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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.

Eric Wofsey
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Twistediso
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  • It is not clear to me that $A^n$ deserves the name "free in a topological sense" when $n$ is infinite. That should mean some sort of universal compactification of $A^{\oplus n}$, and I don't believe $A^n$ is suitably universal (it is too small). In particular, a free module in a topological sense should be a coproduct of copies of $A$ in the category of compact modules, which I would expect to be dual to a product of copies of $A$ (rather than a direct sum of copies of $A$) in the category of discrete modules. – Eric Wofsey Apr 26 '17 at 16:03
  • @EricWofsey You say that you believe a free module in a topological sense should be some coproduct of copies of $A$ in the category of compact modules. If I say consider $\oplus_{i=1}^\infty A,$ then I can of course take the profinite completion of this module. Shouldn't this be some sort of (infinite) coproduct in the category of profinite modules over $A?.$ – Twistediso Apr 29 '17 at 12:57
  • Yes, but that profinite completion will often be much larger than $\prod_{i=1}^\infty A$. For instance, if $A$ is a finite field, then you are completing profinitely with respect to every linear map from $\bigoplus A$ to a finite-dimensional vector space, not just the "finite support" ones that extend continuously to the product. – Eric Wofsey Apr 29 '17 at 13:31
  • @EricWofsey Oh, I thought that the profinite completion of say $\oplus A$ in the case $A$ is a finite field would just be the infinite direct product. If I recall correctly, the profinite completion is $(\oplus_{i=1}^\infty A)/I $ where $I$ is over all those submodules of the infinite direct product is finite. But you're saying this is not the direct product? Or is this the wrong notion of profinite completion? – Twistediso Apr 29 '17 at 13:52
  • @EricWofsey If the profinite completion as I defined it is not the direct product, is there a good way to see this? – Twistediso Apr 29 '17 at 13:53
  • There exist maps $\bigoplus A\to A$ which do not extend to continuous maps $\prod A\to A$ (for instance, the map $\bigoplus A\to A$ given by the identity on each coordinate). But every map from $\bigoplus A$ to a finite module extends to the profinite completion. – Eric Wofsey Apr 29 '17 at 16:53

1 Answers1

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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.

Eric Wofsey
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