Noetherian rings and modules A ring is left-Noetherian if every left ideal of R is finitely generated. Suppose R is a left-Noetherian ring and M is a finitely generated R-module. Show that M is a Noetherian R-module.
I'm thinking we want to proceed by contradiction and try to produce an infinitely generated ideal, but I'm having trouble coming up with what such an ideal will look like.
 A: If $\{x_i\mid 1\leq i\leq n\}$ is a set of generators for $M$, then the obvious map $\phi$ from $R^n$ to $M$ is a surjection. Since $R^n$ is a Noetherian left module, so is $R^n/\ker(\phi)\cong M$.
A: Let $N$ be an $A$-submodule of $M$.
It suffices to prove that $N$ is finitely generated.
We prove this by induction on the number of generators of $M$.
Suppose $M = Ax_1 + \cdots + Ax_n$.
If $n = 1$, $M$ is isomorphic to $A/I$, where $I$ is a left ideal of $A$.
Hence $N$ is finitely generated.
Suppose $n > 1$.
Let $L = Ax_1 + \cdots + Ax_{n-1}$.
There exists the following exact sequence
$$0 \rightarrow N \cap L \rightarrow N \rightarrow M/L.$$
By the induction hypothesis, $N \cap L$ is finitely generated.
Since $M/L$ is generated by the image of $x_n$, the image of $N \rightarrow M/L$ is finitely generated.
Hence $N$ is finitely generated by the following lemma.
Lemma
Let $A$ be a ring.
Let $M$ be an $A$-module.
Let $N$ be an $A$-submodule of $M$.
Suppose $N$ and $M/N$ are finitely generated.
Then $M$ is also finitely generated.
Proof:
Suppose $N$ is generated by $x_1,\dots,x_n$ and $M/N$ is generated by $y_1$ mod $N,\dots,y_m$ mod $N$.
Let $x \in M$.
Then there exist $b_1,\dots,b_m \in A$ such that $x \equiv b_1y_1 + \cdots + b_my_m$ (mod $N$).
Hence there exist $a_1,\dots,a_n \in A$ such that $x - (b_1y_1 + \cdots + b_my_m) = a_1x_1 + \cdots + a_nx_n$.
Hence $M$ is generated by $x_1,\dots,x_n, y_1,\dots,y_m$.
QED
