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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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You might consider doing this. –  Bruno Stonek Jul 20 '12 at 12:56
    
@BrunoStonek So I should post this as an answer? –  Matt N. Jul 20 '12 at 12:58
    
@BrunoStonek Usually, people manage to write more than just "yes" to my questions. –  Matt N. Jul 20 '12 at 12:59
    
I did not express any views on the subject, I just directed you to a meta question that is very related, and to the (only) answer, which was quite well received by the community. If you're worrying about preventing good answers by others by doing what is suggested in that post, then don't worry! There can be multiple answers to one question, no? If someone has something interesting to add, they can always do it, no matter if you posted your intended solution in the question or in an answer. –  Bruno Stonek Jul 20 '12 at 13:01
    
@BrunoStonek Ok : ) –  Matt N. Jul 20 '12 at 13:07

2 Answers 2

up vote 4 down vote accepted

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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But if $N$ is generated by an uncountable set, couldn't you just pick out a countable set and make the exact same argument? I would think you would also have to make some statement about the non-redundancy of generators too, or minimality of the generating set. –  Derek Allums Jul 20 '12 at 13:40
    
I agree with Colin. If feels strange to me to write down a generating set when you are assuming $N$ is not finitely generated. After all, why not just write $N = \langle \cup_{n \in N} n \rangle$. You are right to be worried about redundancies, though that is an issue in your proof as well. Try this: choose elements $n_i \in N$ inductively so that $n_i \not\in \langle n_1, \ldots, n_{i-1}\rangle$... –  Morgan Sherman Jul 20 '12 at 14:28
    
I did indeed assume that it's countably generated, of course without intending to. Thank you! –  Matt N. Jul 21 '12 at 10:30
    
But what do you think of unit3000-21's suggestion to fix my proof? –  Matt N. Jul 21 '12 at 10:31
    
@ClarkKent Take a look at Benjamin's answer below this one. I think that ties up any loose ends. –  Derek Allums Jul 21 '12 at 21:44

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. –  user38268 Jul 20 '12 at 14:53

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