I am studying the text An introduction to Quiver Representations by Derksen. Exercise 1.5.4 asks to prove that if the path algebras $\mathbb{C}Q$ and $\mathbb{C}Q'$ are isomorphic $\mathbb{C}$-algebras then $Q$ and $Q'$ are isomorphic directed graphs if they are acyclic. This I am able to prove(the proof is wrong).
However, will the above result hold if the quivers have cycles? I think the answer should be NO but I am unable to find a counterexample. In any case if the path algebras are isomorphic then $Q$ and $Q'$ will have the same number of edges. This is because they form the free generators and the path algebras are finitely generated algebras. As vector spaces they will now be infinite dimensional which is the reason why I think the underlying quivers might not be isomorphic since it is "easier" for infinite dimensional vectors spaces to be isomorphic and the basis here are the paths in the quivers. But still I don't have a counterexample.
Best Answer
I'll assume that the quiver $Q$ has finitely many vertices and arrows.
Then even if $Q$ has oriented cycles, it is it is still true that the path algebra $\mathbb{C}Q$ determines the quiver $Q$.
(1) In the case of a quiver with no oriented cycles, a quick way to recover the quiver is as follows:
There is a simple module $S_i$ associated with each vertex $i$, and the number of arrows from vertex $i$ to vertex $j$ is $\dim_{\mathbb{C}}\operatorname{Ext}^1_{\mathbb{C}Q}(S_i,S_j)$. [I'm not sure that this is the proof that Derksen was asking for, as the book doesn't seem to assume knowledge of $\operatorname{Ext}$.]
So knowing the simple $\mathbb{C}Q$-modules and the extensions between them lets you recover the quiver.
(2) If the quiver has oriented cycles but no loops (arrows with the target equal to source) then there are more simple modules, but the obvious simples associated to the vertices are the only $1$-dimensional simples, and the same method of recovering the quiver works, if you only consider the $1$-dimensional simples.
(3) If the quiver has loops, then there are more $1$-dimensional simples (consider the representation with $\mathbb{C}$ at vertex $i$, zero at every other vertex, with the loops at vertex $i$ acting by multiplication by arbitrary scalars). But the same method work if we can pick out one $1$-dimensional simple module for each vertex.
There may be a simpler method, but one way to do this is to consider the abelianization of $\mathbb{C}Q$. This is a product of polynomial algebras $\mathbb{C}[x_1,\dots,x_{r_i}]$, one for each vertex $i$, where $r_i$ is the number of loops at vertex $i$. So it has one primitive idempotent for each vertex, and for each of these idempotents we can choose any $1$-dimensional simple module (it doesn't matter which) that is not annihilated by that idempotent. As before, the quiver can then be recovered by considering $\operatorname{Ext}^1_{\mathbb{C}Q}$ between these simple modules.