I would like to know, whether there are a quiver $Q$ and an admissible ideal $I$ such that the quiver algebra $\mathbb{F}_3Q/I$ and the group algebra $\mathbb{F}_3 (C_3\times C_3)$ are Morita equivalent.
($\mathbb{F}_3$ denotes the field with 3 elements)
Thank you very much.
Let's first consider the case of $\mathbb{F}_3(C_3)$. This is isomorphic to $\mathbb{F}_3[x]/(x^3)$, the isomorphism is given by sending $g\mapsto 1+x$, where $g$ is a chosen generator of the cyclic group $C_3$. (compare the answer to this question. This algebra is obviously given by a quiver with relations, namely the 1-loop quiver with relation $x^3=0$.
For $C_3\times C_3$ either one uses the same method or one notes that $\mathbb{F}_3(C_3\times C_3)=\mathbb{F}_3(C_3)\otimes_{\mathbb{F}_3} \mathbb{F}_3(C_3)$ which is therefore isomorphic to $\mathbb{F}_3[x]/(x^3)\otimes_{\mathbb{F}_3} \mathbb{F}_3[y]/(y^3)\cong \mathbb{F}_3[x,y]/(x^3,y^3)$. This is already a quiver with relations, namely the 2-loop quiver (as Aaron suggested) with relations $x^3=0$, $y^3=0$, and $xy=yx$.
