Let $R$ be a commutative Noetherian ring. Prove that an $R$ module $M$ is noetherian iff $M$ is finitely generated.

One way is obvious. The other I have to prove every submodule of $M$ is finitely generated

  • $\begingroup$ Or to prove that a finite direct sum of noetherian modules is noetherian and likewise is a quotient module. $\endgroup$ – user26857 Apr 25 '14 at 14:18

Let $R$ be a commutative noetherian ring.

As user26857 indicated, you can show that over noetherian rings finite direct sums of noetherian modules and quotients of noetherian modules by noetherian submodules are noetherian.

Both statements follow from the following fact.

Fact: Let $0 \to M' \to M \to M'' \to 0$ be an exact sequence of $R$-modules. Then $M$ is noetherian if and only if both $M'$ and $M''$ are noetherian.

Using this fact, we first see that direct sums of noetherian modules are noetherian: Let $M_1, \dots, M_n$ be noetherian $R$-modules. Assume by induction that $\oplus_{i=1}^{n-1} M_i$ is noetherian. Since the sequence $0 \to M_n \to \oplus_{i=1}^{n} M_i \to \oplus_{i=1}^{n-1} M_i \to 0$ is exact, we conclude that $\oplus_{i=1}^{n} M_i$ is noetherian.

Similarly, if $M$ is a noetherian $R$-module and $N$ an $R$-submodule, then $0 \to N \to M \to M/N \to 0$ is exact. It follows that $M/N$ (and $N$) are also noetherian as $R$-modules.

Finally, if $M$ is a finitely generated module over $R$, then $M$ is a quotient of $R^n$ and therefore noetherian.

Proof of the fact: Assume that $M$ is noetherian. An ascending chain of submodules of $M'$ (or $M''$) gives rise to an ascending chain of submodules in $M$ is is therefore stationary. Therefore the sequences in $M'$ and $M''$ become stationary and so $M'$ and $M''$ are noetherian.

Assume now that $M'$ and $M''$ are noetherian. Given an ascending chain of submodules of $M$, the preimages give rise to an ascending chain of submodules in $M'$ and the images to an ascending chain of submodules of $M''$. Both become stationary. Since the sequence is exact, the original sequence in $M$ becomes stationary too.

  • $\begingroup$ So $R^{n}$ is noetherian and there is an injective homomorphism $R^n->M$. So in the exact sequence $0->R^{n}->M->A->0$ what (noetherian module) should $M$ go to ? $\endgroup$ – user137090 Apr 25 '14 at 16:14
  • $\begingroup$ @user26857 So the exact sequence is $0->R->R^n->M->0$. Then $R^n$ is noetherian, so is M? $\endgroup$ – user137090 Apr 25 '14 at 18:40
  • $\begingroup$ If M is finitely generated, then there is a surjection $R^n \to M$, i.e. there is an exact sequence $0 \to I \to R^n \to M \to 0$. Since $R^n$ is noetherian, $M$ is noetherian by the above fact. $\endgroup$ – user44400 Apr 26 '14 at 20:07
  • $\begingroup$ What is $I$? There has to be an injective homomorphism $f$ from $I$ to $R^n$ with $im(f)$ equal to the kernel of the map from $R^n$ to $M$. Thanks $\endgroup$ – user137090 Apr 26 '14 at 21:13
  • 1
    $\begingroup$ You define $I$ to be the kernel of the surjection $R^n \to M$. The map $I \to R^n$ is the natural inclusion. $\endgroup$ – user44400 Apr 28 '14 at 9:04

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