Show that if $M_i/M_{i-1}$ is finitely generated then $M$ is finitely generated. 
Let $0=M_0 \subset M_1 \subset \ldots \subset M_n=M$ be a finite chain of submodules. Show that if $M_i/M_{i-1}$ is finitely generated then $M$ is finitely generated.

This looks really easy but I can't see how to prove it.
Is $M_i/M_{i-1}$ a submodule of $M$? If so, then $M_i/M_{i-1}$ is noetherian and then using induction $M$ is noetherian.
 A: Let $\pi_{i}:M_{i}\to M_{i}/M_{i-1}$ be the quotient map and $\{\pi_{i}(v_{k}^{i})\}_{k=1}^{k=n_{i}}$ a set of generators of $M_{i}/M_{i-1}$. If $x_{i}\in M_{i}$, then there exists $\{\lambda_{i}^{k}\}_{k=1}^{n_{i}}$ in the ring such that
$$ \pi_{i}(x_{i})=\sum_{k=1}^{n_{i}} \lambda_{i}^{k}\pi_{i}(v_{k}^{i})$$
Then,
$$ x_{i}=x_{i-1}+\sum_{k=1}^{n_{i}} \lambda_{i}^{k}v_{k}^{i}$$
for some $x_{i-1}\in M_{i-1}$. Since $M_{n}=M$ and $M_{0}=\{0\}$ we can do this process inductively for any $x\in M$ to show that $\{v_{k}^{i}\}_{i,k}$ is a set of generators for $M$.
A: Let $M$ be a module and $N$ a submodule of $M$. Note that if $N$ and $M/N$ are finitely generated, then $M$ is finitely generated. To see this note that if $x_1,...,x_n$ is a generating set for $N$ and $[y_1],...,[y_m]$ is a generating set for $M/N$, then $x_1,...,x_n,y_1,...,y_m$ is a finite generating set for $M$. (Here I used $[ \ \ \ ]$ to denote the conjugacy class).
Now note that by your assumption 
$$M_1 \cong  M_1/0 \ \ \text{  and} \ \ \ M_2/M_1 $$
are both finitely generated. Hence $M_2$ is finitely generated by what we proved above. We can repeat this argument inductively to prove that $M_n$ is finitely generated.
