This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Homework #2 Solutions pp 254257: 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....
View
Full
Document
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.
 Spring '07
 RyanDaileda

Click to edit the document details