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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.2012-11-28
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    I see. Thanks! I understand now.2012-11-28
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    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.2012-11-28
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    Do you find a solution? The set of proper submodules is inductively ordered is in fact a categorical definition for an object in some category to be finitely generated. And it coincides with the usual definition in a module category over some ring. The proof needs some elementary ordinal number arguments. For details see Categories and Fuctors, Bodo Pareigis.2017-07-12
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    Page 204 if you ask2017-07-12

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