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Unformatted text preview: Math 5125 Friday, November 4 November 4, Ungraded Homework Remarks If M is an Rmodule, then 0 m = = 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 = m = 1 m = ( sr ) m = s ( rm ) = s = 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 2sided 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 and ( 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 Rmodule 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 = r ∈ R such that rm = 0. Then 0 = θ ( rm ) = r ( θ m ) , which shows that θ m ∈ Tor ( N ) as required. Math 5125 Monday, November 7 November 7, Ungraded Homework Exercise 10.2.7 on page 350 Let z be a fixed element of the center of R . Prove that the map m → zm is an Rmodule homomorphism from M to itself. Show that for a commutative ring R the map from R to End R ( M ) given by r → rI is a ring homomorphism (where I is the identity endomorphism). Denote by θ the map m → zm . Then for m , n ∈ M and r ∈ R , we have θ ( m + n ) = z ( m + n ) = zm + zn = θ m + θ n , and because z is in the center of R , we have zr = rz and hence θ ( rm ) = zrm = rzm = r θ m . This proves that θ ∈ End R ( M ) . Now suppose that R is commutative. This means that rI ∈ End R ( M ) for all r ∈ R , because it is the map m → rm for m ∈ M , so we could use the above. Let us denote by φ the map r → rI . Then for r , s ∈ R , we have φ ( r + s )( m ) = ( r + s ) m = rm + sm = φ ( r ) m + φ ( s ) m = ( φ ( r )+ φ ( s ))( m ) for m ∈ M , which shows that φ ( r + s ) = φ ( r )+ φ ( s ) . Also φ ( rs )( m ) = ( rs ) m = r ( sm ) = φ ( r ) φ ( s ) m for all m ∈ M , which shows that φ ( rs ) = φ ( r ) φ ( s ) . This proves that φ is a ring homomor phism 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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