The Stacks project

Proof. Let $f : Y \to X$ be the morphism (29.54.0.1). We get a factorization $Y \to X_{red} \to X$ of $f$ from Schemes, Lemma 26.12.7. By Lemma 29.53.4 we obtain a canonical morphism $X^\nu \to X_{red}$ and that $X^\nu $ is the normalization of $X_{red}$ in $Y$. The lemma follows as $Y \to X_{red}$ is identical to the morphism (29.54.0.1) constructed for $X_{red}$. $\square$

Proof. Minimal primes correspond to irreducible components (Algebra, Lemma 10.26.1), hence we have (1) by assumption. Then $(0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_ t$ because $A$ is reduced (Algebra, Lemma 10.17.2). Then we have $Q(A) = \prod A_{\mathfrak q_ i} = \prod \kappa (\mathfrak q_ i)$ by Algebra, Lemmas 10.25.4 and 10.25.1. This proves (2). Part (3) follows from Algebra, Lemma 10.37.16, or Lemma 29.53.10. Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism

where $f : Y \to X$ is the morphism (29.54.0.1). $\square$

Proof. After replacing $X$ by an affine open neighbourhood of $x$ we may assume that $X$ has finitely many irreducible components and that $x$ is contained in each of them. Then the stalk $(\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A = \mathcal{O}_{X, x}$ in the product $L$ of the residue fields of the minimal primes of $A$. This follows from the construction of the normalization and Lemma 29.53.1. Alternatively, you can use Lemma 29.54.3 and the fact that normalization commutes with localization (Algebra, Lemma 10.36.11). Since $A_{red}$ has finitely many minimal primes (because these correspond exactly to the generic points of the irreducible components of $X$ passing through $x$) we see that $L$ is the total ring of fractions of $A_{red}$ (Algebra, Lemma 10.25.4). Thus our ring is also the integral closure of $A$ in the total ring of fractions of $A_{red}$. $\square$

Proof. Let $f : Y \to X$ be as in (29.54.0.1). The scheme $X^\nu $ is a disjoint union of normal integral schemes because $Y$ is normal and every affine open of $Y$ has finitely many irreducible components, see Lemma 29.53.13. This proves (1). Alternatively one can deduce (1) from Lemmas 29.54.2 and 29.54.3.

The morphism $\nu $ is integral by Lemma 29.53.4. By Lemma 29.53.13 the morphism $Y \to X^\nu $ induces a bijection on irreducible components, and by construction of $Y$ this implies that $X^\nu \to X$ induces a bijection on irreducible components. By construction $f : Y \to X$ is dominant, hence also $\nu $ is dominant. Since an integral morphism is closed (Lemma 29.44.7) this implies that $\nu $ is surjective. This proves (2).

Suppose that $\alpha : X' \to X$ is as in (3). It is clear that $X'$ satisfies the assumptions under which the normalization is defined. Let $f' : Y' \to X'$ be the morphism (29.54.0.1) constructed starting with $X'$. As $\alpha $ is locally birational it is clear that $Y' = Y$ and $f = \alpha \circ f'$. Hence the factorization $X^\nu \to X' \to X$ exists and $X^\nu \to X'$ is the normalization of $X'$ by Lemma 29.53.4. This proves (3).

Let $g : Z \to X$ be a morphism whose domain is a normal scheme and such that every irreducible component dominates an irreducible component of $X$. By Lemma 29.54.2 we have $X^\nu = X_{red}^\nu $ and by Schemes, Lemma 26.12.7 $Z \to X$ factors through $X_{red}$. Hence we may replace $X$ by $X_{red}$ and assume $X$ is reduced. Moreover, as the factorization is unique it suffices to construct it locally on $Z$. Let $W \subset Z$ and $U \subset X$ be affine opens such that $g(W) \subset U$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $W = \mathop{\mathrm{Spec}}(B)$, with $g|_ W$ given by $\varphi : A \to B$. We will use the results of Lemma 29.54.3 freely. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the minimal primes of $A$. As $Z$ is normal, we see that $B$ is a normal ring, in particular reduced. Moreover, by assumption any minimal prime $\mathfrak q \subset B$ we have that $\varphi ^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence if $x \in A$ is a nonzerodivisor, i.e., $x \not\in \bigcup \mathfrak p_ i$, then $\varphi (x)$ is a nonzerodivisor in $B$. Thus we obtain a canonical ring map $Q(A) \to Q(B)$. As $B$ is normal it is equal to its integral closure in $Q(B)$ (see Algebra, Lemma 10.37.12). Hence we see that the integral closure $A' \subset Q(A)$ of $A$ maps into $B$ via the canonical map $Q(A) \to Q(B)$. Since $\nu ^{-1}(U) = \mathop{\mathrm{Spec}}(A')$ this gives the canonical factorization $W \to \nu ^{-1}(U) \to U$ of $\nu |_ W$. We omit the verification that it is unique. $\square$

Proof. We may assume $X$ is reduced, see Lemma 29.54.2. Then the lemma follows either from the local description in Lemma 29.54.3 or from Lemma 29.54.5 part (3) because $\coprod Z_ i \to X$ is integral and locally birational (as $X$ is reduced and has locally finitely many irreducible components). $\square$

Proof. The normalization induces a bijection of irreducible components by Lemma 29.54.5. Let $\eta \in X$ be a generic point of an irreducible component of $X$ and let $\eta ^\nu \in X^\nu $ be the generic point of the corresponding irreducible component of $X^\nu $. Then $\eta ^\nu \mapsto \eta $ and to finish the proof we have to show that $\mathcal{O}_{X, \eta } \to \mathcal{O}_{X^\nu , \eta ^\nu }$ is an isomorphism, see Definition 29.50.1. Because $X$ and $X^\nu $ are reduced, we see that both local rings are equal to their residue fields (Algebra, Lemma 10.25.1). On the other hand, by the construction of the normalization as the normalization of $X$ in $Y = \coprod \mathop{\mathrm{Spec}}(\kappa (\eta ))$ we see that we have $\kappa (\eta ) \subset \kappa (\eta ^\nu ) \subset \kappa (\eta )$ and the proof is complete. $\square$

Proof. Let $V \subset Y$ be an affine open with inverse image $U \subset X$ which is an affine open too. Since $f$ is a birational morphism of integral schemes, the homomorphism $\mathcal{O}_ Y(V) \to \mathcal{O}_ X(U)$ is an injective map of domains which induces an isomorphism of fraction fields. As $Y$ is normal, the ring $\mathcal{O}_ Y(V)$ is integrally closed in the fraction field. Since $f$ is finite (or integral) every element of $\mathcal{O}_ X(U)$ is integral over $\mathcal{O}_ Y(V)$. We conclude that $\mathcal{O}_ Y(V) = \mathcal{O}_ X(U)$. This proves that $f$ is an isomorphism as desired. $\square$

Proof. If $\nu $ is an isomorphism, then $X$ is normal by Lemma 29.54.5. Conversely, suppose $X$ is normal. By Lemma 29.53.6 and Properties, Lemma 28.7.5 we may assume $X$ is integral. By Lemma 29.54.5 the morphism $\nu $ is integral and $X^\nu $ has a unique irreducible component (so it is an integral scheme). We conclude from Lemmas 29.54.7 and 29.54.8. $\square$

Proof. Follows from the definition (Properties, Definition 28.13.1) and Lemma 29.54.3. Namely, in this case the lemma says that $\nu ^{-1}(\mathop{\mathrm{Spec}}(A))$ is the spectrum of the integral closure of $A$ in its field of fractions. $\square$

Proof. Note that a Nagata scheme is locally Noetherian, thus Definition 29.54.1 does apply. The lemma is now a special case of Lemma 29.53.14 but we can also prove it directly as follows. Write $X^\nu \to X$ as the composition $X^\nu \to X_{red} \to X$. As $X_{red} \to X$ is a closed immersion it is finite. Hence it suffices to prove the lemma for a reduced Nagata scheme (by Lemma 29.44.5). Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open. By Lemma 29.54.3 we have $\nu ^{-1}(U) = \mathop{\mathrm{Spec}}(\prod A_ i')$ where $A_ i'$ is the integral closure of $A/\mathfrak q_ i$ in its fraction field. As $A$ is a Nagata ring (see Properties, Lemma 28.13.6) each of the ring extensions $A/\mathfrak q_ i \subset A'_ i$ are finite. Hence $A \to \prod A'_ i$ is a finite ring map and we win. $\square$

Proof. We have to show that $\nu $ is integral, universally injective, and surjective, see Lemma 29.45.5. By Lemma 29.54.5 the morphism $\nu $ is integral. Let $x \in X$ and set $A = \mathcal{O}_{X, x}$. Since $X$ is irreducible we see that $A$ has a single minimal prime $\mathfrak p$ and $A_{red} = A/\mathfrak p$. By Lemma 29.54.4 the stalk $A' = (\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A$ in the fraction field of $A_{red}$. By More on Algebra, Definition 15.106.1 we see that $A'$ has a single prime $\mathfrak m'$ lying over $\mathfrak m_ x \subset A$ and $\kappa (\mathfrak m')/\kappa (x)$ is purely inseparable. Hence $\nu $ is bijective (hence surjective) and universally injective by Lemma 29.10.2. $\square$