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I read that these two proposition are equivalent:

  • $R$ is a noetherian ring;
  • Every submodule of a finitely generated $R-$module is finitely generated.

While I was thinking about it I made the reasoning below, that leads me to an incorrect conclusion, so I must be wrong somewhere. I hope someone can explain me, thank you in advance.

Calling $M$ the $R-$module, we have $M\simeq R/\mathfrak{i_1}\oplus \dots \oplus R/\mathfrak{i_n}$, so it's clear that if $R$ is noetherian, so is $M$; however if $R/\mathfrak{i_1}\oplus \dots \oplus R/\mathfrak{i_n}$ is noetherian for some ideals $\mathfrak{i_1},\dots , \mathfrak{i_n}$, then $R$ is not necessarily noetherian. That means that the fact that $M$ is noetherian doesn't imply that $R$ is noetherian too.

For example, I let the ring of polynomials $\Bbb R[T_1,T_2,\dots ]$ act on the group $\Bbb R^2$, with $T_im=0\; \forall i,\forall \;m\in \Bbb R^2$. I obtain a module which is isomorphic to $\Bbb R[T_1,T_2,\dots ]/(T_1,T_2,\dots)\oplus \Bbb R[T_1,T_2,\dots ]/(T_1,T_2,\dots)$, which is in fact $\Bbb R^2$. This module is noetherian since every sumbodule is a subspace of dimension $1$ but clearly $\Bbb R[T_1,T_2,\dots ]$ is not.

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    $\begingroup$ Not every finitely generated $R$-module can be decomposed as $R/\mathfrak{i_1}\oplus \dots \oplus R/\mathfrak{i_n}$. $\endgroup$ – Angina Seng Jun 30 '19 at 16:16
  • $\begingroup$ $M$ is supposed to be finitely generated $\endgroup$ – Dorian Jun 30 '19 at 16:18
  • $\begingroup$ I don't see what you are asking. In your example, $\Bbb R[T_1,T_2,\ldots]$ is a finitely generated module over itself, but the ideal generated by the $T_i$ is not finitely generated. $\endgroup$ – Angina Seng Jun 30 '19 at 16:18
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    $\begingroup$ If you want a Noetherian module, over a non-Noetherian ring, the easy way to construct one is to take the zero module. (Of course the theorem you cite says nothing about the non-existence of Noetherian modules for non-Noetherian rings.) $\endgroup$ – Angina Seng Jun 30 '19 at 16:39
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    $\begingroup$ It means, that if every submodule of every finitely generated module is finitely generated, then $R$ is Noetherian. $\endgroup$ – Angina Seng Jun 30 '19 at 16:50
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$R$ is an $R-$ module with respect itself so if it has unity then it will be finitely generated then by hypothesis you have that each sub module of $R$ is finitely generated.

You can observe that each ideal $I$ of $R$ is an $R-$ submodule of $R$ so it will be finitely generated, then there exists a finite number of elements $a_1,\dots, a_n$ such that

$I=(a_1,\dots , a_n)$

So $R$ is a noetherian ring.

Now we suppose that $R$ is a noetherian ring. If $f_1,\dots, f_n$ are generators of $M$ then you can define

$\psi: R\times \dots \times R\to M$ such that for each $(r_1,\dots r_n)$ you have that

$\psi(r_1,\dots , r_n):=r_1 f_1+\dots +\dots r_nf_n$

The map is surjective so by first homomorphism theorem you have that

$M\cong (R\times \dots \times R)/ ker(\psi)$

It is easily prove that if $R$ is a noetherian ring then $R\times \dots \times R$ is such that each its submodule is finitely generated.

Each submodule of $(R\times \dots \times R)/ K$ is finetly generated where $K$ is submodule of $R\times \dots \times R$ so you have finished.

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  • $\begingroup$ Thank you what you said is clear, however in my example I don't have a regular module (i.e. a ring considered as a module over itself), I have $\Bbb R ^2$ considered as a $\Bbb R [T_1,T2\dots]-$module $\endgroup$ – Dorian Jun 30 '19 at 16:31

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