The Stacks project

Proof. Let $U$ be an object of $X_{\acute{e}tale}$. Then $U$ is a scheme. Denote $\mathcal{A}|_ U$ the restriction to the Zariski site. Then $\mathcal{A}|_ U$ is a quasi-coherent sheaf of $\mathcal{O}_ U$-algebras hence we can apply Morphisms, Lemma 29.53.1 to find a quasi-coherent subalgebra $\mathcal{A}'_ U \subset \mathcal{A}|_ U$ such that the value of $\mathcal{A}'_ U$ on any affine open $W \subset U$ is as given in the statement of the lemma. If $f : U' \to U$ is a morphism in $X_{\acute{e}tale}$, then $\mathcal{A}|_{U'} = f^*(\mathcal{A}|_ U)$ where $f^*$ means pullback by the morphism $f$ in the Zariski topology; this holds because $\mathcal{A}$ is quasi-coherent (see introduction to Properties of Spaces, Section 66.29 and the references to the discussion in the chapter on descent on schemes). Since $f$ is étale we find that More on Morphisms, Lemma 37.19.1 says that we get a canonical isomorphism $f^*(\mathcal{A}'_ U) = \mathcal{A}'_{U'}$. This immediately tells us that we obtain a sub presheaf $\mathcal{A}' \subset \mathcal{A}$ of $\mathcal{O}_ X$-algebras over $X_{\acute{e}tale}$ which is a sheaf for the Zariski topology and has the right values on affine objects. But the fact that each $\mathcal{A}'_ U$ is quasi-coherent on the scheme $U$ and that for $f : U' \to U$ étale we have $\mathcal{A}'_{U'} = f^*(\mathcal{A}'_ U)$ implies that $\mathcal{A}'$ is quasi-coherent on $X_{\acute{e}tale}$ as well (as this is a local property and we have the references above describing quasi-coherent modules on $U_{\acute{e}tale}$ in exactly this manner). $\square$

Proof. It is immediate from the construction that the formation of the normalization of $X$ in $Y$ commutes with étale base change, i.e., part (1) holds. On the other hand, if $X$ and $Y$ are schemes, then for $U \subset X$ affine open, $f_*\mathcal{O}_ Y(U) = \mathcal{O}_ Y(f^{-1}(U))$ and hence $\nu ^{-1}(U)$ is the spectrum of exactly the same ring as we get in the corresponding construction for schemes. $\square$

Proof. Let $\mathcal{O}' \subset f_*\mathcal{O}_ Y$ be the integral closure of $\mathcal{O}_ X$ as in Definition 67.48.3. The morphism $\nu $ is integral by construction, which proves (1). Assume given a factorization $f = \pi \circ g$ with $\pi : Z \to X$ integral as in (2). By Definition 67.45.2 $\pi $ is affine, and hence $Z$ is the relative spectrum of a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras $\mathcal{B}$. The morphism $g : X \to Z$ corresponds to a map of $\mathcal{O}_ X$-algebras $\chi : \mathcal{B} \to f_*\mathcal{O}_ Y$. Since $\mathcal{B}(U)$ is integral over $\mathcal{O}_ X(U)$ for every affine $U$ étale over $X$ (by Definition 67.45.2) we see from Lemma 67.48.1 that $\chi (\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 67.20.7 this provides us with a unique morphism $h : X' \to Z$. We omit the verification that the diagram commutes.

It is clear that (1) and (2) characterize the factorization $f = \nu \circ f'$ since it characterizes it as an initial object in a category. The morphism $h$ in (2) is integral by Lemma 67.45.12. Given a factorization $g = \pi ' \circ g'$ with $\pi ' : Z' \to Z$ integral, we get a factorization $f = (\pi \circ \pi ') \circ g'$ and we get a morphism $h' : X' \to Z'$. Uniqueness implies that $\pi ' \circ h' = h$. Hence the characterization (1), (2) applies to the morphism $h : X' \to Z$ which gives the last statement of the lemma. $\square$

Proof. This follows from the fact that a subring of a reduced ring is reduced. Some details omitted. $\square$

Proof. By Lemma 67.48.4 and the definitions, we may assume that $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $X' = \mathop{\mathrm{Spec}}(A')$ where $A'$ is the integral closure of $A$ in $\Gamma (Y, \mathcal{O}_ Y)$ and $x'$ corresponds to a minimal prime of $A'$. Choose a surjective étale morphism $V \to Y$ where $V = \mathop{\mathrm{Spec}}(B)$ is affine. Then $A' \to B$ is injective, hence every minimal prime of $A'$ is the image of a minimal prime of $B$, see Algebra, Lemma 10.30.5. The lemma follows. $\square$

Proof. Omitted. $\square$

Proof. The question is étale local on $Y$, hence we may reduce to the case where $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Let $h \in \Gamma (X, \mathcal{O}_ X)$. We have to show that $h$ satisfies a monic equation over $R$. Think of $h$ as a morphism as in the following commutative diagram

Let $Z \subset \mathbf{A}^1_ Y$ be the scheme theoretic image of $h$, see Definition 67.16.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ Y \to Y$ is separated, see Lemma 67.8.9. By Lemma 67.16.3 the morphism $X \to Z$ has dense image on underlying topological spaces. By Lemma 67.40.6 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 67.40.7 to conclude that $Z \to Y$ is universally closed (and even proper). Since $Z \subset \mathbf{A}^1_ Y$, we see that $Z \to Y$ is affine and proper, hence integral by Lemma 67.45.7. Writing $\mathbf{A}^1_ Y = \mathop{\mathrm{Spec}}(R[T])$ we conclude that the ideal $I \subset R[T]$ of $Z$ contains a monic polynomial $P(T) \in R[T]$. Hence $P(h) = 0$ and we win. $\square$

Proof. By Lemma 67.45.7 this is a special case of Lemma 67.48.9. $\square$

Proof. The question is étale local on $Y$, see Lemma 67.48.4. Thus we may assume $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Then $R$ is a Noetherian Nagata ring and we have to show that the integral closure of $R$ in $\Gamma (X, \mathcal{O}_ X)$ is finite over $R$. Since $f$ is quasi-compact we see that $X$ is quasi-compact. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$ (Properties of Spaces, Lemma 66.6.3). Then $\Gamma (X, \mathcal{O}_ X) \subset \Gamma (U, \mathcal{O}_ X)$. Since $R$ is Noetherian it suffices to show that the integral closure of $R$ in $\Gamma (U, \mathcal{O}_ U)$ is finite over $R$. As $U \to Y$ is of finite type this follows from Morphisms, Lemma 29.53.15. $\square$