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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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    @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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    Can you? ${}{}{}$2012-04-13