# HW2_soln - Homework #2 Solutions pp 254-257: 18, 34, 36,...

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Unformatted text preview: Homework #2 Solutions pp 254-257: 18, 34, 36, 50, 54 p 241, #18 We apply the subring test. First of all, S 6 = ∅ since a · 0 = 0 implies 0 ∈ S . Now let x, y ∈ S . Then a ( x- y ) = ax- ay = 0- 0 = 0 and a ( xy ) = ( ax ) y = 0 · y = 0 so that x- y, xy ∈ S . Therefore S is a subring of R . p 242, #38 Z 6 = { , 1 , 2 , 3 , 4 , 5 } is not a subring of Z 12 since it is not closed under addition mod 12: 5 + 5 = 10 in Z 12 and 10 6∈ Z 6 . p 243, #42 Let X = a a b b , Y = c c d d ∈ R . Then X- Y = a- c a- c b- d b- d ∈ R since a- c, b- d ∈ Z . Also XY = ac + ad ac + ad bc + bd bc + bd ∈ R since ac + ad, bc + bd ∈ Z . Since R is clearly nonempty, the subring test implies that R is indeed a subring of M 2 ( Z ). p 254, #4 The zero divisors in Z 20 are 2 , 4 , 5 , 6 , 8 , 10 , 12 , 14 , 15 , 16 and 18, since 2 · 10 = 0 mod 20 4 · 15 = 0 mod 20 6 · 10 = 0 mod 20 8 · 5 = 0 mod 20 12 · 5 = 0 mod 20 14 · 10 = 0 mod 20 16 · 5 = 0 mod 20 18 · 10 = 0 mod 20 and every nonzero element not in this list is a unit. In particular this shows that the zero divisors in Z 20 are precisely the nonzero nonunits. This statement generalizes to everyare precisely the nonzero nonunits....
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## This note was uploaded on 02/29/2012 for the course MATH 4363 taught by Professor Ryandaileda during the Spring '07 term at Trinity University.

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HW2_soln - Homework #2 Solutions pp 254-257: 18, 34, 36,...

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