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Proposition 7.15 in Rouquier's paper (see Publication paper) "Dimension of triangulated categories J. K-theory, 1 (2008) 193-256" as follows, for details please see Rouquier's paper (or arXiv):

``Let $A$ be a noetherian finite dimensional $k$-algebra of global dimension $d\in \mathbf{N}\cup \{\infty\}$. Assume $k$ is perfect. Then, $d$ is the minimal integer $i$ such that $A\mathrm{-perf}=\langle A\rangle_{i+1}$.''

I am trying to give a proof as follows:

  1. If the global dimension is $\infty$, then Proposition 7.15 is hold.

  2. Assume that the global dimension is finite. Denote by $min$ the minimal integer $i$ such that $A\mathrm{-perf}=\langle A\rangle_{i+1}$.

    (a) Proposition 7.4 and Proposition 7.25 implies that $A\mathrm{-perf}=\langle A\rangle_{d+1}$, and this implies that $min\leq d$.

    (b) Suppose that there is an $A$-module $W$ such that $\mathrm{pdim}\,W\geq min+1$. Then Lemma 7.13 implies that $W\notin\langle\mathrm{ads}(A)\rangle_{min+1}$. Then there should be a contradiction ! Hence, for any $A$-modules $W$, $\mathrm{pdim}\,W\leq min$, therefore $d\leq min$.

Note that Lemma 7.13, Proposition 7.4 and Proposition 7.25 are in Rouquier's publication paper, for details please see Rouquier's paper.

Q. What I don't understand is that why there is a contradiction?

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    $\begingroup$ how can it be so that some statements in the text indicate that it's not clear how to prove some other statement in the same text?! $\endgroup$ Aug 18, 2014 at 17:44
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    $\begingroup$ The paper is in arXiv. $\endgroup$ Aug 18, 2014 at 17:45
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    $\begingroup$ It might help if you gave a statement of the Proposition and explained what you do and don't understand about the proof. $\endgroup$ Aug 18, 2014 at 17:49
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    $\begingroup$ If an edit such as Jeremy Rickard suggests is made I will vote to reopen. I think questions like this should be allowed in general, though obviously one has to give some background and point to where the difficulty lies $\endgroup$ Aug 18, 2014 at 23:07
  • $\begingroup$ I am sorry for the first time edit! $\endgroup$ Aug 19, 2014 at 3:38

1 Answer 1

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Since $W$ has finite projective dimension, it is isomorphic in the derived category to a bounded complex of projective $A$-modules. Thus it belongs to $A$-perf = $\langle A \rangle_{min + 1}$. That contradicts Lemma 7.13 as you stated in (b).

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  • $\begingroup$ We know that objects in $A-perf$ are bounded complexes of finite generated projective A-modules, why $W \in A-perf$? $\endgroup$ Aug 20, 2014 at 2:54
  • $\begingroup$ @ParksJonehan There is always a finitely generated (in fact, cyclic) module whose projective dimension is equal to the global dimension of the ring. So $W$ can be chosen to be finitely generated, in which case, if the global dimension is finite and the ring is noetherian, its minimal projective resolution is a bounded complex of finitely generated projectives, so $W$ is perfect. $\endgroup$ Aug 20, 2014 at 18:45

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