Ring Theory Problem Set 1 { Solutions Problem 16.1 Let R be a ring with unity 1. Show that (−1)a = −a for all a 2 R. SOLUTION: We have 1+(−1) = 0 by definition. Multiplying that equation on the right by a, we obtain 1 + (−1) · a = 0 · a = 0 by theorem 16.1, part i. By the distributive law, we obtain the equation 1 · a + (−1) · a = 0 and therefore we have a+(−1)a = 0. We also have a+(−a) = 0. Thus, a+(−1)a = a+(−a). The ring R under addition is a group. The cancellation law in that group implies that −a = (−1)a which is the result we wanted to prove. Problem 16.7 Let F be a field and let a; b 2 F . Assume that a 6= 0, Show that there exists an element x 2 F satisfying the equation ax + b = 0. SOLUTION: Since F is a field and a 6= 0, there exists an element a−1 in F such that aa−1 = 1. Let c = −b. Let x = a−1c. Then x 2 F since both a−1 and c are in F . We have ax + b − = a(a−1c) + b = (aa−1)c + b = 1c + b = c + b = 0 : Hence the element x in F chosen above has the property that ax + b = 0. Problem 16.11 Find all units, zero-divisors, and nilpotent elements in the rings Z ⊕ Z, Z3 ⊕ Z3, and Z4 ⊕ Z6. SOLUTION; In general, if R1 and R2 are rings with unity, then so is R1 ⊕ R2. The unity element is (1R1 ; 1R2 ). An element (a1; a2) in R1 ⊕ R2 is a unit if and only if there is an element (b1; b2) in R1 ⊕ R2 such that (a1; a2)(b1; b2) = (1R1 ; 1R2 ). By definition, (a1; a2)(b1; b2) = (a1b1; a2b2). Therefore, the element (a1; a2) is a unit if and only if there exists elements b1 2 R1 and b2 2 R2 such that a1b1 = 1R1 and a2b2 = 1R2 . This means that (a1; a2) is a unit in R1 ⊕ R2 if and only if a1 is a unit in R1 and a2 is a unit in R2. The units in Z are 1 and -1. The units in Z3 are 1 and 2. The units in Z4 are 1 and 3. The units in Z6 are 1 and 5. Therefore, The units in Z ⊕ Z are (1; 1); (1; −1); (−1; 1), and (−1; −1). The units in Z3 ⊕ Z3 are (1; 1); (1; 2); (2; 1), and (2; 2). The units in Z4 ⊕ Z6 are (1; 1); (1; 5); (3; 1), and (3; 5). Suppose that (a1; a2) is an element of R1 ⊕ R2 and that n is a positive integer. Then n n n we clearly have (a1; a2) = (a1 ; a2 ). The additive identity in R1 ⊕ R2 is (0R1 ; 0R2 ). The n n n equation (a1; a2) = (0R1 ; 0R2 ) is equivalent to the two equations a1 = 0R1 and a2 = 0R2 . Consequently, if (a1; a2) is a nilpotent element of R1 ⊕ R2, then it follows that a1 is a nilpotent element in R1 and a2 is a nilpotent element in R2. The converse is true too. To see this, assume that a1 is a nilpotent element in R1 and a2 is a nilpotent element in R2. e f Then, by definition, there exists positive integers e and f such that a1 = 0R1 and a2 = 0R2 . Let n = ef = fe. Then n is a positive integer and we have an = aef = (ae)f = 0f = 0 and an = afe = (af )e = 0e = 0 1 1 1 R1 R1 2 2 2 R2 R2 n Therefore, (a1; a2) = (0R1 ; 0R2 ) and hence (a1; a2) is a nilpotent element of R1 ⊕ R2. In summary, we have shown that (a1; a2) is a nilpotent element of R1 ⊕ R2 if and only if a1 is a nilpotent element in R1 and a2 is a nilpotent element in R2. The only nilpotent element of Z is 0. The only nilpotent element of Z3 is 0. The nilpotent elements of Z4 are 0 and 2. The only nilpotent element of Z6 is 0. It follows that The only nilpotent element in Z ⊕ Z is (0; 0). The only nilpotent element in Z3 ⊕ Z3 is (0; 0). The nilpotent elements in Z4 ⊕ Z6 are (0; 0) and (2; 0). Suppose that (a1; a2) is an element of R1 ⊕ R2. Then (a1; a2) is a zero-divisor if and only if there exists an element (b1; b2) in R1 ⊕ R2 such that (b1; b2) 6= (0R1 ; 0R2 ) and (a1; a2)(b1; b2) = (0R1 ; 0R2 ) : The second equation just means that a1b1 = 0R1 and a2b2 = 0R2 . Also, (b1; b2) 6= (0R1 ; 0R2 ) means that b1 6= 0R1 or b2 6= 0R2 . Consequently, it follows that if (a1; a2) is a zero-divisor in R1 ⊕ R2, then either a1 is a zero divisor in R1 or a2 is a zero divisor in R2. For the converse, suppose that a1 is a zero-divisor in R1. Then a1b1 = 0R1 for some nonzero element b1 2 R1. It follows that (b1; 0R2 ) 6= (0R1 ; 0R2 ) and (a1; a2)(b1; 0R2 ) = (0R1 ; 0R2 ) : Therefore, (a1; a2) is a zero-divisor in R1 ⊕ R2 . A similar argument shows that if a2 is a zero-divisor in R2, then (a1; a2) is a zero-divisor in R1 ⊕ R2 . In summary, we have shown that (a1; a2) is a zero-divisor in R1 ⊕ R2 if and only if either a1 is a zero divisor in R1 or a2 is a zero divisor in R2. The only zero-divisor in Z is 0. The only zero-divisor in Z3 is 0. The zero-divisors in Z4 are 0 and 2. The zero-divisors in Z6 are 0, 2, 3 and 4. The above remark shows that The set of zero-divisors in Z ⊕ Z is f (a; 0) a 2 Z g [ f (0; b) b 2 Z g. The set of zero-divisors in Z3 ⊕ Z3 is f (a; 0) a 2 Z3 g [ f (0; b) b 2 Z3 g. The set of zero-divisors in Z4 ⊕ Z6 is f (a; b) a 2 Z4; b = 0; 2; 3; or 4 g [ f (a; b) b 2 Z6; a = 0 or 2:g : Problem 16.13, part (a) Show that the multiplicative identity in a ring with unity R is unique. SOLUTION: Suppose that e 2 R and that ea = a = ae for all a 2 R. Suppose also that f 2 R and that fa = a = af for all a 2 R. Then we have f = ef = e Therefore, e = f. Thus, there can only be one element in R satisfying the requirements for the multiplicative identity of the ring R. Problem 16.13, part (b) Suppose that R is a ring with unity and that a 2 R is a unit of R. Show that the multiplicative inverse of a is unique. SOLUTION: Suppose that b; c 2 R and that ab = ba = 1 and that ac = ca = 1. Then we have c = 1c = (ba)c = b(ac) = b1 = b : Hence we have c = b. The multiplicative inverse of a is indeed unique. ADDITIONAL PROBLEMS: A: Prove that if R is a division ring, then the center of R is a field. SOLUTION: First of all, suppose that R is any ring with identity. Let S be the center of R. That is, S = f s 2 R j sr = rs for all r 2 R g : We will show that S is a subring of R. The fact that S is a subgroup of R under addition can be seen as follows. For this purpose, suppose that s1; s2 2 S. Then, for all r 2 R, we have s1r = rs1 and s2r = rs2. Therefore, using the distributive laws for R, we have (s1 + s2)r = s1r + s2r = rs1 + rs2 = r(s1 + s2) for all r 2 R. Therefore, s1 + s2 2 S. Furthermore, letting 0 denote the additive identity of R, we have 0 · r = 0 and r · 0 = 0. Hence 0 · r = r · 0. Therefore, 0 2 S. Finally, suppose that s 2 S. Let t = −s, the additive inverse of s in R. We have s+t = 0. Thus, s + t 2 S. Since s is in S and s + t is in S, it follows that, for all r 2 R, we have sr = rs and (s + t)r = r(s + t). Therefore, we have sr + tr = rs + rt = sr + rt Thus, we have the equation sr+tr = sr+rt. Applying the cancellation law for the underlying additive group of R to that equation, it follows that tr = rt for all r 2 R. Therefore, t 2 S. That is, −s 2 S. This completes the verification that S is a subgroup of R under the operation of addition. To complete the proof that S is a subring of R, we must show that if s1 and s2 are in S, then so is s1s2. So, assume that s1; s2 2 S. Then, for all r 2 R, we have s1r = rs1 and s2r = rs2. Consider s1s2, which is an element of R. Using the associative law for multiplication in R many times, it follows that (s1s2)r = s1(s2r) = s1(rs2) = (s1r)s2 = (rs1)s2 = r(s1s2) for all r 2 R. Therefore, we indeed have s1s2 2 S. We have shown that S is a subring of R. If R is a ring with unity 1, then 1r = r = r1 for all r 2 R.