554-HW-10q - x 2 1 2 Determine the elementary divisors of M and the invariant factors of M 10 If N is a direct summand of M(i.e M = N ⊕ K show N

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10 Find a basis for the submodule of Z (3) generated by (1 , 0 , - 1) , (2 , - 3 , 1) , (0 , 3 , 1) , (3 , 1 , 5). 10 Over the polynomial ring Q [ x ] find a base for the submodule of Q [ x ] (3) generated by (2 x - 1 ,x,x 2 +3) , ( x,x,x 2 ) , and ( x + 1 , 2 x, 2 x 2 - 3). 10 Find a basis for the Z -submodule of Z (3) consisting of the set of all ( x,y,z ) satisyfing x + 2 y + 3 z = 0 and x + 4 y + 9 z = 0. 10 Put this matrix into normal form: 6 2 3 0 2 3 - 4 1 - 3 3 1 2 - 1 2 - 3 5 10 Determine the structure of M = Z (3) /K where K is generated by f 1 = (2 , 1 , - 3) and f 2 = (1 , - 1 , 2). 10 Let D = R [ X ] and assume that M is a direct sum of cyclic D -modules whose order ideals (annihilating ideals) are the ideals generated by the polynomials ( x - 1) 3 , ( x 2 + 1) 2 , ( x - 1)( x 2 + 1) 4 , ( x + 2)(
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Unformatted text preview: x 2 + 1) 2 . Determine the elementary divisors of M and the invariant factors of M . 10 If N is a direct summand of M (i.e. M = N ⊕ K , show N is pure in M . 10 If N is a pure submodule of M and ann( x + N ) = ( d ), prove that w can be chosen in x + N such that w + N = x + N and ann( w ) = ( d ). 10 If N is a pure submodule of a finitely generated module M over a PID D , prove that N is a direct summand of M . You may assume that a finitely generated module of a PID is a direct product of cyclic modules. 1...
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This document was uploaded on 01/25/2012.

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