Given $p,q \in R$, when $(p,q) \subseteq k[p,q]$? 
*

*$k$ is a field of characteristic zero.

*$R$ is a commutative $k$-algebra, which is an integral domain.

*$Q(R)$ is the field of fractions of $R$.

*$p,q \in R^{\times}$.

*$I:=(p,q)$ (= the ideal generated by $p$ and $q$).

*$S:=k[p,q] \subseteq R$ (= the $k$-subalgebra of $R$ generated by $p$ and $q$).


An element of $I$ is of the form $Ap+Bq$, for some $A,B \in R$.

(1) I am looking for an example in which $S \subsetneq R$ and $I=(p,q) \subseteq k[p,q]=S$. (Or an additional condition which guarantees that $I \subseteq S$).

Notice that if $S=R$, then $I \subseteq R=S$.

(2) Is something interesting can be said about the set $Q(I):=\{\frac{u}{v}| 0 \neq v,u \in I\}$? or the set 
  $\frac{I}{S}:=\{\frac{u}{v}| 0 \neq v \in S,u \in I\}$?

It seems that $Q(I)$ is not a fractional ideal, but just a $Q(R)$-submodule of $Q(R)$.
Also, $\frac{I}{S}$ is not a fractional ideal,
and $\frac{I}{v_0}:=\{\frac{i}{v_0}|i \in I\}$ (for some fixed $v_0$) is not a fractional ideal (they are even not submodules).
Any hints are welcome!
 A: Let's suppose $p$ and $q$ are nonunits, since otherwise this is impossible.
You don't ask about one generator, but it helps me think to try small cases:
Let's first see if we can solve this for one generator.
Is it possible to find $R$, a domain, and $p\in R$ with $pR\subseteq k[p] \subsetneq R$?
Assume it is possible. Choose $f\in R\setminus k[p]$, then $fp\in k[p]$, so $fp=r(p)p+c$, with $c$ a constant, but then if $c\ne 0$, $(f-r(p))p = c$, so $c\in pR$, so $pR=R$, contradiction. Otherwise $fp=r(p)p$, so $f=r(p)$ (since $R$ is a domain), contradicting the choice of $f$.
Now with two generators.
Assume we have $pR+qR \subseteq k[p,q] \subsetneq R$.
Let $I=pR+qR$. Let $J=pS+qS$.
Then $I=J$. Why? Well, first $J\subseteq I$, clearly, and if 
$i\in I$, then $i\in S$, so $i=j+c$ for some constant $c\in k$ and $j\in J$. But then $i-j=c$, so $c\in I$, and $I\subsetneq R$, so $c=0$. Thus $i=j\in J$.
On the other hand, if $I=J$, then $I\subseteq k[p,q]$ obviously. 
Thus given $k[p,q]\subsetneq R$, then $I\subseteq k[p,q]$ if and only if $I=J$.
Now consider $k[x^2,x^3]\subsetneq k[x]$.
Observe that $I=x^2k[x]+x^3k[x]=x^2k[x]$, which is all polynomials with no linear or constant terms. Then $J$ contains $x^n$ for all $n\ge 2$, since it contains $x^2$, $x^3$, and by the Chicken McNugget theorem, for $n> 2\cdot 3-2-3=1$, $n=2a+3b$ for some $a,b\ge 0$. Thus $J$ contains all monomials of degree $\ge 2$, and hence $J=I$.
