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How to prove the following conclusion :

For any finite quiver $Q$, an ideal $I$ of $KQ$, contained in $R^2_Q$, is admissible if and only if, for each cycle $\sigma$ in $Q$, there exists $s \geq 1$ such that $ \sigma^s \in I$, where, $R_Q$ is the arrow ideal of the path algebra $KQ$.

This conclusion comes from page 53 of the book named " Elements of the Representation Theory of Associative Algebras Volume1" , but I do not know how to prove it, I need a detailed proof.

For any finite quiver $Q$, a two-sided ideal $I$ of $KQ$ is said to be admissible if there exists $m \geq2$ such that $R^m_Q\subseteq I\subseteq R^2_Q$

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What what? You are presumably asking what is the motivation for us to define admissible ideal in that way... – Mariano Suárez-Alvarez Nov 12 '12 at 6:18
Could you make your question more precise? Do you still ask about the motivation for the definition of admissible ideal or is your question now about what you asked in the comments to my answer. The latter would then be a new question which you should post as such. – Julian Kuelshammer Nov 12 '12 at 11:56
My question in the comments is in fact an abortive attempt to resolve above conclusion. – Aimin Xu Nov 13 '12 at 10:48
Now I finally understand your question. You should include the definition of "admissible" in the question in order to make it even more clear. – Julian Kuelshammer Nov 13 '12 at 18:27
up vote 0 down vote accepted

If I understood this paper correct. They are wrong here. Golod and Shafarevich provided in their paper On Classfield Towers a counterexample to the fact that every finitely generated nil ring is nilpotent.

Rewriting this in quiver terms means that if you take the one-vertex quiver $Q$ with $d\geq 2$ loops, then there exist infinitely many (even homogeneous) linear combinations of paths $f_j$ of degree $\geq 2$ such that the ideal $I$ spanned by the $f_j$ is such that $kQ/I$ is infinite-dimensional, but $kQ^+/I$ is nil, i.e. every element in $kQ^+/I$ is nilpotent. In particular every path to some power lies in $I$ and $kQ/I$ is of course not admissible since $kQ/I$ is infinite-dimensional.

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Thanks, you gave a beautifull counterexample to my question! – Aimin Xu Nov 15 '12 at 11:36

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