I have the following problem: If $\Lambda$ is a hereditary, basic and connected algebra and $e$ is an idempotent of $\Lambda$, how can I prove that $e\Lambda e$ is also hereditary?
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1$\begingroup$ Is this homework? $\endgroup$– Bruce WestburyCommented Dec 1, 2011 at 19:24
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1$\begingroup$ And are you assuming finite dimensional? Over any field or an algebraically closed field? This does smell like homework. $\endgroup$– Benjamin SteinbergCommented Dec 1, 2011 at 19:40
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$\begingroup$ If you are assuming finite dimensional and split basic, here is a hint. Assume $\Lambda$ is the path algebra of an acyclic quiver. Convince yourself that you may assume that $e$ is a sum of primitive idempotents corresponding to vertices. Show that $e\Lambda e$ is a path algebra on a certain subquiver. $\endgroup$– Benjamin SteinbergCommented Dec 1, 2011 at 20:10
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$\begingroup$ Is not a homework is just that Im interested in studying this things and I found that problem. Yes I assume $\Lambda$ is finite dimensional and is over any field. $\endgroup$– AntonioCommented Dec 1, 2011 at 20:41
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$\begingroup$ By basic, do you mean split basic (the radical quotient is a product of copies of the field) or just that the radical quotient is a direct product of division rings? $\endgroup$– Benjamin SteinbergCommented Dec 1, 2011 at 23:04
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1 Answer
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If $\Lambda$ is split basic, then by Gabriel's theorem it is isomorphic to $\Bbbk Q$ where $Q$ is a finite acyclic quiver. Up to isomorphism you can assume $e$ is the sum of empty paths running over some subset $X$ of vertices. Then $e\Lambda e$ is isomorphic to the path algebra on the full (i.e. induced) subquiver on the vertex set $X$. Thus it is hereditary.
Mariano Suárez-Álvarez
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answered Dec 2, 2011 at 1:39