= Solution
An <integral scheme> is nonempty, reduced and irreducible. For an <affine scheme> $X=\operatorname{Spec}R$, 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 $R\ne0$. Therefore
$$
\boxed{\operatorname{Spec}R\text{ is integral}\iff R\text{ is an integral domain}.}
$$
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>,
$$
\boxed{V(1-e)=D(e),\qquad X\setminus V(1-e)=V(e)=D(1-e).}
$$
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
$$
\boxed{\{\text{connected components of }X\}\ \longleftrightarrow\ \{\text{primitive idempotents of }\Gamma(X,\mathcal O_X)\}.}
$$
The printed definition needs both \b[nonzero] and \b[orthogonal]. Literally forbidding every sum of two nonzero <idempotents> gives a false statement in characteristic two. For $X=\operatorname{Spec}(\mathbb F_2\times\mathbb F_2)$, 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.
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