# 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". Commented Aug 19, 2017 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 $k[x_{1},x_{2}]$.