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This question was much simpler, but as I was typing it, it became a chain of questions.

My starting question was

Is $\mathbb{Z}_p$ (obtained by the inverse limit procedure with the directed system $\cdots \to \mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z} \to 0$) the integral closure of $\mathbb{Z}$ in $\mathbb{Q}_p$?

My Intuition: Since $\mathbb{Z}_p$ is uncountable, and "the integral closure of $\mathbb{Z}$ in $\mathbb{R}$"(The Algebraic integers?) is countable, maybe the integral closure of $\mathbb{Z}$ in $\mathbb{Q_p}$ should be countable too?

I was hoping to look for a ring which served $\mathbb{R}$ the same purpose as $\mathbb{Z}$ serves $\mathbb{Q}$. That is being integrally closed. As a side note, I also realized that I do not know any ring whose field of fractions is $\mathbb{R}$. Is it because:

  • such a thing does not exist.
  • such a thing exists only by application of AC.
  • I'm being slow today.

Thanks for the help!

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5  
An integral closure of ${\mathbf Z}$ will be countable. The $p$-adic integers are the topological closure of ${\mathbf Z}$ inside the $p$-adic numbers, but topological closure and integral closure are two very different operations. There is no standard example of a useful ring inside the reals that serves the same purpose for the reals as the integers do for the rationals. –  KCd Dec 9 '12 at 3:48
    
Dear @KCd , can you point out somewhere I can see the proof of the latter point you make? Or maybe just a hint as to how to prove it? Many thanks! –  Ravi Donepudi Dec 9 '12 at 3:52
    
As a matter of terminology, in a nonarchimedean field $k$, one often calls the set $\{a\in k : |a|\leq 1\}$ the ring of integers of $k$. If you were using this definition, then $\mathbb{Z}_p$ is definitely the ring of integers of $\mathbb{Q}_p$. For this reason the statement of the question is a little confusing. –  froggie Dec 9 '12 at 4:01
    
@FortuonPaendrag: Please clarify exactly what it is you are asking for a proof of in your comment. My last sentence in the previous comment is not a theorem. You can't prove there isn't a "standard example"; that's just a matter of experience. –  KCd Dec 9 '12 at 5:56
4  
@FortuonPaendrag: Let $F$ be the field of real algebraic numbers (that is, the real numbers that are algebraic over the rationals). Since the reals are uncountable and $F$ is countable, ${\mathbf R} \not= F$. The extension ${\mathbf R}/F$ is purely transcendental. Let $\{X_i\}$ be a transcendence basis of ${\mathbf R}$ over $F$, so ${\mathbf R} = F(\{X_i\})$. You can use the polynomial ring $F[\{X_i\}]$, which is integrally closed since it's a UFD. –  KCd Dec 9 '12 at 16:27

2 Answers 2

Let me answer the OP's side question:

"As a side note, I also realized that I do not know any ring whose field of fractions is $\mathbb{R}$."

Sure you do: $\mathbb{R}$! But I know what you mean: you are asking whether there is a proper subring $R$ of $\mathbb{R}$ such that $\mathbb{R}$ is the field of fractions of $R$. The answer is yes. (Among the choices you provide, I would lean towards the second. The proof that follows certainly uses AC. Whether it must use AC is not really my bag, but I would guess that it does.) This follows from the following stronger result.

Theorem: For a field $K$, the following are equivalent:
(i) $K$ admits a nontrivial $\mathbb{R}$-valued valuation.
(ii) $K$ is not algebraic over a finite field.

This result is proved in these course notes of mine. Well, at least morally: at the moment I can track down a Remark on p. 6 that it will be proved and the result which is the essential content, Theorem 30b) in $\S$ 1.3:

Extension Theorem If $K$ is a field and $|\cdot|$ is a nontrivial non-Archimedean norm on $K$ -- i.e., such that taking $-\log |\cdot|$ gives a nontrivial $\mathbb{R}$-valued (or "rank one") valuation -- then for any extension field $L/K$, the norm $|\cdot|$ extends to $L$.

From this the above proof follows easily: for (i) $\implies$ (ii) just notice that for every element $x$ of finite order in the multiplicative group of a field $K$ we must have $|x| = 1$ for every norm $|\cdot|$. A field $K$ which is algebraic over a finite field has $K^{\times}$ a torsion group: every element has finite order. For (ii) $\implies$ (i) notice that every field $K$ which is not algebraic over a finite field can be expressed as an extension field of either $\mathbb{Q}$ or $\mathbb{F}_p(t)$, and each of these fields famously carries plenty of nontrivial non-Archimedean norms. Apply the Extension Theorem.

To get an answer to your question: let $|\cdot|$ be a nontrivial non-Archimedean norm on $\mathbb{R}$ and let $R$ be its valuation ring. This is indeed a domain, proper in $\mathbb{R}$, with fraction field $\mathbb{R}$.

Now finally one might ask: okay, such rings exist, but who cares? Surely this is just a curiosity. To that I would say: you must read this four page note of Paul Monsky, published in the MONTHLY in 1970. And you had better hold on to your hat while you read it, and give some attention to your socks as well: they are in grave danger of being knocked off. Monsky's Theorem was introduced to me by Aaron Abrams in 2006. It was in fact much of the motivation for the material on extension of valuations which is included in the notes I linked to (although, sadly, I have not gotten around to writing up the application to Monsky's Theorem).

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The ring of integers of $\mathbb{Q}_p$ cannot be $\mathbb{Z}_p$ because of the countability argument that you mention. However, if $\mathbb{Q}_p$ is given the topology inherited from the $p$-adic norm $|\hspace{1mm}|_p$, then $\mathbb{Z}_p$ is the closure (in the topological sense) of $\mathbb{Z}$ in $\mathbb{Q}_p$. A proof of the latter statement can be found in Juergen Neukirch's "Algebraic Number Theory", Chapter $2$, Proposition $2.3$.

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I did not realize @KCd posted a comment while I was posting this answer. –  Rankeya Dec 9 '12 at 3:49

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