Let $\Bbb Q/q\Bbb Z$, for some positive rational $q$, denote the quotient group of the discrete rationals by the subgroup of integers times $q$.


For any $q\_1, q\_2 \in \Bbb Q^+$ and $n \in \Bbb N^+$ such that $q\_2 = n \cdot q\_1$, we can form a surjective homomorphism $\Bbb Q/q\_2\Bbb Z \to \Bbb Q/q\_1\Bbb Z$.


The set of all such $\Bbb Q/q\Bbb Z$ forms a poset with these surjections, and we can take the inverse limit to get something like a "rational solenoid."


1. Is this group isomorphic to the finite adele ring $\Bbb A\_\Bbb Q^f \cong \Bbb Q \otimes \hat{\Bbb Z}$?
2. Is this group equal to the same thing you'd get if you took the inverse limit of $\Bbb Q/n\Bbb Z$ for a natural number $n$ instead of a positive rational $q$?


For #1, I think it is, because the dual group should be a direct limit of localizations of $\hat{\Bbb Z}$, the group of profinite integers, which I believe is isomorphic to $\Bbb Q \otimes \hat{\Bbb Z}$, which is self-dual. The same reasoning applies for #2.


This question was originally asked on MSE [here](https://math.stackexchange.com/questions/1449189/inverse-limit-of-bbb-q-q-bbb-z-isomorphic-to-finite-adeles). It got upvoted but no answers, so I'm giving it a shot to repost here.



Inverse limit of $\Bbb Q/q\Bbb Z$ isomorphic to finite adeles?
For 1 you can indeed use duality, but you can also give an explicit isomorphism:


First for each rational number $q$ it is easy to see that $\mathbb{A}^{f}/(q \hat{\mathbb{Z}}) \simeq \mathbb{Q}/(q\mathbb{Z})$: indeed, $(q \hat{\mathbb{Z}})$ is open and $\mathbb{Q}$ is dense in the finite adeles, so the image from $\mathbb{Q}$ to this quotient is surjective and its kernel is $q \mathbb{Z}$.


So any finite adele gives you an element of the projective limit, this element is zero if and only if the adele $x$ you started from is in $q \hat{\mathbb{Z}}$ for each $q$ which implies it is zero hence the adele inject into the projective limit.


Conversely, you need to work just a little more but an element $y$ of the projective limit is the data of a compatible familly of $y\_q \in \mathbb{Q}/(q \mathbb{Z})$. pick for each $y$ a lifting of $y\_q$ in $\mathbb{Q}$ the compatiblity of the familly implies that $q\_y$ is a cauchy net (for the divisibility order on rational numbers of course, if you don't like net then $y\_{n!}$ is going to be a cauchy sequence that works equally well) in $\mathbb{A}^f$ which hence converge to an adele $x$ whose images in $\mathbb{Q}/(q \mathbb{Z})$ are the $y\_q$.


For your second question, the answer is also yes: you can either do the exact same proof as above and check that the integer are enough for the arguments to work. Or observe that in your projective limit $\mathbb{N}^\* \subset \mathbb{Q}^\*$ is cofinal for the ordering and hence the projective limit indexed by $\mathbb{N}^\*$ and by $\mathbb{Q}^\*$ are the same.


