Lemma 10.26.1. Let $R$ be a ring.

1. For a prime $\mathfrak p \subset R$ the closure of $\{ \mathfrak p\}$ in the Zariski topology is $V(\mathfrak p)$. In a formula $\overline{\{ \mathfrak p\} } = V(\mathfrak p)$.

2. The irreducible closed subsets of $\mathop{\mathrm{Spec}}(R)$ are exactly the subsets $V(\mathfrak p)$, with $\mathfrak p \subset R$ a prime.

3. The irreducible components (see Topology, Definition 5.8.1) of $\mathop{\mathrm{Spec}}(R)$ are exactly the subsets $V(\mathfrak p)$, with $\mathfrak p \subset R$ a minimal prime.

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. $\square$

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