properties of admissible ideals
Proposition 1. If is finite, then is finite dimensional algebra.
Proof. Let be the arrow ideal in . Since for some , then we have a surjective algebra homomorphism . Thus, it is enough to show, that is finite dimensional. But since is a finite quiver, then there is finitely many paths of length at most . It is easy to see, that these paths form a basis of as vector space over . This completes the proof.
Proposition 2. If is finite, then is a finitely generated ideal.
Proof. Consider the short exact sequence
of modules. It is well known that in such sequences the middle term is finitely generated if the end terms are. Of course is finitely generated, because is finite so there is finite number of paths of length .
On the other hand is an ideal in , which is finite dimensional by proposition 1. Thus is a finite dimensional vector space over . But then it is finitely generated module (see this entry (http://planetmath.org/FiniteDimensionalModulesOverAlgebra) for more details), which completes the proof.
Proof. By proposition 2 there is a finite set of generators of . Generally the don’t have to be relations. On the other hand, if denotes the stationary path in , then it can be easily checked, that every element of the form is either zero or a relation. Also, note that
Since is finite, then this completes the proof.
- 1 I. Assem, D. Simson, A. SkowroÃÆski, Elements of the Representation Theory of Associative Algebras, vol 1., Cambridge University Press 2006, 2007