# Integral domain with fraction field equal to $\mathbb{R}$

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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Well, in the prime case it works: you take $\mathbb{Z}$, complete it with respect to the $p$-adic topology, and then take the field of fractions, and you get $\mathbb{Q}_p$ which is an analog of $\mathbb{R}$. –  the L Feb 2 '12 at 10:45
@Ali Bleybel, I don't understand your comments. Why is completion the only way to proceed? –  the L Feb 2 '12 at 11:09

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.
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 $$x^{-1}=a_1+\cdots+a_nx^{n-1}\in A.$$ –  Pierre-Yves Gaillard Feb 2 '12 at 13:42