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Here's one way to recover $Q$ from $KQ$ when $Q$ is finite but not necessarily acyclic. Consider the one-dimensional $KQ$-modules. Each is associated to some vertex of $Q$, and two such modules $M,N$ …

Actually, his original statement assumes that the quiver has no loops, but quivers with loops are covered as well in V.G. … Kac, "Root systems, representations of quivers and invariant theory" in Invariant Theory (ed. F. Gherardelli) LNM 996 (Springer, 1983), pp 74-108. …

In Nüsken, M. "Unique tensor factorization of algebras", Math Ann. (1999) 315-341 this is proved for $K$ of characteristic zero. As far as I know, it's still open in positive characteristic, although …

Take the quiver $1 \rightarrow 2 \rightrightarrows 3$. Let $V$ be the representation with $V_1=k$, $V_2=k^2$, $V_3=k^2$, with arrows acting by $\pmatrix{0&1}$, $\pmatrix{0&1\\0&0}$ and $\pmatrix{1&0\ …

The quiver $$\begin{array}{ccccc} &&1&\to&2\\ &&\uparrow&&\uparrow&\\ 3&\to&4&&5\\ &&\uparrow&&\\ &&6&& \end{array}$$ is a Dynkin quiver of type $D_6$, so has finite representation type. But if you ad …

For every finite dimensional algebra that does not have finite representation type, there is no bound on the dimension of indecomposable modules. This is known as the first Brauer-Thrall conjecture (n …

This doesn't work for quivers with relations. …

Here's an elementary proof that $2n-2$ is a lower bound. Suppose that $$V_1\xrightarrow{\alpha_1}V_2\xrightarrow{\alpha_2}\dots\xrightarrow{\alpha_{n-2}}V_{n-1}\xrightarrow{\alpha_{n-1}}V_n$$ is a rep …