Let $F$ be a field. Call a unital subring $R\subset F$ dense if there is no subfield $K\subsetneq F$ such that $R\subset K$.
Is there a characteristic zero field $F$ such that the only regular Noetherian dense unital subring $R\subset F$ is $F$ itself?
Welcome new contributor!
You are essentially asking if there is a field $F$ such that the only regular noetherian subring of $F$ with fraction field $F$ is $F$ itself.
Every algebraically closed field is such an example, as well as $\mathbb{R}$.
Indeed, suppose that $F$ is the fraction field of a regular noetherian domain $R$ with $R\neq F$. Then $R$ is not a field and so admits a height-$1$ prime ideal $P$. The ring $R_P$ is a regular local ring of dimension $1$, so it is a discrete valuation ring with fraction field $F$. As a result, we have a (surjective) discrete additive valuation $\nu: F^\times\to \mathbb{Z}$ such that $R_P={\cal O}_\nu$. Take $\pi\in R_P$ with $\nu(\pi)=1$. Then $\pi$ cannot have a square root in $F$ (otherwise that square root $x$ would satisfy $2\nu(x)=1$). Thus, $F$ cannot be algebraically closed.
Using cubic roots instead of square roots shows that $F\ncong \mathbb{R}$.
I think $\mathbb{R}$ is an example. Suppose $R\subsetneq \mathbb{R}$ is a regular noetherian dense subring. Then if $R$ is not complete with respect to standard metric of $\mathbb{R}$, then the quotient field of $R$ is also not complete. Contradiction. So $R$ is complete and if $R$ is not discrete around $0$, then we have a sequence $a_i\in R$ which is converging to $0$, so for any natural number $n$, we can choose $N_n$ such that if $i>N_n$, then $|a_i|<\frac{1}{n}$. Then for any natural number $m$, we have $ma_i\in R$ and we may conclude that $R$ is dense in $\mathbb{R}$. So by completeness, $R=\mathbb{R}$. Otherwise, $R$ must be discrete in the topology in $\mathbb{R}$. Then $R$ is conutable and the quotient field is also countable. Contradiction. So $\mathbb{R}$ would be an example.
