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Unformatted text preview: Math 302 Modern Algebra Spring 2009 Exam 2 Solutions 1. (36 points) Let S be the set of integers with an addition and multiplication defined for all a, b S by a b = a + b 1 and a b = a + b ab. (The +, , and symbols denote ordinary integer addition, subtraction, and multiplication, respec- tively.) It can be shown that S is a commutative ring with 0 S = 1 under these operations. (a) Verify ring axiom R10 Existence of a Multiplicative Identity for S . Proof: We need to show there exists an element 1 S S such that for all a S , a 1 S = a and 1 S a = a. Find a 1 S . The element 1 S must satisfy a 1 S = a for all a S . This means a + 1 S a 1 S = a 1 S (1 a ) = 0 This implies 1 S = 0 or a = 1 . Since this must hold for all integers a , not just a = 1, then 1 S = 0. Choose 1 S = 0. Verify that 1 S satisfies: (i) Domain: 1 S S (ii) Propositional Function: ( a S ) [ a 1 S = a 1 S a = a ] Since 1 S = 0 is an integer, then 1 S S . To prove (ii), suppose a is an arbitrary integer. We need to show a 1 S = a and 1 S a = a. We have a 1 S = a 0 = a + 0 + a 0 = a 1 S a = 0 a = 0 + a + 0 a = a as required. Thus 1 S = 0 is a multiplicative identity for S . (b) Verify ring axiom R11 Zero Factor Property for S . We need to show ( a S ) ( b S ) [ a b = 0 S a = 0 S b = 0 S ] . Proof: Suppose a and b are arbitrary integers in S . Suppose a b = 0 S = 1 . We need to show that a = 0 S = 1 or b = 0 S = 1 . By the hypothesis, a b = a + b ab = 1 a 1 + b ab = 0 ( a 1) b ( a 1) = 0 ( a 1)(1 b ) = 0 Therefore a = 1 = 0 S or b = 1 = 0 S , as required. Thus S satisfies the Zero Factor Property. page 2 Math 302 Exam 2 Solutions 1. (continued) (c) Let R be the set of integers with the ordinary integer addition + and multiplication . Define a function f : R S by f ( r ) = 1 r for each r R ....
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