# 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. $\mathrm{\square}$

Title | path algebra of a disconnected quiver |
---|---|

Canonical name | PathAlgebraOfADisconnectedQuiver |

Date of creation | 2013-03-22 19:16:25 |

Last modified on | 2013-03-22 19:16:25 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 14L24 |