= Solution
{c}
The <spectrum of a commutative ring> is the set
$$
\operatorname{Spec}R=\{\mathfrak p\subsetneq R:\mathfrak p\text{ is a prime ideal}\}.
$$
For 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, $V(0)=\operatorname{Spec}R$ and $V(R)=\varnothing$. For any family of <ideals>,
$$
\bigcap_{\lambda}V(I_\lambda)=V\left(\sum_{\lambda}I_\lambda\right),
$$
since containing the sum is equivalent to containing every summand. For two <ideals>,
$$
V(I)\cup V(J)=V(IJ).
$$
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 \b[these sets define a topology on $\operatorname{Spec}R$].
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