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The take-home exam give me the following problem:

Prove that a module is finitely generated if and only if the union of every chain of proper submodules of $M$ is a proper submodule.

For the if side, I assume $M$ is finitely generated and we have a chain of proper submodules. I tried to construct a chain $A_{i}\subset A_{i+1}$ and argue that because $M$ is noetherian this chain must stop somewhere, and hence obtain a contradiction. But I saw this is totally wrong (see my post: Is every finitely generated module Noetherian?). So I am wondering if this approach is still feasible. Others have pointed out that a submodule of a finitely-generated module is not necessarily finitely generated.

For the only if side, I tried to construct a sequence of submodules by taking generators in the form $$\langle m_{1}\rangle \subset \langle m_{1},m_{2} \rangle\subset \langle m_{1},m_{2},m_{3}\rangle....$$ assuming $M$ has infinitely many generators. Then I argue the union of them must be $M$ while individually they are all proper submodules of $M$. But the professor commented "NO! Otherwise $\mathbb{R}$ will be xxx over $\mathbb{Q}$! I cannot read his text and so I wish to ask in here where I got wrong.

Below is a copy and paste of original proof.I feel quite embarrassed.

For the if side, assume $M$ is finitely generated and we have a chain $\{A_{i}\}$. Then $\bigcup_{i\in I} A_{i}\leftrightarrow A_{1}\bigcup \cup_{i\in I,j\not=1}A_{i}$. By our assumption either $A_{1}\subset \cup_{i\in I,j\not=1}A_{i}$ or $\cup_{i\in I,j\not=1}A_{i}\subset A_{1}$. In the first case $A_{1}=M$ and we had contradiction. In the second case we have a chain of submodules $A_{1}\subset \cup_{i\in I,j\not=1}A_{i}$. I claim that there must be some $A_{2}\in \cup_{i\in I,j\not=1}A_{i}$ such that $A_{1}\subset A_{2}$, for otherwise $\cup_{i\in I,j\not=1}A_{i}$ would be all of elements contained in $A_{1}$, which is a contradiction; so there is at least one $A_{i}$ not contained in $A_{1}$, hence must contain $A_{1}$. We now name it as $A_{2}$. Since $M$ is noetherian, the chain $$A_{1}\subset A_{2}....$$ terminates at some $A_{n}$ contained in the original chain. By assumption $A_{n}$ is a proper ideal of $M$ and we verify $\bigcup_{i\in I}A_{i}=\bigcup_{i\le n}A_{i}=A_{n}$ by our construction.

For the only if side, assume every chain of proper submodules has a union that is a proper ideal of $M$. Then in particular we can form a chain of submodules by $\langle m_{1}\rangle \subset \langle m_{1},m_{2}\rangle...$ where $m_{i}$ are pairwise different. If $M$ is not finitely generated, then we can pick $m_{i}$ infinitely as we desire; and every submodule must be a proper submodule. However their union is $M$ by definition. This showed $M$ must be finitely generated.

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The "xxx" in your professor's comment was probably "finite-dimensional". The problem is that if there is no countable set of generators for $M$, then the union of that sequence will not be $M$. For example, as a vector space over $\mathbb{Q}$, $\mathbb{R}$ has no countable basis. – Brad Nov 28 '12 at 3:28
I see. Thanks! I understand now. – Bombyx mori Nov 28 '12 at 3:30
If $M$ is finitely generated, then as you point out it's not necessarily true that every ascending chain is eventually constant (i.e. $M$ is not necessarily Noetherian). But you can show that if the union of that chain is $M$, then the chain is eventually constant, given that $M$ is finitely generated. – Brad Nov 28 '12 at 3:39

$\,\Longrightarrow\,$: Suppose $\,M=\langle\,x_1,...,x_n\,\rangle\,$ is f.g., and let $\,N_1\subset N_2\subset \ldots\,$ be any ascending chain of proper submodules.

If $\,\bigcup_iN_i=M\,$ then there must exist some $\,N_k\,$ in the chain that contains all the generators $\,x_1,...,x_n\,$ of $\,M\,$, contradicting the fact the chain is of proper submodules. Thus the above union must be a proper submodule.

$\,\Longleftarrow\,$: Take any non-zero $\,x_1\in M\,$, then take $\,x_2\in M-\langle x_1\rangle\,$, then $\,x_3\in M-\langle x_1,x_2\rangle\,$ , etc.

We get this way an ascending chain of proper submodules $\,N_1:=\langle x_1\rangle\subset N_2:=\langle x_1,x_2\rangle\subset\ldots\,$ for which we get $\,\bigcup_iN_i=M\,$ if it continues infinitely (i.e., if $\,M\,$ is not f.g.). This contradicts the assumption and then the chain must be finite and, thus, $\,M\,$ is f.g.

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Your proof of the second part is essentially the same as my proof. I do not think it works when you have uncountable cardinals. – Bombyx mori Dec 1 '12 at 17:14

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