INTEGRAL DOMAINS AND FIELDS

Proposition 8.7. If 0 = 1, the ring contains only one element and is called the trivial ring. All other rings are called nontrivial.

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 property 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

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 integral 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 field is a ring in which the nonzero elements form an abelian group under multiplication. In other words, a field 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 fields, but the integers do not form a field.

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

Proof. Let a · b = 0 in a field 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,

Proof. Let D = {x0, x1, x2, . . . , xn} be a finite integral domain with x0 as 0 and x1 as 1. We have to show that every nonzero element of D has a multiplicative 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

Note that Z is an infinite integral domain that is not a field.

Theorem 8.11. Zn is a field 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 finite, it follows from Theorem 8.10 that Zn is a field.

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 field.

√

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

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:

(i)a + b S.

(ii)−a S.

(iii)a · b S.

(iv)1 S.

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 definition above guarantee that S is closed under addition and multiplication. Condition (iv) shows that 1 S. It follows

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 matrices, 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)

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, defined by f (x) = [x], which maps an integer to its equivalence class modulo n, is a ring morphism from (Z, +, ·) to (Zn, +, ·).