The Stacks project

Proof. Note that if $\mathfrak p \in V(I)$, then $I \subset \mathfrak p$. Hence, clearly $\overline{\{ \mathfrak p\} } = V(\mathfrak p)$. In particular $V(\mathfrak p)$ is the closure of a singleton and hence irreducible. The second assertion implies the third. To show the second, let $V(I) \subset \mathop{\mathrm{Spec}}(R)$ with $I$ a radical ideal. If $I$ is not prime, then choose $a, b\in R$, $a, b\not\in I$ with $ab\in I$. In this case $V(I, a) \cup V(I, b) = V(I)$, but neither $V(I, b) = V(I)$ nor $V(I, a) = V(I)$, by Lemma 10.17.2. Hence $V(I)$ is not irreducible. $\square$