
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). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).