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Math Help - Direct Sums of Modules

  1. #1
    Super Member Bernhard's Avatar
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    Direct Sums of Modules

    I am reading Dummit and Foote to gain an understanding of Modules. In section 10.3 Proposition 5 we read:

    Proposition 5

    Let  N_1 ,  N_{2} , ....., Nk be submodules of the R-module M. Then the following are equivalent:

    (1) The map  \pi :  N_1 X  N_{2} X ...X Nk  \rightarrow  N_1 +  N_{2} + ...+ Nk defined by

     \pi (  a_1 ,  a_2 , ..... ,  a_k ) =  a_1 +  a_2 + .... +  a_k is an isomorphism of R-modules

    i.e.  N_1 +  N_{2} + .... + Nk is isomorphic to  N_1 X  N_{2} X .... X Nk

    (2) Nj  \cap (  N_1 +  N_{2} + Nj-1 + Nj+1 .... + Nk ) = 0 for all j = {1,2, ...... , k}

    (3) Every x  \in  N_1 +  N_{2} + .... + Nk can be written uniquely in the form  a_1 +  a_2 + .... +  a_k

    Proof: To prove (1) implies (2), suppose that for some j that (2) fails to hold and let aj  \in (  N_1 +  N_{2} + Nj-1 + Nj+1 + ....+ Nk)  \cap Nj, with aj not equal to 0 .

    Then aj =  a_1 +  a_2 + .... + aj-1 + aj+1 + .... +  a_k for some  a_i  \in Ni and (  a_1 ,  a_2 , ....., aj-1, -aj, aj+1, ...., [LaTeX ERROR: Convert failed] ) would be a nonzero element of Ker  \pi


    My question is this: Why is a non-zero element of Ker  \pi a contradiction. Surely you can have non-zero elements of Ker  \pi ? WHy not?

    {Since I had real trouble with subscripts in Latex, I have attached the relevant page from Dummit and Foote showing the relevant proposition and proof}

    Peter
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  2. #2
    Super Member girdav's Avatar
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    Re: Direct Sums of Modules

    Since \pi is supposed to be an isomorphism, the kernel is the set which consists of the vector 0.
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