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I recently began to study representation theory of algebras and I found this problem:

Suppose $\Lambda$ is a finite dimensional algebra over an arbitrary field. If $\Lambda$ is hereditary, basic and connected and if $e$ is an idempotent of $\Lambda$, then $e\Lambda e$ is hereditary.

I would like that someone give me a simple proof of it because I have no idea of what to do.

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I will write $A$ instead of $\Lambda$...

If $A$ is hereditary, there is a projective resolution $$0\to P_1\to P_0\to A\to0$$ of $A$ as an $A$-bimodule. The right $A$-module $eA$ is projective, as is the left $A$-module $Ae$, so the functor $$M\longmapsto eA\otimes_AM\otimes_AAe$$ from the category of $A$-bimodules to the category of $eAe$-bimodules is exact. We have, therefore a short exact sequence of $eAe$-bimodules $$0\to eA\otimes_AP_1\otimes_AAe\to eA\otimes_AP_0\otimes_AAe\to eA\otimes_AA\otimes_AAe\to0$$ The $eAe$-bimodule $eA\otimes_AA\otimes_AAe$ is isomorphic to $eAe$, and the $eAe$-bimodules $eA\otimes_AP_1\otimes_AAe$ are projective. This means that the projective dimension of $eAe$ as an $eAe$-bimodule is at most $1$. This implies that $eAe$ is hereditary.


If you know that the algebra is the path algebra $kQ$ of an acyclic quiver and you suppose (as you may) that $e$ is the sum of a set of vertices $S\subseteq Q_0$, then it is easy to construct a quiver $Q'$ such that $ekQe\cong kQ'$. Indeed, the vertices of $Q'$ are the elements of $S$ and the arrows in $Q'$ are the paths in $Q$ from a vertex of $S$ to another which cannot be factored as a product of two such paths.

For example, if $Q$ is

enter image description here

and $S=\{1,3,4\}$, the quiver $Q'$ is

enter image description here

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    Where did you use that A is basic and connected?2011-12-01
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    I did not. ${}{}{}$2011-12-01
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    So those hyphotesis are not necessary?2011-12-01
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    You are asking me if my argument is correct...2011-12-01
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    No, what Im asking you is that if we assume the other hypothesis can we give a simpler proof? Actually your proof is very elegant...2011-12-02
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    I have a question: Since $A$ i hereditary then the resolution of A is of the form $0\rightarrow A\rightarrow A$, so I dont understand why you write the resolution $0\rightarrow P_{1}\rightarrow P{0}\rightarrow A\rightarrow 0$? Another question is that in the part that you write the diagrams I think you are using Gabriels theorem but I think this only works for algebraically closed fields right?2011-12-05
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    Jow, you are mixing two things. On one hand, for every algebra (hereditary or not) $A$ the left $A$-module $A$ is projective, so $0\to A\to A$ is *always* a resolution of left $A$-modules. But as I wrote in my answer, I am considering a resolution $0\to P_1\to P_0\to A$ as an $A$ **bimodule**, which is a rather different thing.2011-12-05
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    As for your second point: I wrote «If you know that the algebra is the path algebra kQ of an acyclic quiver». If you *know* that $A$ is $kQ$ for some $Q$, then what I wrote is true independently of the field. Now, if I only know that $A$ is hereditary, basic and finite dimensional, then I need that $k$ be alg. closed to be sure that $A$ is of the form for some $A$: but I am not using this.2011-12-05
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    Sorry, but I dont know much about this. Tell me if Im wrong, an algebra $A$ is hereditary if and only if it has a resolution $0\rightarrow P_{1}\rightarrow P_{0}\rightarrow A\rightarrow 0$ as an $A$-bimodule?2011-12-05
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    That is true, but non obvious. It is easy to show that an algebra satisfying that condition is hereditary, but the converse it less immediate.2011-12-06
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    Thank you so much for your help and patience, I think I understand a little bit more now. Finally, can you give me a reference for this material, specially this equivalence of hereditary algebras?2011-12-06