Results 1 to 4 of 4

Math Help - Direct Sums of Modules

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Direct Sums of Modules

    I am reading Adkins and Weintraub's book : Algebra: An Approach Through Module Theory. On page 119 (3.1) Theorem states:

    Let M be an R-module and let  M_1 ,  M_2 , .....,  M_n be submodules of M such that

    (1) M =  M_1 +  M_2 + ..... +  M_n , and

    (2) for 1  \leq i  \leq n,

     M_i  \cap (  M_1 +  M_2 + ... +  M_{i-1} +  M_{i+1} + ..... +  M_n) = 0

    Then M  \cong  M_1  \oplus  M_2  \oplus .....  \oplus  M_n

    Proof:

    Let  f_i :  M_i  \rightarrow M be the inclusion map, that is,  f_i (x) = x for all s  \in  M_i and define

    f :  M_1  \oplus  M_2  \oplus .....  \oplus  M_n  \rightarrow M

    by f(  x_1 , ......,  x_n ) =  x_1 + .....+  x_n

    f is an R-Module homomorphism and it follows from condition (1) that f is surjective.

    Now suppose that (  x_1 , ......,  x_n )  \in Ker(f) .

    Then  x_1 + .....+  x_n = 0 so that for 1  \leq i  \leq n, we have

     x_i = -(  x_1 + .....+  x_{i-1} +  x_{i+1} + ....+  x_n).

    Therefore  x_i  \in  M_i  \cap (  M_1 +  M_2 + ... +  M_{i-1} +  M_{i+1} + ..... +  M_n) = 0

    so that (  x_1 , ......,  x_n ) = 0 and f is an isomorphism


    My questions are as follows:

    1. why exactly is f surjective

    2. why exactly does showing that (  x_1 , ......,  x_n )  \in Ker(f) implies (  x_1 , ......,  x_n ) = 0 demonstrate that f is an isomorphism?

    Can anyone help?


    I have attached the relevant page of Adkins and Weintraub since the paragraph at the top of the page may be relevant.


    Peter
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32

    Re: Direct Sums of Modules

    1. Since M_1+\ldots + M_n=\left\{m_1+\ldots+m_n,m_1\in M_1,\ldots,m_n\in M_n\right\}, we have by definition of f: f(M_1\oplus\ldots\oplus M_n)=\left\{m_1+\ldots+m_n,m_1\in M_1,\ldots,m_n\in M_n\right\}=M_1+\ldots + M_n=M
    by hypothesis.
    2. You can show that a morphism is injective if and only if its kernel only contains 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    553
    Thanks
    2

    Re: Direct Sums of Modules

    Thanks girdav

    Can you give me some guidance on how to show that a morphism is injective if and only if its kernel only contains 0

    - or let me know where to find a proof

    Peter
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32

    Re: Direct Sums of Modules

    Let M and M' two modules, and \varphi\colon M\to M' a morphism.
    If \ker\varphi =\left\{0\right\} and if x_1, x_2\in M are such that \varphi(x_1)=\varphi(x_2), then \varphi(x_1-x_2)=0 and x_1-x_2\in\ker\varphi.
    If \varphi is injective, then for x\in \ker\varphi we have \varphi(x)=\varphi(0) hence x=0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Confusion about Direct Limit of R-Modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 4th 2011, 01:54 PM
  2. Direct Sums of Modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 29th 2011, 09:15 AM
  3. Modules, Direct Sums and Annihilators
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 18th 2009, 06:24 PM
  4. Direct sum of free modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 25th 2008, 10:06 PM
  5. Direct Sums
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 11th 2006, 07:46 PM

Search Tags


/mathhelpforum @mathhelpforum