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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 вЂњsij /j вЂќ 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

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
п¬Ѓnite-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 п¬Ѓnd 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 sufп¬Ѓcient 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 (Zв€— , В·), 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 п¬Ѓeld 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 п¬Ѓelds and con-
sider many examples. In later chapters we construct new number systems with
properties similar to the familiar systems.

RINGS

A ring (R, +, В·) is a set R, together with two binary operations + and В· on R
satisfying the following axioms. For any elements a, b, c в€€ R,

(i) (a + b) + c = a + (b + c). (associativity of addition)
(ii) a + b = b + a. (commutativity of addition)
(iii) there exists 0 в€€ R, called the (existence of an additive identity)
zero, such that a + 0 = a.
(iv) there exists (в€’a) в€€ R such that (existence of an additive inverse)
a + (в€’a) = 0.
(v) (a В· b) В· c = a В· (b В· c). (associativity of multiplication)
(vi) there exists 1 в€€ R such that (existence of multiplicative identity)
1 В· a = a В· 1 = a.

Modern Algebra with Applications, Second Edition, by William J. Gilbert and W. Keith Nicholson
ISBN 0-471-41451-4 Copyright п›™ 2004 John Wiley & Sons, Inc.

155
156 8 RINGS AND FIELDS

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

Axioms (i)вЂ“(iv) are equivalent to saying that (R, +) is an abelian group, and
axioms (v) and (vi) are equivalent to saying that (R, В·) is a monoid.
The ring (R, +, В·) is called a commutative ring if, in addition,

(viii) a В· b = b В· a for all a, b в€€ R. (commutativity of multiplication)

The integers under addition and multiplication satisfy all of the axioms above,
so that (Z, +, В·) is a commutative ring. Also, (Q, +, В·), (R, +, В·), and (C, +, В·)
are all commutative rings. If there is no confusion about the operations, we write
only R for the ring (R, +, В·). Therefore, the rings above would be referred to as
Z, Q, R, or C. Moreover, if we refer to a ring R without explicitly deп¬Ѓning 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 veriп¬Ѓed 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 Г— n square matrices with real coefп¬Ѓcients forms a ring
(Mn (R), +, В·), which is not commutative if n > 1, because matrix multiplication
is not commutative.
The elements вЂњevenвЂќ and вЂњoddвЂќ form a commutative ring ({even, odd}, +, В·)
where the operations are given by Table 8.1. вЂњEvenвЂќ is the zero of this ring, and
вЂњoddвЂќ is the multiplicative identity. This is really a special case of the following
example when n = 2.

Example 8.1. Show that (Zn , +, В·) is a commutative ring, where addition and
multiplication on congruence classes, modulo n, are deп¬Ѓned by the equations
[x] + [y] = [x + y] and [x] В· [y] = [xy].

Solution. It follows from Example 4.19, that (Zn , +) is an abelian group.
Since multiplication on congruence classes is deп¬Ѓned in terms of represen-
tatives, it must be veriп¬Ѓed that it is well deп¬Ѓned. Suppose that [x] = [x ] and
[y] = [y ], so that x в‰Ў x and y в‰Ў y mod n. This implies that x = x + kn
and y = y + ln for some k, l в€€ Z. Now x В· y = (x + kn) В· (y + ln) = x В· y +
(ky + lx + kln)n, so x В· y в‰Ў x В· y mod n and hence [x В· y] = [x В· y ]. This
shows that multiplication is well deп¬Ѓned.

TABLE 8.1. Ring of Odd and Even Integers
+ В·
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 deп¬Ѓnitions of addition and mul-
tiplication and from the properties of the integers. The zero is , and the unit
is . The left distributive law is true, for example, because

[x] В· ([y] + [z]) = [x] В· [y + z] = [x В· (y + z)]
= [x В· y + x В· z] by distributivity in Z
= [x В· y] + [x В· z] = [x] В· [y] + [x] В· [z].

Example 8.2. Construct the addition and multiplication tables for the ring
(Z5 , +, В·).

Solution. We denote the congruence class [x] just by x. The tables are given
in Table 8.2.
в€љ в€љ
Example 8.3. Show that (Q( 2), +, В·) is a commutative ring where Q( 2) =
в€љ
{a + b 2 в€€ R|a, b в€€ Q}.
в€љ
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 В· are binary
в€љ
operations on Q( 2). If a, b, c, d в€€ Q, we have
в€љ в€љ в€љ в€љ
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 в€€ Q( 2)

since (a + c) and (b + d) в€€ Q. Also,
в€љ в€љ в€љ в€љ
(a + b 2) В· (c + d 2) = (ac + 2bd) + (ad + bc) 2 в€€ Q( 2)

since (ac + 2bd) and (ad + bc) в€€ Q. в€љ
We now check that axioms (i)вЂ“(viii) of a commutative ring are valid in Q( 2).

(i) Addition of real numbers is associative.
(ii) Addition of real numbers is commutative.
в€љ в€љ
The zero is 0 = 0 + 0 2 в€€ Q( 2).
(iii)
в€љ в€љ в€љ
The additive inverse of a + b 2 is (в€’a) + (в€’b) 2 в€€ Q( 2), since
(iv)
(в€’a) and (в€’b) в€€ Q.

TABLE 8.2. Ring (Z5 , +, В·)
+ В·
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.
в€љ в€љ
(vi) The multiplicative identity is 1 = 1 + 0 2 в€€ Q( 2).
(vii) The distributive axioms hold for real numbers and hence hold for ele-
в€љ
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.

, в€©) is a commutative ring for any set X.
Example 8.4. (P (X),

Solution. The axioms (i)вЂ“(viii) of a commutative ring follow from Proposi-
tions 2.1 and 2.3. The zero is Г˜, and the identity is X.

In the ring above, A в€© 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).

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) = {Г˜, 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, +, В·) is a ring, then for all a, b в€€ R:

a В· 0 = 0 В· a = 0.
(i)
a В· (в€’b) = (в€’a) В· b = в€’(a В· b).
(ii)
(в€’a) В· (в€’b) = a В· b.
(iii)
(в€’1) В· a = в€’a.
(iv)
(в€’1) В· (в€’1) = 1.
(v)

Proof. (i) By distributivity, a В· 0 = a В· (0 + 0) = a В· 0 + a В· 0. Adding
в€’(a В· 0) to each side, we obtain 0 = a В· 0. Similarly, 0 В· a = 0.
(ii) Compute a В· (в€’b) + a В· b = a В· (в€’b + b) = a В· 0 = 0, using (i). Therefore,
a В· (в€’b) = в€’(a В· b). Similarly, (в€’a) В· b = в€’(a В· b).
(iii) We have (в€’a) В· (в€’b) = в€’(a В· (в€’b)) = в€’(в€’(a В· b)) = a В· b by (ii) and
Proposition 3.7.
INTEGRAL DOMAINS AND FIELDS 159

TABLE 8.3. Ring P ({a, b, c})
Г˜ A B C A B C X
Г˜ Г˜ A B C A B C X
Г˜
A A C B X C B A
Г˜
B B C A C X A B
Г˜
C C B A B A X C
Г˜
A A X C B C B A
Г˜
B B C X A C A B
Г˜
C C B A X B A C
Г˜
X X A B C A B C

в€© Г˜ A B C A B C X
Г˜ Г˜ Г˜ Г˜ Г˜ Г˜ Г˜ Г˜ Г˜
Г˜ Г˜ Г˜ Г˜
A A A A A
Г˜ Г˜ Г˜ Г˜
B B B B B
Г˜ Г˜ Г˜ Г˜
C C C C C
Г˜ Г˜
A B C A C B A
Г˜ Г˜
B A C C B A B
Г˜ Г˜
C A B B A C C
Г˜
X A B C A B C X

(iv) By (ii), (в€’1) В· a = в€’(1 В· a) = в€’a.
(v) By (iii), (в€’1) В· (в€’1) = 1 В· 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 В· 1 =
a В· 0 = 0. Therefore, the ring contains only the element 0. It can be veriп¬Ѓed that
this forms a ring with the operations deп¬Ѓned by 0 + 0 = 0 and 0 В· 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 в€’ c) = 0, so b = c. However, this prop-
erty does not hold for all rings. For example, in Z4 , we have  В·  = , and
we cannot always cancel since  В·  =  В· , but  = .
If (R, +, В·) is a commutative ring, a nonzero element a в€€ R is called a zero
divisor if there exists a nonzero element b в€€ R such that a В· 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 В· 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  is a zero divisor. Neither
is (P (X), , в€©), 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 В· b = a В· c, then b = c.

Proof. If a В· b = a В· c, then a В· (b в€’ c) = a В· b в€’ a В· c = 0. Since R is an inte-
gral domain, it has no zero divisors. Since a = 0, it follows that (b в€’ 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 в€€ 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 п¬Ѓeld is a ring in which the nonzero elements form an abelian group
under multiplication. In other words, a п¬Ѓeld is a nontrivial commutative ring R
satisfying the following extra axiom.

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

The rings Q, R, and C are all п¬Ѓelds, but the integers do not form a п¬Ѓeld.

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

Proof. Let a В· b = 0 in a п¬Ѓeld F . If a = 0, there exists an inverse a в€’1 в€€ F
and b = (a в€’1 В· a) В· b = a в€’1 (a В· b) = a в€’1 В· 0 = 0. Hence either a = 0 or b = 0,
and F is an integral domain.

Theorem 8.10. A п¬Ѓnite integral domain is a п¬Ѓeld.

Proof. Let D = {x0 , x1 , x2 , . . . , xn } be a п¬Ѓnite 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 = xiв€’1 , and D is a п¬Ѓeld.
SUBRINGS AND MORPHISMS OF RINGS 161

Note that Z is an inп¬Ѓnite integral domain that is not a п¬Ѓeld.

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

Proof. Suppose that n is prime and that [a] В· [b] =  in Zn . Then n|ab. So
n|a or n|b by EuclidвЂ™s Lemma (Theorem 12, Appendix 2). Hence [a] =  or
[b] = , and Zn is an integral domain. Since Zn is also п¬Ѓnite, it follows from
Theorem 8.10 that Zn is a п¬Ѓeld.
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] =  and [s] =  but
[r] В· [s] = [rs] = . Therefore, Zn has zero divisors and hence is not a п¬Ѓeld.

в€љ
Example 8.12. Is (Q( 2), +, В·) an integral domain or a п¬Ѓeld?
в€љ
Solution. From Example 8.3 we know that Q( 2) is a commutative ring. Let
в€љ
a + bв€љ2 be a nonzero element, so that at least one of a and b is not zero. Hence
в€љ
a в€’ b 2 = 0 (because 2 is not in Q), so we have
в€љ
в€љ
aв€’b 2
1 a b
в€љ= в€љ =2 в€’
в€љ 2.
a в€’ 2b2 a 2 в€’ 2b2
a+b 2 (a + b 2)(a в€’ b 2)
в€љ в€љ в€љ
This is an element of Q( 2), and so is the inverse of a + b 2. Hence Q( 2)
is a п¬Ѓeld (and an integral domain).

SUBRINGS AND MORPHISMS OF RINGS

If (R, +, В·) is a ring, a nonempty subset S of R is called a subring of R if for
all a, b в€€ S:

a + b в€€ S.
(i)
в€’a в€€ S.
(ii)
a В· b в€€ S.
(iii)
1 в€€ S.
(iv)

Conditions (i) and (ii) imply that (S, +) is a subgroup of (R, +) and can be
replaced by the condition a в€’ b в€€ S.

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

Proof. Conditions (i) and (iii) of the deп¬Ѓnition above guarantee that S is closed
under addition and multiplication. Condition (iv) shows that 1 в€€ S. It follows
from Proposition 3.8 that (S, +) is a group. (S, +, В·) satisп¬Ѓes the remaining
axioms for a ring because they hold in (R, +, В·).
162 8 RINGS AND FIELDS

For example, Z, Q, and R are all subrings of C. Let D be the set of n Г— n
real diagonal matrices. Then D is a subring of the ring of all n Г— 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.
в€љ в€љ
Example 8.14. Show that Q( 2) = {a + b 2|a, b в€€ Q} is a subring of R.
в€љ в€љ в€љ
Solution. Let a + b 2, c + d 2 в€€ Q( 2). Then
в€љ в€љ в€љ в€љ
(a + b 2) + (c + d 2) = (a + c) + (b + d) 2 в€€ Q( 2).
(i)
в€љ в€љ в€љ
в€’(a + b 2) = (в€’a) + (в€’b) 2 в€€ Q( 2).
(ii)
в€љ в€љ в€љ в€љ
(a + b 2) В· (c + d 2) = (ac + 2bd) + (ad + bc) 2 в€€ Q( 2).
(iii)
в€љ в€љ
1 = 1 + 0 2 в€€ 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, +, В·) and (S, +, В·) be two rings. The function f : R в†’ S
is called a ring morphism if for all a, b в€€ R:

(i) f (a + b) = f (a) + f (b).
(ii) f (a В· b) = f (a) В· 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, +, В·) and (S, вЉ•, ЕЅ), then the conditions for
f : R в†’ S to be a ring morphism are:

(i) f (a + b) = f (a) вЉ• f (b).
(ii) f (a В· b) = f (a)ЕЅf (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 в€ј S.
=
A ring morphism, f , from (R, +, В·) to (S, +, В·) is, in particular, a group
morphism from (R, +) to (S, +). Therefore, by Proposition 3.19, f (0) = 0 and
f (в€’a) = в€’f (a) for all a в€€ R.
The inclusion function, i: S в†’ R, of any subring S into a ring R is always
a ring morphism. The function f : Z в†’ Zn , deп¬Ѓned by f (x) = [x], which maps
an integer to its equivalence class modulo n, is a ring morphism from (Z, +, В·)
to (Zn , +, В·).
SUBRINGS AND MORPHISMS OF RINGS 163

Example 8.15. If X is a one element set, show that f : P (X) в†’ Z2 is a
ring isomorphism between (P (X), , в€©) and (Z2 , +, В·), where f (Г˜) =  and
f (X) = .

Solution. We can check that f is a morphism by testing all the possibilities
for f (A B) and f (A в€© B). Since the rings are commutative, they are

f (Г˜ Г˜) = f (Г˜) =  = f (Г˜) + f (Г˜)
f (Г˜ X) = f (X) =  = f (Г˜) + f (X)
f (X X) = f (Г˜) =  = f (X) + f (X)
f (Г˜ в€© Г˜) = f (Г˜) =  = f (Г˜) В· f (Г˜)
f (Г˜ в€© X) = f (Г˜) =  = f (Г˜) В· f (X)
f (X в€© X) = f (X) =  = f (X) В· f (X).

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

If f : R в†’ S is an isomorphism between two п¬Ѓnite rings, the addition and
multiplication tables of S will be the same as those of R if we replace each
a в€€ R by f (a) в€€ 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 ), +, ЕЅ ), under addition and composition. Show that the function

f : L(Rn , Rn ) в†’ Mn (R)

TABLE 8.4. Ring P (X )
When X Is a Point
Г˜ Г˜
X X
Г˜ Г˜ Г˜ Г˜ Г˜
X
Г˜ Г˜
X X X X

TABLE 8.5. Ring Z2
+ В·
   
     
     
164 8 RINGS AND FIELDS

is a ring morphism, where f assigns to each linear transformation its standard
matrix, that is, its n Г— n coefп¬Ѓcient matrix with respect to the standard basis
of Rn .

Solution. If О± is a linear transformation from Rn to itself, then
пЈ®пЈ№пЈ® пЈ№ пЈ® пЈ№
a11 x1 + В· В· В· + a1n xn
x1 a11 . . . a1n
пЈЇпЈєпЈЇ пЈє пЈЇ. . пЈє.
О±пЈ° . пЈ» = пЈ°. .
пЈ» and f (О±) = пЈ° . .пЈ»
. . .
. . . . .
an1 x1 + В· В· В· + ann xn
xn an1 . . . ann

Matrix addition and multiplication is deп¬Ѓned so that f (О± + ОІ) = f (О±) + f (ОІ)
and f (О± ЕЅ ОІ) = f (О±) В· f (ОІ). Also, if О№ is the identity linear transformation then
f (О№) is the identity matrix.
Any matrix deп¬Ѓnes 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 в†’ Z4 , deп¬Ѓned by f ([x]24 ) = [x]4 is a ring
morphism.

Proof. Since the function is deп¬Ѓned in terms of representatives of equiva-
lence classes, we п¬Ѓrst check that it is well deп¬Ѓned. If [x]24 = [y]24 ,
then x в‰Ў y mod 24 and 24|(x в€’ y). Hence 4|(x в€’ y) and [x]4 = [y]4 , which
shows that f is well deп¬Ѓned.
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 В· [y]24 ) = f ([xy]24 ) = [xy]4 = [x]4 В· [y]4 .
(iii) f (24 ) = 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, +, В·) and (S, +, В·) are two rings, their product is the ring (R Г— S, +, В·)
whose underlying set is the cartesian product of R and S and whose operations
are deп¬Ѓned component-wise by

(r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 ) and (r1 , s1 ) В· (r2 , s2 ) = (r1 В· r2 , s1 В· s2 ).
NEW RINGS FROM OLD 165

It is readily veriп¬Ѓed that these operations do indeed deп¬Ѓne a ring structure on
R Г— 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 , +, В·) 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 Г— Z3 .

Solution. Let Z2 = {0, 1} and Z3 = {0, 1, 2}. Then Z2 Г— 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 п¬Ѓrst coordinate and modulo 3 in the
second coordinate.

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

Theorem 8.19. Zm Г— 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 в†’ Zm Г— Zn

TABLE 8.6. Ring Z2 Г— 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)

В· (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

deп¬Ѓned by f ([x]mn ) = ([x]m , [x]n ) is a group isomorphism. However, this func-
tion also preserves multiplication because

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

Also, f (mn ) = (m , n ); thus f is a ring isomorphism.
It was shown in the discussion following Corollary 4.33 that if gcd(m, n) =
1, Zm Г— 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 В· m2 В· В· В· mr , where gcd(mi , mj ) = 1 if i = j . Then
Zm1 Г— Zm2 Г— В· В· В· Г— Zmr is a ring isomorphic to Zm .

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

powers of distinct primes. Then Zn в€ј ZpО±1 Г— ZpО±2 Г— В· В· В· Г— ZprО±r as rings.
= 1 2

If R is a commutative ring, we can construct the ring of n Г— n matrices with
entries from R, (Mn (R), +, В·). Addition and multiplication are performed as in
real matrices.
For example, (Mn (Z2 ), +, В·) is the ring of n Г— 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 + В· В· В· + an x n ,

where a0 , a1 , a2 , . . . , an в€€ R and n в€€ N. The element ai is called the coefп¬Ѓcient
of x i in p(x). If the coefп¬Ѓcient of x i is zero, the term 0x i may be omitted, and
if the coefп¬Ѓcient 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 coefп¬Ѓcient of x n is the same in each polynomial for every n 0.
In particular,
a0 + a1 x + a2 x 2 + В· В· В· + an x n = 0

is the zero polynomial if and only if a0 = a1 = a2 = В· В· В· = 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 coefп¬Ѓcients of p(x) are zero, then p(x) is called
the zero polynomial, and its degree is not deп¬Ѓned.
в€љ
For example, 4x 2 в€’ 3 is a polynomial over R of degree 2, ix 4 в€’ (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 coefп¬Ѓcients from the commutative ring
R is denoted by R[x]. That is,

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

This forms a ring (R[x], +, В·) called the polynomial ring with coefп¬Ѓcients 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 deп¬Ѓned by
max(m,n)
p(x) + q(x) = (ai + bi )x i
i=0

and
m+n
p(x) В· q(x) = where ck =
ck x k ai bj .
i+j =k
k=0

With a little effort, it can be veriп¬Ѓed that (R[x], +, В·) satisп¬Ѓes 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 coefп¬Ѓcients 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) В· (3x 2 + 4x + 1) = x 5 + 4x 4 + 4x + 1.

When working in Zn [x], the coefп¬Ѓcients, 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) В· q(x)) = deg p(x) + deg q(x).

Proof. Let deg p(x) = n, deg q(x) = m and let p(x) = a0 + В· В· В· + an x n ,
q(x) = b0 + В· В· В· + bm x m , where an = 0, bm = 0. Then the coefп¬Ѓcient of the high-
est power of x in p(x) В· q(x) is an bm , which is nonzero since R has no zero
divisors. Hence deg(p(x) В· q(x)) = m + n.
168 8 RINGS AND FIELDS

If the coefп¬Ѓcient 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) В· (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) В· 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 coefп¬Ѓcients from R. We deп¬Ѓne induc-
tively R[x1 , . . . , xn ] = R[x1 , . . . , xnв€’1 ][xn ]. For example, consider a polynomial
f in R[x, y] = R[x][y], say

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

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

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

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 inп¬Ѓnite sequence
of elements of R, a0 , a1 , a2 , . . . , by ai . Deп¬Ѓne addition, +, and convolution,
в€—, of two such sequences by

ai + bi = ai + bi

and
ai в€— bi = aj bk = a0 bi + a1 biв€’1 + В· В· В· + ai b0 .
j +k=i

The set of all such sequences forms a commutative ring (R N , +, в€—) 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 в€’ai . Now

( ai в€— bi ) в€— ci = aj bk в€— ci
j +k=i
пЈ« пЈ¶
пЈ­ aj bk пЈё 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 в€— ( bi + ci ) = aj (bk + ck )
j +k=i

= aj bk + aj ck = ai в€— bi + ai в€— ci .
j +k=i j +k=i

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

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

Therefore, (R N , +, в€—) is a commutative ring.
Suppose that aq and br are the п¬Ѓrst 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+rв€’1 + В· В· В· + aq br + aq+1 brв€’1 + В· В· В· + aq+r b
j +k=q+r
=0 + + В· В· В· + aq br + + В· В· В· + 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 п¬Ѓeld because 0, 1, 0, 0, . . . has no inverse.
In fact, for any sequence bi , 0, 1, 0, 0, . . . в€— b0 , b1 , b2 , . . . = 0, b0 , b1 , . . . ,
which can never be the identity in the ring.
A formal power series in x with coefп¬Ѓcients from a commutative ring R is
an expression of the form.
в€ћ
a0 + a1 x + a2 x + В· В· В· = where ai в€€ R.
2
ai x i
i=0

In contrast to a polynomial, these power series can have an inп¬Ѓnite 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 deп¬Ѓned in R[[x]] by
в€ћ в€ћ в€ћ
+ = (ai + bi )x i
i i
ai x bi x
i=0 i=0 i=0
170 8 RINGS AND FIELDS

пЈ« пЈ¶
and
в€ћ в€ћ в€ћ
пЈ­ aj bk пЈё x i .
В· =
ai x i bi x i
j +k=i
i=0 i=0 i=0

It can be veriп¬Ѓed that these formal power series do form a ring, (R[[x]], +, В·),
and that the polynomial ring, R[x], is the subring consisting of those power
series with only a п¬Ѓnite number of nonzero terms. In fact, the ring of sequences
(R N , +, в€—) is isomorphic to the ring of formal power series (R[[x]], +, В·). The
function f : R N в†’ R[[x]] that is deп¬Ѓned by f ( a0 , a1 , a2 , В· В· В· ) = a0 + a1 x +
a2 x2 + В· В· В· is clearly a bijection. It follows from the deп¬Ѓnitions 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 п¬Ѓeld 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 п¬Ѓeld, because п¬Ѓelds cannot contain zero divisors
or pairs of noncommutative elements.

Theorem 8.25. If R is an integral domain, it is possible to construct a п¬Ѓeld 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 В· q в€’1 for suitable p, q в€€ R .

Q is called the п¬Ѓeld of fractions of R (or sometimes the п¬Ѓeld of quotients of R).
Proof. Consider the set R Г— R в€— = {(a, b)|a, b в€€ 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 deп¬Ѓne a relation в€ј on R Г— R в€— by

(a, b) в€ј (c, d) ad = bc in R.
if and only if

We verify that this is an equivalence relation.

(i) (a, b) в€ј (a, b), since ab = ba.
(ii) If (a, b) в€ј (c, d), then ad = bc. This implies that cb = da and hence
that (c, d) в€ј (a, b).
FIELD OF FRACTIONS 171

(iii) If (a, b) в€ј (c, d) and (c, d) в€ј (e, f ), then ad = bc and cf = de. This
implies that (af в€’ be)d = (ad)f в€’ b(ed) = bcf в€’ bcf = 0. Since R has
no zero divisors and d = 0, it follows that af = be and (a, b) в€ј (e, f ).

Hence the relation в€ј is reп¬‚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, deп¬Ѓne addition and multiplication in Q by

a c ac ac
+= В·=
and .
bd bd bd bd

These operations on equivalence classes are deп¬Ѓned in terms of particular repre-
sentatives, so it must be checked that they are well deп¬Ѓned. 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)

=
and therefore , which shows that addition is well deп¬Ѓned.
bd bd
ac ac
Also, acb d = a c bd; thus = , which shows that multiplication is well
bd bd
deп¬Ѓned.
It can now be veriп¬Ѓed that (Q, +, В·) is a п¬Ѓeld. 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
В· + = В· = = В·= +
b d f b df bdf bdf b bd bf
a c ae
= В·+В·.
b d bf

a b
The inverse of any nonzero element is . The remaining axioms for a п¬Ѓeld
b a
are straightforward to check.
r
The ring R is isomorphic to the subring R = r в€€ R of Q by an iso-
1
r a
morphism that maps r to . Any element in the п¬Ѓeld Q can be written as
1 b
a b в€’1
a1
a a b
=В·= 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 п¬Ѓeld of fractions.
172 8 RINGS AND FIELDS

If R is an integral domain, the п¬Ѓeld of fractions of the polynomial ring R[x]
is called the п¬Ѓeld of rational functions with coefп¬Ѓcients 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 (1899вЂ“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 п¬Ѓeld (that is, a п¬Ѓeld that is not necessarily
commutative). On the other hand, in 1937, A. I. MalвЂ™cev (1909вЂ“1967) discov-
ered an example of a domain that cannot be embedded in any skew п¬Ѓeld. The
simplest example of a noncommutative skew п¬Ѓeld is the ring of quaternions
(see Exercise 8.36). It has inп¬Ѓnitely many elements, in agreement with a famous
theorem of J. H. M. Wedderburn, proved in 1905, asserting that any п¬Ѓnite skew
п¬Ѓeld is necessarily commutative.

CONVOLUTION FRACTIONS

We now present an application of the п¬Ѓeld 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 п¬Ѓeld 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 п¬Ѓrst mistrusted by math-
ematicians because they did not have a п¬Ѓrm 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,
Оґ(x), which has the property that
в€ћ
Оґ(x) = 0 if x = 0 Оґ(x) dx = 1.
and
в€’в€ћ

If we use the usual deп¬Ѓnition 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
Оґk (x), where
Оґk (x) = 1/k if 0 x k.
0 otherwise
Each function Оґk (x) vanishes outside the interval 0 k and has the prop-
x
erty that
в€ћ
Оґk (x) dx = 1.
в€’в€ћ
CONVOLUTION FRACTIONS 173

d1 d1/2 d1/4

x x x
Figure 8.1. The Dirac delta вЂњfunctionвЂќ is the limit of Оґk as k tends to zero.

Consider the set, C[0, в€ћ), of real-valued functions that are continuous in the
interval 0 x < в€ћ. We deп¬Ѓne the operations of addition and convolution on this
set so that the algebraic structure (C[0, в€ћ), +, в€—) 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 п¬Ѓeld
of fractions. The Polish mathematician Jan Mikusinski constructed this п¬Ѓeld 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 п¬Ѓeld of fractions.
Deп¬Ѓne addition and convolution of two functions f and g in C[0, в€ћ) by
x
(f + g)(x) = f (x) + g(x) and (f в€— g)(x) = f (t)g(x в€’ 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 в€— bi as at biв€’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 в€’f (x).
Convolution is commutative because
x
(f в€— g)(x) = f (t)g(x в€’ t) dt
0
0
=в€’ f (x в€’ u)g(u) du substituting u = x в€’ t
x
x
= g(u)f (x в€’ u) du = (g в€— f )(x).
0

Convolution is associative because
x
(f в€— (g в€— h))(x) = f (t)(g в€— h)(x в€’ t) dt
0
x xв€’t
= g(u)h(x в€’ t в€’ u) du dt
f (t)
0 0
x x
= g(w в€’ t)h(x в€’ w) dw dt,
f (t)
0 t
174 8 RINGS AND FIELDS

w
x
T

t
x

Figure 8.2

putting u = w в€’ t. This integration is over the triangle in Figure 8.2 so, changing
the order of integration,
x w
(f в€— (g в€— h))(x) = f (t)g(w в€’ t)h(x в€’ w) dt dw,
0 0
x
= (f в€— g)(w)h(x в€’ w) dw = ((f в€— g) в€— h)(x).
0

x
((f + g) в€— h)(x) = (f (t) + g(t))h(x в€’ t) dt
0
x x
= f (t)h(x в€’ t) dt + g(t)h(x в€’ t) dt
0 0
= (f в€— h)(x) + (g в€— h)(x).

If f is a function that is the identity under convolution, then f в€— h = h for
all functions h. If we take h to be the function deп¬Ѓned by h(x) = 1 for all
0 x < в€ћ, then
x
(f в€— h)(x) = f (t) dt = 1 for all x 0.
0

There is no function f in C[0, в€ћ) with this property, although the Dirac delta
вЂњfunctionвЂќ does have this property. Hence (C[0, в€ћ), +, в€—) satisп¬Ѓes 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 в€— g = 0
implies that f = 0 or g = 0. This is a hard result in analysis, which is known as
TitchmarshвЂ™s theorem. Proofs can be found in Erdelyi  or Marchand .
However, we can still construct the п¬Ѓeld 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, +, В·) is also an algebraic object that sat-
isп¬Ѓes all the axioms for an integral domain except for the fact that multiplication
has no identity. The п¬Ѓeld of fractions of 2Z is the set of rational numbers; every
rational number can be written in the form 2r/2s, where 2r, 2s в€€ 2Z.
CONVOLUTION FRACTIONS 175

The п¬Ѓeld of fractions of (C[0, в€ћ), +, в€—) is called the п¬Ѓeld of convolution
fractions, and its elements are sometimes called generalized functions, distri-
butions, or operators. Elements of this п¬Ѓeld 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 в€— g/g and the set C[0, в€ћ). 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 deп¬Ѓned to be the identity of
this п¬Ѓeld under convolution; therefore, Оґ = f/f , for any nonzero function f .
The Heaviside step function illustrated in Figure 8.3 is deп¬Ѓned 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 в€— f )(x) = (f в€— h)(x) = f (t)h(x в€’ t) dt = f (t) dt.
0 0

Hence h в€— f is the integral of f . We can use this to deп¬Ѓne integration of any
generalized function. Take the integral of the convolution fraction f/g to be the
fraction (h в€— f )/g.
Denote the inverse of the Heaviside step function by s, so that s = h/ h в€—
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 п¬Ѓeld
of convolution fractions. It is not exactly the usual differential operator, because
convolution by s and by h must commute, and s в€— h = h в€— 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) в€’ f (0).
0
In fact, if the function f has a derivative,

(s в€— f )(x) = f (x) + f (0)Оґ(x)

where Оґ(x) is the identity in the п¬Ѓeld of convolution fractions. Now, when we
calculate h в€— s в€— f , which is equivalent to integrating s в€— 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 HeavisideвЂ™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  or Marchand .

EXERCISES

8.1. Write out the tables for the ring Z4 .
8.2. Write out the tables for the ring Z2 Г— Z2 .
Which of the systems described in Exercises 8.3 to 8.12 are rings under addition
and multiplication? Give reasons.
в€љ
8.3. {a + b 5|a, b в€€ Z}. 8.4. N.
в€љ в€љ
8.5. {a + b 2 + c 3|a, b, c в€€ Z}.
в€љ
8.6. {a + 3 2b|a, b в€€ Q}.
8.7. All 2 Г— 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, +, Г—), where + is the usual addition and a Г— b = 0 for all a, b в€€ Z.
8.11. The set A = {a, b, c} with tables given in Table 8.7.

TABLE 8.7
+ В·
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
+ В·
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 в€€ R.
(a) Show that (P (X), , в€©) is a boolean ring for any set X.
(b) Show that Z2 and Z2 Г— Z2 are boolean rings.
(c) Prove that if R is boolean, then 2a = 0 for all a в€€ R.
EXERCISES 177

(d) Prove that any boolean ring is commutative.
(e) If (R, в€§, в€Ё, ) is any boolean algebra, show that (R, , в€§) is a boolean
ring where a b = (a в€§ b ) в€Ё (a в€§ b).
(f) If (R, +, В·) is a boolean ring, show that (R, в€§, в€Ё, ) is a boolean algebra
where a в€§ b = a В· b, a в€Ё b = a + b + a В· 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 в€© 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 в€€ R}. 8.17. {a + ib|a, b в€€ Q}.
в€љ
8.18. {a + b в€’7|a, b в€€ Z}. 8.19. {z в€€ C| |z| 1}.
8.20. {a + ib|a, b в€€ Z}.
Which of the rings described in Exercises 8.21 to 8.26 are integral domains and
which are п¬Ѓelds?
Z2 Г— Z2 . 8.22. (P ({a}), , в€©).
8.21.
{a + bi|a, b в€€ Q}. 8.24. Z Г— R.
8.23.
в€љ
{a + b 2|a, b в€€ Z}. 8.26. R[x].
8.25.
8.27. Prove that the set C(R) of continuous real-valued functions deп¬Ѓned on the
real line forms a ring (C(R), +, В·), where addition and multiplication of
two functions f, g в€€ C(R) is given by

(f + g)(x) = f (x) + g(x) and (f В· g)(x) = f (x) В· 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 Г— Z2 . 8.31. (P (X), , в€©).
8.30.
8.32. 8.33. Mn (R).
M2 (Z2 ).
Let (R, +, В·) be a ring in which (R, +) is a cyclic group. Prove that
8.34.
(R, +, В·) is commutative ring.
ab
8.35. Show that S = a, b в€€ R is a subring of M2 (R) isomorphic
в€’b a
to C.
О±ОІ
8.36. Show that H = О±, ОІ в€€ C is a subring of M2 (C), where О±
в€’ОІ О±
is the conjugate of О±. This is called the ring of quaternions and gen-
10 Л†
eralizes the complex numbers in the following way: If I = ,i =
01
i0 01 0i
Л† Л†
,j = , and k = in H, show that every
0 в€’i в€’1 0 i0
178 8 RINGS AND FIELDS

Л† Л† Л†
quaternion q has a unique representation in the form q = aI + bi + cj + d k,
Л† Л† Л† Л† Л†Л†
where a, b, c, d в€€ R. Show further that i 2 = j 2 = k 2 = i j k = в€’I and that
these relations determine the multiplication in H. If 0 = q в€€ H, show that
1 Л† Л† Л†
q в€’1 = 2 q в€— , where q в€— = aI в€’ bi в€’ cj в€’ d k, so that H is
a +b 2 + c2 + d 2
a noncommutative skew п¬Ѓeld.
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 Г— Z to Z Г— 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), +, ЕЅ )
is a ring under addition and composition, where (f + g)(a) = f (a) +
g(a), for f, g в€€ End(A). This is called the endomorphism ring of A.
Describe the endomorphism ring End(Z2 Г— Z2 ). Is it commutative?
8.42.
Prove that 10n в‰Ў 1 mod 9 for all n в€€ 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] в€ј R[y].
=
8.45.
Prove that R[x, y] в€ј R[y, x].
=
8.46.
Let (R, +, В·) be a ring. Deп¬Ѓne the operations вЉ• and ЕЅ on R by
8.47.

r вЉ• s = r + s + 1 and r ЕЅs = r В· s + r + s.

(a) Prove that (R, вЉ•, ЕЅ) is a ring.
(b) What are the additive and multiplicative identities of (R, вЉ•, ЕЅ)?
(c) Prove that (R, вЉ•, ЕЅ) is isomorphic to (R, +, В·).
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 nв€’1 b + В· В· В· + a nв€’k bk + В· В· В· + bn .
1 k

8.49. Let (R, +, В·) be an algebraic object that satisп¬Ѓes all the axioms for a ring
except for the multiplicative identity. Deп¬Ѓne addition and multiplication in
R Г— Z by

(a, n) + (b, m) = (a + b, n + m) and
(a, n) В· (b, m) = (ab + ma + nb, nm).

Show that (R Г— Z, +, В·) is a ring that contains a subset in one-to-one
correspondence with R that has all the properties of the algebraic object
(R, +, В·).
EXERCISES 179

8.50. If R and S are commutative rings, prove that the ring of sequences (R Г—
S)N is isomorphic to R N Г— S N .
8.51. If F is a п¬Ѓeld, show that the п¬Ѓeld of fractions of F is isomorphic to F .
8.52. Describe the п¬Ѓeld of fractions of the ring ({a + ib|a, b в€€ Z}, +, В·).
8.53. Let (S, в€—) be a commutative semigroup that satisп¬Ѓes the cancellation law;
that is, a в€— b = a в€— c implies that b = c. Show that (S, в€—) can be embedded
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