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Unformatted text preview: HOMEWORK 1 SOLUTIONS MATH 121 Problem (10.1.2) . Prove that R × and M satisfy the two axioms in Section 1.7 for a group action of the multiplicative group R × on the set M . Solution. For the rst axiom, we have to check that for r 1 , r 2 ∈ R and m ∈ M , we have r 1 ( r 2 m ) = ( r 1 r 2 ) m. For the second axiom, we have to check that for m ∈ M , 1 m = m . Both of these are included as part of the de nition of a module. Problem (10.1.4) . Let M be the module R n described in Example 3 and let I 1 , I 2 , .. . ,I n be left ideals of R . Prove that the following are submodules of M : (a) { ( x 1 , x 2 , .. . ,x n )  x i ∈ I i } . (b) { ( x 1 , x 2 , .. . ,x n )  x i ∈ R and x 1 + x 2 + ··· + x n = 0 } . Solution. (a) Call the set N . We have to check that N is a subgroup, and that for r ∈ R and x ∈ N , rx ∈ N . Let x = ( x 1 , .. . ,x n ) , y = ( y 1 , .. . ,y n ) ∈ N . Then x + y = ( x 1 + y 1 , .. . ,x n + y n ) ∈ N since the I i 's are ideals and hence closed under sums. Now, let x = ( x 1 , .. . ,x n ) ∈ N and r ∈ R . Then rx = ( rx 1 , .. . ,rx n ) ∈ N, again because the I i 's are ideals. Hence N is a submodule. (b) Call this set P . As in part (a), we check that P is a subgroup and is closed under scalar multiplication. Let x = ( x 1 , .. . ,x n ) , y = ( y 1 , .. . ,y n ) ∈ P and r ∈ R . Then x + y = ( x 1 + y 1 , .. . ,x n + y n ) . From the above, we know that this is in N , so we need to check that ( x 1 + y 1 )+ ··· +( x n + y n ) = 0 . This is true because x 1 + ··· + x n = y 1 + ··· + y n = 0 . Similarly, rx = ( rx 1 , .. . ,rx n ) ∈ N, and ( rx 1 ) + ··· + ( rx n ) = r ( x 1 + ··· + x n ) = 0 , so rx ∈ P . Hence P is a submodule. Date : 10 January, 2011. 1 2 MATH 121 Problem (10.1.9) . If N is a submodule of M , the annihilator of N in R is de ned to be { r ∈ R  rn = 0 for all n ∈ N } . Prove that the annihilator of N in R is a 2sided ideal of R . Solution. Let ann( N ) be the annihilator of N in R . Suppose r ∈ ann( N ) , a ∈ R , and n ∈ N . We have to show that ar, ra ∈ ann( N ) . We have ( ar )( n ) = a ( rn ) = 0 because rn = 0 , and ( ra )( n ) = r ( an ) = rn = 0 for some n ∈ N . Hence ann( N ) is a 2sided ideal of R . Problem (10.1.10) . If I is a right ideal of R , the annihilator of I in M is de ned to be { m ∈ M  am = 0 for all a ∈ I } . Prove that the annihilator of I in M is a submodule of M . Solution. Let ann( I ) be the annihilator of I in M . Suppose that m, m ∈ ann( I ) , r ∈ R , and a ∈ I . Then we have a ( m + m ) = am + am = 0 , so m + m ∈ ann( I ) . Furthermore, a ( rm ) = ( ar ) m = a m = 0 for some a ∈ I . Hence rm ∈ ann( I ) . Thus ann( I ) is a submodule of M ....
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This note was uploaded on 11/08/2011 for the course MATH 321 taught by Professor Ergenc during the Spring '11 term at Boğaziçi University.
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