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I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 . On page 109, the projectively stable category is defined by $$ \underline{mod} A = mod A/\mathcal{P}. $$ The objects of $\underline{mod} A$ is the same as the objects in $mod A$ but the $K$-vector space $\underline{Hom}_A(M, N)$ of morphisms from $M$ to $N$ in $\underline{mod} A$ is defined to be the quotient vector space $$ \underline{Hom}_A(M, N) = Hom_A(M, N)/\mathcal{P}(M, N). $$ Here $\mathcal{P}(M, N)$ is the subset of $Hom_A(M, N)$ consisting of all homomorphisms that factor through a projective $A$-module.

By the definition, we have projective modules in $\underline{mod}A$. But it is said that $\underline{mod}A$ does not have projective modules in the book Representation Theory of Artin Algebras. I am confused.

It is said in Proposition 2.2 on page 110 of the book Elements of the Representation Theory of Associative Algebras: Volume 1 that $M \mapsto Tr M$ induces a $K$-linear duality functor $Tr: \underline{mod}A \to \underline{mod}A^{op}$. If $\underline{mod}A$ has a projective module $P$, then $Tr P = 0$ and $Tr (Tr P) = Tr(0) = 0 \neq P$. This contradicts the fact that $Tr$ is a duality functor. So I think that $\underline{mod}A$ does not have a projective module. Is this true? Thank you very much.

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  • $\begingroup$ I cannot answer your question completely at the moment, but one key point is that in the stable module category all projective modules become isomorphic to the zero module since the identity map factors through a projective module and is therefore modded out. $\endgroup$ – Julian Kuelshammer May 5 '13 at 8:57
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    $\begingroup$ For people who works in stable module category, we tend to say "there is no projective module" in the sense that (1) projective modules are isomorphic to zero, and (2) nothing behaves like a projective object homologically (e.g. higher Ext-groups vanishing, well-defined notion of projective cover/resolution, Schur lemma, splitting an surjection etc.); where we do not mean "the original projective module is not in the class of objects of the stable module category". $\endgroup$ – Aaron May 5 '13 at 19:46
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As I already mentioned in the comment, the projective objects get isomorphic to $0$ since the identity map $P\to P$ obviously factors through a projective module and therefore the identity map is the zero map, which is only true for the zero module.

As you can see from that since every projective module gets "killed" when going to the stable module category you can leave them out before, i.e. you can construct the stable module category as a factor category of the full subcategory of all the non-projective modules. In that sense the stable module category does not contain projective modules.

If I am not mistaken this doesn't mean that the stable module category doesn't have projective objects, e.g. take $A=k[x]/x^2$. Then, up to isomorphism, in the stable module category of $A$ there is only one indecomposable object up to isomorphism (since the indecomposable projective gets isomorphic to $0$) and hence this should be a projective object (in the categorical sense).

For your second question: We have $\operatorname{Tr}\operatorname{Tr}(P)=0$ is correct, but $P\cong 0$, so this is no contradiction to being a duality.

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    $\begingroup$ Isn't every module projective in the stable category? At least in the Frobenius case, the stable category is triangulated, so a morphism is an epimorphism iff it is a split epimorphism. Therefore there is no obstruction for lifting morphisms over any epimorphism. $\endgroup$ – Martin May 5 '13 at 9:57
  • $\begingroup$ @Martin Ah, thanks for this complement. I'm not an expert on triangulated categories. So for self-injective algebras all modules become projective in the stable module category. I guess for non-self-injective algebras this is no longer true in general. For example the simple module corresponding to the source in a linearly oriented $A_3$-quiver seems to stay non-projective. $\endgroup$ – Julian Kuelshammer May 5 '13 at 11:42
  • $\begingroup$ Yes, I think that you are right. Thanks for this example. $\endgroup$ – Martin May 5 '13 at 11:54
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    $\begingroup$ The stable module category is triangulated if and only if every projective object is also injective; so it's only true for (finite dimensional) self-injective algebras. For reference, see Happel's book chapter one and/or Rickard's series of papers. The generalisation of this phenomenon on Gorenstein ring is the stable category of maximal CM-module (or if you prefere a a more trendy term - the singularity category). $\endgroup$ – Aaron May 5 '13 at 18:48
  • $\begingroup$ I should also say that, "all modules become projective" is a wrong concept, there is no formalism of ``projectivity/projective objects" in the stable module category, in the sense that object behave like a projective object homologically; moreover unlike its cousin (the derived categories), the tilting objects (in the triangulated sense) is totally useless. These makes understanding stable module so horribly difficult. $\endgroup$ – Aaron May 5 '13 at 19:33

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