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The Stacks project

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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