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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?

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2 Answers 2

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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.

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  • $\begingroup$ Dear jmc, many thanks! However, could you please add detais? To what is the tensoring functor left adjoint, and why it follows that this functor commutes with colimits? Please give references and, if possible, detailed explanations! $\endgroup$ Mar 24, 2015 at 16:08
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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 $\widehat{\mathbb{Z}}(1)$.

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