As this question demonstrates that the sum of idempotents is idempotent iff every pairwise product is zero, for finite matrices with complex entries.

What additional restrictions do we need to put in for this to remain true in commutative ring theory? Of course the same trivial direction remains trivial, it just that the other direction , I would hazard restricts what kinds of rings we can use.

  • $\begingroup$ An idempotent in a commutative ring $R$ is the same thing as a set of connected components of Spec$(R)$; the idempotent is identically one on this set, and identically zero on its complement. Thus, if each connected component has characteristic zero (e.g. if $R$ is a $\mathbb{Q}$-algebra), then a sum of idempotents can only be an idempotent if their corresponding sets of components are disjoint, i.e. the pairwise products are all zero. $\endgroup$ – Kevin Ventullo Dec 2 '12 at 2:45
  • $\begingroup$ If some connected component has characteristic $p$, then taking the sum of the corresponding idempotent with itself $p+1$ times gives back the idempotent, so the statement is not quite true in this case. $\endgroup$ – Kevin Ventullo Dec 2 '12 at 2:46
  • $\begingroup$ The sum of finite idempotent elements in a commtative ring is an idempotnt if and only if every pairwise product is zero. $\endgroup$ – Ali Dec 2 '12 at 16:03
  • $\begingroup$ @ventullo: great, thanks. I don't really know 'Spec' language, but it seems that this is the right angle to take from what you say. $\endgroup$ – Mozibur Ullah Dec 2 '12 at 22:19
  • $\begingroup$ @taherifar: it seems that your statement is only correct if the characteristic of the ring is zero from what Ventullo says. $\endgroup$ – Mozibur Ullah Dec 2 '12 at 22:21

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