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An integral scheme is nonempty, reduced and irreducible. For an affine scheme $X=\mathrm{Spec}R$, reducedness is equivalent to $\sqrt{(0)}=0$: nilpotent elements vanish in every stalk, and conversely a nonzero element has nonzero localization somewhere. Irreducibility is equivalent to the nilradical being prime. Indeed, if $ab$ lies in every prime, then $V(ab)=V(a)\cup V(b)=X$; irreducibility makes one of these closed subsets all of $X$, so $a$ or $b$ lies in the nilradical. Conversely, if the nilradical is prime, its point is a generic point of the whole spectrum. Thus reducedness and irreducibility together say that zero is a prime ideal, with $R\ne 0$. Therefore

For a global idempotent $e$, its germ in any local ring is either zero or one. To see this, $e(1-e)=0$, and at least one of $e,1-e$ is a unit. Multiplication by that unit forces the other to vanish. On each affine chart,These equalities also hold globally. Hence $V(1-e)$ is a clopen set. Conversely, on a clopen subset $U$ and its complement, glue the sections one and zero. They define a global idempotent $e_U$, and the two constructions are inverse. This proves the idempotent–clopen correspondence.

Assume each connected component is open. Every clopen subset is a union of connected components, since its intersection with a connected component is clopen in that component. Each individual component is also closed, so it defines a nonzero idempotent. A decomposition into two nonempty disjoint clopen subsets corresponds exactly to a sum of two nonzero orthogonal idempotents. A union containing at least two components admits such a decomposition by separating one component from the rest. A single component admits none. Consequently the precise conclusion is

The printed definition needs both nonzero and orthogonal. Literally forbidding every sum of two nonzero idempotents gives a false statement in characteristic two. For $X=\mathrm{Spec}({\mathbb{F}}_2\times {\mathbb{F}}_2)$, the component idempotent $(1,0)$ decomposes as $(1,1)+(0,1)$; both summands are nonzero idempotents but are not orthogonal. In fact each of the three nonzero idempotents in this ring decomposes into the other two, although $X$ has two open connected components. Excluding zero is also necessary: zero corresponds to the empty subset, not to a component. With the standard primitive idempotent definition the preceding proof establishes the intended bijection in every characteristic.