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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.

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  • $\begingroup$ Modules don't have ideals, so I guess you want to construct a left submodule which is not finitely generated. But my hint is to prove directly that any submodule is in fact finitely generated. $\endgroup$
    – Brenin
    Nov 6, 2012 at 0:30

2 Answers 2

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

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  • $\begingroup$ Thanks a lot! I think that makes sense. $\endgroup$
    – Erik
    Nov 9, 2012 at 0:52
  • $\begingroup$ @Erik You need to prove that $R^n$ is Noetherian. $\endgroup$ Nov 9, 2012 at 0:56
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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

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  • $\begingroup$ Thanks for the response. I wasn't sure about the basis step. Why does the isomorphism between M and A/I imply N is finitely generated? $\endgroup$
    – Erik
    Nov 9, 2012 at 0:47
  • $\begingroup$ @Erik Because $A/I$ is Noetherian and $N$ is isomorphic to a submodule of $A/I$. $\endgroup$ Nov 9, 2012 at 0:54

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