﻿ ﻿INTEGRAL DOMAINS - A Book of Abstract Algebra

## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 20. INTEGRAL DOMAINS

Let us recall that an integral domain is a commutative ring with unity having the cancellation property, that is,

if a ≠ 0 and ab = ac then b = c (1)

At the end of Chapter 17 we saw that an integral domain may also be defined as a commutative ring with unity having no divisors of zero, which is to say that

if ab = 0 then a = 0 or b = 0 (2)

for as we saw, (1) and (2) are equivalent properties in any commutative ring.

The system of the integers is the exemplar and prototype of integral domains. In fact, the term “integral domain” means a system of algebra (“domain”) having integerlike properties. However, is not the only integral domain: there are a great many integral domains different from .

Our first few comments will apply to rings generally. To begin with, we introduce a convenient notation for multiples, which parallels the exponent notation for powers. Additively, the sum

a + a + ··· + a

of n equal terms is written as n · a. We also define 0 · a to be 0, and let (-n) · a = −(n · a) for all positive integers n. Then

m · a + n · a = (m + n)·a and m · (n · a) = (mn) · a

for every element a of a ring and all integers m and n. These formulas are the translations into additive notation of the laws of exponents given in Chapter 10.

If A is a ring, A with addition alone is a group. Remember that in additive notation the order of an element a in A is the least positive integer n such that n · a = 0. If there is no such positive integer n, then a is said to have order infinity. To emphasize the fact that we are referring to the order of a in terms of addition, we will call it the additive order of a.

In a ring with unity, if 1 has additive order n, we say the ring has “characteristic n.” In other words, if A is a ring with unity,

the characteristic of A is the least positive integer n such that If there is no such positive integer n, A has characteristic 0.

These concepts are especially simple in an integral domain. Indeed,

Theorem 1 All the nonzero elements in an integral domain have the same additive order.

PROOF: That is, every a ≠ 0 has the same additive order as the additive order of 1. The truth of this statement becomes transparently clear as soon as we observe that

n · a = a + a + · · · + a = 1a + · · · + 1a = (1 + · · · + 1)a = (n · 1)a

hence n · a = 0 iff n · 1 = 0. (Remember that in an integral domain, if the product of two factors is equal to 0, at least one factor must be 0.) ■

It follows, in particular, that if the characteristic of an integral domain is a positive integer n, then

n · x = 0

for every element x in the domain.

Furthermore,

Theorem 2 In an integral domain with nonzero characteristic, the characteristic is a prime number.

PROOF: If the characteristic were a composite number mn, then by the distributive law, Thus, either m · 1 = 0 or n · 1 = 0, which is impossible because mn was chosen to be the least positive integer such that (mn) · 1 = 0. ■

A very interesting rule of arithmetic is valid in integral domains whose characteristic is not zero.

Theorem 3 In any integral domain of characteristic p,

(a + b)p = ap + bp for all elements a and b

PROOF: This formula becomes clear when we look at the binomial expansion of (a + b)p. Remember that by the binomial formula, where the binomial coefficient It is demonstrated in Exercise L of Chapter 17 that the binomial formula is correct in every commutative ring.

Note that if p is a prime number and 0 < k < p, then because every factor of the denominator is less than p, hence p does not cancel out. Thus, each term of the binomial expansion above, except for the first and last terms, is of the form px, which is equal to 0 because the domain has characteristic p. Thus, (a + b)p = ap + bp. ■

It is obvious that every field is an integral domain: for if a ≠ 0 and ax = ay in a field, we can multiply both sides of this equation by the multiplicative inverse of a to cancel a. However, not every integral domain is a field: for example, is not a field. Nevertheless,

Theorem 4 Every finite integral domain is a field.

List the elements of the integral domain in the following manner:

0, 1, a1, a2, …, an

In this manner of listing, there are n + 2 elements in the domain. Take any ai, and show that it is invertible: to begin with, note that the products

ai0, ai1, aia1, aia2, …, aian

are all distinct: for if aix = aiy, then x = y. Thus, there are n + 2 distinct products aix; but there are exactly n + 2 elements in the domain, so every element in the domain is equal to one of these products. In particular, 1 = aix for some x; hence ai is invertible.

OPTIONAL

The integral domain is not a field because it does not contain the quotients m/n of integers. However, can be enlarged to a field by adding to it all the quotients of integers; the resulting field, of course, is , the field of the rational numbers. consists of all quotients of integers, and it contains (or rather, an isomorphic copy of ) when we identify each integer n with the quotient n/1. We say that is the field of quotients of .

It is a fascinating fact that the method for constructing from can be applied to any integral domain. Starting from any integral domain A, it is possible to construct a field which contains A: a field of quotients of A. This is not merely a mathematical curiosity, but a valuable addition to our knowledge. In applications it often happens that a system of algebra we are dealing with lacks a needed property, but is contained in a larger system which has that property—and that is almost as good! In the present case, A is not a field but may be enlarged to one.

Thus, if A is any integral domain, we will proceed to construct a field A* consisting of all the quotients of elements in A; and A* will contain A, or rather an isomorphic copy of A, when we identify each element a of A with the quotient a/1. The construction will be carefully outlined and the busy work left as an exercise.

Given A, let S denote the set of all ordered pairs (a, b) of elements of A, where b ≠ 0. That is,

S = {(a, b): a,bA and b ≠ 0}

In order to understand the next step, we should think of (a, b) as a/b. [It is too early in the proof to introduce fractional notation; nevertheless each ordered pair (a, b) should be thought of as a fraction a/b.} Now a problem of representation arises here, because it is obvious that the quotient xa/xb is equal to the quotient a/b; to put the same fact differently, the quotients a/b and c/d are equal whenever ad = bc. That is, if ad = bc, then a/b and c/d are two different ways of writing the same quotient. Motivated by this observation, we define (a, b) ~ (c, d) to mean that ad = bc, and easily verify that ~ is an equivalence relation on the set 5. (Equivalence relations are explained in Chapter 12.) Then we let [a, b] denote the equivalence class of (a, b), that is,

[a,b] = {(c,d) ∈ S : (c, d) ~ (a, b)}

Intuitively, all the pairs which represent a given quotient are lumped together in one equivalence class; thus, each quotient is represented by exactly one equivalence class.

Let us recapitulate the formal details of our construction up to this point: Given the set S of ordered pairs of elements in A, we define an equivalence relation — in S by letting (a, b) ~ (c, d) iff ad= bc. We let [a, b] designate the equivalence class of (a, b), and finally, we let A* denote the set of all the equivalence classes [a, b]. The elements of A* will be called quotients.

Before going on, observe carefully that

[a,b] = [r,s] iff (a, b)~(r,s) iff as = br (3)

As our next step, we define operations of addition and multiplication in A*:

[a, b] + [c, d] = [ad + bc, bd]

and

[a, b] · [c, d] = [ac, bd]

To understand these definitions, simply remember the formulas We must make certain these definitions are unambiguous; that is, if [a, b] = [r, s] and [c, d] = [t, u], we have equations and we must therefore verify that [ad + bc, bd] = [ru + st, su] and [ac, bd] = [rt, su]. This is left as an exercise. It is also left for the student to verify that addition and multiplication are associative and commutative and the distributive law is satisfied.

The zero element is [0,1], because [a, b] + [0,1] = [a, b]. The negative of [a, b] is [-a, b], for [a, b] + [-a, b] = [0, b2] = [0,1]. [The last equation is true because of Equation (3).] The unity is [1,1], and the multiplicative inverse of [a, b] is [b,a], for [a, b] · [b, a] = [ab, ab] = [1,1]. Thus, A* is a field!

Finally, if A′ is the subset of A* which contains every [a, 1], we let ϕ be the function from A to A′ defined by ϕ(a) = [a, 1]. This function is injective because, by Equation (3), if [a, 1] = [b, 1] then a = b. It is obviously surjective and is easily shown to be a homomorphism. Thus, ϕ is an isomorphism from A to A′ so A* contains an isomorphic copy A′ of A.

EXERCISES

A. Characteristic of an Integral Domain

Let A be a finite integral domain. Prove each of the following:

1 Let a be any nonzero element of A. If n · a = 0, where n ≠ 0, then n is a multiple of the characteristic of A.

2 If A has characteristic zero, n ≠ 0, and n · a = 0, then a = 0.

3 If A has characteristic 3, and 5 · a = 0, then a = 0.

4 If there is a nonzero element a in A such that 256 · a = 0, then A has characteristic 2.

5 If there are distinct nonzero elements a and b in A such that 125 · a = 125 · b, then A has characteristic 5.

6 If there are nonzero elements a and in A such that (a + b)2 = a2 + b2, then A has characteristic 2.

7 If there are nonzero elements a and b in A such that 10a = 0 and 14b = 0, then A has characteristic 2.

B. Characteristic of a Finite Integral Domain

Let A be an integral domain. Prove each of the following:

1 If A has characteristic q, then q is a divisor of the order of A.

2 If the order of A is a prime number p, then the characteristic of A must be equal to p.

3 If the order of A is pm, where p is a prime, the characteristic of A must be equal to p.

4 If A has 81 elements, its characteristic is 3.

5 If A, with addition alone, is a cyclic group, the order of A is a prime number.

C. Finite Rings

Let A be a finite commutative ring with unity.

1 Prove: Every nonzero element of A is either a divisor of zero or invertible. (HINT: Use an argument analogous to the proof of Theorem 4.)

2 Prove: If a ≠ 0 is not a divisor of zero, then some positive power of a is equal to 1. (HINT: Consider a, a2, a3,.... Since A is finite, there must be positive integers n < m such that an = am.)

3 Use part 2 to prove: If a is invertible, then a1 is equal to a positive power of a.

D. Field of Quotients of an Integral Domain

The following questions refer to the construction of a field of quotients of A, as outlined on pages 203 to 205.

1 If [a, b] = [r, s] and [c, d] = [t, u], prove that [a, b] + [c, d] = [r, s] + [t, u].

2 If [a, b] = [r, s] and [c, d] = [t, u], prove that [a, b][c, d] = [r, s][t, u].

3 If (a, b) ~ (c, d) means ad = bc, prove that ~ is an equivalence relation on S.

4 Prove that addition in A* is associative and commutative.

5 Prove that multiplication in A* is associative and commutative.

6 Prove the distributive law in A*.

7 Verify that ϕ: A → A′ is a homomorphism.

E. Further Properties of the Characteristic of an Integral Domain

Let A be an integral domain. Prove parts 1-4:

1 Let aA. If A has characteristic p, and n · a = 0 where n is not a multiple of p, then a = 0.

2 If p is a prime, and there is a nonzero element aA such that p · a = 0, then A has characteristic p.

3 If p is a prime, and there is a nonzero element aA such that pm · a = 0 for some integer m, then A has characteristic p.

4 If A has characteristic p, then the function f(a) = ap is a homomorphism from A to A.

# 5 Let A have order p, where p is a prime. Explain why

A = {0,1,2·1,3·1,...,(p − 1).1}

Prove that A p.

# 6 If A has characteristic p, prove that for any positive integer n,

(a) (a + b)pn = apn + bpn

(b) 7 Let AB where A and B are integral domains. Prove: A has characteristic p iff B has characteristic p.

F. Finite Fields

By Theorem 4, “finite integral domain” and “finite field” are the same.

1 Prove: Every finite field has nonzero characteristic.

2 Prove that if A is a finite field of characteristic p, the function f(a) = ap is an automorphism of A; that is, an isomorphism from A to A. (HINT: Use Exercise E4 above and Exercise F7 of Chapter 18. To show that f is surjective, compare the number of elements in the domain and in the range of f.)

The function f(a) = ap is called the Froebenius automorphism.

3 Use part 2 to prove: In a finite field of characteristic p, every element has a p-th root.

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