# unital path algebras

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

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 ${w_{1},\ldots,w_{n}}$ 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. $\square$

,,$\Leftarrow$” 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. $\square$

Title unital path algebras UnitalPathAlgebras 2013-03-22 19:16:23 2013-03-22 19:16:23 joking (16130) joking (16130) 4 joking (16130) Theorem msc 14L24