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Let $R$ be a ring that satisfies ACC on the set of its left ideals and $M$ be a finite generated $R$-module. Prove that every submodule of $M$ is finitely generated.

I know that if $M$ satisfies ACC condition then every submodules of it are finite generated. I only want to show that $M$ satisfies ACC condition.

Help me some hints.

Thanks a lot.

user109584
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    Hint: Rings meeting the ACC condition are called Noetherian rings. Prove that a ring is Noetherian iff all of its ideals are finitely generated. Then prove that a module meets the ACC condition iff all its submodules are finitely generated. – Camilo Arosemena Serrato Nov 25 '13 at 01:46
  • I've done the first part of you hint. For the second, can you explain more precise? – user109584 Nov 25 '13 at 01:52
  • A module that meets the ACC conditon is called Noetherian module. Is it right? – user109584 Nov 25 '13 at 01:56

1 Answers1

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First try to show following fact: For any exact sequence $A\rightarrow M\rightarrow B$, $M$ satisfies ACC iff $A$ and $B$ satisfy ACC. Use it to prove that finitely generated free modules satisfy ACC.

  • There is an exact sequence $R^{n-1}\rightarrow R^n\rightarrow R$. $R$ satisfy ACC. Assume that $R^{n-1}$ satisfy ACC too then by induction so does $R^n$.

Now notice that module $M$ is finitely generated iff there is a finitely generated free module $F$ and a surjection $f:F\rightarrow M$.

  • If $M$ has generators $\{m_1,...,m_n\}$ then there is a surjection $f:R^n\rightarrow M$ defined on base $f:x_i\mapsto m_i$

Use fact again to prove that $M$ satisfy ACC.

  • There is an exact sequence $\ker(f)\rightarrow R^n\rightarrow M$.

Suppose that there is submodule $N\subset M$ which is not finitely generated. Enumerate its elements then we have sequence of submodules $(m_1)\subset (m_1,m_2)...$ It is easy to see that this sequence will not stabilise.

user52045
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