Basic open set

Remarks

The basic open sets have these properties, for $f, g \in R$:

1. $D(f) \cap D(g) = D(fg)$, since a prime ideal contains $fg$ exactly when it contains $f$ or $g$;
2. each open set $X \setminus V(E)$ equals $\bigcup_{f \in E} D(f)$, so the sets $D(f)$ form a basis of the topology;
3. $D(f) = \emptyset$ exactly when $f$ is nilpotent, and $D(f) = \operatorname{Spec} R$ exactly when $f$ is a unit;
4. the map $\operatorname{Spec} R_f \to \operatorname{Spec} R$ induced by $R \to R_f$ is a homeomorphism onto $D(f)$; see prime ideals of a localization.

Example: in $\operatorname{Spec} \Z$, the set $D(6)$ consists of $(0)$ and all $p\Z$ with $p \ne 2, 3$.

Non-example: not every open set is basic. Let $k$ be a field and $R = k[x, y]$. The set $U = \operatorname{Spec} R \setminus \set{(x, y)}$ is open, because $\set{(x,y)} = V(\set{x, y})$. If $U = D(f)$, then $f \ne 0$ and $f$ is not a unit, since $U$ is neither empty nor everything. Then an irreducible factor $g$ of $f$ generates a prime ideal $(g)$ that contains $f$, and $(g) \ne (x, y)$ because $(x,y)$ is not principal. So $(g) \notin D(f)$ but $(g) \in U$, a contradiction.