showing that an R-module is Noetherian if a submodule is Noetherian and their quotient is Noetherian That is to say, M is a R-module and N is a submodule.  If N and M/N are both Noetherian, then M is Noetherian.  I'm not exactly sure how to begin.  I have this so far: 
Since N is Noetherian it satisfies the A.C.C, so there is an ascending chain of submodules that stabilizes.  Likewise for M/N.  I was thinking of constructing some chains and then combining them somehow (as in a direct sum) to get a chain in M which would also stabilize, however I think there is a better way.  
Thank you for your insight.  
 A: Let $I_n$ be an ascending chain of submodules, $I_n\subset I_{n+1}$ of $M$ and $p:M\rightarrow M/N$ the projection. $p(I_n)$ is an ascending chain of $M/N$, so there exists $n_0: n>m>N_0$ implies $p(I_n)=p(I_m)$.
There exists also $n_0'$ such that $n>m>n_0'$ implies that $I_n\cap N=I_m\cap N$.
Let $n>m>n_0,n_0'$. Let $x\in I_n$, there exists $y\in I_m$ such that $p(y)=p(x)$ since $p(I_n)=p(I_m)$, $p(x-y)=0$, implies that $x-y\in N$, but $x-y\in I_n$ since $y\in I_m\subset I_n$, thus $x-y\in I_n\cap N=I_m\cap N$, thus $x-y\in I_m$. Therefore $x\in I_m$.
A: A proof with the characterization of $M$ being Noetherian if and only if every submodule is finitely generated.
Suppose $L$ is a submodule of $M$. Then $L\cap N$ is finitely generated because $N$ is Noetherian. Also $(L+N)/N$ is finitely generated because $M/N$ is finitely generated. Since
$$
\frac{L+N}{N}\cong \frac{L}{L\cap N}
$$
we know that $L/(L\cap N)$ is finitely generated. Let $\{x_1,\dots,x_m\}$ be set of generators for $L\cap N$ and $\{y_1+(L\cap N),\dots, y_n+(L\cap N)\}$ be a set of generators for $L/(L\cap N)$.
Then $\{x_1,\dots,x_m,y_1,\dots,y_n\}$ is a set of generators for $L$.
A: Suppose $L$ is a submodule of $M$. Then $L\cap N$ is finitely generated because $N$ is Noetherian. Also, $(L+N)/N$ is finitely generated because $M/N$ is finitely generated. We have a short exact sequence 
$0\rightarrow L\cap N\rightarrow L\rightarrow \frac{L+N}{N}\rightarrow 0$ is an short exact sequence. The left and right modules of the short exact sequence, therefore $L$ is finitely generated. Hence, $M$ is Noetherian. 
