Modern Algebra Math 542 Spring 2007 R. Pollack Solutions for HW #5. 1. Which of the following are examples of ring homomorphisms? Explain!

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1 Modern Algebra Math 542 Spring 2007 R. Pollack Solutions for HW #5 1. Which of the following are examples of ring homomorphisms? Explain! (a) φ : R R defined by φ(x) = 2x. This is not a ring homomorphism. For example, φ(1) φ(1) = 2 2 = 4, while φ(1 1) = φ(1) = 2. (b) φ : Z 5 Z 5 defined by φ(x) = x 5. This is a ring homomorphism. Indeed x 5 = x (mod 5) by Fermat s little theorem so this map is simply the identity map. (c) φ : Q Q Q defined by φ((a, b)) = a. (d) φ : C(R, R) R defined by φ(f(x)) = f(1). (( )) a b (e) φ : M 2 (Z) Z defined by φ = a. c d This is not a ring homomorphism. For example, (( ) ( )) (( )) 2 2 φ = φ = 2, 2 2 while φ (( )) (( )) φ = 1 1 = 1 ( ) a b (f) φ : C M 2 (R) defined by φ(a + bi) =. b a

2 2. Let φ : R S be a ring homomorphism. Prove that φ(0) = 0. On the one hand φ(0 + 0) = φ(0) as 0 is the additive identity. On the other hand, φ(0 + 0) = φ(0) + φ(0) since φ is a homomorphism. Thus, φ(0) = φ(0) + φ(0). Adding φ(0) to both sides then gives φ(0) = Let φ : R S be a ring homomorphism. Prove that φ( x) = φ(x). We have φ(0) = φ(x + x) = φ(x) + φ( x) since φ is a homomorphism. By the previous question, φ(0) = 0. Thus, φ(x) + φ( x) = 0. Adding φ(x) to both sides then gives φ( x) = φ(x). 4. Let φ : R S be a ring homomorphism. Recall that φ(r) = {s S : s = φ(r) for some r R}. Prove that φ(r) is a subring of S. Using the subring test, let s 1, s 2 φ(r). Then there exists r 1, r 2 R such that φ(r 1 ) = s 1 and φ(r 2 ) = s 2. We have s 1 s 2 = φ(r 1 ) φ(r 2 ) = φ(r 1 r 2 ) which by definition is in φ(r). The check for multiplication follows the along the same line.

3 5. Let φ : R S be a ring homomorphism. If A R is an ideal, is it true that φ(a) is an ideal of S? Prove this or give a counter-example. This is false. Consider R = Z and S = C. We can then take φ to be the inclusion of Z into C. If we take A = Z, then φ(a) is again Z viewed as a subset of C. Note that Z is not an ideal of C as 1 Z and i C, but 1 i = i / Z. 6. Let φ : R S be a ring homomorphism. Let B S be an ideal. We define φ 1 (B) = {r R : φ(r) B}. Prove this or give a counter- Is it true that φ 1 (B) is an ideal of R? example. This is true. To prove it recall the following defintion. x φ 1 (B) if and only if φ(x) B. Let x, y φ 1 (B). We must show that the x y φ 1 (B) or equivalently that φ(x y) B. Since x, y φ 1 (B), we have φ(x) B and φ(y) B. Thus φ(x y) = φ(x) φ(y) B as B is an ideal. Therefore x y φ 1 (B). Now let x φ 1 (B) and r R. We must check that rx φ 1 (B) or equivalently that φ(rx) B. Since x φ 1 (B), we have φ(x) B. Thus φ(rx) = φ(r)φ(x) B as B is an ideal and φ(r) S. Therefore, rx φ 1 (B). Chapter 15: 21. Determine all ring homomorphisms from Z to Z. Let φ : Z Z be a ring homomorphism. Note that for n Z, φ(n) = nφ(1). Thus φ is completely determined by its value on 1. Since 1 is an idempotent in Z (i.e. 1 2 = 1), we know by question #24 that φ(1) is again idempotent. Let s determine all of the idempotents of Z. To this end, take x Z such that x 2 = x. Thus x 2 x = x(x 1) = 0.

4 Since Z is an integral domain, we deduce that either x = 0 or x = 1. Thus the complete list of idempotents of Z are 0 and 1. Thus φ(1) being idempotent implies that either φ(1) = 0 or φ(1) = 1. In the first case, φ(n) = 0 for all n and in the second case φ(n) = n for all n. Thus, the only ring homomorphisms from Z to Z are the zero map and the identity map. 22. Suppose φ is a ring homomorphism from Z Z to Z Z. What are the possibilities for φ((1, 0))? Note that (1, 0) 2 = (1 2, 0) = (1, 0) and thus (1, 0) is idempotent. By question #24, we then have that φ((1, 0)) is idempotent. So let s determine all idempotents of Z Z. So take (x, y) Z Z such that (x, y) 2 = (x, y). Then x 2 = x and y 2 = y and by the solution to #21, we have that x = 0 or 1 and y = 0 or 1. Thus the only possibilities for φ((1, 0)) are (0, 0),(1, 0),(0, 1) and (1, 1). 23. Determine all ring homomorphisms from Z Z to Z Z. A ring homomorphism from Z Z to Z Z is determined by its values on (1, 0) and (0, 1). Indeed, (a, b) = a(1, 0) + b(0, 1) and thus φ((a, b)) = aφ((1, 0)) + bφ((0, 1)). By question #22 the only possible values for φ((1, 0)) are (0, 0),(1, 0),(0, 1) and (1, 1). Similarly, the only possible values for φ((0, 1)) are these same 4 values. Thus, in total there are at most 16 possible ring homomorphisms from Z Z to Z Z. However, not all of these 16 maps are ring homomorphisms. Let s make a list: 1. φ((1, 0)) = (1, 0), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (a + b, 0). 2. φ((1, 0)) = (0, 1), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (b, a). 3. φ((1, 0)) = (1, 1), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (a + b, a). 4. φ((1, 0)) = (0, 0), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (b, 0). 5. φ((1, 0)) = (1, 0), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (a, b). 6. φ((1, 0)) = (0, 1), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (0, a + b). 7. φ((1, 0)) = (1, 1), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (a, a + b). 8. φ((1, 0)) = (0, 0), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (0, b). 9. φ((1, 0)) = (1, 0), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (a + b, b).

5 10. φ((1, 0)) = (0, 1), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (b, a + b). 11. φ((1, 0)) = (1, 1), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (a + b, a + b). 12. φ((1, 0)) = (0, 0), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (b, b). 13. φ((1, 0)) = (1, 0), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (a, 0). 14. φ((1, 0)) = (0, 1), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (0, a). 15. φ((1, 0)) = (1, 1), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (a, a). 16. φ((1, 0)) = (0, 0), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (0, 0). Note that if φ is going to be a ring homomorphism, it must send (1, 1) to an idempotent (as (1, 1) is clearly idempotent). Looking at the above 16 maps, we have that maps 1,3,6,7,9,10 and 11 don t have this property and thus can t be ring homomorphisms. (For instance, the first map sends (1, 1) to (2, 0) which isn t an idempotent.) This leaves us with the 9 maps: φ((a, b)) = (a, b), (b, a), (b, 0), (0, b), (b, b), (a, 0), (0, a), (a, a) or (0, 0) which one can verify are in fact all ring homomorphisms. 24. Prove that a ring homomorphism carries an idempotent to an idempotent. Let a R be an idempotent. This means that a 2 = a. We need to show that φ(a) is an idempotent. We have and thus φ(a) is idempotent. φ(a) 2 = φ(a 2 ) = φ(a)