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.
On
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
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 $R^n/\ker(\phi)\cong M$.