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It's a common theorem that when $M$ is a finite-free $R$-module of rank $n$, there is a natural isomorphism $M\cong M^{\vee\vee}$, where $M^\vee$ denotes the dual. So $M^{\vee\vee}$ is also free of finite rank $n$.

If we only know that $M$ is free, but not necessarily of finite rank, is it still true that the natural homomorphism is injective?

I was thinking one could take a basis $\{e_i\}_{i\in I}$ for $M$ over $R$, with $\{e^\vee_i\}_{i\in I}$ to be the dual basis of $M^\vee$. (Apparently this part does not follow in the nonfinite case. How can it be made to work?) The argument I'm familiar with is that if $m\in M$ is nonzero, then $m$ has nonzero coordinates relative to $\{e_i\}$, so $e^\vee_i\neq 0$ for some $i$. So the evaluation map $\mathrm{ev}_m(e^\vee_i)\neq 0$, so $\mathrm{ev}_m\neq 0$ in $M^{\vee\vee}$. Here $\mathrm{ev}_m$ is the map sending $\phi$ to $\phi(m)$ for $\phi\in M^\vee$. So the only $m$ for which $\mathrm{ev}_m=0$ in $M^{\vee\vee}$ is $m=0$, and thus the map is injective.

Is this valid, or is there a subtlety that relies on $M$ being finite that I'm not seeing? Thank you.

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    Dear @Pierre-YvesGaillard, sorry for the multiple pings! I just saw your latest comment. Thank you then for the nice concise proof then when $M$ is free of any rank.2012-03-12

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