Units in a ring between an integral domain $R$ and its fraction field.

Question

Let $R$ be an integral domain with fraction field $K$, and let $\mathcal{T}$ be the set of subrings $T$ of $K$ such that $R\subseteq T$.

Call $R\;$unit-complete$\;$if for all $T\in \mathcal{T}$, either $T=R$ or $T$ has a unit element which is not in $R$.

Call $R\;$strongly unit-complete$\;$if for all $T\in \mathcal{T}$, either $T=R$ or $T$ contains ${\large{\frac{1}{r}}}$ for some non-unit element $r\in R$.

It's immediate that strongly unit-complete implies unit-complete.

It's easily shown that if $R$ is a $\text{PID}$, then $R$ is strongly unit-complete.

Proof:

Let $R$ be a PID with fraction field $K$, and let $T$ be a subring of $K$ with $R\subset T$ (proper inclusion).

Fix $t\in T{\setminus}R$, and write $t={\large{\frac{a}{b}}}$, where $a,b\in R$ are such that $b$ is a non-unit of $R$ and $\gcd(a,b)=1$.

From ${\large{\frac{a}{b}}}\in T$, we get $1-{\large{\frac{a}{b}}}\in T$, so ${\large{\frac{b-a}{b}}}\in T$.

From $\gcd(a,b)=1$, we get $\gcd(a,b-a)=1$.

Then since $R$ is a $\text{PID}$, there exist $u,v\in R$ such that $ua+v(b-a)=1$, so $$ \frac{a}{b},\frac{b-a}{b}\in T \implies u\Bigl(\frac{a}{b}\Bigr)+v\Bigl(\frac{b-a}{b}\Bigr)\in T \implies \frac{1}{b}\in T $$ hence $R$ is strongly unit-complete.

Some questions:

$(1)\;$If a noetherian integral domain $R$ is strongly unit-complete, must $R$ be a $\text{UFD}$?

$(2)\;$Is there an integral domain $R$ which is unit-complete but not strongly unit-complete?

Answer 1

Here's a proof that if $R$ is a Dedekind domain with torsion class group, then $R$ is strongly unit-complete. This shows that the answer to question (1) is no, as there are Dedekind domains with torsion class group which are not UFDs, e.g. $\Bbb Z[\sqrt{-5}]$.
Let $K$ be the fraction field of $R$. Let $K \supset T \supsetneq R$ be an overring, then we can find $\frac{a}{b} \in T \setminus R, a,b \in R$ so that $b$ is not a unit in $R$. As the class group of $R$ is torsion, we can find some $n$ such that $(a,b)^n=(c)$. Write $c=\sum_{i=0}^nc_ia^ib^{n-i}$ for some $c_i \in R$, divide by $b^n$ and we get $$\frac{c}{b^n}=\sum_{i=0}^n c_i (\frac{a}{b})^i \in T.$$ Because $b^n \in (a,b)^n=(c)$, we get $b^n=tc$ for some $t \in R$, so we get $$T \ni \frac{c}{b^n}=\frac{1}{t}$$

So it remains to show that $t$ is not a unit in $R$. Suppose $t\in R^\times$, then $(b)^n=(c)=(a,b)^n$. By unique factorization of ideals into prime ideals in $R$, this implies $(a,b)=(b)$, i.e. $b \mid a$, contradicting $\frac{a}{b} \notin R$.