Inspired by the success of my recent question:
A morphism of free modules that is an isomorphism upon dualization.
I thought of the following.
Suppose that $A$ is a profinite, commutative ring, (i.e a ring that is the inverse limit of finite rings). I can then consider modules that are free in a topological sense. Call a topological module $M$ over $A$ profinitely free if $M \cong A^n$ as topological modules for some $n$ which is not necessarily finite. We can take the continuous dual $M^\circ$ of $M$ and get a discrete $A$-module. My questions are:
1. If $M$ is profinitely free, must $M^\circ$ be free? This is easy to see if $A$ is a field.
2. Supposing a positive answer to $1,$ if $M$ is profinitely free, is $(M^\circ)^*$ where the dual is given the pointwise topology, isomorphic to $M?$
If $1$ and $2$ don't hold for any profinite ring $A,$ I would be interested to know for which class of profinite rings it holds for.
The answer to at least one of the questions is no in general (assuming you want the isomorphism in (2) to be given by the canonical map). For instance, let $k$ be a nonzero finite ring and let $A=k^{\mathbb{N}}$ and let $M=A^{\mathbb{N}}$. Consider the map $f:M\to A$ which sends a sequence $(a_n)$ of elements of $A$ to its diagonal $(a_{nn})$ (which is a sequence of elements of $k$). It is easy to see this is a homomorphism. It is also clearly continuous, since the topology on $M=k^{\mathbb{N}\times\mathbb{N}}$ is just the product topology and $f$ is just a projection onto a subproduct (namely the subproduct given by the diagonal in $\mathbb{N}\times\mathbb{N}$).
Now let $N=A^{\oplus\mathbb{N}}$. We have $N\subseteq M^\circ$ in an obvious way, and the induced map $M\to (M^\circ)^*\to N^*$ is an isomorphism. On the other hand, the map $f$ described above is an element of $M^\circ$ that does not come from any element of $N$, so $N$ is a proper submodule of $M^\circ$. If the natural map $M\to (M^\circ)^*$ were an isomorphism, then, that would mean the induced map $(M^\circ)^*\to N^*$ is an isomorphism. By my answer to your previous question, if $M^\circ$ is free, this would imply the inclusion $N\to M^\circ$ is an isomorphism as well. Since this inclusion is not surjective, either $M^\circ$ must not be free or $M\to (M^\circ)^*$ must not be an isomorphism. That is, either the answer to (1) or the answer to (2) is no.
(I don't know whether $M^\circ$ is actually free in this case, though I suspect it isn't. The existence of $f$ shows at least that $M^\circ$ is not free in the obvious way you might expect.)
In order to get a positive answer, I believe what you would want is for there to be a neighborhood $U$ of $0$ in $A$ which does not contain any nontrivial ideal. You can then imitate my answer to your previous question, using $f^{-1}(U)$ in place of $\ker(f)$. You get finitely many $x_1,\dots,x_n$ such that $f^{-1}(U)$ contains all $\alpha$ such that $\alpha(x_i)=0$ for all $i$. But if $\alpha(x_i)=0$ for all $i$, the same is true of any scalar multiple of $\alpha$, so the entire ideal generated by $f(\alpha)$ must be contained in $U$, which implies $f(\alpha)=0$. The rest of the argument is then the same, and shows that the continuous dual of $A^n$ is just $A^{\oplus n}$ for any $n$, which clearly gives positive answers to both of your questions. This argument should work for any topological ring $A$ in which such a $U$ exists.
However, if you are considering only the case that $A$ is profinite, such a $U$ only exists when $A$ is actually finite. Indeed, any neighborhood of $0$ must contain the kernel of a homomorphism from $A$ to a finite ring, which is a nontrivial ideal unless $A$ is finite.