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Unformatted text preview: Algebra , Assignment 10 Solutions 1. Let R be a ring and M R an ideal. Assume M negationslash = R (but do not assume M is maximal). (a) Assume there exists an ideal N with M N R and N negationslash = M,R . Let a N with a negationslash M . Prove that a has no multiplicative inverse in R/M Solution: For any r R , a r = ar . Setting n = ar , as N is an ideal n N . We claim n negationslash = 1. For if n = 1 then n 1 = m M N so 1 = n ( n 1) N and N = R . (b) Assume there does not exist an ideal N with M N R and N negationslash = M,R . Prove that a has an multiplicative inverse in R/M for all a negationslash = 0. Solution: Set M + = { m + ar : m M,r R } Then a M + so M negationslash = M + and M M + and therefore (as there is no intermediate size ideal) M + = R so 1 M + and 1 = m + ar for some r R and so 1 = a r . 2. Let R be a ring and let a,b R . Set M = { ar + bs : r,s R } Prove that M is an ideal. Solution: (a) Identity: 0 = a (0) + b (0) M (b) Closure under Addition. Let , M so that = ar 1 + bs 1 , = ar 2 + bs 2 . Then + = a ( r 1 + r 2...
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This note was uploaded on 04/01/2011 for the course MATH 101 taught by Professor Josephs during the Fall '08 term at NYU.
 Fall '08
 JOSEPHS
 Algebra

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