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Let $C$ be an abelian category with enough projectives and $\underline{C}$ the stable category of $C$ that is obtained by factoring out projective modules. When is the functor $\Omega^1 : \underline{C} \rightarrow \underline{C}$ an equivalence?

We can assume $C$ is a module category of a ring $R$ in case that helps. For $R$ a finite dimensional algebra, this should be true iff $R$ is self-injective.

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  • $\begingroup$ Not sure whether the tag "triangulated-categories" fits too well since the stable category is triangulated iff the algebra is quasi-Frobenius. But maybe it is quasi-Frobenius iff $\Omega^1$ is an equivalence iff the stable category is triangulated? $\endgroup$
    – Mare
    Commented Oct 20, 2018 at 12:41
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    $\begingroup$ Generally speaking, I don't think it's wise to edit a question to ask a new question, after an answer has been given. Usually people start a new question for that sort of thing. But, I'll try to answer. It depends on what you mean by "stable module category" - if you mean "factor out projectives" then this is not triangulated for rings that are not quasi-Frobenius, and indeed $\Omega^1$ is not an equivalence (if it were, that would cause the stable module category to be triangulated, by sec 7.1 of Hovey's book, see also 9.3 in Hovey's cotorsion paper referenced below). $\endgroup$ Commented Oct 20, 2018 at 13:31
  • $\begingroup$ But, if you mean "factor out all modules of finite projective dimension" then the answer is yes. You can work with Gorenstein rings, or indeed any ring, and using either Hovey or the Bravo, Gillespie, Hovey machinery, $\Omega^1$ will be an equivalence. $\endgroup$ Commented Oct 20, 2018 at 13:32
  • $\begingroup$ @David: By your 2nd comment: 'not qF implies not triangulated' (i.e. 'triangulated implies qF') and '$\Omega^1$ is an equivalence implies triangulated'. In combination: '$\Omega^1$ is an equivalence implies qF'. Together with your answer below, you have shown that $\Omega^1$ is an equivalence iff R is qF. $\endgroup$
    – tj_
    Commented Oct 22, 2018 at 18:55
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    $\begingroup$ @Mare: No worries. I'd say now that everything has been answered, it's best to just leave it as is. Cheers. David. $\endgroup$ Commented Oct 22, 2018 at 20:23

2 Answers 2

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There's an elementary proof that if $\Omega$ is a self-equivalence of $\underline{C}$ then $C$ also has enough injectives, and projectives and injectives coincide. In particular, this shows that if $C$ is $\text{Mod-}R$ for a ring $R$, then $R$ is quasi-Frobenius.

For any $C$ with enough projectives, it's straightforward to prove that if $X$ and $Y$ are isomorphic in $\underline{C}$, then $X$ is a summand of $Y\oplus P$ and $Y$ is a summand of $X\oplus Q$ in $C$ for some projectives $P$ and $Q$. In particular $X$ is a subobject of a projective iff $Y$ is, and the only objects isomorphic in $\underline{C}$ to zero are the projectives.

So if $\Omega$ is a self-equivalence of $\underline{C}$ then every object of $C$ must be a subobject of a projective.

If $P$ is projective and $\alpha:P\to X$ a monomorphism, then composing with a monomorphism from $X$ to a projective $Q$ and taking the cokernel, we get a short exact sequence $$0\to P\to Q\to Y\to0.$$ Since $\Omega Y\cong P\cong0$ in $\underline{C}$ and we’re assuming $\Omega$ is a self-equivalence, $Y$ is projective, and the short exact sequence splits. Hence $\alpha$ splits, and so $P$ is injective.

Therefore all projectives are injective, and since every object is a subobject of a projective, there are enough injectives.

Every injective is a subobject, and therefore a summand, of a projective, and so every injective is projective.

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The classical thing to do is take $R$ to be quasi-Frobenius. Then, the stable category is triangulated, hence $\Omega^1$ is an equivalence. This is in Happel's book Triangulated categories in the representation theory of finite dimensional algebras, though I learned it from Mark Hovey's book Model Categories.

Hovey proved a generalization in Cotorsion pairs, model category structures, and representation theory, to the settings of Gorenstein rings $R$, but you have to kill all modules of finite projective dimension, not just the projective modules. Later, Daniel Bravo generalized it to all rings, in his thesis, and the strongest result appears in The stable module category of a general ring, by Bravo, Gillespie, and Hovey. This involves the class of modules admitting a projective resolution by finitely generated projectives.

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  • $\begingroup$ Thanks, can you state the precise theorem in Bravo, Gillespie, and Hovey? It sounds like it is not directly related when they kill more than just the projective modules. So is there are a non-quasi-frobenius ring where $\Omega^1$ is an equivalence of $\underline{C}$? $\endgroup$
    – Mare
    Commented Oct 20, 2018 at 12:27
  • $\begingroup$ I believe the paper is in flux a bit, as it works its way through the publication process. But, in the first version on arxiv, Corollary 6.5 gives the "exact projective model structure", where the trivially cofibrant objects are projective modules (so, they are killed in the homotopy category). Throughout that section, several model structures are given, that all coincide for Gorenstein rings (Prop 6.10), but some involve killing the injectives too. $\endgroup$ Commented Oct 20, 2018 at 12:55

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