Definition and context:

An (associative, unital, not necessarily commutative) ring $R$ is called classical if every regular element of $R$ is a unit. Equivalently, $R$ is its own classical ring of fractions, i.e. Ore localisation with respect to its set of regular elements. These rings feature e.g. in this MO question. (Btw: The connection with depth in Qing Liu's answer might be of its own interest.) They are implicit in any treatment of orders (in the sense of Asano, Goldie, ...) insofar as an order is implicit in a classical ring. The recent name "classical" and some basic facts appear in §11 of Lam's Lectures on Modules and Rings. Another name is "full quotient ring". Standard examples are:

  • von Neumann regular rings
  • right or left self-injective rings ("principally injective" suffices)
  • local rings whose maximal ideal is nil
  • commutative rings with Krull dimension 0

Other instructive examples (commutative, reduced, of positive dimension) are the rings constructed in David Speyer's and Will Sawin's answers to this mildly related MO question.

Classical rings are closed under directed unions. A product of rings is classical iff each factor is.

There is a classification of Noetherian classical rings in: Stafford, J. T. "Noetherian full quotient rings." Proceedings of the London Mathematical Society 3.3 (1982): 385-404.


In the answer to this MO question, Manny Reyes shows that if a commutative $R$ is classical, then so is $M_n(R)$ for all $n$. I wondered if "commutative" is really needed. Also, of the examples above, at least the first two are defined by Morita invariant properties. If I'm not mistaken, it follows from the cited paper by Stafford that "being (two-sided) Noetherian and classical" is Morita invariant. So why not go all in and ask the


1.) Is "being classical" a Morita invariant property?

This was also triggered by another MO question by Bill Cook; namely, if the answer here were yes, an answer there would be "any progenerator". There are natural

Side Questions (and follow-ups in case of a counterexample):

2.) Are there properties P (e.g. commutative) such that if $R$ and $S$ are Morita equivalent and one or both also have property P, then $R$ is classical iff $S$ is? (As said, I think Stafford shows this for P = Noetherian.)

3.) Are there further properties, which are implied by or do imply "classical", and which are Morita invariant? (Like "self-injective" and "von Neumann regular" above.)


  • Stafford's scope in the Noetherian case is broader and his results are stronger, insofar as he investigates the invariance of "having a full quotient ring", not just our special case "being a full quotient ring". It would not be necessary to generalise all of his results (which seems challenging) to give a positive answer to the first question.

  • A positive answer to 1 [or 2] would be equivalent to describing classicality [+ P] of a ring in purely categorical terms of its (say, left) module category. E.g. the co-Hopfian property alluded to in Greg Marks', Chris Leary's and rschwieb's answers to the mentioned question seems like a start.

  • Besides the classical (Ore) ring of fractions, there is, among others, the concept of "maximal ring of quotients", sometimes called "Utumi quotient ring". Here Morita invariance and further analogues of Stafford's results are known in great generality: see this paper, especially prop. 2.5, and ch. X §3 of Stenström's Rings of Quotients. In the latter, invariance is also discussed with respect to Gabriel filters, and people cleverer than me might find further evidence or counter-evidence.

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    $\begingroup$ This is far from being an answer, but your condition is equivalent to every module homomorphism $R\to R$ that is epi and mono in the category of finitely generated projectives being an isomorphism. A related (but strictly stronger), and obviously Morita invariant property is that every mono and epi map in that category is an isomorphism. I wondered if anybody has ever studied that stronger property. Even for finite-dimensional algebras over a field it's sometimes true and sometimes not, in a non-trivial way. $\endgroup$ Mar 19, 2013 at 19:13


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