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I have known the following exercise for some time. Take $p_1,\ldots,p_n$ idempotents in $M_m(K)$ ($K$ field of characteristic zero). Show that $p:=p_1+\cdots+p_n$ is idempotent if and only if $p_ip_j=0$ for all $i\neq j$. The solution I know essentially uses the fact that the rank of an idempotent is equal to its trace.

I am interested in generalizations of this result. In particular, for what other algebras does this hold? Is there an algebra where this fails? I am asking in particular because I don't know how to find a counterexample. Thanks for your help.

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    "the trace of an idempotent is equal to its trace"?2012-04-13
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    @joriki I meant rank, sorry.2012-04-13
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    It's unclear how you would generalize this result to an algebra other than $M_n(K)$; how would you supply the parameter $n$?2012-04-13
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    @Qiaochu: the n's should not be the same n's, sorry.2012-04-13

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This is true for all finite dimensional simple algebras, because each of them becomes a matrix algebra after extension to the algebraic closure of the ground field.

It is also true for direct products of finite dimensional simple algebras, because everything happens in each direct factor separately.

Let now $A$ is an arbitrary finite dimensional algebra over a field and $J$ is its Jacobson radical. Let $p_1,\dots,p_n$ be idempotents of $A$ such that $p=p_1+\cdots+p_n$ is also idempotent Let us denote $\bar p_i$ and $\bar p$ the images of the $p_i$ and of $p$ in $A/J$, which are idempotents there. Since the algebra $A/J$ is a direct product of finite dimensional simple algebras, we know that $\bar p_i\bar p_j=0$ for all $i,j$. This means that $p_ip_j\in J$ for all $i,j$.

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    Yes, thanks. What about general finite-dimensional, say, complex algebras?2012-04-13
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    But this does not mean that $p_ip_j=0$. Can we construct a counterexample continuing in this direction?2012-04-13
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    Can you? ${}{}{}$2012-04-13