0
$\begingroup$

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?

$\endgroup$
11
  • 1
    $\begingroup$ Is this homework? $\endgroup$ Dec 1, 2011 at 19:24
  • 1
    $\begingroup$ And are you assuming finite dimensional? Over any field or an algebraically closed field? This does smell like homework. $\endgroup$ Dec 1, 2011 at 19:40
  • $\begingroup$ 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. $\endgroup$ Dec 1, 2011 at 20:10
  • $\begingroup$ 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. $\endgroup$
    – Antonio
    Dec 1, 2011 at 20:41
  • $\begingroup$ 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? $\endgroup$ Dec 1, 2011 at 23:04

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.