Results 1 to 2 of 2

Thread: Direct Sums of Modules

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    594
    Thanks
    2

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

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

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

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

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

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

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

    Then aj = $\displaystyle a_1 $ + $\displaystyle a_2 $ + .... + aj-1 + aj+1 + .... +$\displaystyle a_k $ for some $\displaystyle a_i $ $\displaystyle \in $ Ni and ( $\displaystyle a_1 $ , $\displaystyle a_2 $ , ....., aj-1, -aj, aj+1, ...., $\displaystyle a_k $ ) would be a nonzero element of Ker $\displaystyle \pi $


    My question is this: Why is a non-zero element of Ker $\displaystyle \pi $ a contradiction. Surely you can have non-zero elements of Ker $\displaystyle \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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32

    Re: Direct Sums of Modules

    Since $\displaystyle \pi$ is supposed to be an isomorphism, the kernel is the set which consists of the vector $\displaystyle 0$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Direct Sums
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Dec 27th 2014, 08:24 AM
  2. Confusion about Direct Limit of R-Modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 4th 2011, 01:54 PM
  3. Direct Sums of Modules
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Aug 30th 2011, 03:09 AM
  4. Modules, Direct Sums and Annihilators
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Feb 18th 2009, 06:24 PM
  5. Direct sum of free modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 25th 2008, 10:06 PM

Search Tags


/mathhelpforum @mathhelpforum