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Is there an interesting ring $S ⊂ ℂ$ such that $ℂ = Q(S)$? I’m thinking no, but how can I prove it?

k.stm
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    The question is unclear, what does "interesting" mean? There are uncountably many subrings with this property. – Martin Brandenburg Dec 14 '14 at 13:30
  • @MartinBrandenburg I should have written “nontrivial”. I wanted to avoid “Yeah, well $S = ℂ$.” but I feared there might be some other nontrivial subrings I overlooked which are equally pointless, so I wrote “interesting”. The answer given by Hurkyl is absolutely satisfactory to me. – k.stm Dec 14 '14 at 15:01
  • Related: http://math.stackexchange.com/questions/1907319/ – Watson Feb 02 '17 at 12:19

2 Answers2

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There is an isomorphism $\mathbf{C} \cong \mathbf{C}_p$ -- i.e. you can extend the p-adic absolute value to the complexes.

$\mathbf{C}_p$ is the field of fractions of its ring of integers: the subring of numbers whose $p$-adic absolute value is less than or equal to $1$.

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    But $\mathbf{C}_p$ is the topological completion of the algebraic closure $\overline{\mathbf{Q}_p}$; doesn't the $p$-adic valuation make no sense on $\mathbf{C}_p$? In other words, I don't think $\mathbf{C}_p$ has a ring of integers. – Sal Dec 14 '14 at 12:51
  • Can we extend this to the reals as well? Is $S' = O_ℂ ∩ ℝ$ a subring with $ℝ = Q(S')$ (where $O_ℂ = {z ∈ ℂ;~|z|_p ≤ 1}$)? – k.stm Dec 14 '14 at 12:51
  • @Sal: Thanks; I should say the absolute value, not the valuation. –  Dec 14 '14 at 12:53
  • @Hurkyl: I'm afraid I don't think that solves anything. – Sal Dec 14 '14 at 12:58
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    @Sal: Then I don't understand the content of your objection; unless I'm overlooking something, it seems clear that the metric extends to the metric completion, and that it remains an ultrametric. –  Dec 14 '14 at 13:04
  • @Sal: Why not? Since the continuation of the $p$-adic absolute value has to be unique for algebraic extensions of $ℚ_p$, it makes sense to define a continued $p$-adic absolute value on the algebraic closure $\overline{ℚ_p}$. Because this value defines the metric on $\overline{ℚ_p}$, it has to be uniformly continuous and can therefore be extended to its own metric completion. The metric on the completion is still an ultrametric and so the ring given by Hurkyl is still a valid ring whose field of fractions must be all of $ℂ_p$. – k.stm Dec 14 '14 at 13:05
  • @Hurkyl: Apologies, you're correct. I was just confused. – Sal Dec 14 '14 at 13:13
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Depends what you mean by "interesting". The smallest examples of rings with $\mathbb{C}$ as their field of fractions would be $$R[\{t_\alpha\}_{\alpha\in A}]$$ where $R$ denotes the ring of algebraic integers, and $\{t_\alpha\}_{\alpha\in A}$ is a (edit: pure) transcendence basis for the extension $\mathbb{C}/F$, where $F$ is the field of algberaic numbers i.e. the field of fractions of $R$.

(Pure transcendence basis means that the $t_\alpha$ are transcendental numbers, algebraically independent, and $\mathbb{C}=F(\{t_\alpha\}_{\alpha\in A})$.)

Of course they are just isomorphic to $R$ with continuum-many variables adjoined. I don't think these rings are terribly useful, but they could arguably be "interesting".

Sal
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  • This construction shouldn't work as is: e.g. $\sqrt{t_\alpha}$ is not in the field of fractions of the ring you construct. You still need to make an algebraic extension of the field of fractions in order to get $\mathbf{C}$. –  Dec 14 '14 at 13:01
  • Grabbed the wrong element to take as a counterexample –  Dec 14 '14 at 13:02
  • @Hurkyl: If I change to pure transcendence bases, that should solve the error you've pointed out, right? – Sal Dec 14 '14 at 13:10