Is there any proper subring of $\mathbb{R}$ with field of fractions equal to $\mathbb{R}$? Can we construct that proper subring? Is it necessarily an integral domain?
Updated: Is there an example to construct the solution?
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Mojee KD
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4The only easy question in that list is the last. A subring of an integral domain is surely an integral domain itself. – Jyrki Lahtonen Aug 14 '15 at 09:45
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1Since the field of fractions only makes sense for integral domains, it must be integral. – sebigu Aug 14 '15 at 09:49
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1Related: http://mathoverflow.net/questions/47103/is-every-field-the-field-of-fractions-of-an-integral-domain – user26857 Aug 14 '15 at 16:33
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Let $(x_i)_{i\in I}$ be a transcendence basis of $\mathbb R$ over $\mathbb Q$, and $S=\mathbb Z[x_i:i\in I]$. Then $Q(S)$, the field of fractions of $S$, is $\mathbb Q(x_i:i\in I)$, and $Q(S)\subset\mathbb R$ is an algebraic field extension. Now let $R$ be the integral closure of $S$ in $\mathbb R$. We have $R\subsetneq\mathbb R$ (why?) and $Q(R)=\mathbb R$ (why?).
user26857
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1The above answer works for arbitrary fields of characteristic zero which are not algebraic over $\mathbb Q$. – user26857 Aug 14 '15 at 16:31
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