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Let $\left(\widehat{\mathbb Z}\right)^\times=\prod_p{\mathbb Z}_p^\times$ be the unit group of the ring $\widehat{\mathbb{Z}}$, which is the profinite completion of $\mathbb Z$. We give it the product topology. For a given prime $p$ let $$ f_p=(1,p)\in\mathbb{Z}_p^\times\ \times\ \prod_{q\ne p}\mathbb{Z}_q^\times. $$ Is the group generated by all $f_p$ dense in $\left(\widehat{\mathbb{Z}}\right)^\times$?

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    $\begingroup$ A subgroup is dense iff it is contained in $\ell\widehat{\mathbf{Z}}$ for no prime $\ell$. Given this, you should be able to answer your question. $\endgroup$
    – YCor
    Commented Mar 18 at 9:32
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    $\begingroup$ Sorry, I meant dense in the unit group, not the ring. I am not sure, whether a similar argument applies for the unit group. $\endgroup$
    – Antonius
    Commented Mar 18 at 10:00
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    $\begingroup$ TeX note: $\widehat{\mathbb Z}^\times$ \widehat{\mathbb Z}^\times puts the \times too high. To get it at the proper level, you can \smash the \widehat{\mathbb Z} to force TeX to forget its height: $\smash{\widehat{\mathbb Z}}^\times$ \smash{\widehat{\mathbb Z}}^\times. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Mar 18 at 13:09
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    $\begingroup$ No, the times is correct, otherwise we have the profinite completion of the unit group of $\mathbb Z$, which is $\{\pm 1\}$. I have put brackets around the ring, to make it clear, what I mean. $\endgroup$
    – Antonius
    Commented Mar 19 at 14:22
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    $\begingroup$ Don't worry, I think there was no danger that you are misunderstood. Similarly $\mathbb{Z}_p^{\times}$ is not the $p$-adic completion of $\{\pm1\}$. $\endgroup$
    – Chris Wuthrich
    Commented Mar 19 at 14:40

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More is true: the generating set $\{f_p:\text{$p$ is a prime}\}$ is dense in $\smash{\widehat{\mathbb Z}}^\times$. To see this, it suffices to show that, for any finite set of primes $S$, the projection of the above generating set onto $\prod_{q\in S}\mathbb{Z}_q^\times$ is dense. Let $t$ be a positive integer whose prime factors are from $S$, and let $a\in(\mathbb{Z}/t\mathbb{Z})^\times$ be a reduced residue class modulo $t$. Then it suffices to show that there is a prime $p\notin S$ such that $p\equiv a\pmod{t}$. Such a prime exists by Dirichlet's theorem on primes in arithmetic progressions, and we are done.

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