3
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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 $$\sup\{d\mid\text{Tor}^A_d(M,S)\neq0\text{ for some simple left module }S\}.$$

$\endgroup$
1
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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." $\endgroup$ Commented Jun 5, 2019 at 19:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .