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Adv Alegbra HW Solutions 84

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

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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 fi 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 ] ∪ [ A C B ] . On the other hand, AB + AC = ( A B ) + ( A C ) = [ ( A B ) ( A C ) ] ∪ [ ( A C ) ( A B ) ] = [ ( A B ) ( A C ) ] ∪ [ ( A C ) ( A B ) ] = [ ( A B A ) ( A B C ) ] ∪ [ ( A C A ) ( A C B ) ] = ( A B C ) ( A C B ), because A A = . (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 . But X = A B A implies that A = X . (iii) If Y X (that is, Y is a proper subset of
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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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