Lemma 29.51.1. Let $X$ be a scheme. Let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras. The subsheaf $\mathcal{A}' \subset \mathcal{A}$ defined by the rule

$U \longmapsto \{ f \in \mathcal{A}(U) \mid f_ x \in \mathcal{A}_ x \text{ integral over } \mathcal{O}_{X, x} \text{ for all }x \in U\}$

is a quasi-coherent $\mathcal{O}_ X$-algebra, the stalk $\mathcal{A}'_ x$ is the integral closure of $\mathcal{O}_{X, x}$ in $\mathcal{A}_ x$, and for any affine open $U \subset X$ the ring $\mathcal{A}'(U) \subset \mathcal{A}(U)$ is the integral closure of $\mathcal{O}_ X(U)$ in $\mathcal{A}(U)$.

Proof. This is a subsheaf by the local nature of the conditions. It is an $\mathcal{O}_ X$-algebra by Algebra, Lemma 10.35.7. Let $U \subset X$ be an affine open. Say $U = \mathop{\mathrm{Spec}}(R)$ and say $\mathcal{A}$ is the quasi-coherent sheaf associated to the $R$-algebra $A$. Then according to Algebra, Lemma 10.35.12 the value of $\mathcal{A}'$ over $U$ is given by the integral closure $A'$ of $R$ in $A$. This proves the last assertion of the lemma. To prove that $\mathcal{A}'$ is quasi-coherent, it suffices to show that $\mathcal{A}'(D(f)) = A'_ f$. This follows from the fact that integral closure and localization commute, see Algebra, Lemma 10.35.11. The same fact shows that the stalks are as advertised. $\square$

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