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I don't know whether this question is a bit too vague for MO or not, so feel free to delete it if you see fit.

The p-adic integer is defined by taking the inverse limit $\ldots \mathbb{Z} / p^2 \rightarrow \mathbb{Z}/p $.One way to see the p-adic integers is to see it as dealing with $ \mathbb{Z} / p, \mathbb{Z} / p^2, \ldots $ at the same time. So $p$-adic integers allow us to see the structure of the ring of integers at the prime $p$. Taking the fractional field we obtain the $ p$-adic rational field $\mathbb{Q}_p$.

This construction is useful in study the arithmetic of the ring (field). For example, in the theory of class field theory, we study the question in $\mathbb{Q}_p$ first and glue them together and do something more to get the solution for $\mathbb{Q}$.

I want to ask why it is not possible for us to construct a structure that will allow us to see the ring of integer at two primes $p,q$ together and see how do they interacts? The analog of the above inverse limit construction seems to still work though not an intergral domain. However, we can still localize where it is possible.

Here is the motivation for the above question. We know that the global question are not solve by simply gluing together solution for local question. I attribute that to the fact that the primes does not play alone but interact with one another. An illustration of this can be seen through the fact that the product of all normalized absolute value is 1. So my question is why not isolate two primes to understand how they are interacting with one another instead of looking at all of them at once. I think there is some complicated issue that will arise from this. Just want to know what they are.

A more particular question may be like this: Let call the construction obtained above $\mathbb{Q}_{p,q} $. Is there something in the same vein of class field theory for this object. What is the obstacle in having such a theory. I am vaguely know that we have a more general Galois theory not only for fields but for rings also.

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    $\begingroup$ It might help if you said what you mean by "interacts". You can invert any collection of primes you want, and you'll get a subring of Q. Taking a profinite completion of this ring will yield a product of p-adic rings for p ranging over the set of primes not inverted. In particular, if you don't invert any primes, you get $\widehat{\mathbb{Z}}$. $\endgroup$
    – S. Carnahan
    Apr 1, 2010 at 16:01
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    $\begingroup$ You probably already realize this, but $\lim_{\infty \leftarrow n} \mathbb{Z}/(pq)^n \cong \mathbb{Z}_p \times \mathbb{Z}_q$. So you don't get any new rings by this construction. $\endgroup$ Apr 1, 2010 at 17:10
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    $\begingroup$ Isn't the answer to this question "the adeles"? By the Chinese remainder theorem, looking at information modulo pq is the same as looking at it mod p, and mod q, separately. This "separateness" is captured by direct products. $\endgroup$ Apr 1, 2010 at 17:44
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    $\begingroup$ The obstacle, to me seems to be the Chinese Remainder Theorem, which demands that when you look at only a finite number of primes, they don't interact. $\endgroup$
    – stankewicz
    Apr 1, 2010 at 18:06
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    $\begingroup$ Here is a specific example in the category of Riemann surfaces: a meromorphic function that's zero on a set with a limit point is identically zero. This is a global obstruction to constructing such functions that don't appear when you only look at finitely many points. $\endgroup$ Apr 2, 2010 at 8:25

3 Answers 3

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As stankewicz said, it is a general principle in number theory that whenever only finitely many primes are involved, they act "independently" in the sense that analyzing what is happening locally at each prime separately is enough to understand what all the finitely many primes are doing. One example of this is the Chinese Remainder Theorem.

Here is another: if you want a version of the integers with two primes $p$ and $q$, start with $\mathbb{Z}$ and invert every prime $\ell \not \in \{p,q\}$. This gives a semilocal ring with $(p)$ and $(q)$ as the nonzero primes. Similarly, you can do this with any two prime ideals $\mathfrak{p}, \mathfrak{q}$ in a Dedekind domain $R$. But the ring you get is not very interesting: it is a semi-local Dedekind domain, hence its class group is trivial, very likely the unit group $R^{\times}$ will be infinitely generated, etc. This domain is the intersection of the two DVRs $R_{\mathfrak{p}}$, $R_{\mathfrak{q}}$, and everything you want to know about it can be reduced to the DVRs. The same with two replaced by any finite set...

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  • $\begingroup$ The answer provided by dke and damiano are very informative, but I think this answer the original question, so I accept this answer. $\endgroup$
    – abcdxyz
    Apr 2, 2010 at 13:36
  • $\begingroup$ May I enquire why the fact that the unit group is infinitely generated makes it not interesting? $\endgroup$
    – abcdxyz
    Apr 2, 2010 at 13:38
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As mentioned in the comments, mimicking the construction of the $p$-adic numbers for two primes doesn't produce much new (although one might be led towards interesting stuff like $\widehat{\mathbb{Z}}$, the adeles, etc. - indeed, the 10-adics are sometimes used to motivate discussion of the $p$-adic numbers).

There is a rather speculative way of 'measuring' the arithmetic interaction between two primes based on absolute i.e. ${\mathbb F}_1$-derivations given in Kurokawa et al's paper Absolute Derivations and Zeta Functions. Here, one is looking at the (not necessarily additive) maps $D:{\mathbb Z}\rightarrow{\mathbb Z}$ satisfying $$D(ab)=D(a)b+aD(b).$$ These are essentially generated by the 'number derivative' for prime $p$, $$\frac{\partial}{\partial p}:mp^n\mapsto \begin{cases} mnp^{n-1}\qquad\mbox{for $n\geq 1$,} \\ 0\qquad\qquad\;\,\mbox{for $n=0$} \end{cases}$$ where $(m,p)=1$.

Now one can ask how two primes $p$ and $q$ are interacting by looking at the commutator of $\partial/\partial p$ and $\partial/\partial q$. In particular, in the paper above, they construct a zeta function out of this map, meromorphically continue it and look at the value at 0. They call this the quantum noncommutativity of $p$ and $q$, which is given by the bizarre-looking formula $$\frac{1}{12pq}\left((q-1)\sum_{k=1}^{\infty}\frac{p^kq^{p^{-k}}}{(1-q^{p^{-k}})^2} - (p-1)\sum_{k=1}^{\infty}\frac{q^kp^{q^{-k}}}{(1-p^{q^{-k}})^2}\right)$$ and somehow measures how $p$ and $q$ interact. What one might do with that is anybody's guess but it's certainly an interesting paper, recommended reading for anyone with an interest in ${\mathbb F}_1$-ideas.

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    $\begingroup$ +1 for sheer trippiness. Does anyone actually claim to understand what's going on here? The only article on $\mathbb{F}_1$ that ever made good sense to me was Henry Cohn's Monthly article (Henry always makes good sense to me). $\endgroup$ Apr 1, 2010 at 18:54
  • $\begingroup$ Who knows what's going on ? I remember Christophe Soulé being both thrilled and mystified by it. The only other place I've seen it mentioned is in Jeff Lagarias' follow-up paper front.math.ucdavis.edu/0404.5350 which essentially suggests (quelle surprise) that to approach the Riemann hypothesis using these ideas would require considerable modification ! $\endgroup$
    – dke
    Apr 1, 2010 at 19:10
  • $\begingroup$ There's now a question on this topic here mathoverflow.net/questions/25974/calculus-on-rationals $\endgroup$
    – j.c.
    May 26, 2010 at 12:42
  • $\begingroup$ The link in dke's comment is broken, here's a replacement: arxiv.org/abs/math/0404350 $\endgroup$
    – David Roberts
    Mar 29, 2022 at 7:39
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This might be slightly off-topic, but I am surprised that no one mentioned Gauss' Quadratic Reciprocity: this is one of the very few fundamental results that allows you to deduce something about one prime, knowing something about a different prime. The intertwining is really surprising and (in some not so remote sense) it opens the way to Brauer-Manin obstructions. In fact, I am not sure that I know of any Brauer-Manin obstruction that may not eventually be reduced to an application of Quadratic Reciprocity.

EDIT: The reason for bringing up the Brauer-Manin obstruction is that it provides "equations" in the space of adelic solutions to a diophantine problem. These relations, while somehow "invisible" from each individual prime, become apparent by considering all the primes at once. Gauss' reciprocity then comes in when trying to make everything explicit and "squeezing" the information from all the primes to an interaction between only two primes. This is the relation to the two primes in the question and to the various inverse limits of Z considered in the other answers.

All of this was implicit in my comment, but I thought that it might be a good idea to expand in this.

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  • $\begingroup$ +1 very good point and not at all off-topic (certainly more on-topic than my answer!) Your use of the word 'intertwining' reminds me that under the 'primes as knots in Spec Z' analogy, mathoverflow.net/questions/4075/… the Legendre symbol is the arithmetic analogue of mod 2 linking number. $\endgroup$
    – dke
    Apr 2, 2010 at 10:26
  • $\begingroup$ Can I deduce from your second last statement that in some sense quadratic reciprocity is the only obstacle from the picture that all the primes behave independently? $\endgroup$
    – abcdxyz
    Apr 2, 2010 at 13:42
  • $\begingroup$ I would definitely not think so: the Brauer-Manin obstruction often reduces (with a significant amount of work!) to an application of Gauss' reciprocity, but I do not know if this is always the case. Moreover, there certainly are obstructions to the existence of rational points that are not explained by the Brauer-Manin obstruction. I would say that this is not enough evidence to conclude that Gauss' reciprocity is the only obstacle! $\endgroup$
    – damiano
    Apr 2, 2010 at 13:58

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