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Let $A$ be an associative ring, and $e\in A$ be an idempotent i.e. $e^2=e.$ It is well-known that $J(eAe)=eJ(A)e,$ where $J(-)$ denotes the Jacobson radical. It seems natural to try to compare $J(eAe)^2$ with $J(A)^2.$ My question is the following: is it always true that $eJ(A)^2e=eJ(A)eJ(A)e?$ Trivial cases when the answer is yes is when $e$ is a central idempotent or $AeA=A.$ The case which I really care about is when $A$ is a finite dimensional algebra over a discrete valuation ring. Thanks in advance!

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up vote 2 down vote accepted

There is an obvious inclusion but not equality in general. For a counter example take a quiver algebra and choose $e$ to correspond to a proper subset of the vertices $I$. Then if you have a directed path from a vertex in $I$ to a vertex not in $I$ and then to a vertex in $I$ this path is an element in one subspace but not the other.

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Very nice!Thanks Bruce. –  zamanjan Dec 1 '10 at 19:48
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