Zariski topology

Remarks

These sets are the closed sets of a topology:

1. $V(\set{0}) = X$ and $V(\set{1}) = \emptyset$;
2. $\bigcap_i V(E_i) = V\left(\bigcup_i E_i\right)$ for any family of subsets $E_i$;
3. $V(I) \cup V(J) = V(IJ)$ for ideals $I, J$, because a prime ideal contains $IJ$ only if it contains $I$ or $J$.

Also $V(E) = V(I)$ for the ideal $I$ generated by $E$, and $V(I) = V(\sqrt I)$. The closure of a point $\mathfrak p$ is $V(\mathfrak p)$, so $\set{\mathfrak p}$ is closed exactly when $\mathfrak p$ is a maximal ideal.

Example: in $\operatorname{Spec} \Z$, the closed sets are $\operatorname{Spec} \Z$ itself and the finite sets of nonzero prime ideals. Indeed $V(n\Z)$ for $n \ne 0$ is the finite set of $p\Z$ with $p \mid n$.

Non-example: $\operatorname{Spec} \Z$ with this topology is not Hausdorff. The point $(0)$ is not closed: its closure is $V(0)$, the whole space. Any nonempty open set contains $(0)$, so no two nonempty open sets are disjoint.

The open sets $D(f) = X \setminus V(\set{f})$ are the basic open sets and form a basis of the topology.