Inverse limit of $\Bbb Q/q\Bbb Z$ isomorphic to finite adeles?

Question

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. It got upvoted but no answers, so I'm giving it a shot to repost here.

Answer 1

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.