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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?

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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.

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    $\begingroup$ It is slightly simpler to just take a minimal resolution of M and tensor it with A/radA: the differentials of the resulting complex are all zero, and the modules nonzero because of Nakayama. $\endgroup$ Commented May 30, 2022 at 23:16
  • $\begingroup$ @MarianoSuárez-Álvarez can you really use Nakayama? the modules here are not finitely generated. $\endgroup$
    – the L
    Commented Nov 23, 2022 at 20:00
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    $\begingroup$ @theL Although it may not technically be Nakayama's lemma, it's still true for finite dimensional algebras that if $M\otimes_AA/\operatorname{rad}A=0$ then $M=0$, even if $M$ is not finitely generated. $\endgroup$ Commented Dec 10, 2022 at 10:35

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