there exists non-zero b ∈ R such that ab = 0R .

If R is non-trivial and has no zero divisors, then it is called an integral

domain. Put another way, a non-trivial ring R is an integral domain if

216 Rings

and only if the following holds: for all a, b ∈ R, ab = 0R implies a = 0R or

b = 0R .

Note that if u is a unit in R, it cannot be a zero divisor (if ub = 0R , then

multiplying both sides of this equation by u’1 yields b = 0R ). In particular,

it follows that any ¬eld is an integral domain.

Example 9.15. Z is an integral domain. 2

Example 9.16. For n > 1, Zn is an integral domain if and only if n is

prime. In particular, if n is composite, so n = n1 n2 with 1 < n1 < n and

1 < n2 < n, then [n1 ]n and [n2 ]n are zero divisors: [n1 ]n [n2 ]n = [0]n , but

[n1 ]n = [0]n and [n2 ]n = [0]n . 2

Example 9.17. Q, R, and C are ¬elds, and hence are also integral domains.

2

Example 9.18. For two non-trivial rings R1 , R2 , an element (a1 , a2 ) ∈

R1 — R2 is a zero divisor if and only if a1 is a zero divisor, a2 is a zero

divisor, or exactly one of a1 or a2 is zero. In particular, R1 — R2 is not an

integral domain. 2

We have the following “cancellation law”:

Theorem 9.3. If R is a ring, and a, b, c ∈ R such that a = 0R and a is not

a zero divisor, then ab = ac implies b = c.

Proof. ab = bc implies a(b ’ c) = 0R . The fact that a = 0 and a is not a

zero divisor implies that we must have b ’ c = 0R , and so b = c. 2

Theorem 9.4. If D is an integral domain, then:

(i) for all a, b, c ∈ D, a = 0D and ab = ac implies b = c;

(ii) for all a, b ∈ D, a | b and b | a if and only if a = bc for some c ∈ D— .

(iii) for all a, b ∈ D with b = 0D and b | a, there is a unique c ∈ D such

that a = bc, which we may denote as a/b.

Proof. The ¬rst statement follows immediately from the previous theorem

and the de¬nition of an integral domain.

For the second statement, if a = bc for c ∈ D— , then we also have b = ac’1 ;

thus, b | a and a | b. Conversely, a | b implies b = ax for x ∈ D, and b | a

implies a = by for y ∈ D, and hence b = bxy. If b = 0R , then the equation

a = by implies a = 0R , and so the statement holds for any c; otherwise,

cancel b, we have 1D = xy, and so x and y are units.

For the third statement, if a = bc and a = bc , then bc = bc , and cancel

b. 2

9.1 De¬nitions, basic properties, and examples 217

Theorem 9.5. The characteristic of an integral domain is either zero or a

prime.

Proof. By way of contradiction, suppose that D is an integral domain with

characteristic m that is neither zero nor prime. Since, by de¬nition, D is

not a trivial ring, we cannot have m = 1, and so m must be composite. Say

m = st, where 1 < s < m and 1 < t < m. Since m is the additive order of

1D , it follows that (s · 1D ) = 0D and (t · 1D ) = 0D ; moreover, since D is an

integral domain, it follows that (s · 1D )(t · 1D ) = 0D . So we have

0D = m · 1D = (st) · 1D = (s · 1D )(t · 1D ) = 0D ,

a contradiction. 2

Theorem 9.6. Any ¬nite integral domain is a ¬eld.

Proof. Let D be a ¬nite integral domain, and let a be any non-zero element

of D. Consider the a-multiplication map that sends b ∈ D to ab, which

is a group homomorphism on the additive group of D. Since a is not a

zero-divisor, it follows that the kernel of the a-multiplication map is {0D },

hence the map is injective, and by ¬niteness, it must be surjective as well.

In particular, there must be an element b ∈ D such that ab = 1D . 2

Theorem 9.7. Any ¬nite ¬eld F must be of cardinality pw , where p is

prime, w is a positive integer, and p is the characteristic of F .

Proof. By Theorem 9.5, the characteristic of F is either zero or a prime,

and since F is ¬nite, it must be prime. Let p denote the characteristic. By

de¬nition, p is the exponent of the additive group of F , and by Theorem 8.42,

the primes dividing the exponent are the same as the primes dividing the

order, and hence F must have cardinality pw for some positive integer w. 2

Of course, for every prime p, Zp is a ¬nite ¬eld of cardinality p. As we

shall see later (in Chapter 20), for every prime p and positive integer w,

there exists a ¬eld of cardinality pw . Later in this chapter, we shall see some

speci¬c examples of ¬nite ¬elds whose cardinality is not prime (Examples

9.35 and 9.47).

Exercise 9.4. Let R be a ring of characteristic m > 0, and let n be any

integer. Show that:

(a) if gcd(n, m) = 1, then n · 1R is a unit;

(b) if 1 < gcd(n, m) < m, then n · 1R is a zero divisor;

(c) otherwise, n · 1R = 0R .

218 Rings

Exercise 9.5. Let D be an integral domain, m ∈ Z, and a ∈ D. Show that

ma = 0D if and only if m is a multiple of the characteristic of D or a = 0D .

Exercise 9.6. For n ≥ 1, and for all a, b ∈ Zn , show that if a | b and b | a,

then a = bc for some c ∈ Z— . Thus, part (ii) of Theorem 9.4 may hold for

n

some rings that are not integral domains.

Exercise 9.7. This exercise depends on results in §8.6. Using the funda-

mental theorem of ¬nite abelian groups, show that the additive group of a

¬nite ¬eld of characteristic p and cardinality pw is isomorphic to Z—w .

p

9.1.3 Subrings

De¬nition 9.8. A subset S of a ring R is called a subring if

(i) S is a subgroup of the additive group R,

(ii) S is closed under multiplication, and

(iii) 1R ∈ S.

It is clear that the operations of addition and multiplication on a ring R

make a subring S of R into a ring, where 0R is the additive identity of S and

1R is the multiplicative identity of S. One may also call R an extension

ring of S.

Some texts do not require that 1R belongs to a subring S, and instead

require only that S contains a multiplicative identity, which may be di¬erent

than that of R. This is perfectly reasonable, but for simplicity, we restrict

ourselves to the case when 1R ∈ S.

Expanding the above de¬nition, we see that a subset S of R is a subring

if and only if 1R ∈ S and for all a, b ∈ S, we have

a + b ∈ S, ’a ∈ S, and ab ∈ S.

If fact, to verify that S is a subring, it su¬ces to show that ’1R ∈ S and

that S is closed under addition and multiplication; indeed, if ’1R ∈ S and S

is closed under multiplication, then S is closed under negation, and further,

1R = ’(’1R ) ∈ S.

Example 9.19. Z is a subring of Q. 2

Example 9.20. Q is a subring of R. 2

Example 9.21. R is a subring of C.

Note that for ± := a+bi ∈ C, with a, b ∈ R, we have ± = ± i¬ a+bi = a’bi

¯

i¬ b = 0. That is, ± = ± i¬ ± ∈ R. 2

¯

9.1 De¬nitions, basic properties, and examples 219

Example 9.22. The set Z[i] of complex numbers of the form a + bi, with

a, b ∈ Z, is a subring of C. It is called the ring of Gaussian integers.

Since C is a ¬eld, it contains no zero divisors, and hence Z[i] contains no

zero divisors. Hence, Z[i] is an integral domain.

Let us determine the units of Z[i]. If ± ∈ Z[i] is a unit, then there exists

± ∈ Z[i] such that ±± = 1. Taking norms, we obtain

1 = N (1) = N (±± ) = N (±)N (± ).

Clearly, the norm of a Gaussian integer is a non-negative integer, and so

N (±)N (± ) = 1 implies N (±) = 1. Now, if ± = a + bi, with a, b ∈ Z, then

N (±) = a2 + b2 , and so N (±) = 1 implies ± = ±1 or ± = ±i. Conversely, it

is clear that ±1 and ±i are indeed units, and so these are the only units in

Z[i]. 2

Example 9.23. Let m be a positive integer, and let Q(m) be the set of

rational numbers of the form a/b, where a and b are integers, and b is

relatively prime to m. Then Q(m) is a subring of Q, since for any a, b, c, d ∈ Z

with gcd(b, m) = 1 and gcd(d, m) = 1, we have

ac ad + bc ac ac

·=,

+= and

bd bd bd bd

and since gcd(bd, m) = 1, it follows that the sum and product of any two

element of Q(m) is again in Q(m) . Clearly, Q(m) contains ’1, and so it follows