# Finitely generated module over Noetherian ring

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.

• Not every finitely generated $R$-module can be decomposed as $R/\mathfrak{i_1}\oplus \dots \oplus R/\mathfrak{i_n}$. – Angina Seng Jun 30 '19 at 16:16
• $M$ is supposed to be finitely generated – Dorian Jun 30 '19 at 16:18
• 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. – Angina Seng Jun 30 '19 at 16:18
• 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.) – Angina Seng Jun 30 '19 at 16:39
• It means, that if every submodule of every finitely generated module is finitely generated, then $R$ is Noetherian. – Angina Seng Jun 30 '19 at 16:50

$$R$$ 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.

• 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 – Dorian Jun 30 '19 at 16:31