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Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one, and all path algebras are basic.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every finitely generated $A$-module has a projective cover), but I don't have a reference for this. If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths). After the discussion in the comments, I am now uncertain whether the same might be true without completion.

The fact that quivers of algebras are determined only up to non-canonical isomorphism is going to prevent you from writing down a specific isomorphism $\Delta\to\Gamma$ given an isomorphism $K\Delta\to K\Gamma$, however.

Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one, and all path algebras are basic.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every finitely generated $A$-module has a projective cover), but I don't have a reference for this. If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths). After the discussion in the comments, I am now uncertain whether the same might be true without completion: since an isomorphism $K\Delta\to K\Gamma$ need not preserve the arrow ideals, one cannot deduce an uncompleted result from the completed one.

The fact that quivers of algebras are determined only up to non-canonical isomorphism is going to prevent you from writing down a specific isomorphism $\Delta\to\Gamma$ given an isomorphism $K\Delta\to K\Gamma$, however.

Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one, and all path algebras are basic.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that the result you want is still true when $\Delta$ and $\Gamma$ are allowed to have cycles, but I don't know a reference for this. I am also fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every simple $A$-module has a projective cover). If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths).

The fact that quivers of algebras are determined only up to non-canonical isomorphism is going to prevent you from writing down a specific isomorphism $\Delta\to\Gamma$ given an isomorphism $K\Delta\to K\Gamma$, however.

Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one, and all path algebras are basic.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every finitely generated $A$-module has a projective cover), but I don't have a reference for this. If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths). After the discussion in the comments, I am now uncertain whether the same might be true without completion.

The fact that quivers of algebras are determined only up to non-canonical isomorphism is going to prevent you from writing down a specific isomorphism $\Delta\to\Gamma$ given an isomorphism $K\Delta\to K\Gamma$, however.

Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that the result you want is still true when $\Delta$ and $\Gamma$ are allowed to have cycles, but I don't know a reference for this. I am also fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every simple $A$-module has a projective cover). If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths).

Benjamin and Theo's answers are already very good, but just to add a little; it is in fact true that any basic finite-dimensional algebra $A$ over a field $K$ determines (up to non-canonical isomorphism) a quiver $Q$ such that $A\cong KQ/I$ for some ideal $I$ contained in that generated by paths of length $2$. (Any finite-dimensional algebra is Morita equivalent to a basic one, and all path algebras are basic.) This is proved in Assem–Simson–Skowroński's Elements of the representation theory of associative algebras, Vol. 1, and implies the result you want when $\Delta$ and $\Gamma$ are acyclic.

The construction (which is essentially the same as Benjamin and Theo's) is to first pick a maximal set of pairwise orthogonal idempotents in $A$ (equivalent to a decomposition of $A$, as a module on one side, into indecomposable projective modules, which are pairwise non-isomorphic by basicness). Let $R$ be the radical, the minimal ideal such that $A/R$ is semi-simple (i.e. isomorphic to $K^n$). Then the vertices of $Q$ correspond to the idempotents, and the number of arrows $i\to j$ is the dimension of $e_j(R/R^2)e_i$ for $e_i$, $e_j$ the corresponding idempotents. (Or possibly of $e_i(R/R^2)e_j$ depending on how you like to compose paths.) Choosing elements of $R$ descending to a basis of $R/R^2$ determines a surjection $KQ\to A$, and one can show that the kernel has the required property.

I am fairly certain that the result you want is still true when $\Delta$ and $\Gamma$ are allowed to have cycles, but I don't know a reference for this. I am also fairly certain that one can get $A\cong KQ/I$ as above, with $A$ determining $Q$, whenever $A$ is semi-perfect (meaning every simple $A$-module has a projective cover). If true, then $\widehat{K\Delta}\cong \widehat{K\Gamma} \implies \Delta\cong\Gamma$, where $\widehat{K\Delta}$ denotes the completion of $K\Delta$ with respect to the arrow ideal (so that one allows formal linear combinations of paths).

The fact that quivers of algebras are determined only up to non-canonical isomorphism is going to prevent you from writing down a specific isomorphism $\Delta\to\Gamma$ given an isomorphism $K\Delta\to K\Gamma$, however.