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Suppose we have a profinite abelian group $A$, then we can define a $\hat{\Bbb{Z}}$-module structure on $A$. However, not every $\hat{\Bbb{Z}}$-module is a profinite group, as one can see for example with $\Bbb Q/\Bbb Z$. So my question is, what is the relation between profinite abelian groups and $\hat{\Bbb{Z}}$-modules? Since this question is quite vague, here are some more precise questions:

  • If we have two profinte abelian groups $A$ and $B$ and a homomorphism of abstract groups $f:A \to B$ is there some relation between the continuity of $f$ and the $\hat{\Bbb{Z}}$-linearity?
  • Does the $\hat{\Bbb{Z}}$-module structure on a profinite abelian group determine its topology?
  • Given a $\hat{\Bbb{Z}}$-module $A$ is there some non-obvious criterion that decides whether the $\hat{\Bbb{Z}}$-module structure comes from a profinite topology on $A$?
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1) Obviously continuity implies $\hat{\mathbf{Z}}$-linear. The converse is false. Indeed, consider $F$ cyclic of order $p$, an infinite set $I$ and a non-continuous homomorphism $F^I\to F$. Then it is clearly a $\hat{Z}$-module homomorphism.

2) For the same reason, the answer is no: all group automorphisms of $F^I$ are $\mathbf{Z}$-module automorphism, and they don't all preserve the topology.

Beware that this example shows that for a profinite abelian group $A$, the topology is not always the inverse limit of the discrete quotients $A/nA$. On the other hand, it it always the inverse limit of the $A/nA$, but the latter can be non-discrete.

If you restrict to abelian groups $A$ such that $A/nA$ is finite for all $n$, things go better and then 1,2 have a positive answer, but even forgetting the $\hat{\mathbf{Z}}$-module structire. Indeed if $A\to B$ is a $\mathbf{Z}$-module homomorphism, then so is the composition $A\to B/nB$ for all $n$. The latter is trivial on the open subgroup $nA$ and hence is continuous. Since this holds for all $n$, continuity follows. Also the topology in this special case is the inverse limit of the discrete topologies on the $A/nA$.


Added: if $A,B$ are profinite abelian groups, then any group homomorphism $A\to B$ (regardless of topology) is a $\mathbf{Z}$-module homomorphism. Indeed, to show that $f(tx)=tf(x)$ it is enough to restrict to the case when $A$ is pro-cyclic, and the previous argument works.

This is not true when $B$ is an arbitrary continuous $\mathbf{Z}$-module, for instance when $A=\mathbf{Z}_p$ and $B$ is the discrete group $\mathbf{Q}/\mathbf{Z}$. Then for group homomorphisms $A\to B$ in this case, continuous is equivalent to being a $\mathbf{Z}$-module homomorphism, and there are plenty of non-continuous homomorphisms.

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