Let $R$ be a commutative ring and denote by $K(R)$ its total ring of fractions, the localization of $R$ with respect to $R\_{\mathrm{reg}}$. For every multiplicative subset $U \subseteq R$ there is a canonical map of $R$-algebras
$$H : K(R)[U^{-1}] \to K(R[U^{-1}]).$$
This may fail to be surjective; see Kleiman's article about Misconceptions about $K\_X$, available [here](http://retro.seals.ch/digbib/fr/view;jsessionid=9363B3CFB982CD263083C1E7C2410C12?rid=ensmat-001:1979:25::372&id=ssearch&id2=browse4&id3=). At least it is true when $R$ is reduced and $\mathrm{Spec}(R)$ has finitely many irreducible components.


**Question.** Is $H$ an isomorphism for every reduced commutative ring $R$?


Actually this is Exercise 3.15a in Eisenbud's book on commutative algebra (page 113), but my gut feeling is that the answer is no in general and that this is another misconception about $K(R)$ *or* Eisenbud forgot to mention the noetherian hypothesis. There is a hint on page 716 which says that $K(R)$ is zero-dimensional; well again this is only clear when $R$ is noetherian.


What I have done so far: The map $H$ is well-defined by the universal properties. One checks that $H$ is injective. It is easy to see that $H$ is surjective iff the following property holds, where $\tau : R \to R[U^{-1}]$ denotes the canonical localization map:


$(\*)$ If $r \in R$ such that $\tau(r) \in R[U^{-1}]$ is regular, then there is some regular $s \in R$ such that $\tau(r) | \tau(s)$.


The general case can be easily reduced to $K(R)$; we arrive at the


**Equivalent question.** Is the class of reduced rings with the property that every regular element is a unit (sometimes called "classical ring", see [MO/42647](https://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor)) stable under localization?


If $R$ is zero-dimensional and reduced, this means that every element is associated to some idempotent element (see Gilmer's *Zero dimensional rings*); the same then must be true for localizations. Hence nothing goes wrong for zero-dimensional rings. This applies, in particular, to artinian rings, and therefore also to finite rings. But I strongly suspect that an *infinite* product of finite nontrivial rings has a chance to be a counterexample, for example $\prod\_{n} \mathbb{Z}/p^n$. This is is a reduced (EDIT: no!) classical ring of positive dimension (thus contradicting Eisenbud's hint). Is the localization at $p$ also classical?


**More general.** What can be said about the structure of reduced classical rings?



Total ring of fractions vs. Localization
A counterexample from one of [my other MO answers](https://mathoverflow.net/questions/33294/linearly-independent-subsets-of-a-free-module) seems to work again. Let $k$ be an algebraically closed field. Let $R$ be the ring of functions $f: k^2 \to k$ such that there exists a polynomial $\overline{f} \in k[x,y]$ with  

1. $f(x,y) = \overline{f}(x,y)$ for all but finitely many $(x,y) \in k^2$ and  

2. $f(0,0) = \overline{f}(0,0)$.


**$R$ is reduced:** If $f(x,y) \neq 0$, then $f(x,y)^n \neq 0$ for all $n$. $\square$


**Every element $f$ of $R$ is either a unit or a zero divisor:** 


**Case 1: $f$ is nowhere zero.** In this case, $\overline{f}$ must lie in $k^{\ast}$, as otherwise $\overline{f}$ vanishes at infinitely many points and $f=\overline{f}$ at all but finitely many of them. So $f^{-1}$ is equal to the nonzero constant $\overline{f}^{-1}$ at all but finitely many points, and $\overline{f}^{-1} \in R$. So, in this case, $f$ is a unit.


**Case 2: $f(x\_0,y\_0)=0$**. Without loss of generality, we may assume that $(x\_0,y\_0) \neq (0,0)$. This is because, if $f(0,0)=0$ then $\overline{f}(0,0)=0$, implying that $\overline{f}$ vanishes at infinitely many points and $f=\overline{f}$ at all but finitely many of them, so we can find some other $(x\_0,y\_0)$ at which $f$ also vanishes. Let $\delta(x,y)$ be $1$ if $(x,y) = (x\_0, y\_0)$ and $0$ otherwise. Then $f \delta=0$ and $\delta \neq 0$, showing that $f$ is a zero divisor. $\square$


The set of functions vanishing at $(0,0)$ is clearly a maximal ideal of $R$; which we will denote $(0,0)$. We claim that $R\_{(0,0)} \cong k[x,y]\_{(0,0)}$. **Proof sketch:** We claim that $f = \overline{f}$ in the localization. To see this, let $g$ vanish at the finitely many points where $f \neq \overline{f}$, but $g(0,0) \neq 0$. Then $fg=\overline{f} g$, and $g$ is invertible in the localization. This shows that $f s^{-1} = \overline{f} \overline{s}^{-1}$ for any $f$ and $s$. $\square$.


Clearly, $k[x,y]\_{(0,0)}$ is not classical. 


I think this construction can clearly be generalized to make classical rings which have any local ring of dimension $\geq 2$ as a localization. 


