A Zariski-closed subset is irreducible exactly when its radical defining ideal is prime, and then it has a unique generic point

Suppose first that $Z$ is irreducible. By [L1], write $Z=V(\mathfrak{a})$ with $\mathfrak{a}$ radical. Let $xy\in \mathfrak{a}$. Then $V(\mathfrak{a})\subseteq V((xy))=V((x))\cup V((y))$ by [L2]. Since $Z$ is irreducible, $Z$ is contained in $V((x))$ or in $V((y))$. By [L1], this means $x\in I(Z)=\mathfrak{a}$ or $y\in I(Z)=\mathfrak{a}$. Therefore $\mathfrak{a}$ is prime.

Suppose conversely that $\mathfrak{a}$ is prime. If $Z=F_1\cup F_2$ with $F_1,F_2$ closed subsets of $Z$, then each $F_i$ is closed in the ambient spectrum, so [L1] gives radical ideals ${\mathfrak{b}}_i$ with $F_i=V({\mathfrak{b}}_i)$. Using $Z=V(\mathfrak{a})$ and [L2], one has $V(\mathfrak{a})=V({\mathfrak{b}}_1)\cup V({\mathfrak{b}}_2)=V({\mathfrak{b}}_1{\mathfrak{b}}_2)$. Since [L1] identifies $\mathfrak{a}$ as the radical defining ideal of $Z$, one has ${\mathfrak{b}}_1{\mathfrak{b}}_2\subseteq \mathfrak{a}$. Primality of $\mathfrak{a}$ then gives ${\mathfrak{b}}_1\subseteq \mathfrak{a}$ or ${\mathfrak{b}}_2\subseteq \mathfrak{a}$, so $V(\mathfrak{a})\subseteq V({\mathfrak{b}}_1)$ or $V(\mathfrak{a})\subseteq V({\mathfrak{b}}_2)$. Hence $Z=F_1$ or $Z=F_2$, and $Z$ is irreducible.

Under either condition, steps 1.1 and 1.2 show that $\mathfrak{a}$ is prime and $Z=V(\mathfrak{a})$. By [L3], the closure of $\{\mathfrak{a}\}$ is $V(\mathfrak{a})=Z$, so $\mathfrak{a}$ is a generic point of $Z$ in the sense of Specialisations, generalisations, and generic points.

If $\eta$ is another generic point of $Z$, then [L3] gives $\overline{\{\eta \}}=Z=\overline{\{\mathfrak{a}\}}$. Fact [L4] now forces $\eta =\mathfrak{a}$. Thus the generic point is unique.

Therefore a nonempty Zariski-closed subset is irreducible exactly when its radical defining ideal is prime, and in that case it has the unique generic point $\mathfrak{a}$.