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I'm a bit confused by the relationship between $V$, $V^*$, and $V^{**}$. If $V$ is finite dimensional, everything works out nicely: $V$ is isomorphic to $V^*$, and $V$ is naturally isomorphic to $V^{**}$.

My question is: what if $V$ is not finite dimensional? Do these isomorphisms still hold?

Furthermore, we can consider a free module $M$ with a basis $X$. If $X$ is finite, say $X=\{x_1,\cdots,x_n\}$, then $\operatorname{Hom}_R(M,R)$ is a free module with the (dual) basis $\{\pi_i:M\to R\}$, where $\pi_i(x_j)=\delta_{ij}$. In this case $M$ is isomorphic to $\operatorname{Hom}_R(M,R)$ as $R$-modules.

Now, if the basis of $M$ is not finite, is $\operatorname{Hom}_R(M,R)$ still free? If it has a basis, what can we say about it?

Any insight is appreciated! Thank you in advance!

Sardines
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  • They can never be isomorphic in the infinite dimensional case because the duel will always have a cardinality that's strictly greater than the original vector space. The same holds for the double duel. – CyclotomicField Nov 29 '22 at 02:19
  • If $V$ is an infinite-dimensional Banach space, the natural map from $V$ to $V^{*}$ is an isometry, but it is possible for the map to not be surjective. Spaces $V$ for which the map is surjective are called reflexive. If $V$ is a Hilbert space, then $V \approx V^$ by the Riesz conjugate linear isomorphism. – Mason Nov 29 '22 at 02:40
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    @Mason given the way that the question is phrased, I think Banach spaces and Hilbert spaces are beside the point: this seems to be a question about algebraic duals of vector spaces/modules without any topology. – Lukas Heger Nov 29 '22 at 02:49

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If $R=\Bbb Z$ and $M=\bigoplus_{n \in\Bbb N}\Bbb Z$, then $\mathrm{Hom}_R(M,R)=\prod_{n \in \Bbb N}\Bbb Z$. One can show that this module is not free, but that's a nontrivial fact. See this question for details.

Although in this case, the dual module is not free, it turns out that the double dual of $M$ is isomorphic to $M$. Namely, we have $$\mathrm{Hom}_{\Bbb Z}(\mathrm{Hom}_{\Bbb Z}(\bigoplus_{n \in \Bbb N} \Bbb Z,\Bbb Z),\Bbb Z)\cong \mathrm{Hom}_{\Bbb Z}(\prod_{n \in \Bbb N}\Bbb Z,\Bbb Z) \cong \bigoplus_{n \in \Bbb N} \Bbb Z$$This is again a nontrivial result and something that cannot happen over a field if the basis is infinite, as CyclotomicField remarked in the comments.

Lukas Heger
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