# path algebra of a disconnected quiver

Let $Q$ be a disconnected quiver, i.e. $Q$ can be written as a disjoint union of two quivers $Q^{\prime}$ and $Q^{\prime\prime}$ (which means that there is no path starting in $Q^{\prime}$ and ending in $Q^{\prime\prime}$ and vice versa) and let $k$ be an arbitrary field.

The path algebra $kQ$ is isomorphic to the product of path algebras $kQ^{\prime}\times kQ^{\prime\prime}$.

Proof. If $w$ is a path in $Q$, then $w$ belongs either to $Q^{\prime}$ or $Q^{\prime\prime}$. Define linear map

 $T:kQ\to kQ^{\prime}\times kQ^{\prime\prime}$

by $T(w)=(w,0)$ if $w\in Q^{\prime}$ or $T(w)=(0,w)$ if $w\in Q^{\prime\prime}$ and extend it linearly to entire $kQ$. We will show that $T$ is an isomorphism of algebras.

If $w,w^{\prime}$ are paths in $Q$, then since $Q^{\prime}$ and $Q^{\prime\prime}$ are disjoint, then each of them entirely lies in $Q^{\prime}$ or $Q^{\prime\prime}$. Now since $Q^{\prime}$ and $Q^{\prime\prime}$ don’t have common vertices it follows that $w\cdot w^{\prime}=w^{\prime}\cdot w=0$. Without loss of generality we may assume, that $w$ is in $Q^{\prime}$ and $w^{\prime}$ is in $Q^{\prime\prime}$. Then we have

 $T(w\cdot w^{\prime})=T(0)=(0,0)=(w,0)\cdot(0,w^{\prime})=T(w)\cdot T(w^{\prime% }).$

If both lie in the same component, for example in $Q^{\prime}$, then

 $T(w\cdot w^{\prime})=(w\cdot w^{\prime},0)=(w,0)\cdot(w^{\prime},0)=T(w)\cdot T% (w^{\prime}).$

Since $T$ preservers multiplication on paths, then $T$ preserves multiplication and thus $T$ is an algebra homomorphism.

Obviously by definition $T$ is 1-1.

It remains to show, that $T$ is onto. Assume that $(a,b)\in kQ^{\prime}\oplus kQ^{\prime\prime}$. Then we can write

 $(a,b)=\sum_{i,j}\lambda_{i,j}(v_{i},w_{j})=\sum_{i,j}\lambda_{i,j}(v_{i},0)+% \sum_{i,j}\lambda_{i,j}(0,w_{j}),$

where $v_{i}$ are paths in $Q^{\prime}$ and $w_{j}$ are paths in $Q^{\prime\prime}$. It can be easily checked, that

 $T\left(\sum_{i,j}\lambda_{i,j}(v_{i}+w_{j})\right)=(a,b).$

Here we consider all $v_{i}$ and $w_{j}$ as paths in $Q$.

Thus $T$ is an isomorphism, which completes the proof. $\square$

Title path algebra of a disconnected quiver PathAlgebraOfADisconnectedQuiver 2013-03-22 19:16:25 2013-03-22 19:16:25 joking (16130) joking (16130) 4 joking (16130) Theorem msc 14L24