# About Loewy length of syzygy

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.

• Any own ideas? Where did you get this question from? Commented Nov 27, 2018 at 18:20
• 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. Commented Nov 27, 2018 at 18:29

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.