Given a finitely generated $R$-module $M$, suppose there is a set of generators $m_1, \cdots m_n$. If we could show that $\{m_i\}$ are linearly independent, does this imply $M$ is free?

I think it should since if $\{m_i\}$ are linearly independent, then the map $$\phi: M \rightarrow R^n \quad \text{ where } \quad m=\sum_i r_im_i \mapsto (r_1, \cdots, r_n)$$ is both injective and surjective.

Is this correct?

  • $\begingroup$ But can't you just reduce the collection of generators into a linearly-independent set anyway? $\endgroup$ – gary Jul 27 '17 at 1:02
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    $\begingroup$ @gary no in general. If it is torsion-free, then yes. I don't know much besides that. $\endgroup$ – Cauchy Jul 27 '17 at 1:45

Yes, an equivalent definition of a free module is that it has a basis, i.e it has a set of generators that are linearly independent.


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