Does every local domain admit an injective homomorphism into a discrete valuation subring of its fraction field? Let $R$ be a local ring which is an integral domain with fraction field $K$. Does there exist a discrete valuation subring $D\le K$ containing $R$?
 A: If the local domain $R$ is noetherian, then the answer is "Yes": let $S$ be the integral closure of $R$ in $K$. Then $S$ in general is not noetherian but by a Theorem of Nagata still has the property that the localisations $S_p$ at prime ideals $p$ of height $1$ are discrete valuation rings.
Extension to address the question, whether the extension $S_p|R$ is local: by the general properties of integral ring extensions one knows that $R$ and $S$ possess the same Krull dimension. Moreover there exists a prime ideal $q$ of $S$ lying over the maximal ideal $m$ of $R$; every such $q$ is itself a maximal ideal.
A commutative ring $R$ is called catenary if for all prime ideals $q\subset q^\prime$ of $R$ all maximal chains of prime ideals starting with $q$ and ending with $q^\prime$ have the same length.
A commutative ring $R$ is called universally catenary if every finitely generated ring extension $S$ of $R$ is catenary.
Assume now that the local ring $R$ has dimension $>1$, is universally catenary and that its integral closure $S$ in the field of fractions $K$ of $R$ is a finitely generated extension of $R$. Then none of the localisations $S_p$, $p$ a prime ideal of height $1$, is a local extension of $R$.
Proof: assume that the localisation $S_p$ is a local extension of $R$. Then $p=pS_P\cap S$ is a prime ideal lying over the maximal ideal $m$ of $R$. Hence $p$ is maximal. All maximal chains starting from $0$ and ending with $p$ by assumption have the same length, namely $\dim(S)=\dim(R)>1$. Hence there exists a prime ideal $p_0\subset p$ and $p$ cannot have height equal to $1$.
A: Let $F$ be a field and take the ring
$$ R = \bigcup_{n=1}^{\infty} F[[ x^{1/n} ]] $$
Its fraction field is the field of Puiseaux series
$$ K = \bigcup_{n=1}^{\infty} F(( x^{1/n} )) $$
$R$ is a valuation ring with value group the nonnegative rationals, so it is not a discrete valuation domain.
However, every nonzero element of $K$ can be written uniquely in the form $u x^a$, where $u \in R^\times$ and $a$ is a rational number. We can conclude that there does not exist any ring satisfying $R \subsetneq D \subsetneq K$.
