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I wonder if there is an integral domain $A\subseteq \mathbb{R}$ which is not a field, and such that the field of fractions of $A$ is equal to $\mathbb{R}$?

Edit: here as a possible direction: it is known that there is no finite field extension $k\subseteq \mathbb{R}$. Given such an $A$, can we produce such a $k$, contradicting this theorem?

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    @Ali Bleybel, I don't understand your comments. Why is completion the only way to proceed?2012-02-02

2 Answers 2

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The field extension $\mathbb{R}/\mathbb{Q}$ posesses a transcendence basis $T$. Consider the polynomial ring $\mathbb{Q}[T]$; it is not a field. The extension $\mathbb{R}/\mathbb{Q}(T)$, $\mathbb{Q}(T)$ the fraction field of $\mathbb{Q}[T]$ is algebraic. Let $A$ be the integral closure of $\mathbb{Q}[T]$ in $\mathbb{R}$. The fraction field of $A$ then equals $\mathbb{R}$, and $A$ is not a field, because by the Lying-over-Theorem there exists a prime ideal of $A$ lying over a given non-zero prime ideal of $\mathbb{Q}[T]$.

Remark: one can show that an integral domain $A$ having $\mathbb{R}$ as field of fractions is not noetherian.

Remark 2: the argument just given essentially shows: if a field $K$ does not contain an integral domain $A$ such that $K$ is the field of fractions of $A$ and $A$ is not a field, then $K$ is an algebraic extension of a finite field.

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    Dear Hagen: Great answer! For those who haven't seen this: If a field $B$ is integral over a domain $A$, then $A$ is a field. Proof: If $0\neq x\in A$, then $x^{-1}\in B$, so $x^{-n}=a_1x^{1-n}+\cdots+a_n,\quad a_i\in A,$ and thus $x^{-1}=a_1+\cdots+a_nx^{n-1}\in A.$2012-02-02
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The question was asked and answered on MathOverflow in 2010, in somewhat greater generality, and by two different methods.