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I allready asked this on MO, but did not get any answer.

Given a finite quiver with relations. When is the path algebra modulo relations hereditary?

If the path algebra is finite dimensional or there are no relations, the answer is well known. What happens if the path algebra is infinite dimensional and there are relations?

I know that the question is everything but precise, however I would appreciate any reference dealing with this topic.

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    $\begingroup$ Do you assume $I \subseteq J^2$ (where $I$ is the ideal generated by the relations and $J$ is the ideal generated by the arrows)? $\endgroup$
    – Dag Oskar Madsen
    Commented Sep 5, 2013 at 10:22
  • $\begingroup$ Note that the path algebra with no relations is always hereditary. $\endgroup$
    – Benjamin Steinberg
    Commented Sep 5, 2013 at 12:54
  • $\begingroup$ Yes, you can assume that $I\subset J^2$ if that helps. $\endgroup$
    – Oliver Straser
    Commented Sep 5, 2013 at 14:00
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    $\begingroup$ Then I believe $J/I$ is not projective, which contradicts $kQ/I$ being hereditary. (In a hereditary ring, submodules of projectives should be projective.) I have to think a bit to formalize the proof. $\endgroup$
    – Dag Oskar Madsen
    Commented Sep 5, 2013 at 17:27
  • $\begingroup$ Here is a proof sketch, details need to be checked and filled in. There is an epimorphism of $kQ/I$-modules $J/IJ \rightarrow J/I \rightarrow 0$ with $J/IJ$ projective. If $J/I$ is projective, then this morphism splits and we get (I think) an isomorphism $J/IJ \simeq J/I$, also as $kQ$-modules. This leads to a contradiction since the finite dimensional spaces $J^m(J/IJ)/J^{m+1}(J/IJ)$ and $J^m(J/I)/J^{m+1}(J/I)$ have different dimensions. $\endgroup$
    – Dag Oskar Madsen
    Commented Sep 6, 2013 at 7:08

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The algebra is $A=kQ/I$ with the assumptions $I \subseteq J^2$ and $I/(JI+IJ) \neq 0$, where $J$ is the ideal in $kQ$ generated by the arrows.

Assume $A$ is hereditary. Then $J/I$ is a projective $A$-module and the exact sequence of $A$-modules $$0 \rightarrow I/IJ \rightarrow J/IJ \rightarrow J/I \rightarrow 0$$ must split. Let $s \colon J/IJ \rightarrow I/IJ$ denote a retraction.

Let $x$ be an element in $I \smallsetminus (JI+IJ)$. Since $I \subseteq J^2$, we can write $$x=p_1 q_1 + \cdots + p_n q_n$$ with $p_i,q_i \in J$ for all $1 \leq i \leq n$. Then $$x+IJ=s(p_1 q_1 + \cdots + p_n q_n+IJ)=p_1 s(q_1) + \cdots + p_n s(q_n) +IJ \in JI+IJ,$$ a contradiction.

Therefore $A$ is not hereditary.

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  • $\begingroup$ I am assuming here the existence of a minimal relation, that is $I/(JI+IJ) \neq 0$. Perhaps this is automatic. $\endgroup$
    – Dag Oskar Madsen
    Commented Sep 6, 2013 at 17:03

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