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# all3 - Math 4124 Wednesday April 20 Eighth Homework...

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Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely 2 4 · 13 2 . To find the number of abelian groups of order 16, we find the number of partitions of 4. The partitions are (4), (3,1), (2,2), (2,1,1), (1,1,1,1) (so for example (3,1) corresponds to the group Z / 8 Z × Z / 2 Z ). Thus there are 5 abelian groups of order 16. Also there are 2 abelian groups of order 9. Therefore the total number of abelian groups of order 2704 is 5 · 2 = 10. 2. Exercises 5.2.2(e) and 5.2.3(e). List the elementary divisors and invariant factors for the abelian groups of order 44100. 44100 = 2 2 · 3 2 · 5 2 · 7 2 . Thus the number of abelian groups of order 44100 is 2 4 = 16. In the table below n will indicate Z / n Z ; thus using this notation there are two abelian groups of order p 2 , namely p 2 and p , p . Elementary Divisors Invariant Factors 2 2 , 3 2 , 5 2 , 7 2 44100 2 2 , 3 2 , 5 2 , 7 , 7 6300,7 2 2 , 3 2 , 5 , 5 , 7 2 8820,5 2 2 , 3 , 3 , 5 2 , 7 2 14700,3 2 , 2 , 3 2 , 5 2 , 7 2 22050,2 2 2 , 3 2 , 5 , 5 , 7 , 7 1260,35 2 2 , 3 , 3 , 5 , 5 , 7 2 2940,15 2 , 2 , 3 , 3 , 5 2 , 7 2 7350,6 2 2 , 3 , 3 , 5 2 , 7 , 7 2100,21 2 , 2 , 3 2 , 5 , 5 , 7 2 4410,10 2 , 2 , 3 2 , 5 2 , 7 , 7 3150,14 2 2 , 3 , 3 , 5 , 5 , 7 , 7 420,105 2 , 2 , 3 2 , 5 , 5 , 7 , 7 630,70 2 , 2 , 3 , 3 , 5 2 , 7 , 7 1050,42 2 , 2 , 3 , 3 , 5 , 5 , 7 2 1470,30 2 , 2 , 3 , 3 , 5 , 5 , 7 , 7 210,210 3. Exercise 7.2.7. Let R be a commutative ring with 1. Prove that the center of the ring M n ( R ) is the set of scalar matrices. First we show that the scalar matrices are in the center of M n ( R ) , so let A be a scalar matrix, say A = rI where r R . Then the i j entry of A is r δ i j , where δ is the Kronecker δ . If B M n ( R ) has entries b i j , then the i j entry of AB is k r δ ik b k j = rb i j , whereas the i j entry of BA is k b ik r δ k j = b i j r . Since R is commutative, rb i j = b i j r for all i , j , hence A commutes with B . It follows that A is in the center of M n ( R ) .

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Conversely suppose A is in the center of M n ( R ) and write A = ( a i j ) . Then by Exercise 7.2.6, we have E pq AE rs = a qr E ps . Also because A is in the center, this is E pq E rs A = δ qr E ps A . Therefore a qr E ps = δ qr E ps A . By considering the case q = r , we now see that a qr = 0 if q = r . Also a qq E 11 = E 11 A , so a qq is independent of q and we deduce that A is a scalar matrix. 4. Exercise 7.3.1. Prove that the rings 2 Z and 3 Z are not isomorphic. Suppose θ : 2 Z 3 Z is an isomorphism. Set x = θ 2. Then x 2 = θ ( 2 2 ) = θ 4 and 2 x = θ 2 + θ 2 = θ 4. Therefore x 2 - 2 x = 0. The only solutions to this are x = 0 and x = 2. But 2 / 3 Z . Therefore θ 2 = 0. But this is not possible because θ is one-to-one, so the result is proven.
Math 4124 Wednesday, April 27 Ninth Homework Solutions 1. Exercise 7.3.10. Decide which of the following are ideals of the ring Z [ x ] . Let I be the relevant subset of Z [ x ] . (a) The set of all polynomials whose constant term is a multiple of 3. Yes. Obviously 0 I and I is an abelian group under addition. Finally suppose f = a 0 + a 1 x + ··· + a m x m I and g = b 0 + b 1 x + ··· + b n x n Z [ x ] . Then 3 divides a 0 and fg = a 0 b 0 + ( a 1 b 0 + a 0 b 1 ) x + ··· + a m b n x m + n . Since 3 divides a 0 b 0 , we see that fg I and we have proven that I is an ideal.

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all3 - Math 4124 Wednesday April 20 Eighth Homework...

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