# Go from finitely generated module to free module

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?

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