Is every integral domain contained in a discrete valuation ring?

Question

Is is true that every integral domain which is not a field is contained in a proper subring of its fraction field which is a DVR?

Answer 1

No, this isn't true. For instance, let $R$ be a valuation ring with value group $\mathbb{Q}$, with fraction field $K$. If $x\in K\setminus R$, then $R[x]=K$, since for every $a\in K$, $ax^{-n}\in R$ for some $n\in\mathbb{N}$ (because $\mathbb{Q}$ is archimedean). So if $S$ is a ring such that $R\subseteq S\subset K$, then $S=R$. Since $R$ is not a DVR, this means there is no such $S$ that is a DVR.

Answer 2

If $V$ is a valuation ring, then every overring of $V$ is also a valuation ring. Moreover, these are localizations of $V$. Suppose that $R$ is an overring of $V$ which is a DVR. Then $R=V_{\mathfrak p}$ with $\operatorname{ht}\mathfrak p=1$. All we need is a valuation ring $V$ such that $V_{\mathfrak p}$ is not noetherian for all height one primes $\mathfrak p$. In particular, every rank one valuation ring which is not a DVR is a counterexample.