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If we have two primes, $p$ and $p'$ and the $p,p'$-adic integers $\mathbb{Z}_p$ and $\mathbb{Z}_{p'}$. Then the integers inject into both of the rings, so we can say that $\mathbb{Z} \subset \mathbb{Z}_p \cap \mathbb{Z}_{p'}$. My question is, are the integers the entire intersection? I assume they are as all the other elements of $\mathbb{Z}_p$ and $\mathbb{Z}_{p'}$ rely on what $p$ and $p'$ are.

The comments have determined that $\mathbb{Z}$ is not the largest ring that embeds into both $\mathbb{Z}_p$ and $\mathbb{Z}_{p'}$. But is $\mathbb{Z}$ the largest ring that injects to all $\mathbb{Z}_p$.

user106209
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    As Henning points out, there is a problem in interpreting the intersection. We can view both $\Bbb{Z}p$ and $\Bbb{Z}{p'}$ as subsets of $\Bbb{C}$, but to get there we need to identify transcendence bases for both the adic-fields, and then pick a way of mapping those to algebraically independent complex numbers. It should be obvious that there is nothing canonical about this process, so the question is left vague. What we can say with certainty is that $$(\Bbb{Z}p\cap\Bbb{Q})\cap(\Bbb{Z}{p'}\cap\Bbb{Q})$$ will be there ($\Bbb{Q}$ is canonical as the prime field). – Jyrki Lahtonen Dec 09 '13 at 19:10
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    (cont'd) So the rational numbers with denominators coprime to $pp'$ will certainly be in the intersection. Some algebraic numbers will also be there. For example, if $n$ is a quadratic residue modulo both $p$ and $p'$ (both odd for simplicity, then $\pm\sqrt{n}$ will be in the intersection et cetera. – Jyrki Lahtonen Dec 09 '13 at 19:13
  • Thank you for this nice answer! Do you know if $\mathbb{Z}$ is the largest rings that embeds into all $\mathbb{Z}$? – user106209 Dec 09 '13 at 19:34
  • Instead of asking a new question when your old one is resolved, please make a new question. Since @HenningMakholm resolved it, you should accept his answer and then make a new one. –  Dec 09 '13 at 19:37
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    All the $\Bbb{Z}_p$ contain transcendental units, so for example $\Bbb{Z}[\pi,\pi^{-1}]$ embeds into all of them. I don't know if this useful in any way :-) – Jyrki Lahtonen Dec 09 '13 at 19:41

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I think this is a case where you need to distinguish between "injects into" and "is a subset of". Each of $\mathbb Z_p$ and $\mathbb Z_{p'}$ is an abstract thing that one can choose to realize formally in several different ways (which we don't normally distinguish between because they produce isomorphic results) -- and which precise formalism we use will determine whether the sets that represent $\mathbb Z_p$ and $\mathbb Z_{p'}$ even intersect at all. Whether they do is more a property of the "uninteresting" choices you make during the construction, than it is a property of the idea of $p$-adic integers per se.

For example, you may choose to realize $\mathbb Z_p$ as a set of sequences of digits $0\le d<p$, and in that case there will certainly be an intersection, such as the number $...1111$. But it will behave differently under the $p$-adic arithmetic than under the $p'$-adic arithmetic.

A better question might be whether there's a larger ring than $\mathbb Z$ that embeds homomorphically into both of $\mathbb Z_p$ and $\mathbb Z_{p'}$.

The answer to that is yes. For example, unless I'm mistaken $\mathbb Z_p$ always contains a subring isomorphic to $\mathbb Z[X]$, with $X$ represented by some element of $\mathbb Z_p$ that is transcendental over $\mathbb Z$. I think that $\sum_{k=0}^\infty p^{f(k)}$ where $f$ is a "sufficiently fast-growing" function will qualify, but even if not, a cardinality argument says that there must be some transcendentals in $\mathbb Z_p$ that can be used.

There's no canonical identification of these copies of $\mathbb Z[X]$, of course.

Note that this also makes $\mathbb Z[X]$ a ring that embeds into all $\mathbb Z_p$s. (It's not the largest such one, of course; you can add continuum many different unknowns and still embed into all of them, given enough choice).

hmakholm left over Monica
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