 Solution

The spectrum of a commutative ring is the setFor an ideal $I\subseteq R$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. The Zariski topology declares precisely these subsets to be closed sets. Using arbitrary subsets instead of ideals gives the same definition, because a prime ideal contains a subset exactly when it contains the ideal generated by that subset.

We verify the closed set axioms directly. First, $V(0)=\mathrm{Spec}R$ and $V(R)=\varnothing$. For any family of ideals,since containing the sum is equivalent to containing every summand. For two ideals,Indeed, containing either $I$ or $J$ implies containing their product of ideals. Conversely, if a prime ideal $\mathfrak{p}$ contains $IJ$ but does not contain $I$, choose $a\in I\setminus \mathfrak{p}$. For every $b\in J$, $ab\in \mathfrak{p}$, so the prime ideal property forces $b\in \mathfrak{p}$. Thus $J\subseteq \mathfrak{p}$.

The displayed identities establish arbitrary intersections and finite unions of closed sets, including their empty cases. Therefore these sets define a topology on $\mathrm{Spec}R$.