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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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This is probably a better way to put the question: is every ring between $k[x,y]$ and $k(x)[y]$ noetherian? – Yashar Nov 8 '12 at 9:22
yes, you can also write $k[x][y]$ instead of $k[x,y]$. – user18119 Nov 8 '12 at 9:24
up vote 3 down vote accepted

Let $p\in k[x]$ be a prime and $T:=\{p^i : i\in\mathbb{N}\}$. Then the ring


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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Thanks. I'm not sure about your last sentence! I write $f_i=p^{-n_i}g_iy$, where $g_i = yh_i + u_i, \ h_i \in k[x,y], \ u_i \in k[x]$. We then have $c = \sum v_ip^{-n_i}u_i$ for some $v_i \in k[x]$. Right? – Yashar Nov 8 '12 at 14:13
Yes. As a consequence of your last equation the exponent $v$ in $c=a/p^v$, $a\in k[x]$ not divisible by $p$, would be bounded from above by the maximum of the $n_i$. But this is impossible since $c$ can be chosen arbitrarily. – Hagen Knaf Nov 8 '12 at 14:33
Right! So, $p$ doesn't have to be prime and any $p \in k[x] \setminus k$ will work. – Yashar Nov 8 '12 at 15:00
In fact you can replace $T^{-1}k[x]$ by any proper overring of $k[x]$ within $k(x)$. Every such ring is a localization of $k[x]$ with respect to a suitable multiplicative subset. It thus cannot be a finitely generated $k[x]$-module, since that would imply that the ring extension is integral. – Hagen Knaf Nov 8 '12 at 15:43
@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 $\mathrm{Frac}(D)[y]$. – user18119 Nov 9 '12 at 0:10

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. – Yashar Nov 8 '12 at 14:15

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