# For $n>1$ and a ring $R$ s.t. $k[x_1,\dots, x_n]\subset R\subset k(x_1,\dots, x_n)$ is $R$ a localization of $k[x_1,\dots, x_n]$?

Let $k$ be a field. Consider $n>1$ and a ring $R$ sitting between $k[x_1,\dots, x_n]$ and $k(x_1,\dots, x_n)=\mathrm{Frac}(k[x_1,\dots, x_n])$.

Q1: Is it true that $R$ is some localization of $k[x_1,\dots, x_n]$ as I cannot come up with a counter-example?

For $n=1$, it is clear as $k[x]$ is a Bezout domain which says everything sitting between $k[x]$ and $k(x)$ is some localization of $k[x]$.

Q2. The reason I am asking this question is that for ring of regular functions $O_{A^n}(U)$ on any open set is finite intersection of $O(D_i)$ where $D_i$ are distinguished open sets/basis covering $U$ and $O(D_i)$. So $O_{A^n}(U)$ is realized as inverse limit of $O(D_i)$. It looks like for most of $U\subset A^n$, I see them as some sort of localization of $O(A^n)$. Is there a counter example to this?

Q3. $O(U)=O(\cup D_i)=\cap O(D_i)$. $\cup D_i$ can be realized as direct limit. $O(\cup D_i)$ are regular functions from $D_i$ to $k$ which is basically $\mathrm{Hom}(-,k)$. So $\mathrm{Hom}$ converts the direct limit to inverse limit. However this conversion is in abelian category. How should I realize this notion? Or is this notion correct?

• As for terminology, from here : "The rings in which every overring is a localization are said to have the QR property; they include the Bézout domains and are a subset of the Prüfer domains". – Watson Aug 19 '17 at 20:49

Here are two examples.

(Example 1): Consider $R := k[x_{1},x_{2},\{x_{1}/x_{2}^{n} \;:\; n > 0\}]$. A basis for $R$ as a $k$-vector subspace of $k(x_{1},x_{2})$ is the collection $\{x_{1}^{i_{1}}x_{2}^{i_{2}}\}_{i_{1} \ge 0 , i_{2} \ge 0} \cup \{x_{1}^{i_{1}}x_{2}^{i_{2}}\}_{i_{1} > 0 , i_{2} < 0}$ of monomials.

Claim: The ring $R$ is not Noetherian (so it is not isomorphic to a localization of $k[x_{1},x_{2}]$).

Proof: For $n > 0$, let $\mathfrak{a}_{n}$ be the ideal of $R$ generated by $x_{1}/x_{2}^{n}$. We show that the inclusion $\mathfrak{a}_{n} \subset \mathfrak{a}_{n+1}$ is not an equality for all $n > 0$. Suppose there exists $r \in R$ such that $r(x_{1}/x_{2}^{n}) = x_{1}/x_{2}^{n+1}$. This implies $r = 1/x_{2}$ using the multiplication law in the fraction field $k(x_{1},x_{2})$; this is a contradiction since $1/x_{2} \not\in R$.

(Example 2): Consider $R := k[x_{1},x_{2},\frac{x_{1}}{x_{2}}] = k[x_{2},\frac{x_{1}}{x_{2}}]$. Since $R$ is abstractly isomorphic to the polynomial ring $k[x,y]$, the canonical inclusion $k^{\times} \to R^{\times}$ is an isomorphism.

We have the following description of the units of a localization of a UFD:

Let $A$ be a UFD, let $W$ be a set of irreducible elements of $A$ that are pairwise coprime, and let $S$ be the multiplicative subset of $A$ generated by $W$. Then the canonical inclusion $$\textstyle A^{\times} \oplus \bigoplus_{w \in W} \mathbb{Z} \to (S^{-1}A)^{\times}$$ is an isomorphism (of abelian groups).

This shows that the inclusion $k[x_{1},x_{2}] \subset R$ is not of the form $k[x_{1},x_{2}] \to S^{-1}(k[x_{1},x_{2}])$ for a nontrivial multiplicative subset $S$ of $k[x_{1},x_{2}]$.