Adv Alegbra HW Solutions 84

Adv Alegbra HW Solutions 84 - B ( Y ) is distinct from the...

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84 3.7 (i) If X is a set, prove that the Boolean group B ( X ) with elements the subsets of X and addition given by symmetric difference, U + V = ( U V ) ( V U ) , is a commutative ring if one de f nes multiplication by UV = U V . Solution. The subset X is one ,for X U = U for every sub- set U X ;1 ±= 0 because X is nonempty; it is easy to check associativity U ( V W ) = ( U V ) W and commutativity U V = V U . The proof of distributivity is longer. A ( B + C ) = A ∩[ ( B C ) ( C B ) ] =[ A ( B C ) ]∪[ A ( C B ) ] =[ A B C 0 ]∪[ A C B 0 ] . On the other hand, AB + AC = ( A B ) + ( A C ) =[ ( A B ) ( A C ) 0 ]∪[ ( A C ) ( A B ) 0 ] =[ ( A B ) ( A 0 C 0 ) ]∪[ ( A C ) ( A 0 B 0 ) ] =[ ( A B A 0 ) ( A B C 0 ) ] ∪[ ( A C A 0 ) ( A C B 0 ) ] = ( A B C 0 ) ( A C B 0 ), because A A 0 = . (ii) Prove that B ( X ) contains exactly one unit. Solution. If A X is a unit in B ( X ) , then there is B X with AB = 1; that is, A B = X .Bu t X = A B A implies that A = X . (iii) If Y ± X (that is, Y is a proper subset of X
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Unformatted text preview: B ( Y ) is distinct from the one in B ( X ) . Conclude that B ( Y ) is not a subring of B ( X ) . Solution. The unit in B ( Y ) is Y . which is distinct from the unit in B ( X ) , namely, X . (iv) Prove that every element U B ( X ) satis f es U 2 = U . Solution. U U = U . 3.8 (i) If R is a domain and a R satis f es a 2 = a , prove that either a = 0 or a = 1. Solution. If a 2 = a , then 0 = a 2 a = a ( a 1 ) . Since R is a domain, either a = 0 or a 1 = 0. (ii) Show that the commutative ring F ( R ) in Example 3.11(i) contains elements f = , 1 with f 2 = f ....
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