Is there a chain of finitely generated modules $M_i$ not satisfying ascending chain condition s.t. $\cup_i M_i$ is finitely generated? It is easy to find such a chain s.t. $\cup_i M_i$ is not finitely generated. So I am curious about the other situation. Can $\cup_i M_i$ be finitely generated?
Thanks in advance.
 A: For the record: you are putting more information in the subject line than in the body. It should be the other way around! The body should be self-contained and include all relevant information, while the subject line gives a rough idea of what the question is about.

First, say $\{M_i\}_{i\in\omega}$  is an ascending chain of modules, and let $M=\cup_{i\in\omega}$ be their union, which is also a module. The chain "stabilizes" if and only if there exists $N\in\omega$ such that for all $m\geq N$, $M_m=M_N$.
First, let us put no conditions on any of the $M_i$. We have the following three examples:

*

*The chain stabilizes and $M$ is finitely generated. Just take your favorite finitely generated module (say, a cyclic one), and let $M_i=M$ for all $i\in\omega$.


*The chain stabilizes and $M$ is not finitely generated. Take your favorite non-finitely generated module (e.g., the free module on the set $\{x_i\}_{i\in\omega}$), and let $M_i=M$ for all $i\in\omega$.


*The chain does not stabilize and $M$ is not finitely generated. Take your favorite non-finitely generated but countably generated module $M$ (e.g., the one mentioned above), let $X=\{m_i\}_{i\in\omega}$ be a generating set for $M$, and let $M_i=\langle x_0,\ldots,x_i\rangle$ for each $i\in\omega$. The chain never stabilizes, and the union is $M$, which is not finitely generated.
However,
Proposition 1. If $M$ is finitely generated, then the chain stabilizes.
Proof. Let $m_1,\ldots,m_k$ be a generating set for $M$. For each $j$, $1\leq j\leq k$, there exists $i_j\in\omega$ such that $m_j\in M_{i_j}$. Let $N=\max\{i_1,\ldots,i_k\}$. Then $m_j\in M_{i_j}\leq M_N$ for all $j$, hence for all $n\geq N$,
$$M = \langle m_1,\ldots,m_k\rangle \leq M_N\leq M_n \leq M,$$
so we have equality throughout and in particular the chain stabilizes. $\Box$
If we also require each $M_i$ to be finitely generated, then example 2 no longer applies. In that case, we have:
Proposition 2. Let $\{M_i\}_{i\in\omega}$ be a chain of finitely generated modules. Then the chain stabilizes if and only if $M=\cup_{i\in\omega}M_i$ is finitely generated.
Proof. If the chain stabilizes, then there exists $N\in\omega$ such that $M=M_N$, and since $M_N$ is finitely generated, so is $M$. The converse is Proposition 1. $\Box$
