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Is my proof of proposition 6.2 on page 75 correct? (it's different from what they do in the book)

Proposition 6.2.: $M$ is a Noetherian $A$ module $\iff$ every submodule of $M$ is finitely generated.

My proof:

$\implies$ Assume $M$ has a submodule $N$ that is not finitely generated. Say, $N = \langle \bigcup_{i=1}^\infty \{ n_i \} \rangle$. Then the following is an increasing chain that is not stationary: $\langle n_1 \rangle \subset \langle n_1, n_2 \rangle \subset \langle n_1, n_2, n_3 \rangle \subset \dots$ Hence $M$ is not Noetherian.

$\Longleftarrow$ Assume $M$ is not Noetherian. Then there is a non-stationary increasing chain of submodules $N_1 \subset N_2 \subset \dots$. Then $N = \bigcup_{k=1}^\infty N_k$ is a submodule. If $N$ was finitely generated, $N_k$ would be stationary. Hence $N$ is a submodule that is not finitely generated.

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    @BrunoStonek Ok : )2012-07-20

2 Answers 2

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The first part is not quite correct: you have assumed $N$ is countably generated.

The second part looks good, though you should make sure you know why the fact that $N$ is finitely generated implies that the sequence is stationary.

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    @unit3000-21 Yes, thanks. I did and I understand it now.2012-07-22
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Your first part is strange. I would rephrase the argument as follows: Suppose that $M$ is a Noetherian $A$ - module and there exists a submodule $N$ that is not finitely generated. Now choose $f_1 \in N$. Since $N$ is not finitely generated, $\langle f_1 \rangle \subsetneqq N$ and so we can choose $f_2$ such that $f_2 \in N - \langle f_1 \rangle$. By the same argument as before, we can choose $f_3 \in N - \langle f_1,f_2\rangle$ such that $\langle f_1,f_2,f_3 \rangle \subsetneqq N.$ We now have a chain of submodules of $N$ (and hence of $M$) such that

$\langle f_1 \rangle \subsetneqq \langle f_1,f_2 \rangle \subsetneqq \ldots \subsetneqq \langle f_1,f_2,f_3 \rangle $

contradicting the ACC.

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    You may want to formalise the choosing process using the axiom of choice though.2012-07-20