``Minimal generating ring" for a field of fractions In this answer and the linked MathOverflow post, it's shown that any field $F$ of characteristic zero contains a proper subring $A$ such that $F$ is the field of fractions of $A$.
However, there is often more than one such $A$. For example, $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$, $\mathbb{Z}[1/2]$, $\mathbb{Z}[3/8]$, etc. But in some sense, $\mathbb{Z}$ is the "true" subring of $\mathbb{Q}$ that generates $\mathbb{Q}$ as a field of fractions.
In particular, $\mathbb{Z} \subset \mathbb{Z}[1/2]$ and $\mathbb{Z} \subset \mathbb{Z}[3/8]$. So my question: for any field $F$ of char zero, is there a minimal proper subring $A \subset F$ such that $F$ is the field of fractions of $A$? Where minimal means that if $A$ and $B$ are proper subrings of $F$ whose fields of fractions are $F$, and if $A$ is minimal, then $A\subset B$.
 A: Minimal such subrings need not exist. For example, consider $\mathbb{Q} (\sqrt{3})$. Then, for any positive integer $n$, the subring $\mathbb{Z} [n \sqrt{3}]$ has $\mathbb{Q} (\sqrt{3})$ as its field of fractions. But we have
$$\mathbb{Z} [\sqrt{3}] \supset \mathbb{Z} [2 \sqrt{3}] \supset \mathbb{Z} [2^2 \sqrt{3}] \supset \mathbb{Z} [2^3 \sqrt{3}] \supset \cdots$$
and the intersection of this descending chain is just $\mathbb{Z}$, which does not have $\mathbb{Q} (\sqrt{3})$ as its field of fractions.
More generally, you do this with any non-trivial finite extension of $\mathbb{Q}$.
A: A counterexample can be found by considering the subrings of $K[X]$ of the form $K[X^n,X^{n+1}]$, $n\ge 1$, which have intersection $K$. 
A: Such a subring doe not necessarily exist. For example, consider the ring $\mathbb Q[x]$. For each $n$, the subring $R_n=\mathbb Q[x^n,x^{n+1}]$ has $\mathbb Q(x)$ as its field of fractions. If there were a minimal $A$ with this property, then we would have to have $A \subseteq \bigcap R_n=\mathbb Q$, a contradiction.
