Let $A$ be a finite dimensional algebra over a field $k,$ $J\subset A$ some ideal of $A$ and $B=A/J$ be a quotient algebra. Assume that $\text{Ext}^n_A(B_A,B_A)=0$ for $n>0$ and $B$ has finite projective dimension as a right $A-$module. It is said in some paper that it is easy to see that the category $B-$mod generates the strict, triangulated subcategory $\mathcal{D}$ of $D^b(A-\text{mod})$ consisting of objects which have cohomology in the subcategory $B-$mod of $A-$mod and also that $D^b(B-\text{mod})\rightarrow D^b(A-\text{mod}) $ factors through $\mathcal{D}.$

I am new in the ground of triangulated categories -- I understand well definitions, notions but I don't have certain intuitions.

Of course, category $B-$mod is abelian, but not triangulated. It seems for me informally that triangulated category generated by $B-$mod consists of objects which have cohomology in the subcategory $B-$mod but I would like to ask you a method, a way to formalize this intuition -- why it is true in a formal way?

I don't see also this factorization through $\mathcal{D}$ -- how we map an arrows of $D^b(B-\text{mod})$ to an arrows of $\mathcal{D}?$


1 Answer 1


Let me try to give a answer I hope will help, but this is not a complete answer though. I assume all the categories are bounded. I will assume that $B-mod$ is thick (maybe this is implied by the condition you give ?).

Denote $T\subset D(A-mod)$ the strict triangulated subcategory of $D(A-mod)$ generated by $B-mod$ and $D\subset D(A-mod)$ the subcategory of object with cohomology in $B-mod$.

First, $T\subset D$: if $x \to y \to z$ is a distinguished triangle, using the long exact sequence in cohomology you get that if two of these objects have cohomology in $B-mod$, so has the third (by thickness of $D(B-mod)$).

Conversely, $D \subset T$: you can proceed by induction on the length of the complex. For example, for a length two complex $X^\bullet := 0 \to M \xrightarrow{f} N \to 0$ in $D$, you have the triangle $Y^\bullet \to X^\bullet \to Z^\bullet$ with $Y^\bullet = 0 \to ker(f) \to 0 \to 0$ and $Z^\bullet = 0 \to 0 \to coker(f) \to 0$. This triangle is distinguished because it is isomorphic to the triangle $$Y^\bullet \to X^\bullet \to Cone$$

No the factorization $D(B-mod) \to D \to D(A-mod)$ is just looking at complex of $B$-modules as complex of $A$-modules with cohomology of $B$-modules.


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