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.

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

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

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

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 $(a,b)\in k{Q}^{\prime }\oplus k{Q}^{\prime \prime }$. 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. $\mathrm{\square }$