Is there any proper subring of $\mathbb{R}$ with field of fractions equal to $\mathbb{R}$?

Question

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?

Answer 1

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?).