z2 — Z3 and Z6 are isomorphic as groups; we now show that

they are isomorphic as rings.

Theorem 8.19. Zm — Zn is isomorphic as a ring to Zmn if and only if

gcd(m, n) = 1.

Proof. If gcd(m, n) = 1, it follows from Theorems 4.32 and 3.20 that the

function

f : Zmn ’ Zm — Zn

TABLE 8.6. Ring Z2 — Z3

+ (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2)

(0, 0) (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2)

(0, 1) (0, 1) (0, 2) (0, 0) (1, 1) (1, 2) (1, 0)

(0, 2) (0, 2) (0, 0) (0, 1) (1, 2) (1, 0) (1, 1)

(1, 0) (1, 0) (1, 1) (1, 2) (0, 0) (0, 1) (0, 2)

(1, 1) (1, 1) (1, 2) (1, 0) (0, 1) (0, 2) (0, 0)

(1, 2) (1, 2) (1, 0) (1, 1) (0, 2) (0, 0) (0, 1)

· (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2)

(0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0)

(0, 1) (0, 0) (0, 1) (0, 2) (0, 0) (0, 1) (0, 2)

(0, 2) (0, 0) (0, 2) (0, 1) (0, 0) (0, 2) (0, 1)

(1, 0) (0, 0) (0, 0) (0, 0) (1, 0) (1, 0) (1, 0)

(1, 1) (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2)

(1, 2) (0, 0) (0, 2) (0, 1) (1, 0) (1, 2) (1, 1)

166 8 RINGS AND FIELDS

de¬ned by f ([x]mn ) = ([x]m , [x]n ) is a group isomorphism. However, this func-

tion also preserves multiplication because

f ([x]mn · [y]mn ) = f ([xy]mn ) = ([xy]m , [xy]n ) = ([x]m [y]m , [x]n [y]n )

= ([x]m , [x]n ) · ([y]m , [y]n ) = f ([x]mn ) · f ([y]mn ).

Also, f ([1]mn ) = ([1]m , [1]n ); thus f is a ring isomorphism.

It was shown in the discussion following Corollary 4.33 that if gcd(m, n) =

1, Zm — Zn and Zmn are not isomorphic as groups, and hence they cannot be

isomorphic as rings.

We can extend this result by induction to show the following.

Theorem 8.20. Let m = m1 · m2 · · · mr , where gcd(mi , mj ) = 1 if i = j . Then

Zm1 — Zm2 — · · · — Zmr is a ring isomorphic to Zm .

Corollary 8.21. Let n = p1 1 p2 2 · · · pr r be a decomposition of the integer n into

±± ±

powers of distinct primes. Then Zn ∼ Zp±1 — Zp±2 — · · · — Zpr±r as rings.

= 1 2

If R is a commutative ring, we can construct the ring of n — n matrices with

entries from R, (Mn (R), +, ·). Addition and multiplication are performed as in

real matrices.

For example, (Mn (Z2 ), +, ·) is the ring of n — n matrices with 0 and 1 entries.

Addition and multiplication is performed modulo 2. This is a noncommutative

2

ring with 2(n ) elements.

If R is a commutative ring, a polynomial p(x) in the indeterminate x over

the ring R is an expression of the form

p(x) = a0 + a1 x + a2 x 2 + · · · + an x n ,

where a0 , a1 , a2 , . . . , an ∈ R and n ∈ N. The element ai is called the coef¬cient

of x i in p(x). If the coef¬cient of x i is zero, the term 0x i may be omitted, and

if the coef¬cient of x i is one, 1x i may be written simply as x i .

Two polynomials f (x) and g(x) are called equal when they are identical, that

is, when the coef¬cient of x n is the same in each polynomial for every n 0.

In particular,

a0 + a1 x + a2 x 2 + · · · + an x n = 0

is the zero polynomial if and only if a0 = a1 = a2 = · · · = an = 0.

If n is the largest integer for which an = 0, we say that p(x) has degree n and

write degp(x) = n. If all the coef¬cients of p(x) are zero, then p(x) is called

the zero polynomial, and its degree is not de¬ned.

√

For example, 4x 2 ’ 3 is a polynomial over R of degree 2, ix 4 ’ (2 + i)x 3 +

3x is a polynomial over C of degree 4, and x 7 + x 5 + x 4 + 1 is a polynomial

over Z2 of degree 7. The number 5 is a polynomial over Z of degree 0; the zero

NEW RINGS FROM OLD 167

polynomial and the polynomials of degree 0 are called constant polynomials

because they contain no x terms.

The set of all polynomials in x with coef¬cients from the commutative ring

R is denoted by R[x]. That is,

R[x] = {a0 + a1 x + a2 x 2 + · · · + an x n |ai ∈ R, n ∈ N}.

This forms a ring (R[x], +, ·) called the polynomial ring with coef¬cients from

R when addition and multiplication of the polynomials

n m

p(x) = and q(x) =

i

bi x i

ai x

i=0 i=0

are de¬ned by

max(m,n)

p(x) + q(x) = (ai + bi )x i

i=0

and

m+n

p(x) · q(x) = where ck =

ck x k ai bj .

i+j =k

k=0

With a little effort, it can be veri¬ed that (R[x], +, ·) satis¬es all the axioms

for a commutative ring. The zero is the zero polynomial, and the multiplicative

identity is the constant polynomial 1.

For example, in Z5 [x], the polynomial ring with coef¬cients in the integers

modulo 5, we have

(2x 3 + 2x 2 + 1) + (3x 2 + 4x + 1) = 2x 3 + 4x + 2

and

(2x 3 + 2x 2 + 1) · (3x 2 + 4x + 1) = x 5 + 4x 4 + 4x + 1.

When working in Zn [x], the coef¬cients, but not the exponents, are reduced

modulo n.

Proposition 8.22. If R is an integral domain and p(x) and q(x) are nonzero

polynomials in R[x], then

deg(p(x) · q(x)) = deg p(x) + deg q(x).

Proof. Let deg p(x) = n, deg q(x) = m and let p(x) = a0 + · · · + an x n ,

q(x) = b0 + · · · + bm x m , where an = 0, bm = 0. Then the coef¬cient of the high-

est power of x in p(x) · q(x) is an bm , which is nonzero since R has no zero

divisors. Hence deg(p(x) · q(x)) = m + n.

168 8 RINGS AND FIELDS

If the coef¬cient ring is not an integral domain, the degree of a product may be

less than the sum of the degrees. For example, (2x 3 + x) · (3x) = 3x 2 in Z6 [x].

Corollary 8.23. If R is an integral domain, so is R[x].

Proof. If p(x) and q(x) are nonzero elements of R[x], then p(x) · q(x) is

also nonzero by Proposition 8.22. Hence R[x] has no zero divisors.

The construction of a polynomial ring can be iterated to obtain the ring of

polynomials in n variables x1 , . . . , xn , with coef¬cients from R. We de¬ne induc-

tively R[x1 , . . . , xn ] = R[x1 , . . . , xn’1 ][xn ]. For example, consider a polynomial

f in R[x, y] = R[x][y], say

f = f0 + f1 y + f2 y 2 + · · · + fn y n ,

where each fi = fi (x) is in R[x]. If we write fi = a0i + a1i x + a2i x 2 · · · for

each i, then

f (x, y) = a00 + a10 x + a01 y + a20 x 2 + a11 xy + a02 y 2 + · · · .

Clearly, we can prove by induction from Corollary 8.22 that R[x1 , . . . , xn ] is

an integral domain if R is an integral domain.

Proposition 8.24. Let R be a commutative ring and denote the in¬nite sequence

of elements of R, a0 , a1 , a2 , . . . , by ai . De¬ne addition, +, and convolution,

—, of two such sequences by

ai + bi = ai + bi

and

ai — bi = aj bk = a0 bi + a1 bi’1 + · · · + ai b0 .

j +k=i

The set of all such sequences forms a commutative ring (R N , +, —) called the

ring of sequences in R. If R is an integral domain, so is R N .

Proof. Addition is clearly associative and commutative. The zero element is

the zero sequence 0 = 0, 0, 0, . . . , and the negative of ai is ’ai . Now

( ai — bi ) — ci = aj bk — ci

j +k=i

«

aj bk cl =

= aj bk cl .

j +k=m j +k+l=i

l+m=i

NEW RINGS FROM OLD 169

Similarly, bracketing the sequences in the other way, we obtain the same result,

which shows that convolution is associative.

Convolution is clearly commutative and the distributive laws hold because

ai — ( bi + ci ) = aj (bk + ck )

j +k=i

= aj bk + aj ck = ai — bi + ai — ci .

j +k=i j +k=i

The identity in the ring of sequences is 1, 0, 0, . . . because

1, 0, 0, . . . — a0 , a1 , a2 , . . . = 1a0 , 1a1 + 0a0 , 1a2 + 0a1 + 0a0 , . . .

= a0 , a1 , a2 , . . . .

Therefore, (R N , +, —) is a commutative ring.

Suppose that aq and br are the ¬rst nonzero elements in the nonzero sequences

ai and bi , respectively. Then the element in the (q + r)th position of their

convolution is

aj bk = a0 bq+r + a1 bq+r’1 + · · · + aq br + aq+1 br’1 + · · · + aq+r b

j +k=q+r

=0 + + · · · + aq br + + · · · + 0 = aq br .

0 0

Hence, if R is an integral domain, this element is not zero and the ring of

sequences has no zero divisors.

The ring of sequences cannot be a ¬eld because 0, 1, 0, 0, . . . has no inverse.

In fact, for any sequence bi , 0, 1, 0, 0, . . . — b0 , b1 , b2 , . . . = 0, b0 , b1 , . . . ,

which can never be the identity in the ring.

A formal power series in x with coef¬cients from a commutative ring R is

an expression of the form.

∞

a0 + a1 x + a2 x + · · · = where ai ∈ R.

2

ai x i

i=0

In contrast to a polynomial, these power series can have an in¬nite number of

nonzero terms.

We denote the set of all such formal power series by R[[x]]. The term formal is

used to indicate that questions of convergence of these series are not considered.

Indeed, over many rings, such as Zn , convergence would not be meaningful.

Motivated by RN , addition and multiplication are de¬ned in R[[x]] by

∞ ∞ ∞

+ = (ai + bi )x i

i i

ai x bi x

i=0 i=0 i=0

170 8 RINGS AND FIELDS

«

and

∞ ∞ ∞

aj bk x i .

· =

ai x i bi x i

j +k=i

i=0 i=0 i=0

It can be veri¬ed that these formal power series do form a ring, (R[[x]], +, ·),

and that the polynomial ring, R[x], is the subring consisting of those power

series with only a ¬nite number of nonzero terms. In fact, the ring of sequences

(R N , +, —) is isomorphic to the ring of formal power series (R[[x]], +, ·). The

function f : R N ’ R[[x]] that is de¬ned by f ( a0 , a1 , a2 , · · · ) = a0 + a1 x +

a2 x2 + · · · is clearly a bijection. It follows from the de¬nitions of addition, mul-

tiplication, and convolution in these rings, that f is a ring morphism.

FIELD OF FRACTIONS

We can always add, subtract, and multiply elements in any ring, but we cannot

always divide. However, if the ring is an integral domain, it is possible to enlarge

it so that division by nonzero elements is possible. In other words, we can con-

struct a ¬eld containing the given ring as a subring. This is precisely what we did

following Example 4.2 when constructing the rational numbers from the integers.

If the original ring did have zero divisors or was noncommutative, it could

not possibly be a subring of any ¬eld, because ¬elds cannot contain zero divisors

or pairs of noncommutative elements.

Theorem 8.25. If R is an integral domain, it is possible to construct a ¬eld Q,

so that the following hold:

(i) R is isomorphic to a subring, R , of Q.

(ii) Every element of Q can be written as p · q ’1 for suitable p, q ∈ R .

Q is called the ¬eld of fractions of R (or sometimes the ¬eld of quotients of R).

Proof. Consider the set R — R — = {(a, b)|a, b ∈ R, b = 0}, consisting of pairs

of elements of R, the second being nonzero. Motivated by the fact that a = d c

b

in Q if and only if ad = bc, we de¬ne a relation ∼ on R — R — by

(a, b) ∼ (c, d) ad = bc in R.

if and only if

We verify that this is an equivalence relation.

(i) (a, b) ∼ (a, b), since ab = ba.

(ii) If (a, b) ∼ (c, d), then ad = bc. This implies that cb = da and hence

that (c, d) ∼ (a, b).

FIELD OF FRACTIONS 171

(iii) If (a, b) ∼ (c, d) and (c, d) ∼ (e, f ), then ad = bc and cf = de. This

implies that (af ’ be)d = (ad)f ’ b(ed) = bcf ’ bcf = 0. Since R has

no zero divisors and d = 0, it follows that af = be and (a, b) ∼ (e, f ).

Hence the relation ∼ is re¬‚exive, symmetric, and transitive.

Denote the equivalence class containing (a, b) by a/b and the set of equiva-

lence classes by Q. As in Q, de¬ne addition and multiplication in Q by

ad + bc

a c ac ac

+= ·=

and .

bd bd bd bd

These operations on equivalence classes are de¬ned in terms of particular repre-

sentatives, so it must be checked that they are well de¬ned. If a/b = a /b and

c/d = c /d , then ab = a b and cd = c d. Hence

(ad + bc)(b d ) = (ab )dd + bb (cd ) = (a b)dd + bb (c d)

= (a d + b c )(bd)

ad + bc ad +bc

=

and therefore , which shows that addition is well de¬ned.

bd bd

ac ac

Also, acb d = a c bd; thus = , which shows that multiplication is well

bd bd

de¬ned.

It can now be veri¬ed that (Q, +, ·) is a ¬eld. The zero is 0/1, and the identity

is 1/1. For example, the distributive laws hold because

cf + de a(cf + de) a(cf + de) b

a c e a ac ae

· + = · = = ·= +

b d f b df bdf bdf b bd bf

a c ae

= ·+·.

b d bf

a b

The inverse of any nonzero element is . The remaining axioms for a ¬eld

b a

are straightforward to check.

r

The ring R is isomorphic to the subring R = r ∈ R of Q by an iso-

1

r a

morphism that maps r to . Any element in the ¬eld Q can be written as

1 b

a b ’1

a1

a a b

=·= where and are in R .

1b 11 1 1

b

If we take R = Z to be integers in the above construction, we obtain the

rational numbers Q as the ¬eld of fractions.

172 8 RINGS AND FIELDS

If R is an integral domain, the ¬eld of fractions of the polynomial ring R[x]

is called the ¬eld of rational functions with coef¬cients in R. Its elements can

be considered as fractions of one polynomial over a nonzero polynomial.

A (possibly noncommutative) ring is called a domain if ab = 0 if and only

if a = 0 or b = 0. Thus the commutative domains are precisely the integral

domains. In 1931, Oystein Ore (1899“1968) extended Theorem 8.25 to a class

of domains (now called left Ore domains) for which a ring of left fractions

can be constructed that is a skew ¬eld (that is, a ¬eld that is not necessarily

commutative). On the other hand, in 1937, A. I. Mal™cev (1909“1967) discov-

ered an example of a domain that cannot be embedded in any skew ¬eld. The

simplest example of a noncommutative skew ¬eld is the ring of quaternions

(see Exercise 8.36). It has in¬nitely many elements, in agreement with a famous

theorem of J. H. M. Wedderburn, proved in 1905, asserting that any ¬nite skew

¬eld is necessarily commutative.

CONVOLUTION FRACTIONS

We now present an application of the ¬eld of fractions that has important implica-

tions in analysis. This example is of a different type than most of the applications

in this book. It can be omitted, without loss of continuity, by those readers not

interested in analysis or applied mathematics.

We construct the ¬eld of fractions of a set of continuous functions, and use

it to explain two mathematical techniques that have been used successfully by

engineers and physicists for many years, but were at ¬rst mistrusted by math-

ematicians because they did not have a ¬rm mathematical basis. One such

technique was introduced by O. Heaviside in 1893 in dealing with electrical

circuits; this is called the operational calculus, and it enabled him to solve par-

tial differential equations by manipulating differential operators as if they were

algebraic quantities. The second such technique is the use of impulse functions

in applied mathematics and mathematical physics. In 1926, when solving prob-

lems in relativistic quantum mechanics, P. Dirac introduced his delta function,

δ(x), which has the property that

∞

δ(x) = 0 if x = 0 δ(x) dx = 1.

and

’∞

If we use the usual de¬nition of functions, no such object exists. However, it

can be pictured in Figure 8.1 as the limit, as k tends to zero, of the functions

δk (x), where

δk (x) = 1/k if 0 x k.

0 otherwise

Each function δk (x) vanishes outside the interval 0 k and has the prop-

x

erty that

∞

δk (x) dx = 1.

’∞

CONVOLUTION FRACTIONS 173

d1 d1/2 d1/4

x x x

Figure 8.1. The Dirac delta “function” is the limit of δk as k tends to zero.

Consider the set, C[0, ∞), of real-valued functions that are continuous in the

interval 0 x < ∞. We de¬ne the operations of addition and convolution on this

set so that the algebraic structure (C[0, ∞), +, —) is nearly an integral domain:

convolution does not have an identity so the structure fails to satisfy Axiom

(vi) of a ring. However, it is still possible to embed this structure in its ¬eld

of fractions. The Polish mathematician Jan Mikusinski constructed this ¬eld of

fractions and called such elements operators or generalized functions. The Dirac

delta function is a generalized function and is in fact the identity for convolution

in the ¬eld of fractions.

De¬ne addition and convolution of two functions f and g in C[0, ∞) by

x

(f + g)(x) = f (x) + g(x) and (f — g)(x) = f (t)g(x ’ t) dt.

0

This convolution of functions is the continuous analogue of convolution of

sequences, as can be seen by writing the ith term of the sequence

i

ai — bi as at bi’t .

t=0

It is clear that addition is associative and commutative, and the zero function

is the additive identity. Also, the negative of f (x) is ’f (x).

Convolution is commutative because

x

(f — g)(x) = f (t)g(x ’ t) dt

0

0

=’ f (x ’ u)g(u) du substituting u = x ’ t

x

x

= g(u)f (x ’ u) du = (g — f )(x).

0

Convolution is associative because

x

(f — (g — h))(x) = f (t)(g — h)(x ’ t) dt

0

x x’t

= g(u)h(x ’ t ’ u) du dt

f (t)

0 0

x x

= g(w ’ t)h(x ’ w) dw dt,

f (t)

0 t

174 8 RINGS AND FIELDS

w

x

T

t

x

Figure 8.2

putting u = w ’ t. This integration is over the triangle in Figure 8.2 so, changing

the order of integration,

x w

(f — (g — h))(x) = f (t)g(w ’ t)h(x ’ w) dt dw,

0 0

x

= (f — g)(w)h(x ’ w) dw = ((f — g) — h)(x).

0

The distributive laws follow because

x

((f + g) — h)(x) = (f (t) + g(t))h(x ’ t) dt

0

x x

= f (t)h(x ’ t) dt + g(t)h(x ’ t) dt

0 0

= (f — h)(x) + (g — h)(x).

If f is a function that is the identity under convolution, then f — h = h for

all functions h. If we take h to be the function de¬ned by h(x) = 1 for all

0 x < ∞, then

x

(f — h)(x) = f (t) dt = 1 for all x 0.

0

There is no function f in C[0, ∞) with this property, although the Dirac delta

“function” does have this property. Hence (C[0, ∞), +, —) satis¬es all the axioms

for a commutative ring except for the existence of an identity under convolution.

Furthermore, there are no zero divisors under convolution; that is, f — g = 0

implies that f = 0 or g = 0. This is a hard result in analysis, which is known as

Titchmarsh™s theorem. Proofs can be found in Erdelyi [36] or Marchand [37].

However, we can still construct the ¬eld of fractions of this algebraic object in

exactly the same way as we did in Theorem 8.25. For example, the even integers

under addition and multiplication, (2Z, +, ·) is also an algebraic object that sat-

is¬es all the axioms for an integral domain except for the fact that multiplication

has no identity. The ¬eld of fractions of 2Z is the set of rational numbers; every

rational number can be written in the form 2r/2s, where 2r, 2s ∈ 2Z.

CONVOLUTION FRACTIONS 175

The ¬eld of fractions of (C[0, ∞), +, —) is called the ¬eld of convolution

fractions, and its elements are sometimes called generalized functions, distri-

butions, or operators. Elements of this ¬eld are the abstract entities f/g, where

f and g are functions. There is a bijection between the set of elements of the

form f — g/g and the set C[0, ∞). It is possible to interpret other convolution

fractions as impulse functions, discontinuous functions, and even differential or

integral operators. The Dirac delta function can be de¬ned to be the identity of

this ¬eld under convolution; therefore, δ = f/f , for any nonzero function f .

The Heaviside step function illustrated in Figure 8.3 is de¬ned by h(x) = 1

if x 0, and h(x) = 0 if x < 0. The function is continuous when restricted

to the nonnegative numbers and, in some sense, is the integral of the Dirac

delta function. Convolution by h acts as an integral operator on any continuous

function because

x x

(h — f )(x) = (f — h)(x) = f (t)h(x ’ t) dt = f (t) dt.

0 0

Hence h — f is the integral of f . We can use this to de¬ne integration of any

generalized function. Take the integral of the convolution fraction f/g to be the

fraction (h — f )/g.

Denote the inverse of the Heaviside step function by s, so that s = h/ h —

h. This element s is not a genuine function, but only a convolution fraction.

Convolution by s acts, in some sense, as a differential operator in the ¬eld

of convolution fractions. It is not exactly the usual differential operator, because

convolution by s and by h must commute, and s — h = h — s must be the identity.

If f (x) is a continuous function, we know from the calculus, that the derivative of

x x

f (t) dt is just f (x); however, 0 f (t) dt is not just f (x) but is f (x) ’ f (0).

0

In fact, if the function f has a derivative,

(s — f )(x) = f (x) + f (0)δ(x)

where δ(x) is the identity in the ¬eld of convolution fractions. Now, when we

calculate h — s — f , which is equivalent to integrating s — f from 0 to x, we

obtain the function f back again.

By repeated convolution with s or h, a generalized function can be differenti-

ated or integrated any number of times, the result being another generalized func-

tion. We can even differentiate or integrate a fractional number of times. These

h

x

Figure 8.3. Heaviside step function.

176 8 RINGS AND FIELDS

operations s and h can be used to explain Heaviside™s operational calculus, in

which differential and integral operators are manipulated like algebraic symbols.

For further information on the algebraic aspects of generalized functions and

distributions, see Erdelyi [36] or Marchand [37].

EXERCISES

8.1. Write out the tables for the ring Z4 .

8.2. Write out the tables for the ring Z2 — Z2 .

Which of the systems described in Exercises 8.3 to 8.12 are rings under addition

and multiplication? Give reasons.

√

8.3. {a + b 5|a, b ∈ Z}. 8.4. N.

√ √

8.5. {a + b 2 + c 3|a, b, c ∈ Z}.

√

8.6. {a + 3 2b|a, b ∈ Q}.

8.7. All 2 — 2 real matrices with zero determinant.

8.8. All rational numbers that can be written with denominator 2.

8.9. All rational numbers that can be written with an odd denominator.

8.10. (Z, +, —), where + is the usual addition and a — b = 0 for all a, b ∈ Z.

8.11. The set A = {a, b, c} with tables given in Table 8.7.

TABLE 8.7

+ ·

a b c a b c

a a b c a a a a

b b c a b a b c

c c a b c a c c

8.12. The set A = {a, b, c} with tables given in Table 8.8.

TABLE 8.8

+ ·

a b c a b c

a a b c a a a a

b b c a b a c b

c c a b c a b c

8.13. A ring R is called a boolean ring if a 2 = a for all a ∈ R.

(a) Show that (P (X), , ©) is a boolean ring for any set X.

(b) Show that Z2 and Z2 — Z2 are boolean rings.

(c) Prove that if R is boolean, then 2a = 0 for all a ∈ R.

EXERCISES 177

(d) Prove that any boolean ring is commutative.

(e) If (R, §, ∨, ) is any boolean algebra, show that (R, , §) is a boolean

ring where a b = (a § b ) ∨ (a § b).

(f) If (R, +, ·) is a boolean ring, show that (R, §, ∨, ) is a boolean algebra

where a § b = a · b, a ∨ b = a + b + a · b and a = 1 + a.

This shows that there is a one-to-one correspondence between boolean

algebras and boolean rings.

8.14. If A and B are subrings of a ring R, prove that A © B is also a subring of R.

8.15. Prove that the only subring of Zn is itself.

Which of the sets described in Exercises 8.16 to 8.20 are subrings of C? Give

reasons.

8.16. {0 + ib|b ∈ R}. 8.17. {a + ib|a, b ∈ Q}.

√

8.18. {a + b ’7|a, b ∈ Z}. 8.19. {z ∈ C| |z| 1}.

8.20. {a + ib|a, b ∈ Z}.

Which of the rings described in Exercises 8.21 to 8.26 are integral domains and

which are ¬elds?

Z2 — Z2 . 8.22. (P ({a}), , ©).

8.21.

{a + bi|a, b ∈ Q}. 8.24. Z — R.

8.23.

√

{a + b 2|a, b ∈ Z}. 8.26. R[x].

8.25.

8.27. Prove that the set C(R) of continuous real-valued functions de¬ned on the

real line forms a ring (C(R), +, ·), where addition and multiplication of

two functions f, g ∈ C(R) is given by

(f + g)(x) = f (x) + g(x) and (f · g)(x) = f (x) · g(x).

Find all the zero divisors in the rings described in Exercises 8.28 to 8.33.

Z4 . 8.29. Z10 .

8.28.

Z4 — Z2 . 8.31. (P (X), , ©).

8.30.

8.32. 8.33. Mn (R).

M2 (Z2 ).

Let (R, +, ·) be a ring in which (R, +) is a cyclic group. Prove that

8.34.

(R, +, ·) is commutative ring.

ab

8.35. Show that S = a, b ∈ R is a subring of M2 (R) isomorphic

’b a

to C.

±β

8.36. Show that H = ±, β ∈ C is a subring of M2 (C), where ±

’β ±

is the conjugate of ±. This is called the ring of quaternions and gen-

10 ˆ

eralizes the complex numbers in the following way: If I = ,i =

01

i0 01 0i

ˆ ˆ

,j = , and k = in H, show that every

0 ’i ’1 0 i0

178 8 RINGS AND FIELDS

ˆ ˆ ˆ

quaternion q has a unique representation in the form q = aI + bi + cj + d k,

ˆ ˆ ˆ ˆ ˆˆ

where a, b, c, d ∈ R. Show further that i 2 = j 2 = k 2 = i j k = ’I and that

these relations determine the multiplication in H. If 0 = q ∈ H, show that

1 ˆ ˆ ˆ

q ’1 = 2 q — , where q — = aI ’ bi ’ cj ’ d k, so that H is

a +b 2 + c2 + d 2

a noncommutative skew ¬eld.

Find all the ring morphisms from Z to Z6 .

8.37.

Find all the ring morphisms from Z15 to Z3 .

8.38.

Find all the ring morphisms from Z — Z to Z — Z.

8.39.

Find all the ring morphisms from Z7 to Z4 .

8.40.

If (A, +) is an abelian group, the set of endomorphisms of A, End(A), con-

8.41.

sists of all the group morphisms from A to itself. Show that (End(A), +, Ž )

is a ring under addition and composition, where (f + g)(a) = f (a) +

g(a), for f, g ∈ End(A). This is called the endomorphism ring of A.

Describe the endomorphism ring End(Z2 — Z2 ). Is it commutative?

8.42.

Prove that 10n ≡ 1 mod 9 for all n ∈ N. Then prove that an integer is

8.43.

divisible by 9 if and only if the sum of its digits is divisible by 9.

8.44. Find the number of nonisomorphic rings with three elements.

Prove that R[x] ∼ R[y].

=

8.45.

Prove that R[x, y] ∼ R[y, x].

=

8.46.

Let (R, +, ·) be a ring. De¬ne the operations • and Ž on R by

8.47.

r • s = r + s + 1 and r Žs = r · s + r + s.

(a) Prove that (R, •, Ž) is a ring.

(b) What are the additive and multiplicative identities of (R, •, Ž)?

(c) Prove that (R, •, Ž) is isomorphic to (R, +, ·).

8.48. Let a and b be elements of a commutative ring. For each positive integer

n, prove the binomial theorem:

n n

(a + b)n = a n + a n’1 b + · · · + a n’k bk + · · · + bn .

1 k

8.49. Let (R, +, ·) be an algebraic object that satis¬es all the axioms for a ring

except for the multiplicative identity. De¬ne addition and multiplication in

R — Z by

(a, n) + (b, m) = (a + b, n + m) and

(a, n) · (b, m) = (ab + ma + nb, nm).

Show that (R — Z, +, ·) is a ring that contains a subset in one-to-one

correspondence with R that has all the properties of the algebraic object

(R, +, ·).

EXERCISES 179

8.50. If R and S are commutative rings, prove that the ring of sequences (R —

S)N is isomorphic to R N — S N .

8.51. If F is a ¬eld, show that the ¬eld of fractions of F is isomorphic to F .

8.52. Describe the ¬eld of fractions of the ring ({a + ib|a, b ∈ Z}, +, ·).

8.53. Let (S, —) be a commutative semigroup that satis¬es the cancellation law;

that is, a — b = a — c implies that b = c. Show that (S, —) can be embedded