TITLE: Can we infer an isomorphism of quivers from an isomorphism of their corresponding path algebras?
QUESTION [7 upvotes]: Given a pair $\Delta, \Gamma$ of quivers and a field $K$ one can construct the corresponding path algebras $K\Delta, K\Gamma$. I came upon a paper claiming (section 3, 2nd paragraph) that an isomorphism of $K\Delta \cong K\Gamma$ of the path algebras induces an isomorphism of the quivers $\Delta \cong \Gamma$. However, no proof is given for this claim, and so I am looking for a reference giving a proof and stating explicitly the required conditions on the quivers or the exact meaning of a quiver isomorphism in this context.

REPLY [2 votes]: 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 $M,N$ are associated to the same vertex if and only if there is some free algebra quotient $F=K\langle x_1,\dots,x_n\rangle$ of $KQ$ such that the actions of $KQ$ on $M$ and $N$ both factor through $F$.
So the vertex set of $Q$ can be recovered from $KQ$. Picking one one-dimensional module $M_v$ for each vertex $v$, the number of arrows from vertex $v$ to vertex $w$ is the dimension of $\text{Ext}^1(M_v,M_w)$.