# 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. $\mathrm{\square}$

Title | unital path algebras |
---|---|

Canonical name | UnitalPathAlgebras |

Date of creation | 2013-03-22 19:16:23 |

Last modified on | 2013-03-22 19:16:23 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 14L24 |