When is an integral domain a localization of a proper subring?

Question

Given an integral domain $R$, is $R$ a localization of a proper subring if and only if there is some unit $u \in R^{\times}$ for which $u^{-1} \notin \mathbb{Z}[u]$, where $\mathbb{Z}[u]$ denotes the (unital) subring of $R$ generated by $u$?

One direction clearly holds: If $R$ is a localization of a proper subring $S$, then any for any nonunit $u$ in $S$ that becomes a unit in $R$, one must have $u^{-1} \notin \mathbb{Z}[u]$.

Conversely, suppose that $u$ is a unit in $R$ for which $u^{-1} \notin \mathbb{Z}[u]$. Then, there is clearly at least one subring of $R$ containing $u$ but not $u^{-1}$, namely $\mathbb{Z}[u]$. But of course, $\mathbb{Z}[u,u^{-1}]$, the subring of $R$ generated by $u$ and $u^{-1}$, is not necessarily all of $R$, so $\mathbb{Z}[u]$ is not the subring we want for $S$.

Perhaps, $S$ should be a subring maximal among those that contain $u$ but not $u^{-1}$, but we need to show whether such a subring actually exists (with Zorn's lemma, maybe) and whether for this subring $S$, $S[u^{-1}]=R$.

Answer 1

Let $F=Frac(\Bbb{Z}/char(R))$,

• If $u$ is transcendental over $F$,

  With the axiom of choice let $X\subset R$ be a transcendence basis of $Frac(R)$ chosen such that $u\in X$.

Note that $F[X]_{(u)}[u^{-1}]=F(X):\ \ $ for $\frac{a(X)}{b(X)}\in F(X)=Frac(F[X])$ there is some $j$ such that $b(X)u^{-j}\in F[X]-u F[X]$ and $u^{-j} \frac{a(X)}{ b(X)u^{-j}}\in F[X]_{(u)}[u^{-1}]$

Any $r\in R$ is algebraic over $F(X)$, it is the root of some $h(t) \in F(X)[t]$,

$u^m h(t) \in F[X]_{(u)}[t]$ whose leading coefficient is $u^{n(r)} a$ with $a\in F[X]_{(u)}^\times$ so $r u^{n(r)}$ is integral over $F[X]_{(u)}$.

Let $$S = \Bbb{Z}/(char(R))[\{ r u^{n(r)} ,r\in R\}]$$

Then $S$ is integral over $F[X]_{(u)}$ so $u^{-1}\not \in S$ and clearly $R=S[u^{-1}]$.

• If $u$ is algebraic over $F$ then $char(R)=0$. Let $O$ be the integral closure of $\Bbb{Z}[u]$ in $\Bbb{Q}(u)$ and $m\subset O$ a maximal ideal containing $u$ (which exists because $u^{-1}\not \in O$) and $(\pi)$ the maximal ideal of $O_m$.

  Take again a transcendence basis $X$. Then $\Bbb{Q}(u)(X) = O_m[X]_{(\pi)}[\pi^{-1}]=O_m[X]_{(\pi)}[u^{-1}]$ and with $n(r)$ constructed as above such that $r\pi^{n(r)}$ thus $r u^{n(r)}$ is integral over $O_m[X]_{(\pi)}$ then $$S = \Bbb{Z}[\{ r u^{n(r)} ,r\in R\}]$$ works (it is integral over $O_m[X]_{(\pi)}$ so $u^{-1}\not \in S$)