Solutions to Assignment 3
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1 Solutions to Assignment 3 Question 1. [Exercises 3.1 # 2] Let R = {0 e b c} with addition multiplication defined by the following tables. Assume associativity distributivity show that R is a ring with identity. Is R commutative? Is R a field? + 0 e b c 0 0 e b c e e 0 c b b b c 0 e c c b e 0 0 e b c e 0 e b c b 0 b b 0 c 0 c 0 c Solution: We assume that the associative laws for addition multiplication hold in T also that the distributive laws hold in T. From the tables it is clear that T is closed under addition multiplication. Also the tables are symmetric about the main diagonal so that the commutative laws hold for both addition multiplication. Again from the tables it is clear that 0 is the additive identity e is the multiplicative identity. Also each element is its own additive inverse e is the only nonzero element in T which has a multiplicative inverse. Therefore with these definitions of addition multiplication T is a commutative ring with identity but is not a field. Question 2. [Exercises 3.1 # 10] Let Z[ i ] denote the set {a + b i a b Z}. Show that Z[ i ] is a subring of C. Solution: Since Z[ i ] C In order to show that Z[ i ] is a subring of the ring C it is enough to show that it is closed under addition multiplication contains the additive identity from C contains additive inverses. Note that if z 1 = a 1 + b 1 i z 2 = a 2 + b 2 i are in Z[ i ] then z 1 + z 2 = (a 1 + a 2 ) + (b 1 + b 2 )i Z[ i ] z 1 z 2 = (a 1 a 2 b 1 b 2 ) + (a 1 b 2 + a 2 b 1 )i Z[ i ] since a 1 b 1 a 2 b 2 Z Z is a ring. Also 0 C = 0 R + 0 R i = 0 Z + 0 Z i Z[ i ] since 0 Z = 0 R. Similarly if z = a + bi Z[ i ] then a + bi + ( a) + ( b)i = (a + ( a)) + (b + ( b))i = 0 Z + 0 Z i = 0 R + 0 R i = 0 C since 0 Z = 0 R so that Z[ i ] contains additive inverses.
2 Question 3. [Exercises 3.1 # 12]. Let T be the ring of continuous functions from R to R. Let S = {f T f(2) = 0}. Is S a subring of T? Solution: Since T is a ring under the operations of pointwise addition multiplication we need only show that S is closed under addition multiplication contains the additive identity additive inverses. Note that if f g S then (f + g)(2) = f(2) + g(2) = = 0 (f g)(2) = f(2) g(2) = 0 0 = 0 so that S is closed under addition multiplication. If 0 T is the additive identity in T then 0 T (t) = 0 for all t R in particular 0 T (2) = 0 so that 0 T S. Also if f S then the additive inverse f T satisfies ( f) + f = 0 T that is ( f)(t) + f(t) = 0 T (t) = 0 for all t R in particular ( f)(2) = 0 f(2) = 0 0 = 0 f S. Therefore S is a subring of T. Question 4. [Exercises 3.1 # 16]. Show that the subset R = { } of Z 18 is a subring. Does R have an identity? Solution: Note that using the addition multiplication from Z 18 the addition multiplication tables for R are given below We see from the tables above that R is subring of Z 18 but has no element that can act as a multiplicative identity since the product of any two elements is either 0 or 9.
3 Question 5. [Exercises 3.1 # 18]. Define a new addition multiplication on Z by a b = a + b 1 a b = a + b ab where the operations on the right-h side of the equal signs are ordinary addition subtraction multiplication. Prove that with the new operations Z is an integral domain. Solution: Clearly Z is closed under addition multiplication with these operations addition multiplication are commutative. To see that addition is associative we have so that a (b c) = (a b) c for all a b c Z. To see that multiplication is associative we have a (b c) = a + (b c) 1 = a + b + c 2 (a b) c = a b + c 1 = a + b + c 2 a (b c) = a + (b c) a(b c) = a + b + c bc a(b + c) + abc = a + b + c (ab + ac + bc) + abc (a b) c = a b + c (a b)c = a + b + c ab (a + b ab)c = a + b + c (ab + ac + bc) + abc so that a (b c) = (a b) c for all a b c Z. To show that the distributive laws hold since multiplication is commutative we only need to prove one of them holds. We have a (b c) = a + (b + c 1) a(b + c 1) = a + (1 a)(b + c 1) (a b) (a c) = a + b ab + a + c ac 1 = a + (1 a)(b + c 1) so that a (b c) = (a b) (a c) for all a b c Z. The additive identity is z = 1 since a z = a + z 1 = a The additive inverse of a Z is given by x = 2 a since a x = a + x 1 = a + 2 a 1 = 1 = z The multiplicative identity is e = 0 since a e = a + e a e = a + 0 a 0 = a Therefore Z with these operations of addition multiplication is a commutative ring with identity.
4 To see that these operations make Z into an integral domain suppose that a b Z a b = z then that is a b = a + b ab = 1 = z (1 a)(1 b) = 1 a b + ab = 0 since Z with the usual operations of addition multiplication is an integral domain this implies that either 1 a = 0 or 1 b = 0 so that either a = 1 = z or b = 1 = z. Question 6. [Exercises 3.1 # 24]. The addition table part of the multiplication table for a three-element ring are given below. Use the distributive laws to complete the multiplication table. + r s t r r s t s s t r t t r s r s t r r r r s r t t r Solution: The completed tables are given below. + r s t r r s t s s t r t t r s r s t r r r r s r t s t r s t The entries were filled in by noting that r is the additive identity in the ring from the addition table we have r = s r = s(s + t) = s s + s t = t + s t so that s t = t = s. Similarly so that t s = t = s. Finally so that t t = s = t. r = r s = (s + t)s = s s + t s = t + t s r = r t = (s + t)t = s t + t t = s + t t Question 7. [Exercises 3.2 # 2]. An element e of a ring is said to be idempotent if e 2 = e. (a) Find four idempotent elements in the ring M 2 (R). (b) Find all idempotents in Z 12. (c) Prove that the only idempotents in an integral domain are O R 1 R.
5 Solution: (a) The matrices ( ) ( ) ( ) ( ) are all idempotent as is easily checked. (b) In Z 12 we have 0 0 = = = = = = = = = = = = 1 there are 4 idempotents in Z 12 namely (c) If R is an integral domain a R is an idempotent then a 2 = a implies that a(a 1 R ) = 0 R since R is an integral domain then either a = 0 R or a 1 R = 0 R that is either a = 0 R or a = 1 R. Question 8. [Exercises 3.2 # 12]. (a) Prove that [ a ] is a unit in Z n if only if (a n) = 1 in Z. (b) Prove that [ a ] is a nonunit in Z n if only if [ a ] is a zero divisor. Solution: (a) Suppose that [ a ] is a unit in Z n then there exists a u Z such that [ a ] [ u ] = [ 1 ] in Z n that is au 1 (mod n) so there exists an integer k such that au = 1 + kn from this it is clear that any common positive divisor of a n must also divide 1 therefore (a n) = 1. Conversely if (a n) = 1 then there exist integers u v such that au + nv = 1 therefore au 1 (mod n) so that [ a ] is a unit in Z n. [ a ] [ u ] = [ 1 ] (b) Suppose that [ a ] is a zero divisor in Z n then [ a ] [ 0 ] there exists b Z such that [ b ] [ 0 ] but [ a ] [ b ] = [ 0 ]. Therefore n a there exists b Z such that n b but n a b. This implies that (a n) > 1 since if (a n) = 1 n a b this implies that n b which is a contradiction. From part (a) since (a n) > 1 then [ a ] is a nonunit in Z n. Conversely if [ a ] [ 0 ] is a nonunit in Z n then (a n) > 1 if we let d = (a n) b = n d then ab = a n d = a d n so that n a n b but n ab so that [ a ] [ 0 ] [ b ] [ 0 ] but [ a ] [ b ] = [ 0 ] so that [ a ] is a zero divisor in Z n.
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