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I have that $e_1,e_2, ..., e_n$ are pairwise orthogonal idempotents in a ring $R$ with unity. Then I consider the sum \begin{equation} S = \sum_{i=1}^{n}Re_i \end{equation}

So, is this a direct sum of left ideals?

It seems quite clear to me that the $Re_i$ are left ideals, but the direct sum is slightly harder for me to show. Let $x\in S$, so I write it like this \begin{equation} x = r_1 + r_2 + ... + r_n \end{equation} where $r_i \in Re_i$. From a result in my book I know that if I can show $$x=0 \Rightarrow r_i = 0 \: \forall i$$ then I have the $S$ is a direct sum.

So suppose $x=0$, but that there are some $r_i$'s which are not $0$, that instead they cancel each other out. So $r_k \in Re_k$ and $r_l \in Re_l$ and suppose $r_k = -r_l$. Since the $R$ is not necessarily commutative, the only way I can see this happening is if $r_k = e_{l}e_{k} = 0$ and $r_l = e_{k}e_{l} = 0$. If this last argument is true, then I have effectively shown that $S$ is a direct sum I think.

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    $\begingroup$ Ask yourself, what is $x\cdot e_k$? $\endgroup$ Sep 27, 2016 at 10:43

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This should be a complete set of pairwise orthogonal idempotents. In that case $1=e_1+e_2+\cdots +e_n$. So, any $x$ can be written as $$x= x.1=x.(e_1+e_2+\cdots +e_n) =x.e_1 + x.e_2+\cdots +x.e_n $$

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