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Let $A$ be a finite dimensional $\mathbb C$-algebra. Let $e_1,\ldots,e_r\in A$ be nonzero idempotents (with $r>0$), i.e. $e_i^2=e_i$. My question is: Can it happen that $e_1+\cdots+e_r=0$? I can't think of a single example.

Note: I do not require the $e_i$ to be central, primitive, or orthogonal.

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Example: $e_1=e_2=\dots=e_r=0.$ –  NikolajK Jan 29 '13 at 12:34
    
Ok, I inserted the appropriate edit ;) –  Jesko Hüttenhain Jan 29 '13 at 12:35
    
Remark: Initially I tried to prove it just from the relation and using sqaring of the whole expression, you get to $\sum_{i\ne j=1}^{r-1}e_ie_j=-2 \sum_{i=1}^{r-1}e_i$. –  NikolajK Jan 29 '13 at 14:06
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@NickKidman: the general statement of the question is false in fields of positive characteristic, for example $1+1+1+1+1=0$ in the field of five elements. So there's no purely algebraic deduction. –  Colin McQuillan Jan 29 '13 at 14:07
    
Hint: Embed $A$ in $L(A)\simeq M_n(\mathbb{C})$. Then the trace of an idempotent is equal to its rank. –  1015 Jan 29 '13 at 14:16

2 Answers 2

up vote 4 down vote accepted

We can WLOG assume that the algebra $A$ is embedded in $\mathrm{M}_n\left(\mathbb C\right)$ for some $n\in\mathbb N$ (because the $A$-module $A$ is faithful and finite-dimensional, so that $A$ is embedded in $\mathrm{End}_{\mathbb C} A \cong \mathrm{M}_n\left(\mathbb C\right)$ for $n=\dim_{\mathbb C}A$). Then, $e_1, e_2, \dots, e_r$ are idempotent matrices, and have idempotent sum (because $0$ is idempotent). According to MathOverflow question #115067, any finite list of idempotent matrices over $\mathbb C$ (or any other field of characteristic $0$) having idempotent sum must be a list of orthogonal idempotents. Hence, your idempotents $e_1, e_2, \dots, e_r$ are orthogonal. Thus, $e_1\left(e_1+e_2+\cdots+e_r\right) = e_1e_1 + e_1e_2 + \cdots + e_1e_r = e_1 + 0 + \cdots + 0 = e_1$. Since $e_1+e_2+\cdots+e_r=0$, this rewrites as $e_1\cdot 0 = e_1$, so that $e_1 = 0$. This contradicts the assumption that $e_1, e_2, \dots, e_r$ are nonzero idempotents.

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Even easier: the trace of a non-zero idempotent matrix is a positive integer, so their sum cannot be zero. –  Colin McQuillan Jan 29 '13 at 12:51

Here is a proof which does not make use of the hypothesis that $A$ is finite-dimensional. It works over any field of characteristic $0$. Since $A$ acts on itself by left multiplication, it suffices to answer this question for idempotent endomorphisms $e_1, e_2, ... e_r$ of some vector space $V$ (not necessarily finite-dimensional). Let $V_1, V_2, ... V_r$ be the images of $e_1, e_2, \dots, e_r$. If $V_1 \cap V_2$ is nonzero, pick direct sum decompositions

$$V_1 \cong V_1' \oplus (V_1 \cap V_2)$$ $$V_2 \cong V_2' \oplus (V_1 \cap V_2)$$

so that $e_1 = e_1' + e_{1 \cap 2}$ and $e_2 = e_2' + e_{1 \cap 2}$ where $e_i'$ is the projection onto $V_i'$ and $e_{1 \cap 2}$ is the projection onto $V_1 \cap V_2$. Now $e_1 + e_2 = e_1' + e_2' + 2 e_{1 \cap 2}$ where all three idempotents appearing in this sum are orthogonal. Next, consider the intersections $V_1' \cap V_3, V_2' \cap V_3, V_1 \cap V_2 \cap V_3$, and continue decomposing the idempotents by choosing direct sum decompositions in this manner. In the end we will have written $e_1 + e_2 + \cdots + e_r$ as a sum, with positive integer coefficients, of nonzero orthogonal idempotents, and over a field of characteristic zero such a sum is clearly nonzero (e.g. because we can pick a nonzero vector in the image of each idempotent, and their sum is not sent to zero).

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