A clopen decomposition of the spectrum comes from a nontrivial idempotent

By [L1], there are radical ideals $I,J⊴R$ such that $C=V(I)$ and $\mathrm{Spec}(R)\setminus C=V(J)$. Since $C$ and its complement are disjoint and cover the spectrum, [A1] gives $V(I+J)=\mathrm{\varnothing }$ and $V(IJ)=\mathrm{Spec}(R)$. If $I+J$ were proper, then either $R$ would be the zero ring, in which case $I+J=R$ anyway, or [L4] would place $I+J$ inside a maximal ideal, hence inside a prime ideal, contradicting $V(I+J)=\mathrm{\varnothing }$. Therefore $I+J=R$, and [L3] gives $I\cap J=IJ$.

Every element of $IJ$ therefore lies in every prime ideal of $R$. By [L2], $IJ\subseteq \mathrm{Nil}(R)$.

Choose $x\in I$ and $y\in J$ with $x+y=1$. Then $x(1-x)=xy\in IJ\subseteq \mathrm{Nil}(R)$ by step 2.1, so $x^n(1-x{)}^n=0$ for some $n\ge 1$. Expanding $1=(x+(1-x){)}^{2n-1}$ shows that every term is divisible by $x^n$ or by $(1-x{)}^n$, so there exist $a,b\in R$ with $1=a{x}^n+b(1-x{)}^n$. Put $e=a{x}^n$. Then $e\in I$, $1-e=b(1-x{)}^n\in J$, and $e(1-e)=ab{x}^n(1-x{)}^n=0$, so $e^2=e$.

If $\mathfrak{p}\in C=V(I)$, then $e\in I\subseteq \mathfrak{p}$, while $1-e\mathrm{\notin }\mathfrak{p}$ because otherwise $1\in \mathfrak{p}$. Thus $\mathfrak{p}\in V(e)=D(1-e)$. Conversely, if $\mathfrak{p}\in V(e)$, then $e\in \mathfrak{p}$. Since $e(1-e)=0$ and $\mathfrak{p}$ is prime, one also has $1-e\mathrm{\notin }\mathfrak{p}$, so $\mathfrak{p}\mathrm{\notin }V(J)=\mathrm{Spec}(R)\setminus C$. Hence $\mathfrak{p}\in C$. Therefore $C=V(e)=D(1-e)$.

The same argument with $1-e$ in place of $e$ gives $\mathrm{Spec}(R)\setminus C=V(1-e)=D(e)$. If $C$ is nonempty and proper, then neither $D(e)$ nor $D(1-e)$ is empty, so $e\mathrm{\ne }0$ and $e\mathrm{\ne }1$.

Thus every clopen subset of the spectrum comes from an idempotent, and a nonempty proper clopen subset comes from a nontrivial idempotent.