## 28.52 Normalization

Next, we come to the normalization of a scheme $X$. We only define/construct it when $X$ has locally finitely many irreducible components. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. Let $X^{(0)} \subset X$ be the set of generic points of irreducible components of $X$. Let

28.52.0.1
\begin{equation} \label{morphisms-equation-generic-points} f : Y = \coprod \nolimits _{\eta \in X^{(0)}} \mathop{\mathrm{Spec}}(\kappa (\eta )) \longrightarrow X \end{equation}

be the inclusion of the generic points into $X$ using the canonical maps of Schemes, Section 25.13. Note that this morphism is quasi-compact by assumption and quasi-separated as $Y$ is separated (see Schemes, Section 25.21).

Definition 28.52.1. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. We define the *normalization* of $X$ as the morphism

\[ \nu : X^\nu \longrightarrow X \]

which is the normalization of $X$ in the morphism $f : Y \to X$ (28.52.0.1) constructed above.

Any locally Noetherian scheme has a locally finite set of irreducible components and the definition applies to it. Usually the normalization is defined only for reduced schemes. With the definition above the normalization of $X$ is the same as the normalization of the reduction $X_{red}$ of $X$.

Lemma 28.52.2. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. The normalization morphism $\nu $ factors through the reduction $X_{red}$ and $X^\nu \to X_{red}$ is the normalization of $X_{red}$.

**Proof.**
Let $f : Y \to X$ be the morphism (28.52.0.1). We get a factorization $Y \to X_{red} \to X$ of $f$ from Schemes, Lemma 25.12.6. By Lemma 28.51.4 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 (28.52.0.1) constructed for $X_{red}$.
$\square$

If $X$ is reduced, then the normalization of $X$ is the same as the relative spectrum of the integral closure of $\mathcal{O}_ X$ in the sheaf of meromorphic functions $\mathcal{K}_ X$ (see Divisors, Section 30.23). Namely, $\mathcal{K}_ X = f_*\mathcal{O}_ Y$ in this case, see Divisors, Lemma 30.25.1 and its proof. We describe this here explicitly.

Lemma 28.52.3. Let $X$ be a reduced scheme such that every quasi-compact open has finitely many irreducible components. Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open. Then

$A$ has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$,

the total ring of fractions $Q(A)$ of $A$ is $Q(A/\mathfrak q_1) \times \ldots \times Q(A/\mathfrak q_ t)$,

the integral closure $A'$ of $A$ in $Q(A)$ is the product of the integral closures of the domains $A/\mathfrak q_ i$ in the fields $Q(A/\mathfrak q_ i)$, and

$\nu ^{-1}(U)$ is identified with the spectrum of $A'$ where $\nu : X^\nu \to X$ is the normalization morphism.

**Proof.**
Minimal primes correspond to irreducible components (Algebra, Lemma 10.25.1), hence we have (1) by assumption. Then $(0) = \mathfrak q_1 \cap \ldots \cap \mathfrak q_ t$ because $A$ is reduced (Algebra, Lemma 10.16.2). Then we have $Q(A) = \prod A_{\mathfrak q_ i} = \prod \kappa (\mathfrak q_ i)$ by Algebra, Lemmas 10.24.4 and 10.24.1. This proves (2). Part (3) follows from Algebra, Lemma 10.36.16, or Lemma 28.51.10. Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism

\[ \mathop{\mathrm{Spec}}\left(\prod \kappa (\mathfrak q_ i)\right) \longrightarrow \mathop{\mathrm{Spec}}(A) \]

where $f : Y \to X$ is the morphism (28.52.0.1).
$\square$

Lemma 28.52.4. Let $X$ be a scheme such that every quasi-compact open has a finite number of irreducible components. Let $\nu : X^\nu \to X$ be the normalization of $X$. Let $x \in X$. Then the following are canonically isomorphic as $\mathcal{O}_{X, x}$-algebras

the stalk $(\nu _*\mathcal{O}_{X^\nu })_ x$,

the integral closure of $\mathcal{O}_{X, x}$ in the total ring of fractions of $(\mathcal{O}_{X, x})_{red}$,

the integral closure of $\mathcal{O}_{X, x}$ in the product of the residue fields of the minimal primes of $\mathcal{O}_{X, x}$ (and there are finitely many of these).

**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 28.51.1. Alternatively, you can use Lemma 28.52.3 and the fact that normalization commutes with localization (Algebra, Lemma 10.35.11). 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 10.24.4). Thus our ring is also the integral closure of $A$ in the total ring of fractions of $A_{red}$.
$\square$

Lemma 28.52.5. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components.

The normalization $X^\nu $ is a disjoint union of integral normal schemes.

The morphism $\nu : X^\nu \to X$ is integral, surjective, and induces a bijection on irreducible components.

For any integral morphism $\alpha : X' \to X$ such that for $U \subset X$ quasi-compact open the inverse image $\alpha ^{-1}(U)$ has finitely many irreducible components and $\alpha |_{\alpha ^{-1}(U)} : \alpha ^{-1}(U) \to U$ is birational^{1} there exists a factorization $X^\nu \to X' \to X$ and $X^\nu \to X'$ is the normalization of $X'$.

For any morphism $Z \to X$ with $Z$ a normal scheme such that each irreducible component of $Z$ dominates an irreducible component of $X$ there exists a unique factorization $Z \to X^\nu \to X$.

**Proof.**
Let $f : Y \to X$ be as in (28.52.0.1). 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 28.51.13. This proves (1). Alternatively one can deduce (1) from Lemmas 28.52.2 and 28.52.3.

The morphism $\nu $ is integral by Lemma 28.51.4. By Lemma 28.51.13 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 28.42.7) 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 (28.52.0.1) constructed starting with $X'$. As $\alpha $ is 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 28.51.4. 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 28.52.2 we have $X^\nu = X_{red}^\nu $ and by Schemes, Lemma 25.12.6 $Z \to X$ 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 28.52.3 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 10.36.12). 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.
$\square$

Lemma 28.52.6. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. Let $Z_ i \subset X$, $i \in I$ be the irreducible components of $X$ endowed with the reduced induced structure. Let $Z_ i^\nu \to Z_ i$ be the normalization. Then $\coprod _{i \in I} Z_ i^\nu \to X$ is the normalization of $X$.

**Proof.**
We may assume $X$ is reduced, see Lemma 28.52.2. Then the lemma follows either from the local description in Lemma 28.52.3 or from Lemma 28.52.5 part (3) because $\coprod Z_ i \to X$ is integral and birational (as $X$ is reduced and has locally finitely many irreducible components).
$\square$

Lemma 28.52.7. Let $X$ be a reduced scheme with finitely many irreducible components. Then the normalization morphism $X^\nu \to X$ is birational.

**Proof.**
The normalization induces a bijection of irreducible components by Lemma 28.52.5. 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 28.48.1. Because $X$ and $X^\nu $ are reduced, we see that both local rings are equal to their residue fields (Algebra, Lemma 10.24.1). 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.
$\square$

Lemma 28.52.8. A finite (or even integral) birational morphism $f : X \to Y$ of integral schemes with $Y$ normal is an isomorphism.

**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.
$\square$

Lemma 28.52.9. Let $X$ be an integral, Japanese scheme. The normalization $\nu : X^\nu \to X$ is a finite morphism.

**Proof.**
Follows from the definition (Properties, Definition 27.13.1) and Lemma 28.52.3. 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.
$\square$

Lemma 28.52.10. Let $X$ be a Nagata scheme. The normalization $\nu : X^\nu \to X$ is a finite morphism.

**Proof.**
Note that a Nagata scheme is locally Noetherian, thus Definition 28.52.1 does apply. The lemma is now a special case of Lemma 28.51.14 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 28.42.5). Let $\mathop{\mathrm{Spec}}(A) = U \subset X$ be an affine open. By Lemma 28.52.3 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 27.13.6) 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.
$\square$

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