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# [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

- Anonymous1 decade agoFavorite 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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