The Stacks project

Proof. This is a subsheaf by the local nature of the conditions. It is an $\mathcal{O}_ X$-algebra by Algebra, Lemma 10.36.7. 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 10.36.12 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 10.36.11. The same fact shows that the stalks are as advertised. $\square$

Proof. Let $\mathcal{O}' \subset f_*\mathcal{O}_ Y$ be the integral closure of $\mathcal{O}_ X$ as in Definition 29.53.3. 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 29.44.1 $\pi $ is affine, and hence $Z$ is the relative spectrum of a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras $\mathcal{B}$. The morphism $g : Y \to Z$ 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 open $U \subset X$ (by Definition 29.44.1) we see from Lemma 29.53.1 that $\chi (\mathcal{B}) \subset \mathcal{O}'$. By the functoriality of the relative spectrum Lemma 29.11.5 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.

From the universal property in (2) we see that $f'$ does not factor through a proper closed subscheme of $X'$. Hence the scheme theoretic image of $f'$ is $X'$. Since $f'$ is quasi-compact (by Schemes, Lemma 26.21.14 and the fact that $\nu $ is separated as an affine morphism) we see that $f'(Y)$ is dense in $X'$. Hence $f'$ is dominant.

Observe that $g$ is quasi-compact and quasi-separated by Schemes, Lemmas 26.21.13 and 26.21.14. Thus the last statement of the lemma makes sense. The morphism $h$ in (2) is integral by Lemma 29.44.14. 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 final assertion of the lemma. $\square$

Proof. By Lemmas 29.53.4 (1) and 29.44.6 the base change $X_2 \times _{X_1} X'_1 \to X_2$ is integral. Note that $f_2$ factors through this morphism. Hence we get a unique morphism $X'_2 \to X_2 \times _{X_1} X'_1$ from Lemma 29.53.4 (2). This gives the arrow $X'_2 \to X'_1$ fitting into the commutative diagram and uniqueness follows as well. $\square$

Proof. Clear from the construction. $\square$

Proof. If $Y \to X'' \to X'$ is the normalization of $X'$ in $Y$, then we can apply Lemma 29.53.4 to the composition $X'' \to X$ to get a canonical morphism $h : X' \to X''$ over $X$. We omit the verification that the morphisms $h$ and $X'' \to X'$ are mutually inverse (using uniqueness of the factorization in 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 29.53.6 we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Choose a finite affine open covering $Y = \bigcup \mathop{\mathrm{Spec}}(B_ i)$. Then $X' = \mathop{\mathrm{Spec}}(A')$ and the morphisms $\mathop{\mathrm{Spec}}(B_ i) \to Y \to X'$ jointly define an injective $A$-algebra map $A' \to \prod B_ i$. Thus the lemma follows from Algebra, Lemma 10.30.5. $\square$

Proof. In terms of integral closures this corresponds to the following fact: Let $A \to B$ be a ring map. Suppose that $B = B_1 \times B_2$. Let $A_ i'$ be the integral closure of $A$ in $B_ i$. Then $A_1' \times A_2'$ is the integral closure of $A$ in $B$. The reason this works is that the elements $(1, 0)$ and $(0, 1)$ of $B$ are idempotents and hence integral over $A$. Thus the integral closure $A'$ of $A$ in $B$ is a product and it is not hard to see that the factors are the integral closures $A'_ i$ as described above (some details omitted). $\square$

Proof. The question is local on $S$, hence we may assume $S = \mathop{\mathrm{Spec}}(R)$ 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 $Z \subset \mathbf{A}^1_ S$ be the scheme theoretic image of $h$, see Definition 29.6.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ S \to S$ is separated, see Schemes, Lemma 26.21.14. By Lemma 29.6.3 the morphism $X \to Z$ is dominant. By Lemma 29.41.7 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 29.41.9 to conclude that $Z \to S$ is universally closed (and even proper). Since $Z \subset \mathbf{A}^1_ S$, we see that $Z \to S$ is affine and proper, hence integral by Lemma 29.44.7. Writing $\mathbf{A}^1_ S = \mathop{\mathrm{Spec}}(R[T])$ 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 29.44.7 this is a special case of Lemma 29.53.11. $\square$

Proof. Let $U \subset X$ be an affine open. Consider the inverse image $U'$ of $U$ in $X'$. Set $V = f^{-1}(U)$. By Lemma 29.53.6 we $V \to U' \to U$ is the normalization of $U$ in $V$. Say $U = \mathop{\mathrm{Spec}}(A)$. Then $V$ is quasi-compact, and hence has a finite number of irreducible components by assumption. Hence $V = \coprod _{i = 1, \ldots n} V_ i$ is a finite disjoint union of normal integral schemes by Properties, Lemma 28.7.5. By Lemma 29.53.10 we see that $U' = \coprod _{i = 1, \ldots , n} U_ i'$, where $U'_ i$ is the normalization of $U$ in $V_ i$. By Properties, Lemma 28.7.9 we see that $B_ i = \Gamma (V_ i, \mathcal{O}_{V_ i})$ is a normal domain. Note that $U_ i' = \mathop{\mathrm{Spec}}(A_ i')$, where $A_ i' \subset B_ i$ is the integral closure of $A$ in $B_ i$, see Lemma 29.53.1. By Algebra, Lemma 10.37.2 we see that $A_ i' \subset B_ i$ is a normal domain. Hence $U' = \coprod U_ i'$ is a finite union of normal integral schemes and hence is normal.

As $X'$ has an open covering by the schemes $U'$ we conclude from Properties, Lemma 28.7.2 that $X'$ is normal. On the other hand, each $U'$ is a finite disjoint union of irreducible schemes, hence every quasi-compact open of $X'$ has finitely many irreducible components (by a topological argument which we omit). Thus $X'$ is a disjoint union of normal integral schemes by Properties, Lemma 28.7.5. It is clear from the description of $X'$ above that $Y \to X'$ is dominant and induces a bijection on irreducible components $V \to U'$ for every affine open $U \subset X$. The bijection of irreducible components for the morphism $Y \to X'$ follows from this by a topological argument (omitted). $\square$

Proof. There is an immediate reduction to the case $S = \mathop{\mathrm{Spec}}(R)$ where $R$ is a Nagata ring by assumption (1). We have to show that the integral closure $A$ of $R$ in $\Gamma (X, \mathcal{O}_ X)$ is finite over $R$. Since $f$ is quasi-compact by assumption (2) we can write $X = \bigcup _{i = 1, \ldots , n} U_ i$ with each $U_ i$ affine. Say $U_ i = \mathop{\mathrm{Spec}}(B_ i)$. Each $B_ i$ is reduced by assumption (5) and has finitely many minimal primes $\mathfrak q_{i1}, \ldots , \mathfrak q_{im_ i}$ by assumption (3) and Algebra, Lemma 10.26.1. We have

the second inclusion by Algebra, Lemma 10.25.2. We have $\kappa (\mathfrak q_{ij}) = (B_ i)_{\mathfrak q_{ij}}$ by Algebra, Lemma 10.25.1. Hence the integral closure $A$ of $R$ in $\Gamma (X, \mathcal{O}_ X)$ is contained in the product of the integral closures $A_{ij}$ of $R$ in $\kappa (\mathfrak q_{ij})$. Since $R$ is Noetherian it suffices to show that $A_{ij}$ is a finite $R$-module for each $i, j$. Let $\mathfrak p_{ij} \subset R$ be the image of $\mathfrak q_{ij}$. As $\kappa (\mathfrak q_{ij})/\kappa (\mathfrak p_{ij})$ is a finitely generated field extension by assumption (4), we see that $R \to \kappa (\mathfrak q_{ij})$ is essentially of finite type. Thus $R \to A_{ij}$ is finite by Algebra, Lemma 10.162.2. $\square$

Proof. This is a special case of Lemma 29.53.14. Namely, (2) holds as the finite type morphism $f$ is quasi-compact by definition and quasi-separated by Lemma 29.15.7. Condition (3) holds because $X$ is locally Noetherian by Lemma 29.15.6. Finally, condition (4) holds because a finite type morphism induces finitely generated residue field extensions. $\square$

Proof. We can replace $X$ by an affine neighbourhood of the image of $x'$. Hence we may assume $X = \mathop{\mathrm{Spec}}(A)$ with $A$ Nagata. By Lemma 29.53.15 the morphism $X' \to X$ is finite. Hence we can write $X' = \mathop{\mathrm{Spec}}(A')$ for a finite $A$-algebra $A'$. By Lemma 29.53.7 after replacing $X$ by $X'$ we reduce to the case described in the next paragraph.

The case $X = X' = \mathop{\mathrm{Spec}}(A)$ with $A$ Noetherian. Let $\mathfrak p \subset A$ be the prime ideal corresponding to our point $x'$. Choose $g \in \mathfrak p$ not contained in any minimal prime of $A$ (use prime avoidance and the fact that $A$ has finitely many minimal primes, see Algebra, Lemmas 10.15.2 and 10.31.6). Set $Z = f^{-1}V(g) \subset Y$; it is a closed subscheme of $Y$. Then $f(Z)$ does not contain any generic point by choice of $g$ and does not contain $x'$ because $x'$ is not in the image of $f$. The closure of $f(Z)$ is the set of specializations of points of $f(Z)$ by Lemma 29.6.5. Thus the closure of $f(Z)$ does not contain $x'$ because the condition $\dim (\mathcal{O}_{X', x'}) = 1$ implies only the generic points of $X = X'$ specialize to $x'$. In other words, after replacing $X$ by an affine open neighbourhood of $x'$ we may assume that $f^{-1}V(g) = \emptyset $. Thus $g$ maps to an invertible global function on $Y$ and we obtain a factorization

Since $X = X'$ this implies that $A$ is equal to the integral closure of $A$ in $A_ g$. By Algebra, Lemma 10.36.11 we conclude that $A_\mathfrak p$ is the integral closure of $A_\mathfrak p$ in $A_\mathfrak p[1/g]$. By our choice of $g$, since $\dim (A_\mathfrak p) = 1$ and since $A$ is reduced we see that $A_\mathfrak p[1/g]$ is a finite product of fields (the product of the residue fields of the minimal primes contained in $\mathfrak p$). Hence $A_\mathfrak p$ is normal (Algebra, Lemma 10.37.16) and the proof is complete. Some details omitted. $\square$