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Suppose $\Lambda$ is an Artin algebra with $\DeclareMathOperator{\rad}{rad}\rad^3\Lambda=0 $, and $ M $ any finite $\Lambda$-module with projective dimension finite. Proof that $\Omega M $ has Loewy length at most 2.

This question is from a paper: On the finitistic global dimension conjecture for Artin algebras - Igusa and Todorov. It is a corollary.

My idea is: we can guarantee that $\rad^3M=0$ because $\rad M=M\rad\Lambda$, and take the projective cover of $M$: $$\Omega M \rightarrow P \rightarrow M \rightarrow 0$$

$ΩM=\ker(f)$ is a submodule of $\rad(P)$ (where $P$ is the first term of projective cover). And thinking use exactness of this for make something.

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    $\begingroup$ Any own ideas? Where did you get this question from? $\endgroup$ Commented Nov 27, 2018 at 18:20
  • $\begingroup$ This is from a paper: On the finitistic global dimension conjecture for Artin algebras. My ideia is: we can guarantee that $ rad^3 M=0 $, and $\Omega M $ is a submodule of $ rad(P)$ (where P is the first term of projective cover). Use exactness of this for make something. $\endgroup$ Commented Nov 27, 2018 at 18:29

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I discuss this question to my professor today, the answer is simple.

Take $M$ a f.g. $\Lambda$-module with $pdim M < \infty$.

Since $Mrad\Lambda = radM$, and we have $\Omega M \subseteq rad P$ (and is a submodule), then $$rad^2\Omega M = rad(rad\Omega M)=rad\Omega M rad\Lambda=\Omega M rad^2\Lambda \leq rad P rad^2\Lambda=Prad^3\Lambda=0$$ Then, the Loewy length of $\Omega M$ is less or equal 2.

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