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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 16.2 Integral Domains and Fields

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Let us briefly recall some definitions. If R is a commutative ring and r is a nonzero element in R, then r is said to be a zero divisor if there is some nonzero element s∈R such that rs=0. A commutative ring with identity is said to be an integral 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 a field.
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Example 16.12.

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If i2=−1, then the set Z[i]={m+ni:m,n∈Z} forms a ring known as the Gaussian 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 in Z[i]. Then α―=a−bi is also a unit since if αβ=1, then α―β―=1. If β=c+di, then
1=αβα―β―=(a2+b2)(c2+d2).
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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.
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Example 16.13.

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The set of matrices
F={(1001),(1110),(0111),(0000)}
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with entries in Z2 forms a field.
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Example 16.14.

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The set Q(2)={a+b2:a,b∈Q} is a field. The inverse of an element a+b2 in Q(2) is
aa2−2b2+−ba2−2b22.
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We have the following alternative characterization of integral domains.

Proof.

Let D be an integral domain. Then D has no zero divisors. Let ab=ac with a≠0. Then a(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=ac implies b=c. Let ab=0. If a≠0, then ab=a0 or b=0. Therefore, a cannot be a zero divisor.
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The following surprising theorem is due to Wedderburn.

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 map λa:D∗→D∗ by λa(d)=ad. This map makes sense, because if a≠0 and d≠0, then ad≠0. The map λa is 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, d must also be a right inverse for a. Consequently, D is a field.
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For any nonnegative integer n and any element r in a ring R we write r+⋯+r (n times) as nr. We define the characteristic of 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 of R by charR.
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Example 16.17.

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For every prime p, Zp is a field of characteristic p. By Proposition 3.4, every nonzero element in Zp has an inverse; hence, Zp is a field. If a is any nonzero element in the field, then pa=0, since the order of any nonzero element in the abelian group Zp is p.

Proof.

If 1 has order n, then n 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 n exists such that n1=0, then the characteristic of R is zero.

Proof.

Let D be an integral domain and suppose that the characteristic of D is n with n≠0. If n is not prime, then n=ab, where 1<a<n and 1<b<n. By Lemma 16.18, we need only consider the case n1=0. Since 0=n1=(ab)1=(a1)(b1) and there are no zero divisors in D, either a1=0 or b1=0. Hence, the characteristic of D must be less than n, which is a contradiction. Therefore, n must be prime.
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