## 37.19 Normalization revisited

Normalization commutes with smooth base change.

Lemma 37.19.1. Let $f : Y \to X$ be a smooth morphism of schemes. Let $\mathcal{A}$ be a quasi-coherent sheaf of $\mathcal{O}_ X$-algebras. The integral closure of $\mathcal{O}_ Y$ in $f^*\mathcal{A}$ is equal to $f^*\mathcal{A}'$ where $\mathcal{A}' \subset \mathcal{A}$ is the integral closure of $\mathcal{O}_ X$ in $\mathcal{A}$.

Proof. This is a translation of Algebra, Lemma 10.147.4 into the language of schemes. Details omitted. $\square$

$\xymatrix{ Y_2 \ar[r] \ar[d]_{f_2} & Y_1 \ar[d]^{f_1} \\ X_2 \ar[r]^\varphi & X_1 }$

be a fibre square in the category of schemes. Assume $f_1$ is quasi-compact and quasi-separated, and $\varphi$ is smooth. Let $Y_ i \to X_ i' \to X_ i$ be the normalization of $X_ i$ in $Y_ i$. Then $X_2' \cong X_2 \times _{X_1} X_1'$.

Proof. The base change of the factorization $Y_1 \to X_1' \to X_1$ to $X_2$ is a factorization $Y_2 \to X_2 \times _{X_1} X_1' \to X_2$ and $X_2 \times _{X_1} X_1' \to X_2$ is integral (Morphisms, Lemma 29.44.6). Hence we get a morphism $h : X_2' \to X_2 \times _{X_1} X_1'$ by the universal property of Morphisms, Lemma 29.53.4. Observe that $X_2'$ is the relative spectrum of the integral closure of $\mathcal{O}_{X_2}$ in $f_{2, *}\mathcal{O}_{Y_2}$. If $\mathcal{A}' \subset f_{1, *}\mathcal{O}_{Y_1}$ denotes the integral closure of $\mathcal{O}_{X_1}$, then $X_2 \times _{X_1} X_1'$ is the relative spectrum of $\varphi ^*\mathcal{A}'$, see Constructions, Lemma 27.4.6. By Cohomology of Schemes, Lemma 30.5.2 we know that $f_{2, *}\mathcal{O}_{Y_2} = \varphi ^*f_{1, *}\mathcal{O}_{Y_1}$. Hence the result follows from Lemma 37.19.1. $\square$

Lemma 37.19.3 (Normalization and smooth morphisms). Let $X \to Y$ be a smooth morphism of schemes. Assume every quasi-compact open of $Y$ has finitely many irreducible components. Then the same is true for $X$ and there is a unique isomorphism $X^\nu = X \times _ Y Y^\nu$ over $X$ where $X^\nu$, $Y^\nu$ are the normalizations of $X$, $Y$.

Proof. By Descent, Lemma 35.16.3 every quasi-compact open of $X$ has finitely many irreducible components. Note that $X_{red} = X \times _ Y Y_{red}$ as a scheme smooth over a reduced scheme is reduced, see Descent, Lemma 35.18.1. Hence we may assume that $X$ and $Y$ are reduced (as the normalization of a scheme is equal to the normalization of its reduction by definition). Next, note that $X' = X \times _ Y Y^\nu$ is a normal scheme by Descent, Lemma 35.18.2. The morphism $X' \to Y^\nu$ is smooth (hence flat) thus the generic points of irreducible components of $X'$ lie over generic points of irreducible components of $Y^\nu$. Since $Y^\nu \to Y$ is birational we conclude that $X' \to X$ is birational too (because $X' \to Y^\nu$ induces an isomorphism on fibres over generic points of $Y$). We conclude that there exists a factorization $X^\nu \to X' \to X$, see Morphisms, Lemma 29.54.5 which is an isomorphism as $X'$ is normal and integral over $X$. $\square$

Lemma 37.19.4 (Normalization and henselization). Let $X$ be a locally Noetherian scheme. Let $\nu : X^\nu \to X$ be the normalization morphism. Then for any point $x \in X$ the base change

$X^\nu \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^ h) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^ h), \quad \text{resp.}\quad X^\nu \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh}) \to \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$

is the normalization of $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^ h)$, resp. $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$.

Proof. Let $\eta _1, \ldots , \eta _ r$ be the generic points of the irreducible components of $X$ passing through $x$. The base change of the normalization to $\mathop{\mathrm{Spec}}(\mathcal{O}_{X, x})$ is the spectrum of the integral closure of $\mathcal{O}_{X, x}$ in $\prod \kappa (\eta _ i)$. This follows from our construction of the normalization of $X$ in Morphisms, Definition 29.54.1 and Morphisms, Lemma 29.53.1; you can also use the description of the normalization in Morphisms, Lemma 29.54.3. Thus we reduce to the following algebra problem. Let $A$ be a Noetherian local ring; recall that this implies the henselization $A^ h$ and strict henselization $A^{sh}$ are Noetherian too (More on Algebra, Lemma 15.45.3). Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be its minimal primes. Let $A'$ be the integral closure of $A$ in $\prod \kappa (\mathfrak p_ i)$. Problem: show that $A' \otimes _ A A^ h$, resp. $A' \otimes _ A A^{sh}$ is constructed from the Noetherian local ring $A^ h$, resp. $A^{sh}$ in the same manner.

Since $A^ h$, resp. $A^{sh}$ are colimits of étale $A$-algebras, we see that the minimal primes of $A$ and $A^{sh}$ are exactly the primes of $A^ h$, resp. $A^{sh}$ lying over the minimal primes of $A$ (by going down, see Algebra, Lemmas 10.39.19 and 10.30.7). Thus More on Algebra, Lemma 15.45.13 tells us that $A^ h \otimes _ A \prod \kappa (\mathfrak p_ i)$, resp. $A^{sh} \otimes _ A \prod \kappa (\mathfrak p_ i)$ is the product of the residue fields at the minimal primes of $A^ h$, resp. $A^{sh}$. We know that taking the integral closure in an overring commutes with étale base change, see Algebra, Lemma 10.147.2. Writing $A^ h$ and $A^{sh}$ as a limit of étale $A$-algebras we see that the same thing is true for the base change to $A^ h$ and $A^{sh}$ (you can also use the more general Algebra, Lemma 10.147.5). $\square$

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