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?



Characteristic zero field with unique regular Noetherian dense unital subring
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 $F\ncong \mathbb{R}$.


