Theorem HW7.1. properties: Prove proposition 68. It states: Let R be a ring. We have the following 1. The ring R only has one additive identity. That is, if 0 R with 0 +b = b+0 = b for every b R, then 0 = 0. 2. The ring R only has one multiplicative identity. That is, if 1 R with 1 b = b1 = b for every b R, then 1 = 1. 3. For every a R, there is only one additive inverse of a. That is, if b R with a+b = b+a = 0, then b = a. 4. For every a R, the additive inverse of the additive inverse of a is a itself. That is, ( a) = a. 5. For every a R, then the additive inverse of a is the product of the additive inverse of 1 and a. That is, ( 1)a = a. In particular, ( 1) 2 = 1. 6. More generally, for any a, b R, then ( a)b = ab = a( b). Note: There are two things to prove here! 7. For every a R, we have a 0 = 0 a = 0. 8. The cancellation law for addition holds. That is, if a, b, c R with a + b = a + c, then b = c. 9. The additive inverse of a sum is the sum of the additive inverses. That is, for a,b R, we have (a + b) = a + b. 10. For a,b,c R, if ab = ba = 1 and ac = 1, then b = c. Proof. 1. Since 0 is an additive identity, we have 0 = 0+0. Since 0 is an additive identity, we have 0 + 0 = 0. By transitivity of equality, we have 0 = 0. 2. Since 1 is a multiplicative identity, we have 1 = 1 1. Since 1 is a multiplicative identity, we have 1 1 = 1. By transitivity of equality, we have 1 = 1. 3. Suppose b and b are both additive inverses of a. Then, b = b + 0 so that b = b as desired. by additive identity = b + (a + b ) as b is an additive inverse of a = (b + a) + b by associativity of addition = 0 + b as b is an additive inverse of a = b by additive identity Page 1 of 5
4. We have that a + a = a + ( a) = 0 as a is the additive inverse of a. This means that a is an additive inverse of a. Since ( a) is the additive inverse of a, we have that a + ( ( a)) = ( a) + ( a) = 0. By the uniqueness of additive inverses (see #3), we have that a = ( a) as desired. 5. By the uniqueness of additive inverses, it suffices to show that ( 1)a + a = a + ( 1)a = 0. We have the first equality by the commutativity of addition, so we show the second holds. We have, a + ( 1)a = 1 a + ( 1) a by multiplicative identity = (1 + ( 1)) a by distributivity = 0 a by additive inverses = 0 by # 7 so that both equalities hold and ( 1)a = a as they are both additive inverses of a. Moreover, ( 1) 2 = ( 1)( 1) so letting a = 1, we see that ( 1) 2 is the additive inverse of 1, which is 1 by part # 4. 6. For the first equality, we have ( a)b = (( 1)a)b by # 5 = ( 1)(ab) by associativity of multiplication = ab by # 5. For the second equality, we show that a( b) is an additive inverse of ab so that ab = a( b) by part # 3. We have, ab+a( b) = a( b)+ab = a( b+b) = a 0 = 0 by commutativity of addition, distributivity, and # 7. Thus, a( b) = ab by the uniqueness of additive inverses (see # 3). 7. We have a 0 = a (0 + 0) = a 0 + a 0 by the distributive property and the fact that 0 is an additive identity. Adding (a 0) to both sides gives 0 = (a 0) + a 0 = (a 0) + a 0 + a 0 = a 0 so that a 0 = 0. To see that 0 a = 0, we repeat the argument. We have 0 a = (0 + 0) a = 0 a + 0 a by the distributive property and the fact that 0 is an additive identity. Adding (0 a) to both sides gives so that 0 a = 0 as well. 0 = (0 a) + 0 a = (0 a) + 0 a + 0 a = 0 a Page 2 of 5
8. Say a + b = a + c. Then, we have b = 0 + b so that b = c as desired. by additive identity = ( a + a) + b by additive inverse of a = a + (a + b) by associativity of addition = a + (a + c) by assumption = ( a + a) + c by associativity of addition = 0 + c by additive inverse of a = c by additive identity 9. We have (a +b) + (a +b) = (a +b)+ (a +b) as (a +b) is an additive inverse of a +b. We also have (a + b) + ( a + b) = ( a + b) + (a + b) by commutativity of addition = a + ( b + a) + b by associativity of addition = a + (a + b) + b by commutativity of addition = ( a + a) + ( b + b) by associativity of addition = 0 + 0 by additive inverses of a and b = 0 by additive identity so that a + b is also an additive inverse of a + b. By the uniqueness of additive inverses (see # 3), we have that a + b = (a + b). Here s an alternate proof: We have (a + b) = ( 1)(a + b) by # 5 = ( 1)a + ( 1)b by distributivity = a + b by # 5 so that (a + b) = a + b as desired. 10. We have b = b 1 by multiplicative identity = b (ac) by assumption that ac = 1 = (ba) c by associativity of multiplication = 1 c by assumption that ba = 1 = c by multiplicative identity Page 3 of 5
so that b = c as desired. Theorem HW 7.2. Let R be a ring. An element e R is called an idempotent if e 2 = e. The trivial idempotents are 0 and 1. (a) Give an example of a ring with non-trivial idempotents. (b) Prove that if e is an idempotent, so is 1 e. (c) Prove that if R has no zero divisors, then the only idempotents of R are the trivial ones. Proof. (a) There ( are) many possible examples here. The one I had in mind is M 2 (Z). There, the 1 0 element is an idempotent since 0 0 ( ) ( ) ( ) 1 0 1 0 1 0 =. 0 0 0 0 0 0 ( ) 1 0 However is neither the additive identity nor the multiplicative identity of M 0 0 2 (Z). (b) Suppose that e 2 = e in a ring R. Then, (1 e) 2 = (1 e) (1 e) = 1 (1 e) e (1 e) by distributivity = 1 e e + ( e) 2 by distributivity = 1 e e + e 2 see below for why ( a) 2 = a 2 for any a R = 1 e e + e by assumption that e 2 = e = 1 e by additive inverses and identity so that (1 e) 2 = 1 e and 1 e is also an idempotent. It remains to show that ( a) 2 = a 2 in any ring. We have ( a) 2 = ( a)( a) = (( 1)a)( a) by Proposition 68, part # 5 = ( 1)(a( a)) by associativity of multiplication = ( 1)( aa) by Proposition 68, part # 6 = ( aa) by Proposition 68, part # 5 = aa = a 2 by Proposition 68, part # 4 so that ( a) 2 = a 2 as desired, thus completing the proof. Page 4 of 5
(c) Suppose that R has no zero divisors, and let e R be an idempotent. Then, e(1 e) = e e 2 = e e = 0 so that e = 0 or 1 e = 0 and e = 1. Thus, the only idempotents in a ring with no zero divisors are 0 and 1. Page 5 of 5