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Does Morita theory hint higher modules for noncommutative ring?

MathOverflow Asked on November 3, 2021

Two possibly noncommutative rings are called Morita equivalent if their left-module categories are equivalent. In the commutative case, Morita equivalence is nothing more than ring isomorphism. Otherwise, there are many known examples where this does not hold.

That means modules alone are not enough to characterize the ring. Are there notions of higher modules or higher structures, such that the corresponding higher Morita equivalence is nothing more than ring isomorphism?

In short, can you find a better notion of modules that faithfully capture their underlying ring?

3 Answers

Although I think my answer "pointed categories" is an important one, there is another way that the question could be interpreted: What is an interesting class of rings which are recoverable up to isomorphism from their categories of modules? And perhaps, if we are not trying to be too universal or functorial, we won't mind if the isomorphism is not unique.

For this, I strongly recommend Morita's original paper introducing his equivalences. (Kiiti Morita, Duality for modules and its applications to the theory of rings with minimum condition, Sci. Rep. Tokyo Kyoiku Daigaku, Sect. A 6, 83-142 (1958). You can find a PDF by googling.) Among the many things in that paper is a theorem saying that rings satisfying a natural minimality condition called "basic" are Morita equivalent iff they are isomorphic (commutative rings, I believe, satisfy this condition) and that every finite-dimensional ring is Morita equivalent to one satisfying this minimality condition. See my answer to a related question.

Answered by Theo Johnson-Freyd on November 3, 2021

As explained for example here

https://doi.org/10.1112/plms/s3-72.2.281

the most direct analogue of Morita equivalence for (connected) graded algebras actually boils down to graded isomorphism.

However there are good reasons why one might actually want equivalence relations that are weaker than isomorphism: Morita, twisting (in the graded case, as in Zhang's work above), derived, and so on. Making some broad classification feasible, for example; this is what noncommutative geometry seeks to do via the point scheme of Artin-Tate-van den Bergh, for example.

Answered by Jan Grabowski on November 3, 2021

Yes. The trick is to use not just categories, but pointed categories, which are categories equipped with a choice of object (the "pointing"). Given any ring $R$, the category $mathrm{Mod}(R)$ is naturally pointed by the rank-1 free module, i.e. $R$-as-an-$R$-module, which I will write as $R_R$. Then it is almost trivial that the pointed category $(mathrm{Mod}(R),R_R)$, up to equivalence, recovers $R$ up to isomorphism.

What's that? I didn't tell you what the morphisms are between pointed categories, so you don't what the equivalences are? Well, you do actually know what the equivalences are: an equivalence of pointed categories $(mathcal{C},C) simeq (mathcal{D},D)$ is an equivalence of categories $F : mathcal{C} oversetsimto mathcal{D}$ together with an isomorphism $f : FC cong D$. I mean, what else could it be? Anything else wouldn't justify the name. But there is actually an interesting question of what are the morphisms which are not equivalences. Surely, a morphism $(mathcal{C},C) to (mathcal{D},D)$ should consist of a functor $F : mathcal{C} to mathcal{D}$ together with a morphism $f$ between $FC$ and $D$. The interesting question is whether $f$ should be: (1) an isomorphism; (2) a morphism $f : FC to D$; (3) a morphism $f : D to FC$. These three options have names: (1) is called a strong pointed functor; (2) is called an oplax pointed functor; and (3) is called a lax pointed functor. It is almost trivial to show that all three options give the same notion of equivalence of pointed categories, but they give different bicategories of pointed categories (and this difference matters in applications).

An advantage of working with pointed categories is that there are plenty of pointed categories which share with $(mathrm{Mod}(R), R_R)$ some of its nice structural properties, but not all, and so are not of that form.

Answered by Theo Johnson-Freyd on November 3, 2021

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