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(A) Give an Example of a Subset of a Ring That Is a Subgroup Under Addition but Not a Subring

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(A) Give an Example of a Subset of a Ring That Is a Subgroup Under Addition but Not a Subring Math 307 Abstract Algebra Homework 9 Sample solution 1. (a) Give an example of a subset of a ring that is a subgroup under addition but not a subring. (b) Give an example of a finite non-commutative ring. Solution. (a) Example 1. Let R = C and S = fix : x 2 Rg. Then 0 2 S and a − b 2 S whenever a; b 2 S. But i · i = −1 2= S. Example 2. Let H = h(2; 3)i 2 Z ⊕ Z. Then H = f(2k; 3k): k 2 Zg is a subgroup under addition. But (2; 3)(2; 3) = (4; 9) 2= H. 4 (b) Let R = M2(Z2). Then there are 2 elements because each entries has two choices. 0 1 Clearly, AB 6= BA if A = Bt = . 0 0 If we consider the subring S of R consisting of matrices with the second row equal zero. Then S has four elements and is not commutative. 2. Show that if m; n are integers and a; b are elements in a ring. Then (ma)(nb) = (mn)(ab). [Note that for positive m, ma = a + ··· + a (m times and (−m)a = m(−a).] Solution. If m or n is zero, then both sides equal 0. If m; n 2 N, then (a + ··· + a)(b + ··· + b) = (ab + ··· + ab) = (mn)(ab): | {z } | {z } | {z } m n mn If m is negative and n is positive, then (ma)(nb) + (jmja)(nb) = ((m + jmj)a)(nb) = 0 so that (ma)(nb) = −(jmjn)(ab) = (mn)(ab). Similarly, if m is positive and n is negative, then (ma)(nb) = (mn)(ab). Finally, if m; n are negative, then (ma)(nb) = (−|mja)(−|njb) = jmnj(ab) = (mn)(ab). 3. Let R be a ring. (a) Suppose a 2 R. Show that S = fx 2 R : ax = xag is a subring. (b) Show that the center of R defined by Z(R) = fx 2 R : ax = xa for all a 2 Rg is a subring. Solution. (a) Note that 0 2 Sa is non-empty. Suppose x; y 2 Sa. Then ax = xa and ay = ya. So, a(x − y) = ax − ay = xa − ya = (x − y)a. So, x − y 2 Sa. Also, a(xy) = (xa)y = (xy)a. So, xy 2 Sa. It follows that Sa is a subring. (b) Note that 0 2 S is non-empty. Suppose x; y 2 Z(R). Then ax = xa and ay = ya. So, a(x − y) = ax − ay = xa − ya = (x − y)a for any a 2 R. So, x − y 2 Z(R). Also, a(xy) = (xa)y = (xy)a for any a 2 R. So, xy 2 Z(R). It follows that Z(R) is a subring. Alternatively, we can show that intersection of subrings is a subring, and use the fact that Z(R) = \a2RSa. 4. Let R be a ring. (a) Prove that R is commutative if and only if a2 − b2 = (a + b)(a − b) for all a; b 2 R. (b) Prove that R is commutative if a2 = a for all a 2 R. (Such a ring is called a Boolean ring.) 1 Solution. (a) If R is commutative, then (a + b)(a − b) = a2 + ab − ba − b2 = a2 − b2 for any a; b 2 R. Suppose (a + b)(a − b) = a2 + ab − ba − b2 = a2 − b2 for any a; b 2 R. Then ab − ba = 0, i.e., ab = ba. (b) Suppose a2 = a for all a 2 R. Then for any a; b 2 R, a2 + b2 = a + b = (a + b)2 = a2 + ab + ba + b2 so that ab + ba = 0. Hence, ab = −ba = (−ba)2 = (ba)2 = ba. 5. Give an example of a Boolean ring with 4 elements. Give an example of a Boolean ring with infinitely many elements. Solution. Let B = Z2 = f0; 1g such that 0 + 0 = 0 and 0 + 1 = 1 + 0 = 1 + 1 = 1, and 00 = 01 = 10 = 0 and 11 = 1. Then B is a Boolean ring with 2 elements, and B ⊕ B is a Boolean ring with 4 elements. 1 Let R = B = f(a1; a2;::: ): ai 2 B for each ig. One can show that R is a ring under the 2 entrywise addition and multiplication operation. Also, (a1; a2; ··· ) = (a1; a2;::: ). So, R is a Boolean ring with infinitely many elements. 6. Show that every nonzero element of Zn is a unit (element with multiplicative inverse) or a zero-divisor. ∗ Solution. Suppose k 2 Zn. If gcd(k; n) = 1, then there is x; y 2 Z such that kx + ny = 1. Thus, x is the inverse of k in Zn. If gcd(k; n) = d > 1. Then n=d 2 Zn is nonzero such that k(n=d) = 0 2 Zn so that k is a zero divisor. The conclusion follows. 7. Show that every nonzero element in Z7[i] = fa + bi : a; b 2 Z7g has a multiplicative inverse. 2 2 2 Solution. Note that in Z7, (1 ; 2 ;:::; 6 ) = (1; 4; 2; 2; 4; 1). Thus, for any nonzero a + ib 2 2 2 Z7[i] we have a + b 2 f1; 2; 4; 3; 5; 6g and is invertible. Thus, we can always find (x + iy) such that (a + ib)(x + iy) = 1 by setting x + iy = (a2 + b2)−1(a − ib) if (a; b) 6= (0; 0). (Extra 5 points.) Show that not every nonzero element in Zp[i] has a multiplicative inverse for a given prime p. Solution. Consider 2+i 2 Z5[i]. One sees that there is no (x+iy)(2+i) = 1, say, by checking all nonzero x + iy 2 Z5[i]. (Extra 5 points.) Determine those prime p such that every nonzero element in Zp[i] has a multiplicative inverse. Solution. From previous discussion, we see that every nonzero element in Zp[i] has a multi- plicative inverse if and only if a2 + b2 is invertible for all nonzero (a; b) 6= (0; 0). [What is the condition on this? We will see later that it is equivalent to the condition that x2 + 1 = 0 has no solution in Zp.] 8. (a) Given an example of a commutative ring without zero-divisors that is not an integral domain. (b) Find two elements a and b in a ring such that a; b are zero-divisors, a + b is a unit. Solution. (a) 2Z. (b) 2; 3 2 Z6. 2 9. (a) Give an example to show that the characteristic of a subring of a ring R may be different from that of R. (b) Show that the characteristic of a subdomain of an integral domain D is the same as that of D. (c) (Extra problem mentioned in class, 3 points) Find a subring R1 of a ring R so that the unity in R is different from that in R1. Solution. (a) Z2 ⊕ Z2 has unity (1; 1), but the subring Z ⊕ f0g has unit (1; 0). (b) Suppose D has unit 1, and a subdomain D^ has unity 1.^ Then 1^1^ = 1^ in D^ and 11^ = 1^ in D. So, 1(1^ − 1)^ = 0 implies that 1^ = 1. (c) Suppose R = Z2 ⊕ Z2 and R1 = Z2 ⊕ f0g. 3.
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