IDEALS AND HOMOMORPHISMS - A Book of Abstract Algebra

A Book of Abstract Algebra, Second Edition (1982)

Chapter 18. IDEALS AND HOMOMORPHISMS

We have already seen several examples of smaller rings contained within larger rings. For example, image is a ring inside the larger ring image, and image itself is a ring inside the larger ring image. When a ring B is part of a larger ring A, we call Ba subring of A. The notion of subring is the precise analog for rings of the notion of subgroup for groups. Here are the relevant definitions:

Let A be a ring, and B a nonempty subset of A. If the sum of any two elements of B is again in B, then B is closed with respect to addition. If the negative of every element of B is in B, then B is closed with respect to negatives. Finally, if the product of any two elements of B is again in B, then B is closed with respect to multiplication. B is called a subring of A if B is closed with respect to addition, multiplication, and negatives. Why is B then called a subring of A? Quite elementary:

If a nonempty subset B ⊆ A is closed with respect to addition, multiplication, and negatives, then B with the operations of A is a ring.

This fact is easy to check: If a, b, and c are any three elements of B, then a, b, and c are also elements of A because BA. But A is a ring, so

image

Thus, in B addition and multiplication are associative and the distributive law is satisfied. Now, B was assumed to be nonempty, so there is an element bB but B is closed with respect to negatives, so −b is also in B. Finally, B is closed with respect to addition; hence b + (−b) ∈ B. That is, 0 is in B. Thus, B satisfies all the requirements for being a ring.

For example, image is a subring of image because the sum of two rational numbers is rational, the product of two rational numbers is rational, and the negative of every rational number is rational.

By the way, if B is a nonempty subset of A, there is a more compact way of checking that B is a subring of A :

B is a subring of A if and only if B is closed with respect to subtraction and multiplication.

The reason is that B is closed with respect to subtraction iff B is closed with respect to both addition and negatives. This last fact is easy to check, and is given as an exercise.

Awhile back, in our study of groups, we singled out certain special subgroups called normal subgroups. We will now describe certain special subrings called ideals which are the counterpart of normal subgroups: that is, ideals are in rings as normal subgroups are in groups.

Let A be a ring, and B a nonempty subset of A. We will say that B absorbs products in A (or, simply, B absorbs products) if, whenever we multiply an element in B by an element in A (regardless of whether the latter is inside Bor outside B), their product is always in B. In other words,

for all bB and xA, xb and bx are in B.

A nonempty subset B of a ring A is called an ideal of A if B is closed with respect to addition and negatives, and B absorbs products in A.

A simple example of an ideal is the set image of the even integers. image is an ideal of image because the sum of two even integers is even, the negative of any even integer is even, and, finally, the product of an even integer by any integer is always even.

In a commutative ring with unity, the simplest example of an ideal is the set of all the multiples of a fixed element a by all the elements in the ring. In other words, the set of all the products

xa

as a remains fixed and x ranges over all the elements of the ring. This set is obviously an ideal because

image

and

y(xa) = (yx)a

This ideal is called the principal ideal generated by a, and is denoted by

a

As in the case of subrings, if B is a nonempty subset of A, there is a more compact way of checking that B is an ideal of A :

B is an ideal of A if and only if B is closed with respect to subtraction and B absorbs products in A.

We shall see presently that ideals play an important role in connection with homomorphism

Homomorphisms are almost the same for rings as for groups.

A homomorphism from a ring A to a ring B is a function f : A → B satisfying the identities

f(xl + x2) = f(x1) + f(x2)

and

f(x1x2) = f(x1)f(x2)

There is a longer but more informative way of writing these two identities:

1. If f(x1) = y1 and f(x2) = y2 then f(x1 + x2) = y1 + y2.

2. If f(x1) = y1 and f(x2) = y2 then f(x1 x2) = y1 y2

In other words, if f happens to carry xl to y1 and x2 to y2, then, necessarily, it must carry x1 + x2 to y1 + y2 and x1x2 to y1y2. Symbolically,

If image and image, then necessarily

image

One can easily confirm for oneself that a function f with this property will transform the addition and multiplication tables of its domain into the addition and multiplication tables of its range. (We may imagine infinite rings to have “nonterminating” tables.) Thus, a homomorphism from a ring A onto a ring B is a function which transforms A into B.

For example, the ring image6 is transformed into the ring image3 by

image

as we may verify by comparing their tables. The addition tables are compared on page 136, and we may do the same with their multiplication tables:

image

If there is a homomorphism from A onto B, we call B a homomorphic image of A. If f is a homomorphism from a ring A to a ring B, not necessarily onto, the range of/is a subring of B. (This fact is routine to verify.) Thus, the range of a ring homomorphism is always a ring. And obviously, the range of a homomorphism is always a homomorphic image of its domain.

Intuitively, if B is a homomorphic image of A, this means that certain features of A are faithfully preserved in B while others are deliberately lost. This may be illustrated by developing further an example described in Chapter 14. The parity ring P consists of two elements, e and o, with addition and multiplication given by the tables

image

We should think of e as “even” and o as “odd,” and the tables as describing the rules for adding and multiplying odd and even integers. For example, even + odd = odd, even times odd = even, and so on.

The function f: imageP which carries every even integer to e and every odd integer to o is easily seen to be a homomorphism from image to P this is made clear on page 137. Thus, P is a homomorphic image of image. Although the ring Pis very much smaller than the ring image, and therefore few of the features of image can be expected to reappear in P, nevertheless one aspect of the structure of image is retained absolutely intact in P, namely, the structure of odd and even numbers. As we pass from image to P, the parity of the integers (their being even or odd), with its arithmetic, is faithfully preserved while all else is lost. Other examples will be given in the exercises.

If f is a homomorphism from a ring A to a ring B, the kernel of f is the set of all the elements of A which are carried by f onto the zero element of B. In symbols, the kernel of f is the set

K = {xA : f(x) = 0}

It is a very important fact that the kernel of f is an ideal of A. (The simple verification of this fact is left as an exercise.)

If A and B are rings, an isomorphism from A to B is a homomorphism which is a one-to-one correspondence from A to B. In other words, it is an injective and surjective homomorphism. If there is an isomorphism from A to Bwe say that A is isomorphic to B, and this fact is expressed by writing

AB

EXERCISES

A. Examples of Subrings

Prove that each of the following is a subring of the indicated ring:

1 {x + imagey : x, yimage} is a subring of image.

2 {x + 21/3 y + 22/3 z : x, y, zimage} is a subring of image.

3 {x2y : x, yimage} is a subring of image.

# 4 Let image(image) be the set of all the functions from image to image which are continuous on (−∞, ∞) and let image(image) be the set of all the functions from image to image which are differentiable on (−∞, ∞). Then image(image) and image(image) are subrings of image(image).

5 Let image(image) be the set of all functions from image to image which are continuous on the interval [0,1]. Then image(image) is a subring of image(image), and image(image) is a subring of image(image).

6 The subset of image2(image) consisting of all matrices of the form

image

is a subring of image2(image).

B. Examples of Ideals

1 Identify which of the following are ideals of image × image, and explain: {(n, n) : nimage}; {(5n, 0) : nimage}; {(n, m) : n + m is even}; {(n, m) : nm is even}; {(2n, 3m) : n, mimage}.

2 List all the ideals of image12.

# 3 Explain why every subring of imagen is necessarily an ideal.

4 Explain why the subring of Exercise A6 is not an ideal.

5 Explain why image(image) is not an ideal of image(image).

6 Prove that each of the following is an ideal of image(image):

(a) The set of all f such that f(x) = 0 for every rational x.

(b) The set of all f such that f(0) = 0.

7 List all the ideals of P3. (P3 is defined in Chapter 17, Exercise D.)

8 Give an example of a subring of P3 which is not an ideal.

9 Give an example of a subring of image3 × image3 which is not an ideal.

C. Elementary Properties of Subrings

Prove parts 1–6:

1 A nonempty subset B of a ring A is closed with respect to addition and negatives iff B is closed with respect to subtraction.

2 Conclude from part 1 that B is a subring of A iff B is closed with respect to subtraction and multiplication.

3 If A is a finite ring and B is a subring of A, then the order of B is a divisor of the order of A.

# 4 If a subring B of an integral domain A contains 1, then B is an integral domain. (B is then called a subdomain of A.)

# 5 Every subring containing the unity of a field is an integral domain.

6 If a subring B of a field F is closed with respect to multiplicative inverses, then B is a field. (B is then called a subfield of F.)

7 Find subrings of image18 which illustrate each of the following:

(a) A is a ring with unity, B is a subring of A, but B is not a ring with unity.

(b) A and B are rings with unity, B is a subring of A, but the unity of B is not the same as the unity of A.

8 Let A be a ring, f : AA a homomorphism, and B = {xA : f(x) = x}. Prove that B is a subring of A.

9 The center of a ring A is the set of all the elements aA such that ax = xa for every xA. Prove that the center of A is a subring of A.

D. Elementary Properties of Ideals

Let A be a ring and J a nonempty subset of A.

1 Using Exercise C1, explain why J is an ideal of A iff J is closed with respect to subtraction and J absorbs products in A.

2 If A is a ring with unity, prove that J is an ideal of A iff J is closed with respect to addition and J absorbs products in A.

3 Prove that the intersection of any two ideals of A is an ideal of A.

4 Prove that if J is an ideal of A and 1 ∈ J, then J = A.

5 Prove that if J is an ideal of A and J contains an invertible element a of A, then J = A.

6 Explain why a field F can have no nontrivial ideals (that is, no ideals except {0} and F).

E. Examples of Homomorphisms

Prove that each of the functions in parts 1–6 is a homomorphism. Then describe its kernel and its range.

1 ϕ : image(image)→image given by ϕ(f) = f(0).

2 h : image × imageimage given by h(x, y) = x.

3 h : imageimage2(image) given by

image

4 h : image × imageimage2(image) given by

image

# 5 Let A be the set image × image with the usual addition and the following “multiplication”:

(a, b) image (c, d) = (ac, bc)

Granting that A is a ring, let f : Aimage2(image) be given by

image

6 h : PcPc given by h(A) = A image D, where D is a fixed subset of C.

7 List all the homomorphisms from image2 to image4; from image3 to image6.

F. Elementary Properties of Homomorphisms

Let A and B be rings, and f : AB a homomorphism. Prove each of the following:

1 f(A) = {f(x): xA} is a subring of B.

2 The kernel of f is an ideal of A.

3 f(0) = 0, and for every aA, f(−a) = −f(a).

4 f is injective iff its kernel is equal to {0}.

5 If B is an integral domain, then either f(l) = 1 or f(l) = 0. If f(l) = 0, then f(x) = 0 for every xA. If f(1) = 1, the image of every invertible element of A is an invertible element of B.

6 Any homomorphic image of a commutative ring is a commutative ring. Any homomorphic image of a field is a field.

7 If the domain A of the homomorphism f is a field, and if the range of f has more than one element, then f is injective. (HINT: Use Exercise D6.)

G. Examples of Isomorphisms

1 Let A be the ring of Exercise A2 in Chapter 17. Show that the function f(x) = x − 1 is an isomorphism from image to A hence imageA.

2 Let image be the following subset of image2(image):

image

Prove that the function

image

is an isomorphism from image to image. [REMARK: You must begin by checking that f is a well-defined function; that is, if a + bi = c + di, then f(a + bi) = f(c + di). To do this, note that if a + bi = c + di then ac = (db)i; this last equation is impossible unless both sides are equal to zero, for otherwise it would assert that a given real number is equal to an imaginary number.]

3 Prove that {(x, x) : ximage} is a subring of image × image, and show {(x, x) : ximage} ≅ image.

4 Show that the set of all 2 × 2 matrices of the form

image

is a subring of image2(image), then prove this subring is isomorphic to image.

For any integer k, let kimage designate the subring of image which consists of all the multiples of k.

5 Prove that image ∉ 2image then prove that 2image ∉ 3image. Finally, explain why if kl, then kimagelimage. (REMEMBER: How do you show that two rings, or groups, are not isomorphic?)

H. Further Properties of Ideals

Let A be a ring, and let J and K be ideals of A.

Prove parts 1-4. (In parts 2-4, assume A is a commutative ring.)

1 If J image K = {0}, then jk = 0 for every jJ and kK.

2 For any aA, Ia = {ax + j + k : xA, jJ, kK} is an ideal of A.

# 3 The radical of J is the set rad J = {aA : anJ for some nimage}. For any ideal J, rad J is an ideal of A.

4 For any aA, {xA : ax = 0} is an ideal (called the annihilator of a).

Furthermore, {xA : ax = 0 for every aA} is an ideal (called the annihilating ideal of A). If A is a ring with unity, its annihilating ideal is equal to {0}.

5 Show that {0} and A are ideals of A. (They are trivial ideals; every other ideal of A is a proper ideal.) A proper ideal J of A is called maximal if it is not strictly contained in any strictly larger proper ideal: that is, if JK, where Kis an ideal containing some element not in J, then necessarily K = A. Show that the following is an example of a maximal ideal: In image(image), the ideal J = {f : f(0) = 0}. [HINT: Use Exercise D5. Note that if gK and g(0) ≠ 0 (that is, gJ), then the function h(x) = g(x) − g(0) is in J hence h(x) − g(x) ∈ K. Explain why this last function is an invertible element of image(image).]

I. Further Properties of Homomorphisms

Let A and B be rings. Prove each of the following:

1 If f : AB is a homomorphism from A onto B with kernel K, and J is an ideal of A such that K image J then f(J) is an ideal of B.

2 If f : AB is a homomorphism from A onto B, and B is a field, then the kernel of f is a maximal ideal. (HINT: Use part 1, with Exercise D6. Maximal ideals are defined in Exercise H5.)

3 There are no nontrivial homomorphisms from image to image. [The trivial homomorphisms are f(x) = 0 and f(x) = x.]

4 If n is a multiple of m, then imagem is a homomorphic image of imagen.

5 If n is odd, there is an injective homomorphism from image2 into image2n.

† J. A Ring of Endomorphisms

Let A be a commutative ring. Prove each of the following:

1 For each element a in A, the function πa defined by πa(x) = ax satisfies the identity πa(x + y) = πa(x) + πa(y). (In other words, πa is an endomorphism of the additive group of A.)

2 πa is injective iff a is not a divisor of zero. (Assume a ≠ 0.)

3 πa is surjective iff a is invertible. (Assume A has a unity.)

4 Let image denote the set {πa : aA} with the two operations

[πa + πb](x) = πa(x) + πb(x) and πa πb = πaπb

Verify that image is a ring.

5 If ϕ : Aimage is given by ϕ(a) = πa, then ϕ is a homomorphism.

6 If A has a unity, then ϕ is an isomorphism. Similarly, if A has no divisors of zero then ϕ is an isomorphism.