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Free $R-module$ over an integral domain R that is not a field has finite length iff it is $0-module$

I try to begin with the most navie case:If $F$ is free $R-module$ on singleton,that is,$R$,I attempt to prove if submodule $M$ is simple,then it must be trivial.But I can't see how to do it..Any hint will be helpful , thanks.

Mugenen
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    $R$ (as an $R$-module) has submodules which are the ideals of $R$. Take a nonunit element $0\neq f$ of $R$. Consider the ideals $(f),(f^2),(f^3),...,(f^n)$. That is a chain of ideals (submodules) of length $n$. None of them are equal since $f$ is not a unit. Hence the length of $R$ is infinite. For the general case take a map $R\rightarrow M$, where $r\mapsto m\cdot r$ for some fixed $m\neq 0$ and use the same trick in the image of $R$ (which will be a quotient of $R$). I believe that should be okay. – Levent Mar 05 '18 at 12:07
  • @Levent thanks,I got it. – Mugenen Mar 05 '18 at 12:15
  • I said it will be a quotient of $R$ but since the module is free, it will be just $R$. – Levent Mar 05 '18 at 12:16
  • @Levent Sorry,I became confused again...Why the series you construct is a composition series? – Mugenen Mar 05 '18 at 12:32
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    It is not a composition series but you also don't need a composition series. Length of a module is the maximum of the lengths of chain of submodules. Even if you define it as the maximum of the length of composition series, no problem. You see a chain of length $n$, if you refine it, the length increases. – Levent Mar 05 '18 at 12:35
  • @Levent The definition of length in my book is that we say an R-module $M$ has length $m$ if it has a finite composition series of length m..So if we want to use the construction to get a contradiction,we must prove the series you constructed could be refine to a composition series which has longer length than the length of $R$.Is my understand right?.. – Mugenen Mar 05 '18 at 13:35
  • So the chain $(f^n)\subseteq (f^{n-1})\subseteq ...\subseteq (f)$ is a chain of submodules of length $n$. You can refine it to a composition series but this composition series necessarily has length larger than $n$. When you increase $n$ you see that composition series should be larger than all natural numbers, hence infinite (refining always increase the length of the chain, by definition). I don't say this chain is the composition series for the module, but the composition series for the module contains it, hence it does not have a finite length. – Levent Mar 05 '18 at 13:52
  • @Levent I find what confuse me is the things about infiniteness..The definition of composition series I learned is finite series..So is the Jordan Holder thm.All series appear in these statement are finite.So when I want to write down the proof rigorously,I always get stuck.For example,we could have series of arbitrary length,and you say it could be refined to compositions series which is not finite,but I can't see why this refinement exist.. – Mugenen Mar 05 '18 at 14:38
  • Refinement exists by Zorn's lemma. But as I said again, you don't have to study infinite composition series to answer this question. It is enough to show that the lengths of chain of submodules is not bounded above. – Levent Mar 05 '18 at 14:41
  • @Levent .And even if we get a such refinement,I can't see why we could use Jordan Holder Thm to get contradiciton(since this the statement of this thm is about two "finite"composition series.) – Mugenen Mar 05 '18 at 14:41
  • Let us continue this discussion in chat. – Levent Mar 05 '18 at 14:45

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