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I have that $R$ is a domain and $M$ is a finitely-generated $R$-module. I am trying to show that there is a free submodule $F$ such that $M/F$ is a torsion $R$-module.

I have the fact: If $X$ is a torsion generating set of $M$ then $M$ is torsion.

I am trying the following strategy. I take $X = \{x_1, ... ,x_n \}$ my generating set of $M$ then I take a maximal linearly independent subset of $X$, say $B = \{x_1, ..., x_m \}$ (up to reordering) with $m \le n$. Then I look at the quotient $M/F$ where $F = <B>$. I know that this quotient is generated by $\{x_{m+1} + <B>, ..., x_n + <B> \}$. Now I want to show that these elements are all torsion and the result will follow. So I suppose it's not torsion so for every non-zero $r \in R$, $rx_i$ is non zero ($i = m+1,..,n$) now I am trying to use this to contradict that B is maximal linearly independent. I'm not quite sure if this is a strategy that will work.

Thanks for any help

Wooster
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1 Answers1

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Notice that $\{ x_1 , ..., x_m , x_{m+1} \}$ is linearly dependent, so that $\sum_1 ^{m+1} r_i x_i =0 $ for some nonzero $r_i$. Since B is linearly independent, you must have that $r_{m+1} \neq 0$, so that $r_{m+1} x_{m+1} \in F$. The same goes for all the other $x_i$ with $i>m$.

Ofir
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  • Thanks for your answer, just struggling to see why the fact $r_{m+1}x_{m+1} \in F$ tells us that $x_{m+1} + $ is torsion? - Never mind, I see it now, because this tells us that $r_{m+1}x_{m+1} + = $ – Wooster Mar 10 '14 at 16:48