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 $A$?

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.