# properties of admissible ideals

Let $Q$ be a quiver, $k$ a field and $I$ an admissible ideal (see parent object) in the path algebra^{} $kQ$. The following propositions^{} and proofs are taken from [1].

Proposition 1. If $Q$ is finite, then $kQ/I$ is finite dimensional algebra.

Proof. Let ${R}_{Q}$ be the arrow ideal in $kQ$. Since ${R}_{Q}^{m}\subseteq I$ for some $m$, then we have a surjective^{} algebra homomorphism $kQ/{R}_{Q}^{m}\to kQ/I$. Thus, it is enough to show, that $kQ/{R}_{Q}^{m}$ is finite dimensional. But since $Q$ is a finite quiver, then there is finitely many paths of length at most $m$. It is easy to see, that these paths form a basis of $kQ/{R}_{Q}^{m}$ as vector space over $k$. This completes^{} the proof. $\mathrm{\square}$

Proposition 2. If $Q$ is finite, then $I$ is a finitely generated^{} ideal.

Proof. Consider the short exact sequence^{}

$$\text{xymatrix}0\text{ar}[r]\mathrm{\&}{R}_{Q}^{m}\text{ar}[r]\mathrm{\&}I\text{ar}[r]\mathrm{\&}I/{R}_{Q}^{m}\text{ar}[r]\mathrm{\&}0$$ |

of $kQ$ modules. It is well known that in such sequences the middle term is finitely generated if the end terms are. Of course ${R}_{Q}^{m}$ is finitely generated, because $Q$ is finite so there is finite number of paths of length $m$.

On the other hand $I/{R}_{Q}^{m}$ is an ideal in $kQ/{R}_{Q}^{m}$, which is finite dimensional by proposition 1. Thus $I/{R}_{Q}^{m}$ is a finite dimensional vector space over $k$. But then it is finitely generated $kQ$ module (see this entry (http://planetmath.org/FiniteDimensionalModulesOverAlgebra) for more details), which completes the proof. $\mathrm{\square}$

Proposition 3. If $Q$ is finite, then there exists a finite set^{} of relations^{} (http://planetmath.org/RelationsInQuiver) $\{{\rho}_{1},\mathrm{\dots},{\rho}_{m}\}$ such that $I$ is generated by them.

Proof. By proposition 2 there is a finite set of generators^{} $\{{a}_{1},\mathrm{\dots},{a}_{n}\}$ of $I$. Generally the don’t have to be relations. On the other hand, if ${e}_{x}$ denotes the stationary path in $x\in {Q}_{0}$, then it can be easily checked, that every element of the form ${e}_{x}\cdot {a}_{i}\cdot {e}_{y}$ is either zero or a relation. Also, note that

$${a}_{i}=\sum _{x,y\in {Q}_{0}}{e}_{x}\cdot {a}_{i}\cdot {e}_{y}.$$ |

Since $Q$ is finite, then this completes the proof. $\mathrm{\square}$

## References

- 1 I. Assem, D. Simson, A. SkowroÃÆski, Elements of the Representation Theory of Associative Algebras, vol 1., Cambridge University Press 2006, 2007

Title | properties of admissible ideals |
---|---|

Canonical name | PropertiesOfAdmissibleIdeals |

Date of creation | 2013-03-22 19:16:48 |

Last modified on | 2013-03-22 19:16:48 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 14L24 |