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sol10

sol10 - Algebra Assignment 10 Solutions 1 Let R be a ring...

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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 α, β M so that α = ar 1 + bs 1 , β = ar 2 + bs 2 . Then α + β = a ( r 1 + r 2 ) + b ( s 1 + s 2 ) M .

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