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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. 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) 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?

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Sorry, I am being dumb. Why is $\prod \mathbb{Z}/p^k$ reduced? Isn't the element $(p^{\lfloor k/2 \rfloor})$ nilpotent? – David Speyer Jun 18 '12 at 18:22
Oh, I see that you correct this in comments to Will's answer. – David Speyer Jun 18 '12 at 18:33
up vote 12 down vote accepted

A counterexample from one of my other MO answers 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.

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Sorry, I've deleted my question in the meantime. Another one: In the definition of $\delta$ we should swap 0 and 1, right? And in the end you mean that $k[x,y]_{(0,0)}$ is not classical, right? – Martin Brandenburg Jun 19 '12 at 12:51
Right on both counts, thanks! – David Speyer Jun 19 '12 at 13:02
Ok. Thank you for this neat counterexample! And indeed we can use "almost polynomial functions on $k^n$" to get examples in dimension $n \geq 2$. – Martin Brandenburg Jun 19 '12 at 13:21

Consider the ring $\prod k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. First some notation. If $f$ is an element, then $f_n$ is the part of it in $k[x,y]/(x^n,x^{n-1}y,...,xy^{n-1},y^n)$. The degree of $f_n$ is the highest power of the ideal $(x,y)$ that it lies in. Since this is not reduced, the ring we will work with is this modulo the ideal of nilpotents. An element $f$ is nilpotent if $f_n^k$ has degree at least $n$ for some fixed $k$, or, equivalently, if $f_n$ has degree at least $n/k$. Since $n/k$ is positive, if $f_n$ is degree $0$ then $f$ cannot be nilpotent.

This will allow us to check that the ring is classical. If $f_n$ has positive degree, then $f$ is a zero-divisor. Take $g$ such that $g_n=1$, $g_m=0$ for $m\neq n$. $g_n$ is degree $0$ so $g$ is not nilpotent. $f_mg_m$ has degree at least $m/n$, so $f_mg_m$ is nilpotent. Thus a regular element is one which is degree $0$ everywhere, and such an element is invertible.

The localization at $y$, however, is not classical. $f y^l=0$ if the degree of $f_n y^l$ is at least $n/k$, equivalently, if the degree of $f_n$ is at least $n/k-l$. This is the kernel of the localization. If $xf$ is in this ideal then $f$ is in the ideal, since $f$ would have degree $n/k-l-1$. So $x$ is regular. We need to check that $x$ is not invertible, which would happen when $xa+b=y^d$ for some $d$ and some $b$ in the ideal. (then the inverse would be $a/y^d$.) Choose some $n$ high enough that the degree of $b_n$ is more than $d$. Then the equation $xa_n+b_n=y^d$ is impossible.

This completes the problem.

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Hm, can you elaborate this a little bit and define the ring explicitly in the answer? – Martin Brandenburg Jun 19 '12 at 9:14
Sure. This sufficient? – Will Sawin Jun 19 '12 at 14:38
Not sure why this isn't getting more upvotes. I think it works, and it is very different from my answer. – David Speyer Jun 19 '12 at 22:42
Thank you, Will. – Martin Brandenburg Jun 20 '12 at 8:25

[I might have misunderstood your equivalent question]. Some hints: a commutative reduced ring is the same as a subring of a direct product of fields (i.e. a ring of field-valued functions, but the co-domain can change with the point); you also wants that never zero functions have an inverse. So typically one considers continuous, of differentiable (or whatever) real valued functions. Then you want to invert also a function f that is sometimes zero; this "kills" the zero set Z of f (the localization is a ring of functions defined outside Z). But then you want that every g which is nonzero outside Z has a inverse, i.e. inverting one particular f should invert every g. Taking the sequences which have limit (i.e. the continuous functions on the Alexandroff compactification of the naturals), suppose that you want to invert f(n)=1/n ; does this invert all positive and converging to zero sequences? Inverting f gives functions that are of polynomial growth when n tends to infinity, so what happens for a g with exponential decrease?

Added: as it was kindly noted, it happens that the ring of continuous functions (on a generic completely regular space) is not a classical ring of quotients; its ring of quotients is obtained considering functions which are definite and continuous on a dense co-zero set (dense open set in the normal case), two such functions being identified when they coincide on a dense co-zero set. But then this ring of quotients is von Neumann regular (and it is naturally identified with the ring of continuous functions on a certain almost P-space associated to the original space); being von Neumann regular there is no hope to obtain the kind of example wanted (as it was implicitly remarked by the requester). However, the idea that "regular" functions can be modified on a ideal of "small" sets (in this case, the sets whose closure has empty interior) still works, and in another answer it was well used (with polynomial functions as regular functions and finite sets as ideal of small sets. One can use other choices of regular functions and/or small sets: meager sets, measure zero sets, ... however, for algebraic geometry, polynomials and finite sets are the most natural choices)

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$1/n$ is not a zero-divisor in the ring you described, despite being $0$ at $\infty$. The only thing that, when multiplied by it, gives the zero sequence, is the zero sequence. – Will Sawin Jun 19 '12 at 15:24

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