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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

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    yes, you can also write $k[x][y]$ instead of $k[x,y]$.2012-11-08

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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 $T$.

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    @Hagen: Nice example ! For any domain $D$ whichi is not a field, imitating your construction, $S=D+yD[y]_f$ with $f\in D$ non-zero and not invertible, should be a non-noetherian sub-$D[y]$-algebra of $\mathrm{Frac}(D)[y]$.2012-11-09
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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 $D$).

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    Thanks, I will look at the paper later.2012-11-08