Consider the free $k$-algebra $k[x_i]_{i \in I}$ indexed by $I$. Then is $Hom_{k-Mod}(k[x_i]_{i \in I},k) \cong k[x_i]_{i \in I}$?
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6This already fails for a one element $I$; the dimension of $k[x]$ as a vector space over $k$ is countable, the dual has uncountable dimension. – egreg Feb 08 '14 at 22:32
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@egreg Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Apr 12 '15 at 11:36
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@JulianKuelshammer I'll do it immediately. – egreg Apr 12 '15 at 12:22
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This already fails for $I$ a singleton: the dimension of $k[x]$ as a vector space over $k$ is countable, while the dual $$ \operatorname{Hom}_{k\text{-Mod}}(k[x],k) $$ has uncountable dimension.
See Slick proof? A vector space has the same dimension as its dual if and only if it is finite dimensional on MathOverflow.
So the result above is general for any set of indeterminates.
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Better, stay on MSE where you can find a proof wtihout the defects of the linked MO proof. – Bill Dubuque Apr 14 '15 at 18:40