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

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.

The question was asked and answered on MathOverflow in 2010, in somewhat greater generality, and by two different methods.