6.2 Ring Homomorphisms 303

6.2 Ring Homomorphisms

We turn our attention now to ring homomorphisms and their relations to ideals and quotient rings.

Definition 6.6 I Ring Homomorphism

If R and Rr are rings, a ring homomorphism from R to Rr is a mapping u: R S Rr such that u(x 1 y) 5u(x) 1u(y) and u(xy) 5u(x)u(y) for all x and y in R.

That is, a ring homomorphism is a mapping from one ring to another that preserves both ring operations. This situation is analogous to the one where a homomorphism from one group to another preserves the group operation, and it explains the use of the term homomorphism in both situations. It is sometimes desirable to use either the term group homomorphism or the term ring homomorphism for clarity, but in many cases, the context makes the meaning clear for the single word homomorphism. If only groups are under consideration, then homomorphism means group homomorphism, and if rings are under consideration, homomorphism means ring homomorphism. Some terminology for a special type of homomorphism is given in the following definition.

Definition 6.7 I Ring Epimorphism, Isomorphism

Let u be a homomorphism from the ring R to the ring Rr. 1. If u is onto, then u is called an epimorphism and Rr is called a homomorphic image of R. 2. If u is a one-to-one correspondence (both onto and one-to-one), then u is an isomorphism.

Example 1 Consider the mapping u: Z S Zn defined by u(a) 5 a .

Since 3 4 u(a 1 b) 5 a 1 b 5 a 1 b 5u(a) 1u(b)

and 3 4 3 4 3 4 u(ab) 5 ab 5 a b 5u(a)u(b)

for all a and b in Z, u is a homomorphism3 4 from3 43 Z4 to Zn. In fact, u is an epimorphism and Zn is a homomorphic image of Z. I 304 Chapter 6 More on Rings

Example 2 Consider u: Z6 S Z6 defined by u a 5 4 a .

It follows from 13 42 3 4 u a 1 b 5 4 a 1 b 5 4 a 1 4 b 13 4 3 42 13 4 3 42 5u a 1u b 3 4 3 4 that u preserves addition. For multiplication, we13 have42 13 42

u a b 5u ab 5 4 ab 5 4ab

and 13 4 3 42 13 42 3 4 3 4 u a u b 5 4 a 4 b 5 16 ab 5 16ab 5 4ab ,

since 16 5 4 in13Z426. 13Thus42 u1 is3 421 a homomorphism.3 42 3 4 3 It can4 be 3 verified4 that u(Z6) 5 0 , 2 , 4 ,and we see that u is neither onto nor one-to-one. I 3 4 3 4 53 4 3 4 3 46 Theorem 6.8 I Images of Zero and Additive Inverses

If u is a homomorphism from the ring R to the ring Rr, then a. u(0) 5 0, and b. u(2r) 5 2u(r) for all r [ R.

p ⇒ q Proof The statement in part a follows from u(0) 5u(0) 1 0 5u(0) 1u(0) 2u(0) 5u(0 1 0) 2u(0) 5u(0) 2u(0) 5 0. (p q) ⇒ r To prove part b, we observe that

¿ u(r) 1u(2r) 5ur 1 (2r) 5u(0) 3 4 5 0. Since the additive inverse is unique in the additive group of Rr, 2u(r) 5u(2r).

Under a ring homomorphism, images of subrings are subrings, and inverse images of subrings are also subrings. This is the content of the next theorem. 6.2 Ring Homomorphisms 305

Theorem 6.9 I Images and Inverse Images of Subrings

Suppose u is a homomorphism from the ring R to the ring Rr. a. If S is a subring of R, then u(S) is a subring of Rr. b. If Sr is a subring of Rr, then u21(Sr) is a subring of R.

(p q) ⇒ r Proof To prove part a, suppose S is a subring of R. We shall verify that the conditions of Theorem 5.3 are satisfied by u(S). The element u(0) 5 0 is in u(S), so u(S) is nonempty. ¿ Let xr and yr be arbitrary elements of u(S). Then there exist elements x, y [ S such that u(x) 5 xr and u(y) 5 yr. Since S is a subring, x 1 y and xy are in S. Therefore,

u(x 1 y) 5u(x) 1u(y) 5 xr 1 yr and

u(xy) 5u(x)u(y) 5 xryr

are in u(S), and u(S) is closed under addition and multiplication. Since 2x is in S and

u(2x) 52u(x) 52xr,

we have 2xr [ u(S), and it follows that u(S) is a subring of Rr. (p q) ⇒ r To prove part b, assume that Sr is a subring of Rr. We have 0 in u21(Sr) since u(0) 5 0, so u21(Sr) is nonempty. Let x [ u21(Sr) and y [ u21(Sr). This implies that u(x) [ Sr and ¿ u(y) [ Sr. Hence u(x) 1u(y) 5u(x 1 y) and u(x)u(y) 5u(xy) are in Sr, since Sr is a subring. Now

u(x 1 y) [ Sr ⇒ x 1 y [ u21(Sr)

and

u(xy) [ Sr ⇒ xy [ u21(Sr).

We also have

u(x) [ Sr ⇒ 2u(x) 5u(2x) [ Sr ⇒ 2x [ u21(Sr),

and u21(Sr) is a subring of R by Theorem 5.3.

Definition 6.10 I Kernel

If u is a homomorphism from the ring R to the ring Rr, the kernel of u is the set

ker u5 x [ R u(x) 5 0 .

5 0 6 306 Chapter 6 More on Rings

Example 3 In Example 1, the epimorphism u: Z S Zn is defined by u(a) 5 a . Now u(a) 5 0 if and only if a is a multiple of n, so 3 4 3 4 ker u5 c, 22n, 2n, 0, n, 2n, c for this u. 5 6 In Example 2, the homomorphism u: Z6 S Z6 defined by u a 5 4 a has kernel given by 13 42 3 4 ker u5 0 , 3 . I

In these two examples, ker u is an ideal53 of 4the3 46domain of u. This is true in general for homomorphisms, according to the following theorem.

Theorem 6.11 I Kernel of a Ring Homomorphism

If u is any homomorphism from the ring R to the ring Rr, then ker u is an ideal of R, and ker u5{0} if and only if u is one-to-one.

p ⇒ q Proof Under the hypothesis, we know that ker u is a subring of R from Theorem 6.9. For any x [ ker u and r [ R,we have u(xr) 5u(x)u(r) 5 0 ? u(r) 5 0, and similarly u(rx) 5 0. Thus xr and rx are in ker u, and ker u is an ideal of R. u ⇐ v Suppose u is one-to-one. Then x [ ker u implies u(x) 5 0 5u(0), and therefore x 5 0. Hence ker u5{0} if u is one-to-one. u ⇒ v Conversely, if ker u5{0}, then u(x) 5u(y) ⇒ u(x) 2u(y) 5 0 ⇒ u(x 2 y) 5 0 ⇒ x 2 y 5 0 ⇒ x 5 y. This means that u is one-to-one if ker u5{0}, and the proof is complete.

Example 4 This example illustrates the last part of Theorem 6.11 and provides a nice example of a ring isomorphism. For the set U 5 {a, b}, the power set of U is p(U) 5 {[, A, B, U}, where A 5 {a} and B 5 {b}. With addition defined by X 1 Y 5 (X c Y) 2 (X d Y) and multiplication by X ? Y 5 X d Y, p(U) forms a ring, as we saw in Example 5 of Section 5.1. Addition and multiplication tables for p(U) are given in Figure 6.1. 6.2 Ring Homomorphisms 307

1[ABU . [ ABU [[ABU [[[[[ AA[ UB A [ A [ A BBU[ A B [[BB UUBA[ U [ ABU I Figure 6.1

The ring R 5 Z2 { Z2 was introduced in Exercises 47 and 48 of Section 5.1. If we write 0 for 0 and 1 for 1 in Z2,the set R is given by R 5 {(0, 0), (1, 0), (0, 1), (1, 1)}. Addition and multiplication tables for R are displayed in Figure 6.2. 3 4 3 4

1 (0, 0) (1, 0) (0, 1) (1, 1) . (0, 0) (1, 0) (0, 1) (1, 1) (0, 0) (0, 0) (1, 0) (0, 1) (1, 1) (0, 0) (0, 0) (0, 0) (0, 0) (0, 0) (1, 0) (1, 0) (0, 0) (1, 1) (0, 1) (1, 0) (0, 0) (1, 0) (0, 0) (1, 0) (0, 1) (0, 1) (1, 1) (0, 0) (1, 0) (0, 1) (0, 0) (0, 0) (0, 1) (0, 1) (1, 1) (1, 1) (0, 1) (1, 0) (0, 0) (1, 1) (0, 0) (1, 0) (0, 1) (1, 1) I Figure 6.2

Consider the mapping u: p(U) S R defined by u([) 5 (0, 0), u(A) 5 (1, 0), u(B) 5 (0, 1), u(U) 5 (1, 1). If each element x in the tables for p(U) is replaced by u(x), the resulting tables agree com- pletely with those in Figure 6.2. Thus u is an isomorphism. We note that the kernel of u con- sists of the zero element in p(U). I

We know now that every kernel of a homomorphism from a ring R is an ideal of R. The next theorem shows that every ideal of R is a kernel of a homomorphism from R. This means that the ideals of R and the kernels of the homomorphisms from R to another ring are the same subrings of R.

Theorem 6.12 I Quotient Ring ⇒ Homomorphic Image

If I is an ideal of the ring R, the mapping u: R S R I defined by u(r) 5 r 1 >I is an epimorphism from R to R I with kernel I.

> 308 Chapter 6 More on Rings

p ⇒ q Proof It is clear that the rule u(r) 5 r 1 I defines an onto mapping u from R to R I and that ker u5I. Since > u(x 1 y) 5 (x 1 y) 1 I 5 (x 1 I) 1 (y 1 I) 5u(x) 1u(y) and u(xy) 5 xy 1 I 5 (x 1 I)(y 1 I) 5u(x)u(y), u is indeed an epimorphism from R to R I.

> The last theorem shows that every quotient ring of a ring R is a homomorphic image of R. A result in the opposite direction is given in the next theorem.

Strategy I In the proof of Theorem 6.13, it is shown that a certain rule defines a mapping f. When the defining rule for a possible mapping is stated in terms of a certain type of representation for the elements, the rule does not define a mapping unless the result is independent of the representation of the elements—that is, unless the rule is well-defined.

Theorem 6.13 I Homomorphic Image ⇒ Quotient Ring

If a ring Rr is a homomorphic image of the ring R, then Rr is isomorphic to a quotient ring of R.

p ⇒ q Proof Suppose u is an epimorphism from R to Rr, and let K 5 ker u. For each a 1 K in R K, define f(a 1 K) by

> f(a 1 K) 5u(a). To prove that this rule defines a mapping, let a 1 K and b 1 K be arbitrary elements of R K. Then

> a 1 K 5 b 1 K ⇔ a 2 b [ K ⇔ u(a 2 b) 5 0 ⇔ u(a) 5u(b) ⇔ f(a 1 K) 5f(b 1 K). This shows that f is well-defined and one-to-one as well. From the definition of f, it follows that f(R K) 5u(R). But u(R) 5 Rr, since u is an epimorphism. Thus f is onto and, consequently, is a one-to-one correspondence from R K to Rr. > > 6.2 Ring Homomorphisms 309

For arbitrary a 1 K and b 1 K in R K,

f (a 1 K) 1 (b 1 K) 5f>(a 1 b) 1 K 5u(a 1 b) 3 4 3 4 5u(a) 1u(b) since u is an epimorphism 5f(a 1 K) 1f(b 1 K) and f (a 1 K)(b 1 K) 5f(ab 1 K) 5u(ab) 3 4 5u(a)u(b) since u is an epimorphism 5f(a 1 K)f(b 1 K). Thus f is an isomorphism from R K to Rr.

As an immediate consequence> of the proof of this theorem, we have the following Fundamental Theorem of Ring Homomorphisms.

Theorem 6.14 I Fundamental Theorem of Ring Homomorphisms

If u is an epimorphism from the ring R to the ring Rr,thenRr is isomorphic to R ker u.

We now see that, in the sense of isomorphism, the homomorphic images of a ring> R are the same as the quotient rings of R. This gives a systematic way to search for all the homomorphic images of a given ring. To illustrate the usefulness of this method, we shall find all the homomorphic images of the ring Z of integers.

Example 5 In order to find all homomorphic images of Z, we shall find all possible ideals of Z and form all possible quotient rings. According to Theorem 6.3, every ideal of Z is a principal ideal. For the trivial ideal (0) 5 {0}, we obtain the quotient ring Z (0), which is isomorphic to Z, since a 1 (0) 5 b 1 (0) if and only if a 5 b. For the other trivial ideal (1) 5 Z, we obtain the quotient ring Z Z, which has only one element and >is isomorphic to {0}. As shown in the proof of Theorem 6.3, any nontrivial ideal I of Z has the form I 5 (n) for some > † positive integer n . 1. For these ideals, we obtain the quotient rings Z (n) 5 Zn. Thus the homomorphic images of Z are Z itself, {0}, and the rings Zn . I > Exercises 6.2 True or False Label each of the following statements as either true or false. 1. A ring homomorphism from a ring R to a ring Rr must preserve both ring operations. 2. If a homomorphism exists from a ring R to a ring Rr, then Rr is called a homomorphic image of R.

†See the paragraph immediately following Example 5 in Section 6.1.