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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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    The only part your wrong is that $\{ e_i^\vee\}$ is NOT a basis. In fact if $M= \bigoplus_{i \in I} R e_i$, then $M^\vee = \prod_{i \in I} R e_i^\vee$.2012-03-11
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    Thanks @user10676. How can the idea be adapted correctly then?2012-03-12
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    Dear Adeal: The natural map $M\to M^{**}$ is injective because the condition $x\neq0$ for $x\in M$ means that there is a coordinate $f\in M^*$ satisfying $f(x)\neq0$. (I of course assume that $M$ is free.)2012-03-12
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    Thanks @Pierre-YvesGaillard. Is that all there is to it then? If so, why is the argument in the finite rank case a little more involved?2012-03-12
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    Dear @Pierre-YvesGaillard, I apologize, I'm just confused with the responses I've gotten. The proof I gave in the body of question is basically the one Keith Conrad gives in his blurb on Dual Modules as [Theorem 4.2](http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/dualmod.pdf) so I'm surprised to hear it's not right. I'm curious, if the proof of injectivity when M is free is so simple, why would the proof be presented as such?2012-03-12
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    Dear Adeal: (Keith Conrad's proof is certainly correct.) Let $f:R^n\to M$ be an isomorphism, and consider the following chain of isomorphisms: $M\to R^n\to R^{n**}\to R^{**n}\to R^n\to M$, the first map being $f^{-1}$, the last being $f$. Claim: the composition is the natural map $\phi_M : M \to M^{ * * }$. Proof: use the fact that $\phi_M$ is compatible (in a sense that can be made precise) with morphisms and finite direct sums. The compatibility with finite direct sums follows from the natural isomorphism $(M\oplus N)^*\simeq M^*\oplus N^*$. (By the way, I up-voted your nice question.)2012-03-12
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    Dear Adeal: PS: The above is the way I would phrase the argument for $M$ free of finite rank. I believe the proof I gave for the injectivity when $M$ is free of any rank is complete and correct.2012-03-12
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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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