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Let $Q$ be a finite quiver without oriented cycles and let $k$ be a field. Then $A = kQ$ is a finite-dimensional path algebra. Let $e_1, \dots, e_n$ be the idempotents corresponding to the vertices $1, \dots, n$ of $Q$. Then the modules $I(i) = D(e_iA)$ are precisely the indecomposable injective modules.

Question: Assume $I(i)$ is not projective. Then there is an Auslander-Reiten sequence $0 \to \tau I(i) \to E(i) \to I(i) \to 0$. How can we describe the module $E(i)$?

In the article "Representations of Wild quivers" by Otto Kerner, the author seems to implicitly say in the proof of theorem 3.8 that $E(i) = \bigoplus I(j) \oplus \bigoplus \tau I(j')$, where $j$ and $j'$ together run through all vertices in the quiver which are adjacent to $i$. Is this correct? Do you know a reference for this?

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Yes, that's correct.

For any finite dimensional algebra, the left minimal almost split morphism from an indecomposable injective $I$ is the natural epimorphism $I\to I/\text{soc }I$. This is Prop. IV.3.5 in Assem, Simson and Skowronski's Elements of the Representation Theory of Associative Algebras, vol. I.

In the case of the path algebra of a quiver, $I(i)/\text{soc }I(i)$ is the direct sum of the $I(j')$, as $j'$ runs over the tails of all arrows into $i$. So the arrows in the Auslander-Reiten quiver starting from $I(i)$ are $I(i)\to I(j')$. None of the $I(j')$ are projective, or they would be proper direct summands of $I(i)$, so there are corresponding arrows $\tau I(j')\to I(i)$. Then there are also the irreducible maps from indecomposable injectives to $I(i)$, which as shown above are maps $I(j)\to I(i)$ where $j$ runs over the heads of all arrows from $i$.

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  • $\begingroup$ Thanks, so all the $I(j)$ and $I(j')$ I mentioned appear in the sink map to $I(i)$, which is in fact all I need. But, out of interest: How can one show that there are no further irreducible maps ending in $I(i)$? $\endgroup$ – Leon Lang Aug 29 '17 at 13:36
  • $\begingroup$ @LeonLang There's a bijection between arrows in the AR quiver from $X$ to non-projective indecomposables $Y$ and arrows from non-injective indecomposables $\tau Y$ to $X$. So once you know all the arrows from $X$, the only "extra" arrows to $X$ will be from injectives. $\endgroup$ – Jeremy Rickard Aug 29 '17 at 15:09
  • $\begingroup$ Thanks. But why are the only injectives with an irreducible map $i(j) \to I(i)$ the ones where we have an arrow $i \to j$ in $Q$? I thought about it and I think that Hom$(I(j), I(i)) \neq 0$ if and only if there is a path $i \rightsquigarrow j$. But I don't know why there is only an irreducible map in it if there is an arrow in it. $\endgroup$ – Leon Lang Aug 29 '17 at 17:49
  • $\begingroup$ @LeonLang That's what Prop. IV.3.5 tells you. The irreducible maps from $I(j)$ are to the indecomposable summands of $I(j)/\text{soc }I(j)$. $\endgroup$ – Jeremy Rickard Aug 29 '17 at 18:37
  • $\begingroup$ Ah, I'm sorry :-) I somehow had the wrong perspective. Thank you very much! :-) $\endgroup$ – Leon Lang Aug 29 '17 at 21:49

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