MATH 113 - WORKSHEET 5 SOLUTIONS In problems (1)-(3), let (A; +) be an abelian group. An endomorphism of A is a group homomorphism φ : A ! A. Let End(A) = Hom(A; A) be the set of endomorphisms of A. Given two endomorphisms of A, φ and , we can add them to obtain a new function (φ + ): A ! A, given by (φ + )(a) = φ(a) + (a). As usual, we can also compose them to obtain a new function (φ ◦ ): A ! A. (1) Show that (End(A); +) is an abelian group. Here + is addition of functions as defined above. This includes checking that if φ, 2 End(A), then (φ + ) 2 End(A). Solution: + is well-defined: Let φ, 2 End(A). We need to show φ+ 2 End(A). Indeed, Let a; b 2 A. Then (φ + )(a + b) = φ(a + b) + (a + b) = φ(a) + φ(b) + (a) + (b) = φ(a) + (a) + φ(b) + (b) = (φ + )(a) + (φ + )(b): Associativity: Let φ, ; θ 2 End(A). Then for all a 2 A, ((φ + ) + θ)(a) = (φ + )(a) + θ(a) = φ(a) + (a) + θ(a) = φ(a) + ( + θ)(a) = (φ + ( + θ))(a): So (φ + ) + θ = φ + ( + θ). Identity: The identity element of End(A) is the zero map z : A ! A, given by z(a) = 0. This is an endomorphism, since z(a + b) = 0 = 0 + 0 = z(a) + z(b). And for all φ 2 End(A), for all a 2 A, (z + φ)(a) = z(a) + φ(a) = 0 + φ(a) = φ(a) = φ(a) + 0 = φ(a) + z(a) = (φ + z)(a): So z + φ = φ = φ + z. 1 Inverses: Let φ 2 End(A). Then −φ : A ! A, given by (−φ)(a) = −φ(a) is the inverse of φ. This is an endomorphism, since (−φ)(a + b) = −φ(a + b) = −(φ(a) + φ(b)) = −φ(a) + −φ(b) = (−φ)(a) + (−φ)(b). And for all a 2 A, (φ + −φ)(a) = φ(a) + −φ(a) = 0 = z(a) = −φ(a) + φ(a) = (−φ + φ)(a): So φ + −φ = z = −φ + φ. Commutativity: Let φ, 2 End(A). Then for all a 2 A,(φ+ )(a) = φ(a)+ (a) = (a) + φ(a) = ( + φ)(a). So φ + = + φ. (2) Given the result of Problem (1), show that (End(A); +; ◦) is a ring with unity. Note that this ring is not commutative in general, since function composition is not commutative in general. Solution: Multiplication is well-defined: This is the fact that the composition of two homomorphisms is a homomorphism. Here's the check: Let φ, 2 End(A). Let a; b 2 A. Then (φ ◦ )(a + b) = φ( (a + b)) = φ( (a) + (b)) = φ( (a)) + φ( (b)) = (φ ◦ )(a) + (φ ◦ )(b). Multiplication is associative: This is just associativity of composition of functions. Distributive laws: Let φ, ; θ 2 End(A). For all a 2 A, (φ ◦ ( + θ))(a) = φ(( + θ)(a)) = φ( (a) + θ(a)) = φ( (a)) + φ(θ(a)) = (φ ◦ )(a) + (φ ◦ θ)(a) = (φ ◦ + φ ◦ θ)(a): So φ ◦ ( + θ) = φ ◦ + φ ◦ θ. Also, ((φ + ) ◦ θ)(a) = (φ + )(θ(a)) = φ(θ(a)) + (θ(a)) = (φ ◦ θ)(a) + ( ◦ θ)(a) = (φ ◦ θ + ◦ θ)(a): So (φ + ) ◦ θ = φ ◦ θ + ◦ θ. Unity: The unity is the identity for function composition, which is the identity function. Of course this is an endomorphism of A. (3) Given a ring with unity R, let A(R) = (R; +) be the abelian group of R under addition. Note that as sets, A(R) = R. Given r 2 R, show that λr : A(R) ! A(R), defined by λr(a) = ra, is an endomorphism. Given the result of Problem (2), show that the map φ : R ! End(A(R)) given by φ(r) = λr is a one-to-one ring homomorphism. This result is the analogue of Cayley's theorem for rings. Cayley's theorem states that every group embeds into the permutation group of some set (in fact, its own underlying set). We have shown that every ring embeds into the endomorphism ring of some abelian group (in fact, its own underlying abelian group). Solution: Let a; b 2 A(R). Then λr(a + b) = r(a + b) = ra + rb = λr(a) + λr(b), so λr 2 End(A(R)). φ is a ring homomorphism: Let r; s 2 R. We need to show that φ(r + s) = φ(r) + φ(s), that is, that λr+s = λr + λs. Indeed, for all a 2 A(R), λr+s(a) = (r + s)a = ra + sa = λr(a) + λs(a). We also need to show that φ(rs) = φ(r)φ(s), that is, that λrs = λr ◦ λs. Indeed, for all a 2 A(R), λrs(a) = rsa = r(λs(a)) = λr(λs(a)) = (λr ◦ λs)(a). φ is one to one: Let r; s 2 R, and suppose that φ(r) = φ(s). Then λr = λs, so in particular r = r1 = λr(1) = λs(1) = s1 = s. (4) Let R be a ring with unity, and let U(R) = fr 2 R j r is a unitg. Show that (U(R); ·) is a group. This includes showing that U(R) is closed under multiplication. If F is a field, conclude that F ∗ = F n f0g is an abelian group. The group U(R) is called the group of units of R and is sometimes denoted R×. Solution: U(R) is closed under multiplication: Let u; v 2 U(R). Then u−1 and v−1 exist in R, and (uv)−1 = v−1u−1, by the usual check, so uv 2 U(R). Associativity: This is just associativity of multiplication in R. Identity: R is a ring with unity, and 1−1 = 1, so 1 2 U(R). Inverses: Let u 2 U(R). Then u−1 exists in R, and moreover u−1 2 U(R), since (u−1)−1 = u. If F is a field, then F ∗ = U(F ), since every nonzero element of F is a unit. Moreover, U(R) is abelian, since multiplication is commutative in F . ∗ (5) Consider the field Zp, p prime. Use Problem (4) (i.e. the fact that Zp is a group) to ∗ p−1 p show that for all a 2 Zp, a = 1, and a = a. This result (and its restatement in (6)) is called Fermat's little theorem. ∗ ∗ Solution: Note that Zp is a finite group of order (p−1). Let a 2 Zp. By Lagrange's theorem, a has order d for some d j (p − 1). So de = p − 1 for some e 2 Z. Then ap−1 = ade = (ad)e = 1e = 1. Then ap = ap−1a = 1a = a. (6) Let p be a prime number. Use Problem (5) to show that if n 2 Z, with p - n, then np−1 ≡ 1 (mod p) and np ≡ n (mod p). Use this result to compute the remainder of 36000002 when divided by 7. ∗ p−1 Solution: Since p - n, n 6= 0 in Zp, so n 2 Zp. Then by Problem (5), n = p−1 p p n = 1 and n = n = n in Zp. These equations translate to the congruences np−1 ≡ 1 (mod p) and np ≡ n (mod p). 36000002 ≡ 36000000 · 32 ≡ (36)1000000 · 9 ≡ 11000000 · 9 ≡ 9 ≡ 2 (mod 7)..