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Let $R$ be an Artin algebra and let $0 \to A \to B \to C \to 0$ be an Auslander-Reiten sequence of finitely generated left $R$-modules. Is it always true that the projective cover of $B$ equals to the direct sum of the projective cover of $A$ and the projective cover of $C$? Thank you very much.

Edit: I would also like to know the following.

Let $A \to B \to C$ ($B$ can be a direct sum of indecomposable modules) be a mesh in an Auslander-Reiten quiver. Is it true that the projective cover of $B$ is isomorphic to the direct sum of the projective cover of $A$ and the projective cover of $C$?

Thank you very much.

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    $\begingroup$ Did you look at any examples? It's true if and only if $A$ is not simple, so you would have found a counterexample by looking at literally any Auslander-Reiten quiver (of a non-semisimple algebra). $\endgroup$ Oct 24, 2020 at 17:26
  • $\begingroup$ @JeremyRickard, thank you very much. Are there some references about this fact? I need to cite this fact in a paper. $\endgroup$ Oct 24, 2020 at 19:16
  • $\begingroup$ The question is equivalent to asking whether the head of $B$ is isomorphic to the direct sum of the heads of $A$ and $C$, which can be detected by applying $\operatorname{Hom}_R(-.S)$ to the sequence for each simple module $S$, and by the definition of an Auslander-Reiten sequence, that gives a short exact sequence if and only if $A\not\cong S$. $\endgroup$ Oct 24, 2020 at 21:00
  • $\begingroup$ @JeremyRickard, thank you very much for your proof. $\endgroup$ Oct 25, 2020 at 8:32
  • $\begingroup$ @JeremyRickard, thank you very much for your help. I have another question. I think that this property is true for the Auslander-Reiten sequence of the algebra $B_{k,n}$ in the post. Does your proof also work for the algebra $B_{k,n}$? $\endgroup$ Oct 25, 2020 at 15:52

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No, take $R=K[x]/(x^2)$ and $C=A=S$ the simple $R$-module. Then $0 \rightarrow S \rightarrow R \rightarrow S \rightarrow 0$ is the Auslander-Reiten sequence of $S$ and the projective cover of $R$ and $S$ is $R$.

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