Bourbaki

Exercise 12, § 8

Algebra, I, 173

Let AA be a ring and (ai)i∈I(a_i)_{i \in I} a finite family of left ideals of AA such that the additive group of AA is the direct sum of the additive groups aia_i. If 1=∑i∈Iei(ei∈ai)1 = \sum_{i \in I} e_i (e_i \in a_i), then ei2=ei,eiej=0e_i^2 = e_i, e_i e_j = 0 if i≠ji \neq j and ai=Aeia_i = Ae_i (write x=x.1x = x . 1 for all x∈Ax \in A). Conversely, if (ei)i∈I(e_i)_{i \in I} is a finite family of idempotents of AA such that eiej=0e_i e_j = 0 for i≠ji \neq j and 1=∑iei1 = \sum_i e_i, then AA is the direct sum of the left ideals AeiAe_i. For the ideals AeiAe_i to be two-sided, it is necessary and sufficient that the eie_i be in the centre of AA.

Solution

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Transcribed from the printed volume. Not translated.