When are dual modules free?

The dual module of a finitely generated module is reflexive, that is, $M^{**}=M$, and reflexives are awfully close to projectives. Specifically, if $R$ is a Noetherian domain, then a module is projective if $Ext^i(M,R)=0$ for all $i>0$, and its reflexive if $Ext^i(M,R)=0$ for $i=1,2$.

It is also worth noting that every reflexive is the dual of some module, specifically of $M^*$. Therefore, your question amounts to "for what rings is every reflexive module free?" In this light, its very similar to the question of when every projective module is free.

From the above Ext criterion, its clear that if the global dimension of $R$ is less than or equal to 2, that being reflexive is the same as being projective. I would go so far as to conjecture the converse is true: that if gldim of R is 3 or more, that there is a non-projective module which is reflexive (and hence it is non-free).

If this conjecture is true, then the answer to your question is "rings with global dimension 2 or less, such that every projective is free". Of course, its not immediately clear what these are, but its a start.

OK, I now have a counter-example. Thanks to the previous answers for showing me where to look.

Let $A = k[x,y,z]$. Let $M$ be the kernel of the map $(x,y,z) : A^3 \to A$ and $N$ the co-kernel of the map $(x,y,z)^T: A \to A^3$. I claim that $M = N^{*}$ but $M$ is not free.

To see that $M=N^{*}$, consider the defining sequence $$0 \to A \to A^3 \to N \to 0.$$ This gives rise to $$0 \to N^{*} \to A^3 \to A.$$ The kernel of the right hand map is $M$ by definition.

Now, let's see that $M$ is not free. We have a graded short exact sequence $$0 \to M \to A^3 \to A[1] \to k[1] \to 0.$$ So the Hilbert series of $M$ is $$\frac{3}{(1-t)^3} - \frac{t^{-1}}{(1-t)^3} + t^{-1} = \frac{(1-t)^3 - 1 + 3t}{t(1-t)^3} = \frac{3t-t^2}{(1-t)^3}.$$ If $M$ were a free module, its Hilbert series would look like $(t^a+t^b)/(1-t)^3$.

$N$ is also not free; I have not figured out whether $N$ is reflexive.

Every dual $T^*$, where $T$ is torsion-free -- and hence every reflexive module -- is a second syzygy, as displayed by dualizing a projective presentation of $T$. On the other hand, it follows from Auslander-Bridger (or see a slightly more readable presentation by Masek (last Corollary in this paper) that if a ring $R$ satisfies S1 and is Gorenstein at the minimal primes, then every second syzygy is reflexive.

Therefore, for reduced rings being a dual of a torsion-free is equivalent to being reflexive is equivalent to being a second syzygy. In particular, as long as the global dimension is at least 3, there are duals that are not projective.

Back to the original question: when can you conclude the dual of $T$ is free? (I'm going to talk only about local rings, so ignore the distinction between free and projective.) Assume $A$ is a regular local ring. If there is a module $N$ such that $\operatorname{Ext}(T,N)=0$ for $i = 1, ..., \operatorname{depth} (N)-2$, then the dual of $T$ is free. In particular, if $N$ has depth less than or equal to $3$ and $\operatorname{Ext}_R^1(T,N)=0$, then the dual of $T$ is free. This is in a recent paper by Jothilingam, but is not hard to prove directly. It's not a condition solely on $A$, but maybe it's useful.