A module that is free in the profinite sense and double dual. 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.
 A: 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.
