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Let $A$ be a finite dimensional quiver algebra such that any two indecomposable modules with the same dimension vector are isomorphic.

Question: In case $A$ has $n$ simple modules and finite global dimension, does $A$ have Loewy length at most $n$?

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Yes.

Let $A=kQ/I$ be a quiver algebra with $n$ simple modules and finite global dimension such that $\text{rad}^nA\neq0$. For a vertex $i$ of the quiver $Q$, I'll use $e_i$ to denote the corresponding primitive idempotent of $A$, and $S_i$ to denote the corresponding simple (right) $A$-module.

Since $\text{rad}^nA\neq0$, there is a path of length $n$ that is nonzero in $A$. Since $Q$ has $n$ vertices, this path must pass through one of them twice. So there is a directed cycle $c$ that is nonzero in $A$.

Suppose $c$ starts and ends at vertex $i$, so $c=e_ice_i\neq0$.

Let $K$ be a submodule of the indecomposable projective $e_iA$ that is maximal subject to $c\not\in K$, and let $X=e_iA/K$. Then $c$ spans the socle of $X$, so $X$ is a module whose head and socle are both isomorphic to $S_i$, and so $\text{rad}X$ and $X/\text{soc}X$ have the same dimension vector. They are also both indecomposable, since they have simple socle and head respectively.

Since $A$ has finite global dimension, the No Loops Conjecture implies that $S_i$ is not a summand of the head of $\text{rad}X$, so $\text{rad}X\not\cong X/\text{soc}X$.

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  • $\begingroup$ What do you mean with $c \neq K$? Probably that c not $\in$ K? $\endgroup$
    – Mare
    Mar 29, 2020 at 9:33
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    $\begingroup$ @Mare Yes, sorry. Fixed. $\endgroup$ Mar 29, 2020 at 9:40
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    $\begingroup$ Thanks, a very nice proof. $\endgroup$
    – Mare
    Mar 29, 2020 at 9:40

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