# A question about Noetherian rings [closed]

Is a ring Noetherian if and only if every submodule of a finitely generated module over it is finitely generated? If so, how does one prove this?

## closed as off-topic by rschwieb, Sahiba Arora, Namaste, TomGrubb, Siong Thye GohAug 2 '17 at 6:03

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• What's your definition of Noetherian ring? – egreg Aug 1 '17 at 9:50
• What about the solutions here and here don't cover your question? Are you having trouble proving $\impliedby$ direction? Did you even look for your question at all before blurting out this problem statement? – rschwieb Aug 1 '17 at 17:26

Let $M$ be a module over a ring $R$. We say that the module $M$ is Noetherian if and only if every submodule of $M$ is finitely generated. This is equivalent to saying that there is no strictly ascending chain of submodules of $M$. From here, it is easy to see that

1. Every submodule and homomorphic image of a Noetherian $R$-module is Noetherian.

2. Conversely, if $M$ is an $R$-module, and $N$ is a submodule of $M$ such that both $N$ and $M/N$ are Noetherian, then so is $M$.

The $R$-module $R$ is Noetherian as a ring if and only if it is Noetherian as a module. Also, if $R$ is Noetherian, then so is the $R$-module $\bigoplus\limits_{i=1}^n R$ for any $n \geq 0$. You have an exact sequence

$$0 \rightarrow \bigoplus\limits_{i=1}^{n-1} R \rightarrow \bigoplus\limits_{i=1}^n R \rightarrow R \rightarrow 0$$

with which you can apply the previous statement and induction.

Now to prove that $R$ is Noetherian if and only if every submodule of a finitely generated $R$-module is finitely generated.

($\Rightarrow$): a finitely generated $R$-module $M$ is a homomorphic image of $\bigoplus\limits_{i=1}^n R$, which is Noetherian. Hence $M$ is Noetherian, i.e. every submodule is finitely generated.

($\Leftarrow$): $R$ is a finitely generated $R$-module.