Let $p\ne\ell$ be two prime numbers, and let $K_{\ell}$ be the maximal extension of $\mathbb Q$ ramified only at $\ell$ and $\infty$ (i.e. it is the fixed subfield of $\overline{\mathbb Q}$ by all the inertia groups at finite places not lying over $\ell).$ Let $\mathfrak p$ be a place of $K_{\ell}$ over $p.$ What is the structure of the decomposition group $D_{\mathfrak p}$ in $Gal(K_{\ell}/\mathbb Q)?$
A priori, it is a quotient of $Frob_p^{\widehat{\mathbb Z}}=D_p/I_p$ by some closed subgroup, where $D_p$ and $I_p$ are in $Gal(\overline{\mathbb Q}/\mathbb Q).$ Is it $\widehat{\mathbb Z}$ itself? Note that when passing to the abelianization $Gal(K_{\ell}/\mathbb Q)^{ab}\simeq\mathbb Z_{\ell}^{\times},$ which is an $\ell$-adic Lie group ($\cong\mathbb F_{\ell}^{\times}\times\mathbb Z_{\ell}$), it has a large kernel in $\widehat{\mathbb Z}.$
Here is an attempt to prove that it is. We have $D_p/I_p=\varprojlim_{(p,n)=1} Gal(\mathbb Q_p(\mu_n)/\mathbb Q_p),$ so to show that $$ D_p/I_p\to Gal(K_{\ell}/\mathbb Q) $$ is injective, it is equivalent to showing that for each $n$ prime to $p,$ there exists a finite Galois subextension $F$ of $K_{\ell}/\mathbb Q,$ such that for some ($\Leftrightarrow$ any) prime $v$ of $F$ over $p,$ we have $F_v=\mathbb Q_p(\mu_n).$ In this case, the map $$ \sigma\mapsto\sigma|_F:Gal(F_v/\mathbb Q_p)\to Gal(F/\mathbb Q), $$ is the inclusion of the decomposition group $D_v.$ One may assume $n$ is a prime power (by factoring it into prime powers), and coprime to $\ell$ (since one knows what to do when $n$ is a power of $\ell:$ take $F=\mathbb Q(\mu_n)$). One idea is to try finding a monic polynomial $f(x)\in\mathbb Z[x]$ of degree $\varphi(n),$ close enough to the cyclotomic polynomial $\Phi_n(x)$ $p$-adically (so that Krasner's lemma applies), such that the discriminant of $f(x)$ is, up to sign, a power of $\ell.$ One problem is how to achieve equality (rather than just an inclusion) when applying Krasner's lemma: this seems to lead to factorisation of $f(x)$ over $\mathbb Q_p.$ The other is how to deal with the discriminant: this doesn't seem to be of local nature.
One can also replace $\ell$ by an arbitrary integer $N,$ and ask the same question for $p\nmid N.$ Of course, if the decomposition group is the full group $\widehat{\mathbb Z}$ for $\ell,$ then it is so for any $N$ divisible by $\ell.$ The question also has a counterpart for function fields, but let me stick with number field here.
