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Unformatted text preview: 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 Z i into the full ring of Gaussian integers: On S := Z R , 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....
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This note was uploaded on 04/01/2008 for the course MATH 535 taught by Professor Brownawell,woodro during the Fall '08 term at Pennsylvania State University, University Park.
 Fall '08
 BROWNAWELL,WOODRO
 Math, Algebra, Integers

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