unital path algebras

Let $Q$ be a quiver and $k$ an arbitrary field.

Proposition. The path algebra $kQ$ is unitary if and only if $Q$ has a finite number of vertices.

Proof.,,$\Rightarrow$” Assume, that $Q$ has an infinite number of vertices and let $1\in kQ$ be an identity. Then we can express $1$ as

$$1=\sum _{i=1}^n{\lambda }_n\cdot w_n$$

where ${\lambda }_n\in k$ and $w_n$ are paths (they form a basis of $kQ$ as a vector space). Since $Q$ has an infinite number of vertices, then we can take a stationary path $e_x$ for some vertex $x$ such that there is no path among $w_1,\mathrm{\dots },w_n$ ending in $x$. By definition of $kQ$ and by the fact that $1$ is an identity we have:

$$e_x=1\cdot e_x=\left(\sum _{i=1}^n{\lambda }_n\cdot w_n\right)\cdot e_x=\sum _{i=1}^n{\lambda }_n\cdot (w_n\cdot e_x)=\sum _{i=1}^n{\lambda }_n\cdot 0=0.$$

Contradiction. $\mathrm{\square }$

,,$\Leftarrow$” If the set $Q_0$ of vertices of $Q$ is finite, then put

$$1=\sum _{q\in Q_0}{e}_q$$

where $e_q$ denotes the stationary path (note that $1$ is well-defined, since the sum is finite). If $w$ is a path in $Q$ from $x$ to $y$, then $e_x\cdot w=w$ and $w\cdot e_y=w$. All other combinations of $w$ with $e_q$ yield $0$ and thus we obtain that

$$1\cdot w=w=w\cdot 1.$$

This completes the proof. $\mathrm{\square }$