Auslander-Reiten theory: exercise $23.b$ of 'Elements of the Representation Theory of Associative Algebras'

I am solving exercise $23.b$ of chapter IV of 'Elements of the representation theory of associative algebras' by Assem, Simson and Skowronski. The question is the following: Consider the following quiver where all arrows point towards vertex $1$. $$\require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} % \hskip 1em % \place{0}{0}{3} \place{0}{10}{2} \place{10}{0}{4} \place{10}{10}{5} \place{5}{5}{1} % \place{1}{6.5}{\dline{3}{3}} \place{6.5}{1}{\dline{3}{3}} \place{6.5}{6.5}{\uline{3}{3}} \place{1}{1}{\uline{3}{3}} \hskip18em\Rule{0em}{14em}{1.5em}$$

We have to construct the portion of the Auslander-Reiten quiver of the path algebra of this quiver containing the indecomposable projective representations. Moreover, we have to show it contains no injectives.

My attempt: $P(1)=S(1)$ and $P(1)=\text{rad}(P(i))$ for $2\leq i\leq 5$, hence the Auslander-Reiten quiver starts in the projective $P(1)$ and then there are $4$ maps, namely $P(1)\rightarrow P(i)$ for $2\leq i\leq 5$, since $S(1)$ is a simple projective, these are all irreducible morphisms starting at $S(1)$. A quick calculation then yields that $\dim\tau^{-1}S(1)=(3,1,1,1); \dim\tau^{-2}S(1)=(5,2,2,2)$ and $\dim\tau^{-3}S(1)=(8,3,3,3)$. Then, by induction one can show that $\dim\tau^{-n}S(1)=(4(n-1),n,n,n,n)$ for all $n\geq 3$.

A quick calculation shows that $\dim I(1)=(1,1,1,1,1)$ and $I(j)=S(j)$ for $j\neq 1$. Thus we know all (indecomposable) injectives. If $0\rightarrow \tau^{-n}S(1)\rightarrow E \rightarrow \tau^{-(n+1)}S(1)\rightarrow 0$ is an almost split exact sequence, then I can show that no summand of $E$ is injective. We can see this by constructing the Auslander-Reiten quiver using the knitting technique, the AR-quiver looks like a horizontal line of meshes with $4$ middle terms. By considering the dimension vectors it follows that no injective is ever reached in this portion of the AR-quiver.

Is the above reasoning correct, is my calculation of this portion of the AR-quiver complete? I'm not 100% familiar with the theory and would like some feedback on whether or not I understand it. Thank you.

• Your reasoning looks ok to me. You haven't mentioned it in your post, but using induction (and the "knitting technique"), you can also compute the dimension vectors of $\tau^{-n} P(i)$ for all $n$ and all $i$; this will tell you right away that no dimension vectors of injective modules will occur in this way. – Pierre-Guy Plamondon Aug 8 '16 at 8:52
• @Pierre-GuyPlamondon: Thank you. I just wanted some confirmation on the reasoning. If I remember correctly I did make a calculation error at some point but that's not too important. – Mathematician 42 Aug 8 '16 at 9:50