Non-open subgroups of finite index of $(\hat Z)^\times$

Question

It is claimed here that $(\hat Z)^\times$ has non-open subgroups of finite index but I couldn't find any example. What are some examples of such subroups? More generally, I would be glad to see other examples of non-open subgroups of finite index in profinite groups.

Answer 1

This is strictly speaking not an answer to the question, but here's a non-constructive proof that $H=\widehat{\Bbb Z}^\times$ has a non-open subgroup of finite index. First we construct a continuous surjection $H \to G=\prod_{i=1}^\infty \Bbb Z/2\Bbb Z$. From the isomorphism of topological rings $\widehat{\Bbb Z} \cong \prod_p \Bbb Z_p$, we obtain an isomorphism of topological groups $\widehat{\Bbb Z}^\times \cong \prod_p \Bbb Z_p^\times$. So it suffices to show that for odd $p$, $\Bbb Z_p^\times$ has an index $2$ open subgroup. ($\Bbb Z_2^\times$ also has one, but I'll stick to the odd case for simplicity) There's an isomorphism $\Bbb (\Bbb Z_p)^\times \cong \Bbb F_p^\times \times (1+p\Bbb Z_p)$, arising from the split short exact sequence $$1 \to 1+p\Bbb Z_p \to \Bbb Z_p^\times \to \Bbb F_p^\times \to 1$$ The map $\Bbb Z_p^\times \to \Bbb F_p^\times$ is the reduction map (using that $\Bbb Z_p/p\Bbb Z_p \cong \Bbb Z/p\Bbb Z$). Proving that it is a split surjection can be done e.g. via Hensel's lemma. $\Bbb F_p^\times$ is discrete and cyclic of order $p-1$ and hence has an open subgroup of index $2$.

So we have reduced the problem to showing that $G=\prod_{i=1}^\infty \Bbb Z/2\Bbb Z$ has a non-open index $2$ subgroup. This is equivalent to showing that there's a linear functional on the $\Bbb F_2$ vector space $G$ (i.e. a linear map $G \to \Bbb F_2$) that is not continuous with respect to the discrete topology on $\Bbb F_2$. Note that $\bigoplus_{i=1}^\infty \Bbb Z/2\Bbb Z \subseteq \prod_{i=1}^\infty \Bbb Z/2\Bbb Z$ is dense. The (algebraic) dual space of $\bigoplus_{i=1}^\infty \Bbb Z/2\Bbb Z$ is isomorphic to $\prod_{i=1}^\infty \Bbb Z/2\Bbb Z$ which has cardinality and hence dimension $\mathfrak{c}$ (of continuum size). The algebraic dual of $\prod_{i=1}^\infty \Bbb Z/2\Bbb Z$ on the other hand has dimension strictly larger than $\mathfrak{c}=\operatorname{dim} \prod_{i=1}^\infty \Bbb Z/2\Bbb Z$, by a general fact about algebraic duals of infinite-dimensional vector spaces (over any field). Thus there exists a discontinuous functional on $G$ and hence a non-open index $2$ subgroup of $G$ and hence a non-open index $2$ subgroup of $H=\widehat{\Bbb Z}^\times$.

Speculation/Question for Set Theorists I don't know much about set theory, but it could be possible that some kind of non-constructiveness is necessary for the proof. At least when we look at $\prod_{i=1}^\infty \Bbb Z/2\Bbb Z$ as a ring and ask for maximal ideals (which correspond to multiplicative functionals), then the existence of non-open examples corresponds to the existence of non-principal ultrafilters. Thus one could be led to speculate that perhaps in a model of set theory without choice where there are no free ultrafilters on $\Bbb N$, it might be possible that there is no example of a discontinuous functional on $\prod_{i=1}^\infty \Bbb Z/2\Bbb Z$.