阿甘 asked in 科學數學 · 1 decade ago

[高等代數]拜託幫我這一題就好了...

Prove that the union of an ascending chain of submodules is a

submodule.

2 Answers

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  • Anonymous
    1 decade ago
    Favorite Answer

    Let {N(i)} be an ascending chain of sub-modules of the given R-module

    M, where R be a commutative ring with identity (anyway, at least it is a

    ring).

    Let N=∪N(i) be the union of all sub-modules of the given ascending

    chain {N(i)}.

    Since the ascending chain would terminates, so there exists an integer k

    such that N(k) =N(k + 1)=N(k + 2)=….

    Therefore, N=∪N(i) =N(k) for some integer k.

    Since N(k) itself a sub-module of M, therefore the union N of an

    ascending chain of sub-modules {N(i)} is of course a sub-module of M.

    Complete the proof.

    2008-01-15 10:11:31 補充:

    The proof based on the definition of ascending chain condition as Frank

    said, this is a trivial problem. It's that proof OK to you ?

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  • 1 decade ago

    照定義做就可以了

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