MATH535_HW7

MATH535_HW7 - Serge Ballif MATH 535 Homework 7 1 This...

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Serge Ballif MATH 535 Homework 7 October 19, 2007 1. This exercise shows (part of) the fact every ring R can be considered as a subring of a ring with identity element. We mimic putting the subring i into the full ring of Gaussian integers: On S := × , define the operations ( m, r ) + ( n, s ) := ( m + n, r + s ) ( m, r )( n, s ) := ( mn, nr + ms + rs ) . a) Show associativity of multiplication. b) Show one of the distributivity laws. c) Find the identity element. d) Give the embedding of R into S and check that it is a homomorphism. (a) ( ( m, r )( n, s ) ) ( o, t ) = ( mn, nr + ms + rs )( o, t ) = ( mno, o ( nr + ms + rs ) + mn ( t ) + ( nr + ms + rs ) t ) = ( mno, onr + oms + ors + mnt + nrt + mst + rst ) = ( mno, no ( r ) + m ( os + nt + st ) + r ( os + nt + st ) = ( m, r )( no, os + nt + st ) = ( m, r ) ( ( n, s )( o, t ) ) . (b) ( ( m, r ) + ( n, s ) ) ( o, t ) = ( m + n, r + s )( o, t ) = ( ( m + n ) o, o ( m + n ) + ( m + n ) t + ( r + s ) t ) = ( mo + no, om + mt + rt + on + nt + st ) = ( mo, om + mt + rt ) + ( no, on + nt + st ) = ( m, r )( o, t ) + ( n, s )( o, t ) . (c) We claim that the element (1 , 0) is the multiplicative identity of S . We verify this claim with the calculation (1 , 0)( m, r ) = (1( m ) , m (0) + 1( r ) + 0( r )) = ( m (1) , 1( r ) + m (0) + r (0) = ( m, r )(1 , 0). (d) Let j : R S be given by j ( r ) = (0 , r ). Claim 1. The map j is a ring homomorphism. Proof. j ( rs ) = (0 , rs ) = (0 , 0( r ) + 0( s ) + rs ) = (0 , r )(0 , s ) = j ( r ) j ( s ). j ( r + s ) = (0 , r + s ) = (0 + 0 , r + s ) = (0 , r ) + (0 , s ) = j ( r ) + j ( s ). Claim 2. The map j is injective. Proof. Suppose that (0 , r ) = j ( r ) = j (0) = (0 , 0). Then (0 , r ) = (0 , 0) is the unique additive identity of S , so r = 0. Hence j is injective. 2. a) Let a 1 , a 2 A , an (additively written) abelian group. Let a 1 , a 2 have relatively prime (finite) orders n 1 , n 2 . Show that a 1 + a 2 has order n 1 n 2 . b) Conclude that if G 1 and G 2 are cyclic groups of relatively prime (finite) orders, then G 1 × G 2 is also cyclic. Page 1 of 4

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Serge Ballif MATH 535 Homework 7 October 19, 2007 (a) We can write multiplication by whole numbers distributively and treat it like regular multiplication since the group is abelian. A calculation shows that ( a 1 + a 2 ) n 1 n 2 = a 1 n 1 n 2 + a 2 n 2
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