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

Home , Center (algebra), Center (ring theory), Unit (ring theory)

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 = {ix : x ∈ R}. Then 0 ∈ S and a − b ∈ S whenever a, b ∈ S. But i · i = −1 ∈/ S. Example 2. Let H = h(2, 3)i ∈ Z ⊕ Z. Then H = {(2k, 3k): k ∈ Z} is a subgroup under addition. But (2, 3)(2, 3) = (4, 9) ∈/ 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 ∈ 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) + (|m|a)(nb) = ((m + |m|)a)(nb) = 0 so that (ma)(nb) = −(|m|n)(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) = (−|m|a)(−|n|b) = |mn|(ab) = (mn)(ab).

3. Let R be a ring. (a) Suppose a ∈ R. Show that S = {x ∈ R : ax = xa} is a subring. (b) Show that the center of R deﬁned by Z(R) = {x ∈ R : ax = xa for all a ∈ R} is a subring.

Solution. (a) Note that 0 ∈ Sa is non-empty. Suppose x, y ∈ Sa. Then ax = xa and ay = ya. So, a(x − y) = ax − ay = xa − ya = (x − y)a. So, x − y ∈ Sa. Also, a(xy) = (xa)y = (xy)a. So, xy ∈ Sa. It follows that Sa is a subring. (b) Note that 0 ∈ S is non-empty. Suppose x, y ∈ Z(R). Then ax = xa and ay = ya. So, a(x − y) = ax − ay = xa − ya = (x − y)a for any a ∈ R. So, x − y ∈ Z(R). Also, a(xy) = (xa)y = (xy)a for any a ∈ R. So, xy ∈ 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) = ∩a∈RSa.

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 ∈ R. (b) Prove that R is commutative if a2 = a for all a ∈ 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 ∈ R. Suppose (a + b)(a − b) = a2 + ab − ba − b2 = a2 − b2 for any a, b ∈ R. Then ab − ba = 0, i.e., ab = ba. (b) Suppose a2 = a for all a ∈ R. Then for any a, b ∈ 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 = {0, 1} 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. ∞ Let R = B = {(a1, a2,... ): ai ∈ B for each i}. 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 ∈ Zn. If gcd(k, n) = 1, then there is x, y ∈ Z such that kx + ny = 1. Thus, x is the inverse of k in Zn. If gcd(k, n) = d > 1. Then n/d ∈ Zn is nonzero such that k(n/d) = 0 ∈ Zn so that k is a zero divisor. The conclusion follows.

7. Show that every nonzero element in Z7[i] = {a + bi : a, b ∈ Z7} 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 Z7[i] we have a + b ∈ {1, 2, 4, 3, 5, 6} 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 ∈ Z5[i]. One sees that there is no (x+iy)(2+i) = 1, say, by checking all nonzero x + iy ∈ 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 multiplicative 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 ∈ 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 ⊕ {0} has unit (1, 0). (b) Suppose D has unit 1, and a subdomain Dˆ has unity 1.ˆ Then 1ˆ1ˆ = 1ˆ in Dˆ and 11ˆ = 1ˆ in D. So, 1(1ˆ − 1)ˆ = 0 implies that 1ˆ = 1.