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# nov04 - -x n = xn = 0 and we deduce that x y ∈ I and-x...

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Math 5125 Friday, November 4 November 4, Ungraded Homework Remarks If M is an R -module, then 0 m = 0 = r 0, and ( - r ) m = - rm for all m M and r R . We shall use these elementary properties without comment. Exercise 10.1.3 on page 343 Assume that rm = 0 for some r R and some m M with m = 0. Prove that r does not have a left inverse (i.e., there is no s R such that sr = 1). Suppose to the contrary that there is an s R such that sr = 1. Then 0 = m = 1 m = ( sr ) m = s ( rm ) = s 0 = 0 which is a contradiction and the result is proven. Exercise 10.1.9 on page 344 If N is a submodule of M , the annihilator of N in R is defined to be { r R | rn = 0 for all n N } . Prove that the annihilator of N in R is a 2-sided ideal of R . Let I denote the annihilator of N in R . First note that 0 I , because 0 n = 0 for all n N . If x , y I , then xn = yn = 0 for all n N , hence ( x + y ) n = xn + yn = 0 + 0 = 0 and
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Unformatted text preview: (-x ) n =-( xn ) = 0, and we deduce that x + y ∈ I and-x ∈ I . We have so far shown that I is an abelian group under addition. Suppose r ∈ R . Then ( rx ) n = r ( xn ) = r = 0 and ( xr ) n = x ( rn ) = 0, and we conclude that rx and xr ∈ I . This completes the proof that I is an ideal of R . Exercise 10.2.8 on page 350 Let φ : M → N be an R-module homomorphism. Prove that φ ( Tor ( M )) ⊆ Tor ( N ) (where Tor ( M ) indicates the torsion elements of M ). Let m ∈ Tor ( M ) . Then there exists 0 6 = r ∈ R such that rm = 0. Then 0 = θ ( rm ) = r ( θ m ) , which shows that θ m ∈ Tor ( N ) as required....
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