Proposition 14

Let $A$ be a ring. For a subset $Y$ of $X = \mathrm{Spec}(A)$ to be irreducible, it is necessary and sufficient that the ideal $\mathfrak{J}(Y)$ be prime.

Writing $p = \mathfrak{J}(Y)$, we note that, for an element $f \in A$, the relation $f \in p$ is equivalent to $Y \subset V(f)$. Suppose that $Y$ is irreducible and let $f, g$ be elements of $A$ such that $fg \in p$. Then

as $Y$ is irreducible and $V(f)$ and $V(g)$ are closed, $Y \subset V(f)$ or $Y \subset V(g)$, hence $f \in p$ or $g \in p$, which proves that $p$ is prime.

Suppose now that $p$ is prime; then $\overline{Y} = V(p)$ (Proposition 11 (ii)) and, as $p$ is prime, $p = \mathfrak{J}(\{p\})$, whence $\overline{Y} = V(\mathfrak{J}(\{p\})) = \{p\}$ (Proposition 11 (ii)). As a set consisting of a single point is irreducible, $Y$ is irreducible (no. 1, Proposition 2).