Éléments de Géométrie Algébrique — English

§23. Japanese rings

The results of this section will be completed in (IV, 7.6) and (7.7).

23.1. Japanese rings

Definition (23.1.1).

We say that an integral ring $A$ is a Japanese ring if, for every finite extension $K^{\prime }$ of its field of fractions $K$, the integral closure $A^{\prime }$ of $A$ in $K^{\prime }$ is an $A$-module of finite type (in other words, a finite $A$-algebra). We say that a ring $A$ is universally Japanese if every integral $A$-algebra of finite type is a Japanese ring.

Translator's note. "Japanese ring" is EGA's vocabulary, following Akizuki's circle and Nagata's early papers. The modern literature usually calls these "Nagata rings" or, with a slightly different condition, "pseudo-geometric rings" (Matsumura). We preserve EGA's term throughout.

It is clear that if $A$ is a Japanese ring, then every ring of fractions $S^{-1}A$ is also a Japanese ring, since (with the preceding notation) $S^{-1}{A}^{\prime }$ is the integral closure of $S^{-1}A$ in $K^{\prime }$, and is evidently an $S^{-1}A$-module of finite type.

An integral Noetherian ring, even a discrete valuation ring, is not necessarily a Japanese ring [30].

Proposition (23.1.2).

Let $A$ be an integral Noetherian ring, $K$ its field of fractions. If, for every finite radicial extension $K^{\prime }$ of $K$, the integral closure of $A$ in $K^{\prime }$ is an $A$-module of finite type, then $A$ is a Japanese ring.

Since $A$ is Noetherian, in order to verify that $A$ is a Japanese ring, it suffices to see that for every finite quasi-Galois extension $L$ of $K$, the integral closure $B$ of $A$ in $L$ is an $A$-module of finite type. Now $L$ is a Galois extension of the greatest radicial extension $K^{\prime }$ of $K$ contained in $L$; and if $A^{\prime }$ is the integral closure of $A$ in $K^{\prime }$, then $B$ is the integral closure of $A^{\prime }$ in $L$. But we know that in a separable extension of $K^{\prime }$, the integral closure of $A^{\prime }$ is an $A^{\prime }$-module of finite type (Bourbaki, Alg. comm., chap. V, §1, n° 6, cor. 1 of prop. 20), whence the proposition.

It follows from (23.1.2) that when $K$ is of characteristic 0, to say that $A$ is a Japanese ring means that its integral closure $A^{\prime }$ is an $A$-module of finite type.

Theorem (23.1.3) (Tate).

Let $A$ be an integral Noetherian ring, $x\ne 0$ an element of $A$. Suppose the following conditions are satisfied:

(i) $A$ is integrally closed.

(ii) $\mathfrak{p}=xA$ is prime and $A$ is separated and complete for the $\mathfrak{p}$-preadic topology.

(iii) $A/xA$ is a Japanese ring.

Then $A$ is a Japanese ring.

We shall use the following lemma:

Lemma (23.1.3.1).

Let $A$ be a ring, $x$ an element of $A$ not a zero-divisor in $A$ and such that $xA=\mathfrak{p}$ is prime; then, for every integer $n>0$, the inverse image of $x^n{A}_{\mathfrak{p}}$ in $A$ is $x^{n}A$.

Indeed, suppose that $b\in A$ is an element such that $b/1=x^{n}a/s$ in $A_{\mathfrak{p}}$, where $a\in A$ and $s\notin \mathfrak{p}$; there exists therefore $s^{\prime }\notin \mathfrak{p}$ such that $s^{\prime }sb=x^{n}as$, whence $s^{\prime }sb\in \mathfrak{p}$, and since $s^{\prime }s\notin \mathfrak{p}$, this entails $b\in \mathfrak{p}$, in other words $b=x{b}^{\prime }$ with $b^{\prime }\in A$; since $x$ is not a zero-divisor, we conclude $s^{\prime }s{b}^{\prime }=x^{n-1}as$ and it suffices to reason by induction on $n$.

To prove the theorem, we may restrict to the case where the field of fractions $K$ of $A$ is of characteristic $p>0$, since $A$ is integrally closed. Let $K^{\prime }$ be a finite radicial extension of $K$, so that there exists a power $q=p^e$ such that $K^{\prime q}\subset K$; by replacing $K^{\prime }$ by a larger radicial extension, we may even suppose that there exists $y\in K^{\prime }$ such that $y^q=x$. Let $A^{\prime }$ be the integral closure of $A$ in $K^{\prime }$; since $A$ is integrally closed, $A^{\prime }$ is the set of $x^{\prime }\in K^{\prime }$ such that $x^{\prime q}\in A^{\prime }\cap K=A$. Set $V=A_{\mathfrak{p}}$; the maximal ideal $\mathfrak{m}=xV$ of $V$ being principal, we know that the Noetherian local ring $V$ is a discrete valuation ring (17.1.4); since $V$ is integrally closed, the same

reasoning as above shows that the integral closure $V^{\prime }$ of $V$ in $K^{\prime }$ is the set of $x^{\prime }\in K^{\prime }$ such that $x^{\prime q}\in V$; we know (Bourbaki, Alg. comm., chap. VI, §8, n° 6, prop. 6, and chap. V, §2, n° 3, lemma 4) that $V^{\prime }$ is a discrete valuation ring, whose maximal ideal ${\mathfrak{m}}^{\prime }$ is the set of $x^{\prime }\in K^{\prime }$ such that $x^{\prime q}\in \mathfrak{m}$, and moreover the residue field $V^{\prime }/{\mathfrak{m}}^{\prime }$ is a finite extension of $V/\mathfrak{m}$ (loc. cit., chap. VI, §8, n° 1, lemma 2). Let us show that, for every integer $n>0$, we have

$${\mathfrak{m}}^{\prime n}\cap A^{\prime }=y^n{A}^{\prime }.(\mathrm{23.1.3.2})$$

Indeed, since $y^q=x$, we have $y^q\in {\mathfrak{m}}^{\prime }$, hence $y^n{A}^{\prime }\subset {\mathfrak{m}}^{\prime n}\cap A^{\prime }$. Conversely, let $x^{\prime }\in {\mathfrak{m}}^{\prime n}\cap A^{\prime }$, and set $x^{\prime }=z^{\prime }{y}^n$ with $z^{\prime }\in K^{\prime }$. We have $x^{\prime q}\in A^{\prime }$; on the other hand, $x^{\prime q}$ is a sum of products $t_1^{\prime }\cdots t_n^{\prime }$ with $t_i^{\prime }\in {\mathfrak{m}}^{\prime }$, hence $t_i^{\prime q}\in \mathfrak{m}$, and we conclude that $x^{\prime q}\in {\mathfrak{m}}^n\cap A$. Now lemma (23.1.3.1) proves that ${\mathfrak{m}}^n\cap A=x^{n}A$, hence we have $z^{\prime q}{x}^n\in x^{n}A$, or again $z^{\prime q}\in A$ since $x\ne 0$, which establishes (23.1.3.2).

Let us show in the second place that on $A^{\prime }$ the xA'-preadic topology is separated; this topology is also the yA'-preadic topology of $A^{\prime }$, and it follows from (23.1.3.2) that this topology is induced by the ${\mathfrak{m}}^{\prime }$-preadic topology of $V^{\prime }$, which is separated since $V^{\prime }$ is a discrete valuation ring.

Let us next prove that $A^{\prime }/x{A}^{\prime }$ is an $A$-module of finite type. Since $x=y^q$ and $y^k{A}^{\prime }/y^{k+1}{A}^{\prime }$ is isomorphic to $A^{\prime }/y{A}^{\prime }$, we may restrict to showing that $A^{\prime }/y{A}^{\prime }$ is an $A$-module of finite type; but formula (23.1.3.2) applied for $n=1$ shows that $A^{\prime }/y{A}^{\prime }$ is a subring of $V^{\prime }/{\mathfrak{m}}^{\prime }$, that is to say of the residue field of $V^{\prime }$, which is a finite extension of the residue field $V/\mathfrak{m}$ of $V$. Now $V/\mathfrak{m}$ is the field of fractions of $A/\mathfrak{p}$, and since $A^{\prime }$ is integral over $A$, $A^{\prime }/y{A}^{\prime }$ is integral over $A/\mathfrak{p}$, hence contained in the integral closure of $A/\mathfrak{p}$ in $V^{\prime }/{\mathfrak{m}}^{\prime }$; since by hypothesis $A/\mathfrak{p}$ is a Noetherian Japanese ring, $A^{\prime }/y{A}^{\prime }$ is an $(A/\mathfrak{p})$-module of finite type, and a fortiori an $A$-module of finite type.

This being so, the xA'-preadic topology of $A^{\prime }$ being separated, $A^{\prime }$ is a subring of its completion Â' for this topology; but since $A^{\prime }/x{A}^{\prime }$ is an $A$-module of finite type and $A$ is separated and complete for the xA-preadic topology, Â' is an $A$-module of finite type by virtue of $(0_I,\mathrm{7.2.9})$, hence so is $A^{\prime }$. Q.E.D.

Corollary (23.1.4).

Let $A$ be a Noetherian integrally closed Japanese ring. Then every ring of formal power series $A[[T_1,\cdots ,T_r]]$ is a Japanese ring.

We know that for every Noetherian integrally closed ring $B$, the ring of formal power series B[[T]] is Noetherian and integrally closed (Bourbaki, Alg. comm., chap. V, §1, n° 4, prop. 14); we may therefore, by induction on $n$, restrict to proving that A[[T]] is a Japanese ring. Now the element $x=T$ verifies all the conditions of (23.1.3), whence the conclusion.

Theorem (23.1.5) (Nagata).

Every complete integral Noetherian local ring $A$ is a Japanese ring.

We know (19.8.8, (ii)) that there exists a subring $B$ of $A$ which is a complete regular local ring, such that $A$ is a finite $B$-algebra; it evidently suffices to prove

that $B$ is a Japanese ring, in other words, we may restrict to the case where $A$ is moreover regular. Let us reason by induction on $n=\dim (A)$, the theorem being evident for $n=0$. So suppose $n>0$, and, denoting by $\mathfrak{m}$ the maximal ideal of $A$, let $x$ be an element of $\mathfrak{m}-{\mathfrak{m}}^2$; we know (17.1.8) that $A/xA$ is a regular ring, and moreover ((17.1.7) and (16.3.4)) that $\dim (A/xA)=n-1$; furthermore, the ideal xA is closed in $A$ $(0_I,\mathrm{7.3.5})$, hence $A/xA$ is complete. By virtue of the induction hypothesis, $A/xA$ is therefore a Japanese ring, and it then follows from (23.1.3) that so is $A$.

Corollary (23.1.6).

Let $A$ be a complete integral Noetherian local ring, $K$ its field of fractions, $K^{\prime }$ a finite extension of $K$; then the integral closure $A^{\prime }$ of $A$ in $K^{\prime }$ is a complete local ring.

We already know by (23.1.5) that $A^{\prime }$ is a finite $A$-algebra, hence a complete Noetherian semi-local ring, and consequently a direct composite of local rings; but since $A^{\prime }$ is integral, it is necessarily a local ring.

Proposition (23.1.7).

Let $A$ be an integral Noetherian local ring, $K$ its field of fractions, $K^{\prime }$ a finite extension of $K$, $A^{\prime }$ the integral closure of $A$ in $K^{\prime }$. Suppose that the completion Â is reduced, and denote by $R$ its total ring of fractions.

(i) If the ring $R{\otimes }_K{K}^{\prime }$ is reduced (which will happen in particular when $K^{\prime }$ is a separable extension of $K$, $R$ being a direct composite of fields, extensions of $K$ (Bourbaki, Alg., chap. VIII, §7, n° 3, cor. 1 of th. 1)), then $A^{\prime }$ is an $A$-module of finite type.

(ii) If in particular $R$ is a separable $K$-algebra, then $A$ is a Japanese ring.

(i) Set for simplicity $A_1=\hat{A}$, $A_1^{\prime }=A^{\prime }{\otimes }_A{A}_1$, $K_1^{\prime }=K^{\prime }{\otimes }_A{A}_1$. Since A_1 is a flat $A$-module $(0_I,\mathrm{7.3.5})$, $A_1^{\prime }$ identifies with a subring of $K_1^{\prime }$ and is evidently integral over A_1. On the other hand, if we set $K_1=K{\otimes }_A{A}_1$, we may write $K_1^{\prime }=K^{\prime }{\otimes }_K{K}_1$; since $A$ is integral and A_1 is a flat $A$-module, every element $\ne 0$ of $A$ is A_1-regular $(0_I,\mathrm{6.3.4})$; since $K=S^{-1}A$, where $S=A-0$, we have $K_1=S^{-1}{A}_1$, and the preceding remark proves that K_1 identifies with a subring of the total ring of fractions $R$ of A_1. Since every $K$-module is flat, $K_1^{\prime }$ identifies with a subring of $K^{\prime }{\otimes }_{K}R$, which by hypothesis is reduced and is a finite $R$-algebra. Denote by ${\mathfrak{q}}_i(1\le i\le n)$ the minimal prime ideals of A_1, by $B_i$ the integral ring $A_1/{\mathfrak{q}}_i$, by $L_i$ the field of fractions of $B_i$, so that $R$ is a direct composite of the $L_i$, and $K^{\prime }{\otimes }_{K}R$ a direct composite of the $K^{\prime }{\otimes }_K{L}_i$; these latter $K$-algebras, being reduced, are direct composites of fields, finite extensions of the $L_i$. Since $B_i$ is a complete integral local ring, Nagata's theorem (23.1.5) proves that the integral closure of $B_i$ in $K^{\prime }{\otimes }_K{L}_i$ is a $B_i$-module of finite type, and a fortiori an A_1-module of finite type; since every element of $K^{\prime }{\otimes }_{K}R$ which is integral over A_1 is also integral over $B_i$ (being annihilated by ${\mathfrak{q}}_i$), we finally conclude that the integral closure of A_1 in $K^{\prime }{\otimes }_{K}R$ is an A_1-module of finite type. But since $A_1^{\prime }$ is contained in this integral closure and A_1 is Noetherian, $A_1^{\prime }$ is also an A_1-module of finite type. Finally, since A_1 is a faithfully flat $A$-module $(0_I,\mathrm{7.3.5})$, we conclude that $A^{\prime }$ is an $A$-module of finite type (Bourbaki, Alg. comm., chap. I, §3, n° 6, prop. 11).

(ii) The hypothesis entails that the $L_i$ are separable extensions of $K$, hence $K^{\prime }{\otimes }_{K}R$ is reduced (Bourbaki, Alg., chap. VIII, §7, n° 3, th. 1), and one may apply (i) to every finite extension $K^{\prime }$ of $K$, which proves our assertion.

23.2. Integral closure of an integral Noetherian local ring

(23.2.1)

In what follows, for every ring $A$, we shall write $A_{red}$ for short to denote the quotient of $A$ by its nilradical (so that if $X=\mathrm{Spec}(A)$, we have X_red = Spec(A_red)).

We shall say that a local ring $A$ is unibranch if the ring $A_{red}$ is integral and if the integral closure of $A_{red}$ is a local ring, which generalizes the definition given in (III, 4.3.6). We shall say that $A$ is geometrically unibranch if it is unibranch and if the residue field of the local ring (which is the integral closure of $A_{red}$) is a radicial extension of that of $A$. It is clear that an integrally closed local ring is geometrically unibranch; it follows from (23.1.6) that a complete integral Noetherian local ring is unibranch.

Lemma (23.2.2) [*].

Let $A$ be an integral ring, $A^{\prime }$ its integral closure; for the canonical morphism $f:\mathrm{Spec}(A^{\prime })\to \mathrm{Spec}(A)$ to be radicial, it is necessary and sufficient that for every $\mathfrak{p}\in \mathrm{Spec}(A)$, $A_{\mathfrak{p}}$ be geometrically unibranch; when this is the case, $f$ is a homeomorphism.

Indeed, for every $\mathfrak{p}\in \mathrm{Spec}(A)$, the integral closure of $A_{\mathfrak{p}}$ is $A_{\mathfrak{p}}^{\prime }$ (Bourbaki, Alg. comm., chap. V, §1, n° 5, prop. 16), and all the prime ideals of $A_{\mathfrak{p}}^{\prime }$ above $\mathfrak{p}{A}_{\mathfrak{p}}$ are maximal (loc. cit., §2, n° 1, prop. 1). To say that $f$ is injective therefore means that for every $\mathfrak{p}\in \mathrm{Spec}(A)$, $A_{\mathfrak{p}}^{\prime }$ is a local ring, that is to say that the $A_{\mathfrak{p}}$ are unibranch. To say that every $A_{\mathfrak{p}}$ is geometrically unibranch then means that $f$ is radicial by virtue of (I, 3.5.8). When this is the case, $f$ is surjective and closed (II, 6.1.10), hence a homeomorphism.

[*] In the remainder of this number, we use notions and results expounded in chap. IV, §§2 and 5; since the results of this number are not used before chap. IV, §6, this does not entail any vicious circle.

Lemma (23.2.3).

Let $A$ be an integral ring, $K$ its field of fractions, $B$ a Noetherian ring, $\varphi :A\to B$ a ring homomorphism making $B$ a flat $A$-module. Let $K^{\prime }$ be an extension of $K$; set $B_1=B_{red}$, and let $R$ be the total ring of fractions of B_1. Then, for every sub-$A$-algebra $A^{\prime }$ of $K^{\prime }$, the canonical homomorphism

  (A' ⊗_A B)_red → (K' ⊗_A R)_red                                             (23.1.8.1)

is injective.

We may consider the canonical homomorphism $h:A^{\prime }{\otimes }_{A}B\to K^{\prime }{\otimes }_{A}R$ as the composite of the following homomorphisms

  A' ⊗_A B  ⟶  K' ⊗_A B  ⟶  K' ⊗_A B_1  ⟶  K' ⊗_A R.
            u            v              w

Since $B$ is a flat $A$-module, $u$ is injective; similarly, $K^{\prime }$ being a flat $K$-module and $K$ a flat $A$-module, $K^{\prime }$ is a flat $A$-module, and $w$ is therefore injective since B_1 is reduced, hence the homomorphism $B_1\to R$ injective. Finally, for the same reason, if $\mathfrak{N}$ is

the nilradical of $B$, the kernel of $v$ is $K^{\prime }{\otimes }_A\mathfrak{N}$, hence it is contained in the nilradical of $K^{\prime }{\otimes }_{A}B$; we conclude immediately that if the image under $h=w\circ v\circ u$ of an element $x\in A^{\prime }{\otimes }_{A}B$ is nilpotent, then $x$ is nilpotent, which proves the lemma.

Proposition (23.2.4).

Let $A$ be an integral ring, $K$ its field of fractions, $B$ a Noetherian ring, ${\mathfrak{q}}_i(1\le i\le n)$ its minimal prime ideals, $\varphi :A\to B$ a ring homomorphism. Suppose that the $B/{\mathfrak{q}}_i$ are Japanese rings, and that $B$ is a flat $A$-module. Let $K^{\prime }$ be a finite extension of $K$, and let $(C_{\lambda })$ be the increasing filtering family of subrings of $K^{\prime }$ which are finite $A$-algebras and admit $K^{\prime }$ for field of fractions; the union $A^{\prime }$ of the $C_{\lambda }$ is therefore the integral closure of $A$ in $K^{\prime }$. Then:

(i) There exists an index $\alpha$ such that for $\lambda \ge \alpha$, the canonical homomorphism

  (C_α ⊗_A B)_red → (C_λ ⊗_A B)_red                                           (23.2.4.1)

is bijective.

(ii) If moreover $B$ is a faithfully flat $A$-module, the morphism $\mathrm{Spec}(A^{\prime }{\otimes }_{A}B)\to \mathrm{Spec}(C_{\alpha }{\otimes }_{A}B)$ is radicial.

(i) Let $B_1=B_{red}$, and let $R$ be the total ring of fractions of B_1; since $A$ is integral and $B$ a flat $A$-module, every element $\ne 0$ of $A$ is $B$-regular $(0_I,\mathrm{6.3.4})$, hence the composite homomorphism $A\to B\to B_1$ extends to a homomorphism $K\to R$; we may then write $K^{\prime }{\otimes }_{A}R=K^{\prime }{\otimes }_{K}K{\otimes }_{A}R$, and since $K^{\prime }{\otimes }_{K}K$ identifies with $K^{\prime }$, we have $K^{\prime }{\otimes }_{A}R=K^{\prime }{\otimes }_{K}R$. Since $R$ is a direct composite of the fields of fractions $L_i$ of the $B/{\mathfrak{q}}_i$, $K^{\prime }{\otimes }_{K}R$ is a direct composite of the $K^{\prime }{\otimes }_K{L}_i$, and consequently $(K^{\prime }{\otimes }_{K}R{)}_{red}$ is a direct composite of a finite number of finite extensions of the $L_i$. By virtue of the hypothesis, the integral closure of $B/{\mathfrak{q}}_i$ in a finite extension of $L_i$ is a $(B/{\mathfrak{q}}_i)$-module of finite type, hence a $B$-module of finite type; we conclude that the integral closure $B^{\prime }$ of $B$ in $(K^{\prime }{\otimes }_{A}R{)}_{red}$ is a $B$-module of finite type. By virtue of (23.2.3), the $(C_{\lambda }{\otimes }_{A}B{)}_{red}$ identify canonically with subrings of $(K^{\prime }{\otimes }_{A}R{)}_{red}$ which are finite $B$-algebras, hence contained in $B^{\prime }$. Since $B$ is Noetherian and $B^{\prime }$ a $B$-module of finite type, the filtering family of the $(C_{\lambda }{\otimes }_{A}B{)}_{red}$ admits a greatest element $(C_{\alpha }{\otimes }_{A}B{)}_{red}$, which proves (i).

(ii) Since $B$ is a faithfully flat $A$-module, it suffices (IV, 2.6.1, (v)) to show that the morphism $\mathrm{Spec}(A^{\prime }{\otimes }_{A}B)\to \mathrm{Spec}(C_{\alpha }{\otimes }_{A}B)$ is radicial, or, what amounts to the same (I, 5.1.6), that the morphism Spec(A' ⊗_A B)_red → Spec(C_α ⊗_A B)_red is radicial; but one has $A^{\prime }{\otimes }_{A}B={\underrightarrow{\lim }}_{\lambda }(C_{\lambda }{\otimes }_{A}B)$, hence (IV, 5.13.2) (A' ⊗_A B)_red = lim⃗_λ (C_λ ⊗_A B)_red, and the conclusion results from (i).

Corollary (23.2.5).

Let $A$ be an integral Noetherian local ring, $K$ its field of fractions, $K^{\prime }$ a finite extension of $K$. Let $(C_{\lambda })$ be the increasing filtering family of subrings of $K^{\prime }$ which are finite $A$-algebras (hence Noetherian semi-local rings (Bourbaki, Alg. comm., chap. IV, §2, n° 5, cor. 3 of prop. 9)) and admit $K^{\prime }$ for field of fractions. Then there exists $\alpha$ such that the homomorphism $({\hat{C}}_{\alpha }{)}_{red}\to ({\hat{C}}_{\lambda }{)}_{red}$ is an isomorphism for $\lambda \ge \alpha$, and if $A^{\prime }$ is the integral closure of $A$ in $K^{\prime }$, the morphism $\mathrm{Spec}({\hat{A}}^{\prime })\to \mathrm{Spec}({\hat{C}}_{\alpha })$ is radicial (cf. (23.2.2)).

We apply (23.2.4) taking $B=\hat{A}$; since the $B/{\mathfrak{q}}_i$ are complete Noetherian local rings, they are Japanese rings (23.1.5) and $B$ is a faithfully flat $A$-module $(0_I,\mathrm{7.3.5})$; moreover one has $C_{\lambda }{\otimes }_{A}B={\hat{C}}_{\lambda }$ (Bourbaki, Alg. comm., chap. III, §3, n° 4, th. 3 and chap. IV, §2, n° 5, cor. 3 of prop. 9).

Corollary (23.2.6).

Under the hypotheses of (23.2.5), the integral closure $A^{\prime }$ of $A$ in $K^{\prime }$ is a semi-local ring; if $\mathfrak{m}$ is the maximal ideal of $A$, then, for every maximal ideal ${\mathfrak{m}}^{\prime }$ of $A^{\prime }$, $A^{\prime }/{\mathfrak{m}}^{\prime }$ is a finite extension of $A/\mathfrak{m}$.

The fact that the homomorphism $({\hat{C}}_{\alpha }{)}_{red}\to ({\hat{C}}_{\lambda }{)}_{red}$ is bijective for $\lambda \ge \alpha$ entails that the number of maximal ideals of $C_{\lambda }$ is constant for $\lambda \ge \alpha$ and that if ${\mathfrak{m}}_{\alpha }$ is a maximal ideal of $C_{\alpha }$ and ${\mathfrak{m}}_{\lambda }$ the unique maximal ideal of $C_{\lambda }$ above ${\mathfrak{m}}_{\alpha }$, the fields $C_{\lambda }/{\mathfrak{m}}_{\lambda }$ and $C_{\alpha }/{\mathfrak{m}}_{\alpha }$ are canonically isomorphic. The conclusion results from the fact that ${\mathfrak{m}}^{\prime }=\underrightarrow{\lim }{\mathfrak{m}}_{\lambda }$ and $A^{\prime }/{\mathfrak{m}}^{\prime }=\underrightarrow{\lim }(C_{\lambda }/{\mathfrak{m}}_{\lambda })$ $(0_{III},\mathrm{10.3.1.3})$.

Proposition (23.2.7) (Y. Mori).

Let $A$ be an integral Noetherian local ring, $K$ its field of fractions, $A^{\prime }$ the integral closure of $A$. Then $A^{\prime }$ is a semi-local Krull ring (Bourbaki, Alg. comm., chap. VII, §1); in other words, there exists a family $(V_{\lambda })$ of discrete valuation rings having $K$ for field of fractions, such that $A^{\prime }=\bigcap V_{\lambda }$ and that every $x\in K$ belongs to all the $V_{\lambda }$ save for a finite number of them. Moreover, there exists a subring $C$ of $K$ which is a finite $A$-algebra, and such that the $V_{\lambda }$ are the integral closures of the rings $C_{{\mathfrak{p}}_{\lambda }}$, where $({\mathfrak{p}}_{\lambda })$ is the family of prime ideals of height 1 of the local ring $C$.

Let us first prove the following lemma:

Lemma (23.2.7.1).

Let $A$, $B$ be two rings, $\varphi :A\to B$ a homomorphism making $B$ a faithfully flat $A$-module. Suppose that $A$ is integral; then, if $C=B_{red}$, the composite homomorphism $\psi :A\to B\to B_{red}=C$ is injective and extends to an injective homomorphism of the field of fractions $K$ of $A$ into the total ring of fractions $R$ of $C$; moreover, if $C^{\prime }$ is the integral closure of $C$ in $R$, then $C^{\prime }\cap K$ is the integral closure of $A$.

Since $A$ is integral and $\varphi$ injective, the intersection of $\varphi (A)$ and the nilradical $\mathfrak{N}$ of $B$ is reduced to 0, and since $C=B/\mathfrak{N}$, $\psi$ is injective. Every $a\ne 0$ in $A$ being a non-zero-divisor in $A$ is no more so in $B$ by flatness $(0_I,\mathrm{6.3.4})$; we deduce that $a$ is also not a zero-divisor in $C$, for if one had $ax\in \mathfrak{N}$ for an $x\notin \mathfrak{N}$ in $B$, one would deduce $a^n{x}^n=0$ for an integer $n$, which contradicts the preceding since $x^n\ne 0$. We may therefore extend $\psi$ to an injective homomorphism of $K$ into $R$. To prove the last assertion, note that it is clear that the integral closure $A^{\prime }$ of $A$ is contained in $C^{\prime }\cap K$. Conversely, let $x\in C^{\prime }\cap K$; $x$ is therefore integral over $C$, and a fortiori over $B$, in other words B[x] is a finite $B$-algebra; moreover, $K$ identifies with a subring of $B{\otimes }_{A}K$, and the subring B[x] of $B{\otimes }_{A}K$ identifies with $B{\otimes }_{A}A[x]$ by flatness; we conclude that A[x] is an $A$-module of finite type (Bourbaki, Alg. comm., chap. I, §3, n° 6, prop. 11), hence that $x\in A^{\prime }$.

This lemma being established, we shall apply it to the canonical injection of $A$ into its completion Â, which is a faithfully flat $A$-module; $B=(\hat{A}{)}_{red}=\hat{A}/\mathfrak{N}$ (where $\mathfrak{N}$ is the nilradical of Â) is a reduced complete Noetherian local ring, whose total ring of

fractions $R$ is therefore a direct composite of a finite number of fields $L_i$; the canonical image $B_i$ of $B$ in $L_i$ is an integral and complete Noetherian local ring, of which $L_i$ is the field of fractions, and the integral closure $B^{\prime }$ of $B$ in $R$ is the direct composite of the $B_i^{\prime }$, where $B_i^{\prime }$ is the integral closure of $B_i$. But by virtue of (23.1.5), $B_i^{\prime }$ is a finite $B_i$-algebra, hence an integrally closed Noetherian local ring. We know (Bourbaki, Alg. comm., chap. VII, §1, n° 3, cor. of th. 2) that $B_i^{\prime }$ is a Krull ring. Now, for every $i$, we have a homomorphism ${\varphi }_i:K\to L_i$, which is injective, and consequently ${\varphi }_i^{-1}(B_i^{\prime })$ is a Krull ring in $K$. But since $A^{\prime }=B^{\prime }\cap K$ by virtue of the lemma and $B^{\prime }={\bigcap }_i{\varphi }_i^{-1}(B_i^{\prime })$, $A^{\prime }$ is the intersection of a finite number of Krull rings and is consequently a Krull ring. To prove the last assertion of the proposition, note that there exists a finite $A$-algebra $C\subset K$ such that the morphism $\mathrm{Spec}(A^{\prime })\to \mathrm{Spec}(C)$ is radicial and a homeomorphism (23.2.4); for every prime ideal ${\mathfrak{p}}^{\prime }\in \mathrm{Spec}(A^{\prime })$, ${\mathfrak{p}}^{\prime }$ is therefore the only prime ideal of $A^{\prime }$ above $\mathfrak{p}={\mathfrak{p}}^{\prime }\cap C$ and $A_{{\mathfrak{p}}^{\prime }}^{\prime }$ the integral closure of $C_{\mathfrak{p}}$; on the other hand, the fact that $\mathrm{Spec}(A^{\prime })\to \mathrm{Spec}(C)$ is a homeomorphism entails that the map ${\mathfrak{p}}^{\prime }\mapsto {\mathfrak{p}}^{\prime }\cap C$ is a bijection from the set of prime ideals of height 1 of $A^{\prime }$ onto the set of prime ideals of height 1 of $C$; whence the conclusion, taking account of Bourbaki, Alg. comm., chap. VII, §1, n° 6, th. 4.

Remarks (23.2.8).

(i) One must take care to note, in the application of (23.2.6) and (23.2.7), that the ring $A^{\prime }$ is not necessarily a Noetherian ring, as an example of Nagata with $\dim (A)=3$ shows [30].

(ii) The conclusions of (23.2.7) are still valid if $A^{\prime }$ is the integral closure of $A$ in a finite extension $K^{\prime }$ of $K$; it suffices in fact to consider a finite $A$-algebra $B$ of which $K^{\prime }$ is the field of fractions; $B$ is a Noetherian semi-local ring, hence an intersection of a finite number of Noetherian local rings $B_{{\mathfrak{m}}_i}$ (${\mathfrak{m}}_i$ maximal ideals of $B$), and its integral closure (which is equal to $A^{\prime }$) is the intersection of the $A_{{\mathfrak{m}}_i}^{\prime }$ (Bourbaki, Alg. comm., chap. II, §3, n° 3, cor. 4 of th. 1) which are the integral closures of the $B_{{\mathfrak{m}}_i}$, hence Krull rings by (23.2.7); consequently $A^{\prime }$ is a Krull ring.

(iii) One can show ([30], 3.3.10) that for every integral Noetherian ring $A$ (not necessarily local), the integral closure of $A$ is a Krull ring.

Corollary (23.2.9).

Let $A$ be an integral Noetherian ring, $A^{\prime }$ its integral closure. Suppose that for every ring $C$ such that $A\subset C\subset A^{\prime }$ and such that $C$ is a finite $A$-algebra, and for every prime ideal $\mathfrak{q}$ of height 1 in $C$, $\mathfrak{q}\cap A$ is a prime ideal of height 1 in $A$. Let $P$ be the set of prime ideals of height 1 in $A^{\prime }$; then one has

  A' = ⋂_{𝔭 ∈ P} A'_𝔭.

Since $A^{\prime }$ is a torsion-free $A$-module, one has $A^{\prime }={\bigcap }_{\mathfrak{m}}{A}_{\mathfrak{m}}^{\prime }$, where $\mathfrak{m}$ runs through the set of maximal ideals of $A$ (Bourbaki, Alg. comm., chap. II, §3, n° 3, cor. 4 of th. 1); it will therefore suffice to prove that $A_{\mathfrak{m}}^{\prime }$ is the intersection of the $A_{\mathfrak{p}}^{\prime }$ for the prime ideals $\mathfrak{p}\subset \mathfrak{m}$ of height 1. Since $A_{\mathfrak{m}}^{\prime }$ is the integral closure of $A_{\mathfrak{m}}$, we see that we may restrict to demonstrating the corollary for $A_{\mathfrak{m}}$. Now there is then, by virtue of (23.2.7), an $A_{\mathfrak{m}}$-algebra

$B$ of finite type contained in $A_{\mathfrak{m}}^{\prime }$ such that $A_{\mathfrak{m}}^{\prime }$ is the intersection of the integral closures of the rings $B_{\mathfrak{q}}$, where $\mathfrak{q}$ runs through the set of prime ideals of height 1 of $B$. But $B$ is of the form $C_{\mathfrak{m}}$, where $C$ is a finite $A$-algebra; hence, for every prime ideal $\mathfrak{q}$ of height 1 in $B$, $\mathfrak{q}\cap A$ is by hypothesis a prime ideal of height 1 in $A$; since the integral closure of $B_{\mathfrak{q}}$ contains that of $A_{\mathfrak{q}\cap A}$ and the latter contains $A_{\mathfrak{m}}^{\prime }$, it indeed follows that $A_{\mathfrak{m}}^{\prime }$ is the intersection of the $A_{\mathfrak{p}}^{\prime }$ for $\mathfrak{p}\in P$ and $\mathfrak{p}\subset \mathfrak{m}$, which completes the proof.

Remarks (23.2.10).

(i) We shall see ((IV, 5.6.3) and (5.6.10)) that the hypothesis of (23.2.9) is satisfied when $A$ is universally catenary, in particular ((IV, 5.10.17) and (5.11.2)) when $A$ is a quotient of a regular ring; finally, it is evidently satisfied if $\mathrm{Spec}(A^{\prime })\to \mathrm{Spec}(A)$ is a homeomorphism, in particular if this morphism is radicial (23.2.2).

(ii) The conclusion of (23.2.9) is no longer necessarily exact when one omits the hypothesis on the finite sub-$A$-algebras of $A^{\prime }$. Indeed, one can define an integral Noetherian local ring $A$, of dimension 2, whose integral closure $A^{\prime }$ is a finite $A$-algebra, but such that there is a prime ideal ${\mathfrak{p}}^{\prime }$ of $A^{\prime }$ of height 1 above the maximal ideal $\mathfrak{m}$ (of height 2) of $A$, and such moreover that for every prime ideal $\mathfrak{p}$ of height 1 in $A$, $A_{\mathfrak{p}}$ is integrally closed (IV, 5.6.11); the intersection of these rings $A_{\mathfrak{p}}$ cannot then be equal to $A^{\prime }$ since $A^{\prime }$ is a Krull ring (Bourbaki, Alg. comm., chap. VII, §1, n° 5, prop. 9).

(To be continued.)