# A property for primitive idempotents

Let $$R$$ be a (commutative) ring (with identity). A nonzero idempotent $$e\in R$$ is called primitive idempotent, whenever it has no decomposition into $$a+b$$ where $$a$$ and $$b$$ are nonzero orthogonal ($$ab=0$$) idempotents in $$R$$. I am looking for a characterization for rings with the property that if $$e$$ is a primitive idempotent of $$R$$ and $$e=r+s$$, where $$r, s\in R$$, then $$(e)\subseteq (r)$$ or $$(e)\subseteq (s)$$ where $$(x)$$ is the ideal generated by $$x$$ in $$R$$.

The property you are looking at is that of a local idempotent.

Let $$R$$ be a commutative ring with $$1$$. A local idempotent of $$R$$ is an idempotent $$e\in R$$ such that $$eRe = eR$$ is a local ring. Local idempotents are always primitive, but primitive idempotents need not be local (e.g., $$1\in\mathbb Z$$ is primitive but not local).

Here is the argument that your property is the same as being local.

Assume that $$e$$ is a local idempotent. Then $$R\cong eR\times (1-e)R$$, where the factor $$eR$$ is a local ring. In coordinates, $$e = (1,0)$$. Suppose that $$e=r+s$$. Writing this in coordinates yields $$(1,0)=(r_1,r_2) + (s_1,s_2)$$. Since $$r_1, s_1\in eR$$, which is local, one of $$r_1$$ or $$s_1$$ is a unit in $$eR$$. Suppose that it is $$r_1$$. Then $$e = (1,0) = (r_1,r_2)*(r_1^{-1},0) \in (r)$$, or $$(e)\subseteq (r)$$. Similarly, if $$s_1$$ is a unit in $$eR$$, then $$e\in (s)$$, or $$(e)\subseteq (s)$$.

Now suppose that $$e$$ is not local, so $$eR$$ is not a local ring. Let $$M, N\subseteq eR$$ be distinct maximal ideals. Then $$M+N=eR$$, so there exist $$m\in M$$ and $$n\in N$$ such that $$m+n=1$$ in $$eR$$. Thus, writing in coordinates in $$R \cong eR\times (1-e)R$$, we have $$(m,0) + (n,0) = (1,0) = e$$. I let $$r = (m,0)$$ and $$s=(n,0)$$, so $$r+s=e$$. But $$e=(1,0)\notin (r)$$, since $$(r) \subseteq M\times \{0\}$$ and $$(1,0)\notin M\times \{0\}$$. Similarly $$e\notin (s)$$.

You have stated I am looking for a characterization for rings with the property that if $$\ldots$$ ETC.'' One (partial?) answer is: those rings whose primitive idempotents are local. I'm not sure how much more you can say than this, but the following paper

W.K. Nicholson.
I-rings.
Trans. Amer. Math. Soc. 207 (1975), 361--373.

proves that primitive idempotents are local in any $$$$semipotent'' ring. A commutative ring is semipotent if every ideal not contained in the Jacobson radical contains a nonzero idempotent.