Let us briefly recall some definitions. If R is a ring and r is a nonzero element in R, then r is said to be a zero divisor if there is some nonzero elements∈R such that rs= 0. A commutative ring with identity is said to be anintegral domain if it has no zero divisors. If an element a in a ring R with identity has a multiplicative inverse, we say that a is a unit. If every nonzero element in a ring R is a unit, then R is called a division ring. A commutative division ring is called afield.
Example 16.12. If i2 =−1, then the set Z[i] ={m+ni:m, n ∈Z} forms a ring known as theGaussian integers. It is easily seen that the Gaussian integers are a subring of the complex numbers since they are closed under addition and multiplication. Let α =a+bi be a unit inZ[i]. Thenα=a−biis also a unit since if αβ= 1, thenαβ= 1. Ifβ=c+di, then
1 =αβαβ= (a2+b2)(c2+d2).
Therefore, a2+b2 must either be 1 or −1; or, equivalently, a+bi = ±1 or a+bi = ±i. Therefore, units of this ring are ±1 and ±i; hence, the Gaussian integers are not a field. We will leave it as an exercise to prove that the Gaussian integers are an integral domain.
16.2. INTEGRAL DOMAINS AND FIELDS 185 Example 16.13. The set of matrices
F = {( 1 0 0 1 ) , ( 1 1 1 0 ) , ( 0 1 1 1 ) , ( 0 0 0 0 )}
with entries in Z2 forms a field.
Example 16.14. The set Q(√2 ) = {a+b√2 : a, b ∈ Q} is a field. The inverse of an elementa+b√2inQ(√2 ) is a a2−2b2 + −b a2−2b2 √ 2.
We have the following alternative characterization of integral domains.
Proposition 16.15 Cancellation Law. LetDbe a commutative ring with identity. Then Dis an integral domain if and only if for all nonzero elementsa∈Dwithab=ac, we have b=c.
Proof. Let D be an integral domain. Then D has no zero divisors. Let ab = ac with a̸= 0. Thena(b−c) = 0. Hence,b−c= 0 and b=c.
Conversely, let us suppose that cancellation is possible in D. That is, suppose that ab=acimplies b=c. Let ab= 0. If a= 0̸ , thenab=a0 orb= 0. Therefore, acannot be a zero divisor.
The following surprising theorem is due to Wedderburn. Theorem 16.16. Every finite integral domain is a field.
Proof. Let D be a finite integral domain and D∗ be the set of nonzero elements of D. We must show that every element in D∗ has an inverse. For each a∈D∗ we can define a mapλa :D∗→D∗ by λa(d) =ad. This map makes sense, because ifa̸= 0 andd̸= 0, then
ad̸= 0. The map λais one-to-one, since for d1, d2 ∈D∗,
ad1=λa(d1) =λa(d2) =ad2
implies d1 = d2 by left cancellation. Since D∗ is a finite set, the map λa must also be
onto; hence, for some d∈ D∗,λa(d) = ad = 1. Therefore,a has a left inverse. Since D is
commutative,dmust also be a right inverse fora. Consequently,D is a field.
For any nonnegative integer n and any element r in a ring R we write r+· · ·+r (n times) as nr. We define the characteristicof a ring R to be the least positive integer n such that nr = 0 for all r ∈ R. If no such integer exists, then the characteristic of R is defined to be 0. We will denote the characteristic ofR by charR.
Example 16.17. For every primep,Zp is a field of characteristic p. By Proposition 3.4, every nonzero element inZp has an inverse; hence,Zp is a field. Ifais any nonzero element
in the field, thenpa= 0, since the order of any nonzero element in the abelian group Zp is
p.
Lemma 16.18. Let R be a ring with identity. If 1 has order n, then the characteristic of R is n.
Proof. If1 has order n, thenn is the least positive integer such that n1 = 0. Thus, for all r∈R,
nr=n(1r) = (n1)r = 0r= 0.
On the other hand, if no positive nexists such that n1 = 0, then the characteristic ofR is zero.
Theorem 16.19. The characteristic of an integral domain is either prime or zero.
Proof. LetD be an integral domain and suppose that the characteristic of D is nwith n̸= 0. Ifn is not prime, then n=ab, where 1< a < n and 1< b < n. By Lemma16.18, we need only consider the case n1 = 0. Since 0 =n1 = (ab)1 = (a1)(b1)and there are no zero divisors inD, eithera1 = 0orb1 = 0. Hence, the characteristic ofDmust be less than n, which is a contradiction. Therefore,n must be prime.