An overring of a polynomial ring, noetherian or not? Let $S$ be a commutative domain and let $k$ be a subfield of $S$. Let $R:=k[x,y] \subseteq S$ be the polynomial ring in two variables $x,y$ and suppose that for every $s \in S$ there exists some $0 \neq r \in k[x]$ such that $rs \in R$, i.e. $S \subseteq (k[x] \setminus \{0\})^{-1}R$. My question: does $S$ have to be noetherian?
Thanks 
 A: In the paper of A. Wadsworth, Pairs of domains where all intermediate domains are Noetherian, it is given the following criterion: for noetherian domains $D\subset E$ the pair $(D[y],E[y])$ is a noetherian pair, that is, all its intermediate subrings are noetherian, if and only if $E$ is a finite integral extension of $D$. 
With $D=k[x]$ and $E=k(x)$ we are very far from a finite integral extension, so your example is not a noetherian pair. Moreover, the proof given by Wadsworth (which is very simple) suggests how to find concrete counterexamples.
Edit. As I said before, Wadsworth's paper suggest how to find examples of non-noetherian rings in a non-noetherian pair: take an intermediate ring $A$ between $D[y]$ and $E[y]$ and $0\neq I\subset A$ a proper ideal such that $A/I$ is not a finitely generated $D[y]$-module. Then the ring $D[y]+I$ is not noetherian. For example $A=E[y]$ and $I=yE[y]$ satisfies the requirement. In this case we get the non-noetherian ring $D+yE[y]=k[x]+yk(x)[y]$. 
The example given by Hagen falls in this class of examples: just take $D\subset E'\subset E$ with $E'$ also not finite over $D$ and use the same construction as above with $E'$ instead of $E$. In his case $E'$ is a ring of fractions of $D$ different from the quotient field (of $D$).
A: Let $p\in k[x]$ be a prime and $T:=\{p^i : i\in\mathbb{N}\}$. Then the ring
$S:=k[x]+T^{-1}k[x][y]y$
is not noetherian: the subset $I:=T^{-1}k[x][y]y$ is an ideal of $S$. It is not finitely generated, since if $f_1,\ldots ,f_r$ were a set of generators, then for every $c\in T^{-1}k[x]$
$cy=\sum\limits_{i=1}^r g_if_i$, $g_i\in S$.
This yields that the coefficients of the monomials of degree $1$ of the polynomials $f_i$ generate $T^{-1}k[x]$ as a $k[x]$-module, which is impossible due to the choice of $T$.
