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



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> Question: In case $A$ has $n$ simple modules and finite global dimension, does $A$ have Loewy length at most $n$?
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On algebras where indecomposable modules are determined by their dimension vectors
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 $\text{rad}X\not\cong X/\text{soc}X$.


