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. $\square$

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

Proof. Consider the short exact sequence

 $\xymatrix{0\ar[r]&R_{Q}^{m}\ar[r]&I\ar[r]&I/R_{Q}^{m}\ar[r]&0}$

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. $\square$

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

Proof. By proposition 2 there is a finite set of generators $\{a_{1},\ldots,a_{n}\}$ 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. $\square$

## References

• 1 I. Assem, D. Simson, A. SkowroÃÆski, Elements of the Representation Theory of Associative Algebras, vol 1., Cambridge University Press 2006, 2007
Title properties of admissible ideals PropertiesOfAdmissibleIdeals 2013-03-22 19:16:48 2013-03-22 19:16:48 joking (16130) joking (16130) 4 joking (16130) Theorem msc 14L24