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Let $M=\langle e_1,\ldots,e_n\rangle$ be a finitely generated $R$-module.

My question is can we construct a free submodule $F$, i.e, isomorphic to $R^s$ for some $s$, finding a subset $S=\{e_1,\ldots,e_s\}$ such that $S$ is a maximal independent subset of $M$, then $S$ generates this free submodule $F$ of $M$ with basis $S$.

I know this is true in vector spaces, but in general free modules?

Thanks in advance

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$\Bbb Z/3\Bbb Z$ is a finitely generated $\Bbb Z$ module, but it does not contain any free $\Bbb Z$ modules $\Bbb Z^s$ for any integer $s>0$.

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  • $\begingroup$ I'm studying this material math.uga.edu/~pete/integral.pdf of Prof. Peter Clark and in the last lines of the page 34, he uses this argument with R integral domain, am I forgetting something? $\endgroup$ – user42912 Mar 18 '14 at 19:33
  • $\begingroup$ @user42912 From that context, we see that there is an additional assumption that $M$ is torsionfree. $\Bbb Z/3\Bbb Z$ is not torsionfree. That is what you omitted from your original post. $\endgroup$ – rschwieb Mar 18 '14 at 20:07
  • $\begingroup$ So should I edit the question? thanks $\endgroup$ – user42912 Mar 18 '14 at 20:08
  • $\begingroup$ @user42912 Well, it is a little late now to supply context now that there are two correct answers to the original question. You may as well ask a new question with the correction. Remember to supply context ahead of time to avoid this :) $\endgroup$ – rschwieb Mar 18 '14 at 20:09
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    $\begingroup$ I've already asked the question with a link to this one, thanks again :) $\endgroup$ – user42912 Mar 18 '14 at 21:04
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In general, the construction you're looking for is not possible, unless you count a trivial solution which I'll explain in a moment. Consider the $\mathbb Z$-module $\mathbb Z/(4)\times\mathbb Z/(4)$ with generating set $S=\{(1,0),(0,1)\}$. No submodule of $S$ is isomorphic to $\mathbb Z$ or $\mathbb Z^2$.

On the other hand, if you allow the "rank $0$ module" $\{0\}$ and define the span of $\varnothing$ to be $\{0\}$ (this is a standard definition) then you can pick the zero submodule and the empty subset of $S$ to make your construction. But that's not very interesting.

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  • $\begingroup$ I'm studying this material math.uga.edu/~pete/integral.pdf of Prof. Peter Clark and in the last lines of the page 34, he uses this argument with $R$ integral domain. $\endgroup$ – user42912 Mar 18 '14 at 19:32

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