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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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    Now I finally understand your question. You should include the definition of "admissible" in the question in order to make it even more clear.2012-11-13

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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!2012-11-15