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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 $R$?

user26857
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2 Answers2

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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 $1$.

Hagen Knaf
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  • Interesting! Do you have a reference for this theorem of Nagata? –  Mar 06 '17 at 06:52
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    You can find it in Nagata's book "Local Rings". To be correct one has to say that the theorem was proved by Mori for local rings $R$ and by Nagata in the general case. – Hagen Knaf Mar 06 '17 at 06:57
  • That's why it's called the Mori-Nagata Theorem. – user26857 Mar 06 '17 at 11:25
  • @HagenKnaf Do you know if the resulting injection of $R$ into this discrete valuation subring $S_p\subset K$ is a local ring homomorphism? –  Mar 06 '17 at 19:32
  • Typically the answer is "no". However the general siatuation is a bit complicated, so I extend my answer to give an explanation. – Hagen Knaf Mar 06 '17 at 22:57
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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 $R \subsetneq D \subsetneq K$.