Total ring of fractions of non-noetherian reduced ring

Question

As is known, see The total ring of fractions of a reduced Noetherian ring is a direct product of fields if the ring $A$ is noetherian and reduced, we can show that the total ring of fractions must be a direct product of fields, however in the case of non-noetherian, what can be said? Must it be krull dimension 0? Must it be a product (of arbitrarily many) fields? I think it must be no for the second question but I don't know many constructions for non-noetherian rings.

Answer 1

To the first question, no.

For a reduced ring $R$, being $0$-dimensional is equivalent to being Von Neumann Regular. It is well known that the total quotient ring of a reduced ring is VNR iff the minimal spectrum of $R$ is compact and $R$ has the property that f.g. ideals consist of zero divisors iff they have nonzero annihilators.

See this nice paper of Matlis, in particular 1.15. This makes it easy to construct examples of reduced rings which don't have $0$-dimensional total ring of fractions.

This also settles the second question, since VNRs can be equivalently characterized as subdirect products of fields.

It would be a little misleading to end the answer there, though. There is a canonical embedding of any reduced ring in a $0$-dimensional 'fraction-like ring.' This is typically approached in one of two equivalent ways: (1) by characterizing the total ring of fractions as a direct limit of the modules $\text{Hom}(d, R)$ where $d$ is a regular element, and then generalizing that to a direct limit of the modules $\text{Hom}(D, R)$ where $D$ is a dense ideal of $R$, or (2) as the injective envelope of $R$. These constructions work also for noncommutative rings, and when $R$ is non-singular (i.e. its elements have non-essential annihilators, equivalent to being reduced for commutative rings) you can actually put a ring structure on these module constructions, and this yields a self-injective Von Neumann Regular ring.

(For commutative rings, the first approach is detailed in the first chapter of these great notes, or for the noncommutative case I'd recommend looking in T.Y. Lam's Lecture's on Modules and Rings. For the injective envelope approach, again Matlis' paper is helpful.)

Answer 2

A commutative Noetherian reduced ring is a Goldie ring, and so it has a semisimple classical ring of quotients. Therefore the ring of quotients is a finite product of fields.

This extends the non-Noetherian case slightly because there are non-Noetherian Goldie rings.

In general, the maximal ring of quotients $Q_{max}(R)$ for a reduced ring is von Neumann regular and self-injective. Since $R$ is commutative, its classical ring of quotients embeds in this ring. There are a couple conditions that can ensure the two rings of quotients are equal: one is when every dense ideal contains a regular element, and a stronger one, I think, is the ACC on annihilator ideals.

Otherwise, the classical ring of quotients may not be so well behaved.