# When $\textrm{Tor}_n^A(-,A/radA)\neq 0$? ($A$ a finite dimensional $K$-algebra)

This question arrise from a proof in paper: Unbounded derived categories and finitistic dimension conjecture - Jeremy Rickard, more spefically Theorem 4.3.

Let $$A$$ be a finite dimensional algebra over a field $$K$$ and $$M$$ a right $$A$$-module with projective dimension $$d$$. So let $$P^{\bullet}$$ be the minimal projective resolution of $$M$$ considered as a complex. Then \begin{align} \textrm{Tor}_d^A(M,A/radA)\neq 0\end{align} that is, $$P^{\bullet}[-d]\otimes_A (A/(radA))$$ has nonzero cohomology in degree zero.($$P^{\bullet}[-d]$$ is the complex shifted $$d$$ times to right)

The question is: Why Tor is nonzero? That is, how justify this statement.

I'm grateful for any help.

For finite dimensional algebras, flat modules are projective, so the projective dimension of $$M$$ is the same as its weak dimension, which is $$\sup\{d\mid \text{Tor}^A_d(M,-)\neq0\}.$$ Since the class of left modules $$X$$ such that $$\text{Tor}^A_d(M,X)=0$$ is closed under coproducts and extensions, and every module is an iterated extension of coproducts of simple modules, this is $$\sup\{d\mid\text{Tor}^A_d(M,S)\neq0\text{ for some simple module }S\},$$ which yields the claim since every simple module is a direct summand of $$A/\text{rad}A$$.
Alternatively, writing $$D$$ for the vector space dual $$D-=\text{Hom}_k(-,k)$$, for any left module $$X$$, $$\text{Hom}_A(M,DX)\cong D(M\otimes_AX).$$ And so, taking derived functors, $$\text{Ext}^d_A(M,DX)\cong D\text{Tor}^A_d(M,X).$$
So $$\text{pd}(M)=\sup\{d\mid\text{Ext}^d_A(M,S)\neq0\text{ for some simple right module }S\}$$ is the same as $$\sup\{d\mid\text{Tor}^A_d(M,S)\neq0\text{ for some simple left module }S\}.$$
It probably cant get better than the author of the article answering himself but here a small addition that also answers the question and gives a direct interpretation of this Tor, namely it counts something. Since we do homological algebra, we can assume that the algebra is basic. As in Jeremy Rickards answer one has $$Tor_d^A(M,A/radA)=DExt_A^d(M,D(A/radA))$$, but $$D(A/radA)=A/radA$$ (as left/right modules) and then the length of $$Ext_A^d(M,D(A/radA))$$ counts how many indecomposable direct summands $$P_d$$ has when $$(P_i)$$ is the minimal projective resolution of $$M$$. When $$M$$ has projective dimension at least $$d$$, one has $$P_d \neq 0$$ which gives the statement.
• Thanks for the explanation @Mare. I have a doubt about how would be an explicit proof for this: "$Extd_A(M,D(A/radA))$ counts how many indecomposable direct summands $P_d$ has when $(P_i)$ is the minimal projective resolution of M." Commented Jun 5, 2019 at 19:23