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Assume that $(R,\mathfrak{m})$ is a commutative local ring of equal characteristic zero. So $R$ contains the field of rationals. The well known $\mathfrak{m}$-adic completion of $R$ provides a complete ring $\hat{R}$ whose coefficient field is isomorphic to the residue field of $R$. Do there exists a topological method (completion) for providing a local Noetheiran extension $S$ of $R$ such that $S$ contains the real numbers and also contains $R$ as a dense subring.

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Why does $R$ contain the field of rationals? I don't see $1/2$ being an element of the localization of $\mathbb{Z}$ by the multiplicative subset $\mathbb{Z}\setminus(2)$. –  Jesko Hüttenhain Jul 29 '12 at 12:40
    
Sorry, by the characteristic zero I meant equal characteristic zero. Then the elements of $R$ of the form $n:=1+\ldots+1$ (n times) are unit. So the localization of $R$ by the multiplicative subset $\{n:n\in \mathbb{Z}\}$ is isomorphic to $R$. –  Aurora Jul 29 '12 at 12:57
    
The obvious algebraic method is to tensor over $\mathbb Q$ or another number field with $\mathbb R$. The existence of this method makes it unlikely a cleverer mathod will produce a more natural-seeming result. –  Will Sawin Jul 29 '12 at 20:08
    
Sawin@ Is $R⨂_Q\mathbb{R}$ again a local Noetherian ring? I am more interested in an extension which contains R as a dense subring. In addition, it is expected for that to be local Noetherian. –  Aurora Jul 29 '12 at 22:02
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@Aurora: Since each of your two comments in effect modified the question, it might be good to edit these modifications into the question itself, so that future readers can find the intended question by simply reading it rather than reading the whole comment thread. –  Andreas Blass Jul 29 '12 at 22:29
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