# More questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$

This question is a follow-up to an older one about a proof of Rickard that the stable module category can be described as a triangulated Verdier quotient. In fact it's a complemented in that this question asks for clarification of every point that the previous question didn't touch upon.

Setup:

Let $$A$$ be a self-injective algebra (so projective = injective for modules) and let $$D^b(A)$$ and $$K^b(A)$$ be the bounded derived category and the full subcategory consisting of the perfect complexes (complexes whose modules are all projective). The singularity category is defined to be the quotient $$D^b_\mathtt{sg}(A) = D^b(A)/K^b(A)$$.

Next let $$\mathtt{stmod}(A)$$ be the stable module category. This is the category whose objects are homomorphisms and whose maps are module homomorphisms modulo those that factor through a projective.

In his paper "Derived categories and stable equivalence" Jeremy Rickard proves that the stable module category and the singularity category are equivalent. The functor $$F\colon\mathtt{stmod}(A) \to D^b_\mathtt{sg}(A)$$ sends a module $$M$$ to the complex which is $$M$$ in degree $$0$$ and $$0$$'s in all other degrees.

Rickard proves in two sentences that this functor is an exact functor.

A distinguished triangle in $$\mathtt{stmod}(A)$$ is a triangle $$X \to Y \to Z \to X[1]$$ coming from a pushout diagram of modules $$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ra{} & X & \ra{} & I & \ra{} & \Omega^{-1} X & \ra{} & 0 \\ & & \da{} & & \da{} & & \da{} & & & \\ 0 & \ras{} & Y & \ras{} & Z & \ras{} & \Omega^{-1} X & \ras{} & 0 \\ \end{array}$$ where $$X \to I$$ is the embedding of $$X$$ into its injective hull. Since short exact sequences of modules give distinguished triangles in the derived category, and since $$FI = 0$$, $$FX \to FY \to FZ \to FX[1]$$ is a distinguished triangle.

Could someone clarify what is going on here? I can understand that $$X \to I \to \Omega^{-1} X \to X[1]$$ and $$Y \to Z \to \Omega^{-1} X \to Y[1]$$ are distinguished triangles in $$D^b(A)$$ and that $$FI = 0$$, but I entirely fail to see how this gets to the conclusion.

Update: This question is now resolved. One uses the defining property of $$D^b(A)$$ to infer that we have distinguished triangles $$X \to I \to \Omega^{-1} X \to X[1]$$ and $$Y \to Z \to \Omega^{-1} X \to Y[1]$$. Passing to the singularity category, $$I$$ gets killed, along us to deduce an isomorphism between $$\Omega^{-1} X$$ and $$X[1]$$, and hence an isomorphism of triangles between $$Y \to Z \to \Omega^{-1} X \to Y[1]$$ and $$Y \to Z \to X[1] \to Y[1]$$. The latter is therefore distinguished, and so by shifting, $$X \to Y \to Z \to X[1]$$ is distinguished too.

Next, it's proved that the functor is full, where he claims the following.

... no non-projective $$A$$-module is isomorphic in the derived category to an object of $$K^b(A)$$.

But this seems unlikely to me. If $$M$$ is a non-projective $$A$$-module then $$M$$ admits a projective resolution $$P^* \to M$$ and if $$P^*$$ can be made finite then it's an object of $$K^b(A)$$ quasi-isomorphic to $$M$$.

Later on, Rickard proves that the functor is essentially surjective. Every object $$X$$ in $$D^b(A) / K^b(A)$$ is isomorphic to a complex of projectives $$P^* = \cdots \to P^r \to P^{r+1} \to \cdots P^s \to 0$$, where $$r < 0$$ and $$P^*$$ has zero homology in degrees less than $$r$$. The previous question already touched upon the fact that $$P^*$$ is, in fact, not in $$D^b(A)$$, so I'll ignore this. The natural map from $$P^*$$ to $$\widetilde{P}^* = \cdots \to P^{r-1} \to P^r \to 0 \to \cdots$$ is an isomorphism in $$D^b(A) / K^b(A)$$ since its mapping cone is a bounded complex of projectives. At this point Rickard writes

... and there is a complex $$Q^* = \cdots \to P^{r-1} P^r \to Q^{r+1} \to \cdots \to Q^0 \to 0$$ which is the projective resolution of some module $$M$$ and whose natural map to $$\widetilde{P}^*$$ is an isomorphism in $$D^b(A) / K^b(A)$$. Thus $$P^* \cong FM$$.

This strikes me as simply rephrasing what has to be proved, in terms of projective resolutions rather than in terms of modules. Could someone clarify what is being done here?

• For your first question, since $I$ is injective, the distinguished triangle $X\to I\to \Omega^{-1}\to X[1]$ in the derived category degenerates in the singularity category to give an isomorphism $\Omega^{-1}\cong X[1]$. We then use this isomorphism in the distinguished triangle $Y\to Z\to \Omega^{-1}/to Y[1]$ and rotate to get the distinguished triangle $X\to Y\to Z/to X[1]$ in the singularity category. Commented Mar 19, 2021 at 12:25
• Thanks @AndrewHubery, that makes sense. I've updated the question to accomodate your resolution. Commented Mar 23, 2021 at 12:52
• For the second question: In the situation at hand, a module is projective or it has infinite projective dimension: The self-injectivity of $A$ allows you to shorten any finite, non-zero projective resolution. Commented Mar 23, 2021 at 12:55
• A nice question
– user822157
Commented Mar 23, 2021 at 13:04
• @Hanno Ah yes, of course. Because any projective $A$-module is also injective, and so if $P' \hookrightarrow P$ is an inclusion then $P'$ is a summand of $P$. Still I believe that some care must be taken: In the last part, Rickard implicitly uses an identification of the derived category of bounded complexes with the category of eventually acyclic complexes, and an infinite resolution still exists in the latter. Commented Mar 23, 2021 at 13:27

First, let's pretend we're working in $$\textbf{D}(A)/\textbf{K}^b(A)$$ instead of $$\textbf{D}^b(A)/\textbf{K}^b(A)$$. Then, the argument shows that within $$\textbf{D}(A)/\textbf{K}^b(A)$$, any object $$X$$ of $$\textbf{D}^b(A)/\textbf{K}^b(A)$$ is isomorphic to a complex $$\ldots\to P^{r-1}\to P^r\to 0$$ with no cohomology below $$r$$. In other words, it's isomorphic to a stalk complex $$M[-r]$$ concentrated in degree $$r$$, for some module $$M$$. This stalk complex is isomorphic to the degre-$$0$$ talk complex $$(\Omega^{-r} M) [0]$$ in $$\textbf{D}^b(A)/\textbf{K}^b(A)$$ which you see by taking a projective resolution $$0\to M\to Q^{r+1}\to \ldots\to Q^0\to \Omega^{-r} M\to 0$$ to the right. This is, I think, the argument Rickard presents.
You can eliminate the transit through $$\textbf{D}(A)/\textbf{K}^b(A)$$ as follows: Given $$X\in \textbf{D}^b(A)$$, consider a bicomplex $$T$$ consisting of projective resolutions of the components of $$X$$. The totalization of this (unbounded!) bicomplex $$T$$ is an unbounded complex of projectives quasi-isomorphic to $$X$$, which you then go on to truncate as described. You can avoid expand+truncate step by already truncating the vertical projective resolutions in $$T$$ in such a way that the totalization of $$T$$ is projective except for the lowest non-zero entry (this will require truncating the projective resolutions at different places). This will give you a bounded complex $$0\to M\to Q^r\to\ldots\to Q^s\to 0$$ quasi-isomorphic to $$X$$, and hence $$X\cong M[-r]$$ in $$\textbf{D}^b(A)/\textbf{K}^b(A)$$ as before.
• Thanks, this makes sense! The fact that $M[-r]$ and $(\Omega^{-r} M)[0]$ are equivalent is I think already immediate from the exactness of $F$, which is proved earlier on. Commented May 31, 2021 at 8:37