= Solution
Choose a separation $\operatorname{Spec}R=U\sqcup V$ into two nonempty <clopen sets>. Both are closed in the <Zariski topology>, so write $U=V(I)$ and $V=V(J)$ for <ideals> $I,J$. Their disjointness gives $V(I+J)=\varnothing$. A proper <ideal> lies in a <maximal ideal>, so this forces $I+J=R$.
Their union is all of the <spectrum of a commutative ring>, and $V(I)\cup V(J)=V(IJ)$. Thus every <prime ideal> contains $IJ$, so
$$
IJ\subseteq\bigcap_{\mathfrak p\in\operatorname{Spec}R}\mathfrak p
=\sqrt{(0)}=0.
$$
The last equality uses that $R$ is a <reduced ring>. For <comaximal ideals>, $I\cap J=IJ$: if $i+j=1$ and $r\in I\cap J$, then $r=ri+rj\in IJ$. Therefore the <Chinese remainder theorem> gives
$$
\boxed{R\cong R/I\times R/J}.
$$
The isomorphism sends $r$ to its two residue classes. Its kernel is $I\cap J=0$; for surjectivity, $a\bmod I$ and $b\bmod J$ are obtained from $aj+bi$, where $i\in I$, $j\in J$ and $i+j=1$. The two factors are nonzero because $U,V$ are nonempty, so $I,J$ are proper.
This is a <disconnected reduced spectrum product decomposition>. The elements $i,j$ also satisfy $ij=0$, $i^2=i$, $j^2=j$, exhibiting the <ring product decomposition by an idempotent>. No Noetherian hypothesis is required.
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