# Noetherian domain whose fraction field is such that some specific proper submodules are projective

Let $$R$$ be a Noetherian domain (which is not a field) with fraction field $$K$$. Suppose every proper $$R$$-submodule of $$K$$ of the form $$R[1/a]$$, where $$a\in R$$, is projective as an $$R$$-module. Then, I can show that $$R$$ has exactly one non-zero prime ideal.

My question is : Does it follow that $$R$$ is normal i.e. integrally closed in $$K$$ ?

• When you write $R[1/a]$, I assume $a\neq 0$ and it is just the localization, that is just inverting $a$. What is your example where your hypothesis is satisfied and $R$ is not a field? – Mohan Dec 1 '18 at 23:24
• @Mohan: Any DVR ... and my aim is to prove that they can only be DVR ... – user521337 Dec 1 '18 at 23:47
• if $R$ is a dvr an $a\neq o$ and not unit, then $R[1/a]$ is the fraction field and it is not projective over $R$. – Mohan Dec 2 '18 at 2:05
• @Mohan: yes, in that case there are no proper submodules of the form $R[1/a]$ and my condition is trivially satisfied ... – user521337 Dec 2 '18 at 2:33

In any $$1$$-dimensional local (not necessarily Noetherian) domain, $$K=R[1/a]$$ for every nonunit $$a \not= 0$$. This is a special case of the following fact:

Let $$R$$ be a domain with fraction field $$K$$, $$R \subsetneq K$$. Then there exists some $$a \not=0$$ in the intersection of the (nonzero) minimal primes of $$R$$ (sometimes called the pseudoradical of $$R$$) iff $$K = R[1/a]$$ for some $$a \in R$$.

Proof of fact: Suppose $$a$$ is in every minimal prime. Let $$S$$ be the multiplicatively closed set $$S = \{a^n\}_{n=0}^{\infty}$$. The saturation of $$S$$ must be $$R \setminus\{0\}$$, otherwise the complement of $$S$$ would contain a prime and therefore absurdly contain $$a$$. Conversely, if $$K = R[1/a]$$, pick an element $$p \in P$$ prime. Noting $$\frac{1}{p} = \frac{b}{a^n}$$ for some $$b \in R$$, it is clear that $$a \in P$$.

Therefore any $$1$$-dimensional Noetherian local domain that is not normal provides a counterexample by trivializing your additional assumption about proper $$R$$-submodules of $$K$$.

An easy example is.....

$$R = k[[T^2, T^3]] \cong k[[X,Y]]/ (X^2 - Y^3)$$.

$$R$$ is 1-dimensional, Noetherian, and local with maximal ideal $$(T^2, T^3)$$. Clearly $$R$$ is not normal (indeed its integral closure is $$k[[T]]$$.