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Unformatted text preview: (x ) n =( xn ) = 0, and we deduce that x + y I andx 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 : M N be an Rmodule homomorphism. Prove that ( 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 = ( rm ) = r ( m ) , which shows that m Tor ( N ) as required....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.
 Fall '07
 PALinnell
 Math, Algebra

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