Adeles and twisted adeles

Question

Let $\mu_n$ denote the group of $n$-th roots of unity in ${\mathbb{C}}$, i.e., $\mu_n=\ker[{\mathbb{C}}^*\overset{n}{\longrightarrow}{\mathbb{C}}^*]$. We set $$ \mu=\varinjlim_n \mu_n\subset {\mathbb{C}}^*,\quad {\widehat{\mathbb{Z}}}(1)=\varprojlim_n \mu_n,\quad {\mathbb{A}}(1)={\widehat{\mathbb{Z}}}(1)\otimes_{{\mathbb{Z}}} {\mathbb{Q}}.$$ I would like to have a canonical isomorphism $$ \mu \overset{\sim}{\longrightarrow} {\mathbb{A}}(1)/{\widehat{\mathbb{Z}}}(1)$$ similar to the canonical isomorphism $$\mathbb{Q}/\mathbb{Z}\to {\mathbb{A}}/{\widehat{\mathbb{Z}}} .$$ Is it possible to construct such a canonical isomorphism?

Answer 1

Doesn't this boil down to:

$$ \mathbb{A}(1)/\hat{\mathbb{Z}}(1) \cong \mathbb{Q}/\mathbb{Z} \otimes \hat{\mathbb{Z}}(1) \cong (\mathrm{colim}\, \mathbb{Z}/n\mathbb{Z}) \otimes \hat{\mathbb{Z}}(1) \cong \mathrm{colim}\, (\mathbb{Z}/n\mathbb{Z} \otimes \hat{\mathbb{Z}}(1)) \cong \mathrm{colim}\, \mu_{n} = \mu$$

Here I use that tensoring commutes with colimits, because it is a left-adjoint: $$ \mathrm{Hom}(A \otimes B, C) \cong \mathrm{Hom}(A, \mathrm{Hom}(B, C)) $$ (This is the universal property of the tensor product.)

It is a general fact that left (resp. right) adjoint functors commute with colimits (resp. limits). See https://en.wikipedia.org/wiki/Adjoint_functors#Limit_preservation.

Answer 2

Explicitly, an element of $\widehat{\mathbb{Z}}(1)$ is a sequence $(\zeta_n)_{n\geq1}$ with $\zeta_1=1$ and $(\zeta_{nr})^r=\zeta_n$. The homomorphism $$\begin{array}{rcl} \widehat{\mathbb{Z}}(1)\otimes\mathbb{Q} & \longrightarrow & \mu\\ (\zeta_n)_{n\geq1}\otimes \frac{a}{b} & \longmapsto & (\zeta_b)^a \end{array}$$ is well-defined and clearly surjective, and is easily seen to have kernel $\widehat{\mathbb{Z}}(1)$.