Classification of rings between a PID and its field of fractions? Let $D$ be a PID and let $\mathrm{Frac}(D)$ be its field of fractions. I want to classify the intermediate rings $D\subseteq R\subseteq \mathrm{Frac}(D)$.
Theorem: Every such ring $R$ is a localization $D[S^{-1}]$ for some subsemigroup $S\subseteq (R,\times,1)$.
Proof: Given $D\subseteq R\subseteq\mathrm{Frac}(D)$ consider the set of denominators $$S:=\left\{ b\in D: \frac{a}{b}\in R \text{ and } \mathrm{gcd}(a,b)=1\right\}.$$
If $b\in S$ then there exists $\frac{a}{b}\in R$ with $\mathrm{gcd}(a,b)=1$ by definition. Since $D$ is a PID there exist $x,y\in D$ such that $1=ax+by$. Then dividing by $b$ gives $$\frac{1}{b}=\frac{a}{b} x + y.$$ Since $R$ is a ring this implies that $\frac{1}{b}\in R$. Now we can see that $S$ is a subsemigroup of $(D,\times,1)$. Indeed, we have $1\in S$, and if $b_1,b_2\in S$ then by the above remarks we have $\frac{1}{b_1},\frac{1}{b_2}\in R$, hence $\frac{1}{b_1b_2}\in R$. It follows that $b_1b_2\in S$. Thus the localization $D[S^{-1}]$ is well-defined.
I claim that $R=D[S^{-1}]$. Indeed, we have $R\subseteq D[S^{-1}]$ by definition. Conversely, consider any $\frac{a}{b}\in D[S^{-1}]$, i.e., with $b\in S$. By the above remarks this implies $\frac{1}{b}\in R$ and hence $\frac{a}{b}=a\frac{1}{b}\in R$. $\square$
Thus we obtain a surjective map from the subsemigroups of $(D,\times,1)$ to the rings between $D$ and $\mathrm{Frac}(D)$. This map is not injective because it is possible to have $D[S^{-1}]=D[T^{-1}]$ for different subsemigroups $S,T$. However, given $S$, there is a largest such subsemigroup $T$ defined by $$T=\left\{a\in D : ab\in S \text{ for some } b\in D\right\}.$$ Such sets $T$ are characterized by the condition that for all $a,b\in D$ we have $$ab\in T \Longleftrightarrow a\in T \text{ and } b\in T.$$ Thus, since $D$ is a UFD, these sets (and therefore the intermediate rings $D\subseteq R\subseteq \mathrm{Frac}(D)$) are in bijection with sets of prime elements.
Finally, my question is this: 
Is there a name for a set $S\subseteq D$ such that $1\in S$ and for all $a,b\in D$ we have $$ab\in S \Longleftrightarrow a\in S \text{ and } b\in S\,\,?$$
I have been unable to find any references. Thanks.
 A: The question is elementary and was already answered in the comments. I'm posting a cw answer so that the question can be ticked as answer (otherwise it remains regularly bumped).
Let $I_D$ be the set of irreducible elements, $I_D/D^\times$ its quotient modulo multiplication by invertible elements, and $\mathcal{P}(X)$ the power set of $X$. 

Then the map $S\mapsto D[S^{-1}]$, from $\mathcal{P}(I_D)$ to the set $\mathcal{L}_D$ of subrings of $\mathrm{Frac}(D)$ containing $D$, is surjective, and induces a bijection from $\mathcal{P}(I_D/D^\times)$ to $\mathcal{L}_D$.

The surjectivity follows from the fact that if $R\in\mathcal{L}_D$ and $a/b\in R$ with $a,b$ coprime, and $s$ is irreducible and divides $b$, then $1/s\in R$. This is an easy consequence of Bézout. 
The injectivity is easy: clearly the set of elements of $D$ that is invertible in $R$ is invariant by multiplication by $D^\times$, and if $S$ is a set of irreducible elements in $D$, then the set of elements in $D$ invertible in $D[S^{-1}]$ is exactly $SD^\times$.
(Such a description does not pass to the case of UFDs: in the polynomial algebra $K[x,y]$, $1/x\notin K[x,y,y/x]$; more generally it fails in an arbitrary non-principal UFD.)
Next, for $D$ an arbitrary domain, one can consider the set of localizations of $D$, that is, the subset $\mathcal{L}'_D$ of $\mathcal{D}$ consisting of those $D[S^{-1}]$ with $S\subset D^\times$. 
Such an $S$ can always be chosen to be "saturated" in the sense that it is a submonoid of $D^\times$ satisfying $ab\in S$ implies $a,b\in S$. (About "unitary" I find it hard to find a worse choice of terminology, since "unitary" is widely used in the sense "contains 1".)
The above description of saturated submonoids of $D^\times$, namely as indexed by subsets of $I_D/D^\times$, holds for arbitrary UFDs. It's actually purely a property of the monoid $(D,\times)$, as is being a UFD ($D$ is a UFD iff $(D,\times)$ has an absorbing element $0$, the complement $D\smallsetminus\{0\}$ is a submonoid, and $D\smallsetminus\{0\}$ is direct product of a group with a free monoid.)
A: you can find such a classification in the following link:
https://www.researchgate.net/publication/283199027_Subrings_of_Q
Ali Jaballah Subrings of Q, Journal of Science and Technology Vol 2(No 2):1-13, 1997.
Abstract:  In this paper we determined all the subrings of Q, the field of rational numbers. We established a one to one correspondence between the set of subrings of Q containing Z, the ring of integers, and the set of subsets of prime numbers. We also studied the rings S between Z and Q that satisfy finiteness conditions on the set of prime ideals or on the set of rings between Z and or between S and Q, and we established equations relating the cardinality of these sets. Finally we have generalized some of these results to principal ideal domains.
