The spectrum of the integers has one generic point, closed points (p), and basic opens D(n)

By [A1], the points of $\mathrm{Spec}(\mathbb{Z})$ are $(0)$ and the prime ideals $(p)$. Since $(0)\subseteq (p)$ for every prime number $p$, fact [L1] shows that every $(p)$ is a specialization of $(0)$. Thus $(0)$ is the unique generic point.

Fact [A1] says that the maximal ideals of $\mathbb{Z}$ are exactly the ideals $(p)$, so [L2] shows that the closed points of the spectrum are exactly the points $(p)$.

For a nonzero integer $n$, fact [L3] says that $D(n)$ consists of the prime ideals that do not contain $n$. The point $(0)$ never contains a nonzero integer, and $(p)$ contains $n$ exactly when $p$ divides $n$. Hence $D(n)=\{(0)\}\cup \{(p):p\nmid n\}$.

Therefore $\mathrm{Spec}(\mathbb{Z})$ has one generic point, closed points $(p)$, and the distinguished opens described above.