The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.25.1. 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 10.16.2. Hence $V(I)$ is not irreducible. $\square$


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