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Introduction: A number system, long known but seldom seen, is (re)introduced for which some elements are torsion and some are torsion-free. A topology question and an analytic number theory question follow background material.

$D=\prod\limits_{p\in\mathbb{P}}\frac{\mathbb{Z}}{p\mathbb{Z}}$ is a commutative profinite ring with identity $\boldsymbol{1}=(1+2\mathbb{Z},1+3\mathbb{Z},1+5\mathbb{Z}\dots)$, using the componentwise presentation of $D$. Alternatively, one can define $D^{\ast}$ as in Section 10 of Hewitt and Ross, where coordinate 0 can take values $0,1$, coordinate 1 can take values $0,1,2$, coordinate 2 can take values $0,1,2,3,4$, coordinate 3 can take values $0,1,2,3,4,5,6$, etc., and an addition using "carrying" is defined and leveraged to define a multiplication, again defining $D^{\ast}$ to be a commutative profinite ring with identity $(1,0,0,\dots)$. In $D^\ast$ one counts as follows: $(0,0,\dots),(1,0,\dots),(0,1,0,\dots),(1,1,0,\dots),(0,2,0,\dots),(1,2,0,\dots),(0,0,1,0,\dots),(1,0,1,0,\dots),(0,1,1,0,\dots),(1,1,1,0,\dots),(0,2,1,0,\dots),(1,2,1,0,\dots),(0,0,2,0,\dots),(1,0,2,0,\dots),(0,1,2,0,\dots),(1,1,2,0,\dots),(0,0,3,0,\dots),(1,0,3,0,\dots),(0,1,3,0,\dots),(1,1,3,0,\dots),(0,2,3,0,\dots),(1,2,3,0,\dots),(0,0,4,0,\dots),(1,0,4,0,\dots),(0,1,4,0,\dots),(1,1,4,0,\dots),(0,2,4,0,\dots),(1,2,4,0,\dots),(0,0,0,1,0,\dots),(1,0,0,0,1,0,\dots),...$

In $D^\ast$, nonnegative integers have finitely many nonzero coordinates. However, for example, $-1=(1,2,4,6,10,\dots)$ has infinitely many nonzero coordinates.

Because a commutative profinite ring is uniquely determined by its $p$-Sylow factors, $D$ and $D^\ast$ are topologically isomorphic; in other words, they are each representations of the same topological ring.

$D$ is topologically generated by $\boldsymbol{1}$; that is, $D=\overline{\mathbb{Z}\boldsymbol{1}}$. $D^\ast$ is topologically generated by $(1,0,0,\dots)$: $D^\ast=\overline{\mathbb{Z}(1,0,0,\dots)}$. Note that the torsion subgroup of $D$ is $S=\bigoplus\limits_{p\in\mathbb{P}}\frac{\mathbb{Z}}{p\mathbb{Z}}$ and $S$ is a dense, non-locally-compact subgroup of $D$. Is it not somewhat fantastical that $D$ (resp. $D^\ast$) has a countable dense torsion subgroup, $S$, as well as a countable dense torsion-free subgroup, $\mathbb{Z}\boldsymbol{1}$?

In the Hewitt and Ross presentation of $D^{\ast}$, it is more evident that a number system is being constructed than in the easier-to-work-with $D$ (at least to me). The order of the prime coordinates in $D^{\ast}$ can be permuted by a permutation of the set of indices $\{ 0,1,2,3,\dots\}$ and the resulting construct is topologically isomorphic to $D^{\ast}$; I have often thought that this simple fact could be used to great effect in number theory, because normally number systems are intrinsically ordered; it is probably a reflection of my lack of expertise, but this flexibility seems to me to be a very big plus. In $D$ the order of the prime coordinates are inconsequential as well, but when you do computations in $D^{\ast}$ with permuted coordinates you see how it might be analytically quite useful (grain of salt); one imagines arguments where one inductively permutes the coordinates for a desired effect.

An example of how elements correspond between the two profinite ring presentations (analogous to the two common presentations of the $p$-adic integers, sans order requirements) the torsion element $(1+2\mathbb{Z},0+3\mathbb{Z},0+5\mathbb{Z},\dots)\in D$ corresponds to the torsion element $(1,\frac{3-1}{2},\frac{5-1}{2},\frac{7-1}{2},\dots)\in D^{\ast}$.

What cannot be represented? Well, in $D$ the element $\boldsymbol{1}$ cannot be divided by a prime $p$ because the equation $px=1$ has no solution in the coordinate/factor $\mathbb{Z}/p\mathbb{Z}$. Similarly, $1/p\notin D^\ast$. However, if one embeds $D$ (equivalently, $D^\ast$) in the compact, connected (divisible) $1$-dimensional solenoid $G=(D \times\mathbb{R})/\mathbb{Z}(\boldsymbol{1},1)$, where the topology on $G$ is induced by a metric and $G$ is topologically isomorphic to $(\mathbb{Q}D \times\mathbb{R})/X(\boldsymbol{1},1)$ for the the divisible, dense, non-locally-compact, incomplete metric subgroup $\mathbb{Q}D$ of $G$ generated by all profinite subgroups, and $X\subset\mathbb{Q}D$ is a dense subgroup algebraically isomorphic to $\sum\limits_{p\in\mathbb{P}}\mathbb{Z}\frac{1}{p}$, the Pontryagin dual of $G$, then elements of $D$ can be divided in $G$, with divisors lying in $\mathbb{Q}D$. Restated, the second representation of $G$ provides a natural setting for $D$ in which one can now divide while preserving the profinite topology on $D$. (There is a subtlety here in that, just as $\mathbb{R}$ is locally compact but its bijective continuous image in $G$ is not locally compact, $\mathbb{Q}D$ is locally compact in the numerator $\mathbb{Q}D\times\mathbb{R}$ (as the union/direct limit of a countable ascending chain of finitely generated profinite abelian groups each with finite index in the next), but its bijective continuous image in $G$ is not locally compact.) Also, $\mathbb{Q}D$ is algebraically isomorphic to $\prod\limits_{p\in\mathbb{P}}\mathbb{Z}(p^\infty)$, which by the way is algebraically isomorphic to $\mathbb{R}/\mathbb{Z}$.

  1. Question What is the non-locally-compact (metric) subspace topology on $\mathbb{Q}D$ inside $G$?

Both $D$ and $D^\ast$ have countably many coordinates, so their cardinality is that of the continuum. If you randomly select an element in $D$ it will be transcendental. Which begs the question, what irrational algebraic numbers lie in $D$? Well, all the algebraic integers lie in $D$ and all the algebraic numbers lie in $\mathbb{Q}D$ - OK...slight exaggeration...for a monic irreducible polynomial in $\mathbb{Z}[x]$ with degree $>1$, there are infinitely many primes for which there is a zero mod $p$ and there are infinitely many primes for which there is no zero mod $p$ - if one stipulates that a $0$ be placed in every coordinate where there is no zero mod $p$ and a zero is inserted in each of the remaining coordinates, then there is a continuum of representations in $D$ for the set of algebraic integers in $\overline{\mathbb{Q}}$ which are zeros of the polynomial. Since an appropriate nonzero integer multiple of any given algebraic number is an algebraic integer, there is also a continuum of representations in $\mathbb{Q}D\subset G$ for each set of algebraic numbers associated with a monic irreducible polynomial in $\mathbb{Q}[x]$. Without going into detail, a combination of an equivalence relation on $\mathbb{Q}D$ and a group action by ${\rm Aut}\,\mathbb{Z}[x]$ can be defined to differentiate between and thus uniquely represent the individual algebraic numbers within a set of zeros of a given monic irreducible polynomial in $\mathbb{Q}[x]$. So, with a pile of grains of salt, the entirety of $\overline{\mathbb{Q}}$ is sitting right there inside of $\mathbb{Q}D\subset G$.

For example, if a prime equals $2$ or is congruent to $1$ modulo $4$, then there is a solution to the equation $x^2 + 1 = 0$ mod $p$. Thus, $(1+2\mathbb{Z},0+3\mathbb{Z},2+5\mathbb{Z},0+7\mathbb{Z},0+11\mathbb{Z},5+13\mathbb{Z},\dots)$ is a representative of $\pm i$ in $D$. One could arbitrarily define $i$ to be the element of $D$ with the minimal representative between $1$ and $p-1$ in each coordinate where $x^2 + 1=0$ has a solution; again, toggling solutions in each such $p$ coordinate produces a continuum of representatives of $\pm i$, which I claim comprises a path component of $G$ (for another day).

$G$ has a Haar measure with total measure $1$ (WLOG). We can integrate all day inside of $G$. For example, $\mathbb{Q}D$ is a countable union of closed sets, a Borel-measurable set. There is a paper by Hewitt and Ritter, Fourier Series on Solenoids, Math. Ann. 257, 61-83 (1981), where they show exactly how to do harmonic analysis in a completely arbitrary solenoid ($1$-dimensional compact, connected abelian group). Our $G$ is a relatively special case.

  1. Question Assuming the construction above, embedding $\overline{\mathbb{Q}}$ in the $1$-dimensional solenoid $\mathbb{Q}D\subset G$, is properly vetted, and knowing from Hewitt and Ritter how to do Fourier analysis in an arbitrary solenoid, is there a reasonable case for formulating a $1$-dimensional version of the classical Riemann conjecture (intrinsically within $G$)?
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With regard to Question 1:

The subspace topology on $\mathbb{Q}D$ inside $G$ is equivalent to the final topology on $\mathbb{Q}D$ coherent with the family of subgroups of $G$ containing $D$ as a subgroup of finite index. The subspace topology on $\mathbb{Q}D$ is equivalent to the topology on $\mathbb{Q}D$ induced by the metric $d$ on $G$, which is the quotient metric of the translation-invariant product metric on $\mathbb{Q}D\times\mathbb{R}$. The restriction of $d$ to any profinite subgroup of $G$ is non-Archimedean. The final topology is usually easy to work with, but it seems that if any serious digging is necessary relating to the topology on $\mathbb{Q}D$, one would need to localize to a single profinite subgroup where a very explicit non-Archimedean metric can be used. One can also view the countable collection of profinite subgroups having finite index over or in $D$ as a sheaf.

Just as the continuous bijective image of $\mathbb{R}$ in $G$ (the path component of $0$) is a dense, non-locally-compact subgroup, the topological group $\mathbb{Q}D$ external to $G$ is locally compact with continuous bijective image (which we also denote) $\mathbb{Q}D$ a dense, non-locally-compact subgroup of $G$. Outside of $G$, $\mathbb{Q}D$ also has the final topology, but the subgroups with finite index over $D$ are open subgroups with the case inside $G$ where they are only closed; the final topology induced by a countable ascending chain of compact open subgroups, each with finite index in the next, is known to coincide with the topological direct limit topology resulting from this directed system of compact subspaces, that multiplication on the direct limit is continuous under the direct limit topology, and that, last but not least, the resulting topology on $\mathbb{Q}D$ is locally compact.

With regard to Question 2:

Yes.

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