 Disconnected reduced spectrum product decomposition

For a reduced ring with a nonempty disconnected spectrum of a commutative ring, choose a separation $V(I)\bigsqcup V(J)$. Disjointness gives $I+J=R$; covering the spectrum gives $IJ\subseteq \sqrt{(0)}=0$. For comaximal ideals, $I\cap J=IJ$. The Chinese remainder theorem then supplies a product of two nonzero quotient rings. Equivalently, if $i+j=1$ with $i\in I$, $j\in J$, then $i,j$ are nontrivial orthogonal idempotents. This connects topology of the spectrum to ring product decomposition by an idempotent without a Noetherian assumption.