Apologizes if this is a basic question, but I am new to the area of finite dimensional algebras. I am reading the paper "Unbounded derived categories and the finitistic dimension conjecture" by Jeremy Rickard (<https://arxiv.org/abs/1804.09801>) and have a question about the proof of Theorem 4.3.


In the proof, $P\_i$ is a projective resolution of the module $M\_i$, and has length $d\_i$.
The author then concludes that $Tor\_{d\_i}^A(M\_i,A/rad(A))$ is non-zero.
My question is: why is this the case? does this follow from some property of finite dimensional algebras? or this a more general fact?



About a recent paper of Rickard on finitistic dimension
I use $M$ instead of $M\_i$ and $d$ instead of $d\_i$.


For a finite dimensional algebra $A$ over a field $K$ we have in general
$D Ext\_A^i(Y,DZ)=Tor\_i^{A}(Y,Z)$ using the duality $D=Hom\_K(-,K)$.
Thus $Tor\_d^{A}(M,A/radA)=D Ext\_A^d(A/rad A, D(M))$ is non-zero since $D(M)$ has injective dimension at least $d$ ($D$ is a duality so this is equivalent to $M$ having projective dimension at least $d$).


Here I used that in general for a simple module $S$ and a module $N$ with minimal injective coresolution $0 \rightarrow N \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots $ we have $Ext\_A^n(S,N)$ being non-zero if and only if the injective envelope $I(S)$ of $S$ is a direct summand of $I^n$. Note that $A/rad A$ is simply the direct sum of all simple $A$-modules.


