﻿ ﻿QUOTIENT RINGS - A Book of Abstract Algebra

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

### Chapter 19. QUOTIENT RINGS

We continue our journey into the elementary theory of rings, traveling a road which runs parallel to the familiar landscape of groups. In our study of groups we discovered a way of actually constructing all the homomorphic images of any group G. We constructed quotient groups of G, and showed that every quotient group of G is a homomorphic image of G. We will now imitate this procedure and construct quotient rings.

We begin by defining cosets of rings:

Let A be a ring, and J an ideal of A. For any element a ∈ A, the symbol J + a denotes the set of all sums j + a, as a remains fixed and j ranges over J. That is,

J + a = {j + a : jJ}

J + a is called a coset of J in A.

It is important to note that, if we provisionally ignore multiplication, A with addition alone is an abelian group and J is a subgroup of A. Thus, the cosets we have just defined are (if we ignore multiplication) precisely the cosets of the subgroup J in the group A, with the notation being additive. Consequently, everything we already know about group cosets continues to apply in the present case—only, care must be taken to translate known facts about group cosets into additive notation. For example, Property (1) of Chapter 13, with Theorem 5 of Chapter 15, reads as follows in additive notation:

We also know, by the reasoning which leads up to Lagrange’s theorem, that the family of all the cosets J + a, as a ranges over A, is a partition of A.

There is a way of adding and multiplying cosets which works as follows:

In other words, the sum of the coset of a and the coset of b is the coset of a + b; the product of the coset of a and the coset of b is the coset of ab.

It is important to know that the sum and product of cosets, defined in this fashion, are determined without ambiguity. Remember that J + a may be the same coset as J + c [by Condition (1) this happens iff c is an element of J + a], and, likewise, J + b may be the same coset as J + d. Therefore, we have the equations

Obviously we must be absolutely certain that J + (a + b) = J + (c + d) and J + ab = J + cd. The next theorem provides us with this important guarantee.

Theorem 1 Let J be an ideal of A. If J + a = J + c and J + b = J + d, then

(i)J + (a + b) = J + (c + d), and

(ii)J + ab = J + cd.

PROOF: We are given that J + a = J + c and J + b = J + d; hence by Condition (2),

acJ and bdJ

Since J is closed with respect to addition, (ac) + (bd) = (a + b) − (c + d) is in J. It follows by Condition (2) that J + (a + b) = J + (c + d), which proves (i). On the other hand, since J absorbs products in A,

and therefore (abcb) + (cbcd) = abcd is in J. It follows by Condition (2) that J + ab = J + cd. This proves (ii). ■

Now, think of the set which consists of all the cosets of J in A. This set is conventionally denoted by the symbol A/J. For example, if J + a, J + b, J + c,… are cosets of J, then

A/J = {J + a, J + b, J + c,…}

We have just seen that coset addition and multiplication are valid operations on this set. In fact,

Theorem 2 A/J with coset addition and multiplication is a ring.

PROOF: Coset addition and multiplication are associative, and multiplication is distributive over addition. (These facts may be routinely checked.) The zero element of A/J is the coset J = J + 0, for if J + a is any coset,

(J + a) + (J + 0) = J + (a + 0) = J + a

Finally, the negative of J + a is J + (–a), because

(J + a) + (J + (–a)) = J + (a + (–a)) = J + 0 ■

The ring A/J is called the quotient ring of A by J.

And now, the crucial connection between quotient rings and homomorphisms :

Theorem 3 A/J is a homomorphic image of A.

Following the plan already laid out for groups, the natural homomorphism from A onto A/J is the function f which carries every element to its own coset, that is, the function f given by

f(x) = J + x

This function is very easily seen to be a homomorphism.

Thus, when we construct quotient rings of A, we are, in fact, constructing homomorphic images of A. The quotient ring construction is useful because it is a way of actually manufacturing homomorphic images of any ring A.

The quotient ring construction is now illustrated with an important example. Let be the ring of the integers, and let ⟨6⟩ be the ideal of which consists of all the multiples of the number 6. The elements of the quotient ring /⟨6⟩are all the cosets of the ideal (6), namely:

We will represent these cosets by means of the simplified notation . The rules for adding and multiplying cosets give us the following tables:

One cannot fail to notice the analogy between the quotient ring /⟨6⟩ and the ring 6. In fact, we will regard them as one and the same. More generally, for every positive integer n, we consider n to be the same as /⟨n⟩. In particular, this makes it clear that n is a homomorphic image of .

By Theorem 3, any quotient ring A/J is a homomorphic image of A. Therefore the quotient ring construction is a way of actually producing homomorphic images of any ring A. In fact, as we will now see, it is a way of producing all the homomorphic images of A.

Theorem 4 Let f : AB be a homomorphism from a ring A onto a ring B, and let K be the kernel of f. Then BA/B.

PROOF: To show that A/K is isomorphic with B, we must look for an isomorphism from A/K to B. Mimicking the procedure which worked successfully for groups, we let ϕ be the function from A/K to B which matches each coset K + x with the element f(x); that is,

ϕ(K + x) = f(x)

Remember that if we ignore multiplication for just a moment, A and B are groups and f is a group homomorphism from A onto B, with kernel K. Therefore we may apply Theorem 2 of Chapter 16: ϕ is a well-defined, bijective function from A/K to B. Finally,

Thus, ϕ is an isomorphism from A/K onto B. ■

Theorem 4 is called the fundamental homomorphism theorem for rings. Theorems 3 and 4 together assert that every quotient ring of A is a homomorphic image of A, and, conversely, every homomorphic image of A is isomorphic to a quotient ring of A. Thus, for all practical purposes, quotients and homomorphic images of a ring are the same.

As in the case of groups, there are many practical instances in which it is possible to select an ideal J of A so as to “factor out” unwanted traits of A, and obtain a quotient ring A/J with “desirable” features.

As a simple example, let A be a ring, not necessarily commutative, and let J be an ideal of A which contains all the differences

abba

as a and b range over A. It is quite easy to show that the quotient ring A/J is then commutative. Indeed, to say that A/J is commutative is to say that for any two cosets J + a and J + b,

(J + a)(J + b) = (J + b)(J + a) that is J + ab = J + ba

By Condition (2) this last equation is true iff abbaJ. Thus, if every difference abba is in J, then any two cosets commute.

A number of important quotient ring constructions, similar in principle to this one, are given in the exercises.

An ideal J of a commutative ring is said to be a prime ideal if for any two elements a and b in the ring,

If abJ then aJ or bJ

Whenever J is a prime ideal of a commutative ring with unity A, the quotient ring A/J is an integral domain. (The details are left as an exercise.)

An ideal of a ring is called proper if it is not equal to the whole ring. A proper ideal J of a ring A is called a maximal ideal if there exists no proper ideal K of A such that JK with JK (in other words, J is not contained in any strictly larger proper ideal). It is an important fact that if A is a commutative ring with unity, then J is a maximal ideal of A iff A/J is a field.

To prove this assertion, let J be a maximal ideal of A. If A is a commutative ring with unity, it is easy to see that A/J is one also. In fact, it should be noted that the unity of A/J is the coset J + 1, because if J + a is any coset, (J + a)(J + 1) = J + a1 = J + a. Thus, to prove that A/J is a field, it remains only to show that if J + a is any nonzero coset, there is a coset J + x such that (J + a)(J + x) = J + 1.

The zero coset is J. Thus, by Condition (3), to say that J + a is not zero, is to say that aJ. Now, let K be the set of all the sums

xa + j

as x ranges over A and j ranges over J. It is easy to check that K is an ideal. Furthermore, K contains a because a = 1a + 0, and K contains every element jJ because j can be written as 0a + j. Thus, K is an ideal which contains J and is strictly larger than J (for remember that aK but aJ). But J is a maximal ideal! Thus, K must be the whole ring A.

It follows that 1 ∈ K, so 1 = xa + j for some xA and jJ. Thus, 1 – xa = jJ, so by Condition (2), J + 1 = J + xa = (J + x)(J + a). In the quotient ring A/J, J + x is therefore the multiplicative inverse of J + a.

The converse proof consists, essentially, of “unraveling” the preceding argument; it is left as an entertaining exercise.

EXERCISES

A. Examples of Quotient Rings

In each of the following, A is a ring and J is an ideal of A. List the elements of A/J, and then write the addition and multiplication tables of A/J.

Example A = 6, J = {0, 3}.

The elements of A/J are the three cosets J = J + 0 = {0,3}, J + 1 = {1,4}, and J + 2 = {2,5}. The tables for A/J are as follows:

1 A = 10, J = {0,5}.

2 A = P3, J = {0, {a}}. (P3 is defined in Chapter 17, Exercise D.)

3 A = 2 × 6; J = {(0,0), (0,2), (0,4)}.

B. Examples of the Use of the FHT

In each of the following, use the FHT (fundamental homomorphism theorem) to prove that the two given groups are isomorphic. Then display their tables.

Example 2 and 6/⟨2⟩.

The following function is a homomorphism from 6 onto 2:

(Do not prove that J is a homomorphism.)

The kernel of f is {0, 2, 4} = (2). Thus:

It follows by the FHT that 26/⟨2⟩.

1 5 and 20/⟨5⟩.

2 3 and 6/⟨3⟩.

3 P2 and P3/K, where K = {0, {c}}. [HINT: See Chapter 18, Exercise E6. Consider the function f(X) = X ∩ {a, b}.]

4 2 and 2 × 2/K, where K = {(0, 0), (0,1)}.

C. Quotient Rings and Homomorphic Images in ()

1 Let ϕ be the function from () to × defined by ϕ(f) = (f(0), f(1)). Prove that ϕ is a homomorphism from () onto × , and describe its kernel.

2 Let J be the subset of () consisting of all f whose graph passes through the points (0,0) and (1,0). Referring to part 1, explain why J is an ideal of (), and ()/J × .

3 Let ϕ be the function from () to (, ) defined as follows:

ϕ(f) = f = the restriction of J to

(NOTE: The domain of f is and on this domain f is the same function as f.) Prove that ϕ is a homomorphism from () onto (, ), and describe the kernel of ϕ. [(, ) is the ring of functions from to .]

4 Let J be the subset of () consisting of all f such that f(x) = 0 for every rational x. Referring to part 3, explain why J is an ideal of () and ()/J().

D. Elementary Applications of the Fundamental Homomorphism Theorem

In each of the following let A be a commutative ring. If aA and n is a positive integer, the notation na will stand for

a + a + ⋯ + a (n terms)

1 Suppose 2x = 0 for every xA. Prove that (x + y)2 = x2 + y2 for all x and y in A. Conclude that the function h(x) = x2 is a homomorphism from A to A. If J = {xA : x2 = 0} and B = {x2 : xA), explain why J is an ideal of A, B is a subring of A, and A/JB.

2 Suppose 6x = 0 for every xA. Prove that the function h(x) = 3x is a homomorphism from A to A. If J = {x : 3x = 0} and B = {3x : xA}, explain why J is an ideal of A, B is a subring of A, and A/JB.

3 If a is an idempotent element of A (that is, a2 = a), prove that the function πa(x) = ax is a homomorphism from A into A. Show that the kernel of πa is Ia, the annihilator of a (defined in Exercise H4 of Chapter 18). Show that the range of πa is ⟨a⟩. Conclude by the FHT that A/Ia = ⟨a⟩.

4 For each aA, let πa be the function given by πa(x) = ax. Define the following addition and multiplication on = {πa : aA}:

πa + πb = πa+b and πa πb = πab

( is a ring; however, do not prove this.) Show that the function ϕ(a) = πa is a homomorphism from A onto . Let I designate the annihilating ideal of A (defined in Exercise H4 of Chapter 18). Use the FHT to show that A/I.

E. Properties of Quotient Rings A/J in Relation to Properties of J

Let A be a ring and J an ideal of A. Use Conditions (1), (2), and (3) of this chapter. Prove each of the following:

# 1 Every element of A/J has a square root iff for every xA, there is some yA such that xy2J.

2 Every element of A/J is its own negative iff x + xJ for every xA.

3 A/J is a boolean ring iff x2xJ for every xA. (A ring S is called a boolean ring iff s2 = s for every sS.)

4 If J is the ideal of all the nilpotent elements of a commutative ring A, then A/J has no nilpotent elements (except zero). (Nilpotent elements are defined in Chapter 17, Exercise M; by M2 and M3 they form an ideal.)

5 Every element of A/J is nilpotent iff J has the following property: for every xA, there is a positive integer n such that xnJ.

# 6 A/J has a unity element iff there exists an element aA such that axxJ and xaxJ for every xA.

F. Prime and Maximal Ideals

Let A be a commutative ring with unity, and J an ideal of A. Prove each of the following:

1 A/J is a commutative ring with unity.

2 J is a prime ideal iff A/J is an integral domain.

3 Every maximal ideal of A is a prime ideal. (HINT: Use the fact, proved in this chapter, that if J is a maximal ideal then A/J is a field.)

4 If A/J is a field, then J is a maximal ideal. (HINT: See Exercise I2 of Chapter 18.)

G. Further Properties of Quotient Rings in Relation to Their Ideals

Let A be a ring and J an ideal of A. (In parts 1–3 and 5 assume that A is a commutative ring with unity.)

# 1 Prove that A/J is a field iff for every element aA, where aJ, there is some bA such that ab – 1 ∈ J.

2 Prove that every nonzero element of A/J is either invertible or a divisor of zero iff the following property holds, where a, xA: For every aJ, there is some xJ such that either axJ or ax – 1 ∈ J.

3 An ideal J of a ring A is called primary iff for all a, bA, if abJ, then either aJ or bnJ for some positive integer n. Prove that every zero divisor in A/J is nilpotent iff J is primary.

4 An ideal J of a ring A is called semiprime iff it has the following property: For every aA, if anJ for some positive integer n, then necessarily aJ. Prove that J is semiprime iff A/J has no nilpotent elements (except zero).

5 Prove that an integral domain can have no nonzero nilpotent elements. Then use part 4, together with Exercise F2, to prove that every prime ideal in a commutative ring is semiprime.

H. Zn as a Homomorphic Image of Z

Recall that the function

f(a) = ā

is the natural homomorphism from onto n. If a polynomial equation p = 0 is satisfied in , necessarily f(p) = f(0) is true in n. Let us take a specific example; there are integers x and y satisfying 11x2 − 8y2 + 29 = 0 (we may take x= 3 and y = 4). It follows that there must be elements and in 6 which satisfy in 6, that is, . (We take and .) The problems which follow are based on this observation.

1 Prove that the equation x2 – 7y2 − 24 = 0 has no integer solutions. (HINT: If there are integers x and y satisfying this equation, what equation will and satisfy in 7?)

2 Prove that x2 + (x + 1)2 + (x + 2)2 = y2 has no integer solutions.

3 Prove that x2 + 10y2 = n (where n is an integer) has no integer solutions if the last digit of n is 2, 3, 7, or 8.

4 Prove that the sequence 3, 8, 13, 18, 23,… does not include the square of any integer. (HINT: The image of each number on this list, under the natural homomorphism from to 5, is 3.)

5 Prove that the sequence 2, 10, 18, 26,… does not include the cube of any integer.

6 Prove that the sequence 3, 11, 19, 27,… does not include the sum of two squares of integers.

7 Prove that if n is a product of two consecutive integers, its units digit must be 0, 2, or 6.

8 Prove that if n is the product of three consecutive integers, its units digit must be 0, 4, or 6.

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