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Thread: Direct Sums of Modules

  1. #1
    Super Member Bernhard's Avatar
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    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 $\displaystyle M_1 $ , $\displaystyle M_2 $ , ....., $\displaystyle M_n $ be submodules of M such that

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

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

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

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

    Proof:

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

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

    by f( $\displaystyle x_1 $ , ......, $\displaystyle x_n $) = $\displaystyle x_1 $ + .....+ $\displaystyle x_n$

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

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

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

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

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

    so that ( $\displaystyle x_1 $ , ......, $\displaystyle 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 ($\displaystyle x_1 $ , ......, $\displaystyle x_n $) $\displaystyle \in $ Ker(f) implies ( $\displaystyle x_1 $ , ......, $\displaystyle 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
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  2. #2
    Super Member girdav's Avatar
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    Re: Direct Sums of Modules

    1. Since $\displaystyle M_1+\ldots + M_n=\left\{m_1+\ldots+m_n,m_1\in M_1,\ldots,m_n\in M_n\right\}$, we have by definition of $\displaystyle f$: $\displaystyle f(M_1\oplus\ldots\oplus M_n)=\left\{m_1+\ldots+m_n,m_1\in M_1,\ldots,m_n\in M_n\right\}=M_1+\ldots + M_n=M$
    by hypothesis.
    2. You can show that a morphism is injective if and only if its kernel only contains $\displaystyle 0$.
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  3. #3
    Super Member Bernhard's Avatar
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    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
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  4. #4
    Super Member girdav's Avatar
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    Re: Direct Sums of Modules

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