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 28.52.3. Let $X$ be a reduced scheme such that every quasi-compact open has finitely many irreducible components. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open. Then

  1. $A$ has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$,

  2. the total ring of fractions $Q(A)$ of $A$ is $Q(A/\mathfrak q_1) \times \ldots \times Q(A/\mathfrak q_ t)$,

  3. the integral closure $A'$ of $A$ in $Q(A)$ is the product of the integral closures of the domains $A/\mathfrak q_ i$ in the fields $Q(A/\mathfrak q_ i)$, and

  4. $\nu ^{-1}(U)$ is identified with the spectrum of $A'$ where $\nu : X^\nu \to X$ is the normalization morphism.

Proof. Minimal primes correspond to irreducible components (Algebra, Lemma 10.25.1), hence we have (1) by assumption. Then $(0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_ t$ because $A$ is reduced (Algebra, Lemma 10.16.2). Then we have $Q(A) = \prod A_{\mathfrak q_ i} = \prod \kappa (\mathfrak q_ i)$ by Algebra, Lemmas 10.24.4 and 10.24.1. This proves (2). Part (3) follows from Algebra, Lemma 10.36.16, or Lemma 28.51.10. Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism

\[ \mathop{\mathrm{Spec}}\left(\prod \kappa (\mathfrak q_ i)\right) \longrightarrow \mathop{\mathrm{Spec}}(A) \]

where $f : Y \to X$ is the morphism (28.52.0.1). $\square$


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