MATH 521A: Abstract Algebra Homework 6 Solutions 1. Let R; S be rings. Consider the embedding map f : R ! R × S given by f : r 7! (r; 0S). Prove that f is a homomorphism. 0 0 0 0 0 We have f(r + r ) = (r + r ; 0S) = (r + r ; 0S + 0S) = (r; 0S) + (r ; 0S) = f(r) + f(r ). We 0 0 0 0 also have f(rr ) = (rr ; 0S) = (rr ; 0S0S) = (r; 0S)(r ; 0S). 2. Let R; S be rings. Consider the projection map f : R × S ! R given by f :(r; s) 7! r. Prove that f is a homomorphism. We have f((r; s) + (r0; s0)) = f((r + r0; s + s0)) = r + r0 = f((r; s)) + f((r0; s0)), and f((r; s)(r0; s0)) = f((rr0; ss0)) = rr0 = f((r; s))f((r0; s0)). 3. We call a ring element x idempotent if x2 = x. Let R; S be rings, and f : R ! S a homomorphism. Suppose x 2 R is idempotent. Prove that f(x) is idempotent. Suppose that x is idempotent, i.e. x = x2. We have f(x) = f(x2) = f(xx) = f(x)f(x), so f(x)2 = f(x). 4. We call a ring element x nilpotent if there is some n 2 N such that xn = 0. Let R; S be rings, and f : R ! S a homomorphism. Suppose x 2 R is nilpotent. Prove that f(x) is nilpotent. n n Recall that f(0R) = 0S. Let x 2 R be nilpotent, i.e. x = 0R. We have f(x) = n f(x)f(x) ··· f(x) = f(xx ··· x) = f(x ) = f(0R) = 0S. 5. Let R; S be rings, and f : R ! S a homomorphism. Define the kernel of f, Kerf = fr 2 R : f(r) = 0Sg. Prove that Kerf is a subring of R. Because f(0R) = 0S, we have 0R 2 Kerf. Suppose that a; b 2 Kerf. Then f(a) = 0S; f(b) = 0S. We have f(a + b) = f(a) + f(b) = 0S + 0S = 0S, so a + b 2 Kerf; this proves additive closure. We also have f(ab) = f(a)f(b) = 0S0S = 0S, so ab 2 Kerf; this proves multiplica- tive closure. By a theorem, −f(a) = f(−a), so 0S = f(a) + f(−a). But f(a) = 0S since a 2 Kerf, so 0S = 0S + f(−a) = f(−a). Hence (−a) 2 Kerf. 6. Let R; S be rings, and f : R ! S a homomorphism. Prove that f is injective (one-to-one) if and only if Kerf = f0Rg. First, suppose that Kerf = f0Rg. Let a; b 2 R such that f(a) = f(b). But then f(a − b) = f(a) − f(b) = 0S, so a − b 2 Kerf. Since Kerf = f0Rg, in fact a − b = 0R, so a − b + b = 0R + b, so a = b. Next, suppose that c 2 Kerf with c 6= 0R. Then f(c) = f(0R) = 0S, so f is not injective. 7. Let R; S be rings, and f : R ! S a homomorphism. Suppose that S1 is a subring of S. −1 Prove that f (S1) = fr 2 R : f(r) 2 S1g is a subring of R. −1 First, 0S 2 S1 since every subring contains zero. Since f(0R) = 0S, we have 0R 2 f (S1). −1 Next, suppose a; b 2 f (S1). Then f(a); f(b) 2 S1. Since S1 is a subring, it is closed so −1 f(a + b) = f(a) + f(b) 2 S1, and f(ab) = f(a)f(b) 2 S1. Hence a + b; ab 2 f (S1). Since S1 is a subring, it contains additive inverses, so −f(a) 2 S1. By a theorem f(−a) = −f(a), −1 so −a 2 f (S1). 8. Let R; S; T be rings, and f : R ! S, g : S ! T two homomorphisms. Prove that g ◦ f : R ! T is a homomorphism. Let a; b 2 R. We have g(f(a + b)) = g(f(a) + f(b) = g(f(a)) + g((f(b)), because f, g are homomorphisms (respectively). Similarly, g(f(ab)) = g(f(a)f(b)) = g(f(a))g(f(b)). 9. Let R; S be rings, and f : R ! S an isomorphism. Let g = f −1, i.e. for all r 2 R, g(f(r)) = r and for all s 2 S, f(g(s)) = s. Prove that g : S ! R is an isomorphism. First, g is a bijection because the inverse of a bijection is a bijection (or we can prove it if we like). We have g(a + b) = g(f(g(a)) + f(g(b))) = g(f(g(a) + g(b))) = g(a) + g(b), where we use the homomorphism property of f for the second equality. Similarly, we have g(ab) = g(f(g(a))f(g(b))) = g(f(g(a)g(b))) = g(a)g(b). a 2b 10. Let S = f( b a ): a; b 2 Zg, which is a subringp of M2;2(Zp) (two-by-two matrices with integer entries). Prove that S is isomorphic to [ 2] = fa + b 2 : a; b 2 g, a subring of . Z p Z pR The notation gives a big hint; define f : S ! [ 2] via f :( a 2b ) 7! a + b 2. We Z pb a p a 2b a0 2b0 a+a0 2(b+b0) 0 0 0 check f ( ) + 0 0 = f 0 0 = (a + a ) + (b + b ) 2 = (a + b 2) + (a + p b a b a b+b a+a b0 2) = f (( a 2b )) + f a0 2b0 . We also have f ( a 2b ) a0 2b0 = f aa0+2bb0 2ab0+2a0b = b a pb0 a0 p p b a b0 a0 a0b+b0a 2bb0+aa0 (aa0 + 2bb0) + (ab0 + ba0) 2 = (a + b 2)(a0 + b0 2) = f (( a 2b )) f a0 2b0 . To prove in- p b a p b0 a0 jection, suppose f (( a 2b )) = f a0 2b0 . Then a + b 2 = a0 + b0 2, which rearranges as p b a b0 a0 a − a0 = (b0 − b) 2. If b0 = b, then a = a0 and ( a 2b ) = a0 2b0 . If b0 6= b, then we divide by p b a b0 a0 0 b − b and discover thatp 2p is rational, a contradiction.p Surjection is \obvious" due to the a 2b definitions: let a + b 2 2 Z[ 2], then f (( b a )) = a + b 2. 11. Recall the ring from HW4 #6: R has ground set Z and operations ⊕; defined as: a ⊕ b = a + b − 1; a b = a + b − ab Prove that R is isomorphic to Z. The hard part of this problem is finding the isomorphsim, which is f : R ! Z via f(x) = 1 − x. We first prove a bijection; if f(x) = f(x0) then 1 − x = 1 − x0 so x = x0. Also, for a 2 Z we have f(1 − a) = 1 − (1 − a) = a. Now, f(a ⊕ b) = f(a + b − 1) = 1 − (a + b − 1) = 2 − a − b = (1 − a) + (1 − b) = f(a) + f(b), and f(a b) = f(a + b − ab) = 1 − (a + b − ab) = 1 − a − b + ab = (1 − a)(1 − b) = f(a)f(b). 12. Recall the ring from HW4 #7: R has ground set Z and operations ⊕; defined as: a ⊕ b = a + b − 1; a b = ab − a − b + 2 Prove that R is isomorphic to Z. The hard part of this problem is finding the isomorphsim, which is f : R ! Z via f(x) = x − 1. We first prove a bijection; if f(x) = f(x0) then x − 1 = x0 − 1 so x = x0. Also, for a 2 Z we have f(a + 1) = (a + 1) − 1 = a. Now, f(a ⊕ b) = f(a + b − 1) = a + b − 2 = (a − 1) + (b − 1) = f(a) + f(b), and f(a b) = f(ab − a − b + 2) = ab − a − b + 2 − 1 = ab − a − b + 1 = (a − 1)(b − 1) = f(a)f(b)..