# Auslander-Reiten sequence and projective covers

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.

• 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). Oct 24, 2020 at 17:26
• @JeremyRickard, thank you very much. Are there some references about this fact? I need to cite this fact in a paper. Oct 24, 2020 at 19:16
• 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 $A\not\cong S$. Oct 24, 2020 at 21:00
• @JeremyRickard, thank you very much for your proof. Oct 25, 2020 at 8:32
• @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 $B_{k,n}$? Oct 25, 2020 at 15:52

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 $$R$$.