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There are a few relevant posts

Error in proof that submodules of f.g. modules are f.g.

Specific proof that any finitely generated $R$-module over a Noetherian ring is Noetherian.

Finitely generated modules over a Noetherian ring are Noetherian

I'm trying to straighten out noetherianess and finite generation in my brain. So, say $M$ is a finitely generated left $R$-module. That is

$$M=Rm_1 + \dots + Rm_k$$

Then it feels to me that every submodule ought to be f.g. since for any chain

$$0=N_0 \subset N_1 \subset \dots \subset N_n \subset \dots $$

We have $M = \bigcup N_j$ and so each $m_i$ has to appear in $N_i$ for some $i$. We can use this to conclude finite generation. Then, since every submodule if f.g. we can conclude that $M$ is noetherian. This is incorrect but I'm having trouble understanding why despite going through other posts.

Then, knowing the above is incorrect, it would suffice to show that, in particular,

$$R + R$$

is noetherian (then it would follow for any finite sum by induction). So, given any chain of submodules (which will look like the sum of submodules of $R$) we will have something like

$$0 \subset N_1 + N'_1 \subset \dots $$

and so we can first stabilize the chain

$$N_1 \subset \dots $$

since it is a chain of submodules in $R$ and similarly for the $N'$ and so the whole chain above must stabilize. Then, given this, there is clearly a map from $R \to M$ which is an isomorphism which would give the result?

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    $\begingroup$ There are rings which are not Noetherian. In that case $M=R$ is finitely generated but has non-finitely generated submodules. $\endgroup$ Oct 27, 2018 at 13:46
  • $\begingroup$ That makes sense. I know that a counter example exists I just don't see the flaw in the logic above? $\endgroup$
    – RhythmInk
    Oct 27, 2018 at 13:47

1 Answer 1

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$$M=Rm_1 + \dots + Rm_k$$

Then it feels to me that every submodule ought to be f.g. since for any chain

$$0=N_0 \subset N_1 \subset \dots \subset N_n \subset \dots $$

Wrong

We have $M = \bigcup N_j$ and so each $m_i$ has to appear in $N_i$.

We do not know that the union of any chain will be the entire module we are looking at. So we aren't able to apply that argument.

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