# hw8 - Math 113 Homework 8 solutions Fraleigh 19.14 If A and...

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Math 113 Homework # 8 solutions Fraleigh 19.14. If A and B are matrices such that the image of B is contained in the kernel of A , then AB = 0. This observation allows us to find that 1 2 2 4 2 2 - 1 - 1 = 0 0 0 0 . Fraleigh 19.17. (a) False, n Z is a subring of Z which has no zero divisors. (b) True, proved in class and the book. (c) False, n Z has characteristic zero since it is a subring of Z . (d) False, n Z has no multiplicative identity for n > 1. (e) True, because if a ring is isomorphic to an integral domain then it is an integral domain. (f) True. If it were finite then m · 1 = n · 1 for some m 6 = n , so ( m - n ) · 1 = 0, so the characteristic divides m - n . (g) False: (1 , 0) · (0 , 1) = (0 , 0). (h) True. Suppose a is a zero divisor and has a multiplicative inverse. Then a 6 = 0, there exists b 6 = 0 with ab = 0, and there exists c with ac = 1. Then 0 = 0 c = ( ab ) c = ( ac ) b = 1 b = b , contradiction. (i) False, n Z is not a domain when n > 1 because it has no multiplica- tive identity. (j) False, Z is not a field. Fraleigh 21.2. F = { a + b 2 | a, b Q } . By a previous problem, F is a field, and F contains D . On the other hand every element of F is a quotient of two elements of D , because p/q + ( r/s ) 2 = ( ps + qr 2) / ( qs ). Fraleigh 21.4. (a) True; this is essentially the definition of Q . (b) False. If R were a field of quotients of Z , then there would be an injective ring homomorphism f : Z R , such that every element of R is the quotient of two elements of the image of f . As explained in class, an injective ring homomorphism f : Z R must be the usual inclusion map. But not every element of R is a quotient of two elements of

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