power series over noetherian rings may 10, 2010heinzer/teaching/math690/power.pdf · 24.4. more...

Power Series Over Noetherian Rings May 10, 2010

William Heinzer

Christel Rotthaus

Sylvia Wiegand

Department of Mathematics, Purdue University, West Lafayette,

Indiana 47907

E-mail address: [email protected]

Department of Mathematics, Michigan State University, East

Lansing, MI 48824-1027


Department of Mathematics, University of Nebraska, Lincoln, NE

68588-0130


1991 Mathematics Subject Classification. Primary 13A15, 13B35, 13F25;Secondary 13B02, 13B22, 13B40, 13C15, 13H05

Key words and phrases. power series, Noetherian integral domain,completion, generic fiber, flatness, prime spectra.

Abstract. In this monograph the authors gather together results and exam-ples from their work of the past two decades related to power series rings andto completions of Noetherian integral domains.

A major theme is the creation of examples that are appropriate inter-sections of a field with a homomorphic image of a power series ring over aNoetherian domain. This technique goes back to work of Akizuki in the 1930sand Nagata in the 1950s. In certain circumstances, such an intersection iscomputable as a directed union, and the Noetherian property for the associ-ated directed union is equivalent to a flatness condition. This flatness criterionsimplifies the analysis of several classical examples and yields new examplessuch as

• A catenary Noetherian local integral domain of any specified dimensionbigger than one that has geometrically regular formal fibers and is notuniversally catenary.

• A three-dimensional non-Noetherian unique factorization domain B suchthat the unique maximal ideal of B has two generators; B has preciselyn prime ideals of height two, where n is an arbitrary positive integer;and each prime ideal of B of height two is not finitely generated but allthe other prime ideals of B are finitely generated.

• A two-dimensional Noetherian local domain that is a birational extensionof a polynomial ring in three variables over a field yet fails to have Cohen-Macaulay formal fibers. This also demonstrates that Serre’s conditionS1 need not lift to the completion. It is related to an example of Ogoma.

Another theme is an analysis of extensions of integral domains R → Shaving trivial generic fiber, that is, every nonzero prime ideal of S has a nonzerointersection with R. Motivated by a question of Hochster and Yao, we presentin Chapters

weiersec29,

ppssec30 and

tgfsec31 results about

• The height of prime ideals maximal in the generic fiber of certain exten-sions involving mixed power series/polynomial rings.

• The prime ideal spectrum of a power series ring in one variable over aone-dimensional Noetherian domain.

• The dimension of S if R → S is a local map of complete local domainshaving trivial generic fiber.

A third theme relates to the questions:• What properties of a Noetherian domain extend to a completion?• What properties of an ideal pass to its extension in a completion?• What properties extend for a more general multi-adic completion?

We give an example of a three-dimensional regular local domain R having aprime ideal P of height two with the property that the extension of P to thecompletion of R is not integrally closed.

All of these themes are relevant to the study of prime spectra of Noether-ian rings and of the induced spectral maps associated with various extensionsof Noetherian rings. We describe the prime spectra of various extensions in-volving power series.

Contents

Preface xi

Chapter 1. Introduction May 17, 2009 1

Chapter 2. Tools March 21, 2010 52.1. Conventions and terminology 52.2. Basic theorems 82.3. Flatness 8Exercises 10

Chapter 3. More Tools April 1, 2010 133.1. Introduction to ideal-adic completions 133.2. Basic theorems about completions 143.3. Excellence, definition and basic properties 15Exercises 17

Chapter 4. First Examples of the Construction March 12, 2010 214.1. Elementary examples 214.2. Historical examples 224.3. Motivating examples 25Exercises 26

Chapter 5. Describing the Construction March 25, 2010 295.1. Two construction methods and a picture 295.2. Universality 31Exercises 33

Chapter 6. Two Approximations and their ConnectionMarch 7, 2010 35

6.1. Approximations for the Inclusion Construction 356.2. Approximations for the Homomorphic Image Construction 396.3. The Inclusion Construct is a Homomorphic Image Construct 416.4. Basic properties of the approximations 42Exercises 44

Chapter 7. Revisiting the Motivating Examples March 20, 2010 457.1. Properties of the motivating examples 457.2. Residual algebraic independence 48Exercises 53

Chapter 8. Building Noetherian domains March 31, 2010 558.1. Flatness and the Noetherian property 55

v

vi CONTENTS

8.2. The Inclusion Version 58

Chapter 9. Examples where flatness holds April 3, 2010 619.1. A general example 619.2. Transfer to the intersection of two ideals 649.3. Regular maps and geometrically regular formal fibers 66Exercises 67

Chapter 10. Introduction to the insider construction May 10, 2010, 16, 17(insideintro) 69

10.1. The Nagata example 6910.2. Other examples inside the general example 70

Chapter 11. The flat locus of a polynomial ring extension Feb 6, 2010, 16, 17(flatpoly) 73

11.1. Flatness criteria 7311.2. The Jacobian ideal and the smooth and flat loci 7511.3. Applications to polynomial extensions 79Exercises 82

Chapter 12. Height-one primes and limit-intersecting elements May 8, 2010 8312.1. The limit-intersecting condition 8312.2. Height-one primes in extensions of integral domains 85Exercises 87

Chapter 13. Insider construction details, May 3, 2010, 16, 17 (insidecon) 8913.1. Describing the construction 8913.2. The flat locus of the insider construction 9013.3. The non-flat locus of the insider construction 91

Chapter 28. Integral closure under extension to the completion (intclsec),May 10, 2010 18 233

28.1. Integral closure under ring extension 23328.2. Extension to the completion 23628.3. Comments and Questions 238

Chapter 15. Catenary local rings with geometrically normal formal fibersApril 21, 2010 101

15.1. Introduction 10115.2. Geometrically normal formal fibers 10215.3. A method for constructing examples 10615.4. Examples that are not universally catenary 10715.5. Regular maps and geometrically regular formal fibers 109Exercises 110

Chapter 16. Non-finitely generated prime ideals in subrings of power seriesrings, June 5, 2009(nonfgsec), 17 111

16.1. Introduction 11116.2. Background and Notation 11216.3. Non-Noetherian examples 11416.4. A more general construction 114

CONTENTS vii

16.5. A More General Construction 115

Chapter 17. Examples using the insider construction, Dec. 2009(insidepssec)16 117

17.1. Introduction 11717.2. Background and Notation 11717.3. Examples 118

Chapter 18. The first Noetherian theorem April 23 2009 12118.1. Introduction 121

Chapter 19. Properties of ring extensions 12319.1. Independence properties for elements 12319.2. Related properties for extensions 123

Chapter 20. Idealwise algebraic independence May 17 2009 12520.1. Introduction 12520.2. Introduction 12920.3. Background 13020.4. 2. Idealwise independence, weakly flat and PDE extensions 13420.5. 3. Primary independence 13920.6. Residual algebraic independence 14320.7. Idealwise independence and flatness 14820.8. Composition, base change and polynomial extensions 15220.9. Passing to the Henselization 15620.10. Summary diagram for the independence concepts 159

Chapter 21. Intermediate rings between a local domain and its completionMay 17 2009 161

21.1. Background 16221.2. Outline 16321.3. Background material 16421.4. Intersections and directed unions 16521.5. Limit-intersecting elements 16821.6. Some examples 17521.7. Corrigendum to “Intermediate rings ...”, 182

Chapter 22. Noetherian rings between a semilocal domain and its completionMay 17, 2009 183

22.1. Introduction 18322.2. Preserving Noetherian under directed unions of localized polynomial

extensions 18422.3. Applications and a classical construction of Noetherian rings 190

Chapter 23. A non-Noetherian limit-intersecting example 19523.1. An example where B = A is non-Noetherian 195

Chapter 24. Approximating discrete valuation rings by regular local rings(appdvrsec), May 17, 2009 12 205

24.1. Introduction 20524.2. Expressing a DVR as a directed union of regular local rings 206

viii CONTENTS

24.3. Proof of Theorem12.1.224.3 207

24.4. More general valuation rings as directed unions of RLRs 210

Chapter 25. Intermediate rings between a local domain and its completion II(intIIsec), 13, May 17 2009 213

25.1. Introduction 21325.2. Flatness criteria 21425.3. Cohen-Macaulay formal fibers and Ogoma’s example 21725.4. Related examples not having Cohen-Macaulay fibers 219

Chapter 26. Building Noetherian and non-Noetherian integral domains usingpower series 225

26.1. Noetherian results 22526.2. Flatness, approximations 22526.3. Explicit constructions 226

Chapter 27. What properties extend to an ideal-adic completion? (wprop,25)May 17, 2009 229

27.1. Associated graded rings of ideals under completion 23027.2. The Sk condition and completion 231

Chapter 28. Integral closure under extension to the completion (intclsec),May 10, 2010 18 233

28.1. Integral closure under ring extension 23328.2. Extension to the completion 23628.3. Comments and Questions 238

Chapter 29. Generic formal fibers of mixed power series/polynomial ringsMay 17 2009 (weiersec), 19 241

29.1. Introduction and Background 24129.2. Variations on a theme of Weierstrass 24329.3. Weierstrass implications for the ring B = K[[X ]] [Y ](X,Y ) 25029.4. Weierstrass implications for the ring C = K[Y ](Y )[[X ]] 25329.5. Subrings of the power series ring K[[z, t]] 25529.6. Weierstrass implications for the localized polynomial ring A 25729.7. Generic fibers of power series ring extensions 260

Chapter 30. Mixed polynomial/power series rings and relations among theirspectra (ppssec), 20, May 17, 2009 263

30.1. Introduction and Background 26330.2. Trivial generic fiber (TGF) extensions and prime spectra 26530.3. Spectra for two-dimensional mixed polynomial/power series rings 26730.4. Higher dimensional mixed power series/polynomial rings 271

Chapter 31. Extensions of local domains with trivial generic fiber May 172009(tgfsec), 21 277

31.1. Introduction 27731.2. General remarks about TGF extensions 27831.3. TGF-complete extensions with finite residue field extension 27931.4. The case of transcendental residue extension 283

CONTENTS ix

Chapter 32. Multi-ideal-adic Completions of Noetherian Rings (multsec), 22Jan. 18, 2009 287

32.1. Introduction 28732.2. Basic mechanics for the multi-adic construction 28832.3. Preserving Noetherian under multi-adic completion 29132.4. Preserving excellence or Henselian under multi-adic completion 294

Chapter 33. Outline/Plan May 17 2009 297

Chapter 34. AMS Monograph Series Sample 299This is an unnumbered first-level section head 299This is a Special Section Head 29934.1. This is a numbered first-level section head 299Exercises 30034.2. Some more list types 301

Bibliography 303

Index 307

Preface

The authors have had as a long-term project the creation of examples usingpower series to analyze and distinguish several properties of commutative ringsand their spectra. This monograph is our attempt to expose the results that havebeen obtained in this endeavor, to put these results in better perspective and toclarify their proofs. We hope in this way to assist current and future researchers incommutative algebra in utilizing the techniques described here.

William Heinzer, Christel Rotthaus, Sylvia Wiegand

xi

CHAPTER 1

Introduction May 17, 2009

introOver the past sixty years, important examples of Noetherian domains have

been constructed using power series, homomorphic images and intersections. Thebasic idea is that, starting with a typical Noetherian integral domain R such as apolynomial ring in several indeterminates over a field, we look for more unusualNoetherian and non-Noetherian extension rings inside a homomorphic image S ofthe completion R of R. The constructed ring A has the form A := L ∩ S, where Lis a field between the fraction field of R and the total quotient ring of S.

We may consider R as a homomorphic image of a power series ring over R.(Completions are discussed in Section

3.13.1 of Chapter

3tools2.) We have the following

major goals:

(1) To construct new examples of Noetherian rings, continuing a traditionthat goes back to Akizuki and Schmidt in the 1930s and Nagata in the1950s.

(2) To construct new non-Noetherian integral domains that illustrate recentadvances in ideal theory.

(3) To study birational extensions of Noetherian local domains; this is relatedto Zariski’s constructions of valuation rings that are birational approxi-mations of Noetherian local domains.

(4) To consider the fibers of an extension R → R∗, where R is a Noether-ian domain and R∗ is the completion of R with respect to an ideal-adictopology, and to relate these fibers with birational extensions of R∗.

These objectives form a complete circle, since (4) is used to accomplish (1).The development of this popular technique to create interesting new rings from

well-known ones goes back to work of Akizuki inA[7], Schmidt in

Sc[87] and Nagata

inN1[70]. This work is continued by Ferrand-Raynaud

FR[27], Lequain

Leq[61], Nishimura

Ni[73], Rotthaus

R1[81],

R2[82], Heitmann

H1[54], Ogoma

O1[76],

O2[77], Brodmann-Rotthaus

BR1[11],

BR2[12], and Weston

W[98]. The present authors have been captivated by this

technique and have been examining it for a number of years. In several chaptersof this monograph (See Chapters

fex4,

constr5,

appcom6,

noehomsec8,

intsec21,

noeslocsec22,

insidepssec17) we present a reorganized

development of our previous work on this technique.In his 1935 paper Akizuki gives an example in characteristic zero of a one-

dimensional Noetherian local domain that is analytically ramifiedA[7], while Schmidt

gives such an example in positive characteristicSc[87, pp. 445-447]. An example due

to Nagata is given inN2[72, Example 3, pp. 205-207]. (See also

N2[72, (32.2), p. 114].)

In Chapterconstr5 (

4.3.14.8) we consider another example constructed by Nagata. This

is the first occurence of a two-dimensional regular local domain containing a fieldof characteristic zero that fails to be a Nagata domain. The example also fails tosatisfy one of the conditions in the definition of an excellent ring. For the definition

1

2 1. INTRODUCTION MAY 17, 2009

and information on Nagata rings and excellent rings see Sections3.022.1 and

3.123.3. We

describe in Example4.3.34.10 a construction due to Rotthaus of a Nagata domain that

is not excellent.An interesting construction first introduced by Ray Heitmann in

H1[54, page 126]

addresses the question of how ring-theoretically to adjoin a transcendental powerseries in an element a to the ring R. We make use of this construction of Heitmannin Chapter

noeslocsec22 in a local or semilocal context. We consider this construction in the

more general context of an arbitrary Noetherian integral domain in Chapter ???.We apply the construction, and its natural generalization to finitely many elements,to exhibit Noetherian extension domains of R inside the (a)-adic completion R∗ ofR. Heitmann in

H2[55] and

H3[56] answers the question of what complete Noetherian

local rings are the completion of a Noetherian local domain, and what completelocal rings are the completion of a Noetherian local unique factorization domain.

In the foundational work of Akizuki, Nagata and Rotthaus (and indeed inmost of the papers cited above) the description of the constructed ring A as anintersection A := L ∩ S, where L is a field between the fraction field of R andthe total quotient ring of S, is not explicitly stated. Instead A is defined as adirect limit or directed union of subrings. The fact that in certain circumstancesthe intersection domain A may be computable as a directed union is crucial toour development of this technique; the description of A as an intersection is oftenunfathomable. There is a natural direct limit domain B associated with A andwe examine conditions when A = B. This motivates our formulation of the limit-intersecting property that we discuss in Chapter

intsec21. A primary goal of our study

is to determine for a given R, S and L whether A is Noetherian.

An important observation related to that goal is that the Noetherian prop-erty for the associated direct limit ring B is equivalent to a flatness condition cf.Chapter

noeslocsec22, Theorem

8.2.1222.12 and Chapter

noehomsec8, Theorem

11.3.28.3.

While it took only about a page (using ideas similar to those in (5.1)) forNagata

N2[72, page 210] to establish the Noetherian property of his example, for the

example of RotthausR1[81, pages 112-118], the proof of the Noetherian property took

7 pages. In Chapternoehomsec8 we discuss results that give the Noetherian property more

quickly for many of the known examples.A special case of this construction is D := L ∩ R, where L is a purely tran-

scendental extension of the quotient field K of the Noetherian domain R; sayL = K(τ1, τ2, . . . , τn), where τ1, τ2, . . . , τn ∈ R are algebraically independent overK. (See Section

4.65.2)) Several questions arise in connection with this setup.

In Chapteridwisec20, we define a first “independence” condition: for (R,m) a nor-

mal excellent local domain with quotient field K, the elements τ1, . . . , τn, whichare algebraically independent over K, are said to be idealwise independent, ifK(τ1, . . . , τn)∩ R equals the localized polynomial ring R[τ1, . . . , τn](m,τ1,...,τn). Theconsideration of idealwise independent elements leads to other related flatness con-ditions, namely two more concepts of independence over R for algebraically inde-pendent elements τ1, . . . , τn of m. A summary of the analysis and properties relatedto the idealwise independence definition is given in Chapter

idwisec20.

Topics we consider include the following:

(1) Various flatness properties of maps of rings as in Chaptersidwisec20 and

intsec21.

1. INTRODUCTION MAY 17, 2009 3

(2) Preservation of various properties of rings and ideals under passage tocompletion.

(3) The catenary and universally catenary property of rings.(4) Excellent rings and geometrically regular and geometrically normal formal

fibers.(5) Examples of non-Noetherian local rings having a Noetherian completion.(6) Schematic diagrams of the partially ordered set of prime ideals of a ring.(7) Prime ideals and the fibers of a map as in Chapter

weiersec29.

(8) The approximation of a DVR with higher dimensional regular local rings.(9) Trivial generic fiber extensions as in the Hochster-Yao question in Chap-

tersweiersec29,

ppssec30, and

tgfsec31.

(10) For subrings R ⊂ C of a ring B as in Chapternoeslocsec22, we consider conditions

in order that B is excellent implies C is excellent.(11) What domains lie between a Noetherian integral domain and its field of

fractions , as in Chapternoehomsec8

We remark that the idealwise construction in Chapteridwisec20 is of a different nature

to the construction in Chapterbuildsec18. The idealwise construction demonstrates that

the intersection domain can sometimes be small or large, depending on Idealwiseconsiderations. There are various flatness properties related to this construction.

CHAPTER 2

Tools March 21, 2010

3toolsIn this chapter we review conventions and terminology, state several basic the-

orems and review the concept of flatness of modules and maps.

2.1. Conventions and terminology3.02

We generally follow the notation of MatsumuraM[68]. We also reference Dummit

and FooteDF[22] for more elementary concepts. Thus by a ring we mean a commuta-

tive ring with identity, and a ring homomorphism R→ S maps the identity elementof R to the identity element of S. The set of prime ideals of a ring R is denotedSpecR. The set SpecR is naturally a partially ordered set with respect to inclusion.For an ideal I of a ring R, let

V(I) = P ∈ SpecR | I ⊆ P .The Zariski topology on SpecR is obtained by defining the closed subsets to be thesets of the form V(I) as I varies over all the ideals of R.

We use Z to denote the ring of integers, N for the positive integers, N0 the non-negative integers, Q the rational numbers, R the real numbers and C the complexnumbers.

Regular elements. An element r of a ring R is said to be a zerodivisor, ifthere exists a nonzero element a ∈ R such that ar = 0, and r is a regular elementif r is not a zerodivisor. The total ring of fractions of the ring R, denoted Q(R),is the localization of R at the multiplicatively closed set of regular elements, thusQ(R) := a/b | a, b ∈ R and b is a regular element . There is a natural embeddingR → Q(R) of a ring R into its total ring of fractions Q(R).

An integral domain, sometimes called a domain or an entire ring, is a nonzeroring in which every nonzero element is a regular element. If R is a subring of anintegral domain S and S is a subring of Q(R), we say S is birational over R, or abirational extension of R.

Local rings. If a ring R (not necessarily Noetherian) has a unique maximalideal m, we say R is local and write (R,m) to denote that R is local with maximalideal m. If (R,m) and (S,n) are local rings, a ring homomorphism f : R→ S is alocal homomorphism if f(m) ⊆ n.

Let (R,m) be a local ring. A subfield k of R is said to be a coefficient field forR if the composite map k → R→ R/m defines an isomorphism of k onto R/m.

If (R,m) is a subring of a local ring (S,n), then S is said to dominate R ifm = n ∩R, or equivalently, if the inclusion map R → S is a local homomorphism.The local ring (S,n) is said to birationally dominate (R,m) if S is an integraldomain that dominates R and S is contained in the field of fractions of R.

5

6 2. TOOLS MARCH 21, 2010

Jacobson radical. The Jacobson radical J (R) of a ring R is the intersectionof all maximal ideals of R. An element z of R is in J (R) if and only if 1 + zr is aunit of R for all r ∈ R.

If z is a nonunit of R, then 1 + zR := 1 + zr | r ∈ R is a multiplicativelyclosed subset of R that does not contain 0. Let (1+zR)−1R denote the localizationR(1+zR) of R at the multiplicatively closed set 1 + zR, cf.

M[68, Section 4]. If P is a

prime ideal of R and P ∩ (1 + zR) = ∅, then (P + zR)∩ (1 + zR) = ∅. Therefore zis in every maximal ideal of (1 + zR)−1R, so z ∈ J ((1 + zR)−1R).

An extension ring S of a ring R is of finite type over R if S is finitely generatedas an R-algebra. An extension ring S of R is essentially of finite type over R if Sis a localization at some multiplicatively closed subset of an extension ring of R offinite type. We also say that S is essentially finitely generated in this case.

If P is a prime ideal of a ring R and e is a positive integer, the eth symbolicpower of P , denoted P (e), is defined as

P (e) := a ∈ R | ab ∈ P e for some b ∈ R \ P .An integral domain R is a valuation domain if for each element a ∈ Q(R) \R,

we have a−1 ∈ R. A valuation domain R is called a discrete valuation ring (DVR)if R is Noetherian and not a field, or equivalently, if R is a principal ideal domain(PID) and not a field.

3.02.1 Remark 2.1. If R is a valuation domain with field of fractions K and F is asubfield of K, then R ∩ F is again a valuation domain and has field of fractions FN2[72, (11.5)]. If R is a DVR and the field F is not contained in R, then R ∩ F isagain a DVR

N2[72, (33.7)].

An integral domain R is said to be integrally closed provided it satisfies thefollowing condition: for every monic polynomial f(x) in the polynomial ring R[x],if a ∈ Q(R) is a root of f(x), then a ∈ R. An ideal of a ring R is said to be integrallyclosed provided it satisfies the following condition: for every monic polynomial f(x)in the polynomial ring R[x], if a ∈ Q(R) is a root of f(x), then a ∈ R. The ring Ris a normal ring if for each P ∈ SpecR the localization RP is an integrally closeddomain

M[68, page 64]. If R is a Noetherian normal ring and p1, . . . ,pr are the

minimal primes of R, then R is isomorphic to the direct product R/p1×· · ·×R/prand each R/pi is an integrally closed domain. Since a nontrivial direct product isnot local, a normal Noetherian local ring is a normal domaiin.

The Krull dimension, or briefly dimension, of a ring R, denoted dimR, is n ifthere exists a chain P0 ( P1 ( · · · ( Pn of prime ideals of R and there is no suchchain of length greater than n. We say that dimR = ∞ if there exists a chain ofprime ideals of R of length greater than n for each n ∈ N.

A local ring (R,m) is a regular local ring (RLR) if R is Noetherian and mcan be generated by dimR elements. If (R,m) is a regular local ring then R is anintegral domain, and the function ordR : R \ 0 → N0 defined by ordR(a) = nif a ∈ mn \mn+1 satisfies the following two properties: (i) ordR(ab) = ordR(a) +ordR(b), and (ii) ordR(a + b) ≥ minordR(a), ordR(b), for a, b ∈ R \ 0. Wedefine a valuation v on Q(R) \ 0, by defining v(a/b) = ordR(a)− ordR(b) for alla, b ∈ R \ 0. The associated valuation domain

V := q ∈ Q(R) \ 0 | v(q) ≥ 0 ∪ 0

2.1. CONVENTIONS AND TERMINOLOGY 7

is a DVR that birationally dominates R. If x ∈m \m2, then V = R[m/x]xR[m/x],where m/x = y/x | y ∈m.

We record the following definitions.

3.02.2 Definitions 2.2. Let R be a commutative ring.

(1) R is called a Nagata ring if R is Noetherian and for every P ∈ SpecR andevery finite extension field L ofQ(R/P ), the integral closure ofR/P in L isfinitely generated as a module over R/P . In Nagata’s book

N2[72] a Nagata

ring is called pseudo-geometric. It is clear from the definition that ahomomorphic image of a Nagata ring is again a Nagata ring. An importantresult proved by Nagata is that a polynomial ring in finitely many variablesover a Nagata ring is again a Nagata ring

N2[72, Theorem 36.5, page 132].

(2) An integral domain R is said to be a Krull domain if there exists a familyF = Vλλ∈Λ of DVRs of its field of fractions Q(R) such that• R =

⋂λ∈Λ Vλ, and

• Every nonzero element of Q(R) is a unit in all but finitely many ofthe Vλ.

A unique factorization domain (UFD) is a Krull domain, and a Noetherianintegral domain is a Krull domain if and only if it is integrally closed. Anintegral domain R is a Krull domain if and only if it satisfies the followingthree properties:• RP is a DVR for each prime ideal P of R of height one.• R =

⋂ RP | P is a height-one prime .• Every nonzero element of R is contained in only finitely many height-

one primes of R.If R is a Krull domain, then F = RP | P is a height-one prime is theunique minimal set of DVRs satisfying the properties in the definition ofa Krull domain.

(3) Let R be a Krull domain and let R → S be an inclusion map of R into aKrull domain S. The extension R → S satisfies the PDE condition (“pasd’eclatement”, or in English “no blowing up”) provided that for everyheight-one prime ideal Q in S, the height of Q∩R is at most one (cf.

F[26,

page 30]).(4) A local ring (R,m) is Henselian provided the following holds: for every

monic polynomial f(x) ∈ R[x] satisfying f(x) ≡ g0(x)h0(x) modulo m[x],where g0 and h0 are monic polynomials in R[x] such that

g0R[x] + h0R[x] + m[x] = R[x],

there exist monic polynomials g(x) and h(x) in R[x] such that f(x) =g(x)h(x) and such that both

g(x) − g0(x) and h(x) − h0(x) ∈ m[x].

In other words, if f(x) factors modulo m[x] into two comaximal factors,then this factorization can be lifted back to R[x]. Alternatively, Henselianrings are rings for which the conclusion to Hensel’s Lemma holds

M[68,

Theorem 8.3].

18.2.0 Definitions 2.3. Let I be an ideal of a ring R.

8 2. TOOLS MARCH 21, 2010

(1) An element r ∈ R is integral over I if there exists a monic polynomialf(x) = xn +

∑ni=1 aix

n−i such that f(r) = 0 and such that ai ∈ Ii foreach i with 1 ≤ i ≤ n.

(2) If J is an ideal contained in I and JIn−1 = In, then J is said to be areduction of I.

(3) The integral closure I of I is the set of elements of R integral over I.(4) If I = I, then I is said to be integrally closed.(5) The ideal I is said to be normal if In is integrally closed for every n ≥ 1.

2.2. Basic theorems3.2

Theorem3.2.012.4 is a famous result proved by Krull that is now called the Krull

intersection theorem.

3.2.01 Theorem 2.4 (KrullM[68, Theorem 8.10]). Let I be an ideal of a Noetherian

ring R.(1) If I is contained in the Jacobson radical J (R) of R, then

⋂∞n=1 I

n = 0,and for each finite R-module M , we have

⋂∞n=1 I

nM = 0.(2) If I is a proper ideal of a Noetherian integral domain, then

⋂∞n=1 I

n = 0.

To prove that a ring is Noetherian, it suffices by the following well-known resultof Cohen to prove that every prime ideal of the ring is finitely generated.

3.2.1 Theorem 2.5 (CohenCo[18]). If each prime ideal of the ring R is finitely gen-

erated, then R is Noetherian.

An interesting result proved by Nishimura is

3.2.2 Theorem 2.6 (NishimuraNi[73, Theorem, page 397]). Let R be a Krull domain.

If R/P is Noetherian for each P ∈ SpecR with htP = 1, then R is Noetherian.

3.2.21 Remark 2.7. It is observed inHRS[33, Lemma 1.5] that the conclusion of The-

orem3.2.22.6 still holds if it is assumed that R/P is Noetherian for all but at most

finitely many of the height one primes P of R.

Theorem3.2.32.8 is useful for describing the maximal ideals of a power series ring

R[[x]]. It is related to the fact that an element f = a0 + a1x + a2x2 + · · · ∈ R[[x]]

with the ai ∈ R is a unit of R[[x]] if and only if a0 is a unit of R.

3.2.3 Theorem 2.8 (N2[72, Theorem 15.1]). Let R[[x]] be the formal power series ring

in a variable x over a commutative ring R. There is a one-to-one correspondencebetween the maximal ideals m of R and the maximal ideals m∗ of R[[x]] where m∗

corresponds to m if and only if m∗ is generated by m and x.

As an immediate corollary of Theorem3.2.32.8, we have

3.2.4 Corollary 2.9. The element x is in the Jacobson radical J (R[[x]]) of thepower series ring R[[x]]. In the formal power series ring S := R[[x1, . . . , xn]], theideal (x1, . . . , xn)S is contained in the Jacobson radical J (S) of S .

2.3. Flatness3.3

The concept of flatness was introduced by Serre in the 1950’s in an appendixto his paper

S[89]. Mumford writes in

Mu[69, page 424]: “The concept of flatness is

a riddle that comes out of algebra, but which technically is the answer to manyprayers.”

2.3. FLATNESS 9

flat Definitions 2.10. For R a ring, an R-module M is flat over R if tensoringwith M preserves exactness of every exact sequence of R-modules. The R-moduleM is said to be faithfully flat over R if, for every sequence of R-modules

S : 0 −−−−→ M1 −−−−→ M2,

S is exact if and only if S ⊗RM is exact.A ring homomorphism φ : R→ S is said to be a flat homomorphism if S is flat

as an R-module.

Flatness is preserved by several standard ring constructions as we record inRemarks

remflat2.11. There is an interesting elementwise criterion for flatness that is

stated as item (2) of Remarksremflat2.11.

remflat Remarks 2.11. The following facts are useful for understanding flatness. Weuse these facts to obtain the results in Chapters

noehomsec8 and

insidepssec17.

(1) Since localization at prime ideals commutes with tensor products, themodule M is flat as an R-module ⇐⇒ MQ is flat as an RQ-module, forevery prime ideal Q of R.

(2) An R-module M is flat over R if and only if for every m1, . . . ,mn ∈ Mand a1, . . . , an ∈ R such that

∑aimi = 0, there exist a positive integer k,

a subset bijni=1,kj=1⊆ R, and elements m′

1, . . . ,m′k ∈M such that mi =∑k

j=1 bijm′j for each i and

∑ni=1 aibij = 0 for each j

M[68, Theorem 7.6] or

[Theorem 1]M1.(3) A finitely generated module over a local ring is flat if and only if it is free

M1[66, Proposition 3.G].

(4) If the ring S is a localizaton of R, then S is flat as an R-moduleM1[66,

(3.D), page 19].(5) Let S be a flat R-algebra. Then S is faithfully flat over R ⇐⇒ for every

proper ideal J of R one has JS 6= SM1[66, Theorem 3, page 28] or

M[68,

Theorem 7.2].(6) If the ring S is a flat R-algebra, then every regular element of R is regular

on SM1[66, (3.F)].

(7) Let S be a faithfully flat R-algebra and let I be an ideal of R. ThenIS ∩R = I

M[68, Theorem 7.5].

(8) Let R be an integral domain with field of fractions K and let S be afaithfully flat R-algebra. By item (5), every nonzero element of R isregular on S and so K naturally embeds in the total quotient ring Q(S)of S. By item (6), all ideals in R extend and contract to themselves withrespect to S, and thus R = K ∩ S. In particular, if S ⊆ K, then R = SM1[66, page 31].

(9) If φ : R → S is a flat homomorphism of rings, then φ satisfies the going-down theorem

M1[66, (5.D), page 33]. This implies for each P ∈ SpecS that

the height of P is greater than or equal to the height in R of φ−1(P ).(10) Let R→ S be a flat homomorphism of rings and let I and J be ideals of

R. Then (I ∩ J)S = IS ∩ JS. If J is finitely generated, then (I :R J)S =IS :S JS

M[68, Theorem 7.4] or

M1[66, (3.H) page 23].

(11) Consider the following short exact sequence of R-modules:

0 −−−−→ A −−−−→ B −−−−→ C −−−−→ 0 .

If A and C are flat over R, then so is B.

10 2. TOOLS MARCH 21, 2010

(12) If S is a flatR-algebra andM is a flat S-module, thenM is a flatR-moduleM[68, page 46].

We use Remarkremflat22.12.3 in Chapter

motiv7.

remflat2 Remark 2.12. Let R be an integral domain.(1) Every flat R-module M is torsionfree, i.e., if r ∈ R, x ∈ M and rx = 0,

then r = 0 or x = 0M1[66, (3.F), page 21]

(2) Every finitely generated torsionfree module over a PID is free, cf.DF[22,

Theorem 5. page 462].(3) Every torsionfree module over a PID is flat. This follows from item (2)

and Remarkremflat2.11.2.

(4) Every injective homomorphism of R into a field is flat. This follows fromRemarks

remflat2.11.12 and

remflat2.11.4.

In Chapterptools23 we discuss other tools we will be using involving ideal-adic com-

pletions and properties of excellent rings.

Exercises(1) Let V be a local domain with nonzero principal maximal ideal yV . Prove that

V is a DVR if⋂∞n=1 y

nV = (0).Comment: It is not being assumed that V is Noetherian, so it needs to beestablished that V has dimension one.

(2) Prove as stated in Remark3.02.12.1 that if R is a valuation domain with field of

fractions K and F is a subfield of K, then R ∩ F is again a valuation domainand has field of fractions F ; also prove that if R is a DVR and the field F isnot contained in R, then R ∩ F is again a DVR.

(3) Prove that a unique factorization domain is a Krull domain.(4) Let R[[x]] be the formal power series ring in a variable x over a commutative

ring R.(i) Prove that a0 + a1x + a2x

2 + · · · ∈ R[[x]], where the ai ∈ R, is a unit ofR[[x]] if and only if a0 is a unit of R.

(ii) Prove that x is contained in every maximal ideal of R[[x]].(iii) Prove Theorem

3.2.32.8 that the maximal ideals m of R are in one-to-one cor-

respondence with the maximal ideals m∗ of R[[x]], where m∗ correspondsto m if and only if m∗ is generated by m and x.

(5) Let R be a commutative ring. Prove items (4)-(7) of Remarksremflat2.11.

(6) Let f : A→ B be a ring homomorphism and let P be a prime ideal of A. Provethat there exists a prime ideal Q in B that contracts in A to P if and only ifthe extended ideal f(P )B contracts to P .

(7) Let f : A → B be an injective ring homomorphism and let P be a minimalprime of A.(i) Prove that there exists a prime ideal Q of B that contracts in A to P .(ii) Deduce that there exists a minimal prime Q of B that contracts in A to P .Suggestion: Consider the multiplicatively closed set A \ P in B.

EXERCISES 11

(8) Let R be a Krull domain having the property that R/P is Noetherian for allbut at most finitely many of the P ∈ SpecR with htP = 1. Prove that R isNoetherian.

Suggestion: By Nishimura’s result Theorem3.2.22.6 and Cohen’s result Theo-

rem3.2.12.5, it suffices to prove each P ∈ SpecR with htP > 1 is finitely gener-

ated. Let P1, . . . , Pn be the height-one prime ideals of R for which R/Pi mayfail to be Noetherian, and choose a ∈ P \ (P1 ∪ · · · ∪ Pn). Observe thataR = q

(e1)1 ∩ · · · ∩ q(es)

s , where q1, . . . , qs are height-one prime ideals of R notin the set P1, . . . , Pn. Prove that R/aR is Noetherian and deduce that P isfinitely generated.

CHAPTER 3

More Tools April 1, 2010

ptools2In this chapter we discuss ideal-adic completions. We also state several results

concerning complete local rings and review the concept of excellence.

3.1. Introduction to ideal-adic completions3.1

3.1.1 Definitions 3.1. Let R be a commutative ring with identity. A filtration onR is a decreasing sequence In∞n=0 of ideals of R. Associated to a filtration thereis a well-defined completion R∗ that may be defined to be the inverse limit 1

(3.13.1.0.1) R∗ = lim←−

n

R/In

and a canonical homomorphism ψ : R → R∗ Nor[75, Chapter 9]. If

⋂∞n=0 In = (0),

then ψ is injective and R may be regarded as a subring of R∗ Nor[75, page 401]. In the

terminology of Northcott, a filtration In∞n=0 is said to be multiplicative if I0 = Rand InIm ⊆ In+m, for all m ≥ 0, n ≥ 0

Nor[75, page 408].

A well-known example of a multiplicative filtration on R is the I-adic filtrationIn∞n=0, where I is a fixed ideal of R such that

⋂∞n=0 I

n = (0). In this case we sayR∗ := lim←−

n

R/In is the I-adic completion of R. If the canonical map R→ R∗ is an

isomorphism, we say that R is I-adically complete.

3.1.2 Remark 3.2. In the special case where I = zR is a principal ideal with z ∈ Rhaving the property that

⋂∞n=1 z

nR = (0), then the z-adic completion of R is theinverse limit

(3.13.1.0.2) R∗ := lim←−

n

R/znR.

The ring R∗ also has the form

R∗ =R[[y]]

(y − z)R[[y]],

where y is an indeterminate over R. This follows fromN2[72, (17.5)], provided that

the ideal (y − z)R[[y]] is closed in the topology on R[[y]] defined by the powersof the ideal (y, z)R[[y]] = (y − z, y)R[[y]]. By Corollary

3.2.42.9, the element y is

in the Jacobson radical of R[[y]]. Hence the image of y in R[[y]](y−z)R[[y]] is in the

Jacobson radical of this quotient ring. Since this quotient ring is Noetherian, wehave

⋂∞n=1

ynR[[y]](y−z)R[[y]] = (0). It follows that the ideal (y − z)R[[y]] is closed. Thus

the elements of R∗ are power series in z with coefficients in R, but without theuniqueness of expression as power series that occurs in the formal power series ringR[[y]]. If R is already complete in its (z)-adic topology, then R = R∗, but often it

1We refer to Appendix A ofM[68] for the definition of direct and inverse limits.

13

14 3. MORE TOOLS APRIL 1, 2010

is the case for a Noetherian integral domain R that there exist elements of R∗ thatare transcendental over the field of fractions of R. If, for example, the Noetherianintegral domain R has countable cardinality, then a simple cardinality argumentimplies that for each nonzero nonunit z of R there exist infinitely many elementsin the (z)-adic completion R∗ of R that are algebraically independent over R.

We reserve the notation R for the situation whereR is a local ring with maximalideal m and R is the m-adic completion of R. For a local ring (R,m), we say thatR is “the” completion of R. If m is generated by elements a1, . . . , an, then R isrealizable by taking the a1-adic completion R∗

1 of R, then the a2-adic completionR∗

2 of R∗1, . . ., and then the an-adic completion of R∗

n−1.We use the following definitions.

3anstuff Definition 3.3. A Noetherian local ring R is said to be(1) analytically unramified if the completion R is reduced, i.e., has no nonzero

nilpotent elements;(2) analytically irreducible if the completion R is an integral domain;(3) analytically normal if the completion R is an integrally closed (i.e., normal)

domain.

If R has any one of the properties of Definition3anstuff3.3, then R is reduced. If R

has either of the last two properties, then R is an integral domain.

3.2. Basic theorems about completions3.38

3.38.0 Remark 3.4. Let I be an ideal of a ring R.(1) If R is I-adically complete, then I is contained in the Jacobson radical

J(R), where J(R) is the intersection of the maximal ideals of RM[68, The-

orem 8.2] orM1[66, 24.B, pages 73-74].

(2) If R is a Noetherian ring, then the I-adic completion R∗ of R is flat overR

M1[66, Corollary 1, page 170].

(3) If R is Noetherian then the I-adic completion R∗ of R is faithfully flatover R ⇐⇒ for each proper ideal J of R we have JR∗ 6= R∗.

(4) If R is a Noetherian ring and I ⊆ J(R), then the I-adic completion R∗ isfaithfully flat over R

M1[66, Theorem 56, page 172].

In Theorem 8 of Cohen’s famous paperCo[18] on the structure and ideal theory

of complete local rings a result similar to Nakayama’s lemma is obtained with-out the usual finiteness condition of Nakayama’s lemma. As formulated in

M[68,

Theorem 8.4], the result is:

3.38.1 Theorem 3.5. (A version of Cohen’s Theorem 8) Let I be an ideal of a ring Rand let M be an R-module. Assume that R is complete in the I-adic topology and⋂∞n=1 I

nM = (0). If M/I is generated over R/I by elements w1, . . . , ws and wi isa preimage in M of wi for 1 ≤ i ≤ s, then M is generated over R by w1, . . . , ws.

Let K be a field and let R = K[[x1, . . . , xn]] be a formal power series ring inn variables over K. It is well-known that there exists a K-algebra embedding of Rinto the formal power series ring K[[y, z]] in two variables over K

ZSII[106, page 219].

We observe in Corollary3.38.23.6 restrictions on such an embedding.

3.38.2 Corollary 3.6. Let (R,m) be a complete local ring and assume that the mapϕ : (R,m)→ (S,n) is a local homomorphism.

3.3. EXCELLENCE, DEFINITION AND BASIC PROPERTIES 15

(1) If mS is n-primary and S/n is finite over R/m, then S is a finitelygenerated R-module.

(2) If mS = n and R/m = S/n, then ϕ is surjective.(3) If R = K[[x1, . . . , xn]] is a formal power series ring in n > 2 variables

over the field K and S = K[[y, z]] is a formal power series ring in twovariables over K, then ϕ(m)S is not n-primary.

We record in Remarks3.38.43.8 several consequences of Cohen’s structure theorems

for complete local rings. We use the following definitions.

3.38.3 Definitions 3.7. Let (R,m) be a local ring.(1) (R,m) is said to be equicharacteristic if R has the same characteristic as

its residue field R/m.(2) A subfield k of R is a coefficient field of R if the canonical map of R →

R/m restricts to an isomorphism of k onto R/m.

3.38.4 Remarks 3.8.M[68, Theorem 28.3 and p. 222] and

N2[72, (32.1), (32.2) and Ex.

1 on page 122] andKatz[59].

(1) Every equicharacteristic complete Noetherian local ring has a coefficientfield.

(2) If k is a coefficient field of a complete Noetherian local ring (R,m) andx1, . . . , xn are generators of m, then every element of R can be expandedas a power series in x1, . . . , xn with coefficients in k. Thus R is a homo-morphic image of a formal power series ring in n variables over k.

(3) Every complete Noetherian local ring (whether equicharacteristic or not)is a homomorphic image of a regular local ring.

(4) If (R,m) is a complete Noetherian local domain, then the integral closureof R in a finite algebraic field extension is a finite R-module.

(5) If a Noetherian local ring R is analytically unramified, then the integralclosure of R is a finite R-module.

(6) Let (R,m) be a one-dimensional Noetherian local domain.(i) The integral closure R of R is a finite R-module if and only if the

completion R of R is reduced, i.e., if and only if R is analyticallyunramified.

(ii) The minimal primes of R are in one-to-one correspondence with themaximal ideals of R.

3.3. Excellence, definition and basic properties3.12

Let f : A → B be a ring homomorphism. The map f can always be factoredas the composite of the surjective map A → f(A) followed by the inclusion mapf(A) → B. This is often helpful for understanding the relationship of A and B. IfJ is an ideal of B, then f−1(J) is an ideal of A called the contraction of J to Awith respect to f . If Q is a prime ideal of B, then P := f−1(Q) is a prime ideal ofA. Thus associated with the ring homomorphism f : A→ B, there is a well-definedmap f∗ : SpecB → SpecA of topological spaces, where for Q ∈ SpecB we definef∗(Q) = f−1(Q) = P ∈ SpecA.

Given a ring homomorphism f : A → B and an ideal I of A, the ideal f(I)Bis called the extension of I to B with respect to f . For P ∈ SpecA, the extensionideal f(P )B is, in general, not a prime ideal of B. The fiber over P in SpecB is the


set of all Q ∈ SpecB such that f∗(Q) = P . Exercise 6 below asserts that the fiberover P is empty if and only if P is the contraction of the extended ideal f(P )B.In general, the fiber over P in SpecB is the spectrum of the ring S−1(B/f(P )B),where S is the multiplicatively closed set A \ P . Notice that a prime ideal Q of Bcontracts to P in A if and only if f(P ) ⊆ Q and Q∩ S = ∅. This describes exactlythe prime ideals of S−1(B/f(P )B).

3.20 Definitions 3.9. Let P and Q be prime ideals of a ring A.(1) If P ( Q, we say that the inclusion P ( Q is saturated if there is no prime idealof A strictly between P and Q.(2) A possibly infinite chain of prime ideals · · · ( Pi ⊆ Pi+1 ( · · · is called saturatedif every inclusion Pi ( Pi+1 is saturated.(3) A ring A is catenary provided for every pair of prime ideals P ( Q of A, everychain of prime ideals from P to Q can be extended to a saturated chain and everytwo saturated chains from P to Q have the same number of inclusions.(4) A ring A is universally catenary provided every finitely generated A-algebra iscatenary.(5) A ring A is said to be equidimensional if dimA = dimA/P for every minimalprime P of A.

Theoremratliff3.10 is a well-known result of Ratliff, cf,

M[68, Theorem 31.7].

ratliff Theorem 3.10. A Noetherian local domain A is universally catenary if andonly if its completion A is equidimensional.

A sharper result also due to Ratliff relating the universally catenary propertyto properties of the completion is Theorem

15.2.13.11.

15.2.1 Theorem 3.11.Ra[80, Theorem 2.6] A Noetherian local ring (R,m) is univer-

sally catenary if and only if the completion of R/p is equidimensional for everyminimal prime ideal p of R.

ucathom Remark 3.12. Every Noetherian local ring that is a homomorphic image of aregular local ring, or even a homomorphic image of a Cohen-Macaulay local ring,is universally catenary, cf.

M[68, Theorem 17.9, page 137].

3.388 Definition 3.13. Let f : A → B be a ring homomorphism, let P ∈ SpecAand set S = A\P . The fiber over P with respect to the map f is said to be regularif the ring S−1(B/f(P )B) := C is a Noetherian regular ring, i.e., C is a Noetherianring with the property that its localization at every prime ideal is a regular localring.

For P ∈ SpecA, let k(P ) denote the field S−1(A/P ), where S = A \ P .

3.389 Definition 3.14. With notation as in Definition3.3883.13, the fiber over P with

respect to the map f is said to be geometrically regular if for every finite extensionfield F of k(P ) the ring C ⊗A F is regular. The map f : A → B is said to havegeometrically regular fibers if for each P ∈ SpecA the fiber over P is geometricallyregular.

3.41 Definition 3.15. A homomorphism f : A→ B of Noetherian rings is said tobe regular if it is flat with geometrically regular fibers.

EXERCISES 17

3.415 Example 3.16. Let x be an indeterminate over a field k of characteristic zero,and let

A := k[x(x− 1), x2(x− 1)](x(x−1),x2(x−1)) ⊂ k[x](x) =: B.

The inclusion map f : A → B has geometrically regular fibers, but f is not flat.Hence f is not a regular morphism.

We present in Chapterflatpoly11 examples of maps of Noetherian rings that are regular,

and other examples that are flat but fail to be regular.

The formal fibers of a Noetherian local ring as in Definition3.393.17 play an im-

portant role in the concept of excellence of a Noetherian ring.

3.39 Definition 3.17. Let (R,m) be a Noetherian local ring and let R be the m-adic completion of R. The formal fibers of R are the fibers of the canonical inclusionmap R → R.

3.42 Definition 3.18. A Noetherian ring A is called a G-ring if for each prime idealP of A the map of AP to its PAP -adic completion is regular, or, equivalently, theformal fibers of AP are geometrically regular for each prime ideal P of A.

3.43 Definition 3.19. A Noetherian ring A is excellent if(i) A is universally catenary,(ii) A is a G-ring, and(iii) for every finitely generated A-algebra B, the set Reg(B) of primes P of

B for which BP is a regular local ring is an open subset of SpecB.

3.435 Remark 3.20. The class of excellent rings includes the ring of integers as wellas all fields and all complete Noetherian local rings

M[68, page 260]. All Dedekind

domains of characteristic zero are excellentM1[66, (34.B)]. The usefulness of the con-

cept of excellent rings is enhanced by the fact that the class of excellent rings isstable under the ring-theoretic operations of localization and passage to a finitelygenerated algebra

G[30, Chap. IV],

M1[66, (33.G) and (34.A)].

3.436 Remark 3.21. In Corollary11.5.39.11 of Chapter

flatholds9, we prove that the 2-dimensional

Noetherian local ring B constructed in Example11.4.59.8 has the property that the map

f : B → B has geometrically regular fibers. This ring B of Example11.4.59.8 is also an

example of a catenary ring that is not universally catenary. Thus the property ofhaving geometrically regular formal fibers does not imply that a Noetherian localring is excellent.

Exercises

(1) (D[21]) Let R be an integral domain and let f ∈ R[[x]] be a nonzero nonunit

of the formal power series ring R[[x]]. Prove that the principal ideal fR[[x]] isclosed in the (x)-adic topology, that is, fR[[x]] =

⋂m≥0(f, x

m)R[[x]].

Suggestion: Reduce to the case where c = f(0) is nonzero. Then f is a unitin the formal power series ring Rc[[x]]. If g ∈ ⋂m≥0(f, x

m)R[[x]], then g = fh

for some h ∈ Rc[[x]], say h =∑n≥0 hnx

n, with hn ∈ Rc. Let m ≥ 1. Asg ∈ (f, xm)R[[x]], g = fq+xmr, for some q, r ∈ R[[x]]. Thus g = fh = fq+xmr,hence f(h− q) = xmr. As f(0) 6= 0, h− q = xms, for some s ∈ Rc[[x]]. Henceh0, h1, . . . , hm−1 ∈ R.


(2) Let R be a commutative ring and let f =∑

n≥0 fnxn ∈ R[[x]] be a power series

having the property that its leading form fr is a regular element of R, that is,ord f = r, so f0 = f1 = · · · = fr−1 = 0, and fr is a regular element of R. Asin the previous exercise, prove that the principal ideal fR[[x]] is closed in the(x)-adic topology.

(3) (EH[23]) Let R be a commutative ring and let P be a prime ideal of the power

series ring R[[x]]. Let P (0) denote the ideal in R of constant terms of elementsof P .(i) If P (0) is a finitely generated ideal of R, prove that P is a finitely generated

ideal of R[[x]].(ii) If x 6∈ P and P (0) is generated by n elements ofR, prove that P is generated

by n elements of R[[x]].(iii) IfR is a PID, prove that every prime ideal ofR[[x]] of height one is principal.

(4) Let f : A → B be as in Example3.4153.16.

(i) Prove as asserted in the text that f has geometrically regular fibers but isnot flat.

(ii) Prove that the inclusion map of C := k[x(x − 1)(x(x−1)) → k[x](x) = B isflat and has geometrically regular fibers. Deduce that the map C → B isa regular map.

(5) Let φ : (R,m) → (S,n) be an injective local map of the Noetherian local ring(R,m) into the Noetherian local ring (S,n). Let R = lim←−

n

R/mn denote the

m-adic completion of R and let S = lim←−n

S/nn denote the n-adic completion

of S.(i) Prove that there exists a map φ : R→ S that extends the map φ : R → S.(ii) Prove that φ is injective if and only if for each positive integer n there

exists a positive integer sn such that nsn ∩R ⊆mn.(iii) Prove that φ is injective if and only if for each positive integer n the ideal

mn is closed in the topology on R defined by the ideals nn ∩Rn∈N, i.e.,the topology on R that defines R as a subspace of S.

Suggestion: For each n ∈ N, we have mn ⊆ nn ∩ R. Hence there existsa map φn : R/mn → R/(nn ∩ R) → S/nn, for each n ∈ N . The familyof maps φnn∈N determine a map φ : R → S. Since R/nn is Artinian, thedescending chain of ideals mn + (ns ∩ R)s∈N stabilizes, and mn is closed inthe subspace topology if and only if there exists a positive integer sn such thatnsn ∩ R ⊆ mn. This holds for each n ∈ N if and only if the m-adic topologyon R is the subspace topology from S.

(6) Let (R,m), (S,n) and (T,q) be Noetherian local rings. Assume there existinjective local maps f : R → S and g : S → T , and let h := gf : R → T bethe composite map. For f : R → S and g : S → T and h : R → T as in theprevious exercise, prove that h = gf .

(7) Let (R,m) and (S,n) be Noetherian local rings such that S dominates R andthe m-adic completion R of R dominates S.(i) Prove that R is a subspace of S.

EXERCISES 19

(ii) Prove that R is an algebraic retract of S, i.e., R → S and there exists asurjective map π : S → R such that π restricts to the identity map on thesubring R of S.

(8) Let k be a field and let R be the localized polynomial ring k[x]xk[x], and thusR = k[[x]]. Let n ≥ 2 be a positive integer. If chark = p > 0, assume that n isnot a multiple of p.(i) Prove that there exists y ∈ k[[x]] such that yn = 1 + x.(ii) For y as in (i), let S := R[yx] → k[[x]]. Prove that S is a local ring integral

over R with maximal ideal (x, yx)S. By the previous exercise, R = k[[x]]is an algebraic retract of S.

(iii) Prove that the integral closure S of S is not local. Indeed, if the fieldk contains a primitive n-th root of unity, then S has n distinct maximalideals. Deduce that R 6= S, so R is a nontrivial algebraic retract of S.

CHAPTER 4

First Examples of the Construction March 12,

2010

fex

The basic idea of the standard form of Construction4.4.15.3 of the next chapter is

to start with a well understood Noetherian domain R and then construct a newring A as follows: take an ideal-adic completion R∗ of R, and then intersect R∗

with an appropriate field L containing R. Thus A := L ∩ R∗. This is made moreexplicit in Section

4.45.1 of Chapter

constr5.

In this chapter we illustrate the construction with several examples.

4.1. Elementary examples4.1

We first consider examples where R is a polynomial ring over the field Q ofrational numbers. In the case of one variable the situation is well understood:

4.1.1 Example 4.1. Let y be a variable over Q, let R := Q[y], and let L be asubfield of the field of fractions of Q[[y]] such that Q(y) ⊆ L. Then the intersectiondomain A := L ∩ Q[[y]] is a rank-one discrete valuation domain (DVR) with fieldof fractions L (cf. Remark

3.02.12.1) and maximal ideal yA and y-adic completion A∗ =

Q[[y]]. For example, if we work with our favorite transcendental function ey andput L = Q(y, ey), then A is a DVR having residue field Q and field of fractions Lof transcendence degree 2 over Q.

The integral domain A of Example4.1.14.1 is perhaps the simplest example of a

Noetherian local domain on an algebraic function field L/Q of two variables that isnot essentially finitely generated over its ground field Q, i.e., A is not the localizationof a finitely generated Q-algebra. However A does have a nice description as aninfinite nested union of localized polynomial rings in 2 variables over Q, cf. Example4.1.1r6.6. Thus in a certain sense there is a good description of the elements of theintersection domain A in this case.

The two-dimensional (two variable) case is more interesting. The followingtheorem of Valabrega

V[96] is useful in considering this case.

4.1.2 Theorem 4.2. (Valabrega) Let C be a DVR, let y be an indeterminate overC, and let L be a subfield of Q(C[[y]]) such that C[y] ⊂ L. Then the integraldomain D = L∩C[[y]] is a two-dimensional regular local domain having completionD = C[[y]], where C is the completion of C.

Exercise 4 of this chapter outlines a proof for Theorem4.1.24.2. Applying Val-

abrega’s Theorem4.1.24.2, we see that the intersection domain is a two-dimensional

regular local domain with the “right” completion in the following two examples:

4.1.3 Example 4.3. Let x and y be indeterminates over Q and let C be the DVRQ(x, ex)∩Q[[x]]. Then A1 := Q(x, ex, y)∩C[[y]] is a two-dimensional regular localdomain with maximal ideal (x, y)A1 and completion Q[[x, y]].

21

22 4. FIRST EXAMPLES OF THE CONSTRUCTION MARCH 12, 2010

This example is generalized in Theorem11.4.19.3, as is explained in Remark

11.4.29.5.2.

4.1.4 Example 4.4. Let x and y be indeterminates over Q and let C be the DVRQ(x, ex) ∩ Q[[x]] as in Example

4.1.34.3. Then A2 := Q(x, y, ex, ey) ∩ C[[y]] is a

two-dimensional regular local domain with maximal ideal (x, y)A2 and completionQ[[x, y]].

4.1.5 Remarks 4.5. (1) There is a significant difference between the integral do-mains A1 of Example

4.1.34.3 and A2 of Example

4.1.44.4. As is shown in Theorem

11.4.19.3,

the two-dimensional regular local domain A1 of Example4.1.34.3 is, in a natural way,

a nested union of 3-dimensional regular local domains. It is possible therefore todescribe A1 rather explicitly. On the other hand, the two-dimensional regular localdomain A2 of Example

4.1.44.4 contains, for example, the element ex−ey

x−y . As discussedin Example

4.7.137.3, the associated nested union domain B naturally associated with A2

is 3-dimensional and non-Noetherian. Notice that the 2-dimensional regular localring A1 is on an algebraic function field in three variables over Q, while A2 is onan algebraic function field in four variables. Since the field Q(x, ex, y) is containedin the field Q(x, ex, y, ey), the local ring A1 is dominated by the local ring A2.

(2) It is shown in Chapteridwisec20 Theorem

6.3.920.30 and Corollary

6.3.1020.33 that if we go

outside the range of Valabrega’s theorem, that is, if we take more general subfieldsL of the field of fractions of Q[[x, y]] such that Q(x, y) ⊆ L, then the intersectiondomain A = L∩Q[[x, y]] can be, depending on L, a localized polynomial ring in n ≥3 variables over Q or even a localized polynomial ring in infinitely many variablesover Q. In particular, A = L ∩ Q[[x, y]] need not be Noetherian. Theorem

4.2.117.2

describes possibilities for the intersection domain A in this setting.

4.2. Historical examples4.3

Related to singularities of algebraic curves, there are classical examples of one-dimensional Noetherian local domains (R,m) such that the m-adic completion Ris not an integral domain, that is, R is analytically reducible, cf. Section

3.22.2. We

demonstrate this in Example4.3.04.6.

4.3.0 Example 4.6. Let X and Y be variables over Q and consider the localizedpolynomial ring

S : = Q[X,Y ](X,Y )Q[X,Y ] and the quotient ring R : =S

(X2 − Y 2 − Y 3)S.

Since the polynomial X2 − Y 2 − Y 3 is irreducible in the polynomial ring Q[X,Y ],the ring R is a one-dimensional Noetherian local domain Let x and y denote theimages in R of X and Y , respectively. The principal ideal yR is primary for themaximal ideal m = (x, y)R, and so the m-adic completion R is also the y-adiccompletion of R. Thus

R =Q[X ][[Y ]]

(X2 − Y 2((1 + Y )).

Since 1 + Y has a square root (1 + Y )1/2 ∈ Q[[Y ]], we see that X2 − Y 2(1 + Y )factors in Q[X ][[Y ]] as

X2 − Y 2(1 + Y ) = (X − Y (1 + Y )1/2) · (X + Y (1 + Y )1/2).

4.2. HISTORICAL EXAMPLES 23

Thus R is not an integral domain. Since the polynomial Z2 − (1 + y) ∈ R[Z] hasx/y as a root and x/y 6∈ R, the integral domain R is not normal, cf. Section

3.022.1.

The birational integral extension R := R[xy ] has two maximal ideals,

m1 := (m,x

y− 1)R = (

x− yy

)R and m2 := (m,x

y+ 1)R = (

x+ y

y)R.

To see, for example, that m1 = (x−yy )R, it suffices to show that m ⊂ (x−yy )R.

It is obvious that x − y ∈ (x−yy )R. We also clearly have x2−y2

y2 ∈ (x−yy )R, and

x2 − y2 = y3. Hence y3

y2 = y ∈ (x−yy )R, and so m1 is principal and generatedby x−y

y . Similarly, the maximal ideal m2 is principal and is generated by x+yy .

Thus R = R[xy ] is a PID, and hence is integrally closed. To better understand thestructure of R and R, it is instructive to extend the homomorphism

ϕ : S −→ S

(X2 − Y 2 − Y 3)S= R.

Let X1 := X/Y and S′ := S[X1]. Then S′ is a regular integral domain and themap ϕ can be extended to a map ψ : S′ → R[xy ] such that ψ(X1) = x

y . The kernelof ψ is a prime ideal of S′ that contains X2 − Y 2 − Y 3. Since X = Y X1, and Y 2

is not in ker ψ, we see that ker ψ = (X21 − 1− Y )S′. Thus

ψ : S′ −→ S′

(X21 − 1− Y )S′ = R[

x

y] = R.

Notice that X21 − 1− Y is contained in exactly two maximal ideals of S′, namely

n1 : = (X1 − 1, Y )S′ and n2 : = (X1 + 1, Y )S′.

The rings S1 := S′n1

and S2 := S′n2

are two-dimensional RLRs that are localquadratic transformations of S, and the map ψ localizes to define maps

ψn1 : S1 → S1

(X21 − 1− Y )S1

= Rm1 and ψn2 : S2 → S2

(X21 − 1− Y )S2

= Rm2 .

4.3.01 Remark 4.7. Examples given by AkizukiA[7] and Schmidt

Sc[87], provide one-

dimensional Noetherian local domains R such that the integral closure R is notfinitely generated as an R-module; equivalently, the completion R of R has nonzeronilpotents, cf.

N2[72, (32.2) and Ex. 1, page 122] and

Katz[59, Corollary 5].

If R is a normal one-dimensional Noetherian local domain, then R is a rank-one discrete valuation domain (DVR) and it is well-known that the completion ofR is again a DVR. Thus R is analytically irreducible. Zariski showed (cf.

ZSII[106,

pages 313-320]) that the normal Noetherian local domains that occur in algebraicgeometry are analytically normal, cf. Section

3.123.3. In particular, the normal local

domains occuring in algebraic geometry are analytically irreducible.This motivated the question of whether there exists a normal Noetherian local

domain for which the completion is not a domain. Nagata produced such examplesin

N1[70]. He also pinpointed sufficient conditions for a normal Noetherian local

domain to be analytically irreducibleN2[72, (37.8)].

In Example4.3.14.8, we present a special case of a construction of Nagata

N1[70],

N2[72, Example 7, pages 209-211] of a two-dimensional regular local domain A thatis not excellent and a two-dimensional normal Noetherian local domain D that isanalytically reducible. These concepts are defined in Section

3.123.3 and Section

3.13.1.


Nagata constructs A as a nested union of subrings. He proves that A is Noetherianwith completion A = Q[[x, y]]. Since the completion of a Noetherian local ring is afaithfully flat extension, it follows (cf. Remark

remflat2.11) that A is also an intersection

as defined in Example4.3.14.8.

Example 4.8. (Nagata) Let x and y be algebraically independent over Q and4.3.1let R be the localized polynomial ring R = Q[x, y](x,y). Then the completion of Ris R = Q[[x, y]]. Let τ ∈ xQ[[x]] be an element that is transcendental over Q(x, y),e.g., τ = ex − 1. Let ρ := y + τ and f := ρ2 = (y + τ)2. Now define

A : = Q(x, y, f) ∩Q[[x, y]] and D : =A[z]

(z2 − f)A[Z],

where z is an indeterminate. It is clear that A is a quasilocal integral domain.Nagata proves that f is a prime element of A and that A is a two-dimensionalregular local domain with completion A = Q[[x, y]], cf. Proposition

16.1n10.2. He also

shows that D is a normal Noetherian local domain.

4.3.2 Remarks 4.9. (1) The integral domain D in Example4.3.14.8 is analytically re-

ducible. This is because the element f factors as a square in the completion A ofA. Thus

D =Q[[x, y, z]]

(z − (y + τ))(z + (y + τ),

which is not an integral domain.(2) The two-dimensional regular local domain A of Example

4.3.14.8 is not a Na-

gata ring and therefore is not excellent (for the definition of a Nagata ring, seeDefinition

3.02.22.2.1, and for the definition of excellence, see Section

3.123.3). To see that

A is not a Nagata ring, notice that A has a principal prime ideal generated byf that factors in A = Q[[x, y]], namely f is the square of the prime element ρ ofA. Therefore the one-dimensional local domain A/fA has the property that itscompletion A/fA has a nonzero nilpotent element. This implies that the integralclosure of the one-dimensional Noetherian domain A/fA is not finitely generatedover A/fA. Hence A is not a Nagata ring. Moreover, the map A → A = Q[[x, y]]is not a regular morphism, cf. Section

3.123.3.

The existence of examples such as the normal Noetherian local domain D ofExample

4.3.14.8 naturally motivated the question: Is a Nagata domain necessarily

excellent? Rotthaus shows inR1[81] that the answer is “no” as described below.

In Example4.3.34.10, we present a special case of the construction of Rotthaus.

InR1[81] the ring A is constructed as a direct limit. The fact that A is Noetherian

implies that A is also an intersection as defined in Example4.3.34.10. To see this, we

use that A is Noetherian implies that its completion A is a faithfully flat extension,and then we apply Remark

remflat2.11.8.

4.3.3 Example 4.10. (Rotthaus) Let x, y, z be algebraically independent over Q andlet R be the localized polynomial ring R = Q[x, y, z](x,y,z). Let τ1 =

∑∞i=1 aiy

i ∈Q[[y]] and τ2 =

∑∞i=1 biy

i ∈ Q[[y]] be power series such that y, τ1, τ2 are alge-braically independent over Q, for example, τ1 = ey − 1 and τ2 = ey

2 − 1. Letu := x+ τ1 and v := z + τ2. Define

A := Q(x, y, z, uv) ∩ (Q[x, z](x,z)][[y]]).

4.3. MOTIVATING EXAMPLES 25

4.3.31 Remark 4.11. The integral domain A of Example4.3.34.10 is a Nagata domain

that is not excellent. Moreover, A is a 3-dimensional regular local domain suchthat the formal fibers of A are geometrically reduced, but are not geometricallyregular. Rotthaus shows in

R1[81] that A is Noetherian and that the completion A

of A is Q[[x, y, z]], so A is a 3-dimensional regular local domain. Since u, v are partof a regular system of parameters of A, it is clear that (u, v)A is a prime ideal ofheight two. It is shown in

R1[81], that (u, v)A ∩ A = uvA. Thus uvA is a prime

ideal and A(u,v) bA/uvA(u,v) bA is a non-regular formal fiber of A. Therefore A is notexcellent.

4.3. Motivating examples4.2

We give a family of examples such that for certain values of the parameters,one obtains an example of a 3-dimensional local Krull domain (B,n) such that Bis not Noetherian, n = (x, y)B is 2-generated and the n-adic completion B of B isa two-dimensional regular local domain.

Let k be a field let x and y be indeterminates over k, and let R be the localizedpolynomial ring R := k[x, y](x,y). If σ, τ ∈ R = k[[x, y]] are algebraically indepen-dent over R, then the polynomial ring R[t1, t2] in the two variables t1, t2 over R, canbe identified with a subring of R by means of an R-algebra isomorphism mappingt1 → σ and t2 → τ . The structure of the local domain A = k(x, y, σ, τ)∩ R dependson the residual behavior of σ and τ with respect to prime ideals of R. Theorem

4.2.14.12

illustrates this in the special case where σ ∈ k[[x]] and τ ∈ k[[y]]. This introducesa technique that is further developed in Chapters

constr5,

appcom6,

motiv7 and

noehomsec8.

4.2.1 Theorem 4.12. Let k be a field, let x and y be indeterminates over k, and let

σ :=∞∑i=0

aixi ∈ k[[x]] and τ :=

∞∑i=0

biyi ∈ k[[y]]

be formal power series that are algebraically independent over the fields k(x) andk(y), respectively. Then A := k(x, y, σ, τ) ∩ k[[x, y]] is a two-dimensional regularlocal domain with maximal ideal (x, y)A and completion A = k[[x, y]]. The inter-section ring A has a subring B with the following properties:

• The local ring A birationally dominates the local ring B.• The ring B is the directed union of a chain of four-dimensional regular

local domains that are localized polynomial rings in four variables over kand are birationally dominated by both A and B.• The local ring B is a Krull domain with maximal ideal n = (x, y)B.• The dimension of B is either 2 or 3.• The local ring B is Hausdorff in the topology defined by the powers of n.• The n-adic completion B of B is canonically isomorphic to k[[x, y]].

Moreover the following statements are equivalent:

(1) The ring B is Noetherian.(2) The dimension of B is 2.(3) The rings B and A are equal.(4) Every finitely generated ideal of B is closed in the n-adic topology on B.(5) Every principal ideal of B is closed in the n-adic topology on B.


Depending on the choice of σ and τ , B may or may not be Noetherian. In particularthere exist certain values for σ and τ such that B 6= A and other values such thatB = A.

To establish the asserted properties of the ring A of Theorem4.2.14.12, we use the

following consequence of the useful result of Valabrega stated as Theorem4.1.24.2 above.

4.2.2 Proposition 4.13. With the notation of Theorem4.2.14.12, the ring A is a two-

dimensional regular local domain with maximal ideal (x, y)A and completion A =k[[x, y]].

Proof. The ring C = k(x, σ) ∩ k[[x]] is a rank-one discrete valuation domainwith completion k[[x]], and the field k(x, y, σ, τ) = L is an intermediate field betweenthe fields of fractions of the rings C[y] and C[[y]]. Hence, by Theorem

4.1.24.2, A =

L ∩C[[y]] is a regular local domain with completion k[[x, y]]. 4.2.11 Remark 4.14. In examining properties of subrings of the formal power se-

ries ring k[[x, y]] over the field k, we use that the subfields k((x)) and k((y)) ofthe field Q(k[[x, y]]) are linearly disjoint over k as defined for example in

ZSI[105,

page 109]. It follows that if α1, . . . , αn ∈ k[[x]] are algebraically independent overk and β1, . . . , βm ∈ k[[y]] are algbraically independent over k, then the elementsα1, . . . , αn, β1, . . . , βm are algebraically independent over k.

We return to the proof of Theorem4.2.14.12 in Chapter

motiv7.

Exercises(1) Prove that the intersection domain A of Example

4.1.14.1 is a DVR with field of

fractions L and (y)-adic completion A∗ = Q[[y]].(2) Let R be an integral domain with field of fractions K. Let F be a subfield of K

and let S := F ∩R. For each principal ideal aS of S, prove that aS = aR ∩ S.(3) Let R be a local domain with maximal ideal m and field of fractions K. Let

F be a subfield of K and let S := F ∩ R. Prove that S is local with maximalideal m ∩ S, and thus conclude that R dominates S. Give examples to showthat either of R or S may be Noetherian while the other is not Noetherian.

(4) Assume the notation of Theorem4.1.24.2. Thus y is an indeterminate over the

DVR C and D = C[[y]] ∩ L, where L is a subfield of the field of fractions ofC[[y]] with C[y] ⊂ L. Let x be a generator of the maximal ideal of C and letR := C[y](x,y)C[y]. Observe that R is a two-dimensional RLR with maximalideal (x, y)R and that C[[y]] is a two-dimensional RLR with maximal ideal(x, y)D[[y]] that dominates R. Let m := (x, y)C[[y]] ∩D.(i) Using Exercise 2, prove that

C ∼= R

yR→ D

yD→ C[[y]]

yC[[y]]∼= C.

(ii) Deduce that C ∼= DyD , and that m = (x, y)D.

(iii) Let k := CxC denote the residue field of C. Prove that D

xD is a DVR andthat

k[y] → R

xR→ D

xD→ C[[y]]

xC[[y]]∼= k[[y]].

EXERCISES 27

(iv) For each positive integer n, prove that

R

(x, y)nR∼= D

(x, y)nD∼= C[[y]]

(x, y)nC[[y]].

Deduce that R = D = C[[y]], where C is the completion of C.(v) Let P be a prime ideal of D such that x 6∈ P . Prove that there exists b ∈ P

such that b(D/xD) = yr(D/xD) for some positive integer r, and deducethat P ⊂ (b, x)D.

(vi) For a ∈ P , observe that a = c1b + a1x, where c1 and a1 are in D. Sincex 6∈ P , deduce that a1 ∈ P and hence a1 = c2b+ a2x, where c2 and a2 arein D. Conclude that P ⊂ (b, x2)D. Continuing this process, deduce that

bD ⊆ P ⊆∞⋂n=1

(b, xn)D.

(vii) Extending the ideals to C[[y]], observe that

bC[[y]] ⊆ PC[[y]] ⊆∞⋂n=1

(b, xn)C[[y]] = bC[[y]],

where the last equality is because the ideal bC[[y]] is closed in the topologydefined by the ideals generated by the powers of x on the Noetherian localring C[[y]]. Deduce that P = bD.

(viii) Conclude by Theorem3.2.12.5 thatD is Noetherian and hence a two-dimensional

regular local domain with completion D = C[[y]].(5) Let k be a field and let f ∈ k[[x, y]] be a formal power series of order r ≥ 2.

Let f =∑∞

n=r fn, where fn ∈ k[x, y] is a homogeneous form of degree n. Ifthe leading form fr factors in k[x, y] as fr = α · β, where α and β are coprimehomogeneous polynomials in k[x, y] of positive degree, prove that f factors ink[[x, y]] as f = g · h, where g has leading form α and h has leading form β.Suggestion. Let G =

⊕n≥0Gn represent the polynomial ring k[x, y] as a

graded ring obtained by defining deg x = deg y = 1. Notice that Gn hasdimension n+ 1 as a vector space over k. Let degα = a and deg β = b. Thena + b = r and for each integer n ≥ r + 1, we have dim(α · Gn−a) = n − a + 1and dim(β ·Gn−b) = n− b + 1. Since α and β are coprime, we have

(α ·Gn−a) ∩ (β ·Gn−b) = fr ·Gn−r.Conclude that α · Gn−a + β · Gn−b is a subspace of Gn of dimension n + 1and hence that Gn = α ·Gn−a + β ·Gn−b. Let ga := α and hb := β. Sincefr+1 ∈ Gr+1 = α ·Gr+1−a + β ·Gr+1−b = ga ·Gb+1 + hb ·Ga+1, there existforms hb+1 ∈ Gb+1 and ga+1 ∈ Ga+1 such that fr+1 = ga·hb+1+hb·ga+1. SinceGr+2 = ga ·Gb+2 + hb ·Ga+2, there exist forms hb+2 ∈ Gb+2 and ga+2 ∈ Ga+2

such that fr+2 − ga+1 · hb+1 = ga · hb+2 + hb · ga+2. Proceeding by induction,assume for a positive integer s that there exist forms ga, ga+1, . . . , ga+s andhb, hb+1, . . . , hb+s such that the power series f−(ga+· · ·+ga+s)(hb+· · ·+hb+s)has order greater than or equal to r + s+ 1. Using that

Gr+s+1 = ga ·Gb+s+1 + hb ·Ga+s+1,


deduce the existence of forms ga+s+1 ∈ Ga+s+1 and hb+s+1 ∈ Gb+s+1 such thatthe power series f − (ga + · · · + ga+s+1)(hb + · · · + hb+s+1) has order greaterthan or equal to r + s+ 2.

(6) Let k be a field of characteristic zero. Prove that both

xy + z3 and xyz + x4 + y4 + z4

are irreducible in the formal power series ring k[[x, y, z]]. Thus there does notappear to be any natural generalization to the case of three variables of theresult in the previous exercise.

CHAPTER 5

Describing the Construction March 25, 2010

constrWe discuss a universal technique that yields the examples of the previous section

and also leads to more examples. There are two versions of the construction thatwe use. At first they appear to be quite different:

• The Inclusion Construction4.4.15.3 defines an intersection A := R∗∩L, where

R∗ is an ideal-adic completion of a Noetherian integral domain R and Lis a subfield of the total quotient ring of R∗ that is generated over Q(R)by elements τ1, . . . , τs of R∗ that are algebraically independent over R.• The Homomorphic Image Construction

4.4.25.4 defines an intersection of a ho-

momorphic image of an ideal-adic completion R∗ of a Noetherian integraldomain R with the field of fractions of R.

The two versions are explained in more detail below. As shown in Remarks4.4.45.7.2,

Construction4.4.25.4 is a generalization of Construction

4.4.15.3.

We begin with the setting for both constructions. This setting will also be usedin later chapters.

setconstr Setting 5.1. Let R be a Noetherian integral domain with field of fractionsK := Q(R), let z ∈ R be a nonzero nonunit, and let R∗ denote the (z)-adiccompletion of R.

zadic Remark 5.2. Section3.13.1 contains material on the (z)-adic completion R∗ of a

Noetherian integral domain R with respect to a nonzero nonunit z of R. Remark3.1.23.2

of Chapter3tools2 implies that R∗ has the form

R∗ =R[[y]]

(y − z)R[[y]],

where y is an indeterminate over R. It is natural to ask for conditions in order thatR∗ is an integral domain, or equivalently, that (y − z)R[[y]] is a prime ideal. Theelement y − z obviously generates a prime ideal of the polynomial ring R[y]. Ourassumption that z is a nonunit of R implies that (y− z)R[[y]] is a proper ideal. Weconsider in Exercise 3 of this chapter examples where (y − z)R[[y]] is prime andexamples where it is not prime.

5.1. Two construction methods and a picture4.4

With R, z and R∗ as in Settingsetconstr5.1, we describe two methods for the construc-

tion associated with R∗. The Inclusion Construction results in a domain containedin a power series extension of R, whereas the Homomorphic Image Constructionresults in a domain that is birational over R and is contained in a homomorphicimage of a power series extension of R.

29

30 5. DESCRIBING THE CONSTRUCTION MARCH 25, 2010

4.4.1 Construction 5.3. Inclusion Construction: Let τ1, . . . , τs ∈ zR∗ be alge-braically independent elements over R. We define A to be the intersection domainA := K(τ1, . . . , τs) ∩R∗.

4.4.2 Construction 5.4. Homomorphic Image Construction: Let I be an ideal ofR∗ having the property that P ∩ R = (0) for each P ∈ SpecR∗ that is associatedto I. Define A := K ∩ (R∗/I).

4.4.21 Note 5.5. The condition in (4.4.25.4), that P ∩ R = (0) for every prime ideal P

of R∗ that is associated to I, implies that the field of fractions K of R embeds inthe total quotient ring Q(R∗/I) of R∗/I. To see this, observe that the canonicalmap R → R∗/I is an injection and that regular elements of R remain regular aselements of R∗/I. In this connection, cf. Exercise 1.

4.4.3 Picture 5.6. The diagram below shows the relationship among these rings.

Q(R∗) R∗ Q(R∗/I)

R∗ L = K(τi) R∗/I K

A = R∗ ∩ L A = (R∗/I) ∩K

R R

(4.4.15.3) A := L ∩R∗ (

4.4.25.4) A := K ∩ (R∗/I)

4.4.4 Remarks 5.7. (1) Chaptersfex4,

intsec21 and

noeslocsec22 feature the Inclusion Construction

4.4.15.3

that realizes the intersection domain A := K(τ1, . . . , τs) ∩ R∗, while Chapternoehomsec8

features the Homomorphic Image Construction4.4.25.4. Iterations of Construction

4.4.15.3

yield the examples of Sections4.14.1,

4.34.2 and

4.24.3. Nagata’s Example

4.3.14.8 fits the format

of Construction4.4.15.3, since

A := Q(x, y, ρ2) ∩Q[[x, y]] = Q(x, y, ρ2) ∩Q[x][[y]].

(2) Construction4.4.25.4 includes Construction

4.4.15.3 as a special case. To see this,

let R, z, R∗ and τ1, . . . , τs ∈ zR∗ be as in Construction4.4.15.3. Let t1, . . . , ts be

indeterminates over R, define S := R[t1, . . . , ts], let S∗ be the z-adic completionof S and let I denote the ideal (t1 − τ1, . . . , ts − τs)S∗. Consider the followingdiagram, where D is the intersection domain result of Construction

4.4.25.4, when R

and R∗ there are replaced by S and S∗, and where λ is the R-algebra isomorphismof S into R∗ that maps ti → τi for i = 1, . . . , s. The map λ naturally extends to ahomomorphism of S∗ onto R∗ with the kernel of this extension being the ideal I.Thus there is an induced isomorphism of S∗/I onto R∗ that we also label as λ.

5.2. UNIVERSALITY 31

4.4.54.4.5 (4.4.45.7.2.1)

S := R[t1, . . . , ts] −−−−→ D := K(t1, . . . , ts) ∩ (S∗/I) −−−−→ S∗/I

λ

y λ

y λ

yR −−−−→ R[τ1, . . . , τs] −−−−→ A := K(τ1, . . . , τs) ∩R∗ −−−−→ R∗.

Since λ maps D isomorphically onto A, we see that Construction4.4.25.4 includes as a

special case Construction4.4.15.3.

(3) Many of the classical examples of interesting Noetherian local domains canbe obtained with Construction

4.4.15.3 where R is a regular local domain. Construc-

tion4.4.15.3, however, is not sufficient to obtain certain other types of rings such as

Ogoma’s celebrated exampleO1[76] of a normal non-catenary Noetherian local do-

main, for a homomorphic image of a regular local ring is universally catenary.Ogoma’s example can be realized, however, as an intersection with the homomor-phic image of a completion as described in Construction

4.4.25.4.

5.2. Universality4.6

In this section we describe how the methods of Section4.45.1 can be regarded as

universal for the construction of certain Noetherian local domains. Consider thefollowing general question.

4.6.1 Question 5.8. Let k be a field and let L/k be a finitely generated field exten-sion. What are the Noetherian local domains (A,n) such that

(1) L is the field of fractions A, and(2) k is a coefficient field for A?

Recall from Section3.022.1, that k is a coefficient field of (A,n) if the composite map

k → A→ A/n defines an isomorphism of k onto A/n.

In relation to Question4.6.15.8, we observe in Theorem

4.6.155.9 the following general

facts.

4.6.15 Theorem 5.9. Let (A,n) be a Noetherian local domain having a coefficientfield k. There exist a Noetherian local subring (R,m) of A and an ideal I of thecompletion R of R such that

(1) The local ring R is essentially finitely generated over k.(2) If Q(A) = L is finitely generated over k, then R has field of fractions L.(3) The field k is a coefficient field for R.(4) The local ring A dominates R and mA = n.(5) The inclusion map ϕ : R → A extends to a surjective homomorphism

ϕ : R → A of the m-adic completion R of R onto the n-adic completionA of A.

(6) The ideal I can be taken to be ker(ϕ), and then(a) R/I ∼= A, so R/I dominates A, and(b) P ∩ A = (0) for every P ∈ Ass(R/I), so the field of fractions Q(A)

of A embeds in the total ring of quotients Q(R/I) of R/I, and(c) A = Q(A) ∩ (R/I).

32 5. DESCRIBING THE CONSTRUCTION MARCH 25, 2010

Proof. Since A is Noetherian, there exist elements t1, . . . , tn ∈ n such that(t1, . . . , tn)A = n. For item 2, we may assume that L = k(t1, . . . , tn), since everyelement of Q(A) has the form a/b, where a, b ∈ n. For the existence of the integraldomain (R,m) and for item 1, we set T := k[t1, . . . , tn] and p := n ∩ T . LetR := Tp and m := n ∩ R. Then k ⊆ R ⊆ A, mA = n and R is essentially finitelygenerated over k and has k as a coefficient field. Thus we have established items1- 4. Even without the assumption that Q(A) is finitely generated over k, thereis a relationship between R and A that is realized by passing to completions. Letϕ be the inclusion map R → A. The map ϕ extends to a map ϕ : R → A, andby Corollary

3.38.23.6.2, the map ϕ is surjective. Let I := ker ϕ. Then R/I = A. This

implies item 5. The assertions in item 6 follow from that fact that A is a Noetherianlocal domain and A ∼= R/I. Applying Remarks

3.38.03.4, we have R/I is faithfully flat

over A, and by Remarkremflat2.11.6 the nonzero elements of A are regular on R/I. The

following commutative diagram, where the vertical maps are injections, displaysthe relationships among these rings:

(5.1)

Rϕ−−−−→ A ∼= R/I −−−−→ Q(R/I)x x x

k⊆−−−−→ R

ϕ−−−−→ A := Q(A) ∩ (R/I) −−−−→ Q(A) .

This completes the proof of Theorem4.6.155.9.

In relation to Question4.6.15.8, we summarize as follows.

4.6.2 Summary 5.10. Every Noetherian local domain (A,n) having a coefficient fieldk, and having the property that the field of fractions L of A is finitely generated overk is realizable as an intersection L ∩ R/I, where I is an ideal in the completion Rof a Noetherian local domain R essentially finitely generated over k and Q(R) = L.

4.6.3 Remark 5.11. Summary4.6.25.10 yields a “poor man’s” classification of the Noe-

therian local domains A with the stated properties. A drawback with (4.6.25.10), how-

ever, is that it is not true for each R,L, I as in (4.6.25.10) that L∩ (R/I) is Noetherian

(cf. ChapterintIIsec25, Section

13.425.4 below). To make the classification more satisfying an

important goal is to identify the ideals I of R and fields L such that L ∩ (R/I) isNoetherian.

If we start with a Noetherian local domain R that is essentially finitely gener-ated over a field k, then there often exist ideals I in the completion R of R suchthat the intersection domain Q(R) ∩ (R/I) is an interesting Noetherian local do-main that birationally dominates R. This technique may be used to describe theexample given by Nagata

N1[70] or

N2[72, Example 7, page 209], and other examples

given inBR1[11],

BR2[12],

H1[54],

O1[76],

O2[77],

R1[81],

R2[82], and

W[98].

In order to give a more precise criterion for the intersection domain A to beNoetherian, we restrict in Constructions

4.4.15.3 and

4.4.25.4 to a completion R∗ of R with

respect to a nonzero nonunit z of R. This sometimes permits an explicit descriptionof the intersection via the approximation methods of Chapter

appcom6. In Chapter

noehomsec8 we

give necessary and sufficient conditions for A to be computable as a nested unionof subrings of a specific form. We also prove that the Noetherian property for theassociated nested union is equivalent to flatness of a certain map.

EXERCISES 33

Exercises

(1) Let A be a Noetherian integral domain and let A → B be a map to a Noetherianextension ring B. For an ideal I of B, prove that the following are equivalent:(i) The induced map A → B/I is injective and each nonzero element of A is

regular on B/I.(ii) The field of fractions Q(A) of A naturally embeds in the total quotient ringQ(B/I) of B/I.

(iii) For each prime ideal P of B that is associated to I we have P ∩A = (0).(2) Let A := k[x] → k[x, y] =: B be a map of the polynomial ring k[x] into the

polynomial ring k[x, y] and let I = xyB. Prove that the induced map A→ B/Iis injective, but the field of fractions Q(A) of A does not embed into the totalquotient ring Q(B/I) of B/I.

(3) Let z be a nonzero nonunit of a Noetherian integral domain R, let y be anindeterminate, and let R∗ = R[[y]]

(z−y)R[[y]] be the (z)-adic completion of R.

(i) If z = ab, where a, b ∈ R are nonunits such that aR+ bR = R, prove thatthere exists a factorization

z − y = (a+ a1y + · · · ) · (b + b1y + · · · ) = (∞∑i=0

aiyi) · (

∞∑i=0

biyi),

where the ai, bi ∈ R, a0 = a and b0 = b.(ii) If R is a principal ideal domain, prove that R∗ is an integral domain if and

only if zR has prime radical.(4) Let A be a Noetherian integral domain and let A → B be a map to a Noetherian

extension ring B. Let I be an ideal of B that satisfies the equivalent conditionsof Exercise 1 and let C := Q(A) ∩ (B/I).(i) Prove that C = a/b | a, b ∈ A and a ∈ I + bB .(ii) Assume that J ⊆ I is an ideal of B that also satisfies the equivalent con-

ditions of Exercise 1, and let D := Q(A) ∩ (R/J). Prove that D ⊆ C.

CHAPTER 6

Two Approximations and their Connection

March 7, 2010

appcomOur goal in this chapter is to describe in detail the Inclusion Construction

4.4.15.3

and the Homomorphic Image Construction4.4.25.4. The first difficulty we face is identi-

fying precisely what we have constructed—because, while the form of the exampleas an intersection as given in Constructions

4.4.15.3 and

4.4.25.4 of Chapter

constr5 is wonderfully

concise, sometimes it is difficult to fathom.Using the notation of Setting

setconstr5.1, the approximation methods in this chapter

describe elements in the constructed intersection domain A of Constructions4.4.15.3

and4.4.25.4. These are useful in describing A. We discuss two approximation meth-

ods corresponding to the two construction methods of Chapterconstr5. We use this

approximation in Chaptermotiv7 to complete the motivating examples of Section

4.24.3.

Associated with each of the Constructions4.4.15.3 and

4.4.25.4 there is a computable

subring B of the intersection domain A that approximates A in each of the con-structions. In the case of Construction

4.4.15.3 we obtain an “approximation” domain

B that is a nested union of localizations of polynomial rings in s variables over R.In the case of Construction

4.4.25.4, the “approximation” domain B is a nested union

of birational extensions of R that are essentially finitely generated R-algebras. Ineach case B is contained in the corresponding intersection domain A.

6.1. Approximations for the Inclusion Construction4.45

We give an explicit description of the approximating domain B of the previousparagraph. We use the last parts, the endpieces, of the power series τ1, . . . , τs.

4.2.3 Endpiece Notation 6.1. Let R be a Noetherian integral domain and let z ∈R be a nonzero nonunit. Let R∗ be the z-adic completion of R, cf. Remark

3.1.23.2.

Each γ ∈ R∗ has an expansion as a power series in z over R,

γ :=∞∑i=0

cizi, where ci ∈ R.

For each positive integer n we define the nth endpiece γn of γ with respect to thisexpansion:

4.2.3.14.2.3.1 (4.2.36.1.1) γn :=

∞∑i=n+1

cizi−n.

For each n ∈ N, we have the relation

4.2.3.24.2.3.2 (4.2.36.1.2) γn = cnz + zγn+1.

35

36 6. TWO APPROXIMATIONS AND THEIR CONNECTION MARCH 7, 2010

In Construction4.4.15.3, the elements τ1, . . . τs ∈ zR∗ are algebraically independent

over the field of fractions Q(R) of R. Hence

U0 := R[τ1, . . . , τs] ⊆ R∗ ∩ Q(R[τ1, . . . , τs]) =: A,

and U0 is a polynomial ring in s variables overR. Each τi ∈ zR∗ has a representationτi :=

∑∞j=1 rijz

j, where the rij ∈ R. For each positive integer n, we associate withthis representation of τi the nth endpiece,

4.2.3.214.2.3.21 (4.2.36.1.3) τin :=

∞∑j=n+1

rijzj−n.

We define

4.2.3.224.2.3.22 (4.2.36.1.4) Un := R[τ1n, . . . , τsn] and Bn := (1 + zUn)−1Un

Then Un is a polynomial ring in s variables over R, and z is in every maximalideal of Bn, so z ∈ J (Bn), cf. Section

3.022.1 Using (

4.2.3.24.2.36.1.2), we have for each positive

integer n a birational inclusion of polynomial rings Un ⊆ Un+1. We also haveUn+1 ⊆ Un[1/z]. By Remark

3.38.03.4.1, the element z is in the Jacobson radical of R∗.

Hence the localization Bn of Un is also a subring of A and Bn ⊆ Bn+1. We define

11.2.011.2.0 (4.2.36.1.5) U :=

∞⋃n=1

Un =∞⋃n=1

R[τ1n, . . . , τsn] and B :=∞⋃n=1

Bn.

It follows that B = (1 + zU)−1U and

11.2.0111.2.01 (4.2.36.1.6) B ⊆ A := R∗ ∩ Q(R[τ1, . . . , τs]).

Remark 6.2. With the notation and setting of (4.2.36.1), the representation

τi =∞∑j=1

rijzj

of τi as a power series in z with coefficients in R is not unique. Indeed, sincez ∈ R, it is always possible to modify finitely many of the coefficients rij in thisrepresentation. It follows that the endpiece τin is also not unique. However, as weobserve in Proposition

4.5.226.3 the rings U and Un are uniquely determined by the τi.

4.5.22 Proposition 6.3. Assume the notation and setting of (4.2.36.1). Then the ring U

and the rings Un are independent of the representation of the τi as power series inz with coefficients in R.

Proof. For 1 ≤ i ≤ s, assume that τi and ωi = τi have representations

τi =∞∑j=1

aijzj and ωi =

∞∑j=1

bijzj,

where each aij , bij ∈ R. We define the nth-endpieces τin and ωin as in (4.2.36.1):

τin =∞∑

j=n+1

aijzj−n and ωin =

∞∑j=n+1

bijzj−n.

Then we have

τi =∞∑j=1

aijzj =

n∑j=1

aijzj + znτin =

∞∑j=1

bijzj =

n∑j=1

bijzj + znωin = ωi.

6.1. APPROXIMATIONS FOR THE INCLUSION CONSTRUCTION 37

Therefore, for 1 ≤ i ≤ s and each positive integer n,

znτin − znωin =n∑j=1

bijzj −

n∑j=1

aijzj , and so τin − ωin =

∑nj=1(bij − aij)zj

zn.

Thus∑nj=1(bij − aij)zj ∈ R is divisible by zn in R∗ . Since znR is closed in the

(z)-adic topology on R, we have znR = R ∩ znR∗. It follows that zn divides thesum

∑nj=1(bij − aij)zj in R. Therefore τin − ωin ∈ R. Thus the rings Un and

U =⋃∞n=1 Un are independent of the representation of the τi.

4.211 Remark 6.4. Assume the setting of Construction4.4.15.3.

(1) Assume that R is a Noetherian local domain with maximal ideal m. Weobserve in this case that B is the directed union of the localized polynomial rings.Cn := (Un)Pn , where Pn := (m, τ1n, . . . , τsn)Un. It is clear that Cn ⊆ Cn+1,and Pn ∩ (1 + zUn) = ∅ implies that Bn ⊆ Cn. We show that Cn ⊆ Bn+1: Letad ∈ Cn, where a ∈ Un and d ∈ Un \ Pn. Then d = d0 +

∑si=1 τinbi, where d0 ∈ R

and each bi ∈ Un. Notice that d0 /∈ m since d /∈ Pn, and so d−10 ∈ R. Thus

dd0 = 1 +∑s

i=1 τinbid−10 ∈ (1 + zUn+1) since each τin ∈ zUn+1 by (

4.2.36.1.2). Hence

ad = ad0

dd0∈ Bn+1, and so B =

⋃∞n=1Bn =

⋃∞n=1 Cn.

(2) For the special case where R is a polynomial ring k[x] over a field k, andR∗ is the (x)-adic completion of R, a proof similar to that for item (1) shows thatB =

⋃Cn, where Cn = (Un)Pn and Pn := (x, τ1n, . . . , τsn)Un.

4.2li Definition 6.5. Using the notation in (4.2.36.1.3) and (

4.2.36.1.5), we say Construc-

tion4.4.15.3 is limit-intersecting over R with respect to the τi if B = A. In this case,

we refer to the sequence of elements τ1, . . . τs ∈ zR∗ as limit-intersecting over R, orbriefly, as limit-intersecting for A.

Example4.1.1r6.6 illustrates the limit-intersecting property. Notice that Example

4.1.14.1

is a special case of Example4.1.1r6.6 .

4.1.1r Example 6.6. Let R = k[x] where k is a field and x is an indeterminate. Let

τ1 :=∞∑j=1

r1jxj , . . . , τs :=

∞∑j=1

rsjxj ∈ xk[[x]]

be algebraically independent over k(x). By Remark3.02.12.1, A := k(x, τ1, . . . , τs)∩k[[x]]

is a DVR with (x)-adic completion A = A∗ = k[[x]]. The elements τ1, . . . , τs arelimit-intersecting over R, that is, A = B, where B is as defined in (

11.2.04.2.36.1.5). This

follows from Theorem11.4.1a9.2. We give the following direct proof.

Proof. Let Cn = k[x, τ1n, . . . , τsn](x,τ1n,...,τsn). By Remark4.2116.4, it suffices to

prove that A =⋃∞n=1 Cn. Notice that Cn is a RLR of dimension s+ 1, and Cn ⊆

Cn+1 ⊆ A with A birationally dominating Cn+1 and Cn+1 birationally dominatingCn. Since Cn is a regular local ring, the function that associates to each nonzerof ∈ Cn the nonnegative integer e := ordCn(f), where f is in the e-th power ofthe maximal ideal of Cn, but not in the (e + 1)-th power, defines a valuation, cf.Section

3.022.1.

Let a ∈ A. Using that A is birational over Cn and that Cn is a UFD, we canwrite a = fn/gn, where fn, gn ∈ Cn are relatively prime. Notice that a ∈ Cn ifand only if ordCn(gn) = 0. Let ordA denote the order valuation associated to theone-dimensional regular local ring A. Since a ∈ A and since ordA is a valuation,


we have ordA(a) = ordA(fn) − ordA(gn) ≥ 0. Since A dominates Cn, we haveordA(gn) ≥ 0 and ordA(gn) = 0 if and only if ordCn(gn) = 0. These statementshold for each of the rings Cn.

To show that a ∈ ⋃∞n=1 Cn and hence that

⋃∞n=1 Cn = A, it suffices to show

that ordCm(gm) = 0 for some positive integer m. If ordA(gn) > 0, it sufficesto show that with the representation a = fn+1

gn+1, where fn+1, gn+1 ∈ Cn+1 are

relatively prime, then ordA(gn+1) < ordA(gn). By (4.2.3.14.2.36.1.1) we see that the maximal

ideal (x, τ1n, . . . , τsn)Cn of Cn is contained in xCn+1. Thus there exist elementsf ′n+1, g

′n+1 in Cn+1 such that

fn = xf ′n+1 and gn = xg′n+1. Hence a =

fn+1

gn+1=

fngn

=xf ′

n+1

xg′n+1

=f ′n+1

g′n+1

.

Since Cn+1 is a UFD and fn+1 and gn+1 are relatively prime in Cn+1, we see thatg′n+1 = bgn+1 for some b ∈ Cn+1. Since A dominates Cn+1 we have

ordA gn = ordA x + ordA g′n+1 = 1 + ordA g′n+1 ≥ 1 + ordA gn+1.

This completes the proof that τ1, . . . , τs are limit-intersecting over R.

Example4.1.2r6.7 extends and generalizes Example

4.1.1r6.6 to a situation where R is a

polynomial ring in several variables over k[x].

4.1.2r Example 6.7. Let m be a positive integer and let R := k[x, y1, . . . , ym] be apolynomial ring in independent variables x, y1, . . . , ym over a field k. Let

τ1 :=∞∑j=1

r1jxj , . . . , τs :=

∞∑j=1

rsjxj ∈ xk[[x]]

be algebraically independent over k(x). Then C := k(x, τ1, . . . , τs) ∩ k[[x]] is aDVR with (x)-adic completion C = C∗ = k[[x]] by Remark

3.02.12.1. Let R∗ denote the

(x)-adic completion of R, and let

A := k(x, τ1, . . . , τs, y1, . . . , ym) ∩ R∗.

Let Un := k[x, y1, . . . , ym, τ1n, . . . , τsn], where the τin are the nth endpieces of theτi. Theorem

11.4.1a9.2 implies that

A = B :=∞⋃n=1

Bn, where Bn := (1 + xUn)−1Un.

Thus the elements τ1, . . . , τs are limit-intersecting over R.

4.5.3 Remark 6.8. The limit-intersecting property depends on the choice of the el-ements τ1, . . . , τs ∈ zR∗. For example, if R is the polynomial ring Q[z, y], thenR∗ = Q[y][[z]]. Let s = 1, and let τ1 = τ := ez − 1 ∈ zR∗. Then τ is alge-braically independent over Q(z, y). Let U0 = R[τ ]. Example

4.1.2r6.7 shows that τ is

limit-intersecting. On the other hand, the element yτ is not limit-intersecting. Forif U ′

0 := R[yτ, ], then Q(U0) = Q(U ′0) and the intersection domain

A = Q(U0) ∩ R∗ = Q(U ′0) ∩ R∗

is the same for τ and yτ . However the approximation domain B′ associated toU ′

0 does not contain τ . Indeed, τ 6∈ R[yτ ][1/z]. Hence B′ is properly containedin the approximation domain B associated to U0. We have B′ ( B = A and thelimit-intersecting property fails for the element yτ .

6.2. APPROXIMATIONS FOR THE HOMOMORPHIC IMAGE CONSTRUCTION 39

6.2. Approximations for the Homomorphic Image Construction4.55

Applying the notation from Settingsetconstr5.1 to the Homomorphic Image Construc-

tion4.4.25.4, we again approach A using a sequence of “computable rings” over R, but

we use the fronts of the power series involved, rather than the ends. The com-putable rings that are so obtained are not localizations of polynomial rings overR; instead they are localizations of finitely generated birational extensions withinthe fraction field of R. Nevertheless, as we explain below, there is a sequence ofcomputable rings associated to the Homomorphic Image Construction

4.4.25.4.

A goal of the computations here is to develop machinery for a proof that flatnessof a certain extension implies that the integral domainA of the Homomorphic ImageConstruction

4.4.25.4 is Noetherian and is a localization of a subring of R[1/z]. This

goal is realized in Theorem11.1.18.1 of Chapter

noehomsec8.

4.5.4 Frontpiece Notation 6.9. For Construction4.4.25.4 one no longer has an ap-

proximation of A by a nested union of polynomial rings over R. Indeed, in Con-struction

4.4.25.4 the extension R ⊆ A is birational. However, there is an analogous

approximation. We are given an ideal I of R∗ with the property that P ∩R = (0)for each P ∈ Ass(R∗/I). Since I ⊂ R∗, each γ ∈ I has an expansion as a powerseries in z over R,

γ :=∞∑i=0

aizi, where ai ∈ R.

For each positive integer n we define the nth frontpiece γn of γ with respect to thisexpansion:

γn :=n∑j=0

ajzj

zn.

Thus with I := (σ1, . . . , σt)R∗, then for each σi we have

σi :=∞∑j=0

aijzj, where the aij ∈ R,

and the nth frontpiece σin of σi is

(4.5.46.9.1) σin :=

n∑j=0

aijzj

zn.

For the Homomorphic Image Construction4.4.25.4, we obtain approximating rings

as follows: We define

(4.5.46.9.2) Un := R[σ1n, . . . , σtn], and Bn := (1 + zUn)−1Un.

The rings Un and Bn are subrings of K. We observe in Proposition11.256.11 that they

may also be considered to be subrings of R∗/I. First we show in Proposition4.5.4c6.10

that the approximating rings Un and Bn are nested.

4.5.4c Proposition 6.10. With the setting of Frontpiece Notation4.5.46.9, for each inte-

ger n ≥ 0 and for each integer i with 1 ≤ i ≤ t, we have(1) σin = −zai,n+1 + zσi,n+1.(2) (z, σi)R∗ = (z, ai0)R∗ and hence (z, I)R∗ = (z, a10, . . . , at0)R∗.(3) (z, σi)R∗ = (z, znσin)R∗ and hence (z, I)R∗ = (z, znσ1n, . . . , z

nσtn)R∗.Thus R ⊆ U0 and we have Un ⊆ Un+1 and Bn ⊆ Bn+1 for each positive integer n.


Proof. For item 1, by Definition4.5.46.9.1, we have σi,n+1 :=

∑n+1j=0

aijzj

zn+1 . Thus

zσi,n+1 =n+1∑j=0

aijzj+1

zn+1=

n∑j=0

aijzj

zn+ zai,n+1 = σin + zai,n+1.

For item 2, by definition

σi :=∞∑j=0

aijzj = ai0 + z

( ∞∑j=1

aijzj−1).

For item 3, observe that

σi =∞∑j=0

aijzj = znσin + zn+1

( ∞∑j=n+1

aijzj−n−1

).

The asserted inclusions follow from this.

11.25 Proposition 6.11. Assume the setting of Frontpiece Notation4.5.46.9 and let n be

a positive integer. As an element of the total quotient ring of R∗/I the nth frontpieceσin is the negative of the nth endpiece of σi defined in Endpiece Notation

4.2.36.1, that

is, for σi :=∑∞j=0 aijz

j, where each aij ∈ R,

σin = −∞∑

j=n+1

aijzj

zn= −

∞∑j=n+1

aijzj−n (mod I).

It follows that σin ∈ K ∩ (R∗/I), and so Un and Bn are subrings of R∗/I.

Proof. Let π denote the natural homomorphism from R∗ onto R∗/I. Usingthat the restriction of π to R is the identity map on R, we have

σi = znσin +∞∑

j=n+1

aijzj =⇒ znσin = σi −

∞∑j=n+1

aijzj

=⇒ π(znσin) = π(σi)− π(∞∑

j=n+1

aijzj)

=⇒ znσin = −znπ(∞∑

j=n+1

aijzj−n).

Therefore znσin ∈ znπ(R∗) = zn(R∗/I). Since z is a regular element of R∗/I, wehave σin = −π(

∑∞j=n+1 aijz

j−n), is an element of R∗/I.

4.5.5 Definition 6.12. We define

11.2.111.2.1 (4.5.56.12.1) U := ∪∞n=1Un, B := ∪∞n=1Bn = (1 + zU)−1U, A := K ∩ (R∗/I).

By Remark3.38.03.4.1, the element z is in the Jacobson radical of R∗. Hence B ⊆ A.

Construction4.4.25.4 is said to be limit-intersecting if B = A.

6.3. THE INCLUSION CONSTRUCT IS A HOMOMORPHIC IMAGE CONSTRUCT 41

6.3. The Inclusion Construct is a Homomorphic Image Construct4.56

Even though they appear different, Proposition11.256.11 shows for each of the power

series in the ideal I that the approximation in Frontpiece Notation4.5.46.9 is, modulo

the ideal I, the negative of the approximation in Endpiece Notation4.2.36.1.

In view of Remark4.4.45.7.2 of Chapter

constr5, Construction

4.4.25.4 includes Construc-

tion4.4.15.3 as a special case. When an Inclusion Construction

4.4.15.3 is translated to

a Homomorphic Image Construction4.4.25.4 under the correspondence described in

(4.4.45.7.2), Proposition

11.276.14 shows that the approximating rings using frontpieces as

in (4.5.46.9) correspond to the approximating rings using endpieces as in (

4.2.36.1).

We revise the notation slightly so that we can view the Inclusion Constructionas a Homomorphic Image Construction as in Remark

4.4.45.7.2:

11.26 Revised Notation 6.13. LetR, z, R∗ and τ1, . . . , τs be as in Construction4.4.15.3.

Thus

τi :=∞∑j=1

rijzj where rij ∈ R.

Let t1, . . . , ts be indeterminates over R, define S := R[t1, . . . , ts], let S∗ be the(z)-adic completion of S and let I denote the ideal (t1 − τ1, . . . , ts − τs)S∗. Noticethat S∗/I ∼= R∗ implies that P ∩ S = (0) for each prime ideal P ∈ Ass(S∗/I).Thus we are in the setting of the Homomorphic Image Construction and can defineD := Q(R)(t1, . . . , ts) ∩ (S∗/I). Let σi := ti − τi, for each i with 1 ≤ i ≤ s.

As in (4.5.46.9), define σin to be the nth frontpiece for σi over S. We consider

(4.4.54.4.45.7.2.1), the same diagram as before, where λ is the R-algebra isomomorphism

that maps ti → τi for i = 1, . . . , s:

S := R[t1, . . . , ts] −−−−→ D := Q(R)(t1, . . . , ts) ∩ (S∗/I) −−−−→ S∗/I

λ

y λ

y λ

yR −−−−→ R[τ1, . . . , τs] −−−−→ A := Q(R)(τ1, . . . , τs) ∩R∗ −−−−→ R∗.

As before, λ maps D isomorphically onto A via λ(ti) = τi for every i with 1 ≤ i ≤ s.Define Vn, Cn, V, C using Frontspiece Notation

4.5.46.9 and Equation

11.2.14.5.56.12.1 over S,

that is, with S and S∗ in place of R and R∗. Define Un, Bn, U,B using EndpieceNotation

4.2.36.1 and Equations

4.2.36.1.4 and Equation

11.2.04.2.36.1.5 over R. Then

(6.1)

Vn : = S[σ1n, . . . , σsn] = R[t1, . . . , ts][σ1n, . . . , σsn],

Cn : = (1 + zVn)−1Vn,

V : =∞⋃n=1

Vn, C :=∞⋃n=1

Cn = (1 + zV )−1V

U : =∞⋃n=1

Un =∞⋃n=1

R[τ1n, . . . , τsn] and B :=∞⋃n=1

Bn = (1 + zU)−1U.

11.27 Proposition 6.14. With the setting of Revised Notation11.266.13, the R-algebra

isomorphism λ has the following properties:

λ(σin) = τin, λ(Vn) = Un, λ(Cn) = Bn, λ(V ) = U, λ(C) = B,

for all i with 1 ≤ i ≤ s and all n ∈ N.


Proof. We have elements rij ∈ R so that

τi :=∞∑j=1

rijzj, τin :=

∞∑j=n+1

rijzj−n

σi := ti − τi = ti −∞∑j=1

rijzj, σin :=

ti −∑n

j=1 rijzj

zn

=⇒ λ(σin) =τi −

∑nj=1 rijz

j

zn

= τin.

The remaining statements of Proposition11.276.14 now follow.

11.27r Remark 6.15. With the setting of Revised Notation11.266.13, since Un is a poly-

nomial ring over R in the variables τ1n, . . . τsn, Proposition11.276.14 implies that Vn is

a polynomial ring over R in the variables σ1n, . . . , σsn. Thus

Vn := S[σ1n, . . . , σsn] = R[t1, . . . , ts][σ1n, . . . , σsn] = R[σ1n, . . . , σsn].

6.4. Basic properties of the approximations

The following general lemma is used in Theorem11.2.46.17 to establish more detailed

descriptions for our setting.

11.2.3 Lemma 6.16. Let S be a subring of a ring T and let x ∈ S be a regular elementof T such that

⋂∞n=1 x

nT = (0). The following conditions are equivalent.(1) (i) xS = xT ∩ S, and (ii) S/xS = T/xT .(2) For each positive integer n we have:

xnS = xnT ∩ S, S/xnS = T/xnT and T = S + xnT .(3) The rings S and T have the same (x)-adic completion.(4) (i) S = S[1/x] ∩ T , and (ii) T [1/x] = S[1/x] + T .

Proof. To see that item (1) implies item (2), observe that

xnT ∩ S = xnT ∩ xS = x(xn−1T ∩ S),

so the equality xn−1S = xn−1T ∩S implies the equality xnS = xnT ∩S. MoreoverS/xS = T/xT implies T = S + xT = S + x(S + xT ) = · · · = S + xnT , soS/xnS = T/xnT for every n ∈ N. Therefore (1) implies (2).

It is clear that item (2) is equivalent to item (3).To see that item (2) implies (4i), let s/xn ∈ S[1/x] ∩ T with s ∈ S and n ≥ 0.

Item (2) implies that s ∈ xnT ∩ S = xnS and therefore s/xn ∈ S. To see (4ii), lettxn ∈ T [1/x] with t ∈ T and n ≥ 0. Item (2) implies that t = s + xnt1 for somes ∈ S and t1 ∈ T . Therefore t

xn = sxn + t1. Thus (2) implies (4).

It remains to show that item (4) implies item (1). To see that (4) implies (1i),let t ∈ T and s ∈ S be such that xt = s. Then t = s/x ∈ S[1/x] ∩ T = S, by (4i).Thus xt ∈ xS. To see that (4) implies (1ii), let t ∈ T . Then t

x = sxn + t′, for some

n ∈ N, s ∈ S and t′ ∈ T by (4ii). Thus t = sxn−1 + t′x. Hence t − t′x = s

xn−1 ∈S[1/x] ∩ T = S, by (4ii).

Theorem11.2.46.17 relates to the analysis of the homomorphic image construction.

11.2.4 Theorem 6.17. With the setting of Frontpiece Notation4.5.46.9 and Definition

4.5.56.12,

we have for each positive integer n

6.4. BASIC PROPERTIES OF THE APPROXIMATIONS 43

(1) The ideals of R containing zn are in one-to-one inclusion preserving cor-respondence with the ideals of R∗ containing zn. In particular, we have(I, z)R∗ = (a10, . . . , at0, z)R∗ and

(I, z)R∗ ∩ R = (a10, . . . , at0, z)R∗ ∩ R = (a10, . . . , at0, z)R.

(2) The ideal (a10, . . . , at0, z)R equals z(R∗/I)∩R under the identification ofR as a subring of R∗/I.

(3) zn(R∗/I)∩A = znA, zn(R∗/I)∩U = znU, zn(R∗/I)∩B = znB.(4) U/znU = B/znB = A/znA = R∗/(znR∗ + I). The rings A, U and B all

have (z)-adic completion R∗/I, that is, A∗ = U∗ = B∗ = R∗/I.(5) R[1/z] = U [1/z], U = R[1/z]∩B = R[1/z]∩A = R[1/z]∩ (R∗/I) and the

integral domains R, U , B and A all have the same field of fractions K.

Proof. The first assertion of (1) follows because R/znR is canonically iso-morphic to R∗/znR∗. The next assertion of (1) follows from part 2 of Proposi-tion

4.5.4c6.10. If γ =

∑ti=1 σiβi + zτ ∈ (I, z)R∗ ∩R, where τ, βi ∈ R∗, then write each

βi = bi + zβ′i, where bi ∈ R, β′

i ∈ R∗. Thus γ−∑ti=1 ai0bi ∈ zR∗ ∩R = zR, and so

γ ∈ (a10, . . . , at0, z)R.Since z(R∗/I) = (z, I)R∗/I, we have (a10, . . . , at0, z)R ⊆ z(R∗/I) ∩ R. The

reverse inclusion in item (2) follows from (I, z)R∗ = (a10, . . . , at0, z)R∗.The first assertion of item (3) follows from the definition of A as (R∗/I)∩K. To

see that z(R∗/I)∩U ⊆ zU , let g ∈ z(R∗/I)∩U . Then g ∈ Un, for some n, impliesg = r0 + g0, where r0 ∈ R, g0 ∈ (σ1n, . . . , σtn)Un. Also σin = −zai,n+1 + zσi,n+1,and so g0 ∈ zUn+1 ⊆ z(R∗/I). Now r0 ∈ (z, σ1, . . . , σt)R∗ = (I, z)R∗. Thus byitem (1), r0 ∈ (a10, . . . , at0, z)R. Also each ai0 = σi− z

∑∞j=1 aijz

j−1 ∈ zU becauseai0 = zσi1 − zai1.

Thus r0 ∈ zU , as desired. This prove that zn(R∗/I) ∩ U = znU . SinceB = (1 + zU)−1U , we also have zn(R∗/I) ∩B = znB..

Item (4) follows from item (3) and Lemma11.2.36.16.

For (5): if g ∈ U , then g ∈ Un, for some n. Clearly each σin ∈ R[1/z], and sog ∈ R[1/z]. To see that U = R[1/z] ∩ B, apply Lemma

11.2.36.16 with S = U, B = T .

Similarly we see that U = R[1/z] ∩A, since

R[1/z] ∩ A = R[1/z] ∩ (Q(R) ∩ (R∗/I)) = R[1/z] ∩ (R∗/I).

It is clear that the integral domains R, U , B and A all have the same field offractions K.

11.2.5 Remark 6.18. We note the following implications from Theorem11.2.46.17 .

(1) Item (5) of Theorem11.2.46.17 implies that the definitions in (

11.2.14.5.56.12.1) of B and

U are independent of(a) the choice of generators for I, and(b) the representation of the generators of I as power series in z.

(2) Item (5) of Theorem11.2.46.17 implies that the rings U = R[1/z]∩ (R∗/I) and

B = (1 + zU)−1U are uniquely determined by z and the ideal I of R∗.(3) Since z is in the Jacobson radical of B, item (4) of Theorem

11.2.46.17 implies

that if b ∈ B is a unit of A, then b is already a unit of B.(4) We have the following diagram displaying the relationships among the

rings. Recall that B = (1 + zU)−1U .


Q(R) Q(U) Q(B) Q(A)⊆−−−−→ Q(R∗/I)x x x x x

R[1/z] U [1/z] ⊆−−−−→ B[1/z] ⊆−−−−→ A[1/z] ⊆−−−−→ (R∗/I)[1/z]x x x x xR

⊆−−−−→ U = ∪ Un ⊆−−−−→ B⊆−−−−→ A

⊆−−−−→ R∗/I.

We record in Corollary11.2.516.19 several basic properties of the approximations

for Inclusion Construction4.4.15.3 that follow from Theorem

11.2.46.17, Remark

11.2.56.18 and

Proposition4.5.226.3.

11.2.51 Corollary 6.19. With the setting of Endpiece Notation4.2.36.1, we have for each

positive integer n(1) znR∗ ∩A = znA, znR∗ ∩B = znB and znR∗ ∩ U = znU .(2) R[1/z] = U [1/z] and U = R[1/z] ∩B = R[1/z] ∩A.(3) R/znR = U/znU = B/znB = A/znA = R∗/znR∗ .(4) R∗ = U∗ = B∗ = A∗.(5) The definitions in (

11.2.04.2.36.1.5) of B and U are independent of the representa-

tions given in Notation4.2.36.1 for the τi as power series in R∗.

Exercises(1) Prove item (2) of Remark

4.2116.4, that is, with R a polynomial ring k[x] over

a field k and R∗ = k[[x]] the (x)-adic completion of R, show that with thenotation of Construction

4.4.15.3 we have B =

⋃Cn, where Cn = (Un)Pn and

Pn := (x, τ1n, . . . , τsn)Un. Thus B is a DVR that is the directed union of abirational family of localized polynomial rings in n+ 1 indeterminates.

(2) Assume the setting of Frontpiece Notation4.5.46.9 and Definition

4.5.56.12. If J is a

proper ideal of B, prove that JB∗ is a proper ideal of B∗, where B∗ is the(z)-adic completion of B.

(3) Assume the setting of Frontpiece Notation4.5.46.9, and let W denote the set of

elements ofR∗ that are regular onR∗/I. Prove that the natural homomorphismπ : R∗ → R∗/I extends to a homomorphism π : W−1R∗ →W−1(R∗/I).

CHAPTER 7

Revisiting the Motivating Examples March 20,

2010

motiv

In this chapter we continue the proof of Theorem4.2.14.12 stated in Chapter

fex4.

We give specific examples that illustrate possible outcomes of the construction inChapter

constr5.

7.1. Properties of the motivating examples4.7

We begin by fixing notation. We also include several remarks concerning theintegral domains that are used in the proof of Theorem

4.2.14.12.

4.7.n Notation and Remarks 7.1. Let k be a field, let x and y be indeterminatesover k, and let

σ :=∞∑i=0

aixi ∈ k[[x]] and τ :=

∞∑i=0

biyi ∈ k[[y]]

be formal power series that are algebraically independent over the fields k(x) andk(y), respectively. Let R := k[x, y](x,y), and let σn, τn be the nth endpieces of σ, τdefined as in (

4.2.36.1). Define

Cn := k[x, σn](x,σn), Dn := k[y, τn](y,τn) Bn := k[x, y, σn, τn](x,y,σn,τn)

(4.77.1.1)

C := k(x, σ) ∩ k[[x]] = lim−→(Cn) =∞⋃n=1

Cn ;

D := k(y, τ) ∩ k[[y]] = lim−→(Dn) =∞⋃n=1

Dn

B := lim−→(Bn) =∞⋃n=1

Bn. A := k(x, y, σ, τ) ∩ k[[x, y]]

The relations (4.2.3.24.2.36.1.2) imply for each positive integer n the inclusions

Cn ⊆ Cn+1, Dn ⊆ Dn+1, and Bn ⊆ Bn+1.

Moreover, for each of these inclusions we have also birational domination of thelarger ring over the smaller, and the local rings Bn, Cn, Dn are all dominated byk[[x, y]]. By Remark

3.02.12.1, the rings C andD are rank-one discrete valuation domains;

as in Example4.1.1r6.6, they are the asserted directed unions. The rings Bn are four-

dimensional regular local domains that are essentially finitely generated over k.Thus B is the directed union of a chain of birational four-dimensional regular local

45

46 7. REVISITING THE MOTIVATING EXAMPLES MARCH 20, 2010

rings. By Proposition4.2.24.13, the ring A is a two-dimensional regular local domain

with maximal ideal (x, y)A and completion A = k[[x, y]].

We prove in Theorem4.2.117.2 other parts of Theorem

4.2.14.12.

4.2.11 Theorem 7.2. Assume the setting of Notation4.7.n7.1. Then B is a local Krull

domain with maximal ideal n = (x, y)B, the dimension of B is either 2 or 3, and Bis Hausdorff in the topology defined by the powers of n. The n-adic completion B ofB is canonically isomorphic to k[[x, y]], and the following statements are equivalent:

(1) B is Noetherian.(2) dimB = 2.(3) B = A.(4) Every finitely generated ideal of B is closed in the n-adic topology on B.(5) Every principal ideal of B is closed in the n-adic topology on B.

Proof. We have for each n ∈ N that

xB ∩Bn = (x, σn)Bn and yB ∩Bn = (y, τn)Bn,

where (x, σn)Bn and (y, τn)Bn are height-two prime ideals of the four-dimensionalregular local domain Bn. Therefore the unique maximal ideal of B is (x, y)B.Also, B(xB) and B(yB) are rank-one discrete valuation domains, since each is thecontraction to the field k(x, y, σ, τ) of the (x)-adic or the (y)-adic valuations ofk[[x, y]]. Moreover, B is birationally dominated by the two-dimensional regularlocal domain A = k(x, y, σ, τ) ∩ k[[x, y]].

Since B is dominated by k[[x, y]], B is Hausdorff in the topology defined by thepowers of n. Since n is finitely generated, B is Noetherian

N2[72, 31.7]. Therefore B

is a 2-dimensional regular local domain that canonically maps onto k[[x, y]]. Thiscanonical surjection must have kernel (0), and so we have B = k[[x, y]].

To see that B is a Krull domain, observe that, if q is a height-one prime of Bn,then q is contained in the union (x, σn)Bn ∪ (y, τn)Bn if and only if q ⊆ xB ∪ yB,and if q is not contained in xB ∪ yB, then Bn+1 ⊆ (Bn)q for each n ∈ N. Itfollows that if q is not contained in xB ∪ yB, then B ⊆ (Bn)q. Moreover, thecanonical map Spec(Bn+1) → Spec(Bn) restricts to a biregular correspondence ofthe height-one primes of Bn+1 not contained in xB∪yB with the height-one primesof Bn not contained in xB ∪ yB. It follows that, if Hn is the multiplicative systemHn := Bn \ ((xB ∩Bn) ∪ (yB ∩Bn)), then

(B0)H0 = · · · = (Bn)Hn = · · · = B[1xy

].

Since xB and yB are principal height-one prime ideals of B, we have

xiBxB ∩ B = xiB and yjByB ∩ B = yjB

for all positive integers i, j, and so B = B[ 1xy ]∩BxB ∩ByB. In particular, it follows

that B is a Krull domain. Since B is a birational extension of the 3-dimensionalNoetherian domain C[y, τ ], the dimension of B is at most 3.

Since the maximal ideal of B is finitely generated, it follows from Nishimura’sTheorem

3.2.22.6 that B is Noetherian if dimB = 2. On the other hand, since n =

(x, y)B, it is clear that if B is Noetherian, then B is a two-dimensional regular local

7.1. PROPERTIES OF THE MOTIVATING EXAMPLES 47

domain with completion k[[x, y]]. Since the completion of a Noetherian local ringis a faithfully flat extension, if B is Noetherian we have

B = k(x, y, σ, τ) ∩ k[[x, y]]

and hence B = A. Therefore statements (1), (2) and (3) are equivalent. If Bis Noetherian, then every ideal in B is closed in the n-adic topology by Krull’sintersection theorem, cf. Theorem

3.2.012.4. Statement (4) obviously implies statement

(5). To complete the proof of Theorem4.2.117.2, it suffices to show that statement (5)

implies that A = B. Since A birationally dominates B, we have B = A if and onlyif bA ∩ B = bB for every element b ∈ n. The principal ideal bB is closed in then-adic topology on B if and only if bB = bB ∩ B. Since B = A and bA = bA ∩A,it follows that every principal ideal bB is closed implies that B = A.

Depending on the choice of σ and τ , the ring B may fail to be Noetherian.Example

4.7.137.3 shows that in the setting of Theorem

4.2.117.2 the ring B can be strictly

smaller than A := k(x, y, σ, τ) ∩ k[[x, y]].4.7.13 Example 7.3. Let τ ∈ k[[y]] be defined to be σ(y), that is, set bi := ai for

every i ∈ N. We then have that θ := σ−τx−y ∈ A. Indeed,

σ − τ = a1(x− y) + a2(x2 − y2) + · · ·+ an(xn − yn) + · · · ,so θ = σ−τ

x−y ∈ k[[x, y]] ∩ k(x, y, σ, τ) = A. As a specific example, one may takek := Q and set σ := ex− 1 and τ := ey − 1. The ring B is a localization of the ringU :=

⋃n∈N k[x, y, σn, τn], where σn and τn are the endpieces defined in (

4.2.3.14.2.36.1.1).

4.7.13.1 Claim 7.4. The element θ is not in B.

Proof. If θ is an element of B, then

σ − τ ∈ (x− y)B ∩ U = (x− y)U.Let S := k[x, y, σ, τ ] and let Un := k[x, y, σn, τn] for each positive integer n. Wehave

U =⋃n∈N

Un ⊆ S[1xy

] ⊆ S(x−y)S,

where the last inclusion is because xy 6∈ (x− y)S. Thus θ ∈ B implies that

σ − τ ∈ (x− y)S(x−y)S ∩ S = (x− y)S,but this contradicts the fact that x, y, σ, τ are algebraically independent over k.

Therefore σ−τx−y 6∈ B, so B ( A and (x− y)B ( (x− y)A ∩B. Since an ideal of

B is closed in the n-adic topology if and only if the ideal is contracted from B andsince B = A, the principal ideal (x − y)B is not closed in the n-adic topology onB. This establishes the existence of a non-Noetherian three-dimensional quasilocalKrull domain B having a two-generated maximal ideal such that B birationallydominates a four-dimensional regular local domain. We discuss this example inmore detail in Example

4.7.13c8.11 of Chapter

noehomsec8.

On the other hand, the ring B may be Noetherian. To complete the proof ofTheorem

4.2.14.12, we establish in Example

7.6.10.17.13 below with k = Q that the elements

σ := ex − 1 and τ := e(ey−1) − 1 give an example where B = A. The property

of σ and τ we use to prove A = B is that for each height-one prime ideal Q of


R = Q[[x, y]] with Q ∩ R 6= (0) the images of σ and τ in R/Q are algebraicallyindependent over R/(Q ∩R). In Example

7.6.10.17.13 we use this to deduce that B = A.

4.7.14 Remark 7.5. In the setting of Notation4.7.n7.1, It seems natural to consider the

ring compositum of C = k[[x]] and D = k[[y]]. We outline in Exercise 2 of thischapter a proof of Kunz that the subring C[D] of k[[x, y]] is not Noetherian.

7.2. Residual algebraic independence

Recall that an extension of Krull domains R → S satisfies the condition PDE(“pas d’eclatement”, or in English “no blowing up”) provided that ht(Q∩R) ≤ 1 foreach prime ideal of height one Q in S, cf. Definition

3.02.22.2. The motivating example

leads us to consider in this section a related property as in the following definition.

7.6.1.0 Definition 7.6. Let R → S denote an extension of Krull domains. An ele-ment ν ∈ S is residually algebraically independent with respect to S over R if ν isalgebraically independent over R and for every height-one prime ideal Q of S suchthat Q∩R 6= 0, the image of ν in S/Q is algebraically independent over R/(Q∩R).

7.6.1r Remark 7.7. If (R,m) is a two-dimensional regular local domain, or more gen-erally an analytically normal Noetherian local domain of dimension at least two, itis naturally to consider the extension of Krull domains R → R and ask about theexistence of an element ν ∈ R that is residually algebraically independent with re-spect to R over R. If R has countable cardinality, so for example, if R = Q[x, y](x,y)is a localized polynomial ring over the field Q, then a cardinality argument impliesthe existence of an element ν ∈ R that is residually algebraically independent withrespect to R over R.

We show in Proposition6.3.420.25 and Theorem

6.4.420.37 of Chapter

idwisec20, that if ν ∈ m

is residually algebraically independent with respect to R over R, then the inter-section ring R ∩ Q(R[ν]) is the localized polynomial ring R[ν](m,ν). Therefore thisintersection ring is Noetherian, but has dimension one more than the dimension ofR and hence is not a subspace of R.

The existence of a residually algebraically independent element is importantfor completing the proof of the motivating example as we demonstrate below.

Setting 7.8. Let R := Q[x, y](x,y) be the localized polynomial ring in twovariables x and y over the field Q of rational numbers. Then the completion of Rwith respect to the maximal ideal m := (x, y)R is R := Q[[x, y]], the formal powerseries ring in the variables x and y.

7.6.1.1 Theorem 7.9. Let σ ∈ xQ[[x]] and τ ∈ yQ[[y]] be such that the following twoconditions are satisfied:

(i) σ is algebraically independent over Q(x) and τ is algebraically independentover Q(y).

(ii) trdegQ Q(y, τ, ∂nτ∂yn n∈N) > r := trdegQ Q(x, σ, ∂nσ

∂xn n∈N).

Following the notation of (4.7.n7.1), let C = k(x, σ)∩k[[x]] and let T = C[y](x,y). Then

ν := σ+ τ is residually algebraically independent over T . Hence ν is also residuallyalgebraically independent over Q[x, y](x,y).

7.2. RESIDUAL ALGEBRAIC INDEPENDENCE 49

Before proving Theorem7.6.1.17.9, we establish the existence of elements σ and τ

satisfying properties (i) and (ii) of Theorem7.6.1.17.9. Let σ = ex − 1 ∈ Q[[x]] and

choose for τ a hypertranscendental element in Q[[y]]. Recall that a power seriesτ =

∑∞i=0 biy

i ∈ Q[[y]] is called hypertranscendental over Q[y] if the set of allpartial derivatives ∂nτ

∂yn n∈N is infinite and algebraically independent over Q(y).(Two examples of hypertranscendental elements are the Gamma function and theRiemann Zeta function.1) Thus there exist elements σ, τ that satisfy the conditionsof Theorem

7.6.1.17.9. Another way to obtain such elements is to set σ = ex − 1 and

τ = e(ey−1) − 1. In this case, the conditions of Theorem

7.6.1.17.9 follow from

Ax[9].

In either case, Theorem7.6.1.17.9 implies that σ + τ is residually algebraically inde-

pendent, and, by Remark7.6.1r7.7, we have the following corollary.

7.6.2.2 Corollary 7.10. There exist elements σ ∈ Q[[x]] and τ ∈ Q[[y]] such thatthe element σ + τ ∈ (x, y)Q[[x, y]] is residually algebraically independent overQ[x, y](x,y). Therefore by Remark

7.6.1r7.7, the three-dimensional localized polynomial

ring Q[x, y, σ + τ ](x,y,σ+τ) is the intersection Q(x, y, σ + τ) ∩ R.

Proof. We begin the proof of Theorem7.6.1.17.9. The ring T := C[y](x,y) is between

R and its completion R and has completion T = R:

R = Q[x, y](x,y) −→ T = C[y](x,y) −→ R = T = Q[[x, y]].

We display the relationships among these rings.

R = Q[[x, y]]

T := C[y](x,y)

R := Q[x, y](x,y)

C := Q(x, σ) ∩Q[[x]]

= ∪ Q[x, σn](x,σn)

The rings of the example

Let P be a height-one prime ideal of R, let bars (for example, x), denote images

in R = R/P and set P := P ∩R and P1 := P ∩ T . Assume that P1 6= 0. It is easyto see for P1 = xT or P1 = yT that the image σ + τ of σ + τ in R := Q[[x, y]]/P isalgebraically independent over T := T/P1. Thus we assume that xy 6∈ P .

In the following commutative diagram, we identify Q[[x]] with Q[[x]] and Q[[y]]with Q[[y]] etc.

1The exponential function is, of course, far from being hypertranscendental.


Q[[y]] R = R/P Q[[x]]ψy ψx

T = T/P1

φy φxQ[y](y) R = R/P Q[x](x)

All morphisms in the diagram are injective and R is finite over both of the ringsQ[[x]] and Q[[y]]. Also R is algebraic over Q[x](x), since trdegQQ(R) = 1.

Let L denote the field of fractions of R. We may consider Q(y, τ, ∂nτ∂yn n∈N)

and Q(x, σ, ∂nσ∂xn n∈N) as subfields of L, where

trdegQ(Q(y, τ, ∂nτ

∂ynn∈N) > trdegQ(Q(x, σ, ∂

nσ

∂xnn∈N).

Let d denote the partial derivative map ∂∂x on Q((x)). Since the extension L of

Q((x)) is finite and separable, d extends uniquely to a derivation d : L → L. LetH denote the algebraic closure in L of the field Q(x, σ, ∂nσ

∂xn n∈N). Let p(x, y) ∈Q[[x, y]] be a prime element generating P . Claim

7.6.5.47.11 asserts that the images of H

and R under d are inside H and (1/p′)R, respectively, as shown in Picture7.6.5.47.11.1.

7.6.5.4 Claim 7.11. With the notation above: (i) d(H) ⊆ H.(ii) There exists a polynomial p(x, y) ∈ Q[[x]][y] with pQ[[x, y]] = P and p(y) = 0.

(iii) d(y) 6= 0 and p′(y)d(y) ∈ R, where p′(y) := ∂p(x,y)∂y .

(iv) For every element λ ∈ R, we have p′(y)d(λ) ∈ R, and so d(R) ⊆ (1/p′)R.

L := Q(R)d

L

R := Q[[x, y]]d 1

p′(y)R

Q[[x]] ∼= Q[[x]]d := ∂

∂x Q[[x]]

Hd

H :=(Q(x, σ, ∂nσ

∂xn ∞n=1))a ∩ L

Q[x, σ, ∂nσ∂xn ∞n=1]

dQ[x, σ, ∂nσ

∂xn ∞n=1]

Q1Q

Q

Picture7.6.5.47.11.1 The image of subrings of L via the extension d of d := ∂

∂x .


Proof. For item (i), since d maps Q(x, σ, ∂nσ∂xn n∈N) into itself, d(H) ⊆ H .

For item (ii), since x and y are not contained in P , the element p(x, y) is regularin y as a power series in Q[[x, y]] (in the sense of Zariski-Samuel

ZSII[106, p.145]). Thus

byZSII[106, Corollary 1, p.145] the element p(x, y) can be written as:

p(x, y) = ε(x, y)(yn + bn−1(x)yn−1 + . . .+ b0(x)),

where ε(x, y) is a unit of Q[[x, y]] and each bi(x) ∈ Q[[x]]. Hence P is also generatedby

p(y) := ε−1p = yn + cn−1yn−1 + · · ·+ c0,

where the ci ∈ Q[[x]].For item (iii), we observe that

p′(y) = nyn−1 + cn−1(n− 1)yn−2 + · · ·+ c1.

Since p(y) is the minimal polynomial of y over the field Q((x)), we have p′(y) 6= 0and 0 = p(y) := yn + cn−1y

n−1 + · · ·+ c1y + c0. Now

0 = d(p(y)) = d(yn + cn−1yn−1 + · · ·+ c1y + c0)

= nyn−1d(y) + cn−1(n− 1)yn−2d(y) + d(cn−1)yn−1 + · · ·+ c1d(y) + d(c1)y + d(c0)

= d(y)(nyn−1 + cn−1(n− 1)yn−2 + · · ·+ c1)+

+ d(cn−1)yn−1 + · · ·+ d(c1)y + d(c0)

= d(y)(p′(y)) +n−1∑i=0

d(ci)yi

=⇒ d(y)(p′(y)) =− ( n−1∑i=0

d(ci)yi)

and d(y) =−1p′(y)

n−1∑i=0

d(ci)yi.

In particular, p′(y)d(y) ∈ R and d(y) 6= 0, as desired for item (iii).For item (iv), we observe that every element λ ∈ R has the form:

λ = en−1(x)yn−1 + · · ·+ e1(x)y + e0(x), where ei ∈ Q[[x]].

Therefore:

d(λ) = d(y)[(n− 1)en−1(x)yn−2 + · · ·+ e1(x)

]+n−1∑i=0

d(ei(x))yi.

The sum expression on the right is in R, and as established above, p′(y)d(y) ∈ R,

and so p′(y)d(λ) ∈ R.

The next claim asserts an expression for d(τ ) in terms of the partial derivative∂τ∂y of τ with respect to y.

7.6.6.6 Claim 7.12. d(τ) = d(y)∂τ∂y .


Proof. For every m ∈ N, we have τ =∑m

i=0 biyi + ym+1τm, where the bi ∈ Q

and τm ∈ Q[[y]] is defined as in (4.2.3.14.2.36.1.1). Therefore

d(τ ) = d(y) · ( m∑i=1

ibiyi−1)

+ d(y)(m+ 1)ymτm + ym+1d(τm).

Thus

p′(y)d(τ ) = p′(y)d(y) ·m∑i=1

ibiyi−1 + ym(p′(y)d(y)(m+ 1)τm + yp′(y)d(τm)).

Since τ =∑∞i=0 biy

i with bi ∈ Q, we have

∂τ

∂y=

m∑i=1

ibiyi−1 + ym

∞∑i=m+1

ibiyi−m−1.

Thus ∂τ∂y −

∑mi=1 ibiy

i−1 ∈ ymR and hence

p′(y)d(τ )− p′(y)d(y)∂τ∂y∈ ym(R)

for every m ∈ N. Since p′(y) 6= 0, we have d(τ) = d(y)∂τ∂y .

Completion of proof of Theorem7.6.1.17.9. Recall that R is algebraic over Q[x](x), and

so y is algebraic over Q[x](x). Also T := C[y](x,y), where C := Q(x, σ) ∩ Q[[x]].Thus T ⊆ H , and so H is the algebraic closure of the field Q(T , ∂nσ

∂xn n∈N) in L.Therefore τ 6∈ H if and only if τ , or equivalently ν = σ + τ , is transcendental overH . By hypothesis, the transcendence degree ofH/Q is r. Since d(H) ⊆ H , if τ werein H , then ∂nτ

∂yn ∈ H for all n ∈ N. This implies that the field Q(y, τ, ∂nτ∂yn n∈N) is

contained in H . This contradicts our hypothesis that trdegQ Q(y, τ, ∂nτ∂yn n∈N) > r.

Therefore ν is residually algebraically independent over T . This completes the proofof Theorem

7.6.1.17.9.

7.6.10.1 Example 7.13. Let σ ∈ Q[[x]] and τ ∈ Q[[y]] satisfy the conditions of Theo-rem

7.6.1.17.9. Then with the setting of Notation

4.7.n7.1 we have B = A. In particular, if

σ = ex − 1 and τ = e(ey−1) − 1, then B = A.

Proof. By Theorem4.2.117.2, it suffices to prove that B is Noetherian. Recall

that T := C[y](x,y), where C := Q(x, σ) ∩ Q[[x]], a DVR, cf. Notation andRemarks

4.7.n7.1. In order to show B is Noetherian we show that the morphism

φy : T [τ ] −→ C[[y]][1/y] is flat, cf Definitionflat2.10. Since flatness is a local property,

cf.remflat2.11.1, it suffices to show for each prime ideal P of C[[y]] with y 6∈ P that

the induced map φP : T [τ ]P∩T [τ ] −→ C[[y]]P is flat. If ht(P ∩ T [τ ]) ≤ 1, thenT [τ ]P∩T [τ ] is a field or a DVR and φP is flat by (

remflat22.12.2) since C[[y]]P is torsion-

free over T [τ ]P∩T [τ ]. If ht(P ∩ T [τ ]) were greater than one, then P ∩ T 6= (0).But then Theorem

7.6.1.17.9 implies that the image of ν = σ + τ in R/P R is transcen-

dental over T/(P ∩ T ), and this implies that ht(P ∩ T [τ ]) ≤ 1. Thus we musthave ht(P ∩ T [τ ]) ≤ 1. Therefore φy : T [τ ] −→ C[[y]][1/y] is flat. We prove inCorollary


noehomsec8 that this flatness implies B is Noetherian and hence

A = B.

EXERCISES 53

Exercises

(1) Let x and y be indeterminates over a field k and let R be the 2-dimensionalRLR obtained by localizing the mixed power series-polynomial ring k[[x]][y] atthe maximal ideal (x, y)k[[x]][y].(i) For each height-one prime ideal P of R different from xR, prove that R/P

is a one-dimensional complete local domain.(ii) For each nonzero prime ideal Q of R = k[[x, y]] prove that Q ∩ R 6= (0).

Conclude that the generic formal fiber of R is zero-dimensional.Suggestion. For part (ii), use Theorem

3.38.13.5. For more information about the

dimension of the formal fibers, seeM1.5[67] and

R3.5[84].

(2) (Kunz) Let L/k be a field extension with L having infinite transcendence degreeover k. Prove that the ring L ⊗k L is not Noetherian. Deduce that the ringk[[x]]⊗k k[[x]], which has k((x)) ⊗k k((x)) as a localization is not Noetherian.Suggestion. Let xλλ∈Λ be a transcendence basis for L/k and consider thesubfield F = k(xλ) of L. The ring L ⊗k L is faithfully flat over its subringF ⊗kF , and if F ⊗kF is not Noetherian, then L⊗kL is not Noetherian. Henceit suffices to show that F ⊗k F is not Noetherian. The module of differentialsΩ1F/k is known to be infinite dimensional as a vector space over F

Kunz[60, 5.4],

and Ω1F/k∼= I/I2, where I is the kernel of the map F ⊗k F → F , defined by

sending a⊗ b 7→ ab. Thus the ideal I of F ⊗k F is not finitely generated.

CHAPTER 8

Building Noetherian domains March 31, 2010

noehomsecIn this chapter we complete the proof of Theorem

4.2.14.12 concerning the motivat-

ing examples. We establish that flatness of a certain map implies that the integraldomain obtained via Constructions

4.4.15.3 or

4.4.25.4 is Noetherian. In particular, we prove

the following theorem.

11.1.1 Theorem 8.1. Let R be a Noetherian integral domain with field of fractionsK. Let z be a nonzero nonunit of R and let R∗ denote the (z)-adic completion of R.Let I be an ideal of R∗ having the property that p∩R = (0) for each p ∈ Ass(R∗/I).The inclusion map R → (R∗/I)[1/z] is flat if and only if A := K ∩ (R∗/I) is bothNoetherian and realizable as a localization of a subring of R[1/z].

In the statement of Theorem11.1.18.1, (R∗/I)[1/z] is the localization of R∗/I at the

multiplicative system generated by z, and the intersectionK∩(R∗/I) is taken insidethe total quotient ring Q(R∗/I) of R∗/I. As discussed in Note

4.4.215.5, the hypothesis

of Theorem11.1.18.1 implies that K embeds in Q(R∗/I).

Theorem11.1.18.1 is stated using the Homomorphic Image Construction

4.4.25.4. In

Section11.338.2, we state in Theorem

11.3.258.8 the corresponding inclusion version of this

result. The intersection domain A constructed using the Inclusion Construction4.4.15.3

is both Noetherian and computable as a localization of an infinite directed unionU = ∪∞n=0Un of polynomial extension rings ofR if and only if the map U0 → R∗[1/z]is flat.

Theorem11.1.18.1 is implied by the Noetherian Example Theorem

11.3.28.3 that is proved

in Section11.38.1. We make use of the approximation ring B of A defined in Sec-

tion4.556.2. The ring B is an infinite nested union of “computable rings” — while

not localized polynomial rings over R, they are localizations of finitely generatedbirational extensions of R. In Section

11.38.1 we also prove a crucial lemma relating

flatness and the Noetherian property.

8.1. Flatness and the Noetherian property11.3

We use Lemma11.3.18.2 in order to prove the Noetherian Example Theorem

11.3.28.3.

This lemma is crucial for our proof of Theorem11.1.18.1. We thank Roger Wiegand for

the proof given for Lemma11.3.18.2. For an introduction to flatness see Section

3.32.3 of

Chapter3tools2.

11.3.1 Lemma 8.2. Let S be a subring of a ring T and let z ∈ S be a regular elementof T . Assume that the equivalent conditions of Lemma

11.2.36.16 are satisfied, that is

zS = zT ∩ S and S/zS = T/zT . Then(1) T [1/z] is flat over S ⇐⇒ T is flat over S.(2) If T is flat over S, then D := (1 + zS)−1T is faithfully flat over C :=

(1 + zS)−1S.

55

56 8. BUILDING NOETHERIAN DOMAINS MARCH 31, 2010

(3) If T is Noetherian and T is flat over S, then C = (1 + zS)−1S is Noe-therian.

(4) If T and S[1/z] are both Noetherian and T is flat over S, then S is Noe-therian

Proof. For (1), if T flat over S, then by transitivity of flatness, cf. Re-mark

remflat2.11.12, the ring T [1/z] is flat over S. For the converse, consider the exact

sequence (using (11.2.36.16.3))

0 −−−−→ S = S[1/z]∩ T α−−−−→ S[1/z]⊕ T β−−−−→ T [1/z] = S[1/z] + T −−−−→ 0,

where α(b) = (b,−b) for all b ∈ S and β(c, d) = c + d for all c ∈ S[1/z], d ∈ T .Since the two end terms are flat S-modules, the middle term S[1/z] ⊕ T is alsoS-flat. Therefore the direct summand T is S-flat.

Since S → T is flat, the embedding C = (1 + zS)−1S → (1 + zS)−1T = D isflat. Since zC is in the Jacobson radical of C and C/zC = S/zS = T/zT = D/zD,each maximal ideal of C is contained in a maximal ideal of D, and so D is faithfullyflat over C. This establishes item (2). If also T is Noetherian, then D is Noetherian,and since D is faithfully flat over C, it follows that C is Noetherian, and thus (3)holds.

For (4), let J be an ideal of S. Since C is Noetherian by part (3) and S[1/z] isNoetherian by hypothesis, there exists a finitely generated ideal J0 ⊆ J such thatJ0S[1/z] = JS[1/z] and J0C = JC. To show J0 = J , it suffices to show for eachmaximal ideal m of S that J0Sm = JSm. If z 6∈ m, then Sm is a localizationof S[1/z], so J0Sm = JSm, while if z ∈ m, then Sm is a localization of C, soJCSm = J0Sm. Therefore J = J0 is finitely generated. It follows that S isNoetherian.

Theorem11.3.28.3 gives precise conditions for the ring B to be Noetherian.

11.3.2 Noetherian Example Theorem 8.3. Homomorphic Image Version Let Rbe a Noetherian integral domain with field of fractions K. Let z be a nonzerononunit of R and let R∗ denote the (z)-adic completion of R. Let I be an ideal ofR∗ having the property that p∩R = (0) for each p ∈ Ass(R∗/I). As in (

4.5.46.9.2), let

U := ∪∞n=1Un, B := ∪∞n=1Bn = (1 + zU)−1U, and A := K ∩ (R∗/I).

The following statements are equivalent:(1) The extension R → (R∗/I)[1/z] is flat.(2) The ring B is Noetherian.(3) The extension B → R∗/I is faithfully flat.(4) The ring A := K ∩ (R∗/I) is Noetherian and A = B.(5) The ring U is Noetherian(6) The ring A is both Noetherian and a localization of a subring of R[1/z].

Proof. For (1) =⇒ (2), if R → (R∗/I)[1/z] is flat, by factoring throughU [1/z] = R[1/z] → (R∗/I)[1/z], we see that U → (R∗/I)[1/z] andB → (R∗/I)[1/z]are flat. By (

11.3.18.2.2) where we let S = U and T = R∗/I, the ring B is Noetherian.

For (2) =⇒ (3), if B is Noetherian, then, (11.2.46.17.4) implies that B∗ = R∗/I

is faithfully flat over B (cf. (remflat2.11.4)). Since z is in the Jacobson radical of B for

each proper ideal J of B we have JB∗ 6= B∗ M[68, Theorem 8.14]. This implies that

B = A is Noetherian, so (3) =⇒ (4).

8.1. FLATNESS AND THE NOETHERIAN PROPERTY 57

For (4) =⇒ (5), the composite embedding

U → B = A → B∗ = A∗ = R∗/I

is flat because B is a localization of U and A is Noetherian, so A∗ is faithfully flatover A. Thus by (

11.3.18.2.1) and (

11.3.18.2.3), where again we let S = U and T = R∗/I, we

have S[1/z] = U [1/z] = R[1/z] is Noetherian, and hence U is Noetherian.If U is Noetherian, then the localization B of U is Noetherian, so B = A is a

localization of U , a subring of R[1/z], so (5) =⇒ (6).For (6) =⇒ (1): since A is a localization of a subring D of R[1/z], we have

A := Γ−1D, where Γ is a multiplicatively closed subset of D. Now R ⊆ A =Γ−1D ⊆ Γ−1R[1/z] = A[1/z] and so A[1/z] is a localization of R. Since A isNoetherian, A → A∗ = R∗/I is flat. Thus A[1/z] → (R∗/I)[1/z] is flat. It followsthat R → (R∗/I)[1/z] is flat.

11.3.21 Corollary 8.4. Let R, I and z be as in Theorem11.3.28.3. If dim(R∗/I) = 1, then

R → (R∗/I)[1/z] is flat and therefore the equivalent conditions of Theorem11.3.28.3 all

hold.

Proof. Since z is in the Jacobson radical of R∗/I, if dim(R∗/I) = 1, thendim(R∗/I)[1/z] = 0, and our hypotheses then imply that every prime of (R∗/I)[1/z]contracts to (0) in R. Therefore (R∗/I)[1/z] is a vector space over K and hence aflat R-module.

11.3.3 Remark 8.5. With R, I and z as in the Noetherian Example Theorem11.3.28.3:

(1) There are examples where A is Noetherian and yet B 6= A, so B in theseexamples is non-Noetherian (see Theorem

4.2.117.2 and Example

4.7.13c8.11 below).

(2) There are examples where A = B is non-Noetherian (see Theorem8.4.423.4

and Remark11.3.268.10 below).

(3) A necessary and sufficient condition that A = B is that A is a localizationof R[1/z]∩A. Indeed, Theorem

11.2.46.17.5 implies that R[1/z]∩A = U and, by

Definition11.2.14.5.56.12.1, B = (1 + zU)−1U . Therefore the condition is sufficient.

On the other hand, if A = Γ−1U , where Γ is a multiplicatively closedsubset of U , then by Remark

11.2.56.18.3, each y ∈ Γ is a unit of B, and so

Γ−1U ⊆ B and A = B.(4) We discuss in Chapter

flatholds9 a family of examples where R → (R∗/I)[1/z] is

flat.

Theorem11.3.48.6 extends the range of applications of the Noetherian Example The-

orem11.3.28.3.

11.3.4 Theorem 8.6. Let R be a Noetherian integral domain with field of fractionsK. Let z be a nonzero nonunit of R and let R∗ denote the (z)-adic completion of R.Let I be an ideal of R∗ having the property that p∩R = (0) for each p ∈ Ass(R∗/I).Also assume that I is generated by a regular sequence of R∗. If R → (R∗/I)[1/z]is flat, then for each n ∈ N we have

(1) Ass(R∗/In) = Ass(R∗/I),(2) R canonically embeds in R∗/In, and(3) R → (R∗/In)[1/z] is flat.

Proof. Let I = (σ1, . . . , σr)R∗ where σ1, . . . , σr is a regular sequence in R∗.Then the sequence σ1, . . . , σr is quasi-regular in the sense of

M[68, Theorem 16.2,


page 125]. It follows that In/In+1 is a free R∗/I-module for each positive integern. Consider the exact sequence

11exact11exact (8.1) 0 −−−−→ In/In+1 −−−−→ R∗/In+1 −−−−→ R∗/In −−−−→ 0.

Proceeding by induction, assume Ass(R∗/In) = Ass(R∗/I). Since

Ass(R∗/In+1) ⊆ Ass(In/In+1) ∪ Ass(R∗/In)

and In/In+1 is a free (R∗/I)-module, it follows that Ass(R∗/In+1) = Ass(R∗/I).Thus R canonically embeds in R∗/In for each n ∈ N.

That R → (R∗/In)[1/z] is flat for every n ∈ N now follows by induction on nand considering the exact sequence obtained by tensoring over R the short exactsequence (

11exact8.1) with R[1/z].

Example 8.7. Let R = k[x, y] be the polynomial ring in the variables x, yover a field k and let R∗ = k[y][[x]] be the (x)-adic completion of R. Fix anelement τ ∈ xk[[x]] such that x and τ are algebraically independent over k, anddefine the k[[x]]-algebra homomorphism φ : k[y][[x]] → k[[x]], by setting φ(y) = τ .Then ker(φ) = (y − τ)R∗ =: I. Notice that φ(R) = k[x, τ ] ∼= R because x andτ are algebraically independent over k. Hence I ∩ R = (0). Also I is a primeideal generated by a regular element of R∗, and (I, x)R∗ is a maximal ideal ofR∗. Corollary

11.3.218.4 and Theorem

11.3.48.6 imply that for each positive integer n, the

intersection ring An := (R∗/In) ∩ k(x, y) is a one-dimensional Noetherian localdomain having (x)-adic completion R∗/In. Since x generates an ideal primaryfor the unique maximal ideal of R∗/In, the ring R∗/In is also the completion ofAn with respect to the powers of the unique maximal ideal of An. The ring A1

is a DVR since R∗/I is a DVR. For n > 1, the completion of An has nonzeronilpotent elements and hence the integral closure of An is not a finitely generatedAn-module, cf. Remarks

3.38.43.8. The inclusion In+ ( In and the fact that An has

completion R∗/In imply that An+1 ( An for each n ∈ N. Hence the rings An forma proper descending chain

A1 ⊃ A2 ⊃ · · · ⊃ An ⊃ · · · ⊃ R

of one-dimensional local birational extensions of R.

8.2. The Inclusion Version11.33

As remarked in (4.4.45.7.2), the Inclusion Construction

4.4.25.4 is a special case of the

Homomorphic Images Construction. Thus we obtain the following result analogousto the Noetherian ExampleTheorem

11.3.28.3 for Inclusion Constructions.

11.3.25 Noetherian Example Theorem 8.8. Inclusion Version Let R be a Noe-therian integral domain with field of fractions K. Let z be a nonzero nonunitof R and let R∗ denote the (z)-adic completion of R. Let τ1, . . . , τs ∈ zR∗ bealgebraically independent elements over K. As in Equations

4.2.36.1.5 and

4.2.36.1.6 of

Notation4.2.36.1, define

A := K(τ1, . . . , τs) ∩R∗, U := ∪∞n=1Un, and B := ∪∞n=1Bn = (1+zU)−1U.

The following statements are equivalent:(1) The extension U0 := R[τ1, . . . τs] → R∗[1/z] is flat.(2) The ring B is Noetherian.(3) The extension B → R∗ is faithfully flat.

8.2. THE INCLUSION VERSION 59

(4) The ring A is Noetherian and A = B.(5) The ring U is Noetherian(6) The ring A is Noetherian and is a localization of a subring of U0[1/z].(7) The ring A is Noetherian and is a localization of a subring of U [1/z].

Proof. For the proof of Theorem11.3.258.8, use the identifications in (

11.266.13) and

(11.276.14).

Remark 8.9. Innoehom[39] a proof is given for Theorem

11.3.258.8 that is an adaptation of

a proof given by Heitmann inH1[54, page 126]. Heitmann considers the case where

there is one transcendental element τ and defines the corresponding extension U tobe a simple PS-extension of R for z. Heitmann proves in this case that a certainmonomorphism condition on a sequence of maps is equivalent to U being NoetherianH1[54, Theorem 1.4].

In Chaptersflatholds9 and

noeslocsec22, for an element z in the Jacobson radical of R we study the

condition in Noetherian Example Theorem11.3.258.8 that the embedding U0 → R∗[1/z]

is flat.

11.3.26 Remark 8.10. Examples where A = B is not Noetherian show that it is possi-ble for A to be a localization of U and yet for A, and therefore also U , to fail to beNoetherian. Thus the equivalent conditions of the Noetherian Example Theorem11.3.258.8 are not implied by the property that A is a localization of U .

4.7.13c Example 8.11. Recall that in Example4.7.137.3 we may take k = Q and σ := ex−1

and τ := ey − 1. We have θ := σ−τx−y is in A := Q[[x, y]] ∩ Q(x, y, σ, τ) and not in

B :=⋃n∈N Q[x, y, σn, τn](x,y,σn,τn), where σn and τn are the endpieces defined in

(4.2.3.14.2.36.1.1). Then A is Noetherian and B ( A. Set R :=

⋃n∈N Q[x, y, σn](x,y,σn).

Then R = Q[y](y)[[x]] ∩ Q(x, y, σ) is an excellent two-dimensional regular localdomain, cf. Theorem

11.4.1a9.2 of Chapter

flatholds9. Inside the (y)-adic completion of R, we

define U :=⋃n∈NR[τn] as in (

11.2.04.2.36.1.5). The ring B is the localization of U at the

multiplicative system 1+ yU , and the rings B and U are not Noetherian. It followsthat A is not a localization of U by Theorem

11.3.258.8.

Proof. By Claim4.7.13.17.4, θ 6∈ B. If U were Noetherian, then B would be Noe-

therian. But the maximal ideal of B is (x, y)B, so if B were Noetherian, then itwould be a regular local domain with completion Q[[x, y]]. Since the completion ofa local Noetherian ring is a faithfully flat extension, and since the field of fractionsof B is Q(x, y, σ, τ), then B would equal A.

That A is Noetherian follows from Valabrega’s Theorem4.1.24.2. If A were a

localization of U , then A would be a localization of B. But each of A and Bhas a unique maximal ideal and the maximal ideal of A contains the maximal idealof B. Therefore B ( A implies that A is not a localization of B.

The following diagram displays the situation concerning possible implicationsamong certain statements for the Inclusion Construction

4.4.15.3 and the approxima-

tions in Section4.456.1:


R∗[1/a] is flat over U0 = R[τ1, . . . , τs] U Noetherian

A is a localization of U A Noetherian

11.3.25r Remark 8.12. In connection with the flatness property, if U0 := R[τ1, . . . , τs] →R∗[1/z] is flat, then for each P ∈ SpecR∗[1/z] one has that htP ≥ ht(P ∩ U0). Itis shown in Chapter

intIIsec25 that conversely this height inequality in certain contexts

implies flatness.

CHAPTER 9

Examples where flatness holds April 3, 2010

flatholdsWe continue the notation of the preceding chapters: thus R is a Noetherian

integral domain with field of fractions K, z is a nonzero nonunit of R, and R∗ isthe (z)-adic completion of R. We consider R∗ as a power series ring in z over R inthe sense of Remark

3.1.23.2. Assume that I is an ideal of R∗ such that p∩R = (0) for

each p ∈ Ass(R∗/I).In Sections

11.49.1 and

11.4a9.2 we present several examples of the Homomorphic Image

Construction4.4.25.4 where the flatness condition of Theorem

11.3.28.3 holds; that is, the

map R → (R∗/I)[1/z] is flat. We also describe several of these examples using theInclusion Constuction

4.4.15.3. The Inclusion Constuction

4.4.15.3 is in certain ways more

transparent.In Section

11.59.3 we investigate special properties of the intersection domain

A := K ∩ (R∗/I) from the Homomorphic Image Construction4.4.25.4, such as hav-

ing geometrically regular formal fibers. We present in Example11.4.59.8 an example

of a Noetherian local domain that is not universally catenary, but has geometri-cally regular formal fibers. The Homomorphic Image Construction

4.4.25.4 permits the

construction of such examples, whereas the Inclusion Construction4.4.15.3 does not.

9.1. A general example11.4

In view of Theorem11.3.28.3, it is natural to ask about the existence of ideals I

of a (z)-adic completion R∗ such that R → (R∗/I)[1/z] is flat. General ExampleTheorem

11.4.19.3 gives a general construction of such examples in the form of the Homo-

morphic Image Construction4.4.25.4. First we prove the General Example Theorem

11.4.1a9.2

for the Inclusion Construction4.4.15.3; we use it for many of our examples and it is

vital to the Insider Construction in Chaptersinsideintro10 and

insidecon13. The proof is easier for this

form. The isomorphisms obtained from Notation11.266.13 and Proposition

11.276.14 yield

the corresponding result for the Homomorphic Image Construction4.4.25.4.

We present the setting used for the versions of the two General Example The-orems together. For convenience we also include the definitions of the intersec-tion and approximation domains corresponding to the two constructions from Sec-tions

4.45.1,

4.456.1 and

4.556.2.

11.4.1set Setting and Notation 9.1. Let x be an indeterminate over a field k. Let rbe a nonnegative integer and s a positive integer. Assume that τ1, . . . , τs ∈ xk[[x]]are algebraically independent over k(x) and let y1, . . . , yr and t1, . . . , ts be additionalindeterminates. We define the following rings:

R := k[x, y1, . . . , yr], R∗ = k[y1, . . . , yr][[x]], V = k(x, τ1, . . . , τs) ∩ k[[x]].Notice that R∗ is the (x)-adic completion of R and V is a DVR, cf. Remark

3.02.12.1.

61

62 9. EXAMPLES WHERE FLATNESS HOLDS APRIL 3, 2010

We use the base ring R to define as in Construction4.4.15.3 and Section

4.456.1

Aincl := k(x, y1, . . . , yr, τ1, . . . τs) ∩R∗, Bincl := (1 + xUincl)−1Uincl,

where Uincl :=⋃n∈NR[τ1n, . . . τsn], each τin is the nth endpiece of τi and each

τin ∈ R∗, for 1 ≤ i ≤ s. By Corollary11.2.516.19, the ring R∗ is the (x)-adic completion

of each of the rings Aincl, Bincl and Uincl.Set S := k[x, y1, . . . , yr, t1, . . . , ts], let S∗ be the (x)-adic completion of S and

let σi := ti− τi, for each i. We define I := (t1− τ1, . . . , ts− τs)S∗ = (σ1, . . . , σs)S∗.With S as the base ring, we define as in Construction

4.4.25.4 and Section

4.556.2

Ahom := k(x, y1, . . . , yr, t1, . . . , ts) ∩ (S∗/I) and Bhom := (1 + xU)−1U,

where U :=⋃n∈N S[σ1n, . . . , σsn], and each σin is the nth frontpiece of σi and each

σin ∈ Q(S) ∩ (S∗/I), for 1 ≤ i ≤ s, cf. Proposition11.256.11. By Theorem

11.2.46.17, the

ring S∗/I is the (x)-adic completion of each of the rings Ahom, Bhom and U .

Proposition11.276.14 and Notation

11.266.13 imply the following isomorphisms

Uincl∼= U, Aincl

∼= Ahom, Bincl∼= Bhom R∗ ∼= S∗/I.

11.4.1a General Example Theorem 9.2. Inclusion Version. Assume Setting andNotation

11.4.1set9.1. Then

(1) The canonical map α : R[τ1, . . . τs] → R∗[1/x] is flat.(2) Bincl = Aincl is Noetherian of dimension r + 1 and is the localization

(1 + xV [y1, . . . , yr])−1V [y1, . . . , yr] of the polynomial ring V [y1, . . . , yr]over the DVR V . Thus Aincl is a regular integral domain.

(3) Bincl is a directed union of localizations of polynomial rings in r + s + 1variables over k.

(4) If k has characteristic zero, then the ring Bincl = Aincl is excellent.

Proof. The mapk[x, τ1, . . . τs] → k[[x]][1/x]

is flat by Remark3.38.03.4.2 since k[[x]][1/x] is a field. It follows that

k[x, τ1, . . . τs]⊗k k[y1, . . . , yr] → k[[x]][1/x]⊗k k[y1, . . . , yr]is flat. We also have k[[x]][1/x]⊗k k[y1, . . . , yr] ∼= k[[x]][y1, . . . , yr][1/x] and

k[x, τ1, . . . τs]⊗k k[y1, . . . , yr] ∼= k[x, y1, . . . , yr][τ1, . . . τs].

Hence the natural inclusion map

k[x, y1, . . . , yr][τ1, . . . τs]β→ k[[x]][y1, . . . , yr][1/x]

is flat. Also k[[x]][y1, . . . , yr] → k[y1, . . . , yr][[x]] is flat since it is the map taking aNoetherian ring to an ideal-adic completion, cf. Remark

3.38.03.4.2. Therefore

k[[x]][y1, . . . , yr][1/x]δ→ k[x, y1, . . . , yr][[x]][1/x]

is flat. It follows that the map

k[x, y1, . . . , yr][τ1, . . . τs]δβ→ R∗[1/x] = k[y1, . . . , yr][[x]][1/x]

is flat. Thus the Noetherian Example Theorem11.3.258.8 implies items (1) - (3).

If k has characteristic zero, then V is excellent and hence item (4) followsfrom item (2) since excellence is preserved under localization of a finitely generatedalgebra, cf. Remark

3.4353.20.

9.1. A GENERAL EXAMPLE 63

We now state and prove the homomorphic image version of the theorem.

11.4.1 General Example Theorem 9.3. Homomorphic Image Version. AssumeSetting and Notation

11.4.1set9.1. Then

(1) The canonical map α : S → (S∗/I)[1/x] is flat.(2) Bhom = Ahom is Noetherian of dimension r+1 and is a localization of the

polynomial ring V [y1, . . . , yr] over the DVR V . Thus Ahom is a regularintegral domain.

(3) Bhom is a directed union of localizations of polynomial rings in r + s+ 1variables over k.

(4) If k has characteristic zero, then Bhom = Ahom is excellent.

Proof. The ideal I := (t1 − τ1, . . . , ts − τs)S∗ is a prime ideal of S∗ andS∗/I ∼= k[y1, . . . , yr][[x]]. The fact that τ1, . . . , τs are algebraically independentover k(x) implies that I ∩ S = (0).

We identify the Homomorphic Image Construction4.4.25.4 for the ring S with

the Inclusion Construction4.4.15.3 for the ring R as in the following diagram slightly

modified from Chapterconstr5, where λ is the R-algebra isomomorphism that maps

ti → τi for i = 1, . . . , s, and K := Q(R).

4.4.554.4.55 (11.4.19.3.1)

S := R[t1, . . . , ts] −−−−→ Ahom := K(t1, . . . , ts) ∩ (S∗/I) −−−−→ S∗/I

λ

y λ

y λ

yR −−−−→ R[τ1, . . . , τs] −−−−→ Aincl := K(τ1, . . . , τs) ∩R∗ −−−−→ R∗.

In view of the identifications displayed in this diagram, the items of Theorem11.4.19.3

follow from the corresponding items of Theorem11.4.1a9.2.

11.4.11 Corollary 9.4. If we adjust Setting and Notation11.4.1set9.1 so that the base rings

are the regular local rings

R := k[x, y1, . . . , yr](x,y1,...,yr) and S := k[x, y1, . . . , yr, t1, . . . , ts](x,y1,...,yr,t1,...,ts),

then the conclusions of Theorems11.4.1a9.2 and

11.4.19.3 are still valid, and Aincl = Bincl and

Ahom = Bhom are Noetherian regular local rings.

11.4.2 Example 9.5. Let S be as in Corollary11.4.119.4. Then t1−τ1, . . . , ts−τs is a regular

sequence in S∗. Hence, with the ideal I = (t1−τ1, . . . , ts−τs)S∗ as in Corollary11.4.119.4,

then Theorem11.3.48.6 implies that S → (S∗/In)[1/x] is flat for each positive integer

n. Using In in place of I, Theorem11.4.19.3 implies the existence for every r and n in

N of a Noetherian local domain A having dimension r + 1 such that the (x)-adiccompletion A∗ of A has nilradical n with nn−1 6= (0).

Here are some more specific examples to which Theorems11.4.19.3 and

11.4.1a9.2 and

Corollary11.4.119.4 apply. Example

11.4.39.6 shows that the dimension of U can be greater

than the dimension of Bhom.

11.4.3 Examples 9.6. Assume the setting and notation of (11.4.1set9.1).

(1) Let S := k[x, t1, . . . , ts], that is, there are no y variables, and let S∗ denotethe (x)-adic completion of S. Then by Theorem

11.4.19.3,

(S∗/I) ∩ Q(S) = Ahom = Bhom


is the DVR obtained by localizing U at the prime ideal xU . In this exampleS[1/x] = U [1/x] has dimension s + 1 and so dimU = s + 1, while dim(S∗/I) =dimAhom = dimBhom = 1.

(2) Essentially the same example as in item (1) can be obtained by usingTheorem

11.4.1a9.2 as follows. Let R = k[x], then R∗ = k[[x]] and

Aincl = k(x, τ1, . . . , τs) ∩ k[[x]] and Aincl = Bincl,

by Theorem11.4.1a9.2. In this case Uincl is a directed union of polynomial rings over k,

Uincl =∞⋃n=1

k[x][τ1n, . . . , τsn],

where the τin are endpieces as in Section4.456.1. By Proposition

11.256.11, the endpieces

are related to the frontpieces of the homomorphic image construction.(3) Applying Corollary

11.4.119.4, one can modify Example

11.4.39.6.1 by taking S to be

the (s + 1)-dimensional regular local domain k[t1, . . . , ts, x](t1,...,ts,x). In this caseSx = Ux has dimension s, while we still have S∗/I ∼= k[[x]]. Thus dimU = s + 1and dim(S∗/I) = 1 = dimAhom = dimBhom.

One can also obtain a local version using the inclusion construction with R =k[x](x) and applying Theorem

11.4.1a9.2. We again have R∗ = k[[x]].

With S as in either (11.4.39.6.1) or (

11.4.39.6.3), the domains Bn constructed from S as in

Theorem11.3.28.3 are (s+1)-dimensional regular local domains dominated by k[[x]] and

having k as a coefficient field. In either case, since (S∗/I)x is a field, S → (S∗/I)xis flat, so by Theorem

11.3.28.3, the family Bnn∈N is a directed union of (s + 1)-

dimensional regular local domains whose union B is Noetherian, and is, in fact aDVR.

(4) In the notation of (11.4.1set9.1) with the adjustment of Corollary

11.4.119.4, let r = 1 and

y1 = y. Thus S = k[x, y, t1, . . . , ts](x,y,t1,...,ts). Then S∗/I ∼= k[y](y)[[x]]. By Theo-rem

11.4.19.3.1, the extension S → (S∗/I)[1/x] is flat. Let V = k[[x]] ∩ k(x, τ1, . . . , τs).

Then V is a DVR and (S∗/I) ∩ Q(S) ∼= V [y](x,y) is a 2-dimensional regular localdomain that is the directed union of (s+ 2)-dimensional regular local domains.

9.2. Transfer to the intersection of two ideals11.4a

We use the homomorphic image construction and assume the notation of The-orem

11.3.28.3. In Theorem

11.4.49.71 we show that in certain circumstances the flatness,

Noetherian and computability properties associated with ideals I1 and I2 of R∗ asdescribed in Theorem

11.3.28.3 transfer to their intersection I1∩I2. We use Theorem

11.4.49.7

in Section11.59.3 to show in Theorem

11.5.29.10 that the property of regularity of the generic

formal fibers also transfers in certain cases to the intersection domain associatedwith the ideal I1 ∩ I2.

11.4.4 Theorem 9.7. Assume the notation of Theorem11.3.28.3 and let I1 and I2 be two

ideals of R∗ such that p ∈ Ass(R∗/Ii) implies p ∩ R = (0) for i = 1, 2. Assumethat R → (R∗/Ii)[1/z] is flat, for i = 1, 2, and that zn ∈ I1 + I2 for some positiveinteger n. For i = 1, 2, let Ai := K∩(R∗/Ii) and let A := K∩(R∗/(I1∩I2)). Then

(1) The ideal I := I1 ∩ I2 satisfies the conditions of Theorem11.3.28.3 and the map

R → (R∗/I))[1/z] is flat. The (z)-adic completion A∗ of A is R∗/I, andthe (z)-adic completion A∗

i of Ai is R∗/Ii, for i = 1, 2.

1A generalization of Theorem11.4.49.7 is given in Theorem

15.3.615.9.

9.2. TRANSFER TO THE INTERSECTION OF TWO IDEALS 65

(2) The ring A∗[1/z] ∼= (A∗1)[1/z] ⊕ (A∗

2)[1/z]. If Q ∈ Spec(A∗) and z 6∈ Q,then (A∗)Q is a localization of either A∗

1 or A∗2.

(3) We have A ⊆ A1∩A2, and (A1)[1/z]∩(A2)[1/z] ⊆ AP for each P ∈ SpecAwith z /∈ P . Thus we have A[1/z] = (A1)[1/z] ∩ (A2)[1/z].

Proof. By Theorem11.3.28.3, the (z)-adic completion A∗

i of Ai is R∗/Ii. SinceAss(R∗/(I1 ∩ I2)) ⊆ Ass(R∗/I1) ∪ Ass(R∗/I2), the condition on assassinators ofTheorem

11.3.28.3 holds for the ideal I1 ∩ I2. The natural R-algebra homomorphism

π : R∗ → (R∗/I1) ⊕ (R∗/I2) has kernel I1 ∩ I2. Further, the localization of πat z is onto because (I1 + I2)R∗[1/z] = R∗[1/z]. Thus (R∗/(I1 ∩ I2))[1/z] ∼=(R∗/I1)[1/z]⊕ (R∗/I2)[1/z] = (A∗

1)[1/z]⊕ (A∗2)[1/z] is flat over R. Therefore A is

Noetherian and A∗ = R∗/I is the (z)-adic completion of A.If Q ∈ Spec(A∗) and z /∈ Q, then A∗

Q is a localization of

A∗[1/z] ∼= (A∗1)[1/z]⊕ (A∗

2)[1/z].

Every prime ideal of (A∗1)[1/z] ⊕ (A∗

2)[1/z] has the form either (Q1)A∗1[1/z] ⊕

(A∗2)[1/z] or (A∗

1)[1/z] ⊕ Q2A∗2[1/z], where Qi ∈ Spec(A∗

i ). It follows that A∗Q

is a localization of either A∗1 or A∗

2.Since R∗/Ii is a homomorphic image of R∗/(I1 ∩ I2), the intersection ring

A ⊆ Ai for i = 1, 2. Let P ∈ SpecA with z /∈ P . Since A∗ = R∗/I is faithfully flatover A, there exists P ∗ ∈ Spec(A∗) with P ∗ ∩ A = P . Then z /∈ P ∗ implies A∗

P∗

is either (A∗1)P∗

1or (A∗

2)P∗2, where P ∗

i ∈ Spec(A∗i ). By symmetry, we may assume

A∗P∗ = (A∗

1)P∗1. Let P1 = P ∗

1 ∩ A1. Since AP → A∗P∗ and (A1)P1 → (A∗

1)P∗1

arefaithfully flat, we have

AP = A∗P∗ ∩K = (A∗

1)P∗1∩K = (A1)P1 ⊇ (A1)[1/z].

It follows that (A1)[1/z] ∩ (A2)[1/z] ⊆ AP . Thus we have

(A1)[1/z] ∩ (A2)[1/z] ⊆⋂AP | P ∈ SpecA and z /∈ P = A[1/z].

Since A[1/z] ⊆ (Ai)[1/z], for i = 1, 2, we have A[1/z] = (A1)[1/z] ∩ (A2)[1/z]. The ring B of Example

11.4.59.8 is a two-dimensional Noetherian local domain that

birationally dominates a three-dimensional regular local domain and is such thatB is not universally catenary. The completion of B has two minimal primes oneof dimension one and one of dimension two and this implies B is not universallycatenary, cf. Theorem

ratliff3.10. It follows that B is not a homomorphic image of a

regular local ring, for every homomorphic image of a regular local ring, or even ofa Cohen-Macaulay local ring, is universally catenary, cf. Remark

ucathom3.12. We present

in Chaptercatsec15 other examples of Noetherian local domains of various dimensions

that are not universally catenary and that have properties similar to those of theNoetherian local domain B of Example

11.4.59.8.

11.4.5 Example 9.8. Let k be a field of characteristic zero2 and let x, y, z be inde-terminates over k. Let R = k[x, y, z](x,y,z), let K denote the field of fractionsof R, and let τ1, τ2, τ3 ∈ xk[[x]] be algebraically independent over k(x, y, z). LetR∗ denote the (x)-adic completion of R. We take the ideal I to be the intersec-tion of the two prime ideals Q := (z − τ1, y − τ2)R∗, which has height 2, andP := (z − τ3)R∗, which has height 1. Then R∗/P and R∗/Q are examples of the

2The characteristic zero assumption implies that the intersection rings A1 and A2 as con-structed below are excellent cf. (

11.4.19.3.4).


form considered in Examples11.4.39.6. Thus (R∗/P )[1/x] and (R∗/Q)[1/x] are both flat

over R. Here R∗/P ∼= k[y](y)[[x]]; the ring V := k[[x]] ∩ k(x, τ3) is a DVR, andA1 := (R∗/P ) ∩K ∼= V [y](x,y) is a two-dimensional regular local domain that is adirected union of three-dimensional RLRs, while A2 := (R∗/Q) ∩K is a DVR.

Since τ1, τ3 ∈ xk[[x]], the ideal (z − τ1, z − τ3)R∗ has radical (x, z)R∗. Hencethe ideal P + Q is primary for the maximal ideal (x, y, z)R∗, so, in particular, Pis not contained in Q. Therefore the representation I = P ∩Q is irredundant andAss(R∗/I) = P,Q. Since P ∩R = Q∩R = (0), the ring R injects into R∗/I. LetA := K ∩ (R∗/I).

By (11.4.49.7.1), the inclusion R → (R∗/I)[1/x] is flat. By Theorem

11.3.28.3, the ring

A is Noetherian and is a localization of a subring of R[1/x]. The map A → A ofA into its completion factors through the map A → A∗ = R∗/I. Since R∗/I hasminimal primes P/I and Q/I with dimR∗/P = 2 and dimR∗/Q = 1, and since Ais faithfully flat over A∗ = R∗/I, the ring A is not equidimensional. It follows thatA is not universally catenary, cf. Theorem

ratliff3.10.

9.3. Regular maps and geometrically regular formal fibers11.5

We show in Corollary11.5.39.113 that the ring A = B of Example

11.4.59.8 has geometri-

cally regular formal fibers. Another example of a Noetherian local domain that isnot universally catenary but has geometrically regular formal fibers is given in

G[30,

(18.7.7), page 144] using a gluing construction; also seeGreco[29, (1.1)].

11.5.1 Proposition 9.9. Assume the notation and setting of Theorem11.3.28.3. Also as-

sume for each P ∈ Spec(R∗/I) with z /∈ P , that the map ψP : RP∩R −→ (R∗/I)Pis regular. Then

(1) B is Noetherian and the map B −→ B∗ = R∗/I is regular.(2) If R is semilocal with geometrically regular formal fibers and z is in the

Jacobson radical of R, then B has geometrically regular formal fibers.

Proof. Since flatness is a local property, the map ψz : R −→ (R∗/I)[1/z] isflat. By Theorem

11.3.28.3 and (

11.2.46.17.4), the ringB is Noetherian with (z)-adic completion

B∗ = R∗/I. Hence B −→ B∗ is flat.For Q ∈ Spec(B), let k(Q) denote the field of fractions of B/Q and let Q0 =

Q ∩R.Case 1: z ∈ Q: Then R/Q0 = B/Q = B∗/QB∗ and the ring B∗

QB∗/QB∗QB∗ =

B∗ ⊗B k(Q) = BQ/QBQ is trivially geometrically regular over k(Q).Case 2: z /∈ Q: Let k(Q) ⊆ L be a finite algebraic field extension. We show thering B∗⊗B L is regular. Let W ∈ Spec(B∗⊗B L) and let W ′ = W ∩ (B∗⊗B k(Q)).The prime W ′ corresponds to a prime ideal P ∈ Spec(B∗) with P ∩ B = Q. Byassumption the map

RQ0 −→ (R∗/I)P = B∗P

is regular. Since z /∈ Q it follows that RQ0 = UQ∩U = BQ and that k(Q0) = k(Q).Thus the ring B∗

P ⊗BQ L is regular. Therefore (B∗ ⊗B L)W , which is a localizationof this ring is regular.

For part (2), since R has geometrically regular formal fibers, so has R∗ byR3[83].

Hence the map θ : B∗ = R∗/I −→ B = (R∗/I) is regular. ByM[68, Thm. 32.1 (i)]

3A generalization of Corollary11.5.39.11 is given in Corollary

15.5.315.19

EXERCISES 67

and part (1) above, it follows that B has geometrically regular formal fibers, thatis, the map B −→ B is regular.

11.5.2 Theorem 9.10. Assume the setting of Theorem11.3.28.3 for two ideals I1 and I2 of

R∗, i.e., if P ∈ Ass(R∗/Ii), then P ∩R = (0), for i = 1, 2. Also assume(1) R is semilocal with geometrically regular formal fibers and z is in the

Jacobson radical of R.(2) (R∗/I1)[1/z] and (R∗/I2)[1/z] are flat R-modules and the ideal I1 + I2 of

R∗ contains some power of z.(3) For i = 1, 2, Ai := K ∩ (R∗/Ii) has geometrically regular formal fibers.

Then A := K ∩ (R∗/(I1 ∩ I2)) = B has geometrically regular formal fibers.

Proof. Let I = I1 ∩ I2. Since R has geometrically regular formal fibers,by (

11.5.19.9.2), it suffices to show for W ∈ Spec(R∗/I) with z 6∈ W that RW0 −→

(R∗/I)W is regular, where W0 := W ∩ R. As in (11.4.49.7), we have (R∗/I)[1/z] =

(R∗/I1)[1/z] ⊕ (R∗/I2)[1/z]. It follows that (R∗/I)W is a localization of eitherR∗/I1 or R∗/I2. Suppose (R∗/I)W = (R∗/I1)W1 , where W1 ∈ Spec(R∗/I1). ThenRW0 = (A1)W1∩A1 and (A1)W1∩A1 −→ (R∗/I1)W1 is regular. A similar argumentholds if (R∗/I)W is a localization of R∗/I2. Thus in either case, RW0 −→ (R∗/I)Wis regular.

11.5.3 Corollary 9.11. The ring A = B of Example11.4.59.8 has geometrically regular

formal fibers, that is, the map φ : A → A is regular.

Proof. By the definition of R and the observations given in (11.4.59.8), the hy-

potheses of (11.5.29.10) are satisfied.

Exercises(1) Describe Example

11.4.39.6.4 in terms of the Inclusion Construction. In particular,

determine the appropriate base ring R for this construction.(2) For the rings A and A∗ of Example

11.4.59.8, prove that A∗ is universally catenary.

CHAPTER 10

Introduction to the insider construction May 10,

2010, 16, 17 (insideintro)

insideintroWe describe in this chapter and Chapter

insidecon13 a version of the Inclusion Con-

struction4.4.15.3 that we call the Insider Construction. We present in this chapter

several examples using this Insider Construction including two classical examples.The Insider Construction makes it easy to create examples. It is then easier thanit is with Construction

4.4.15.3 to determine whether a given example is or is not Noe-

therian. The Insider Construction is described in more generality and with moredetails in Chapter

insidecon13.

For the examples considered in this chapter, we begin with a base ring that isa localized polynomial ring in two or three variables over a field. We construct two“insider” integral domains inside an “outside” ring Aout, where Aout is constructedfrom General Example Theorem

11.4.1a9.2 with the Inclusion Construction

4.4.15.3. By Theo-

rem11.4.1a9.2, this intersection domain Aout is equal to an approximation integral domain

Bout that is the nested union of localized polynomial rings from Section4.456.1. The

two insider integral domains contained in Aout are: Ains, an intersection of a fieldwith a power series ring as in Construction

4.4.15.3, and Bins, a nested union of localized

polynomial rings that “approximates” Ains as in Section4.456.1.

As we describe in Proposition16.1.pg10.6, the condition that the insider approxima-

tion domain Bins is Noetherian is related to flatness of a map of polynomial ringscorresponding to the extension Bins → Bout. Flatness of this map implies thatthe insider approximation domain Bins is Noetherian and therefore that the insiderintersection domain Ains equals Bins and so Ains is also Noetherian. We apply thisobservation in Proposition

16.1n10.2 and Example

16.1r10.7 to conclude that certain con-

structed rings are Noetherian. We use Theorem16.3.913.5 from Chapter

insidecon13 to show that

other constructed rings are not Noetherian.We present more details concerning flatness of polynomial extensions in Chap-

terflatpoly11.

10.1. The Nagata example16.1gn

In Proposition16.1n10.2 we adapt the example of Nagata given as Example

4.3.14.8 in

Chapterfex4, to fit the insider construction.

16.1.1g Setting 10.1. Let k be a field, let x and y be indeterminates over k, and set

R : = k[x, y](x,y) and R∗ : = k[y](y)[[x]].

The power series ring R∗ is the xR-adic completion of R. Let τ ∈ xk[[x]] betranscendental over k(x). We define the intersection domain Aτ corresponding toτ by Aτ := k(x, y, τ) ∩R∗.

69

7010. INTRODUCTION TO THE INSIDER CONSTRUCTION MAY 10, 2010, 16, 17 (INSIDEINTRO)

Let f be a polynomial in R[τ ] that is algebraically independent over Q(R),for example, f = (y + τ)2. The intersection domain corresponding to f is Af =Q(R[f ]) ∩R∗. In the language of the introduction to this chapter, Aτ is the “out-sider” intersection domain Aout and Af , which is contained in Aτ , is the “insider”intersection domain Ains.

Recall from Section4.456.1 and Remark

4.2116.4 that there are natural “approximating”

outsider and insider domains associated to Aτ and Af , namely,

Bτ : =⋃n∈N

k[x, y, τn](x,y,τn) and Bf : =⋃n∈N

k[x, y, fn](x,y,τn),

where the τn are the nth endpieces of τ and the fn are the nth endpieces of f .The rings Bτ and Bf are nested unions of localized polynomial rings over k in 3variables. (These approximating domains are the Cn of Remark

4.2116.4.)

By Corollary11.4.119.4, the extension T := R[τ ]

ψ→ R∗[1/x] is flat, where ψ is the

inclusion map, and the ring Aτ is Noetherian and is equal to the “computable”nested union Bτ . Let S := R[f ] ⊆ R[τ ] and let ϕ be the embedding ϕ : S :=R[f ]

ϕ→ T = R[τ ]. Put α := ψ ϕ : S → R∗[1/x]. Then we have the following

commutative diagram:

R∗[1/x]

R ⊆ S := R[f ] T := R[τ ]

ψ

α:=ψϕ(16.1gn10.1.1)

ϕ

In Proposition16.1n10.2, we present a different proof of the result of Nagata de-

scribed in Example4.3.14.8.

16.1n Proposition 10.2. With the notation of Setting16.1.1g10.1, let f := (y + τ)2. In

the Nagata Example4.3.14.8, the ring Af = Bf is Noetherian with completion k[[x, y]].

Therefore Af is a two-dimensional regular local domain.

Proof. Since T = R[τ ] is a free S-module with free basis 〈1, y + τ〉, the map

Sϕ→ T is flat. As mentioned above, the map T

ψ→ R∗[1/x] is flat. Therefore as

displayed in Diagram16.1gn10.1.1, the map S

α→ R∗[1/x] is flat. By Theorem

11.3.258.8 and

Remark4.2116.4, Bf is Noetherian and Bf = Af .

16.1nr Remark 10.3. In the example of Nagata it is required that the field k be ofcharacteristic different from 2. This assumption is not necessary for showing thatthe domain Af of Proposition

16.1n10.2 is a two-dimensional regular local domain.

10.2. Other examples inside the general example16.1ng

To describe other examples, we modify Setting16.1.1g10.1 as follows.

16.2.2g Setting 10.4. Let k be a field, let x, y, z be indeterminates over k, and set

R : = k[x, y, z](x,y,z) and R∗ : = k[y, z](y,z)[[x]].

The power series ring R∗ is the xR-adic completion of R. Let σ and τ in xk[[x]] bealgebraically independent over k(x). We define the intersection domains Aτ and

10.2. OTHER EXAMPLES INSIDE THE GENERAL EXAMPLE 71

Aσ,τ corresponding to τ and σ, τ as follows:

Aτ := k(x, y, z, τ) ∩R∗ and Aσ,τ := k(x, y, z, σ, τ) ∩R∗

In the following examples we define f to be an element of R[σ, τ ] that is alge-braically independent over Q(R). The intersection domain corresponding to f isAf = Q(R[f ])∩R∗. In the language of the beginning of this chapter, Aτ or Aσ,τ isthe “outsider” intersection domain Aout, and Af , which is contained in Aτ or Aσ,τ ,is the “insider” intersection domain Ains.

As in Setting16.1.1g10.1 for the Nagata Example, there are natural “approximating”

domains associated to Aτ , Aσ,τ and Af , respectively, namely, Bτ , Bσ,τ and Bf ,respectively. The rings Bτ , Bσ,τ and Bf are nested unions of localized polynomialrings over k in 4 or 5 variables.

16.2.2gr Remark 10.5. With Setting16.2.2g10.4, let T := R[σ, τ ] and let S = R[f ], where f

is a polynomial in R[σ, τ ] that is algebraically independent over Q(R). Notice thatthe inclusion map ψ : T → R∗[1/x] is flat. Let ϕ : S → T denote the inclusionmap from S to T . Then we have the following commutative diagram:

R∗[1/x]

R ⊆ S := R[f ] T = R[σ, τ ]

ψ

α:=ψϕ(16.2.2gr10.5.1)

ϕ

We note the following:

(1) Lemma16.3.711.18 implies the map ϕ is flat if and only if the nonconstant

coefficients of f as a polynomial in R[σ, τ ] generate the unit ideal of R.(2) If ϕ is flat then α is the composition of two flat homomorphisms and so

α is also flat.(3) To conclude that α is flat, it suffices to show that ϕx : S → T [1/x] is flat,

for the inclusion map T [1/x] → R∗[1/x] is flat.(4) Theorem

16.3.913.5.1 implies that ϕx : S → T [1/x] is flat if and only if the

nonconstant coefficients of f generate the unit ideal of R[1/x].

Proposition16.1.pg10.6 is a preliminary result regarding flatness; there is a more

extensive discussion in Chapterinsidecon13:

16.1.pg Proposition 10.6. With the notation of Remark16.2.2gr10.5, let ϕx : S → T [1/x]

denote the composition of ϕ : S → T with the localization map from T to T [1/x].If ϕx : S → T [1/x] is flat, then the approximating domain Bf from Section

4.456.1 is

Noetherian and is equal to the intersection domain Af .

Proof. In Diagram‘16.2.2gr10.5.1, the map ψ is flat. Thus the map ψ′ : T [1/x] →

R∗[1/x] is flat. If ϕx : S → T [1/x] is flat, then α = ψ′ ϕx in Diagram16.2.2gr10.5.1

is the composition of two flat maps. Thus α is flat, and hence by Theorem11.3.258.8,

the approximating domain Bf from Section4.456.1 is Noetherian and is equal to the

intersection domain Af .

We use Proposition16.1.pg10.6 to show the Noetherian property for the following

example of RotthausR1[81], Example

4.3.34.10 of Chapters

fex4.

7210. INTRODUCTION TO THE INSIDER CONSTRUCTION MAY 10, 2010, 16, 17 (INSIDEINTRO)

16.1r Example 10.7. (Rotthaus) With the Setting16.2.2g10.4, let f := (y+ σ)(z + τ) and

consider the insider domain Af ⊆ Aσ,τ , the outsider domain. By Proposition16.1.pg10.6,

the associated nested union domain Bf is Noetherian and is equal to Af .

16.1nn Examples 10.8. (1) With the Setting16.2.2g10.4, let f := yσ + zτ . We show in

Examples16.3.1013.6 that the map R[f ] → R[σ, τ ][1/x] is not flat and that Af = Bf , i.e.,

A is “limit-intersecting”, cf. Definition4.2li6.5, but is not Noetherian. Thus we have a

situation where the intersection domain equals the approximation domain, but isnot Noetherian. This gives a simpler example of such behavior than the examplegiven in Section

8.423.1. ????

(2) The following is a related even simpler example: again with the notation ofSetting

16.2.2g10.4, let f := yτ + zτ2 ∈ R[τ ] ⊆ Aτ . Then the constructed approximation

domain Bf is not Noetherian by Theorem16.3.913.5. Moreover, Bf is equal to the

intersection domain Af := R∗ ∩ k(x, y, z, f) by Corollary16.3.413.4.

In dimension two (the two variable case), an immediate consequence of Val-abrega’s Theorem

4.1.24.2 is the following.

16.3.11g Theorem 10.9. (Valabrega) Let x and y be indeterminates over a field k andlet R = k[x, y](x,y). Then R = k[[x, y]] is the completion of R. If L is a fieldbetween the fraction field of R and the fraction field of k[y] [[x]], then A = L∩ R isa two-dimensional regular local domain with completion R.

Example16.1nn10.8 shows that the dimension three analog to Valabrega’s result fails.

With R = k[x, y, z](x,y,z) the field L = k(x, y, z, f) is between k(x, y, z) and thefraction field of k[y, z] [[x]], but L ∩ R = L ∩R∗ is not Noetherian.

16.4nn Example 10.10. The following example is given in Section13.425.4. With the

notation of Setting16.2.2g10.4, let f = (y + σ)2 and g = (y + σ)(z + τ). It is shown in

ChapterintIIsec25 that the intersection domain A := R∗∩k(x, y, z, f, g) properly contains

its associated approximation domain B and that both A and B are non-Noetherian.

CHAPTER 11

The flat locus of a polynomial ring extension Feb

6, 2010, 16, 17 (flatpoly)

flatpolyLet R be a Noetherian ring, let n be positive integer and let z1, . . . , zn be

indeterminates over R. In this chapter we examine the flat locus of a polynomialring extension ϕ of the form

(flatpoly11.01) S := R[f1, . . . , fm]

ϕ→ R[z1, . . . , zn] =: T,

where the fj are polynomials in R[z1, . . . , zn] that are algebraically independentover R1. We are motivated to examine the flat locus of the extension ϕ by theflatness condition in the Insider Construction of Chapter

insideintro10.

We discuss in Section13.2p11.1 a general result on flatness. Then in Section

16.pj11.2

we consider the Jacobian ideal of the map ϕ : S → T of (flatpoly11.01) and describe the

nonsmooth and nonflat loci of this map. In Section16.3p11.3 we discuss applications to

polynomial extensions.

11.1. Flatness criteria13.2p

Theorem13.2.0p11.1 is a general result on flatness involving a trio of Noetherian local

rings.

13.2.0p Theorem 11.1. Let (R,m), (S,n) and (T, `) be Noetherian local rings, andassume there exist local maps:

R −→ S −→ T,

such that(i) R→ T is flat and T/mT is Cohen-Macaulay.(ii) R→ S is flat and S/mS is a regular local ring.

Then the following statements are equivalent:(1) S → T is flat.(2) For each prime ideal w of T , we have ht(w) ≥ ht(w ∩ S).(3) For each prime ideal w of T such that w is minimal over nT , we have

ht(w) ≥ ht(n).

Proof. The implications (2) =⇒ (3) is obvious and the implication (1) =⇒ (2)is clear by

M[68, Theorem 9.5]. To prove (3) =⇒ (1), we show that (3) implies

(a) mS ⊗S T ∼= mT .(b) The map S/mS −→ T/mT is faithfully flat.

1In general for a commutative ring T and a subring R, we say that elements f1, . . . , fm ∈ Tare algebraically independent over R if, for indeterminates t1, . . . , tm over R, the only polynomialG(t1, . . . , tm) ∈ R[t1, . . . , tm] with G(f1, . . . , fm) = 0 is the zero polynomial.

73

7411. THE FLAT LOCUS OF A POLYNOMIAL RING EXTENSION FEB 6, 2010, 16, 17 (FLATPOLY)

ByM[68, Theorem 22.3], conditions (a) and (b) imply flatness of the map S → T

and thus prove that (3 =⇒ (1).Proof of (a): Since R → S is flat, we have mS ∼= mR⊗R S. Therefore

mS ⊗S T ∼= (m⊗R S)⊗S T ∼= m⊗R T ∼= mT,

where the last isomorphism follows because the map R→ T is flat.Proof of (b): By assumption, T/mT is Cohen-Macaulay and S/mS is regular.

Thus in view ofM[68, Theorem 23.1], it suffices to show:

dim(T/mT ) = dim(S/mS) + dim(T/nT ).

Let w be a prime ideal of T such that nT ⊆ w and w is minimal over nT . Since`∩S = n, also w∩S = n and by

M[68, Theorem 15.1], ht(w) ≤ ht(n). By assumption

ht(w) ≥ ht(n) and therefore ht(w) = ht(n). This equality holds for every minimalprime divisor w of nT and therefore ht(n) = ht(nT ). Since n is generated up toradical by ht(n) elements , we have

dim(T/nT ) = dim(T )− ht(nT )

= dim(T )− ht(w)

= dim(T )− ht(n)

Since T and S are flat over R, we have ht(mT ) = ht(mS) = ht(m)M[68, Theorem

15.1]. Therefore

dim T/nT = dim(T )− ht(n)

= dim(T )− ht(mT )− (ht(n)− ht(mS))

= dim(T/mT )− dim(S/mS)).

Therefore S/mS −→ T/mT is faithfully flat and hence by the equivalence of items(1) and (3) of

M[68, Theorem 22.3], S −→ T is faithfully flat. This completes the

proof of Theorem13.2.0p11.1

In Theorem13.2.1p11.2 we present a result closely related to Theorem

13.2.0p11.1 with a

Cohen-Macaulay hypothesis on all the fibers of R→ T and a regularity hypothesison all the fibers of R→ S.

13.2.1p Theorem 11.2. Let (R,m), (S,n) and (T, `) be Noetherian local rings, andassume there exist local maps:

R −→ S −→ T,

such that

(i) R→ T is flat with Cohen-Macaulay fibers,(ii) R→ S is flat with regular fibers.

Then the following statements are equivalent:

(1) S → T is flat with Cohen-Macaulay fibers.(2) S → T is flat.(3) For each prime ideal w of T , we have ht(w) ≥ ht(w ∩ S).(4) For each prime ideal w of T such that w is minimal over nT , we have

ht(w) ≥ ht(n).

11.2. THE JACOBIAN IDEAL AND THE SMOOTH AND FLAT LOCI 75

Proof. The implications (1) =⇒ (2) and (3) =⇒ (4) are obvious and theimplication (2) =⇒ (3) is clear by

M[68, Theorem 9.5]. By Theorem

13.2.0p11.1, item (4)

implies that S → T is flat.To show Cohen-Macaulay fibers for S → T , it suffices to show for each Q ∈

SpecT with P := Q ∩ S that TQ/PTQ is Cohen-Macaulay. Let Q ∩ R = q. Bypassing to R/q ⊆ S/qS ⊆ T/qT , we may assume Q ∩ R = (0). Let htP = n.Since R → SP has regular fibers and P ∩ R = (0), the ideal PSP is generated byn elements. Moreover, faithful flatness of the map SP → TQ implies that the idealPTQ has height n. Since TQ is Cohen-Macaulay, a set of n generators of PSP formsa regular sequence in TQ. Hence TQ/PTQ is Cohen-Macaulay

M[68, Theorems 17.4

and 17.3]. Since flatness is a local property, the following two corollaries are immediate

from Theorem13.2.1p11.2; see also

P[79, Theoreme 3.15].

16.2.3p Corollary 11.3. Let T be a Noetherian ring and let R ⊆ S be Noetheriansubrings of T . Assume that R → T is flat with Cohen-Macaulay fibers and thatR→ S is flat with regular fibers. Then S → T is flat if and only if, for each primeideal P of T , we have ht(P ) ≥ ht(P ∩ S).

As a special case of Corollary16.2.3p11.3, we have:

16.2.4p Corollary 11.4. Let R be a Noetherian ring and let z1, . . . , zn be indetermi-nates over R. Assume that f1, . . . , fm ∈ R[z1, . . . , zn] are algebraically independentover R Then

(1) ϕ : S := R[f1, . . . , fm] → T := R[z1, . . . , zn] is flat if and only if, for eachprime ideal P of T , we have ht(P ) ≥ ht(P ∩ S).

(2) For Q ∈ SpecT , ϕQ : S → TQ is flat if and only if for each prime idealP ⊆ Q of T , we have ht(P ) ≥ ht(P ∩ S).

Proof. Since S and T are polynomial rings over R, the maps R→ S and R→T are flat with regular fibers. Hence both assertions follow from Corollary

16.2.3p11.3.

11.2. The Jacobian ideal and the smooth and flat loci16.pj

We use the following definitions as inSwan[94].

smooth Definition 11.5. Let R be a ring. An R-algebra A is said to be quasi-smoothover R if for every R-algebra B and ideal N of B with N2 = 0, every R-algebrahomomorphism g : A → B/N lifts to an R-algebra homomorphims f : A → B.Thus in the commutative diagram below where the maps from R→ A and R→ Bare the canonical ring homomorphisms that define A and B as R-algebras and themap B → B/N is the canonical quotient ring map

R −−−−→ Ay g

yB −−−−→ B/N

if A is quasi-smooth over R, then there exists an R-algebra homomorphism f :A→ B that preserves commutativity of the diagram. If A is finitely presented andquasi-smooth over R, then A is said to be smooth over R. If A is essentially finitelypresented and quasi-smooth over R, then A is said to be essentially smooth over R.


Let ϕ : S → T be as in (flatpoly11.01).

16.2.5 Definitions and Remarks 11.6. (1) The Jacobian ideal J of the extensionS → T is the ideal of T generated by the m ×m minors of the m × n matrix Jdefined as follows:

J :=(∂fi∂zj

)1≤i≤m, 1≤j≤n

.

(2) For the extension ϕ : S → T , the nonflat locus of ϕ is the set F , where

F := Q ∈ Spec(T ) | the map ϕQ : S → TQ is not flat .We also define the set Fmin and the ideal F of T as follows:

Fmin := minimal elements of F and F :=⋂Q | Q ∈ F.

ByM[68, Theorem 24.3], the set F is closed in the Zariski topology on SpecT .

HenceF = V(F ) := P ∈ Spec T | F ⊆ P .

Thus the set Fmin is a finite set and is equal to the set Min(F ) of minimal primesof the ideal F of T .

Since a flat homomorphism satisfies the going-down theorem cf.Remark3.38.03.4.9,

Corollary16.2.4p11.4 implies that

(i) Fmin ⊆ Q ∈ Spec T | htQ < ht(Q ∩ S), and(ii) If Q ∈ Fmin, then every prime ideal P ( Q satisfies htP ≥ ht(P ∩ S).

16.2.6 Example and Remarks 11.7. (1) Let k be a field, let x and y be indetermi-nates over k and set f = x, g = (x− 1)y. Then k[f, g]

ϕ→ k[x, y] is not flat.

Proof. For the prime ideal P := (x−1) ∈ Spec(k[x, y]), we see that ht(P ) = 1,but ht(P ∩ k[f, g]) = 2; thus the extension is not flat by Corollary

16.2.4p11.4.

(2) The Jacobian ideal J of f and g in (1) is given by:

J = (det

(∂f∂x

∂f∂y

δg∂x

∂g∂y

)) = (det

(1 0y x− 1

)) = (x − 1)k[x, y].

(3) In the example of item (1), the nonflat locus is equal to the set of primeideals Q of k[x, y] that contain the Jacobian ideal (x − 1)k[x, y], thus J = F .

(4) One can also describe the example of item (1) by taking the base ring Rto be the polynomial ring k[x] rather than the field k. Then both T = R[y] andS = R[g] are polynomial rings in one variable over R with g = (x − 1)y. TheJacobian ideal J is the ideal of T generated by ∂g

∂y = x − 1, so is the same as initem (1).

16.2.65 Remark 11.8. A homomorphism f : R → Λ of Noetherian rings is said to beregular, cf. Definition

3.413.15, if f is flat and has geometrically regular fibers. In the

case where Λ is a finitely generated R-algebra, the map f is regular if and only ifit is smooth as can be seen by taking Λ = A in

Swan[94, Corollary 1.2].

We record in Theorem16.2.711.9 observations about smoothness and flatness that

follow from well-known properties of the Jacobian.

11.2. THE JACOBIAN IDEAL AND THE SMOOTH AND FLAT LOCI 77

16.2.7 Theorem 11.9. Let R be a Noetherian ring, let z1, . . . , zn be indeterminatesover R, and let f1, . . . , fm ∈ R[z1, . . . , zn] be algebraically independent over R.Consider the embedding ϕ : S := R[f1, . . . , fm] → T := R[z1, . . . , zn]. Let J denotethe Jacobian ideal of ϕ, and let F and Fmin be as in (

16.2.511.6). Then

(1) Q ∈ SpecT does not contain J ⇐⇒ ϕQ : S → TQ is essentially smooth.Thus J defines the nonsmooth locus of ϕ.

(2) If Q ∈ SpecT does not contain J , then ϕQ : S → TQ is flat. Thus J ⊆ F .(3) Fmin ⊆ Q ∈ SpecT | J ⊆ Q and ht(Q ∩ S) > htQ.(4) Fmin ⊆ Q ∈ SpecT | J ⊆ Q, htQ < dimS and ht(Q ∩ S) > htQ.(5) ϕ is flat ⇐⇒ for every Q ∈ Spec(T ) such that J ⊆ Q and ht(Q) < dimS,

then ht(Q ∩ S) ≤ ht(Q).(6) If htJ ≥ dimS, then ϕ is flat.

Proof. For item 1, we show that our definition of the Jacobian ideal J givenin (

16.2.511.6) agrees with the description of the smooth locus of an extension given in

E[24],

Swan[94, Section 4].

To see this, let u1, . . . , um be indeterminates over R[z1, . . . , zn] and identify

R[z1, . . . , zn] withR[u1, . . . , um][z1, . . . , zn]

(ui − fii=1,...,m).

Since u1, . . . , um are algebraically independent, the ideal J generated by the minorsof J is the Jacobian ideal of the extension ϕ by means of this identification. Wemake this more explicit as follows.

Let B := R[u1, . . . , um, z1, . . . , zn] and I = (fi − uii=1,...,m)B. Consider thefollowing commutative diagram

S := R[f1, . . . , fm] −−−−→ T := R[z1, . . . , zn]

∼=y ∼=

yS1 := R[u1, . . . , um] −−−−→ T1 := B/I

Define as inE[24],

Swan[94, Section 4]

H = HT1/S1 := the radical of Σ∆(g1, . . . , gs)[(g1, . . . , gs) :B I],

where the sum is taken over all s with 0 ≤ s ≤ m, for all choices of s polynomialsg, . . . , gs from I = (f1−u1, . . . , fm−um)B, where ∆ := ∆(g1, . . . , gs) is the ideal

of T ∼= T1 generated by the s× s-minors of(∂gi

∂zj

), and ∆ = T if s = 0.

To establish (16.2.711.9.1), we show that H = rad(J). Since ui is a constant with

respect to zj , we have(∂(fi−ui)∂zj

)=(∂fi

∂zj

). Thus J ⊆ H .

For g1, . . . , gs ∈ I, the s× s-minors of(∂gi

∂zj

)are contained in the s× s-minors

of(∂fi

∂zj

). Thus it suffices to consider s polynomials g1, . . . , gs from the set f1 −

u1, . . . , fm − um. Now f1 − u1, . . . , fm − um is a regular sequence in B. Thusfor s < m, [(g1, . . . , gs) :B I] = (g1, . . . , gs)B. Thus the m × m-minors of

(∂fi

∂zj

)generate H up to radical, and so H = rad(J).

Hence byE[24] or

Swan[94, Theorem 4.1], TQ is essentially smooth over S if and only

if Q does not contain J .


Item (2) follows from item (1) because essentially smooth maps are flat. Inview of Corollary

16.2.4p11.4 and (

16.2.511.6.2), item (3) follows from item (2).

If htQ ≥ dimS, then ht(Q ∩ S) ≤ dimS ≤ htQ. Hence the set Q ∈ SpecT :J ⊆ Q and ht(Q ∩ S) > htQ is equal to the set Q ∈ SpecT : J ⊆ Q, htQ <dimS and ht(Q ∩ S) > htQ. Thus item (3) is equivalent to item (4).

The ( =⇒ ) direction of item (5) is clearM[68, Theorem 9.5]. For the ( ⇐= )

direction of item (5) and for item (6), it suffices to show Fmin is empty, and thisholds by item (4).

16.2.8 Remarks 11.10. (1) For ϕ as in Theorem16.2.711.9, it would be interesting to

identify the set Fmin = Min(F ). In particular we are interested in conditions forJ = F and/or conditions for J ( F . Example

16.2.611.7 is an example where J = F ,

whereas Examples16.4.311.13 contains several examples where J ( F .

(2) If R is a Noetherian integral domain, then the zero ideal is not in Fmin and soF 6= 0.(3) In view of Theorem

16.2.711.9.3, we can describe Fmin precisely as

Fmin = Q ∈ SpecT | J ⊆ Q, ht(Q ∩ S) > htQ and ∀P ( Q, ht(P ∩ S) ≤ ht(P ).(4) Item (3) of Theorem

16.2.711.9 implies that for each prime ideal Q of Fmin there exist

prime ideals P1 and P2 of S with P1 ( P2 such that Q is minimal over both P1Tand P2T .

Corollary16.4.211.11 is immediate from Theorem

16.2.711.9.

16.4.2 Corollary 11.11. Let k be a field, let z1, . . . , zn be indeterminates over k andlet f, g ∈ k[z1, . . . , zn] be algebraically independent over k. Consider the embeddingϕ : S := k[f, g] → T := k[z1, . . . , zn]. Assume that the associated Jacobian ideal Jis nonzero. 2 Then

(1) Fmin ⊆ minimal primes Q of J with ht(Q ∩ S) > htQ = 1 .(2) ϕ is flat ⇐⇒ for every height-one prime ideal Q ∈ SpecT such that

J ⊆ Q we have ht(Q ∩ S) ≤ 1.(3) If htJ ≥ 2, then ϕ is flat.

16.4.2r Remark 11.12. In the case where k is algebraically closed, another argumentcan be used for Corollary

16.4.211.11.3: Each height-one prime ideal Q ∈ SpecT has

the form Q = hT for some polynomial h ∈ T . If ϕ is not flat, then there existsa prime ideal Q of T of height one, such that ht(Q ∩ S) = 2. Then Q ∩ S hasthe form (f − a, g − b)S, where a, b ∈ k. Thus f − a = f1h and g − b = g1hfor some polynomials f1, g1 ∈ T . Now the Jacobian ideal J of f, g is the same asthe Jacobian ideal of f − a, g − b, and an easy computation shows that J ⊆ hT .Therefore htJ ≤ 1.

16.4.3 Examples 11.13. Let k be a field of characteristic different from 2 and letx, y, z be indeterminates over k.

(1) With f = x and g = xy2 − y, consider S := k[f, g]ϕ→ T := k[x, y]. Then

J = (2xy−1)T . Since ht((2xy−1)T ∩S) = 1, ϕ is flat by Corollary16.4.211.11.2. But ϕ

is not smooth, since J defines the nonsmooth locus and J 6= T , cf. Theorem16.2.711.9.1.

Here we have J ( F = T .

2This is automatic if the field k has characteristic zero.

11.3. APPLICATIONS TO POLYNOMIAL EXTENSIONS 79

(2) With f = x and g = yz, consider S := k[f, g]ϕ→ T := k[x, y, z]. Then

J = (y, z)T . Since htJ ≥ 2, ϕ is flat by Corollary16.4.211.11.3. Again ϕ is not smooth

since J 6= T .(3) The examples given in items (1) and (2) may also be described by taking

R = k[x]. In item (1), we then have S := R[xy2 − y] → R[y] =: T . The JacobianJ = (2xy − 1)T is the same but is computed now as just a derivative. In item (2),we have S := R[yz] → R[y, z] =: T . The Jacobian J = (y, z)T is now computed bytaking the partial derivatives ∂(yz)

∂y and ∂(yz)∂z .

(4) Let R = k[x] and S = R[xyz] → R[y, z] =: T . Then J = (xz, xy)T . Thus Jhas two minimal primes xT and (y, z)T . Notice that xT ∩S = (x, xyz)S is a primeideal of S of height two, while (y, z)T ∩ S has height one. Therefore J ( F = xT .

(5) Let R = k[x] and S = R[xy + xz] → R[y, z] =: T . Then J = xT .(6) Let R = k[x] and S = R[xy+ z2] → R[y, z] =: T . Then J = (y, z)T . Hence

S → T is flat but not regular.(7) Let R = k[x] and S = R[xy + z] → R[y, z] =: T . Then J = T . Hence

S → T is a regular map.

16.4.5 Corollary 11.14. With the notation of Theorem16.2.711.9, we have

(1) If Q ∈ Fmin, then Q is a nonmaximal prime of T .(2) Fmin ⊆ Q ∈ SpecT : J ⊆ Q, dim(T/Q) ≥ 1 and ht(Q ∩ S) > htQ.(3) ϕ is flat ⇐⇒ ht(Q ∩ S) ≤ ht(Q) for every nonmaximal Q ∈ Spec(T )

with J ⊆ Q.(4) If dimR = d and htJ ≥ d+m, then ϕ is flat.

Proof. For item (1), suppose Q ∈ Fmin is a maximal ideal of T . Then htQ <ht(Q∩S) by Theorem

16.2.711.9.3. By localizing at R \ (R∩Q), we may assume that R

is local with maximal ideal Q∩R := m. Since Q is maximal, T/Q is a field finitelygenerated over R/m. By the Hilbert Nullstellensatz

M[68, Theorem 5.3], T/Q is

algebraic over R/m and htQ = ht(m)+n. It follows that Q∩S = P is maximal inS and htP = ht(m)+m. The algebraic independence hypothesis for the fi impliesthat m ≤ n, and therefore that htP ≤ htQ. This contradiction proves item (1).Item (2) follows from Theorem

16.2.711.9.3 and item (1).

Item (3) follows from Theorem16.2.711.9.5 and item (1), and item (4) follows from

Theorem16.2.711.9.6.

As an immediate corollary to Theorem16.2.711.9 and Corollary

16.4.511.14, we have:

16.4.7 Corollary 11.15. Let R be a Noetherian ring, let z1, . . . , zn be indetermi-nates over R and let f1, . . . , fm ∈ R[z1, . . . , zn] be algebraically independent overR. Consider the embedding ϕ : S := R[f1, . . . , fm] → T := R[z1, . . . , zn], let J bethe Jacobian ideal of ϕ and let F be the (reduced) ideal that describes the nonflatlocus of ϕ as in (

16.2.511.6.2). Then J ⊆ F and either F = T , that is, ϕ is flat, or

dim(T/Q) ≥ 1, for each Q ∈ Spec(T ) that is minimal over F .

11.3. Applications to polynomial extensions16.3p

Proposition16.4.411.16 considers behavior of the extension ϕ : S → T with respect

to prime ideals of R.


16.4.4 Proposition 11.16. Let R be a commutative ring, let z1, . . . , zn be indetermi-nates over R, and let f1, . . . , fm ∈ R[z1, . . . , zn] be algebraically independent overR. Consider the embedding ϕ : S := R[f1, . . . , fm] → T := R[z1, . . . , zn].

(1) If p ∈ SpecR and ϕpT : S → TpT is flat, then pS = pT∩S and the imagesfi of the fi in T/pT ∼= (R/p)[z1, . . . , zn] are algebraically independent overR/p.

(2) If ϕ is flat, then for each p ∈ Spec(R) we have pS = pT∩S and the imagesfi of the fi in T/pT ∼= (R/p)[z1, . . . , zn] are algebraically independent overR/p.

Proof. Item (2) follows from item (1), so it suffices to prove item (1). Assumethat TpT is flat over S. Then pT 6= T and it follows from

M[68, Theorem 9.5] that

pT ∩ S = pS. If the fi were algebraically dependent over R/p, then there existindeterminates t1, . . . , tm and a polynomial G ∈ R[t1, . . . , tm] \ pR[t1, . . . , tm] suchthat G(f1, . . . , fm) ∈ pT . This implies G(f1, . . . , fm) ∈ pT ∩ S. But f1, . . . , fmare algebraically independent over R and G(t1, . . . , tm) 6∈ pR[t1, . . . , tm] impliesG(f1, . . . , fm) 6∈ pS = pT ∩ S, a contradiction.

16.4.8 Proposition 11.17. Let R be a Noetherian integral domain containing a fieldof characteristic zero. Let z1, . . . , zn be indeterminates over R and let f1, . . . , fm ∈R[z1, . . . , zn] be algebraically independent over R. Consider the embedding ϕ : S :=R[f1, . . . , fm] → T := R[z1, . . . , zn]. Let J be the associated Jacobian ideal and letF be the reduced ideal of T defining the nonflat locus of ϕ . Then

(1) If p ∈ SpecR and J ⊆ pT , then ϕpT : S → TpT is not flat. Thus we alsohave F ⊆ pT .

(2) If the embedding ϕ : S → T is flat, then for every p ∈ Spec(R) we haveJ * pT .

Proof. Item (2) follows from item (1), so it suffices to prove item (1). Letp ∈ SpecR with J ⊆ pT , and suppose ϕpT is flat. Let fi denote the image of fi inT/pT . Consider

ϕ : S := (R/p)[f1, . . . , fm]→ T := (R/p)[z1, . . . , zn].

By Proposition16.4.411.16, f1, . . . , fm are algebraically independent over R := R/p.

Since the Jacobian ideal commutes with homomorphic images, the Jacobian idealof ϕ is zero. Thus for each Q ∈ SpecT the map ϕQ : S → TQ is not smooth. Buttaking Q = (0) gives TQ is a field separable over the field of fractions of S andhence ϕQ is a smooth map. This contradiction completes the proof.

Theorem16.3.711.18 follows from

P[79, Proposition 2.1] in the case of one indetermi-

nate z, so in the case where T = R[z].

16.3.7 Theorem 11.18. Let R be a Noetherian integral domain, let z1, . . . , zn be in-determinates over R, and let T = R[z1, . . . zn]. Suppose f ∈ T \ R. Then thefollowing are equivalent:

(1) R[f ]→ T is flat.(2) For each prime ideal q of R, we have qT ∩R[f ] = qR[f ].(3) For each maximal ideal q of R, we have qT ∩R[f ] = qR[f ].(4) The nonconstant coefficients of f generate the unit ideal of R.(5) R[f ]→ T is faithfully flat.

11.3. APPLICATIONS TO POLYNOMIAL EXTENSIONS 81

Proof. Since f ∈ T \ R and R is an integral domain, the ring R[f ] is apolynomial ring in the indeterminate f over R. Thus the map R → R[f ] is flatwith regular fibers. Hence Corollary

16.2.4p11.4 implies that R[f ] → T is flat ⇐⇒

for each Q ∈ SpecT we have htQ ≥ ht(Q ∩ R[f ]. Let q := Q ∩ R. We haveht q = ht qR[f ] = ht qT . Thus R[f ] → T is flat implies for each q ∈ SpecR thatqT ∩ R[f ] = qR[f ]. Moreover, if P := Q ∩ R[f ] properly contains qR[f ], thenhtP = 1 + ht q, while if Q properly contains qT , then htQ ≥ 1 + ht q. Therefore(1) ⇐⇒ (2) follows from Corollary

16.2.4p11.4. It is obvious that (2) =⇒ (3).

(3) =⇒ (4): Let a ∈ R be the constant term of f . If the nonconstantcoefficients of f are contained in a maximal ideal q of R, then f − a ∈ qT ∩ R[f ].Since R is an integral domain, the element f − a is transcendental over R andf − a 6∈ qR[f ] since R[f ]/qR[f ] is isomorphic to the polynomial ring (R/q)[x].Therefore qT ∩R[f ] 6= qR[f ] if the nonconstant coefficients of f are in q.

(4) =⇒ (2): Let q ∈ SpecR and consider the map16.3.7516.3.75 (11.1)

R[f ]⊗R R/q = R[f ]/qR[f ]ϕ−−−−→ T ⊗R R/q = T/qT ∼= (R/q)[z1, . . . , zn].

Since the nonconstant coefficients of f generated the unit ideal of R, the imageof f in (R/q)[z1, . . . , zn] has positive degree. This implies that ϕ is injective andqT ∩R[f ] = qR[f ].

This completes a proof that items (1), (2), (3) and (4) are equivalent. To showthat these equivalent statements imply (5), it suffices to show for P ∈ Spec(R[f ])that PT 6= T . Let q = P ∩R, and let κ(q) denote the field of fractions of R/q. Letf denote the image of f in R[f ]/qR[f ]. Then R[f ]/qR[f ] ∼= (R/q)[f ], a polynomialring in one variable over R/q. Tensoring the map ϕ of equation

16.3.7511.1 with κ(q) gives

an embedding of the polynomial ring κ(q)[f ] into κ(q)[z1, . . . , zn]. The image of Pin κ(q)[f ] is either zero or a maximal ideal of κ(q)[f ]. In either case, its extensionto κ(q)[z1, . . . , zn] is a proper ideal. Therefore PT 6= T . It is obvious that (5) =⇒(1), so this completes the proof of Theorem

16.3.711.18.

Remark 11.19. A different proof that (4) =⇒ (1) in Theorem16.3.711.18 is as

follows: Let v be another indeterminate and consider the commutative diagram

R[v] −−−−→ T [v] = R[z1, . . . , zn, v]

π

y π′y

R[f ]ϕ−−−−→ R[z1,...,zn,v]

(v−f(z1,...,zn)) .

where π maps v → f and π′ is the canonical quotient homomorphism. ByM1[66,

Corollary 2, p. 152] orM[68, Theorem 22.6 and its Corollary, p. 177], ϕ is flat if the

coefficients of f − v generate the unit ideal of R[v]. Moreover, the coefficients off − v as a polynomial in z1, . . . , zn with coefficients in R[v] generate the unit idealof R[v] if and only if the nonconstant coefficients of f generate the unit ideal of R.For if a ∈ R is the constant term of f and a1, . . . , ar are the nonconstant coefficientsof f , then (a1, . . . , ar)R = R clearly implies that (a− v, a1, . . . , ar)R[v] = R[v]. Onthe other hand, if (a − v, a1, . . . , ar)R[v] = R[v], then setting v = a implies that(a1, . . . , ar)R = R.

We observe in Proposition16.3.811.20 that one direction of Theorem

16.3.711.18 holds for

more than one polynomial: see alsoP[79, Theorem 3.8] for a related result concerning

flatness.


16.3.8 Proposition 11.20. Let z1, . . . , zn be indeterminates over an integral domainR. Let f1, . . . , fm be polynomials in R[z1, . . . , zn] := T that are algebraically inde-pendent over Q(R). If the inclusion map ϕ : S := R[f1, . . . fm] → T is flat, thenthe nonconstant coefficients of each of the fi generate the unit ideal of R.

Proof. The algebraic independence of the fi implies that the inclusion mapR[fi] → R[f1, . . . , fm] is flat, for each i with 1 ≤ i ≤ m. If S −→ T is flat,then so is the composition R[fi] −→ S −→ T , and the statement follows fromTheorem

16.3.711.18

Exercises(1) Let k be an algebraically closed field of characteristic zero and let T denote

the polynomial ring k[x]. Let f ∈ T be a polynomial of degree d ≥ 2 and letS := k[f ].

(i) Prove that the map S → T is free and hence flat.(ii) Prove that the prime ideals Q ∈ SpecT for which S → TQ is not a

regular map are precisely the primes Q such that the derivative dfdx ∈ Q.

(iii) Deduce that S → T is not smooth.(2) With S = k[x, xy2 − y] → T = k[x, y] and J = (2xy − 1)T as in Exam-

ples16.4.311.13.1, prove that ht(J ∩ S) = 1.

Suggestion. Show that J ∩ S ∩ k[x] = (0) and use that, for A an integraldomain, prime ideals of the polynomial ring A[y] that intersect A in (0) are inone-to-one correspondence with prime ideals of K[y], where K = (A \ 0)−1Ais the field of fractions of A.

(3) Let z1, . . . , zn be indeterminates over a ring R, and let T = R[z1, . . . zn]. Sup-pose f ∈ T \ R. Modify the proof of (3) implies (4) of Theorem

16.3.711.18 to prove

that qT ∩R[f ] = qR[f ] for each maximal ideal q of R implies that the noncon-stant coefficients of f generate the unit ideal of R without the assumption thatthe ring R is an integral domain.Suggestion. Assume that the nonconstant coefficients of f are contained in amaximal ideal q of R. Observe that one may assume that f as a polynomialin R[z1, . . . , zn] has zero as its constant term and that the ring R is local withmaximal ideal q. Let M be a monomial in the support of f of minimal totaldegree and let b ∈ R denote the coefficient of M for f . Then b is nonzero, butf ∈ qR[f ] implies that b ∈ qb and this implies, by Nakayama’s lemma, thatb = 0.

CHAPTER 12

Height-one primes and limit-intersecting elements

May 8, 2010

flatconLet z be a nonzero nonunit of a Noetherian integral domain R and let R∗

denote the (z)-adic completion of R. Assume that R and R∗ are integrally closeddomains. As in Construction

4.4.15.3, we consider in this chapter the structure of an

intermediate ring A between R and R∗ of the form A := Q(R)(τ1, τ2, . . . τs) ∩ R,where τ1, τ2, . . . , τs ∈ zR∗ are algebraically independent elements over R. If theintersection ring A can be expressed as a directed union B of localized polynomialextension rings of R as in Section

4.456.1, then the computation of A is easier.

Recall that τ1, τ2, . . . , τs are called limit-intersecting for A if the ring A is such adirected union, cf. Definition

4.2li6.5. In Theorem

7.5.5fc12.4 we give criteria for τ1, τ2, . . . , τs

to be limit-intersecting for A.We use Theorem

7.5.5fc12.4 to establish in Examples

16.3.1013.6 a family of examples where

the approximating ring B is equal to the intersection ring A, but this ring is notNoetherian. In Chapter

intsec21 we consider two stronger forms of the limit-intersecting

condition that are useful for constructing examples and for determining if A isNoetherian or excellent.

12.1. The limit-intersecting condition7.5fc

We consider the following two properties of an extension S → T of integraldomains involving height-one prime ideals.

7.3.4fc Definitions 12.1. Let S → T be an extension of integral domains.(1) We say that the extension S → T is weakly flat, or that T is weakly flat

over S, if every height-one prime ideal P of S with PT 6= T satisfiesPT ∩ S = P .

(2) We say that the extension S → T is height-one preserving, or that T isa height-one preserving extension of S, if for every height-one prime idealP of S with PT 6= T there exists a height-one prime ideal Q of T withPT ⊆ Q.

6.2.9fc Remarks 12.2. Let S → T be an extension of Krull domains.(a) We have T ∩Q(S) = S ⇐⇒ each height-one prime of S is the contraction

of a height-one prime of T . If these equivalent conditions hold for theextension S → T , then T is height-one preserving and weakly flat over S(cf.

N2[72, (33.5) and (33.6)]).

(b) If S → T is flat, then S → T is height-one preserving, weakly flat andsatisfies PDE (cf. Definition

7.6.1.07.6 and

B[10, Chapitre 7, Proposition 15,

page 19]).

83

84 12. HEIGHT-ONE PRIMES

(c) If U is a multiplicative system in S consisting of units of T , then S →T is height-one preserving (respectively weakly flat, respectively satisfiesPDE) ⇐⇒ U−1S → T is height-one preserving (respectively weakly flat,respectively satisfies PDE).

(d) If each height-one prime of S is the radical of a principal ideal, so, in par-ticular, if S is a UFD, then the extension S → T is height-one preserving.For if P is a height-one prime of S and P is the radical of the principalideal xS, then PT 6= T if and only if xT is a proper principal ideal ofT and every proper principal ideal in a Krull domain is contained in aheight-one prime. Hence if PT 6= T , then PT is contained in a height-oneprime of T .

The weakly flat property is important for our work, and so it is fortunate thatit holds in the situations we consider.

6.2.14fc Proposition 12.3. Let S → T be an extension of Krull domains with PT 6= Tfor every height-one prime ideal P of S. Then S = Q(S) ∩ T ⇐⇒ T is weaklyflat over S.

Proof. The assertion that S = Q(S) ∩ T implies T is weakly flat over S isRemark (

6.2.9fc12.2)(a). A direct proof of this assertion involving primary decomposition

of principal ideals goes as follows: Let P be a height-one prime ideal of S, let a bea nonzero element in P and consider an irredundent primary decomposition

aT = Q1 ∩ · · · ∩Qsof the principal ideal aT in the Krull domain T . Since T is a Krull domain, each Qiis primary for a height-one prime ideal Pi of T . The fact that S = Q(S)∩T impliesthat aS = Q(S) ∩ aT . Thus, after renumbering, there is an integer t ∈ 1, . . . , ssuch that the ideal

Q1 ∩ . . . ∩Qt ∩ Sis the P -primary component of the ideal aS. Hence for at least one integer i ∈1, . . . , t we must have that Pi ∩ S = P . Therefore T is weakly flat over S.

For the converse, assume that T is weakly flat over S and that PT 6= T foreach height-one prime ideal P of S. We have S → D := Q(S) ∩ T . It followsthat PD ∩ S = P and SP = (S \ P )−1D, and hence D ⊆ SP for each height-oneprime ideal P of S. Since S =

⋂SP | P is a height-one prime of S, we haveS = D.

We show in Theorem7.5.5fc12.4 that τ1, . . . , τs are limit-intersecting if and only if

the extensionR[τ1, . . . , τs] → R∗[1/z]

is weakly flat.

7.5.5fc Theorem 12.4. Let R be a Noetherian integral domain and let z ∈ R be anonzero nonunit. Let R∗ denote the (z)-adic completion of R and let τ1, . . . , τs ∈ R∗

be algebraically independent over R. Assume that R and R∗ are integrally closeddomains, and let the rings A and B be as defined in Section

4.456.1. Then the following

statements are equivalent:(1) The elements τ1, . . . , τs are limit-intersecting in z over R, i.e., the inter-

mediate rings A and B are equal.(2) The extension R[τ1, . . . , τs] → R∗[1/z] is weakly flat.

12.2. HEIGHT-ONE PRIMES IN EXTENSIONS OF INTEGRAL DOMAINS 85

(3) The extension B → R∗[1/z] is weakly flat.(4) The extension B → R∗ is weakly flat.

Proof. (2)⇒(1): Since A and B are Krull domains with the same field offractions and B ⊆ A it suffices to show that every height-one prime ideal p of Bis the contraction of a (height-one) prime ideal of A. By Corollary

11.2.516.19.3 each

height-one prime of B containing zB is the contraction of a height-one prime of A.Let p be a height-one prime of B that does not contain zB. Consider the prime

ideal q = R[τ1, . . . , τs] ∩ p. Since B[1/z] is a localization of the ring R[τ1, . . . , τs],we see that Bp = R[τ1, . . . , τs]q and so q has height one in R[τ1, . . . , τs]. The limit-intersecting hypothesis implies qR∗ ∩ R[τ1, . . . , τs] = q and there is a height-oneprime ideal w of R∗ with w ∩ R[τ1, . . . , τs] = q. This implies that w ∩ B = pand thus also (w ∩ A) ∩ B = p. Hence every height-one prime ideal of B is thecontraction of a prime ideal of A. Since A is birational over B, this prime ideal of Ais necessarily of height one since the localization of A at this prime ideal birationallydominates a DVR and hence is equal to this DVR.

(3)⇐⇒ (4): By Corollary11.2.516.19.3 each height-one prime of B containing zB is

the contraction of a height-one prime of R∗. If p is a height-one prime of B withz 6∈ p, then pR∗ ∩B = p if and only if pR∗[1/z]∩B = p.

(1)⇒(4): If B = A = Q(B) ∩ R∗, then by Proposition6.2.14fc12.3, the extension

B → R∗ is weakly flat.(4)⇒(2): If B → R∗ is weakly flat, so is the localization B[1/z] → R∗[1/z].

Since B[1/z] is a localization of R[τ1, . . . , τs], item (2) follows.

12.2. Height-one primes in extensions of integral domains

We observe in Proposition6.2.10fc12.5 that a weakly flat extension of Krull domains

is height-one preserving.

6.2.10fc Proposition 12.5. If φ : S → T is a weakly flat extension of Krull domains,then φ is height-one preserving. Moreover, for every height-one prime ideal P of Swith PT 6= T there is a height-one prime ideal Q of T with Q ∩ S = P .

Proof. Let P ∈ SpecS with htP = 1. By assumption PT ∩S = P . Thereforethe ideal PT of T is contained in an ideal Q′ of T that is maximal with respect tonot meeting the multiplicative system S \ P of T . It follows that Q′ is a primeideal of T and Q′ ∩ S = P . Let a be a nonzero element of P and let Q ⊆ Q′ be aminimal prime divisor of aT . Then Q has height one and (0) 6= Q ∩ S ⊆ P , andthus Q ∩ S = P .

The height-one preserving condition does not imply weakly flat as we demon-strate in Example

7.3.6fc12.6.

7.3.6fc Example 12.6. Let x and y be variables over a field k, let R = k[[x]][y](x,y)and let C = k[[x, y]]. There exists an element τ ∈ n = (x, y)C that is algebraicallyindependent over Q(R). For any such element τ , let S = R[τ ](m,τ). Since R is aUFD, the ring S is also a UFD and the local inclusion map ϕ : S → C is height-onepreserving. There exists a height-one prime ideal P of S such that P ∩ R = 0.Since the map S → C is a local map, we have PC 6= C. Because ϕ is height-one preserving, there exists a height-one prime ideal Q of C such that PC ⊆ Q.

86 12. HEIGHT-ONE PRIMES

Since C is the m-adic completion R of R and the generic formal fiber of R is zero-dimensional, dim(C ⊗R Q(R)) = 0. Hence Q ∩ R 6= 0. We have P ⊆ Q ∩ S andP ∩R = (0). It follows that P is strictly smaller than Q ∩ S, so Q ∩ S has heightgreater than one. Therefore the extension ϕ : S → C is not weakly flat.

7.5.6 Remarks 12.7. (1) By Proposition6.2.10fc12.5, an injective map of Krull domains

that is weakly flat is also height-one preserving. Thus the equivalent conditions ofTheorem

7.5.5fc12.4 imply that B → R∗ is height-one preserving.

(2) If the ring B in Theorem7.5.5fc12.4 is Noetherian, then A = B and the equivalent

conclusions of Theorem7.5.5fc12.4 hold.

(3) In Chapternoeslocsec22, (

7.4.421.12), we give an example of a three-dimensional regular

local domain R dominating Q[x, y, z](x,y,z) and having completion Q[[x, y, z]], suchthat there exists an element τ in the (y)-adic completion of R that is residuallylimit-intersecting in y over R but fails to be primarily limit-intersecting in y overR. In particular, the rings A and B constructed using τ are equal, yet A and Bare not Noetherian. We also show in Chapter

noeslocsec22, (

8.2.1222.12) that if R is a Noetherian

semilocal domain, then τ1, . . . , τs ∈ yR∗ are primarily limit-intersecting in y overR if and only if B is Noetherian. If this holds, we also have B = A.

7.3.3fc Question 12.8. Let (C,n) be a complete Noetherian local domain that dom-inates a quasilocal Krull domain (D,m). Assume that the inclusion map D → Cis height-one preserving, and that τ ∈ n is algebraically independent over D. Doesit follow that the local inclusion map ϕ : S := D[τ ](m,τ) → C is height-onepreserving?

Discussion 12.9. If D has torsion divisor class group, then S also has torsiondivisor class group and by item (d) of Remark

6.2.9fc12.2, the extension S → C is

height-one preserving, so the answer to Question7.3.3fc12.8 is affirmative in this case. To

consider the general case, let P be a height-one prime ideal of S that is not assumedto be the radical of a principal ideal. One may then consider the following cases:Case (i): If ht(P ∩ D) = 1, then P = (P ∩ D)S. Since D → C is height-onepreserving, (P ∩D)C ⊆ Q, for some height-one prime ideal Q of C. Then PC =(P ∩D)SC ⊆ Q as desired.Case (ii): Suppose P ∩ D = (0). Let U denote the multiplicative set of nonzeroelements of D. Let t be an indeterminate over D and let S1 = D[t](m,t). Considerthe following commutative diagram where the map from S1 to S is the D-algebraisomorphism taking t to τ and λ is the extension mapping C[[t]] onto C.

U−1S1⊆−−−−→ U−1C[t](n,t)

∪x ∪

xD

⊆−−−−→ S1 = D[t](m,t)⊆−−−−→ C[t](n,t)

⊆−−−−→ C[[t]]

=

y ∼=y λ

yD

⊆−−−−→ S = D[τ ](m,τ)ϕ−−−−→ C.

Under the above isomorphism of S with S1, the prime ideal P correponds to a

height-one prime ideal P1 of S1 such that P1∩D = (0). Since U−1S1 is a localizationof a polynomial ring in one variable over a field, the extended ideal P1U

−1S1 is a

EXERCISES 87

principal prime ideal. Therefore P1 is contained in a proper principal ideal ofU−1C[t](n,t).

However, in the above diagram it can happen that the inclusion map

U−1S1 → U−1C[t](n,t)may fail to be faithfully flat. As an example to illustrate this, let

D := k[x, y = ex − 1](x,y) → k[[x]] =: C

and let P = (xt − y)D[t]. Then P extends to the whole ring in U−1C[t](n,t) sincet− y

x is a unit of U−1C[t](n,t).

ExercisesLet T = k[x, y, z] be a polynomial ring in the 3 variables x, y, z over a field k,

and consider the subring S = k[xy, xz, yz] of T .(1) Prove that the field extension Q(T )/Q(S) is algebraic with [Q(T ) : Q(S)] = 2.(2) Deduce that xy, xz, yz are algebraically independent over k, so S is a polyno-

mial ring in 3 variables over k.(3) Prove that the extension S → T is height-one preserving, but is not weakly

flat.(4) Prove that T ∩ Q(S) = S[x2, y2, z2] is a Krull domain that properly contains

S.(5) Prove that the map S → T [ 1

xyz ] is flat.(6) Prove that S[ 1

xyz ] = T [ 1xyz ]. (Notice that S[ 1

xyz ] is not a localization of S sincexyz is not in Q(S).)

CHAPTER 13

Insider construction details, May 3, 2010, 16, 17

(insidecon)

insideconIn this chapter we continue the development of the Insider Construction begun

in Chapterinsideintro10. In Section

17.2’?? we expand the notation for the Insider Construction

to the case where the base ring R is a Noetherian domain that is not necessarily apolynomial ring over a field. As before we first construct an “outside” Noetheriandomain Aout using the Inclusion Construction

4.4.15.3. We require that this intersection

domain Aout is equal to its corresponding approximation integral domain Bout

that is the nested union of localized polynomial rings from Section4.456.1. Then we

construct inside Aout = Bout two “insider” integral domains: Ains, an intersectionof a field with a power series ring as in Construction

4.4.15.3, and Bins, a nested union

of localized polynomial rings that “approximates” Ains as in Section4.456.1. We show

that Bins is Noetherian and equal to Ains if a certain map of polynomial rings overR is flat.

In Section17.2213.1, we describe background and notation for the construction.

Theorem16.3.213.2 in Section

16.2n13.2 gives necessary and sufficient conditions for the in-

tegral domains constructed with the insider construction to be Noetherian andequal. In Section

16.2213.3, we use the analysis of flatness for polynomial extensions

from Chapterflatpoly11 to obtain a general flatness criterion for the Insider Construction.

This yields examples where the constructed domains A and B are equal and arenot Noetherian.

In Chapterintclsec28, we use the Insider Construction to establish the existence for

each integer d ≥ 3 and each integer h with 2 ≤ h ≤ d − 1 of a d-dimensionalregular local domain (A,n) having a prime ideal P of height h with the propertythat the extension PA is not integrally closed. In Chapter

insidepssec17, we use the Insider

Construction to obtain, for each positive integer n, an explicit example of a 3-dimensional quasilocal unique factorization domain B such that the maximal idealof B is 2-generated, B has precisely n prime ideals of height two, and each primeideal of B of height two is not finitely generated.

13.1. Describing the construction17.22

We use the following setting and details for the Insider Construction in thischapter. This setting includes Noetherian domains that are not necessarily localand thus generalizes Settings

16.1.1g10.1 and

16.2.2g10.4.

16.1.1 Insider Construction Details 13.1. Generalized version LetR be a Noe-therian integral domain. Let z be a nonzero nonunit of R and let R∗ be the (z)-adiccompletion of R. The intersection domain of Construction

4.4.15.3 and the correspond-

ing approximation domain of Section4.456.1 are inside R∗. Let τ1, . . . , τn ∈ zR∗ be

algebraically independent over the field of fractions K of R and let τ abbreviate

89

90 13. INSIDER CONSTRUCTION DETAILS, MAY 3, 2010, 16, 17 (INSIDECON)

the list τ1, . . . , τn. We define the intersection domain corresponding to τ to be thering Aτ := K(τ1, . . . , τn)∩R∗. The Noetherian Example Theorem

11.3.258.8 implies that

Aτ is simultaneously Noetherian and computable as a nested union Bτ of certainassociated localized polynomial rings over R using τ if and only if the extension

T := R[τ ] := R[τ1, . . . , τn]ψ→ R∗[1/z] is flat. Moreover, if this flatness condition

holds, then Aτ is a localization of a subring of T [1/z] and Aτ [1/z] is a localizationof T .

We assume that ψ : T → R∗[1/z] is flat, so that the intersection domain Aτis Noetherian and computable, and we take the outside domain to be Aout :=Aτ = Bτ so that Bout = Aout. Then we construct new “insider” examples insideAout = Aτ as follows: We choose elements f1, . . . , fm of T := R[τ ], considered aspolynomials in the τi with coefficients in R and abbreviated by f . Assume thatf1, . . . , fm are algebraically independent over K; thus m ≤ n. As above we defineAf := K(f1, . . . , fm) ∩ R∗ to be the intersection domain corresponding to f . Welet Bf be the approximation domain corresponding to f , or corresponding to Af ,obtained using the fi as in Section

4.456.1. Sometimes we refer to Bf as the nested

union domain corresponding to Af . We define Ains := Af , and Bins := BfSet S := R[f ] := R[f1, . . . , fm], let ϕ be the embedding

(16.1.113.1.1) ϕ : S : = R[f ]

ϕ→ T := R[τ ],

and let ψ be the inclusion map: R[τ ] → R∗[1/z]. Put α := ψ ϕ : S → R∗[1/a].

We show in Chapterinsideintro10 for the special case where R is a localized polynomial

ring over a field in two or three variables that if ϕx : S → T [1/x] is flat, then Afis Noetherian and is equal to the corresponding approximating domain Bf fromSection

4.456.1. In Section

16.2n13.2 we make a more thorough analysis of conditions for

Af to be Noetherian and equal to Bf . In Section16.2213.3, we present examples where

Bf = Af is not Noetherian.

13.2. The flat locus of the insider construction16.2n

We assume Insider Construction Details16.1.113.1 for this discussion and refer the

reader to Section4.456.1 for details concerning the approximation domains Bτ and Bf

corresponding to the intersection domains Aτ and Af , respectively, of (16.1.113.1).

Theorem11.3.258.8 is the basis for our construction of examples.

LetF := ∩P ∈ Spec(T ) |ϕP : S → TP is not flat .

Thus, as in (16.2.511.6.2), the ideal F defines the nonflat locus of the map ϕ : S → T .

For Q∗ ∈ Spec(R∗[1/z]), we consider flatness of the localization map ϕQ∗∩T :

(16.3.213.2.1) ϕQ∗∩T : S −−−−→ TQ∗∩T

16.3.2 Theorem 13.2. With the notation of (16.1.113.1) we have

(1) For Q∗ ∈ Spec(R∗[1/z]), the map αQ∗ : S → (R∗[1/z])Q∗ is flat if andonly if the map ϕQ∗∩T in (

16.3.213.2.1) is flat.

(2) The following are equivalent:(i) The ring A is Noetherian and A = B.(ii) The ring B is Noetherian.

13.3. THE NON-FLAT LOCUS OF THE INSIDER CONSTRUCTION 91

(iii) For every maximal Q∗ ∈ Spec(R∗[1/z]), the map ϕQ∗∩T in (16.3.213.2.1)

is flat.(iv) FR∗[1/z] = R∗[1/z].

(3) The map ϕz : S → T [1/z] is flat if and only if FT [1/z] = T [1/z]. More-over, either of these conditions implies B is Noetherian and B = A. Itthen follows that A[1/z] is a localization of S.

Proof. For item (1), we have αQ∗ = ψQ∗ϕQ∗∩T : S → TQ∗∩T → (R∗[1/a])Q∗ .Since the map ψQ∗ is faithfully flat, the composition αQ∗ is flat if and only if ϕQ∗∩Tis flat

M1[66, page 27]. For item (2), the equivalence of (i) and (ii) is part of Theo-

rem11.3.258.8. The equivalence of (ii) and (iii) follows from item (1) and Theorem

11.3.258.8.

For the equivalence of (iii) and (iv), we use FR∗ 6= R∗ ⇐⇒ F ⊆ Q∗ ∩ T , forsome Q∗ maximal in Spec(R∗)[1/a]) ⇐⇒ the map in (

16.3.213.2.1) fails to be flat. The

first statement of Item (3) follows from the definition of F and the fact that thenonflat locus of ϕ : S → T is closed. Theorem

11.3.258.8 implies the final statement of

item (3).

13.3. The non-flat locus of the insider construction16.22

To examine the map α : S → R∗[1/z] in more detail, we use the followingterminology.

16.3.3 Definition 13.3. For an extension of Noetherian rings ϕ : A′ → B′ and ford ∈ N, we say that ϕ : A′ → B′, satisfies LFd if for each P ∈ Spec(B′) withht(P ) ≤ d, the composite map A′ → B′ → B′

P is flat.

16.3.4 Corollary 13.4. With the notation of (16.1.113.1), we have ht(FR∗[1/a]) > 1 ⇐⇒

ϕ : S → R∗[1/a] satisfies LF1. These equivalent conditions imply B = A.

Proof. The first equivalence follows from the definition of LF1 and LF1 im-plies weakly flat, so Theorem

7.5.5fc12.4 implies that B = A.

16.3.9 Theorem 13.5. With the notation of (16.1.113.1), if m = 1, that is, there is only

one polynomial f1 = f , S := R[f ] and T := R[τ1, . . . , τn], then(1) The map S → T [1/a] is flat ⇐⇒ the nonconstant coefficients of f

generate the unit ideal in R[1/a],(2) Either of the conditions in (1) implies the insider approximation domain

Bf is Noetherian and is equal to the insider intersection domain Af .(3) Bf is Noetherian and Bf = Af ⇐⇒ for every prime ideal Q∗ in R∗ with

a 6∈ Q∗, the nonconstant coefficients of f generate the unit ideal in Rq,where q := Q∗ ∩R.

(4) If the nonconstant coefficients of f generate an ideal L of R[1/a] of heightd, then the map S → R∗[1/a] satisfies LFd−1, but not LFd.

Proof. Item (1) follows from Theorem16.3.711.18 for the ring R[1/a] with zi = τi.

By Theorems11.3.258.8 and

16.3.213.2, the first condition in item (1) implies item (2).

For item (3), suppose the nonconstant coefficients of f generate the unit ideal ofRq. Then by Theorem

16.3.711.18, Rq[f ] → Rq[τ1, . . . , τn] is flat. Since Rq[τ1, . . . , τn]→

R∗Q∗ is flat, the map Rq[f ] → R∗

Q∗ is also flat. For the other direction, we mayassume that f has constant term zero. Suppose there exists Q∗ ∈ SpecR∗ witha 6∈ Q∗ such that the coefficients of f are in qRq, where q = Q∗∩R. Then f ∈ qR∗

Q∗

92 13. INSIDER CONSTRUCTION DETAILS, MAY 3, 2010, 16, 17 (INSIDECON)

and qR∗Q∗ 6= R∗

Q∗ . If R[f ] → R∗Q∗ is flat, then we have qR∗

Q∗ ∩R[f ] = qR[f ]. Thisimplies f ∈ qR[f ], a contradiction.

For item (4), if Q∗ ∈ Spec(R∗[1/a]), then the map S → (R∗[1/a])Q∗ is notflat if and only if L ⊆ Q∗. By hypothesis there exists a prime ideal Q∗ of height dcontaining L, but no such prime ideal of height less than d.

We return to Example16.1nn10.8 and establish a more general result.

16.3.10 Examples 13.6. Let k be a field and let x, y1, . . . , yd be indeterminates overk. Let R be either

(1) The polynomial ring R := k[x, y1, . . . , yd] with (x)-adic completion R∗ =k[y1, . . . , yd][[x]], or

(2) The localized polynomial ring R := k[x, y1, . . . , yd](x,y1,...,yd) with (x)-adiccompletion R∗.

With the notation of (16.1.113.1), and setting z = x and m = 1, assume that n and s are

each greater than or equal to d. Let f1 = f := y1τ1+ · · ·+ydτd. By Theorem16.3.913.5.4,

the map ϕx : S := R[f ] → T [1/x] satisfies LFd−1, but fails to satisfy LFd becausethe ideal L = (y1, . . . , yd)R[1/x] of nonconstant coefficients of f has height d. Ford ≥ 2, the map ϕx : S → T [1/x] satisfies LF1 and hence by Corollary

16.3.413.4, we have

Af = Bf , that is Af is ”limit-intersecting”. However, since ϕx does not satisfyLFd, the map ϕx is not flat and thus Af is not Noetherian.

CHAPTER 14

Integral closure under extension to the completion

(intclsec), May 10, 2010 18intclsec

Using the insider construction described in Chapterinsidecon13, we present in this

chapter an example of a 3-dimensional regular local domain (A,n) having a height-two prime ideal P with the property that the extension PA of P to the n-adiccompletion A of A is not integrally closed. The ring A in the example is a 3-dimensional regular local domains that is a nested union of 5-dimensional regularlocal domains.

More generally, we use this same technique to establish the existence for eachinteger d ≥ 3 and each integer h with 2 ≤ h ≤ d − 1 of a d-dimensional regularlocal domain (A,n) having a prime ideal P of height h with the property that theextension PA is not integrally closed. A regular local domain having a prime idealwith this property is necessarily not excellent.

We discuss in Section18.228.1 conditions in order that integrally closed ideals of a

ring R extend to integrally closed ideals of R′, where R′ is an R-algebra. In par-ticular, we consider under what conditions integrally closed ideals of a Noetherianlocal ring A extend to integrally closed ideals of the completion A of A.

14.1. Integral closure under ring extension18.2

For properties of integral closure of ideals, rings and modules we refer toSH[95].

In particular, we use the following definition.

18.2.0 Definitions 14.1. Let I be an ideal of a ring R.(1) An element r ∈ R is integral over I if there exists a monic polynomial

f(x) = xn +∑n

i=1 aixn−i such that f(r) = 0 and such that ai ∈ Ii for

each i with 1 ≤ i ≤ n.(2) If J is an ideal contained in I and JIn−1 = In, then J is said to be a

reduction of I.(3) The integral closure I of I is the set of elements of R integral over I.(4) If I = I, then I is said to be integrally closed.(5) The ideal I is said to be normal if In is integrally closed for every n ≥ 1.

18.2.01 Remarks 14.2. We record the following facts about an ideal I of a ring R.(1) An element r ∈ R is integral over I if and only if I is a reduction of

the ideal L = (I, r)R. To see this equivalence, observe that for a monicpolynomial f(x) as in Definition

18.2.028.1.1, we have

f(r) = 0 ⇐⇒ rn = −n∑i=1

airn−i ∈ ILn−1 ⇐⇒ Ln = ILn−1.

93

9414. INTEGRAL CLOSURE UNDER EXTENSION TO THE COMPLETION (INTCLSEC), MAY 10, 2010 18

(2) It is well known that I is an integrally closed idealSH[95, Corollary 1.3.1].

(3) An ideal is integrally closed if and only if it is not a reduction of a properlybigger ideal.

(4) A prime ideal is always integrally closed. More generally, a radical idealis always integrally closed.

(5) Let a, b be elements in a Noetherian ring R and let I := (a2, b2)R. Theelement ab is integral over I. If a, b are a regular sequence, then ab 6∈ Iand the ideal I is not integrally closed. More generally, if h ≥ 2 anda1, . . . , ah are a regular sequence in R and I := (ah1 , . . . , a

hh)R, then I is

not integrally closed.

Our work in this chapter was motivated by the following questions:

18.2.02 Questions 14.3.(1) Craig Huneke: “Does there exist an analytically unramified Noetherian lo-

cal ring (A,n) that has an integrally closed ideal I for which the extensionIA to the n-adic completion A is not integrally closed?”

(2) Sam Huckaba: “If there is such an example, can the ideal of the examplebe chosen to be a normal ideal?”

Related to Question18.2.0228.3.1, we construct in Example

18.3.128.7 a 3-dimensional reg-

ular local domain A having a height-two prime ideal I = P = (f, g)A such thatIA is not integrally closed. Thus the answer to Question

18.2.0228.3.1 is “yes”. This

example also shows that the answer to Question18.2.0228.3.2 is again “yes”. Since f, g

form a regular sequence and A is Cohen-Macaulay, the powers Pn of P have noembedded associated primes and therefore are P -primary

M1[66, (16.F), p. 112],

M[68,

Ex. 17.4, p. 139]. Since the powers of the maximal ideal of a regular local domainare integrally closed, the powers of P are integrally closed. Thus the Rees alge-bra A[Pt] = A[ft, gt] is a normal domain while the Rees algebra A[ft, gt] is notintegrally closed.

Remarks 14.4. Without the assumption that A is analytically unramified,there exist examples even in dimension one where an integrally closed ideal of Afails to extend to an integrally closed ideal in A. If A is reduced but analyticallyramified, then the zero ideal of A is integrally closed, but its extension to A is notintegrally closed.

As we remark in Chapterintro1, examples exhibiting this behavior have been known

for a long time. An example in characteristic zero of a one-dimensional Noetherianlocal domain that is analytically ramified is given by Akizuki in his 1935 paper

A[7].

An example in positive characteristic is given by F.K. SchmidtSc[87, pp. 445-447].

Another example due to Nagata is given inN2[72, Example 3, pp. 205-207]. (See also

N2[72, (32.2), p. 114].)

Let R be a commutative ring and let R′ be an R-algebra. In Remark18.2.128.6 we

list cases where extensions to R′ of integrally closed ideals of R are again integrallyclosed. In this connection we use the following definition.

18.2.10 Definition 14.5. The R-algebra R′ is said to be quasi-normal if R′ is flat overR and the following condition (NR,R′) holds: If C is any R-algebra and D is aC-algebra in which C is integrally closed, then also C ⊗R R′ is integrally closed inD ⊗R R′.

14.1. INTEGRAL CLOSURE UNDER RING EXTENSION 95

18.2.1 Remarks 14.6. Let R be a commutative ring and let R′ be an R-algebra.

(1) ByL[62, Lemma 2.4], if R′ satisfies (NR,R′) and I is an integrally closed

ideal of R, then IR′ is integrally closed in R′.(2) Assume that R and R′ are Noetherian rings and that R′ is a flat R-

algebra. Let I be an integrally closed ideal of R. The flatness of R′ overR implies every P ′ ∈ Ass(R′/IR′) contracts in R to some P ∈ Ass(R/I)M[68, Theorem 23.2]. Since a regular map is quasi-normal, if the mapR → R′

P ′ is regular for each P ′ ∈ Ass(R′/IR′), then IR′ is integrallyclosed.

(3) Principal ideals of an integrally closed domain are integrally closed.(4) In general, integral closedness of ideals is a local condition. If R′ is an R-

algebra that is locally normal in the sense that for every prime ideal P ′ ofR′, the local ring R′

P ′ is an integrally closed domain, then the extensionto R′ of every principal ideal of R is integrally closed by item (3). Inparticular, if (A,n) is an analytically normal Noetherian local domain,then every principal ideal of A extends to an integrally closed ideal of A.

(5) Let (A,n) be a Noetherian local ring and let A be the n-adic completionof A. Since A/q ∼= A/qA for every n-primary ideal q of A, the n-primaryideals of A are in one-to-one inclusion preserving correspondence with then-primary ideals of A. It follows that an n-primary ideal I of A is areduction of a properly larger ideal of A if and only if IA is a reductionof a properly larger ideal of A. Therefore an n-primary ideal I of A isintegrally closed if and only if IA is integrally closed.

(6) If R is an integrally closed domain, then for every ideal I and element x ofR we have xI = xI. If (A,n) is analytically normal and also a UFD, thenevery height-one prime ideal of A extends to an integrally closed idealof A. In particular if A is a regular local domain, then PA is integrallyclosed for every height-one prime P of A. If (A,n) is a 2-dimensional localUFD, then every nonprincipal integrally closed ideal of A has the form xI,where I is an n-primary integrally closed ideal and x ∈ A. In particular,this is the case if (A,n) is a 2-dimensional regular local domain. In viewof item (5), it follows that every integrally closed ideal of A extends to anintegrally closed ideal of A in the case where A is a 2-dimensional regularlocal domain.

(7) If (A,n) is an excellent local ring, then the map A → A is quasi-normalby

G[30, (7.4.6) and (6.14.5)], and in this case every integrally closed ideal

of A extends to an integrally closed ideal of A.(8) Let (A,n) be a Noetherian local domain and let Ah denote the Henseliza-

tion of A. Every integrally closed ideal of A extends to an integrally closedideal of Ah. This follows because Ah is a filtered direct limit of etale A-algebras

L[62, (iii), (i), (vii) and (ix), pp. 800- 801]. Since the map from

A to its completion A factors through Ah, every integrally closed ideal ofA extends to an integrally closed ideal of A if and only every integrallyclosed ideal of Ah extends to an integrally closed ideal of A.


14.2. Extension to the completion18.3

We use results from Chaptersnoehomsec8,

flatholds9 and

insidecon13 In the construction of the following

example.

18.3.1 Construction of Example 14.7. Let k be a field of characteristic zero andlet x, y and z be indeterminates over k. Let R := k[x, y, z](x,y,z) and let R∗ be the(x)-adic completion of R. Thus R∗ = k[y, z](y,z)[[x]], the formal power series ringin x over the 2-dimensional regular local ring k[y, z](y,z).

Let α and β be elements of xk[[x]] that are algebraically independent over k(x).Set

f = (y − α)2, g = (z − β)2, and A = k(x, y, z, f, g) ∩ R∗.

Let T := R[α, β] and S := R[f, g]. Notice that S and T are both polynomialrings in 2 variables over R and that S is a subring of T . Indeed, with u := y − αand v := z − β, we have T = R[u, v] and S = R[u2, v2]. Let

Aout := R∗ ∩ k(x, y, z, α, β).

By Corollary11.4.119.4, the map T → R∗[1/x] is flat and Aout is a 3-dimensional regular

local domain that is a directed union of 5-dimensional regular local domains. AlsoAout has (x)-adic completion R∗

By Theorem16.3.213.2, the ring A is Noetherian and equal to its approximation

domain B provided that the map S → T [1/x] is flat. In our situation, the ring Tis a free S-module with 1, y − α, z − β, (y − α)(z − β) as a free module basis.Therefore the map S → T [1/x] is flat. The following commutative diagram whereall the labeled maps are the natural inclusions displays this situation:

(14.1)

B = A = R∗ ∩ Q(S)γ1−−−−→ Aout = R∗ ∩Q(T )

γ2−−−−→ R∗ = A∗

δ1

x δ2

x ψ

xS = R[, f, g]

ϕ−−−−→ T = R[α, β] T

In order to better understand the structure of A, we recall some of the detailsof the approximation domain B associated to A.

18.3.2 Approximation Technique 14.8. With k, x, y, z, f, g, R and R∗ as in Con-struction

18.3.128.7, we have

f = y2 +∞∑j=1

bjxj , g = z2 +

∞∑j=1

cjxj ,

where bj,∈ k[y] and cj ∈ k[z]. The rth endpieces for f and g are the sequencesfr∞r=1, gr∞r=1 of elements in R∗ defined for each r ≥ 1 by:

fr :=∞∑j=r

bjxj

xrand gr :=

∞∑j=r

cjxj

xr.

For each integer r ≥ 1, the ring Br is the polynomial ring k[x, y, z, fr, gr]localized at the maximal ideal generated by (x, y, z, fr − br, gr − cr), and the ap-proximation domain B =

⋃∞r=1Br.

14.2. EXTENSION TO THE COMPLETION 97

18.3.3 Theorem 14.9. Let A be the ring constructed in (18.3.128.7) and let P = (f, g)A,

where f and g are as in (18.3.128.7). Then

(1) The ring A = B is a 3-dimensional regular local domain that is a nestedunion of 5-dimensional regular local domains.

(2) The ideal P is a height-two prime ideal of A.(3) The (xA)-adic completion A∗ of A is R∗ = k[y, z](y,z)[[x] and PA∗ is not

integrally closed.(4) The completion A of A is R = k[[x, y, z]] and PA is not integrally closed.

Proof. Notice that the polynomial ring T = R[α, β] = R[y − α, z − β] is afree module of rank 4 over the polynomial subring R[f, g] since f = (y − α)2 andg = (z − β)2. Hence the extension

R[f, g] → R[α, β][1/x]

is flat. Thus item (1) follows from Theorem16.3.213.2.

For item (2), it suffices to observe that P has height two and that, for eachpositive integer r, Pr := (f, g)Br is a prime ideal of Br. We have f = xf1 + y2 andg = xg1 + z2. It is clear that (f, g)k[x, y, z, f, g] is a height-two prime ideal. SinceB1 is a localized polynomial ring over k in the variables x, y, z, f1 − b1, g1 − c1, wesee that

P1B1[1/x] = (xf1 + y2, xg1 + z2)B1[1/x]

is a height-two prime ideal of B1[1/x]. Indeed, setting f = g = 0 is equivalent to set-ting f1 = −y2/x and g1 = −z2/x. Therefore the residue class ring (B1/P1)[1/x] isisomorphic to a localization of the integral domain k[x, y, z][1/x]. Since B1 is Cohen-Macaulay and f, g form a regular sequence, and since (x, f, g)B1 = (x, y2, z2)B1 isan ideal of height three, we see that x is in no associated prime of (f, g)B1 (see, forexample

M[68, Theorem 17.6]). Therefore P1 = (f, g)B1 is a height-two prime ideal.

For r > 1, there exist elements ur ∈ k[x, y] and vr ∈ k[x, z] such that f =xrfr + urx + y2 and g = xrgr + vrx + z2. An argument similar to that givenabove shows that Pr = (f, g)Br is a height-two prime of Br. Therefore (f, g)B is aheight-two prime of B = A.

For items 3 and 4, R∗ = B∗ = A∗ by Construction18.3.128.7 and it follows that A =

k[[x, y, z]]. To see that PA∗ = (f, g)A∗ and PA = (f, g)A are not integrally closed,observe that ξ := (y − α)(z − β) is integral over PA∗ and PA since ξ2 = fg ∈ P 2.On the other hand, y − α and z − β are nonassociate prime elements in the localunique factorization domains A∗ and A. An easy computation shows that ξ 6∈ PA.Since PA∗ ⊆ PA, this completes the proof.

In Example18.3.428.10, we observe that this same technique may be used to construct

a d-dimensional regular local domain (A,n) having a prime ideal P of height h withthe property that the extension PA is not integrally closed for each integer d ≥ 3and each integer h with 2 ≤ h ≤ d− 1

18.3.4 Example 14.10. Let k be a field of characteristic zero and for an integer n ≥ 2let x, y1, . . . , yn be indeterminates over k. Let R denote the d := n+ 1 dimensionalregular local ring obtained by localizing the polynomial ring k[x, y1, . . . , yn] at themaximal ideal generated by (x, y1, . . . , yn). Let h be an integer with 2 ≤ h ≤ nand let τ1, . . . , τh ∈ xk[[x]] be algebraically independent over k(x). For each i with


1 ≤ i ≤ h, define fi = (yi − τi)h, and set ui = yi − τi. Consider the rings

S := R[f1, . . . , fh] and T := R[τ1, . . . , τh] = R[u1, . . . , uh].

Notice that S and T are polynomial rings in h variables over R and that T is afinite free integral extension of S. Let R∗ denote the (x)-adic completion of R, anddefine

A := R∗ ∩ Q(S) and Aout := R∗ ∩ Q(T ).The following commutative diagram where the labeled maps are the natural

inclusions displays this situation:

(14.2)

B = A = R∗ ∩ Q(S)γ1−−−−→ Aout = R∗ ∩Q(T )

γ2−−−−→ R∗ = A∗

δ1

x δ2

x ψ

xS = R[, f1, . . . , fh]

ϕ−−−−→ T = R[τ1, . . . , τh] T

Since the map S → T is flat, Theorem16.3.213.2 implies that A is a d-dimensional

regular local ring and is equal to its approximation domain B. Let P := (f1, . . . , fh)A.An argument similar to that given in Theorem

18.3.328.9 shows that P is a prime ideal of

A of height h. We have yi−τi ∈ A∗. Let ξ =∏hi=1(yi−τi). Then ξh = f1 · · · fh ∈ P h

implies ξ is integral over PA∗. Since y1−τ1, . . . , yh−τh is a regular sequence in A∗,we see that ξ 6∈ PA∗. Therefore the extended ideals PA∗ and PA are not integrallyclosed

14.3. Comments and Questions

In connection with Theorem18.3.328.9 it is natural to ask the following question.

18.4.1 Question 14.11. For P and A as in Theorem18.3.328.9, is P the only prime of A

that does not extend to an integrally closed ideal of A?

Com Comments 14.12. In relation to Example18.3.128.7 and to Question

18.4.128.11, con-

sider the following commutative diagram, where the labeled maps are the naturalinclusions:

(14.3)

B = A = R∗ ∩ Q(S)γ1−−−−→ Aout = R∗ ∩Q(T )

γ2−−−−→ R∗ = A∗

δ1

x δ2

x ψ

xS = R[, f, g]

ϕ−−−−→ T = R[α, β] T

Let γ = γ2 · γ1. Referring to the diagram above, we observe the following:(1) Theorem

16.3.213.2 implies that A[1/x] is a localization of S. Furthermore, by

Proposition11.4.19.3 of Chapter

noehomsec8, Aout is excellent. Notice, however, that A

is not excellent since there exists a prime ideal P of A such that PA isnot integrally closed. The excellence of Aout implies that if Q∗ ∈ SpecA∗

and x /∈ Q∗, then the map ψQ∗ : T → A∗Q∗ is regular

G[30, (7.8.3 v)].

(2) Let Q∗ ∈ SpecA∗ be such that x /∈ Q∗ and let q′ = Q∗ ∩ T . ByM[68,

Theorem 32.1] and Item 1 above, if ϕq′ : S → Tq′ is regular, then γQ∗ :A→ A∗

Q∗ is regular.

14.3. COMMENTS AND QUESTIONS 99

(3) Let I be an ideal of A. Since A′ and A∗ are excellent and both havecompletion A, Remark

18.2.128.6.3 shows that the ideals IA′, IA∗ and IA are

either all integrally closed or all fail to be integrally closed.(4) The Jacobian ideal of ϕ : S := k[x, y, z, f, g] → T := k[x, y, z, α, β] is the

ideal of T generated by the determinant of the matrix

J :=

( ∂f∂α

∂g∂α

∂f∂β

∂g∂β

).

Since the characteristic of the field k is zero, this ideal is (y−α)(z− β)T .

In Proposition18.4.328.13, we relate the behavior of integrally closed ideals in the

extension ϕ : S → T to the behavior of integrally closed ideals in the extensionγ : A→ A∗.

18.4.3 Proposition 14.13. With the setting of Theorem18.3.328.9 and Comment

Com28.12.2,

let I be an integrally closed ideal of A such that x 6∈ Q for each Q ∈ Ass(A/I). LetJ = I ∩ S. If JT is integrally closed (resp. a radical ideal) then IA∗ is integrallyclosed (resp. a radical ideal).

Proof. Since the map A → A∗ is flat, x is not in any associated prime ofIA∗. Therefore IA∗ is contracted from A∗[1/x] and it suffices to show IA∗[1/x] isintegrally closed (resp. a radical ideal). Our hypothesis implies I = IA[1/x] ∩ A.By Comment

Com28.12.1, A[1/x] is a localization of S. Thus every ideal of A[1/x] is

the extension of its contraction to S. It follows that IA[1/x] = JA[1/x]. ThusIA∗[1/x] = JA∗[1/x].

Also by CommentCom28.12.1, the map T → A∗[1/x] is regular. If JT is integrally

closed, then Remark18.2.128.6.7 implies that JA∗[1/x] is integrally closed. If JT is a

radical ideal, then the regularity of the map T → A∗[1/x] implies the JA∗[1/x] isa radical ideal.

Proposition 14.14. With the setting of Theorem18.3.328.9 and Comment

Com28.12,

let Q ∈ SpecA be such that QA (or equivalently QA∗) is not integrally closed. Then

(1) Q has height two and x 6∈ Q.(2) There exists a minimal prime Q∗ of QA∗ such that with q′ = Q∗ ∩ T , the

map ϕq′ : S → Tq′ is not regular.(3) Q contains f = (y − α)2 or g = (z − β)2.(4) Q contains no element that is a regular parameter of A.

Proof. By Remark18.2.128.6.6, the height of Q is two. Since A∗/xA∗ = A/xA =

R/xR, we see that x 6∈ Q. This proves item 1.By Remark

18.2.128.6.7, there exists a minimal prime Q∗ of QA∗ such that γQ∗ :

A→ A∗Q∗ is not regular. Thus item 2 follows from Comment

Com28.12.2.

For item 3, let Q∗ and q′ be as in item 2. Since γQ∗ is not regular it is notessentially smooth

G[30, 6.8.1]. By

insideps[45, (2.7)], (y−α)(z−β) ∈ q′. Hence f = (y−α)2

or g = (z − β)2 is in q′ and thus in Q. This proves item 3.Suppose w ∈ Q is a regular parameter for A. Then A/wA and A∗/wA∗ are

two-dimensional regular local domains. By Remark18.2.128.6.6, QA∗/wA∗ is integrally

closed, but this implies that QA∗ is integrally closed, which contradicts our hy-pothesis that QA∗ is not integrally closed. This proves item 4.


Question 14.15. In the setting of Theorem18.3.328.9 and Comment

Com28.12, let Q ∈

SpecA with x /∈ Q and let q = Q ∩ S. If QA∗ is integrally closed, does it followthat qT is integrally closed?


Com28.12, if a

prime ideal Q of A contains f or g, but not both, and does not contain a regularparameter of A, does it follow that QA∗ is integrally closed ?

In Example18.3.128.7, the three-dimensional regular local domain A contains height-

one prime ideals P such that A/P A is not reduced. This motivates us to ask:

Question 14.17. Let (A,n) be a three-dimensional regular local domain andlet A denote the n-adic completion of A. If for each height-one prime P of A, theextension PA is a radical ideal, i.e., the ring A/P A is reduced, does it follow thatPA is integrally closed for each P ∈ SpecA?

Remark 14.18. A problem analogous to that considered here in the sense thatit also deals with the behavior of ideals under extension to completion is addressedby Loepp and Rotthaus in

LR[64]. They construct nonexcellent Noetherian local

domains to demonstrate that tight closure need not commute with completion.

CHAPTER 15

Catenary local rings with geometrically normal

formal fibers April 21, 2010

catsecIn this chapter1 we consider the catenary property in a Noetherian local ring

(R,m) having geometrically normal formal fibers. We prove that the HenselizationRh of R is universally catenary, and relate the catenary and universally catenaryproperties of R to the fibers of the map R → Rh. We present for each integer n ≥ 2an example of a catenary Noetherian local integral domain of dimension n that hasgeometrically regular formal fibers and is not universally catenary.

15.1. Introduction15.1

Recall that a ring R is catenary if, for every pair of comparable prime idealsP ⊂ Q of R, every saturated chain of prime ideals from P to Q has the same length,cf. Definitions

3.203.9. The ring R is universally catenary if every finitely generated

R-algebra is catenary. A Noetherian local ring (R,m) with m-adic completion Rhas geometrically normal (respectively geometrically regular) formal fibers if for eachprime P ofR and for each finite algebraic extension k′ of the field k(P ) := RP /PRP ,the ring R ⊗R k(P )⊗k(P ) k

′ is normal (respectively regular), cf. Definition3.3893.14.

In this chapter we investigate the catenary property in Noetherian local ringshaving geometrically normal formal fibers. In Example

15.4.215.11 we apply a technique

developed in previous chapters to construct, for each integer n ≥ 2, an exampleof a catenary Noetherian local integral domain of dimension n with geometricallyregular formal fibers that is not universally catenary.

Let (R,m) be a Noetherian local ring. We denote the Henselization of R byRh. Recall that (R,m) is formally equidimensional, or in other terminology quasi-unmixed, provided all the minimal primes of the m-adic completion R have thesame dimension. A theorem of Ratliff ralating the universal catenary property toproperties of the completion, cf. Theorem


ptools23, is crucial for our work.

In Section15.215.2 we present results concerning conditions for a Noetherian local

ring (R,m) to be universally catenary. The theorem of Ratliff mentioned aboveleads to our observation in Proposition

15.2.215.1 that a Henselian Noetherian local ring

having geometrically normal formal fibers is universally catenary.Assume that (R,m) is a Noetherian local integral domain having geometrically

normal formal fibers. Corollary15.2.315.4 implies that the Henselization Rh of R is

universally catenary; moreover, if the integral closure R of R is local, then R is

1Material in this chapter comes from a paper we wrote that is included in a volume dedicatedto Shreeram S. Abhyankar in celebration of his seventieth birthday. In his mathematical workRam has opened up many avenues. In this chapter we are pursuing one of these related to powerseries and completions.

101

102 15. CATENARY RINGS AND NORMAL FIBERS

universally catenary. Theorem15.2.615.6 asserts that R is universally catenary if and

only if the set Γ is empty, where

Γ := P ∈ Spec(Rh) | dim(Rh/P ) < dim(R/P ∩R)).We also prove that the subset Γ of SpecRh is stable under generalization in thesense that if Q ∈ Γ and P ∈ SpecRh is such that P ⊆ Q, then P ∈ Γ. Thus Γsatisfies a strong “going down” property.

In Theorem15.2.715.7 we prove for R as above that R is catenary but is not uni-

versally catenary if and only if the set Γ is nonempty and each prime ideal P in Γhas dimension one. It follows in this case that Γ is a finite subset of the minimalprimes of Rh. Thus, as we observe in Corollary

15.2.815.8, if R is catenary but not uni-

versally catenary, this is signaled by the existence of dimension one minimal primesof the m-adic completion R of R. If R is catenary, each minimal prime of R havingdimension different from the dimension of R must have dimension one.

In Section15.315.3 we present examples to illustrate the results of Section

15.215.2. We

apply a construction involving power series, homomorphic images and intersectionsdeveloped in previous chapters, cf. Chapter

noehomsec8.

The construction begins with a Noetherian integral domain that may be takento be a polynomial ring in several indeterminates over a field. In Theorem

15.3.615.9 we

extend this construction by proving that in certain circumstances it is possible totransfer the flatness, Noetherian and computability properties of integral domainsassociated with ideals I1, . . . , In of R to the integral domain associated with theirintersection I = I1 ∩ · · · ∩ In.

We apply these concepts in Examples15.4.115.10 -

15.4.315.12 to obtain Noetherian local

domains that are not universally catenary. In Remark15.4.415.13, we specifiy precisely

which of these rings are catenary. These domains illustrate the results of Section15.215.2, because in Section

15.515.5 we prove that they have geometrically regular formal

fibers.We would like to thank M. Brodmann and R. Sharp for raising a question on

catenary/universally catenary rings that motivated our work in this chapter.

15.2. Geometrically normal formal fibers15.2

Throughout this section (R,m) is a Noetherian local ring. We use Theo-rem

15.2.13.11, the result of Ratliff mentioned above, that relates the universally catenary

property to properties of the completion in order to prove:

15.2.2 Proposition 15.1. Let (R,m) be a Henselian Noetherian local ring havinggeometrically normal formal fibers.

(1) For each prime ideal P of R, the extension PR to the m-adic completionof R is also a prime ideal.

(2) The ring R is universally catenary.

Proof. Item (2) follows from item (1) and Theorem15.2.13.11. In order to prove

item (1), observe that the completion of R/P is R/P R, and R/P is a NoetherianHenselian local integral domain having geometrically normal formal fibers. Hence,by passing from R to R/P , to prove item (1), it suffices to prove that if R isa Henselian Noetherian local integral domain having geometrically normal formalfibers, then the completion R of R is an integral domain.

15.2. GEOMETRICALLY NORMAL FORMAL FIBERS 103

Since R has normal formal fibers, the completion R of R is reduced. Hencethe integral closure R of R is a finitely generated R-module

N2[72, (32.2)]. Moreover,

since R is Henselian, R is localN2[72, (43.12)].

The completion R of R is R ⊗R RN2[72, (17.8)]. Since the formal fibers of R

are geometrically normal, the formal fibers of R are also geometrically normal. Itfollows that R is normal

M[68, Corollary, page 184], and hence an integral domain

because R is local. Since R is a flat R-module, R is a subring of R. Therefore R isan integral domain.

15.2.5 Remark 15.2. Let (R,m) be a Noetherian local domain. An interesting resultproved by Nagata

N2[72, (43.20)] establishes the existence of a one-to-one correspon-

dence between the minimal primes of the Henselization Rh of R and the maximalideals of the integral closure R of R. Moreover, if a maximal ideal m of R corre-sponds to a minimal prime q of Rh, then the integral closure of the Henselian localdomain Rh/q is the Henselization of Rm

N2[72, Ex. 2, page 188],

N3[71]. Therefore

ht(m) = dim(Rh/q).

15.2.55 Remark 15.3. Let (R,m) be a Noetherian local ring, let R denote the m-adiccompletion of R, and let Rh denote the Henselization of R. It is well known thatthe canonical map R → Rh is a regular map with zero-dimensional fibers, and thatR is also the completion of Rh with respect to its unique maximal ideal mh = mRh.The following statements are equivalent:

(1) The map R → R has (geometrically) normal fibers.(2) The map Rh → R has (geometrically) normal fibers.

Let P be a prime ideal of R and let U = R \ P . Then PRh = P1 ∩ · · · ∩ Pn,where the Pi are the minimal prime ideals of PRh. Then PR = (∩ni=1Pi)R. SinceR is faithfully flat over Rh, finite intersections distribute over this extension, andPR = ∩ni=1(PiR). Let S = U−1(R/P R) denote the fiber over P in R and letqi = PiS. The ideals q1, . . . , qn of S intersect in (0) and are pairwise comaximalbecause for i 6= j, (Pi+Pj)∩U 6= ∅. Therefore S ∼=∏n

i=1(S/qi). Since a Noetherianring is normal if and only if it is a finite product of normal Noetherian domains,the fiber over P in R is normal if and only if the the fiber over each of the Pi in Ris normal.

15.2.3 Corollary 15.4. Let R be a Noetherian local domain having geometricallynormal formal fibers. Then

(1) The Henselization Rh of R is universally catenary.(2) If the integral closure R of R is again local, then R is universally catenary.

In particular, if R is a normal Noetherian local domain having geometrically normalformal fibers, then R is universally catenary.

Proof. For item (1), the Henselization Rh of R is a Noetherian local ringhaving geometrically normal formal fibers, so Proposition

15.2.215.1 implies that Rh is

universally catenary. For item (2), if the integral closure of R is local, then, byRemark

15.2.515.2, the Henselization Rh has a unique minimal prime. Since Rh is uni-

versally catenary, the completion R is equidimensional, and hence R is universallycatenary.


Theorem15.2.415.5 relates the catenary property of R to the height of maximal ideals

in the integral closure of R.

15.2.4 Theorem 15.5. Let (R,m) be a Noetherian local domain of dimension d andlet R be the integral closure of R. If R contains a maximal ideal m with ht(m) =r 6∈ 1, d, then there exists a saturated chain of prime ideals in R of length ≤ r.Hence R is not catenary.

Proof. Since R has only finitely many maximal idealsN2[72, (33.10)], there

exists b ∈ m such that b is in no other maximal ideal of R. Let R′ = R[b] and letm′ = m∩R′. Notice that m is the unique prime ideal of R that contains m′. By theGoing Up Theorem

M[68, (9.3)], htm′ = r. Since R′ is a finitely generated R-module

and is birational over R, there exists a nonzero element a ∈m such that aR′ ⊆ R.It follows that R[1/a] = R′[1/a]. The maximal ideals of R[1/a] have the formPR[1/a], where P ∈ Spec(R) is maximal with respect to not containing a. Sincethere are no prime ideals strictly between P and m

M[68, (13.5)], if htP = h, then

there exists in R a saturated chain of prime ideals through P of length h+1. Thusto show R is not catenary, it suffices to establish the existence of a maximal idealof R[1/a] having height different from d− 1. Since R[1/a] = R′[1/a], the maximalideals of R[1/a] correspond to the prime ideals P ′ in R′ maximal with respect tonot containing a. Since htm′ > 1, there exists c ∈ m′ such that c is not in anyminimal prime of aR′ nor in any maximal ideal of R′ other than m′. Hence thereexist prime ideals of R′ containing c and not containing a. Let P ′ ∈ Spec(R′) bemaximal with respect to c ∈ P ′ and a 6∈ P ′. Then P ′ ⊂m′, so htP ′ ≤ r−1 < d−1.It follows that there exists a saturated chain of prime ideals of R of length ≤ r, andhence R is not catenary.

15.2.6 Theorem 15.6. Let (R,m) be a Noetherian local integral domain having geo-metrically normal formal fibers and let Rh denote the Henselization of R. Considerthe set

Γ := P ∈ Spec(Rh) | dim(Rh/P ) < dim(R/(P ∩R)).The following statements then hold.

(1) For p ∈ Spec(R), the ring R/p is not universally catenary if and only ifthere exists P ∈ Γ such that p = P ∩R.

(2) The set Γ is empty if and only if R is universally catenary.(3) If p ⊂ q are prime ideals in R and if there exists Q ∈ Γ with Q ∩R = q,

then there also exists P ∈ Γ with P ∩R = p.(4) If Q ∈ Γ, then each prime ideal P of Rh such that P ⊆ Q is also in Γ,

that is, the subset Γ of SpecRh is stable under generalization.

Proof. The map of R/p to its m-adic completion R/pR factors throughRh/pRh. Since R → R has geometrically normal fibers, so does the map Rh → R.Proposition

15.2.215.1 implies that each prime ideal P of Rh extends to a prime ideal

PR. Therefore, by Theorem15.2.13.11, the ring R/p is universally catenary if and only if

Rh/pRh is equidimensional if and only if there does not exist P ∈ Γ with P∩R = p.This proves items (1) and (2). To prove item (3), observe that if R/p is universallycatenary, then R/q is also universally catenary

M[68, Theorem 31.6].

It remains to prove item (4). Let P ∈ SpecRh be such that P ⊂ Q, andlet ht(Q/P ) = n. Since the fibers of the map R → Rh are zero-dimensional, the

15.2. GEOMETRICALLY NORMAL FORMAL FIBERS 105

contraction to R of an ascending chain of primes

P = P0 ( P1 ( · · · ( Pn = Q

of Rh is a strictly ascending chain of primes from p := P ∩R to q := Q∩R. Henceht(q/p) ≥ n. Since Rh is catenary, we have

dim(Rh/P ) = n+ dim(Rh/Q) < n+ dim(R/q) ≤ dim(R/p),

where the strict inequality is because Q ∈ Γ. Therefore P ∈ Γ.

15.2.7 Theorem 15.7. Let (R,m) be a Noetherian local integral domain having geo-metrically normal formal fibers and let Γ be defined as in Theorem

15.2.615.6. The ring

R is catenary but not universally catenary if and only if

(i) the set Γ is nonempty, and(ii) dimRh/P = 1, for each prime ideal P ∈ Γ.

If these conditions holds, then each P ∈ Γ is a minimal prime of Rh, and Γ is afinite nonempty open subset of SpecRh.

Proof. Assume that R is catenary but not universally catenary. By Theo-rem

15.2.615.6, the set Γ is nonempty and there exist minimal primes P of Rh such that

dim(Rh/P ) < dim(Rh). By Remark15.2.515.2, if a maximal ideal m of R corresponds

to a minimal prime P of Rh, then ht(m) = dim(Rh/P ). Since R is catenary, The-orem

15.2.415.5 implies that the height of each maximal ideal of the integral closure R

of R is either one or dim(R). Therefore dim(Rh/P ) = 1 for each minimal primeP of Rh for which dim(Rh/P ) 6= dim(Rh). Item (4) of Theorem

15.2.615.6 implies each

P ∈ Γ is a minimal prime of Rh and dimRh/P = 1.For the converse, assume that Γ is nonempty and each prime P ∈ Γ has di-

mension one. Then R is not universally catenary by item (2) of Theorem15.2.615.6 and

by item (4) of Theorem15.2.615.6, each prime ideal in Γ is a minimal prime of Rh and

therefore lies over (0) in R. To show R is catenary, it suffices to show for eachnonzero nonmaximal prime ideal p of R that ht(p) + dim(R/p) = dim(R)

M[68,

Theorem 31.4]. Let P ∈ Spec(Rh) be a minimal prime of pRh. Since Rh is flatover R with zero-dimensional fibers, ht(p) = ht(P ). Let Q be a minimal primeof Rh with Q ⊆ P . Then Q 6∈ Γ. For by assumption every prime of Γ has di-mension one, so if Q were in Γ, then Q = P . But P ∩ R = p, which is nonzero,and Q ∩ R = (0). Therefore Q 6∈ Γ and hence dim(Rh/Q) = dim(Rh). SinceRh is catenary, it follows that ht(P ) + dim(Rh/P ) = dim(Rh). Since P 6∈ Γ, wehave dim(R/p) = dim(Rh/P ). Therefore ht(p) + dim(R/p) = dim(R) and R iscatenary.

15.2.8 Corollary 15.8. If R has geometrically normal formal fibers and is catenarybut not universally catenary, then there exist in the m-adic completion R of Rminimal prime ideals q such that dim(R/q) = 1.

Proof. By Theorem15.2.715.7, each prime ideal Q ∈ Γ has dimension one and

is a minimal prime of Rh. Moreover, QR := q is a minimal prime of R. Sincedim(Rh/Q) = 1, we have dim(R/q) = 1.


15.3. A method for constructing examples15.3

In Theorem15.3.615.9 we show in certain circumstances that the flatness, Noether-

ian and computability properties associated with ideals I1, . . . , In of the (z)-adiccompletion R∗ of R as described in Theorem

11.3.28.3 transfer to the integral domain

associated with their intersection. In Section15.515.5, we show that the property of

regularity of formal fibers also transfers in certain cases to the integral domain as-sociated with their intersection. Theorem

15.3.615.9 is a generalization of Theorem

11.4.49.7.

15.3.6 Theorem 15.9. Assume that R is a Noetherian integral domain with field offractions K, z ∈ R is a nonzero nonunit, R∗ is the (z)-adic completion of R, andI1, . . . , In are ideals of R∗ such that, for each i ∈ 1, . . . , n, each associated primeof R∗/Ii intersects R in (0). Also assume the map R → (R∗/Ii)[1/z] is flat foreach i and that the localizations at z of the Ii are pairwise comaximal; that is, forall i 6= j, (Ii + Ij)R∗[1/z] = R∗[1/z]. Let I := I1 ∩ · · · ∩ In, A := K ∩ (R∗/I) and,for i ∈ 1, 2, . . . n, let Ai := K ∩ (R∗/Ii). Then

(1) Each associated prime of R∗/I intersects R in (0).(2) The map R → (R∗/I))[1/z] is flat, so the ring A is Noetherian and is

equal to its associated approximation ring B. The (z)-adic completionA∗ of A is R∗/I, and the (z)-adic completion A∗

i of Ai is R∗/Ii, fori ∈ 1, . . . , n.

(3) The ring A∗[1/z] ∼= (A∗1)[1/z] ⊕ · · · ⊕ (A∗

n)[1/z]. If Q ∈ Spec(A∗) andz 6∈ Q, then (A∗)Q is a localization of one of the A∗

i .(4) We have A ⊆ A1 ∩ · · · ∩An and ∩ni=1(Ai)[1/z] ⊆ AP for each P ∈ SpecA

with z /∈ P . Thus we have A[1/z] = ∩ni=1(Ai)[1/z].

Proof. By Theorem11.3.28.3, the (z)-adic completion A∗

i of Ai is R∗/Ii. SinceAss(R∗/I) ⊆ ⋃ni=1 Ass(R∗/Ii), the condition on assassinators of Theorem

11.3.28.3 holds

for the ideal I. The natural R-algebra homomorphism π : R∗ →⊕ni=1(R

∗/Ii) haskernel I. Further, the localization of π at z is onto because (Ii + Ij)R∗[1/z] =R∗[1/z] for all i 6= j. Thus (R∗/I)[1/z] ∼= ⊕n

i=1(R∗/Ii)[1/z] =

⊕ni=1(A

∗i )[1/z] is

flat over R. Therefore A is Noetherian and is equal to its associated approximationring B, and A∗ = R∗/I is the (z)-adic completion of A.

If Q ∈ Spec(A∗) and z /∈ Q, then A∗Q is a localization of

A∗[1/z] ∼= (A∗1)[1/z]⊕ · · · ⊕ (A∗

n)[1/z].

Every prime ideal of⊕n

i=1(A∗i )[1/z] has the form (Qi)A∗

i [1/z] ⊕⊕

j 6=i(A∗j )[1/z],

where Qi ∈ Spec(A∗i ). It follows that A∗

Q is a localization of A∗i for some i ∈

1, . . . , n.Since R∗/Ii is a homomorphic image of R∗/I, the intersection ring A ⊆ Ai for

each i. Let P ∈ SpecA with z /∈ P . Since A∗ = R∗/I is faithfully flat over A, thereexists P ∗ ∈ Spec(A∗) with P ∗∩A = P . Then z /∈ P ∗ implies A∗

P∗ = (A∗i )P∗

i, where

P ∗i ∈ Spec(A∗

i ) for some i ∈ 1, . . . , n. Let Pi = P ∗i ∩ Ai. Since AP → A∗

P∗ and(Ai)Pi → (A∗

i )P∗i

are faithfully flat, we have

AP = A∗P∗ ∩K = (A∗

i )P∗i∩K = (Ai)Pi ⊇ (Ai)[1/z].

It follows that⋂ni=1(Ai)[1/z] ⊆ AP . Thus we have

n⋂i=1

(Ai)[1/z] ⊆⋂AP | P ∈ SpecA and z /∈ P = A[1/z].

15.4. EXAMPLES THAT ARE NOT UNIVERSALLY CATENARY 107

Since A[1/z] ⊆ (Ai)[1/z], for each i, we have A[1/z] =⋂ni=1(Ai)[1/z].

15.4. Examples that are not universally catenary15.4

Example11.4.59.8 is an example of a two-dimensional Noetherian local domain A

having geometrically regular formal fibers such that the completion A of A has twominimal primes one of dimension one and one of dimension two. Thus A is notuniversally catenary. We generalize this example in the following.

We construct in Example15.4.115.10 a two-dimensional Noetherian local domain

having geometrically normal formal fibers such that the completion has any desiredfinite number of minimal primes of dimensions one and two.

15.4.1 Example 15.10. Let r and s be positive integers and let R be the localizedpolynomial ring in three variables R := k[x, y, z](x,y,z), where k is a field of char-acteristic zero and the field of fractions of R is K := k(x, y, z). Then the (x)-adiccompletion of R is R∗ := k[y, z](y,z)[[x]]. Let τ1, . . . , τr,β1, β2, . . . , βs, γ ∈ xk[[x]] bealgebraically independent power series over k(x). Define

Qi := (z−τi, y−γ)R∗, for i ∈ 1, . . . , r and Pj := (z−βj)R∗, for j ∈ 1, . . . , s.We apply Theorem

15.3.615.9 with Ii = Qi for 1 ≤ i ≤ r, and Ir+j = Pj for 1 ≤ j ≤ s.

Then the Ii satisfy the comaximality condition at the localization at x. Let I :=I1 ∩ · · · ∩ Ir+s and let A := K ∩ (R∗/I). For J an ideal of R∗ containing I, let Jdenote the image of J in R∗/I. Then, for each i with 1 ≤ i ≤ r, dim( (R∗/I)/Qi) =dim(R∗/Qi) = 1 and, for each j with 1 ≤ j ≤ s, dim( (R∗/I)/Pj) = 2. Thus A∗

contains r minimal primes of dimension one and s minimal primes of dimension two.Since A∗ modulo each of its minimal primes is a regular local ring, the completionA of A also has precisely r minimal primes of dimension one and s minimal primesof dimension two.

The integral domain A birationally dominates R and is birationally dominatedby each of the Ai. Corollary

15.5.315.19 implies that A has geometrically regular formal

fibers. Since dim(A) = 2, A is catenary.

We show in Example15.4.215.11 that for every integer n ≥ 2 there is a Noetherian

local domain (A,m) of dimension n that has geometrically regular formal fibersand is catenary but not universally catenary.

15.4.2 Example 15.11. Let R = k[x, y1, . . . , yn](x,y1,...,yn) be a localized polynomialring of dimension n+ 1 where k is a field of characteristic zero. Let σ, τ1, . . . , τn ∈xk[[x]] be n+ 1 algebraically independent elements over k(x) and consider in R∗ =k[y1, . . . , yn](y1,...,yn)[[x]] the ideals

I1 = (y1 − σ)R∗ and I2 = (y1 − τ1, . . . , yn − τn)R∗.

Then the ringA = k(x, y1, . . . , yn) ∩ (R∗/(I1 ∩ I2))

is the desired example. The completion A of A has two minimal primes I1A havingdimension n and I2A having dimension one. By Corollary

15.5.315.19, A has geomet-

rically regular formal fibers. Therefore the Henselization Ah has precisely twominimal prime ideals P,Q which may be labeled so that PA = I1A and QA = I2A.Thus dim(Ah/P ) = n and dim(Ah/Q) = 1. By Theorem

15.2.715.7, A is catenary but

not universally catenary.


In Example15.4.315.12 we construct for each positive integer t and specified non-

negative integers n1, . . . , nt with n1 ≥ 1, a t-dimensional Noetherian local domainA that has geometrically regular formal fibers and birationally dominates a t+ 1-dimensional regular local domain such that the completion A of A has, for each rwith 1 ≤ r ≤ t, exactly nr minimal primes prj of dimension t + 1 − r. Moreover,each A/prj is a regular local ring of dimension t+ 1− r. If ni > 0 for some i 6= 1,then A is not universally catenary and is not a homomorphic image of a regularlocal domain. It follows from Remark

15.2.515.2 that the derived normal ring A of A has

exactly nr maximal ideals of height t+ 1− r for each r with 1 ≤ r ≤ t.15.4.3 Example 15.12. Let t be a positive integer and let nr be a nonnegative integer

for each r with 1 ≤ r ≤ t. Assume that n1 ≥ 1. We construct a t-dimensionalNoetherian local domain A that has geometrically regular formal fibers such thatA has exactly nr minimal primes of dimension t+ 1− r for each r. Let x, y1 . . . , ytbe indeterminates over a field k of characteristic zero.

Let R = k[x, y1, . . . , yt](x,y1,...,yt), let R∗ = k[y1, . . . , yt] [[x]](x,y1,...,yt) denotethe (x)-adic completion of R and let K denote the field of fractions of R. For everyr, j, i ∈ N such that 1 ≤ r ≤ t, 1 ≤ j ≤ nr and 1 ≤ i ≤ r, choose elements τrji ofxk[[x]] so that the set

⋃τrji is algebraically independent over k(x).For each r, j with 1 ≤ r ≤ t and 1 ≤ j ≤ nr, define the prime ideal Prj :=

(y1−τrj1, . . . , yr−τrjr) of height r in R∗. Notice that R∗/Prj is a regular local ringof dimension t+ 1− r. Theorem

11.4.19.3 implies that the extension R → (R∗/Prj)[1/x]

is flat, and that the intersection domain Arj := K ∩ (R∗/Prj) is a regular local ringof dimension t+ 1− r that has (x)-adic completion R∗/Prj.

Let I :=⋂Prj be the intersection of all the prime ideals Prj . Since the τrji ∈

xk[[x]] are algebraically independent over k(x), the sum of any two of these idealsPrj and Pmi, where we may assume r ≤ m, has radical (x, y1, . . . , ym)R∗, andthus (Prj + Pmi)R∗[1/x] = R∗[1/x]. It follows that the representation of I as theintersection of the Prj is irredundant and Ass(R∗/I) = Prj | 1 ≤ r ≤ t, 1 ≤ j ≤nr. Since each Prj ∩ R = (0), we have R → R∗/I, and the intersection domainA := K ∩ (R∗/I) is well defined.

By Theorem15.3.615.9, the map R → (R∗/I)[1/x] is flat, A is Noetherian and A is a

localization of a subring of R[1/x]. Since R∗/Prj is a regular local ring of dimensiont+1− r, the minimal primes of A all have the form prj := PrjA and each A/prj isa regular local ring of dimension t+ 1 − r. The ring A birationally dominates the(t+ 1)-dimensional regular local domain R and all the stated properties hold oncewe prove Corollary

15.5.315.19.

15.4.4 Remark 15.13. By Theorem15.2.715.7, the ring A constructed in Example

15.4.315.12

is catenary if and only if each minimal prime of A has dimension either one or t.By taking nr = 0 for r 6∈ 1, t in Example

15.4.315.12, we obtain additional examples

of catenary Noetherian local domains A of dimension t having geometrically reg-ular formal fibers for which the completion A has precisely nt minimal primes ofdimension one and n1 minimal primes of dimension t.

15.4.45 Remark 15.14. Let (A,n) be a Noetherian local domain constructed as inExample

15.4.315.12 and let Ah denote the Henselization of A. Since each minimal

prime of A is the extension of a minimal prime of Ah and also the extension of aminimal prime of A∗, the minimal primes of Ah and A∗ are in a natural one-to-onecorrespondence. Let P be the minimal prime of Ah corresponding to a minimal

15.5. REGULARITY 109

prime p of A∗. Since the minimal primes of A∗ extend to pairwise comaximal primeideals of A∗[1/x], for each prime ideal Q ⊃ P of Ah with x 6∈ Q, the prime idealP is the unique minimal prime of Ah contained in Q. Let q := Q ∩ A. We havehtq = htQ, and either dim(A/q) > dim(Ah/Q) or else every saturated chain ofprime ideals of A containing q has length less than dimA.

In this connection, we ask:

15.4.46 Question 15.15. Let (A,n) be a Noetherian local domain constructed as inExample

15.4.315.12. If A is not catenary, does it follow that the set

Γ := P ∈ Spec(Ah) | dim(Ah/P ) < dim(A/(P ∩A))is infinite?

15.4.5 Remark 15.16. We thank L. Avramov for suggesting we consider the depthof the rings constructed in Example

15.4.315.12. The catenary rings that arise from this

construction all have depth one. However, Example15.4.315.12 can be used to construct,

for each integer t ≥ 3 and integer d with 2 ≤ d ≤ t−1, an example of a noncatenaryNoetherian local domain A of dimension t and depth d having geometrically regularformal fibers. The (x)-adic completion A∗ of A has precisely two minimal primes,one of dimension t and one of dimension d. To establish the existence of such anexample, with notation as in Example

15.4.315.12, we set m = t− d+ 1 and take nr = 0

for r 6∈ 1,m and n1 = nm = 1. Let

P1 := P11 = (y1 − τ111)R∗ and Pm := Pm1 = (y1 − τm11, . . . , ym − τm1m)R∗.

Consider A∗ = R∗/(P1 ∩ Pm) and the short exact sequence

0 −→ P1

P1 ∩ Pm −→R∗

P1 ∩ Pm −→R∗

P1−→ 0.

Since P1 is principal and not contained in Pm, we have P1 ∩ Pm = P1Pm andP1/(P1 ∩ Pm) ∼= R∗/Pm. It follows that depthA = depthA∗ = depth(R∗/Pm)= d ; cf.

Kap[58, page 103, ex 14] or

BH[14, Prop. 1.2.9, page 11]. By Remark

15.2.515.2, the

derived normal ring A of A has precisely two maximal ideals one of height t andone of height d.

15.5. Regular maps and geometrically regular formal fibers15.5

We show in Corollary15.5.315.19 that each of the rings A constructed in Examples

15.4.115.10,

15.4.215.11 and

15.4.315.12 has geometrically regular formal fibers.

15.5.1 Proposition 15.17. Let R, z, R∗, A and I be as in Theorem11.3.28.3. Assume that

for each P ∈ Spec(R∗/I) with z /∈ P , the map ψP : RP∩R −→ (R∗/I)P is regular.Then

(1) A is Noetherian and the map A −→ A∗ = R∗/I is regular.(2) If R is semilocal with geometrically regular formal fibers and z is in the

Jacobson radical of R, then A has geometrically regular formal fibers.

Proof. Since flatness is a local property (and regularity of a map includesflatness), the map ψz : R −→ (R∗/I)[1/z] is flat. By Theorem

11.3.28.3, the intersection

ring A is Noetherian with (z)-adic completion A∗ = R∗/I. Hence A −→ A∗ is flat.Let Q ∈ Spec(A), let k(Q) denote the field of fractions of A/Q and let Q0 =

Q ∩R.


Case 1: z ∈ Q. Then R/Q0 = A/Q = A∗/QA∗ and the ring A∗QA∗/QA∗

QA∗ =A∗ ⊗A k(Q) = AQ/QAQ is trivially geometrically regular over k(Q).

Case 2: z /∈ Q. Let k(Q) ⊆ L be a finite algebraic field extension. We showthe ring A∗⊗AL is regular. Let W ∈ Spec(A∗⊗AL) and let W ′ = P ∩(A∗⊗Ak(Q)).The prime W ′ corresponds to a prime ideal P ∈ Spec(A∗) with P ∩ A = Q. Byassumption the map

RQ0 −→ (R∗/I)P = A∗P

is regular. Since z /∈ Q it follows that RQ0 = UQ∩U = AQ and that k(Q0) = k(Q).Thus the ring A∗

P ⊗AQ L is regular. Therefore (A∗ ⊗A L)W , which is a localizationof this ring, is regular.

For part (2), since R has geometrically regular formal fibers, so has R∗ byR3[83].

Hence the map θ : A∗ = R∗/I −→ A = (R∗/I) is regular. ByM1[66, Thm. 32.1 (i)]

and part (1) above, it follows that A has geometrically regular formal fibers, thatis, the map A −→ A is regular.

15.5.2 Proposition 15.18. Assume that R, K, z and R∗ are as in Theorem11.3.28.3. For

a positive integer n, let I1, . . . , In be ideals of R∗ such that each associated primeof R∗/Ii intersects R in (0), for i = 1, . . . , n. Let I := I1 ∩ · · · ∩ In. Assume that

(1) R is semilocal with geometrically regular formal fibers and z is in theJacobson radical of R.

(2) Each (R∗/Ii)[1/z] is a flat R-module and, for each i 6= j, the idealsIi(R∗)[1/z] and IjR∗[1/z] are comaximal in (R∗)[1/z].

(3) For i = 1, . . . , n, Ai := K∩(R∗/Ii) has geometrically regular formal fibers.Then A := K ∩ (R∗/I) is equal to its approximation domain B, and has geometri-cally regular formal fibers.

Proof. Since R has geometrically regular formal fibers, by (15.5.115.17.2), it suffices

to show for W ∈ Spec(R∗/I) with z 6∈W that RW0 −→ (R∗/I)W is regular, whereW0 := W ∩R. As in Theorem

15.3.615.9, we have

(R∗/I)[1/z] = (R∗/I1)[1/z] ⊕ · · · ⊕ (R∗/In)[1/z].

It follows that (R∗/I)W is a localization of R∗/Ii for some i ∈ 1, . . . , n. If(R∗/I)W = (R∗/Ii)Wi , where Wi ∈ Spec(R∗/Ii), then RW0 = (Ai)Wi∩Ai and(Ai)Wi∩Ai −→ (R∗/Ii)Wi is regular. Thus RW0 −→ (R∗/I)W is regular.

15.5.3 Corollary 15.19. The rings A of Examples15.4.115.10,

15.4.215.11 and

15.4.315.12 have ge-

ometrically regular formal fibers, that is, the map φ : A→ A is regular.

Proof. By the definition of R and the observations given in (15.5.115.17), the hy-

potheses of (15.5.215.18) are satisfied.

Exercises(1) Let (R,m) be a catenary Noetherian local domain having geometrically normal

formal fibers. If R is not universally catenary, prove that R has depth one.Suggestion: Use Theorem

15.2.715.7 and

BH[14, Prop. 1.2.13].

CHAPTER 16

Non-finitely generated prime ideals in subrings of

power series rings, June 5, 2009(nonfgsec), 17nonfgsec

???alertObserve similarity with Chapterinsidepssec17 ?? Chapter 16. This Chapter may

have a little more and be more polished. ???

Given a power series ring R∗ over a Noetherian integral domain R and anintermediate field L between R and the total quotient ring of R∗, the integraldomain A = L ∩ R∗ often (but not always) inherits nice properties from R∗ suchas being Noetherian. For certain fields L it is possible to approximate A using alocalization B of a particular nested union of polynomial rings over R associated toA; if B is Noetherian then B = A. If B is not Noetherian, we can sometimes identifywhich prime ideals of B are not finitely generated. We have obtained in this way, foreach positive integer n, a three-dimensional quasilocal unique factorization domainB such that the maximal ideal of B is two-generated, B has precisely n prime idealsof height two, each prime ideal of B of height two is not finitely generated and allthe other prime ideals of B are finitely generated. We examine the structure of themap SpecA → SpecB for this example. We also present a generalization of thisexample to higher dimensions.

16.1. Introduction

In this section we analyze the prime ideal structure of particular non-Noetherianintegral domains arising from the general construction developed in several previouschapters.

Briefly, with this technique two types of integral domains are constructed: (1)the intersection of a power series ring with a field yields an integral domain A asin the abstract, and (2) an approximation of the domain A by a nested union B oflocalized polynomial rings has the second form described in the abstract. Classicalexamples such as those of Akizuki

A[7] and Nagata

N2[72, pages 209-211] are created

using the second (nested union) description of this construction. As we observe inChapter

nonnoesec26, it is also possible to realize these classical examples as the intersection

domains of the first description.In Chapter

nonnoesec26 we observe that in certain applications of this technique flatness

of a map of associated polynomial rings implies that the constructed domains areNoetherian and that A = B. In Chapters

nonnoesec26 and

insidepssec17, we apply this observation

to the construction of examples of both Noetherian and non-Noetherian integraldomains.

111

112 16. NON-FINITELY GENERATED PRIME IDEALS

We present in this chapter several examples constructed using this technique ofbuilding non-Noetherian integral domains inside a power series ring for which theprime ideal structure can be described explicitly.

In Section17.316.3, we analyze and provide more details about an “insider” ex-

ample B constructed in Chapterinsidepssec17. For each positive integer n, this construction

produces an example of a 3-dimensional quasilocal unique factorization domainthat is a generalized local ring in the sense of Cohen

Co[18] and has as its completion

a 2-dimensional regular local domain. Moreover, B is not catenary and has pre-cisely n prime ideals of height two. The associated intersection domain A for thisconstruction is a 2-dimensional regular local domain. In Section

17.416.4, we examine

the prime spectrum map SpecA → SpecB and describe three types of height-oneprime ideals of B by how they relate to primes of A.

In Section17.416.4 we establish the existence for each positive integer n ≥ 2 and

each positive integer t of a non-Noetherian integral domain B such that:(a) dimB = n+ 1.(b) The maximal ideal of B is generated by n elements.(c) B has exactly t prime ideals of height n and each of these primes is not

finitely generated.(d) B is a factorial domain.(e) The completion of B is a regular local domain of dimension n.(f) B is a birational extension of the localized polynomial ring over a field in

n+ 1 variables.

16.2. Background and Notation17.2

We recall some details from Sectioninsidepssec17.3 of Chapter

insidepssec17 concerning the insider

adaptation of the construction utilized in this chapter 1.We use the following setting throughout this section.

17.2.1 Setting 16.1. Let k be a field and let x, y1, . . . , ys be indeterminates over k;for convenience, we abbreviate the yi by y. (Often we assume k to have charac-teristic zero. This ensures excellence of the constructed domains in the simplestapplications.) Let R := k[x, y1, . . . , ys](x,y1,...,ys) = k[x, y](x,y) and let R∗ be the(x)-adic completion of R, that is, R∗ := k[y](y)[[x]], a power series ring in x. Wewrite Q(R),Q(R∗), etc, for the total rings of quotients of R,R∗, etc, respectively.Let τ1, . . . , τn be elements of xk[[x]] which are algebraically independent over k(x);we abbreviate them by τ . Let T := R[τ ]. We define the intersection domain Aτcorresponding to τ by Aτ := k(x, y, τ ) ∩R∗ = Q(T ) ∩R∗.

Next we select elements f1, . . . , fm of R[τ ] which are algebraically independentover Q(R); we abbreviate them by f . The intersection domain corresponding to fis Af = Q(R[τ ]) ∩R∗.

In order to obtain Aτ as a nested union of polynomial rings over k in s+m+1variables, we recall some of the details of the construction.2

17.2.2 Approximation Technique 16.2. With k, x, y, s, R andR∗ as in Setting17.2.116.1,

let ρ1, . . . , ρm be elements of R∗ which are algebraically independent over Q(R);

1For more detail see Chaptersnonnoesec26 and

insidepssec17.

2Again more details may be found in Chapterinsidepssec17.

16.2. BACKGROUND AND NOTATION 113

we abbreviate them by ρ. (Thus the intersection domain Aρ corresponding to ρ

is Aρ := k(x, y, ρ) ∩ R∗ = Q(R[ρ]) ∩ R∗.) Write each ρi :=∑∞

j=1 bijxj , with the

bij ∈ R. There are natural sequences ρir∞r=0 of elements in R∗, called the rth

endpieces for the ρi, which “approximate” the ρi, defined by:

For each i ∈ 1, . . . , n and r ≥ 0, ρir :=∞∑

j=r+1

(bijxj)/xr.

Now, for each r ≥ 0, define Ur := R[ρ1r, . . . , ρnr] and set Br to be Ur localizedat the multiplicative system 1 + xUr. Then set U := ∪∞r=0Ur and Bρ = ∪∞r=0Br.Thus U is a nested union of polynomial rings over R and Bρ is a nested union oflocalized polynomial rings over R; clearly U ⊆ Bρ ⊆ Aτ . We refer to Bρ as thenested union domain corresponding to ρ. The definition of the Ur (and hence alsoof Br, U and B) is independent of the representation of the ρi as power series withcoefficients in R (see Proposition

8.2.322.3 in Chapter

noeslocsec22).

For the two sequences of power series in Setting17.2.116.1, namely τ and f , we

have the approximation sequences τ i and fi

and the nested union domains Bτand Bf , which are localizations of

⋃∞r=0R[τ1r, . . . , τnr] and of

⋃∞r=0R[f1r, . . . , fmr]

respectively. Clearly Bf ⊆ Bτ are quasilocal domains with Bτ dominating Bf .

Recall that Proposition11.4.19.3 from Chapter

noehomsec8 implies the following:

17.2.3 Proposition 16.3. Assume the set-up as in Setting17.2.116.1. Then T → R∗[1/x]

is flat, Aτ = Bτ and Aτ is a Noetherian regular local ring; moreover, if char k = 0,then Aτ is excellent.

17.2.4 Insider Construction Details 16.4. Let τ = τ1, . . . , τn and f = f1, . . . , fmbe as in (

17.2.116.1) and (

17.2.216.2). We assume the constant terms in R = k[x, y] of the fi

are zero. Let S := R[f ]. The inclusion map S → T is an injective R-algebrahomomorphism, and m ≤ n.

Let A be the intersection domain for f , i.e. A := Af = Q(S)∩R∗. Let B := Bfbe the nested union domain associated to the f1, . . . , fm, as in (

17.2.216.2). We consider

the inclusion maps φ : S → T and ψ : T → R∗[1/x] as shown in the followingdiagram.

R∗[1/x]

R ⊆ S := R[f ] T := R[τ ](17.2.4.1) ψ

α:=ψϕ

ϕ

??? alertThe next theorem refers to 2.2 of Chapterbuildsec18 which isn’t there. The

diagram might be helpful. ???

17.2.5 Theorem 16.5. ( See also Chaptersbuildsec18, Theorem 2.2 (not there) but diagram

below Remark9.1.2?? and

insidepssec17 (???) Theorem

16.3.213.2.) Assume notation as in (

17.2.416.4).

(1) B is Noetherian and B = A if and only if the map α : S → R∗[1/x] in(17.2.4.1) is flat.

(2) For Q∗ ∈ Spec(R∗[1/x]), the localization αQ∗ : S → (R∗[1/x])Q∗ of themap α in (2.4.1) is flat if and only if the localization ϕQ∗∩T of the mapϕ in (17.2.4.1) is flat.


(3) The following are equivalent:(i) B is Noetherian and A = B.(ii) B is Noetherian.(iii) The localized map ϕQ∗∩T from (2.4.1) is flat for every maximal

ideal Q∗ ∈ Spec(R∗[1/x]).

Proposition17.2.616.6 is helpful for testing whether the map φ of (17.2.4.1) is flat.

Proposition 16.6.insideps[45, Proposition 2.4]17.2.6

Let R be a Noetherian integral domain and let x1, . . . , xn be indeterminatesover R. Assume that f1, . . . , fm ∈ R[x1, . . . , xn] are algebraically independent overR. Then

(1) ϕ : S := R[f1, . . . , fm] → T := R[x1, . . . , xn] is flat if and only if, for eachprime ideal P of T , we have ht(P ) ≥ ht(P ∩ S).

(2) For Q ∈ SpecT , ϕQ : S → TQ is flat if and only if for each prime idealP ⊆ Q of T , we have ht(P ) ≥ ht(P ∩ S).

(3) If ϕx : S → T [1/x] is flat, then B is Noetherian and B = A.

16.3. Non-Noetherian examples17.3

We refer to Examples16.5.117.2. Here are some further notes about these examples.

???alertThe propositions in the Venice paper have a little more than what isstated in the Fez paper. Maybe we should combine them.

16.4. A more general construction17.4

We consider the ring B =⋃∞r=0Br of the previous section in the case where

n = 1 and q = p1 = y. Thus f = yτ , R = k[x, y](x,y), Br = R[yτr](x,y,yτr) and B =⋃∞r=0R[yτr](x,y,yτr). This ring B has exactly one prime ideal Q = (y, yτr∞r=0)B

of height 2. Moreover, Q is not finitely generated and is the only prime ideal of Bthat is not finitely generated. We also have Q = yA ∩B, and Q ∩Br = (y, yτr)Brfor each r ≥ 0.

If q is a heightIf q is -one prime of B, then B/q is Noetherian if and only if qis not contained in Q. This is clear since Q is the unique prime of B that is notfinitely generated and a ring is Noetherian if each prime ideal of the ring is finitelygenerated.

The height-one primes q of B may be separated into three types as follows:

Type I The primes q 6⊆ Q, such as xB. As mentioned above, B/q is Noetherian.These primes are contracted from A. To see this, consider q = gB where g 6∈ Q.Then gA is contained in a height one prime P of A. Then g ∈ (P ∩ B) \ Q soP ∩ B 6= Q. Since mBA = mA, we have P ∩ B 6= mB. Therefore P ∩ B is aheight-one prime containing q, so q = P ∩B and Bq = AP .

There are infinitely many primes q of type I, because every element of mB \Qis contained in a prime q of type I. Thus mB ⊆ Q ∪ ⋃q of type I. Since mB

is not the union of finitely many strictly smaller prime ideals, there are infinitelymany primes q of type I.

If q is a height-one prime of B not of type I, then B = B/q has precisely threeprime ideals. These prime ideals form a chain: (0) ⊂ Q ⊂ (x, y)B = mB.

16.5. A MORE GENERAL CONSTRUCTION 115

Type II The primes q ⊂ Q, where q has height one and is contracted from a primep of A = k(x, y) ∩ R∗, for example, the prime y(y + τ)B. For q of this type, B/qis dominated by the one-dimensional Noetherian local domain A/p. Thus B/q is anon-Noetherian generalized local ring in the sense of Cohen.

For q of Type II, the maximal ideal of B/q is not principal. This follows becausea quasilocal generalized local domain having a principal maximal ideal is a DVRN2[72, (31.5)].

There are infinitely many height-one primes of type II, for example, y(y+xnτ)Bfor each n ∈ N. For q of type II, the DVR Bq is birationally dominated by Ap.Hence Bq = Ap and the ideal

√qA = p ∩ yA.

Type III The primes q ⊂ Q, where q has height one and is not contracted fromA, for example, the prime yB and the prime (y + xnyτ)B for n ∈ N. Since theelements y and y+xnyτ are in mB and are not in m2

B and since B is a UFD, theseelements are necessarily prime. Also the prime ideals (y+xnyτ)B and (y+xmyτ)Bare distinct for n and m distinct positive integers, for if W is a prime ideal of Bthat contains them both and if m > n, then xm−nyτ ∈ W . But x 6∈ Q and Qand mB are the only primes of B containing (y, yτ)B. For q of type III, we have√qA = yA.

If q = yB or q = (y + xnyτ)B, then the image mB of mB in B/q is principal.It follows that the intersection of the powers of mB is Q/q and B/q is not ageneralized local ring. For if P is a principal prime ideal of a ring and P ′ is a primeideal properly contained in P , then P ′ is contained in the intersection of the powersof P .

Kap[58, page 7, ex. 5].

Another way to examine the height-one primes of BObserve that B[1/x] is a localization of the polynomial ring k[x, y, f ] while

A[1/x] is a localization of k[x, y, τ ]. Thus the embedding B[1/x] → A[1/x] is alocalization of the map:

φ : k[x, y, f ] −→ k[x, y, τ ]

which is defined by φ(f) = yτ . Obviously the nonflat locus of φ is defined by theideal (y) of k[x, y, τ ].

If q ∈ B is a height one prime ideal which is not contained in Q, then theextension qA is not contained in yA and for every minimal prime p ⊆ A over qAthe induced map Bq → Ap is flat. In particular, q is not lost in A. If q is a height-one prime of B which is contained in Q then q is of type II or III according to theminimal prime divisors of qA: If there is a minimal prime p ∈ SpecA of qA whichis different from (y), then the map: Bq → Ap is flat, and q is not lost in A. In thiscase, q is of type II. Note that yA is also a minimal prime divisor of qA. If yA isthe only prime divisor of qA, then q is lost in A. In this case the map BQ → A(y)

is a localization.We remark that B/yB is a rank 2 valuation ring. This can be seen directly

or else one may applyHS[52, Prop. 3.5(iv)]. If we do a similar construction with a

prime contained in m2 (instead of y), for example, f = (x2 + y2)τ (over Q), thenB/(x2 + y2) has a two-generated maximal ideal and cannot be a valuation ring.

16.5. A More General Construction17.5

In Theorem17.5.116.7, we obtain a generalization of the construction that gives the

examples considered above.


17.5.1 Theorem 16.7. Let n ≥ 2 be a positive integer. For each positive integer t,there exists a non-Noetherian integral domain B such that:

(a) dimB = n+ 1.(b) The maximal ideal of B is generated by n elements.(c) B has exactly t prime ideals of height n and each of these primes is not

finitely generated.(d) B is a factorial domain.(e) The completion of B is a regular local domain of dimension n.(f) B is a birational extension of the localized polynomial ring over a field in

n+ 1 variables.

Proof. Let k be a field and let x, y1, . . . , yn−1 be variables over k. Define:

R = k[x, y1, . . . , yn−1](x,y1,...,yn−1)

and let τ1, . . . , τn−1 ∈ xk[[x]] be algebraically independent overQ(R). For 1 ≤ i ≤ t,let pi = y1 − xi. Set q = p1p2 . . . pt, and consider the element

f = qτ1 + y2τ2 + . . . yn−1τn−1.

The map:

R[f ] −→ R[τ1, . . . , τn−1]has as its Jacobian ideal: J = (q, y2, . . . , yn−1) which is an ideal of height n − 1that is the intersection of t prime ideals of height n− 1: J = Q1 ∩ . . . ∩Qt, whereQi = (pi, y2, . . . , yn−1).

If we construct B as usual as the union of localized polynomial rings of di-mension n + 1, then by Theorem

16.3.913.5, B is not Noetherian ( or use (

17.2.516.5.1) and

an argument similar to (16.3.313.3.1)Check?? ). If n > 2 then J has height > 1 and

it follows from Theorem7.5.521.18 or Theorem

14.6.326.3 that B = A = Q(R[f ]) ∩ R∗. It

follows from Theorem7.4.521.13 that B is factorial.

Moreover, if we consider our usual set up:

R[f ] −→ R[τ1, . . . , τn−1][1/x] −→ R∗[1/x]the preimage in R[f ] of JR∗[1/x] is the ideal L = (J, f) which has height n. Theideal L corresponds to the intersection of t prime ideals in B of height n, these arethe only primes in B of height n and they are not finitely generated.

17.5.2 Remark 16.8. It would be interesting to know whether for B as in Theorem17.5.116.7, the prime ideals of B of height n are the only prime ideals of B that are notfinitely generated.

17.5.3 Remark 16.9. With the insider construction given in (17.2.116.1)-(

17.2.416.4), if the di-

mension of Bf is greater than that of R∗, then Bf is not catenary. One way tosee this is that in Spec(Bf ) there is always a saturated chain of prime ideals thatincludes (x) and this chain has length equal to dimR∗, while if dimBf > dimR∗,then there exists a saturated chain of prime ideals in Bf of length greater thandimR∗.

CHAPTER 17

Examples using the insider construction, Dec.

2009(insidepssec) 16insidepssec

???alertObserve similarity with Chapternonfgsec16 ???


In this chapterGiven a power series ring R∗ over a Noetherian integral domain R and given

a subfield L of the total quotient ring of R∗ with R ⊆ L, we construct subrings Aand B of L such that B is a localization of a nested union of polynomial rings overR and B ⊆ A := L ∩ R∗. We show in certain cases that flatness of a related mapon polynomial rings is equivalent to the Noetherian property for B. Moreover if Bis Noetherian, then B = A. We use this construction to obtain for each positiveinteger n an explicit example of a 3-dimensional quasilocal unique factorizationdomain B such that the maximal ideal of B is 2-generated, B has precisely n primeideals of height two, and each prime ideal of B of height two is not finitely generated.

???? In Section16.4?? we continue the analysis of the flatness of (

16.1.2??) and the

nonflat locus. We discuss results ofP[79],

Wang[97] and others.

We present for each positive integer n an insider example B such that:(1) B is a 3-dimensional quasilocal unique factorization domain,(2) B is not catenary,(3) the maximal ideal of B is 2-generated,(4) B has precisely n prime ideals of height two,(5) Each prime ideal of B of height two is not finitely generated,(6) For every non-maximal prime P of B the ring BP is Noetherian.

17.2. Background and Notation16.2

We recall Theorem11.3.258.8 with some details for the approximation to the intersec-

tion domain Aτ of (16.1.113.1).

16.2.2 Theorem 17.1. Let R be a Noetherian integral domain with fraction field K.Let a be a nonzero nonunit of R. Let τ1, . . . , τn ∈ aR[[a]] = aR∗ be algebraicallyindependent over K, abbreviated by τ . Let Uτ and Bτ be as in (

16.2.1??). Then the

following statements are equivalent:(1) Aτ := K(τ) ∩R∗ is Noetherian and Aτ = Bτ .(2) Uτ is Noetherian.(3) Bτ is Noetherian.

117

118 17. EXAMPLES INTEGRAL DOMAINS

(4) R[τ ]→ R∗a is flat.

17.3. Examples16.5

16.5.1 Examples 17.2. For each positive integer n, we present an example of a 3-dimensional quasilocal unique factorization domain B such that

(1) B is not catenary,(2) the maximal ideal of B is 2-generated,(3) B has precisely n prime ideals of height two,(4) Each prime ideal of B of height two is not finitely generated,(5) For every non-maximal prime P of B the ring BP is Noetherian.

The notation for this construction is a localized version of the notation ofInsider Construction Details

16.3.5??, with s = 1. Thus k is a field, R = k[x, y](x,y) is

a 2-dimensional regular local ring and R∗ = k[y](y)[[x]] is the (x)-adic completionof R. Let τ =

∑∞j=1 cjx

j ∈ xk[[x]] be algebraically independent over k(x). Letpi ∈ R− xR be such that piR∗ are n distinct prime ideals. For example, we couldtake pi = y − xi. Let q = p1 · · · pn. We set f := qτ and consider the injectiveR-algebra homomorphism S = R[f ] → R[τ ] = T .

LetB be the nested union domain associated to f as in (16.2.1??). If τr =

∑∞j=r+1

cjxj

xr

is the rth endpiece of τ , then ρr := qτr is the rth endpiece of f . For each r ∈ N,let Br = R[ρr](x,y,ρr). Then each Br is a 3-dimensional regular local ring andB =

⋃∞r=1Br.

The map α : S → R∗x is not flat since piR∗

x is a height-one prime and piR∗x∩S =

(pi, f)S is of height two. By Theorem16.2.217.1, B is not Noetherian. By Theo-

remintsec21.4.5, B is a quasilocal unique factorization domain. Moreover, by Theo-

remintsec21.4.4, for each t ∈ N, xtB = xtR∗ ∩ B and R/xtR = B/xtB = R∗/xtR∗.

It follows that the maximal ideal of B is (x, y)B. If P ∈ SpecB is such thatP ∩ R = (0), then because the field of fractions K(f) of B has transcendence de-gree one over the field of fractions K of R, ht(P ) ≤ 1 and hence because B is aUFD, P is principal.

16.5.1.1 Claim 17.3. Let I be an ideal of B and let t ∈ N. If xt ∈ IR∗, then xt ∈ I.Proof. There exist elements b1, . . . , bs ∈ I such that IR∗ = (b1, . . . , bs)R∗. If

xt ∈ IR∗, there exist αi ∈ R∗ such that

xt = α1b1 + · · ·+ αsbs.

We have αi = ai + xt+1λi, where ai ∈ B and λi ∈ R∗. Thus

xt[1− x(b1λ1 + · · ·+ bsλs)] = a1b1 + · · ·+ asbs ∈ B.Since xtR∗ ∩B = xtB, γ := 1− x(b1λ1 + · · ·+ bsλs) ∈ B. Moreover, γ is invertiblein R∗ and hence also in B. It follows that xt ∈ I.

To examine more closely the prime ideal structure of B, it is useful to considerthe inclusion map B → A := R∗ ∩K(f) and the map SpecA→ SpecB.

16.5.2 Proposition 17.4. With the notation of Example16.5.117.2 and A = R∗ ∩ K(f),

we have

17.3. EXAMPLES 119

(1) A is a two-dimensional regular local domain with maximal ideal mA =(x, y)A.

(2) mA is the unique prime of A lying over mB = (x, y)B, the maximal idealof B.

(3) If P ∈ SpecB is nonmaximal, then ht(PR∗) ≤ 1 and ht(PA) ≤ 1. Thusevery nonmaximal prime of B is contained in a nonmaximal prime of A.

(4) If P ∈ SpecB and xq 6∈ P , then htP ≤ 1.(5) If P ∈ SpecB, htP = 1 and P ∩R 6= 0, then P = (P ∩R)B.

Proof. By Proposition16.3.11?? (the result of Valabrega) A := R∗ ∩K(f) is a two-

dimensional regular local domain having the same completion as R and R∗. Thisproves item 1. Since B/xB = A/xA = R∗/xR∗, mA = (x, y)A is the unique primeof A lying over mB = (x, y)B. Thus item 2 holds and also item 3 if x ∈ P . To see(3), it remains to consider P ∈ SpecB with x 6∈ P . By Claim

16.5.117.2.1, for all t ∈ N,

xt 6∈ PR∗. Thus ht(PR∗) ≤ 1. Since A → R∗ is faithfully flat, ht(PA) ≤ 1.For (4), we see by (3) that ht(PA) ≤ 1. Let Q ∈ SpecA be a height-one prime

ideal such that P ⊆ Q. Since xq 6∈ P , we have BP = SP∩S = TQ∩T = AQ, whereS = R[f ] and T = R[τ ]. Thus ht(P ) ≤ 1. For (5), if x ∈ P , then P = xB andthe statement is clear. Assume x 6∈ P . Since Bx is a localization of (Br)x, we have(P ∩R)Br = P ∩Br for all r ∈ N. Thus P = (P ∩R)B.

We observe that the DVRs BxB and AxA are equal. Moreover, A is the nestedunion

⋃∞r=1R[τn](x,y,τn) of 3-dimensional regular local domains. Since A is a two-

dimension regular local domain each nonmaximal prime of A is principal. If pAis a height-one prime of A with pA 6∈ p1A, . . . pnA, then ApA = BpA∩B andht(pA ∩B) = 1. We observe in Claim

16.5.2.117.5 that piA ∩B has height two and is not

finitely generated.

16.5.2.1 Claim 17.5. Let pi be one of the prime factors of q. Then piB is prime in B.Moreover

(1) piB and Qi := (pi, ρ1, ρ2, . . .)B = piA ∩B are the only primes of B lyingover piR in R,

(2) Qi is of height two and is not finitely generated.

Proof. We use that B =⋃∞r=1Br, where Br = R[ρr](x,y,ρr) is a 3-dimensional

regular local ring. For each r ∈ N, piBr is prime in Br. Hence piB is a height-oneprime ideal of B, for i = 1, . . . , n. Since ρr = qτr, piA ∩ Br = (pi, ρr)Br is aheight-two prime ideal of the 3-dimensional regular local domain Br. ThereforeQi := (pi, ρ1, ρ2, . . .)B = piA ∩ B is a nested union of prime ideals of height two,so ht(Qi) ≤ 2. Since piB is a nonzero prime ideal properly contained in Qi,ht(Qi) = 2. Moreover x 6∈ (pi, ρr)Br for each r, so x 6∈ Qi. Hence for each r ∈ N,ρr+1 6∈ (pi, ρr)B and Qi is not finitely generated.

Since x 6∈ Qi and B[1/x] is a localization of the Noetherian domain Bn[1/x],we see that BQi is Noetherian. Since the Qi are the only prime ideals of B of heighttwo and B is a UFD, BP is Noetherian for every non-maximal prime P .

This completes the presentation of Examples16.5.117.2. With regard to the bi-

rational inclusion B → A and the map SpecA → SpecB, we remark that thefollowing holds: Each Qi contains infinitely many height-one primes of B that arethe contraction of primes of A and infinitely many that are not. Among the primesthat are not contracted from A are the piB. In the terminology of

ZSII[106, page 325],

120 17. EXAMPLES INTEGRAL DOMAINS

P is not lost in A if PA ∩B = P . Since piA ∩B = Qi properly contains piB, piBis lost in A. Since (x, y)B is the maximal ideal of B and (x, y)A is the maximalideal of A and B is integrally closed, a version of Zariski’s Main Theorem

Peskine[78],

Ev[25],

implies that A is not essentially finitely generated as a B-algebra.

CHAPTER 18

The first Noetherian theorem April 23 2009

buildsec???(Note: the file is eno11.tex There is not very much of it here.)???Suppose a is a nonzero nonunit of a Noetherian integral domain R. We apply

this construction, and its natural generalization to finitely many elements, to exhibitNoetherian extension domains of R inside the (a)-adic completion R∗ of R. Supposeτ1, . . . , τs ∈ aR∗ are algebraically independent over K, the field of fractions of R.Starting with U0 := R[τ1, . . . , τs], there is a natural sequence of nested polynomialrings Un between R and A := K(τ1, . . . , τs) ∩ R∗. It is not hard to show that ifU := ∪∞n=0Un is Noetherian, then A is a localization of U and R∗[1/a] is flat overU0. We prove, conversely, that if R∗[1/a] is flat over U0, then U is Noetherian andA := K(τ1, . . . , τs)∩R∗ is a localization of U . Thus the flatness of R∗[1/a] over U0

implies the intersection domain A is Noetherian.


Let a be a nonzero nonunit of a Noetherian integral domain R. The (a)-adiccompletion R∗ of R is isomorphic to the ring R[[x]]/(x − a) N2

[72, (17.5), page 55].Thus elements of the (a)-adic completion may be regarded as formal power seriesin a. Of course if R is already complete in its (a)-adic topology, then R = R∗, butoften it is the case that there are elements of R∗ that are transcendental over thefield of fractions K of R.

Suppose τ1, . . . , τs ∈ aR∗ are algebraically independent the field of fractionsK of R. Starting with U0 := R[τ1, . . . , τs], there is a natural sequence of nestedpolynomial rings Un between R and A := K(τ1, . . . , τs)∩R∗. It is not hard to showthat if U := ∪∞n=0Un is Noetherian, then A is a localization of U and R∗[1/a] is flatover U0. We prove, conversely, that if R∗[1/a] is flat over U0, then U is Noetherianand A := K(τ1, . . . , τs) ∩ R∗ is a localization of U . Thus the flatness of R∗[1/a]over U0 implies the intersection domain A is Noetherian.

Let R be a Noetherian integral domain with field of fractions K and let a bea nonzero nonunit of R. We are interested in the structure of certain intermediateintegral domains between R and R∗ := (R, (a)) = R[[x]]/(x − a), the (a)-adiccompletion of R. We are particularly interested in domains of the form A := L∩R∗,where L is an intermediate field between K and the total ring of fractions of R∗.It is often difficult to compute this intersection ring A. Thus we seek conditions inorder that A be realizable as a localization of a directed union of polynomial ringextensions of R.

This intersection construction inside the completion of R with respect to a prin-cipal ideal yields interesting Noetherian rings which are directed unions of localizedpolynomial rings, as we see in this chapter. By contrast, taking the analogousconstruction inside the completion with respect to a maximal ideal, even of an ex-cellent local domain seems less likely to give Noetherian intersection domains. In

121

122 18. THE FIRST NOETHERIAN THEOREM APRIL 23 2009

Chapteridwisec20 we show for a countable excellent local domain (R,m) of dimension

at least two that there exist infinitely many algebraically independent elementsτ1, τ2, . . . in the m-adic completion R of R such that the corresponding intersec-tion domain is a localized polynomial ring in infinitely many variables over R; thatis, R ∩K(τ1, τ2, . . . ) = R[τ1, τ2, . . . ](m,τ1,τ2,... ).

In Chapternoeslocsec22 we consider, in the case where R is a Noetherian semilocal

integral domain and a is an element of the Jacobson radical of R, the conditionthat the embedding U0 → R∗[1/a] is flat. Our goal in this chapter is to examineflatness of the embedding U0 → R∗[1/a] in a more general context, and to provethe following theorem.

9.1.1 Theorem 18.1. Let R be a Noetherian domain, let a ∈ R be a nonzero nonunit,and let τ1, . . . , τs be elements of the (a)-adic completion R∗ of R that are alge-braically independent over R. Then the following conditions are equivalent:

(1) The ring R∗[1/a] is flat over U0 = R[τ1, . . . , τs].(2) The directed union U := ∪∞n=0Un is Noetherian. 1

Moreover, the integral domain A := K(τ1, . . . , τs) ∩ R∗ is a localization of U , andhence is Noetherian, if these equivalent conditions hold.

9.1.12 Remark 18.2. An example given in Chapternoeslocsec22.4 shows that it is possible for

A to be a localization of U and yet for A, and therefore also U , to fail to beNoetherian. Thus the equivalent conditions of Theorem

9.1.118.1 are not implied by the

property that A is a localization of U .

We present in (9.2.518.3) an example that is a modification of Example

4.2.14.12 to show

that A being Noetherian does not imply that U is Noetherian.

???ALERT Fill In the following example from Noehom wherever it appears.???

9.2.5 Example 18.3.

The following diagram displays the situation concerning possible implicationsbetween A being a localization of U and A or U being Noetherian:

R∗[1/a] is flat over U0 = R[τ1, . . . , τs] U Noetherian

A is a localization of U A Noetherian

1Heitmann inH1[54, page 126] considers the case where there is one transcendental element

τ and defines the corresponding extension U to be a simple PS-extension of R for a. Heitmannproves in this case that a certain monomorphism condition on a sequence of maps is equivalentto U being Noetherian

H1[54, Theorem 1.4].

CHAPTER 19

Properties of ring extensions

propIn this chapter we consider various properties related to flatness.

19.1. Independence properties for elements

Let (R,m) be an normal Noetherian local domain with field of fractions K andcompletion (R, m). Elements x1, . . . , xn ∈ m that are algebraically independentover R are said to be idealwise algebraic independent over R if R ∩K(x1, . . . , xn)is equal to R[x1, . . . , xn] localized at the ideal (m, x1, . . . , xn)R[x1, . . . , xn].

19.2. Related properties for extensions

Let S → T be an extension of Krull domains. We say that T is a height-onepreserving extension of S if for every height-one prime ideal P of S with PT 6= Tthere exists a height-one prime ideal Q of T with PT ⊆ Q.)

Let S → T be an extension of Krull domains. We say that T is weakly flat overS if every height-one prime ideal P of S with PT 6= T satisfies PT ∩ S = P .

123

CHAPTER 20

Idealwise algebraic independence May 17 2009idwisec


Many examples in commutative algebra have been constructed using the fol-lowing principle: Let k be a field, let S = k[x1, . . . , xn](x1,...,xn) be a localizedpolynomial ring over k, and let a be an ideal in the completion S of S such thatthe associated primes of a are in the generic formal fiber of S; that is, p ∩ S = (0)for each p ∈ Ass(S/a). Then S embeds in S/a, the fraction field Q(S) of S embedsin the total ring of fractions of S/a; and for certain choices of a, the intersectionD = Q(S) ∩ (S/a) is a Noetherian local domain with completion D = S/a.

Examples that may be realized by this method include Nagata’s first examplesof non-excellent rings

N1[70], Ogoma’s celebrated counterexample to Nagata’s cate-

nary conjectureO1[76],

O2[77], examples of Rotthaus and Brodmann

R1[81],

R2[82],

BR1[11],

BR2[12],

and examples of Nishimura and WestonNi2[74],

W[98]. In fact, we know of no example of

a Noetherian reduced local ring that contains and is of finite transcendence degreeover a coefficient field that is not realizable using this principle, cf. Section

4.65.2. We

conjecture that all Noetherian reduced local ringsD that contain a coefficient field kand that are of finite transcendence degree over k relate to an ideal a in the genericformal fiber of the localization of a polynomial ring S = k[x1, . . . , xn](x1,...,xn), insuch a way that D is a direct limit of etale extensions of such an intersection ringQ(S) ∩ (S/a) as above.

The key to these examples is usually the behavior of the formal fibers of thedomain D. A major problem in this setting is to identify and classify ideals inthe formal fiber of S according to the properties of the intersection domain D =Q(S) ∩ (S/a). The goal of this chapter is to study the significance of the choice ofthe ideal a in this construction.

In many of the examples mentioned above, the expression D = Q(S) ∩ (S/a)may be interpreted so that D is an intersection of the completion of a Noetherianlocal domain R with a subfield. In this chapter we consider this intersection form.More precisely we use the following setting throughout this chapter:

6.1.1s Setting 20.1. Let (R,m) be an excellent normal local domain with field offractions K and completion (R, m). Suppose that τ1, . . . , τn are elements of mwhich are algebraically independent over R and that t1, . . . , tn are indeterminatesover K.

For S = R[τ1, . . . , τn](m.τ1,...,τn) and L the fraction field of S, we consider inthis chapter intermediate rings of the form A = L ∩ R. It is immediate that:

(1) The completion of R[τ1, . . . , τn](m.τ1,...,τn) is isomorphic to R[[t1, . . . , tn]],(2) For a = (t1 − τ1, t2 − τ2, . . . , tn − τn) in R[[t1, . . . , tn]], R ∩ a = 0,(3) R[[t1, . . . , tn]]/a ∼= R,

125

126 20. IDEALWISE ALGEBRAIC INDEPENDENCE MAY 17 2009

Thus, with Sn = R[t1, . . . , tn](m.t1,...,tn) and a = (t1 − τ1, t2 − τ2, . . . , tn − τn) inR[[t1, . . . , tn]], we have Q(Sn) ∩ (Sn/a) = L ∩ R. That is, the expression from theabove paragraphs now has the form L ∩ R, where L is a field between K and thefraction field of R.

For elements x1, . . . , xn ∈ m algebraically independent over R, we are some-times able to prove that R ∩ K(x1, . . . , xn) is equal to R[x1, . . . , xn] localized atthe ideal (m, x1, . . . , xn)R[x1, . . . , xn]. We introduce in this chapter the conceptsof primary, residual and idealwise algebraic independence

Let S → T be an extension of Krull domains. We say that T is a height-onepreserving extension of S if for every height-one prime ideal P of S with PT 6= Tthere exists a height-one prime ideal Q of T with PT ⊆ Q.


Suppose A is a Noetherian local intermediate ring dominating R (in the sensethat the maximal ideal n of A intersects R in m) and dominated by R. The localinjective morphisms R → A → R imply the existence of a canonical surjectionπ : A R, where A is the n-adic completion of A. In this setting it is well knownthat A is a topological subspace of R, i.e., π is an isomorphism, if and only if everyideal of A is closed in the topology on A defined by the powers of m. Since A isassumed to be Noetherian, π an isomorphism implies R is faithfully flat over A,and hence aR ∩ A = aA for each principal ideal aA of A, so A = L ∩ R where Lis the field of fractions of A. Thus A = R implies A = L ∩ R and there can beat most one Noetherian. Without the assumption that A is Noetherian there areexamples where A = R and A is non-Noetherian with the same fraction field as R,see for example

B[10, Chap. III, pages 119-120, Ex. 14]. A with A = R for each

intermediate field L between K and the fraction field K of R.On the other hand, if L is any intermediate field between K and K, then the

ring A = L ∩ R is a quasilocal domain dominating R and dominated by R. It iseasily seen that such a ring A is Hausdorff in the topology defined by the powersof its maximal ideal, and again the injective local morphisms R → A → R implythe existence of a canonical surjection π : A R, where A is the completion of A.This leads us to the question:

6.1.2 Question 20.2. What subdomains A of R have the form L∩ R, where L is anintermediate field between K and K?

In considering Question6.1.220.2, we have come to realize that it is quite broad,

and that the explicit determination of L ∩ R is computationally challenging evenfor relatively simple examples of R and L. We have also discovered that for manyexcellent normal local domains R of dimension at least two, there exist intermediatefields L between K and K such that A = L∩ R fails to be a subspace of R. It canhappen thatA = L∩R is an excellent normal local domain of dimension greater thanthat of R, or even that A fails to be Noetherian. In order to exhibit such exampleswe concentrate in this paper on elements a1, . . . an ∈ m which are algebraicallyindependent over K and satisfy certain additional independence conditions. Wecontinue the study of domains of the form L ∩ R in Chapter

intsec21.

20.1. INTRODUCTION 127

6.1.3 Results 20.3. Here are some specific results related to the general question ofthe structure of L ∩ R.

(1) For a1, . . . an arbitrary elements of R and L = K(a1, . . . , an), the subdo-main A = L∩R is a normal quasilocal birational extension ofR[a1, . . . , an].If the ai are algebraic over R, then the structure of A is well understood;A is an etale extension of R with completion A = R (see

R3.7[85]). But if

the ai are not algebraic over R, the situation is more complicated and thestructure of A depends on the residual behavior of the ai modulo variousprime ideals of R.

(2) For a specific example of an excellent normal local domain to illustratethese ideas, we refer to R = k[x, y](x,y), the localization of the polynomialring over a field k at the maximal ideal generated by the indeterminatesx and y. For this example R = k[[x, y]]. A result of Valabrega

V[96,

Proposition 3] implies that, for R = k[x, y](x,y), if L is a field between K

and the fraction field F of k[x][[y]] or if L is a field between F and K, thenA = L ∩ R is a two-dimensional regular local domain with completion R.

(3) On the other hand, again considering R = k[x, y](x,y), it is well-knownthat, for each positive integer n, the formal power series ring in n variablesover k can be embedded in R = k[[x, y]]; in fact k[[x, y]] contains infinitelymany analytically independent elements

Abhy[2],

AM[5],

AHW[6]. However, if a formal

power series ring Sn in n variables over k is embedded in R = k[[x, y]],and if R ⊆ Sn (so, in particular, if Sn = L ∩ R for some field L betweenK and K), then Sn = R, so that n = 2. More generally, if A is a localNoetherian ring with completion A and B is a complete local ring suchthat B dominates A and A dominates B, then a well-known theorem ofCohen ( cf.

M[68, (8.4)]) implies that B = A.

(4) An example of Nagata shows the existence of a 3-dimensional regular localdomain D with completion D a formal power series ring in 3 variablesover a field k of characteristic p > 0 for which there exists an intermediatefield L between the fraction fields of D and D such that A = D ∩ L isnon-Noetherian. In this example, D is not excellent and L is a finitepurely inseparable extension of the fraction field of D. A discussion ofthis example is given on pages 31-32 of

HRS[33].

(5) An example of Ogoma shows the existence of a four-dimensional excellentregular local domain, indeed a domain D obtained as a localization of apolynomial ring in 4 variables over a countable field, for which there existsa field L contained in the fraction field of D and generated by two elementsover the fraction field of D such that A = D ∩ L is not Noetherian. Adiscussion of this example is given on pages 32-34 of

HRS[33].

(6) InHR[32], an excellent normal local domain R is said to have the Noetherian

intermediate rings property, (NIR), if for each subfield L of K contain-ing K, the quasilocal ring L ∩ R is Noetherian and has completion R.Interesting examples of excellent regular local rings satisfying (NIR) areconstructed in

R3.5[84] and by Shelburne in

Shb[92]. Shelburne shows the ex-

istence for each positive integer d ≥ 3 of an excellent local domain R


containing a field of characteristic p > 0 such that dim(R) = d, the di-mension of the generic formal fiber of R is 0, and R is properly containedin its completion R. In his examples, R has, in fact, infinite transcendencedegree over R.

(7) For a subfield L of K containing K, the following construction consideredin

HRS[33] is sometimes useful for obtaining information about A = L ∩ R.

Suppose (S,n) is an excellent normal local domain dominating R and qis a prime ideal of S such that S/q ∼= R and q ∩ S = (0). Then L,the fraction field of S, embeds in the fraction field of S/q and with thisidentification, A = L ∩ (S/q) = L ∩ R is a quasilocal domain birationallydominating S. In §3 we present examples of this type where A = S.

(8) Recent work of Heitmann inH2[55] and

H3[56] and Loepp in

Lo[63] shows the

richness of the structure of the local domains with a given completion. InH2[55], Heitmann shows that a complete local ring T is the completion ofa local unique factorization domain (UFD) if (i) T has depth at least 2,and (ii) no nonzero element of the prime subring of T is a zero divisor ofT . In

H3[56], Heitmann proves that very often a complete local ring is the

completion of a local ring having an isolated singularity. In particular,for T satisfying (i) and (ii) he shows the existence of a local UFD allof whose proper localizations are regular that has completion T . Hisconstruction is adopted by his student Loepp to obtain more examples ofstrange phenomena that can occur in passing from a local (Noetherian)ring to its completion. A central step in Heitmann’s construction involvespassing from a subring R of T to a bigger ring R′ by adjoining a kind ofindependent element, similar to the residually algebraically independentelements defined below and studied in §4 ??? of this chapter. Theseresidually algebraically independent elements play an important role inhis construction. Certain relations from T become satisfied in R′ (definedas a limit), but by using the residually algebraically independent elements,he is able to control the correspondence between the height-one primeideals of R and those of R′.

ALERT We refer to the following theorems:

6.2.720.12

6.2.1020.15

6.2.1420.19

6.3.320.24

6.6.120.56

6.6.320.58

6.4.420.37

6.5.620.52

6.3.120.22

PDE-extensions.We show for each positive integer n the existence of a subfield Fn of the fraction

field of k[[x, y]] such that k[[x, y]] ∩ Fn is a regular local ring of dimension n. Wealso show the existence of a subfield F∞ of k((x, y)) such that k[[x, y]] ∩ F∞ is aninfinite dimensonal non-Noetherian integral domain.

Similarities here to material in TGF paper Section 3 ???locally flat in height k, denoted LFk.Passing to the Henselization.The idealwise independence conditions do not yield “new” Noetherian rings,

since when these conditions hold, the intersection domain is essentially a localizedpolynomial ring over the original ring. The extension of these conditions to “limit

20.2. INTRODUCTION 129

intersecting” conditions where the intersection domain is a directed union of local-ized polynomial rings over the original ring does lead to many new examples andis quite fruitful. ————————————-

HERE IS THE FILE OF IDEA57ABSTRACT: Suppose (R,m) is an excellent normal local domain with field

of fractions K and completion (R, m). In this paper we consider three concepts ofindependence over R for elements a1, . . . , an ∈ m which are algebraically indepen-dent over K. The first of these, idealwise independence, is that K(a1, . . . , an) ∩ Requals the localized polynomial ring R[a1, . . . , an](m,a1,...,an). If R is countable withdim(R) > 1, we show the existence of an infinite sequence of elements a1, a2, . . .of m that are idealwise independent over R and therefore the existence of a non-Noetherian domain A = K(a1, a2, . . . ) ∩ R such that A is a localized polynomialring in infinitely many variables over R. We relate these three concepts of indepen-dence to flatness conditions of extensions of Krull domains, establish implicationsamong them, and draw some conclusions concerning their equivalence in specialsituations. We also investigate their stability under change of base ring.

20.2. Introduction6.1.i

Over the past forty years many examples in commutative algebra have been con-structed using the following principle: Let k be a field, let S = k[x1, . . . , xn](x1,...,xn)

be a localized polynomial ring over k, and let a be an ideal in the completion S ofS such that the associated primes of a are in the generic formal fiber of S; that is,p∩ S = (0) for each p ∈ Ass(S/a). Then S embeds in S/a, the fraction field Q(S)of S embeds in the fraction field of S/a; and for certain choices of a, the intersectionD = Q(S) ∩ (S/a) is a local Noetherian domain with completion D = S/a.

Examples constructed by this method include Nagata’s first examples of non-excellent rings

N2[72], Ogoma’s celebrated counterexample to Nagata’s catenary con-

jectureO1[76],

O2[77], examples of Rotthaus and Brodmann

R1[81],

R2[82],

BR1[11],

BR2[12], and

examples of Nishimura and WestonNi[73],

W[98]. In fact all examples we know of local

Noetherian reduced rings which contain and are of finite transcendence degree overa coefficient field may be realized using this principle. 1

The key to these examples is usually the behavior of the formal fibers of thedomain D. A major problem in this setting is to identify and classify ideals inthe formal fiber of S according to the properties of the intersection domain D =Q(S) ∩ (S/a). The goal of this paper is to study the significance of the choice ofthe ideal a in this construction.

In many of the examples mentioned above, the expression D = Q(S) ∩ (S/a)may be interpreted so that D is an intersection of the completion of a local Noe-therian domain R with a subfield. In this paper we consider this latter form. Moreprecisely we use the following setting throughout this paper:

Setting 20.4. Let (R,m) be an excellent normal local domain with field offractions K and completion (R, m). Suppose that τ1, . . . , τn are elements of m

1We conjecture that all local Noetherian reduced rings D which contain a coefficient fieldk and which are of finite transcendence degree over k relate to an ideal a in the generic formalfiber of the localization of a polynomial ring S = k[x1, . . . , xn](x1,...,xn), in such a way that D is

a direct limit of etale extensions of such an intersection ring Q(S) ∩ (bS/a) as above.


which are algebraically independent over R and that t1, . . . , tn are indeterminatesover K. For S = R[τ1, . . . , τn](m.τ1,...,τn) and L the fraction field of S, we considerin this paper intermediate rings of the form A = L ∩ R. It is immediate that:

(1) The completion of R[τ1, . . . , τn](m.τ1,...,τn) is isomorphic to R[[t1, . . . , tn]],(2) For a = (t1 − τ1, t2 − τ2, . . . , tn − τn) in R[[t1, . . . , tn]], R ∩ a = 0,(3) R[[t1, . . . , tn]]/a ∼= R,

Thus, with Sn = R[t1, . . . , tn](m.t1,...,tn) and a = (t1 − τ1, t2 − τ2, . . . , tn − τn) inR[[t1, . . . , tn]], we have Q(Sn) ∩ (Sn/a) = L ∩ R. That is, the expression from theabove paragraphs now has the form L ∩ R, where L is a field between K and thefraction field of R.

Before we proceed with our summary of this paper, we give questions, motiva-tion and background information on the study of L∩ R, where L is a field betweenK and the fraction field of R.

20.3. Background6.1.1

Suppose A is a local Noetherian intermediate ring dominating R (in the sensethat the maximal ideal n of A intersects R in m) and dominated by R. The localinjective morphisms R → A → R imply the existence of a canonical surjectionπ : A R, where A is the n-adic completion of A. In this setting it is well knownthat A is a topological subspace of R, i.e., π is an isomorphism, if and only if everyideal of A is closed in the topology on A defined by the powers of m. Since A isassumed to be Noetherian, π an isomorphism implies R is faithfully flat over A,and hence aR ∩ A = aA for each principal ideal aA of A, so A = L ∩ R where Lis the field of fractions of A. Thus A = R implies A = L ∩ R and there can be atmost one Noetherian 2 A with A = R for each intermediate field L between K andthe fraction field K of R.

On the other hand, if L is any intermediate field between K and K, then thering A = L ∩ R is a quasilocal domain dominating R and dominated by R. It iseasily seen that such a ring A is Hausdorff in the topology defined by the powersof its maximal ideal, and again the injective local morphisms R → A → R implythe existence of a canonical surjection π : A R, where A is the completion of A.This leads us to the question:

What subdomains A of R have the form L∩ R, where L is an intermediate fieldbetween K and K?

In considering this question, we have come to realize that it is quite broad,and that the explicit determination of L ∩ R is computationally challenging evenfor relatively simple examples of R and L. We have also discovered that for manyexcellent normal local domains R of dimension at least two, there exist intermediatefields L between K and K such that A = L∩ R fails to be a subspace of R. It canhappen thatA = L∩R is an excellent normal local domain of dimension greater thanthat of R, or even that A fails to be Noetherian. In order to exhibit such exampleswe concentrate in this paper on elements a1, . . . an ∈ m which are algebraically

2Without the assumption that A is Noetherian there are examples where bA = bR and A is

non-Noetherian with the same fraction field as bR, see for exampleB[10, Chap. III, pages 119-120,

Ex. 14].

20.3. BACKGROUND 131

independent over K and satisfy certain additional independence conditions. Thestudy of domains of the form L ∩ R is continued in Chapter

intsec21.

Here are some specific results related to the general question of the structureof L ∩ R.

(1.1) For a1, . . . an arbitrary elements of R and L = K(a1, . . . , an), the subdo-main A = L∩ R is a normal quasilocal birational extension of R[a1, . . . , an]. If theai are algebraic over R, then the structure of A is well understood; A is an etaleextension of R with completion A = R (see

R3.7[85]). But if the ai are not algebraic

over R, the situation is more complicated and the structure of A depends on theresidual behavior of the ai modulo various prime ideals of R.

(1.2) For a specific example of an excellent normal local domain to illustratethese ideas, we refer to R = k[x, y](x,y), the localization of the polynomial ring overa field k at the maximal ideal generated by the indeterminates x and y. For thisexample R = k[[x, y]]. A result of Valabrega

V[96, Proposition 3] implies that, for

R = k[x, y](x,y), if L is a field between K and the fraction field F of k[x][[y]] or ifL is a field between F and K, then A = L ∩ R is a two-dimensional regular localdomain with completion R.

(1.3) On the other hand, again considering R = k[x, y](x,y), it is well-knownthat, for each positive integer n, the formal power series ring in n variables over kcan be embedded in R = k[[x, y]]; in fact k[[x, y]] contains infinitely many analyti-cally independent elements

Abhy0[1],

AM[5],

AHW[6]. However, if a formal power series ring Sn in

n variables over k is embedded in R = k[[x, y]], and if R ⊆ Sn (so, in particular, ifSn = L∩ R for some field L between K and K), then Sn = R, so that n = 2. Moregenerally, if A is a local Noetherian ring with completion A and B is a completelocal ring such that B dominates A and A dominates B, then a well-known theoremof Cohen ( cf.

M[68, (8.4)]) implies that B = A.

(1.4) An example of Nagata shows the existence of a 3-dimensional regular localdomain D with completion D a formal power series ring in 3 variables over a fieldk of characteristic p > 0 for which there exists an intermediate field L between thefraction fields of D and D such that A = D∩L is non-Noetherian. In this example,D is not excellent and L is a finite purely inseparable extension of the fraction fieldof D. A discussion of this example is given on pages 31-32 of

HRS[33].

(1.5) An example of Ogoma shows the existence of a four-dimensional excellentregular local domain, indeed a domain D obtained as a localization of a polynomialring in 4 variables over a countable field, for which there exists a field L containedin the fraction field of D and generated by two elements over the fraction field ofD such that A = D∩L is not Noetherian. A discussion of this example is given onpages 32-34 of

HRS[33].

(1.6) InHR[32], an excellent normal local domain R is said to have the Noetherian

intermediate rings property, (NIR), if for each subfield L of K containing K, thequasilocal ring L ∩ R is Noetherian and has completion R. Interesting examplesof excellent regular local rings satisfying (NIR) are constructed in

R3.5[84] and by

Shelburne inShb[92]. 3

3We remark that Shelburne inShb[92] has provided examples answering Question 2.8 of

HR[32].

He shows existence for each positive integer d ≥ 3 of an excellent local domain R containing afield of characteristic p > 0 such that dim(R) = d, the dimension of the generic formal fiber of


(1.7) For a subfield L of K containing K, the following construction consideredin

HRS[33] is sometimes useful for obtaining information about A = L ∩ R. Suppose

(S,n) is an excellent normal local domain dominating R and q is a prime ideal ofS such that S/q ∼= R and q ∩ S = (0). Then L, the fraction field of S, embedsin the fraction field of S/q and with this identification, A = L ∩ (S/q) = L ∩ R isa quasilocal domain birationally dominating S. In §3 we present examples of thistype where A = S.

(1.8) Recent work of Heitmann inH2[55] and

H3[56] and Loepp in

Lo[63] shows the

richness of the structure of the local domains with a given completion. InH2[55],

Heitmann shows that a complete local ring T is the completion of a local uniquefactorization domain (UFD) if (i) T has depth at least 2, and (ii) no nonzero ele-ment of the prime subring of T is a zero divisor of T . In

H3[56], Heitmann proves that

very often a complete local ring is the completion of a local ring having an isolatedsingularity. In particular, for T satisfying (i) and (ii) he shows the existence of alocal UFD all of whose proper localizations are regular that has completion T . Hisconstruction is adopted by his student Loepp to obtain more examples of strangephenomena which can occur in passing from a local (Noetherian) ring to its com-pletion. A central step in Heitmann’s construction involves passing from a subringR of T to a bigger ring R′ by adjoining a kind of independent element, similar tothe residually algebraically independent elements defined below and studied in §4 ofthis article. These residually algebraically independent elements play an importantrole in his construction. Certain relations from T become satisfied in R′ (defined asa limit), but by using the residually algebraically independent elements, he is ableto control the correspondence between the height-one prime ideals of R and thoseof R′.

We now summarize the results of the present paper.

Summary 20.5. Summary of this paper: In §2 we begin our analysis withidealwise independence as defined in the abstract. We observe that τ1, . . . , τn ∈mR

are idealwise independent over R if and only if the extension R[τ1, . . . , τn] → R isweakly flat in the sense of the definition given in §2. We also show in §2 that asufficient condition for τ1, . . . , τn to be idealwise independent over R is that theextension R[τ1, . . . , τn] → R satisfies PDE (“pas d’eclatement”, or in English “noblowing up”). At the end of §2 we display in a schematic diagram the relationshipsbetween these concepts and some others, for extensions of Krull domains.

In §3 and §4 we present two methods for obtaining idealwise independent ele-ments over a countable ring R. The method in §3 is to find elements τ1, . . . , τn ∈ m,the maximal ideal of R, so that (1) τ1, . . . , τn are algebraically independent over thefraction field of R, and (2) for every prime ideal P of S = R[τ1, . . . , τn](m,τ1,...,τn)

with dim(S/P ) = n, the ideal PR is m-primary. In this case, we say that τ1, . . . , τnare primarily independent over R. If R is countable and dim(R) > 2, we show in(4.5) the existence over R of idealwise independent elements that fail to be primarilyindependent.

For every countable excellent normal local domain R of dimension at least two,we prove in Theorem 3.9 the existence of an infinite sequence τ1, τ2, . . . of elements

R is 0, and R is properly contained in its completion bR. In his examples, bR has, in fact, infinitetranscendence degree over R.

20.3. BACKGROUND 133

of R which are primarily independent over R. It follows that A = K(τ1, τ2, . . . )∩ Ris an infinite-dimensional non-Noetherian quasilocal domain. Thus, for the exampleR = k[x, y](x,y) with k a countable field, and for every positive integer n or n =∞,there exists an extension An = Ln ∩ R of R such that dim(An) = dim(R) + n. Inparticular, the canonical surjection An → R has a nonzero kernel.

In §4 we define τ ∈mR to be residually algebraically independent over R if τ isalgebraically independent over the fraction field of R and for each height-one primeideal P of R such that P∩R 6= 0, the image of τ in R/P is algebraically independentover R/(P ∩ R). We extend the concept of residual algebraic independence to afinite or infinite number of elements τ1, . . . , τn ∈ mR and observe the equivalenceof residual algebraic independence to the extension R[τ1, . . . , τn] → R satisfyingPDE.

We show that primary independence =⇒ residual algebraic independence=⇒ idealwise independence. For R of dimension two, we show that primaryindependence is equivalent to residual algebraic independence, but as remarkedabove, if R has dimension greater than two, then primary independence is strongerthan residual algebraic independence. We show in (4.7) and (4.9) the existence ofidealwise independent elements that fail to be residually algebraically independent.

In §5 we describe the three concepts of idealwise independence, residual al-gebraic independence, and primary independence in terms of certain flatness con-ditions on the embedding φ : R[τ1, . . . , τn](m,τ1,...,τn) → R. In §6 we investigatethe stability of these independence concepts under base change, composition andpolynomial extension. We prove in (6.10) the existence of uncountable excellentnormal local domains R such that R contains infinite sets of primarily independentelements.

We show in §7 that both residual algebraic independence and primary inde-pendence hold for elements over the original ring R exactly when they hold overthe Henselization Rh of R (7.2). Also idealwise independence descends from theHenselization to the ring R. If R is Henselian of dimension two, then all threeconcepts of independence are equivalent for one element τ ∈ m (Corollary 7.6).

The following diagram summarizes some relationships between the indepen-dence concepts for one element τ of m, over a local normal excellent domain (R,m).In the diagram we use “ind.” and “resid.” to abbreviate “independent” and “resid-ually algebraic”.


R Henseliandim(R) = 2

τ primarily ind.

τ resid. ind.

τ idealwise ind.

In §8 we include a diagram which displays many more relationships among theindependence concepts and other related properties.

20.4. 2. Idealwise independence, weakly flat and PDE extensions

First we describe the setting of the idealwise independent concept and we es-tablish notation to be used throughout the paper.

6.2.1 Setting and Notation 20.6. Let (R,m) be an excellent normal local do-main of dimension d with field of fractions K and completion (R, m = mR), andlet t1, . . . , tn, . . . be indeterminates over R. Suppose that τ1 = τ, τ2, . . . , τn, . . . ∈ mare algebraically independent over K. For each n ≥ 0, we consider the followinglocalized polynomial rings:

Sn = R[t1, . . . , tn](m,t1,...,tn),Rn = R[τ1, . . . , τn](m,τ1,...,τn),S∞ = R[t1, . . . , tn, . . . ](m,t1,...,tn,... ) andR∞ = R[τ1, . . . , τn, . . . ](m,τ1,...,τn,... ).Of course, Sn is R-isomorphic to Rn and S∞ is R-isomorphic to R∞ with

respect to the R-algebra homomorphism taking ti → τi for each i. When workingwith a particular n or ∞, we sometimes define S to be Rn or R∞. If n = 0, takeRn = R = Sn.

20.4. 2. IDEALWISE INDEPENDENCE, WEAKLY FLAT AND PDE EXTENSIONS 135

The completion Sn of Sn is R[[t1, . . . , tn]], and we have the following commu-tative diagram:

Sn = R[t1, . . . , tn](m,t1,...,tn)⊆−−−−→ Sn = R[[t1, . . . , tn]]

∼=y λ

yR

⊆−−−−→ S = Rn = R[τ1, . . . , τn](m,τ1,...,τn)ϕ−−−−→ R.

Here the first vertical isomorphism is the R-algebra map taking ti → τi, the restric-

tion of the R-algebra surjection λ : Sn → R where ker(λ) = (t1−τ1, . . . , tn−τn)Sn =p; note that p ∩ Sn = (0).

The central definition of this paper is the following:

6.2.2 Definition 20.7. Let (R,m) and τ1, . . . , τn ∈ m be as in the setting of (2.1).We say that τ1, . . . , τn are idealwise independent overR provided R∩K(τ1, . . . , τn) =Rn. Similarly, an infinite sequence τi∞i=1 of algebraically independent elements ofmR is idealwise independent over R if R ∩K(τi∞i=1) = R∞.

6.2.3 Remarks 20.8. (1) A subset of an idealwise independent set τ1, . . . , τn overR is also idealwise independent over R. For example, to see that τ1, . . . , τm areidealwise independent over R for m ≤ n, let K denote the quotient field of R andobserve that

R∩K(τ1, . . . , τm) = R ∩K(τ1, . . . , τn) ∩K(τ1, . . . , τm)

=R[τ1, . . . , τn](m,τ1,...,τn) ∩K(τ1, . . . , τm) = R[τ1, . . . , τm](m,τ1,...,τm).

(2) Idealwise independence is a strong property of the elements τ1, . . . , τn and ofthe embedding morphism ϕ : Rn → R. As we stated in the introduction, it is oftendifficult to compute R ∩ L when L is an intermediate field between the quotientfields of R and R. In order for R ∩ L to be an intermediate localized polynomialring Rn, there can be no new quotients in R other than those in ϕ(Rn); that is, iff/g ∈ R and f, g ∈ Rn, then f/g ∈ Rn. This does not happen, for example, if oneof the τi is in the completion of R with respect to a principal ideal; in particular,if dim(R) = 1, then there do not exist idealwise independent elements over R.

The following example illustrates Remark 2.3.2:

6.2.4 Example 20.9. Let R = Q[x, y](x,y), the localized ring of polynomials in twovariables over the rational numbers. The elements τ1 = ex−1, τ2 = ey−1, and ρ =ex−ey= τ1 − τ2 of R = Q[[x, y]] belong to completions ofR with respect to principalideals (and so are not idealwise independent). If S = R2 = Q[x, y, τ1, τ2](x,y,τ1,τ2)and L is the quotient field of S, then the elements (ex − 1)/x, (ey − 1)/y, and(ex − ey)/(x − y) are certainly in L ∩ R but not in S. A result of ValabregaV[96, Proposition 3] implies that L ∩ R is a two-dimensional regular local ring withcompletion R.

In the remainder of this section we discuss some properties of extensions ofKrull domains related to idealwise independence. (A diagram near the end of thissection displays the relationships among these properties.) We start by defining aproperty which we prove in (2.7) is satisfied by the extension φ : Rn −→ R:


6.2.5 Definition 20.10. Let A → B be an extension of Krull domains. We say thatB is a height-one preserving extension of A if for every height-one prime ideal P ofA with PB 6= B there exists a height-one prime ideal Q of B with PB ⊆ Q.

6.2.6 Remark 20.11. If A → B is an extension of Krull domains, and if A is fac-torial, or more generally, if every height-one prime ideal of A is the radical of aprincipal ideal, then B is a height-one preserving extension of A. This is clear fromthe fact that every minimal prime divisor of a principal ideal in a Krull domain isof height one.

6.2.7 Proposition 20.12. Let (R,m) and τ1, . . . , τn ∈ m be as in the setting of(2.1). Then the embedding

φ : Rn = R[τ1, . . . , τn](m,τ1,...,τn) −→ R

is a height-one preserving extension.

Proof. Consider the commutative diagram of (2.1):

Sn = R[t1, . . . , tn](m,t1,...,tn)⊆−−−−→ Sn = R[[t1, . . . , tn]]

∼=y λ

yR

⊆−−−−→ S = Rn = R[τ1, . . . , τn](m,τ1,...,τn)ϕ−−−−→ R.

Let P ⊆ Rn be a prime ideal of height one. Under the above isomorphism of Rn

with Sn, P correponds to a height-one prime ideal P0 of Sn. The extended idealP0Sn is reduced and each of its minimal prime divisors is of height one.

The minimal prime divisors Q in R of the ideal PR are in 1−1 correspondencewith the minimal prime divisors Q0 in Sn of the ideal

J = (P0, t1 − τ1, . . . , tn − τn)Sn = (P0, p)Sn

via Q0/p = Q. Since p = ker(λ) is a prime ideal of height n with p∩Sn = (0), eachQ0 is of height n+1 and each Q is of height one. Therefore Rn → R is a height-onepreserving extension.

The concept of idealwise independence is naturally related to other ideal-theoretic properties. If A → B is an extension of domains and F is the fractionfield of A, then it is well known and easily seen that A = B∩F ⇐⇒ each principalideal of A is contracted from B. For an extension A → B of Krull domains, thecondition that A = B ∩ F , where F is the fraction field of A, is related to thefollowing concepts.

6.2.8 Definition 20.13. Let A → B be an extension of Krull domains.(a) We say that B is weakly flat over A if every height-one prime ideal P of A

with PB 6= B satisfies PB ∩A = P .(b) The extension A → B is said to satisfy PDE ( “pas d’eclatement”, or “no

blowing up”) if for every height-one prime ideal Q in B, the height of Q ∩ A is atmost one (cf.

F[26, page 30]).

6.2.9 Remarks 20.14. Let A → B be an extension of Krull domains and let F bethe fraction field of A.

20.4. 2. IDEALWISE INDEPENDENCE, WEAKLY FLAT AND PDE EXTENSIONS 137

(a) We have B ∩ F = A ⇐⇒ each height-one prime of A is the contraction ofa height-one prime of B. If this holds, then B is height-one preserving and weaklyflat over A (cf.

N2[72, (33.5) and (33.6)]).

(b) If A → B is flat, then A → B is height-one preserving, weakly flat andsatisfies PDE (cf.

B[10, Chapitre 7, Proposition 15, page 19]).

(c) If S is a multiplicative system in A consisting of units of B, then A → Bis height-one preserving (respectively weakly flat, respectively satisfies PDE) ⇐⇒S−1A → B is height-one preserving (respectively weakly flat, respectively satisfiesPDE).

6.2.10 Proposition 20.15. If φ : A → B is a weakly flat extension of Krull domains,then φ is height-one preserving. Moreover, for every height-one prime ideal P of Awith PB 6= B there is a height-one prime ideal Q0 of B with Q0 ∩A = P .

Proof. Let P ∈ Spec(A) with ht(P ) = 1. By assumption PB ∩ A = P .Therefore the ideal PB of B is contained in an ideal Q of B that is maximal withrespect to not meeting the multiplicative system A − P . It follows that Q is aprime ideal of B and Q ∩ A = P . Let a ∈ P − (0) and let Q0 ⊆ Q be a minimalprime divisor of aB. Then Q0 has height one and (0) 6= Q0 ∩ A ⊆ P , and thusQ0 ∩A = P .

6.2.11 Proposition 20.16. Let A → B be an extension of Krull domains which isheight-one preserving and satisfies PDE. Then B is weakly flat over A.

Proof. Let P ∈ Spec(A) with ht(P ) = 1. Then PB is contained in a primeideal Q of B of height one. The PDE hypothesis on A → B implies that Q∩A hasheight one. It follows that Q ∩A = P and thus PB ∩A = P .

6.2.12 Corollary 20.17. Let (R,m) and τ1, . . . , τn ∈ mR be as in the setting of(2.1). Let S = R[τ1, . . . , τn](m,τ1,...,τn). If S → R satisfies PDE, then R is weaklyflat over S.

Proof. This is immediate from (2.7) and (2.11). 6.2.13 Example 20.18. Without assuming that the extension is height-one preserv-

ing, it can happen that an extension A → B of Krull domains satisfies PDE and yetB fails to be weakly flat over A. This is the case, for example, if A = k[x, y, z, w] =k[X,Y, Z,W ]/(XY − ZW ), where k is a field and X,Y, Z,W are indeterminatesover k, and B = A[x/z]. Since x/z = w/y, B = k[y, z, x/z] is a polynomial ring inthree variables over k and the height-one prime ideal P = (y, z)A extends in B toa prime ideal of height two. Another way to describe this example is to let r, s, tbe indeterminates over a field k, let A = k[r, s, rt, st] ⊂ k[r, s, t] = B. Then A → Bsatisfies PDE since B is an intersection of localizations of A, but P = (r, s)A is aheight-one prime of A such that PB is a height-two prime of B, so B is not weaklyflat over A.

6.2.14 Proposition 20.19. Let A → B be an extension of Krull domains with PB 6=B for every height-one prime ideal P of A and let F denote the fraction field of A.Then B is weakly flat over A ⇐⇒ A = F ∩ B. Moreover, in this setting, theseequivalent conditions imply that A → B is height-one preserving.

Proof. The assertion that A = F ∩ B implies B is weakly flat over A isRemark (2.9)(a). A direct proof of this assertion involving primary decomposition


of principal ideals goes as follows: Let P be a height-one prime ideal of A, leta ∈ P − (0), and consider an irredundent primary decomposition

aB = Q1 ∩ · · · ∩Qsof the principal ideal aB in the Krull domain B. Since B is a Krull domain, eachQi is primary for a height-one prime ideal Pi of B. The fact that A = F ∩B impliesthat aA = F ∩ aB. Thus, after renumbering, there is an integer t ∈ 1, . . . , s suchthat the ideal

Q1 ∩ . . . ∩Qt ∩Ais the P -primary component of the ideal aA. Hence for at least one integer i ∈1, . . . , t we must have that Pi ∩A = P . Therefore B is weakly flat over A.

Conversely, if B is weakly flat over A, then since PB 6= B, we have PB∩A = Pfor each height-one prime ideal P of A. It follows that A → D = F ∩ B andPD ∩A = P , so AP = (A − P )−1D, and D ⊆ AP for each height-one prime idealP of A. Since A =

⋂AP : P is a height-one prime of A, we have A = D.The last assertion follows by (2.9a) or (2.10).

Let A → B be an extension of Krull domains, F the quotient field of A, Q ∈Spec(B), ht(Q) = 1, P ∈ Spec(A), ht(P ) = 1. The following diagram illustrates(2.5)-(2.14):

(2.9b)

(2.11)

(2.9a)

(2.8a)

(2.5)

(2.8b)

(2.11)

(2.14)

(2.9a)

(2.10)

(2.10)

A → B flat

A → B ht-1 pres. (2.5),PDE (2.8b) and PB 6= B,∀P

A → B w.f. (2.8a)and PB 6= B,∀P

A → B ht-1 pres. (2.5),and PDE (2.8b)

B ∩ F = A ∀P, ∃Q|Q ∩A = P

A → B w.f. (2.8a) PB 6= B =⇒ PB ∩A = P

A → B PDE (2.8b)PB 6= B =⇒ ∃Q|Q ∩A = P

∀Q, ht(Q ∩A) ≤ 1 A → B ht-1 pres. (2.5) PB 6= B =⇒ ∃Q|PB ⊆ Q

The relationships between properties for extensions of Krull domains

20.5. 3. PRIMARY INDEPENDENCE 139

6.2.15 Remark 20.20. The condition in (2.14) that PB 6= B for all height-one primeideals P of A holds if A → B are quasilocal Krull domains with B dominating A,and so it holds for Rn → R as in (2.1).

Summarizing from (2.12) and (2.14), we have the following implications amongthe concepts of weakly flat, PDE and idealwise independence in the setting of (2.1):

6.2.16 Theorem 20.21. Let (R,m) and τ1, . . . , τn ∈mR be as in the setting of (2.1).Then:

(1) τ1, . . . , τn are idealwise independent over R ⇐⇒ R[τ1, . . . , τn] → R isweakly flat.

(2) R[τ1, . . . , τn] → R satisfies PDE =⇒ R[τ1, . . . , τn] → R is weakly flat.Moreover, in view of part (c) of (2.9), these assertions also hold with R[τ1, . . . , τn]replaced by its localization R[τ1, . . . , τn](m,τ1,...,τn).

In order to demonstrate idealwise independence we develop in the next twosections the concepts of primary independence and residual algebraic independence,each of which implies idealwise independence.

20.5. 3. Primary independence6.3

In this section we introduce primary independence, a concept we show to bestronger than idealwise independence (in (3.4) and (4.5) ). We construct infinitelymany primarily independent elements over any countable excellent normal localdomain of dimension at least two (in (3.9) ).

6.3.1 Definition 20.22. Let (R,m) be an excellent normal local domain. We saythat τ1, . . . , τn ∈ mR, which are algebraically independent over the fraction fieldof R, are primarily independent over R, provided that, for every prime ideal P ofS = R[τ1, . . . , τn](m,τ1,...,τn) such that dim(S/P ) ≤ n, the ideal PR is mR-primary.A countably infinite sequence τi∞i=1 of elements of mR is primarily independentover R if, for each n, τ1, . . . , τn are primarily independent over R.

6.3.2 Remarks 20.23. (1) Referring to the diagram, notation and setting of (2.1),primary independence of τ1, . . . , τn as defined in (3.1) is equivalent to the statementthat for every prime ideal P of S with dim(S/P ) ≤ n, the ideal λ−1(PR) = PSn+ker(λ) is primary for the maximal ideal of Sn.

(2) A subset of a primarily independent set is again primarily independent. Forexample, if τ1, . . . , τn are primarily independent over R, to see that τ1, . . . , τn−1 areprimarily independent, let P be a prime ideal of Rn−1 with dim(Rn−1/P ) ≤ n− 1.Then PRn is a prime ideal of Rn with dim(Rn/PRn) ≤ n, and so PR is primaryfor the maximal ideal of R.

6.3.3 Lemma 20.24. Let (R,m) be an excellent normal local domain of dimension atleast 2, let n be a positive integer, and let S = Rn = R[τ1, . . . , τn](m,τ1,...,τn), whereτ1, . . . , τn are primarily independent over R. Let P be a prime ideal of S such thatdim(S/P ) ≥ n+ 1. Then (1) PR is not mR-primary, and (2) PR ∩ S = P .

Proof. For the first statement, suppose that dim(S/P ) ≥ n+ 1 and that PRis primary for mR. Then, referring to the diagram in (2.1), λ−1(PR) = PSn+ker(λ) is primary for the maximal ideal of S, and hence the maximal ideal of S/P S


is the radical of an n-generated ideal, a contradiction because Sn/P Sn ∼= (S/P ) isthe completion of S/P , and dim(S/P ) ≥ n+ 1 implies that dim(S/P ) ≥ n+ 1.

For the second assertion, note that if dim(S/P ) = n + 1, and P < (PR ∩ S),then dim(S/(PR ∩ S) ) ≤ n, which implies that PR = (PR ∩ S)R is primary formR, a contradiction to the first assertion of the lemma. Thus we have PR∩S = Pfor each P such that dim(S/P ) = n+ 1.

If dim(S/P ) > n + 1, then P is an intersection of prime ideals P ′ of S suchthat dim(S/P ′) = n+ 1, say P = ∩P ′∈IP ′. Using the result for P ′, we have

P ⊆ PR ∩ S = (∩P ′∈IP ′)R ∩ S ⊆ ∩P ′∈I(P ′R ∩ S) = ∩P ′∈IP ′ = P.

6.3.4 Proposition 20.25. Let (R,m) be an excellent normal local domain of dimen-sion at least 2, let n be a positive integer, and let S = R[τ1, . . . , τn](m,τ1,...,τn), whereτ1, . . . , τn are primarily independent over R. Then S = L∩ R, where L is the frac-tion field of S. Thus τ1, . . . , τn are idealwise independent elements of R over R. Ifτi∞i=1 is a countably infinite sequence of primarily independent elements of mRover R, then τi∞i=1 are idealwise independent over R.

Proof. Let P be a height-one prime of S. Since S is catenary, dim(S/P ) ≥n+ 1. By (3.3.2), PR∩S = P . Therefore R is weakly flat over S and by (2.16) wehave S = L ∩ R.

6.3.5 Proposition 20.26. Let (R,m) and τ1, . . . , τn ∈ mR be as in (2.1). LetRn = R[τ1, . . . , τn](m,τ1,...,τn)

∼= Sn = R[t1, . . . , tn](m,t1,...,tn), where t1, . . . , tn areindeterminates over R. Then τ1, . . . , τn are primarily independent over R if andonly if one of the equivalent statements (1), (2) or (3) holds:

(1) For each prime ideal P of Sn such that dim(Sn/P ) ≥ n and each primeideal P of Sn minimal over PSn, the images of t1 − τ1, . . . , tn − τn inSn/P generate an ideal of height n in Sn/P .

(2) For each prime ideal P of Sn with dim(Sn/P ) ≥ n and each nonnegativeinteger i ≤ n, every prime ideal Q of Sn minimal over (P, t1−τ1, . . . , ti−1−τi−1)Sn fails to contain ti − τi.

(3) For each ideal P of Sn such that dim(Sn/P ) = n, the images of t1 −τ1, . . . , tn−τn in Sn/P Sn generate an ideal primary for the maximal idealof Sn/P Sn.

Proof. It is clear that (1) and (2) are equivalent, that (1) and (2) imply (3)and that (3) is equivalent to the primary independence of τ1, . . . , τn over R. Itremains to observe that (3) implies (1). For this, let P be a prime ideal of Snsuch that dim(Sn/P ) = n+ h, where h ≥ 0. There exist s1, . . . , sh ∈ Sn so that ifI = (P, s1, . . . , sh)Sn, then for each minimal prime Q of I we have dim(Sn/Q) = n.Item (3) implies that the images of t1 − τ1, . . . , tn − τn in Sn/QSn generate anideal primary for the maximal ideal of Sn/QSn. It follows that the images oft1 − τ1, . . . , tn − τn in Sn/ISn generate an ideal primary for the maximal ideal ofSn/ISn, and therefore that the images of s1, . . . , sh, t1 − τ1, . . . , tn− τn in Sn/P Snare a system of parameters for the (n + h)-dimensional local ring Sn/P Sn. LetP be a minimal prime of PSn. Then dim(Sn/P ) = n + h, and the images of

20.5. 3. PRIMARY INDEPENDENCE 141

s1, . . . , sh, t1 − τ1, . . . , tn − τn in the complete local domain Sn/P are a system ofparameters. It follows that the images of t1 − τ1, . . . , tn − τn in Sn/P generate anideal of height n in Sn/P .

6.3.6 Corollary 20.27. With the notation of (2.1) and (3.5) assume that τ1, . . . , τnare primarily independent over R.

(1) Let I be an ideal of Sn such that dim(S/I) = n. Then the ideal (I, t1 −τ1, . . . , tn − τn)Sn is primary to the maximal ideal of Sn.

(2) Let P ∈ Spec(Sn) be a prime ideal with dim(Sn/P ) > n. Then the idealW = (P, t1 − τ1, . . . , tn− τn)Sn has ht(W ) = ht(P ) + n and W ∩Sn = P .

Proof. Part (1) is an immediate corollary of (3.5.3) and it follows from (3.5.1)that ht(W ) = ht(P )+n. Let λn be the restriction to Sn of the canonical homomor-phism λ : Sn → R from (2.1) so that λn : Sn

∼=→→Rn. Then dim(Rn/λn(P )) > n,and so by (3.3.2), λn(P )R ∩Rn = λn(P ). Now

W ∩ Sn = λ−1(λn(P )R) ∩ λ−1n (Rn) = λ−1

n (λn(P )R ∩Rn) = λ−1n (λn(P )) = P.

To prove the existence of primarily independent elements, we use the follow-

ing prime avoidence lemma over a complete local ring (cfBu[15, Lemma 3],

WW[101,

Lemma 10]). We also use this result in two constructions given in Section 4.

6.3.7 Lemma 20.28. Let (T,n) be a complete local ring of dimension at least 2, andlet t ∈ n − n2. Assume that I is an ideal of T containing t, and that U is acountable set of prime ideals of T each of which fails to contain I. Then thereexists an element a ∈ I ∩ n2 such that t− a 6∈ ⋃Q : Q ∈ U.

Proof. Let Pi∞i=1 be an enumeration of the prime ideals of U . We mayassume that there are no containment relations between the primes of U . Choosef1 ∈ n2∩I so that t−f1 6∈ P1. Then choose f2 ∈ P1∩n3∩I so that t−f1−f2 6∈ P2.Note that f2 ∈ P1 implies t− f1 − f2 /∈ P1. Successively, by induction, choose

fn ∈ P1 ∩ P2 ∩ · · · ∩ Pn−1 ∩ nn+1 ∩ Iso that t− f1 − . . . − fn 6∈

⋃ni=1Pi for each positive integer n. Then

f1 + · · ·+fn∞n=1 is a Cauchy sequence in T which converges to an element a ∈ n2.Now

t− a = (t− f1 − . . . − fn) + (fn+1 + . . . ),where (t− f1 − . . . − fn) /∈ Pn, (fn+1 + . . . ) ∈ Pn. Therefore t− a /∈ Pn, for all n,and t− a ∈ I.

6.3.8 Remark 20.29. Let A → B be an extension of Krull domains. If α is a nonzerononunit of B and α is outside every height-one prime Q of B such that Q∩A 6= (0),then αB ∩ A = (0). In particular, such an element α is algebraically independentover the fraction field of A.

6.3.9 Theorem 20.30. Let (R,m) be a countable excellent normal local domain ofdimension at least 2. Then:

(1) There exists τ ∈mR which is primarily independent over R.(2) If τ1, . . . , τn−1 ∈ mR are primarily independent over R, then there exists

τn ∈mR such that τ1, . . . , τn−1, τn are primarily independent over R.


(3) Thus there exists an infinite sequence τ1, . . . , τn, . . . ∈ mR of elementswhich are primarily independent over R.

Proof. The proof for part (2) also establishes part (1) and part (3). Toprove (2), let t1, . . . , tn be indeterminates over R, and let the notation be as in thesetting of (2.1). Thus we have Sn−1

∼= Rn−1, under the R-algebra isomorphismtaking ti → τi. Let n denote the maximal ideal of Sn. We show the existenceof a ∈ n2 such that, if λ denotes the R-algebra surjection Sn → R with kernel(t1− τ1, . . . , tn−1− τn−1, tn− a)Sn, then τ1, . . . , τn−1 together with the image τn oftn under the map λ are primarily independent over R.

Since Sn is countable and Noetherian we can enumerate as Pj∞j=1 the prime

ideals of Sn such that dim(Sn/Pj) ≥ n. Let I = (t1 − τ1, . . . , tn−1 − τn−1)Sn−1,and let U be the set of all prime ideals of Sn = R[[t1, . . . , tn]] minimal over idealsof the form (Pj , I)Sn for some Pj ; then U is countable and n /∈ U since (Pj , I)Snis generated by n− 1 elements over PjSn and dim(Sn/PjSn) ≥ n. By Lemma 3.7with the ideal I of that lemma taken to be n, there exists an element a ∈ n2 sothat tn− a is outside Q, for every prime ideal Q ∈ U . Let τn ∈ R denote the imageof tn under the R-algebra surjection λ : Sn → R with kernel (I , tn − a)Sn. Thekernel of λ is also generated by (I , tn − τn)Sn. Therefore the setting will be as inthe diagram of (2.1) after we establish Claim

6.3.9c120.31.

6.3.9c1 Claim 20.31. (I , tn − τn)Sn ∩ Sn = (0).

Proof. of Claim 1 Since τ1, . . . , τn−1 are primarily independent, I∩Sn−1 = (0)and ISn ∩ Sn = (0). Let R′

n = Rn−1[tn](max(Rn−1)tn)). Consider the diagram:

Sn = Sn−1[tn](max(Sn−1),tn)⊂−−−−→ Sn = Sn−1[[tn]]

∼=y λ1

yR′n = Rn−1[tn](max(Rn−1),tn)

⊂−−−−→ R[[tn]] ∼= (Sn−1/I)[[tn]],

where λ1 : Sn → Sn/(ISn) is the canonical projection.For Q a prime ideal of Sn, we have Q ∈ U ⇐⇒ λ1(Q) = P , where P is a

prime ideal of R[[tn]] ∼= (Sn−1/I)[[tn]] minimal over λ1(Pj)R[[tn]] for some primeideal Pj of Sn such that dim(Sn/Pj) ≤ n. Since tn − a is outside every Q ∈ U ,tn − λ1(a) = λ1(tn − a) is outside every prime ideal P of R[[tn]], such that P isminimal over λ1(Pj)R[[tn]]. Since Sn is catenary and dim(Sn) = n + dim(R), aprime ideal Pj of Sn is such that dim(Sn/Pj) ≥ n ⇐⇒ ht(Pj) ≤ dim(R). SupposeP is a height-one prime ideal of R[[tn]] such that P ∩ R′

n = P 6= (0). Then P is aminimal prime ideal of PR[[tn]]. But also P = λ1(Q), whereQ is a height-one primeof Sn and dim(Sn/Q) = n + dim(R) − 1 ≥ n. Therefore Q ∈ Pj∞j=1. Hence bychoice of a, we have tn−λ1(a) /∈ P . By Remark 3.8, (tn−λ1(a))R[[tn]])∩R′

n = (0).Hence (I , tn − τn)Sn ∩ Sn = (0).

6.3.9c2 Claim 20.32. Let P be a prime ideal of Sn such that dim(Sn/P ) = n. Thenthe ideal (P, I, tn − τn)Sn is n-primary.

Proof. of Claim6.3.9c220.32 Let Q = P ∩ Sn−1. Either QSn = P , or QSn < P . If

QSn = P , then dim(Sn−1/Q) = n−1 and the primary independence of τ1, . . . , τn−1


implies that (Q, I)Sn−1 is primary for the maximal ideal of Sn−1. Therefore(Q, I, tn − τn)Sn = (P, I, tn − τn)Sn is n-primary in this case. On the other hand,if QSn < P , then dim(Sn−1/Q) = n. Let Q′ be a minimal prime of (Q, I)Sn−1. By(3.5), dim(Sn−1/Q

′) = 1, and hence dim(Sn/Q′Sn) = 2. The primary independenceof τ1, . . . τn−1 implies that Q′∩Sn−1 = Q. Therefore Q′Sn−1[[tn]]∩Sn = QSn < P ,so P is not contained in Q′Sn. Therefore dim(Sn/(P, I)Sn)) = 1 and our choice ofa implies that (P, I, tn − τn)Sn is n-primary.

This completes the proof of Theorem 3.9. 6.3.10 Corollary 20.33. Let (R,m) be a countable excellent normal local domain of

dimension at least 2, and let K denote the fraction field of R. Then there existτ1, . . . , τn, . . . ∈ mR such that A = K(τ1, τ2, . . .) ∩ R is an infinite-dimensionalquasilocal (non-Noetherian) domain. In particular, for k a countable field, thelocalized polynomial ring R = k[x, y](x,y) has such extensions inside R = k[[x, y]].

Proof. By (3.9.3), there exist τ1, . . . , τn, . . . ∈ mR which are primarily inde-pendent over R. It follows that A = K(τ1, τ2, . . .) ∩ R is an infinite-dimensionalquasilocal domain. In particular, A is not Noetherian.

20.6. Residual algebraic independence6.4

We introduce in this section a third concept, that of residual algebraic in-dependence. Residual algebraic independence is a stronger notion than idealwiseindependence, but is weaker than primary independence. In (4.5) we show thatover every countable normal excellent local domain (R,m) of dimension at leastthree there exists an element residually algebraically independent over R, but notprimarily independent over R. In (4.7) and (4.9) we show the existence of idealwiseindependent elements that fail to be residually algebraically independent.

6.4.1 Definition 20.34. Let (R, m) be a complete normal local domain and let Abe a Krull subdomain of R such that A → R satisfies PDE.

(1) An element τ ∈ m is residually algebraically independent with respect toR over A provided that τ is algebraically independent over the fractionfield of A and for each height-one prime P of R such that P ∩ A 6= (0),the image of τ in R/P is algebraically independent over the fraction fieldof A/(P ∩A).

(2) Elements τ1, . . . τn ∈ m are said to be residually algebraically independentover A if for each 0 ≤ i < n, τi+1 is residually algebraically independentover A[τ1, . . . , τi].

(3) An infinite sequence τi∞i=1 of elements of m is residually algebraicallyindependent over A, if τ1, . . . τn are residually algebraically independentover A for each positive integer n.

The following result shows the equivalence of residual algebraic independencefor τ over A to the PDE property for A[τ ] → R.

6.4.2 Proposition 20.35. Let (R,m),τ ∈ mR be as in the setting of (2.1) and letA be a Krull subdomain of R such that A → R satisfies PDE. Then τ is residuallyalgebraically independent with respect to R over A ⇐⇒ A[τ ] → R satisfies PDE.


Proof. Assume A[τ ] → R does not satisfy PDE. Then there exists a primeideal P of R of height one such that ht(P ∩ A[τ ]) ≥ 2. Now ht(P ∩ A) = 1,since PDE holds for A → R. Thus, with p = P ∩ A, we have pA[τ ] < P ∩ A[τ ];that is, there exists f(τ) ∈ P ∩ A[τ ] − pA[τ ], or equivalently there is a nonzeropolynomial f(x) ∈ (A/(P ∩ A) )[x] so that f(τ ) = 0 in A[τ ]/(P ∩ A[τ ]), where τdenotes the image of τ in R/P . This means that τ is algebraic over the quotientfield of A/(P ∩A). Hence τ is not residually algebraically independent with respectto R over A.

For the converse, assume that A[τ ] → R satisfies PDE and let P be a height-oneprime of R such that P∩A = p 6= 0. SinceA[τ ] → R satisfies PDE, P∩A[τ ] = pA[τ ]and A[τ ]/(pA[τ ]) canonically embeds in R/P . Since the image of τ in A[τ ]/pA[τ ]is algebraically independent over A/p, it follows that τ is residually algebraicallyindependent with respect to R over A.

6.4.3 Theorem 20.36. Let (R,m) and τ1, . . . , τn ∈mR be as in the setting of (2.1).The following statements are equivalent:

(1) The elements τ1, . . . , τn are residually algebraically independent with re-spect to R over R.

(2) For each 1 ≤ i ≤ n, if P is a height-one prime ideal of R such thatP ∩R[τ1, . . . , τi−1] 6= 0, then ht(P ∩R[τ1, . . . , τi]) = 1.

(3) R[τ1, . . . , τn] → R satisfies PDE and is weakly flat.(4) R[τ1, . . . , τn] → R satisfies PDE.

Proof. The equivalence of (1) and (2) and of (1) and (4) follows from (4.2).By (2.16) and part (c) of (2.9), (3) and (4) are equivalent.

6.4.4 Theorem 20.37. Let (R,m) and τimi=1 ⊆ m be as in the setting of (2.1),where dim(R) ≥ 2 and m is either a positive integer or m =∞.

(1) If τimi=1 is primarily independent over R, then τimi=1 is residually al-gebraically independent over R.

(2) If τimi=1 is residually algebraically independent over R, then τimi=1 isidealwise independent over R.

(3) If dim(R) = 2, then τimi=1 is primarily independent over R if and onlyif it is residually algebraically independent over R.

Proof. To prove (1), it suffices by (4.3) to show that for each positive integern ≤ m, if τ1, . . . , τn are primarily independent over R, then R[τ1, . . . , τn] → Rsatisfies PDE. Let S = R[τ1, . . . , τn](m,τ1,...,τn) and let the notation be as in thediagram of (2.1).

Let P be a height-one prime ideal of R with P = P∩R 6= (0). Consider the idealW = (P, t1− τ1, . . . , tn− τn)Sn. Using diagram (2.1) we see that λ(W ) = PR ⊆ P .By Corollary 3.6.2, ht(W ) = ht(P )+n. But W ⊆ (P , t1−τ1, . . . , tn−τn) = λ−1(P )and thus

1 + n ≤ ht(P ) + n = ht(W ) ≤ ht(λ−1(P )) ≤ ht(P ) + n = 1 + n.

Therefore ht(P ) = 1.The proof of (2) follows from (4.3) and (2.16).In view of (1), to prove (3), we assume that dim(R) = 2 and n ≤ m is a

positive integer such that τ1, . . . , τn are residually algebraically independent over


R. Let S = R[τ1, . . . , τn](m,τ1,...,τn). By (4.2), S → R satisfies PDE. Let P be aprime ideal of S such that dim(S/P ) ≤ n. Since dim(S) = n+ 2 and S is catenary,it follows that ht(P ) ≥ 2. To show τ1, . . . , τn are primarily independent over R, itsuffices to show that PR is primary for the maximal ideal of R. Since dim(R) = 2,this is equivalent to showing P is not contained in a height-one prime of R, andthis last statement holds since S → R satisfies PDE.

6.4.5 Proposition 20.38. If (R,m) is a countable excellent normal local domainof dimension at least 3, then there exists an element τ ∈ mR which is residuallyalgebraically independent over R, but not primarily independent over R.

Proof. We modify the proof of (3.9). Let t be an indeterminate over R andset S1 = R[t](m,t) so that S1 = R[[t]]. Let Q0 be a height-three prime of S1 thatcontains t and is such that Q0 = Q0 ∩ S1 also has height three. Using Lemma3.7 with I = Q0, there exists a ∈ Q0 ∩ n1

2, where n1 is the maximal ideal of S1,so that t − a ∈ Q0 but t − a is not in any of the other height-three prime idealsof S1 that are minimal over a height-three prime of S1. Let λ be the surjectionS1 → R with kernel (t − a)S1. Then the image τ ∈ mR of t under λ : S1 → Ris not primarily independent because the prime ideal λ(Q0) in S = R[τ ](m,τ) isof height three and is the contraction to S of the prime ideal λ(Q0 of R. Since(t − τ)S1 = (t − a)S1 ⊆ Q0, λ(Q0 is of height two. Therefore τ is not primarilyindependent.

We prove that τ is residually algebraically independent over R: If P is a height-one prime ideal of R with P ∩R 6= 0, then the height of P ∩R is 1 and so the heightof P ∩ S is at most 2. Also λ−1(P ) = Q in S1 has height two—since it’s generatedby the inverse images of the generators of P and ker(λ) = (t− a)S1.

Suppose that the height of P ∩ S = 2. Then under the R-isomorphism ofS1 to S taking t to τ , P ∩ S corresponds to a height-two prime P of S1. Wehave P ⊆ Q ∩ S1 and since S1 is flat over S1, the height of Q ∩ S1 is at mosttwo, so we have P = Q ∩ S1. Let n1 denote the maximal ideal of S1, and chooseb ∈ n1−(P ∪Q0) and a prime ideal Q1 in S1 minimal over (Q, b)S1. Since b /∈ Q, wesee that ht(Q1) = 3 and ht(Q1 ∩S1) = 3, because it properly contains P = Q∩S1.We have:

In S1:Q0 6= Q1 ∩ S1 (ht 3) ⊆−−−−→ Q1min←−−−− (b, Q)S1 (ht. 3)

∪|x ∪|

xIn S1 : P = Q ∩ S1(ht 2)

⊆−−−−→ Q = λ−1(P )⊇←−−−− (P , (t− a))S1 (ht 2)

∼=y λ

yIn S : P ∩ S (ht 2) ⊆−−−−→ P (ht 1 in R).

But then Q1 is minimal over a height-three prime (Q1 ∩ S1) of S1 and t− a ∈ Q1.This implies that Q1 = Q0, and so Q1 ∩ S1 = Q0 ∩ S1 = Q0, a contradiction sinceb 6∈ Q0. We conclude that ht(P ∩ S) = 1 and that τ is residually algebraicallyindependent over R.


6.4.6 Example 20.39. The following construction, similar to that in (4.5), showsthat condition (2) in Definition 4.1 is stronger than

(2′) For each height-one prime ideal P of R with P ∩ R 6= 0, the images ofτ1, . . . , τn in R/P are algebraically independent over R/(P ∩R).

6.4.6c Construction 20.40. Let R be a countable excellent local unique factoriza-tion domain (UFD) of dimension two, for example R = Q[x, y](x,y). As in (3.9),construct τ1 ∈ mR primarily independent over R (or equivalently residually alge-braically independent in this context). Let S2 = R[t1, t2](m,t1,t2), let n denote themaximal ideal of S2, and let U = prime ideals Q of S2 minimal over some idealof form (P, t1 − τ1)S2 where P is a prime ideal of S2 with dim(S2/P ) ≥ 2 andP 6= (t1, t2)S2. Note that all the prime ideals in U have height at most 3 and theideal I = (t1, t2, t1− τ1)S2 is not contained in any prime ideal in U . By Lemma 3.7,we can choose a ∈ n2 ∩ I so that t2 − a /∈

⋃Q : Q ∈ U . Let τ2 be the image oft2 under the R-algebra surjection λ : S2 → R with kernel (t1 − τ1, t2 − a)S2; thenker(λ) has height two. As before, set Ri = R[τ1, τi](m,τ1,τi), for i = 1, 2.

6.4.6cc1 Claim 20.41. τ1, τ2 do not satisfy (2) of Definition 4.1.

Proof. (of Claim6.4.6cc120.41) Let Q be a prime ideal of S2 which is minimal over

(t1, t2, t1−τ1, t2−a)S2. Then by the choice of a, Q is minimal over (t1, t2, t1−τ1)S2.Therefore ht(Q) ≤ 3 and Q ⊇ ker(λ). Let P = λ(Q) in R; then ht(P ) ≤ 1. In factht(P ) = 1 since 0 6= τ1 = λ(t1) ∈ P . Since τ1 is residually algebraically independentover R, ht(P ∩ R1) ≤ 1. But τ1 ∈ P ∩ R1, so ht(P ∩ R1) = 1 and P ∩ R = (0).Now also τ2 = λ(t2) ∈ P ; thus τ1, τ2 ∈ P ∩ R2, so ht(P ∩ R2) ≥ 2. Thus (2) failsby (4.2).

6.4.6cc2 Claim 20.42. τ1, τ2 satisfy (2′) above.

Proof. (of Claim6.4.6cc220.42) Suppose P is a height-one prime ideal of R with

P ∩ R 6= (0) and let Q = λ−1(P ). Then ht(Q) = 3 and ht(P ∩ R) = 1. By theresidual algebraic independence of τ1 over R, ht(P ∩R1) = 1, and so ht(P ∩R2) ≤ 2.If ht(P ∩R2) = 1, we are done. Suppose ht(P ∩R2) = 2. We have:

Q ∩ S1⊆−−−−→ Q ∩ S2

⊆−−−−→ Q = λ−1(P )⊆−−−−→ S2

∼=y ∼=

y λ

y λ

yP ∩R ⊆−−−−→ P ∩R1

⊆−−−−→ P ∩R2⊆−−−−→ P

⊆−−−−→ R.

Thus Q∩S2 = P is a prime ideal of height 2, and ht(Q∩S1) = 1. Also P 6= (t1, t2)S2

because (t1, t2)S2 ∩ R = (0). But this means that Q ∈ U since Q is minimal over(P, t1 − τ1)S2 where P is a prime of S2 with dim(S2/P ) = 2 and P 6= (t1, t2)S2.This contradicts the choice of a and establishes that (2′) holds.

Following a suggestion of the referee, we present in (4.7) a method to obtain anidealwise independent element that fails to be residually algebraically independent.

6.4.7 Proposition 20.43. Let (R,m) be a countable excellent local UFD of dimen-sion at least two. Assume there exists a height-one prime P of R such that P iscontained in at least two distinct height-one primes P and Q of R. Also assume


that P is not the radical of a principal ideal in R. Then there exists τ ∈ mR thatis idealwise independent but not residually algebraically independent over R.

Proof. Let t be an indeterminate over R and set S1 = R[t](m,t) so that S1 =R[[t]]. Let n1 denote the maximal ideal of S1.

Using Lemma 3.7 with I = (P , t)S1 and U = p ∈ Spec(S1) |p 6= I, ht(p) ≤2, and p minimal over p ∩ S1, there exists a ∈ (P , t)S1 ∩ n1

2, such that t − a /∈⋃p |p ∈ U, but t − a ∈ (P , t)S1. That is, if t − a ∈ p, for some prime idealp 6= (P , t)S1 of S1 with ht(p) ≤ 2, then ht(p) > ht(p∩S1). Let λ be the surjectionS1 → R with kernel (t − a)S1. By construction, (t − a)S1 ∩ S1 = (0). Thereforethe restriction of λ to S1 maps S1 isomorphically onto S = R[τ ](m,τ), where λ(t) =τ ∈mR is algebraically independent over the fraction field of R.

That τ is not residually algebraically independent over R follows because theprime ideal λ((P, t)S1) = (P, τ)S has height two and is the contraction to S ofthe prime ideal λ((P , t)S1) = P of R. Since (t − τ)S1 = (t − a)S1 ⊆ (P , t)S1,λ((P , t)S1) has height one and equals P . Therefore τ is not residually algebraicallyindependent over R.

Our choice of t− a insures that each height-one prime q other than P of R hasthe property that ht(q ∩ S) ≤ 1. We show that τ is idealwise independent over Rby showing each height-one prime of S is the contraction of a height-one prime ofR. Let φ : S1 → S denote the restriction of λ. For q a height-one prime of S, letq1 := φ−1(q) denote the corresponding height-one prime of S1. Then (q1, t− a)S1

is an ideal of height two. Let w1 be a height-two prime of S1 containing (q1, t− a).If q1 is not contained in (P , t)S1, then by the choice of t− a, w1 ∩S1 has height atmost one. Therefore w1 ∩ S1 = q1. Let w = λ(w1). Then w is a height-one primeof R and w ∩ S = q.

Therefore each height-one prime q of S such that q1 := φ−1(q) is not containedin (P , t)S1 is the contraction of a height-one prime of R. Since λ((P , t)S1) ∩ S =(P, τ)S, it remains to consider height-one primes q of S such that q ⊆ (P, τ)S.By construction we have PS = Q ∩ S. Let q be a height-one prime of S suchthat q 6= PS and q ⊆ (P, τ)S. Since R is a UFD, S is a UFD and q = fS foran element f ∈ q. Since P is not the radical of a principal ideal, there exists aheight-one prime q 6= P of R such that f ∈ q. Since ht(q ∩ S) ≤ 1, we haveq ∩ S = fS = q. Therefore τ is idealwise independent over R.

6.4.8 Remark 20.44. A specific example of a countable excellent local UFD havinga height-one prime P satisfying the conditions in (4.7) is R = k[x, y, z](x,y,z), wherek is the algebraic closure of the field Q and z2 = x3 + y7. That R is a UFD isshown in

Sam[88, page 32]. Since z − xy is an irreducible element of R, the ideal

P = (z − xy)R is a height-one prime of R. It is observed inHL[31, pages 300-301]

that in the completion R of R there exist distinct height-one primes P and Q lyingover P . Moreover, the blowup of P has a unique exceptional prime divisor and thisexceptional prime divisor is not on the blowup of an m-primary ideal. ThereforeP is not the radical of a principal ideal of R.

In (4.9) we present an alternative method to obtain idealwise independent ele-ments that are not residually algebraically independent.


6.4.9 Proposition 20.45. Let (R,m) be a countable excellent local UFD of dimen-sion at least two. Assume there exists a height-one prime P0 of R such that P0 iscontained in at least two distinct height-one primes P and Q of R. Also assumethat the Henselization (Rh,mh) of R is a UFD. Then there exists τ ∈ mR that isidealwise independent but not residually algebraically independent over R.

Proof. Since R is excellent, P := P ∩Rh and Q := Q∩Rh are distinct height-one primes of Rh with PR = P , and QR = Q. Let x ∈ Rh be such that xRh = P .Theorem 3.9 implies there exists y ∈mR that is primarily independent and henceresidually algebraically independent over Rh.

We show that τ = xy is idealwise independent but not residually algebraicallyindependent over R. Since x is nonzero and algebraic over R, xy is algebraicallyindependent over R. Let S = R[xy](m,xy). Then S is a UFD and P ∩S = xR∩S ⊇(P0, xy)S has height at least two in S. Therefore by (4.3), xy is not residuallyalgebraically independent over R.

Since y is idealwise independent over Rh, every height-one prime of the polyno-mial ring Rh[y] contained in the maximal ideal n = (mh, y)Rh[y] is the contractionof a height-one prime of R. To show xy is idealwise independent over R, it sufficesto show every prime element w ∈ (m, xy)R[xy] is such that wR[xy] is the con-traction of a height-one prime of Rh[y] contained in n. If w 6∈ (P, xy)Rh[xy], thenthe constant term of w as a polynomial in Rh[xy] is in mh − P . Thus w ∈ n andw 6∈ xRh[y]. Since Rh[xy][1/x] = Rh[y][1/x] and xRh[y] ∩Rh[xy] = (x, xy)Rh[xy],it follows that there is a prime factor u of w in Rh[xy] such that u ∈ n − xRh[y].Then uRh[y] is a height-one prime of Rh[y] and uRh[x] ∩Rh[xy] = uRh[xy]. SinceRh[xy] is faithfully flat over R[xy], it follows that uRh[y] ∩R[xy] = wR[xy].

We have QRh[xy] = QRh[y]∩Rh[xy] and QRh[xy]∩R[xy] = P0R[xy]. Thus itremains to show, for a prime element w ∈ (m, xy)R[xy] such that w ∈ (P, xy)Rh[xy]and wR[xy] 6= P0R[xy], that wR[xy] is the contraction of a height-one prime of Rh

contained in n. Since (P, xy)Rh[xy] ∩ R[xy] = (P0, xy)R[xy], it follows that w isa nonconstant polynomial in R[xy] and the constant term w0 of w is in P0. Inthe polynomial ring Rh[y] we have w = xnv, where v 6∈ xRh[y]. If v0 denotes theconstant term of v as a polynomial in Rh[y], then xnv0 = w0 ∈ P0 ⊆ R impliesxnv0 ∈ Q ⊆ Rh. Since x ∈ Rh − Q, we must have v0 ∈ Q and hence v ∈ n. Alsov 6∈ xRh[y] implies there is a height-one prime ideal v of Rh[y] with v ∈ v and x 6∈ v.Then, since Rh[y]v is a localization of Rh[xy], v ∩Rh[xy] is a height-one prime ofRh[xy] that is contained in (mh, xy)Rh[xy]. It follows that v ∩ Rh[xy] = wRh[xy]which completes the proof of (4.9).

6.4.10 Remark 20.46. For a specific example of (4.9), take R to be the localizedpolynomial ring in two variables over a countable field k where k has characteristicnot equal to 2, say R = k[s, t](s,t). Then P0 = (s2 − t2 − t3)R is a height-one primeof R and P0R = (s2−t2−t3)k[[s, t]] is the product of two distinct height-one primesof R.

20.7. Idealwise independence and flatness6.5

This section contains more results relating idealwise independence, residualalgebraic independence, and primary independence. We describe all three notionsin terms of flatness of certain localizations of the canonical embedding φ : Rn =

20.7. IDEALWISE INDEPENDENCE AND FLATNESS 149

R[τ1, . . . , τn](m,τ1,...,τn) → R. We start with an easy characterization of weakly flatand PDE morphisms.

6.5.1 Proposition 20.47. Let φ : A −→ B be an injective morphism of Krull do-mains.

(1) φ is weakly flat ⇐⇒ for every height-one prime ideal P ∈ Spec(A)such that PB 6= B there is a height-one prime ideal Q ∈ Spec(B) withP ⊆ Q ∩A such that the induced morphism on the localizations

φQ : AQ∩A −→ BQ

is faithfully flat.(2) φ satisfies PDE if and only if for every prime ideal Q ∈ Spec(B) with

ht(Q) = 1 the induced morphism on the localizations


is faithfully flat.

Proof. We use in both (1) and (2) that for each height-one prime P ∈ Spec(A)the induced morphism φP : AP −→ (A − P )−1B is flat (a domain extension of aDVR is always flat); and φP is faithfully flat ⇐⇒ P (A − P )−1B 6= (A − P )−1Bwhich is equivalent to the existence of a prime in B lying over P in A.

For the proof of (1), to see ( ⇐= ), we use that φQ a faithfully flat morphismimplies φQ satisfies the going-down property (see (5.5.1)). Hence Q∩A is of heightone, so P = Q ∩A, and thus PB ∩ A = P . For ( =⇒ ), suppose P ∈ Spec(A) hasheight one and φ is weakly flat. Then (2.10) implies the existence of Q ∈ Spec(B)of height one such that Q ∩A = P . Since BQ is a localization of (A − P )−1B, wesee that φQ is faithfully flat.

For the proof of (2), ( =⇒ ) is clear by the remark above, and ( ⇐= ) followsfrom the fact that a faithfully flat morphism satisfies the going-down property.

6.5.2 Corollary 20.48. Let (R,m) and τ1, . . . , τn ∈ m be as in the setting of (2.1),and let φ : Rn = R[τ1, . . . , τn](m,τ1,...,τn) → R denote the canonical embedding.Then:

(1) ”(1)” τ1, . . . , τn are idealwise independent over R if and only if for everyheight-one prime ideal P of Rn there is a prime ideal Q ⊆ R with Q∩Rn =P such that the induced morphism of the localizations

φ bQ : (Rn)P −→ R bQis faithfully flat.

(2) ”(2)” τ1, . . . , τn are residually algebraic independent over R if and onlyif for every height-one prime ideal Q ⊆ R the induced morphism of thelocalizations

φ bQ : (Rn) bQ∩Rn−→ R bQ

is faithfully flat.

In order to describe primary independence in terms of flatness of certain local-izations of the embedding φ : Rn −→ R, we introduce the following definition:

6.5.3 Definition 20.49. Let φ : A −→ B be an injective morphism of commutativerings and let k ∈ N be an integer with 1 ≤ k ≤ d = dim(B) where d is an integer or


d =∞. Then φ is called locally flat in height k - LFk for short - if for every primeideal Q ∈ Spec(B) with ht(Q) ≤ k the induced morphism on the localizations


is faithfully flat.

The following proposition is an immediate consequence of (5.1):

6.5.4 Proposition 20.50. Let φ : A −→ B be an injective morphism of Krull do-mains. Then φ satisfies PDE if and only if φ satisfies LF1.

6.5.5 Remarks 20.51. We use the following results on flatness.(1) Let φ : A −→ B be an injective morphism of commutative rings. Suppose

that φ satisfies LFk. Then for every Q ∈ Spec(B) with ht(Q) ≤ k we have thatht(Q ∩A) ≤ ht(Q)

M1[66, Theorem 4, page 33].

(2) Let A be a Noetherian ring, I an ideal in A, and M an I-adically ideal-separated A-module. Then M is A-flat ⇐⇒ (i) M/IM is (A/I)-flat, and (ii)I ⊗AM ∼= IM

M[68, part (1) ⇐⇒ (3) of Theorem 22.3].

6.5.6 Theorem 20.52. Let (R,m) and τ1, . . . , τn ∈ m be as in the setting of (2.1).Suppose that dim(R) = d. Then:

(1) The elements τ1, . . . , τn are residually algebraically independent over R⇐⇒ φ : Rn = R[τ1, . . . , τn](m,τ1,...,τn) −→ R satisfies LF1.

(2) The elements τ1, . . . , τn are primarily independent over R ⇐⇒ φ : Rn =R[τ1, . . . , τn](m,τ1,...,τn) −→ R satisfies LFd−1.

Proof. For (1) apply (5.4) and (4.2). To prove =⇒ in (2), let Q ∈ Spec(R)with ht(Q) ≤ d− 1. Put Q = Q ∩Rn and P = Q ∩R = Q ∩R. We show that theinduced morphism

φ bQ : (Rn)Q −→ R bQ

is faithfully flat. By (5.5.2), we have to verify two conditions:

(1) ”(a)” The morphism φ bQ : (Rn/PRn)Q −→ (R/P R) bQ is faithfully flat.

(2) ”(b)” P (Rn)Q ⊗(Rn)QR bQ ∼= PR bQ.

Proof. (of (a)) We observe that the ring (Rn/PRn)Q is a localization ofthe polynomial ring k(P )[τ1, . . . , τn] where k(P ) = RP /PRP . Hence the ring(Rn/PRn)P is regular and so is the ring (R/P R) bQ, since R is excellent. In par-

ticular, the ring (R/P R) bQ is Cohen-Macaulay, andM[68, Theorem 23.1] applies.

Therefore we need to show only the following dimension formula:

dim(R/P R) bQ = dim(Rn/PRn)Q + dim(R/QR) bQ.

Since QR is contained in Q and ht(Q) ≤ d− 1, primary independence implies thatdim(Rn/Q) > n. (If dim(Rn/Q) ≤ n, then QR is mR-primary.)

By Corollary 3.6.2, every minimal prime divisor W ∈ Spec(R) of QR hasht(W ) = ht(Q). Let W ∈ Spec(R) be a minimal prime divisor of QR contained in

20.7. IDEALWISE INDEPENDENCE AND FLATNESS 151

Q. Then:

dim(R/QR) bQ = dim(R bQ)− ht(QR bQ)

= dim(R bQ)− ht(W )

= dim(R bQ)− ht(Q(Rn)Q)

= dim(R bQ)− ht(PR bQ)− (ht(Q(Rn)Q)− ht(P (Rn)Q))

= dim( (R/P R) bQ)− dim( (Rn/PRn)Q).

Proof. of (b) Since RP −→ (R− P )−1(Rn) is a flat extension we have that

P (Rn)Q ∼= PRP ⊗RP (Rn)Q.

Therefore

P (Rn)Q ⊗(Rn)QR bQ ∼= (PRP ⊗RP (Rn)Q)⊗(Rn)Q

R bQ ∼= PRP ⊗RP R bQ ∼= PR bQwhere the last isomorphism is implied by the flatness of the canonical morphismRP −→ R bQ.

For ⇐= of (2), let P ∈ Spec(Rn) be a prime ideal with dim(Rn/P ) ≤ n.Suppose that PR is not m-primary and let Q ⊇ PR be a minimal prime divisor ofPR. Then ht(Q) ≤ d−1. Put Q = Q∩Rn, then LFd−1 implies that the morphism

φ bQ : (Rn)Q −→ R bQis faithfully flat. Hence by going-down (5.5.1), ht(Q) ≤ d− 1. But P ⊆ Q and Rnis catenary, so d− 1 ≥ ht(Q) ≥ ht(P ) ≥ d, a contradiction.

6.5.7 Remark 20.53. The results above yield a different proof of statements (1)and (3) of Theorem 4.4, that primarily independent elements are residually alge-braically independent and that in dimension two, the two concepts are equivalent.Considering again our basic setting from (2.1), with d = dim(R), Theorem 5.6equates the LFd−1 condition on the extension Rn = R[τ1, . . . , τn](m,τ1,...,τn) −→ R,to the primary independence of the τi. Also Proposition 5.4 and Theorem 4.3yield that residual algebraic independence of the τi is equivalent to the extensionRn = R[τ1, . . . , τn](m,τ1,...,τn) −→ R satisfying LF1. Clearly LFi =⇒ LFi−1, fori > 1, and if d = dim(R) = 2, then LFd−1 = LF1.

6.5.8 Remark 20.54. In the setting of (2.1), if τ1, . . . , τn are primarily independentover R and dim(R) = d, then φ : Rn −→ R satisfies LFd−1, but not LFd, that is,φ fails to be faithfully flat. (Faithful flatness would imply going-down and hencedim(Rn) ≤ d = dim(R).)

6.5.9 Remark 20.55. By a modification of Example 2.13, it is possible to obtain,for each integer d ≥ 2, a local injective morphism φ : (A,m) −→ (B,n) of normallocal Noetherian domains with B essentially of finite type over A, φ(m)B = n,and dim(B) = d such that φ satisfies LFd−1, but fails to be faithfully flat overA. Let k be a field and let x1, . . . , xd, y be indeterminates over k. Let A bethe localization of k[x1, . . . , xd, x1y, . . . , xdy] at the maximal ideal generated by(x1, . . . , xd, x1y, . . . , xdy), and let B be the localization of A[y] at the prime ideal


(x1, . . . , xd)A[y]. Then A is an d+1-dimensional normal local domain and B is an d-dimensional regular local domain birationally dominating A. For any nonmaximalprime Q of B we have BQ = AQ∩A. Hence φ : A −→ B satisfies LFd−1, but φ isnot faithfully flat since dim(B) < dim(A). However, this example of a local LFk-morphism which fails to be faithfully flat also fails to be height-one preserving. AsProposition 2.7 shows the morphisms studied in this paper are automatically height-one preserving, and we believe that this condition is central for our investigations.We do not have an example of a local algebra extension essentially of finite typewhich is both LFk and height-one preserving, but fails to be faithfully flat.

20.8. Composition, base change and polynomial extensions6.6.

In this section we investigate idealwise independence, residual algebraic inde-pendence, and primary independence under polynomial ring extensions and local-izations of these polynomial extensions.

We start with a more general situation. Let

C

A B

ψ→ψ

φ→φ→

be a commutative diagram of commutative rings and injective morphisms. We seein (6.1) that many of the properties of injective morphisms we consider are stableunder composition of morphisms:

6.6.1 Proposition 20.56. Let φ : A −→ B and ψ : B −→ C be injective morphismsof commutative rings.

(1) If φ and ψ satisfy LFk, then ψφ satisfies LFk.(2) If C is Noetherian, ψ is faithfully flat and the composite map ψφ satisfies

LFk, then φ satisfies LFk.(3) Let A,B and C be Krull domains. Assume that for each height-one prime

Q of B,QC 6= C. If φ and ψ are height-one preserving (respectively weaklyflat), then ψφ is height-one preserving (respectively weakly flat).

Proof. The first statement follows from the fact that a flat morphism satisfiesgoing-down

M1[66, Theorem 4, page 33]. For (2), since C is Noetherian and ψ is

faithfully flat, B is Noetherian. Let Q ∈ Spec(B) with ht(Q) = d ≤ k. We showφQ : AQ∩A −→ BQ is faithfully flat. By localization of B and C at B − Q, wemay assume that B is local with maximal ideal Q. Since C is faithfully flat overB, QC 6= C. Let Q′ ∈ Spec(C) be a minimal prime of QC. Since C is Noetherianand B is local with maximal ideal Q, we have ht(Q′) ≤ d and Q′ ∩ B = Q. Sincethe composite map ψφ satisfies LFk, the composite map

AQ′∩A = AQ∩AφQ→ −→BQ = BQ′∩B

ψQ′→ −→CQ′

is faithfully flat. This and the faithful flatness of ψQ′ : BQ′∩B −→ CQ′ implies thatφQ is faithfully flat

M1[66, (4.B) page 27].

For (3), let P be a height-one prime of A such that PC 6= C. Then PB 6= Bso if φ and ψ are height-one preserving then there exists a height-one prime Q of

20.8. COMPOSITION, BASE CHANGE AND POLYNOMIAL EXTENSIONS 153

B such that PB ⊆ Q. By assumption, QC 6= C (and ψ is height-one preserving),so there exists a height-one prime Q′ of C such that QC ⊆ Q′. Hence PC ⊆ Q′.

Now if φ and ψ are weakly flat, by (2.10) there exists a height-one prime Q ofB such that Q ∩A = P . Again by assumption, QC 6= C, thus weakly flatness of ψimplies QC ∩B = Q. Now

P ⊆ PC ∩A ⊆ QC ∩A = QC ∩B ∩A = Q ∩A = P.

6.6.2 Remarks 20.57. If in (6.1.3) the Krull domains B and C are quasilocal and ψis a local morphism, then clearly QC 6= C for each height-one prime Q of B.

If a morphism λ of Krull domains is faithfully flat, then λ is a height-onepreserving, weakly flat morphism which satisfies condition LFk for every integerk ∈ N. Thus if φ : A −→ B and ψ : B −→ C are injective morphisms of Krulldomains, such that one of φ or ψ is faithfully flat and the other is weakly flat(respectively height-one preserving or satisfies LFk), then the composition ψφ isagain weakly flat (respectively height-one preserving or satisfies LFk). Moreover, ifthe morphism ψ is faithfully flat, we also obtain the following converse to (6.1.3) :

6.6.3 Proposition 20.58. Let φ : A −→ B and ψ : B −→ C be injective morphismsof Krull domains. Suppose that the morphism ψ is faithfully flat. If ψφ is height-onepreserving (respectively weakly flat), then φ is height-one preserving (respectivelyweakly flat).

Proof. Suppose that P is a height-one prime ideal of A such that PB 6= B.Since ψ is faithfully flat, PC 6= C, so if ψφ is height-one preserving, then thereexists a height-one prime ideal Q′ of C containing PC. Now Q = Q′∩B has heightone by going-down for flat extensions, and PB ⊆ Q′ ∩ B = Q, so φ is height-onepreserving. The proof of the weakly flat statement is similar, using (2.10).

Next we consider a commutative square of commutative rings and injectivemorphisms:

A′ φ′−−−−→ B′

µ

x ν

xA

φ−−−−→ B

6.6.4 Proposition 20.59. In the diagram above, suppose that µ and ν are faithfullyflat. Let k ∈ N. Then:

(1) (Ascent) Suppose B′ = B⊗AA′, or a localization of B⊗AA′, and ν is thecanonical morphism associated with this tensor product. If φ : A −→ Bsatisfies LFk, then φ′ : A′ −→ B′ satisfies LFk.

(2) (Descent) If B′ is Noetherian and φ′ : A′ −→ B′ satisfies LFk, thenφ : A −→ B satisfies LFk.

(3) (Descent) Suppose that the rings A,A′, B and B′ are Krull domains. Ifφ′ : A′ −→ B′ is height-one preserving (respectively weakly flat), thenφ : A −→ B is height-one preserving (respectively weakly flat).

Proof. For (1), assume that φ satisfies LFk and let Q′ ∈ Spec(B′) withht(Q′) ≤ k. Put Q = (ν)−1(Q′), P ′ = (φ′)−1(Q′), and P = µ−1(P ′) = φ−1(Q) and


consider the commutative diagrams:

A′ φ′−−−−→ B′ A′

P ′φ′

Q′−−−−→ B′Q′

µ

x ν

x µP ′x νQ′

xA

φ−−−−→ B APφQ−−−−→ BQ

The flatness of ν implies that ht(Q) ≤ k and so by assumption, φQ is faithfullyflat. The ring B′

Q′ is a localization of BQ ⊗AP A′P ′ and BQ is faithfully flat over

AP implies B′Q′ is faithfully flat over A′

P ′ .For (2), by (6.1.1) or ????

6.1.1s20.1 , φ′µ = νφ satisfies LFk. Now by (6.1.2), φ

satisfies LFk.Item (3) follows immediately from the assumption that µ and ν are faithfully

flat morphisms and hence going-down holdsM1[66, Theorem 4, page 33].

Next we examine the situation for polynomial extensions.

6.6.5 Proposition 20.60. Let (R,m) and τimi=1 ⊆ m be as in the setting of (2.1),where m is either an integer or m =∞, and the dimension of R is at least 2. Letz be an indeterminate over R. Then:

(1) τimi=1 is residually algebraically independent over R ⇐⇒ τimi=1 isresidually algebraically independent over R[z](m,z).

(2) If τimi=1 is idealwise independent over R[z](m,z), then τimi=1 is idealwiseindependent over R.

Proof. Let n ∈ N be an integer with n ≤ m. SetRn = R[τ1, . . . , τn](m,τ1,...,τn).Let φ : Rn −→ R and µ : Rn −→ Rn[z] be the inclusion maps. We have the follow-ing commutative diagram:

Rn[z](max(Rn),z)φ′

−−−−→ R′ = R[z]( bm,z)

µ

x µ′x

Rnφ−−−−→ R

The ring R′ is a localization of the tensor product R ⊗Rn Rn[z] and Proposition6.4 applies. Thus, for (1), φ satisfies LF1 if and only if φ′ satisfies LF1. Sincethe inclusion map ψ of R′ = R[z]( bm,z) to its completion R[[z]] is faithfully flat, weobtain equivalences:

φ satisfies LF1 ⇐⇒ φ′ satisfies LF1 ⇐⇒ ψφ′ satisfies LF1.(2) If the τi are idealwise independence over R[z](m,z), the morphism ψφ′ is

weakly flat. Thus φ′ is weakly flat and the statement follows by (6.1). We also obtain:

6.6.6 Proposition 20.61. Let A → B be an extension of Krull domains such thatfor each height-one prime P ∈ Spec(A) we have PB 6= B, and let Z be a (possiblyuncountable) set of indeterminates over A. Then A → B is weakly flat if and onlyif A[Z] → B[Z] is weakly flat.

Proof. Let F denote the fraction field of A. By (2.14), the extension A → Bis weakly flat if and only if F ∩ B = A. Thus the assertion follows from F ∩ B =A ⇐⇒ F (Z) ∩B[Z] = A[Z].

20.8. COMPOSITION, BASE CHANGE AND POLYNOMIAL EXTENSIONS 155

6.6.7 Remark 20.62. It would be interesting to know whether the converse of (6.5.2)is true. It is unclear that a localization of a weakly flat morphism is again weaklyflat, that isDoes there exist a weakly flat morphism φ : A −→ B of Krull domains and aheight-one prime P ∈ Spec(A) such that PB has a minimal prime divisor Q withht(Q) > 1?

If so, the map A −→ BQ fails to be weakly flat. Note that if P is the radicalof a principal ideal, then each minimal prime divisor of PB is of height one.

6.6.8 Remark 20.63. Primary independence never lifts to polynomial rings. Tosee that τ ∈ mR fails to be primarily independent over R[z](m,z), observe thatmR[z](m,z) is a dimension-one prime ideal that extends to mR[[z]], which also hasdimension one and is not (m, z)-primary in R[[z]]. Alternatively, in the languageof locally flat morphisms, if the elements τimi=1 ⊆ m are primarily independentover R, then (6.1) implies that the morphism

φ′ : Rn[z](max(Rn),z) −→ R[[z]]

satisfies condition LFd−1, where d = dim(R). For τimi=1 to be primarily in-dependent over R[z](m,z), however, the morphism φ′ has to satisfy LFd, sincedimR[z](m,z) = d + 1. Using (6.1) again this forces φ : Rn −→ R to satisfycondition LFd and thus φ is flat, which can happen only if n = 0. This is an in-teresting phenomenon; the construction of primarily independent elements involvesall parameters of the ring R.

In the remainder of this section we consider localizations of polynomial ex-tensions so that the dimension does not increase. Theorem 6.9 gives a methodto obtain residually algebraically independent and primarily independent elementsover an uncountable excellent local domain. In (6.9) we make use of the fact thatif A is a Noetherian ring and Z is a set of indeterminates over A, then the ringA(Z) obtained by localizing the polynomial ring A[Z] at the multiplicative systemof polynomials whose coefficients generate the unit ideal of A is again a Noetherianring

GH[28, Theorem 6].

6.6.9 Theorem 20.64. Let (R,m) and τimi=1 ⊆ m be as in the setting of (2.1),where m is either an integer or m = ∞, and dim(R) = d ≥ 2. Let Z be aset (possibly uncountable) of indeterminates over R and let R(Z) = R[Z](mR[Z]).Then:

(1) τimi=1 is primarily independent over R ⇐⇒ τimi=1 is primarily inde-pendent over R(Z).

(2) τimi=1 is residually algebraically independent over R ⇐⇒ τimi=1 isresidually algebraically independent over R(Z).

(3) If τimi=1 is idealwise independent over R(Z), then τimi=1 is idealwiseindependent over R.

Proof. Let n ∈ N be an integer with n ≤ m, put Rn = R[τ1, . . . , τn](m,τ1,...,τn)

and let n denote the maximal ideal of Rn. Let φ : Rn −→ R and µ : Rn −→Rn(Z) = Rn[Z]nRn[Z] be the inclusion maps. We have the following commutative


diagram:

Rn(Z)φ′

−−−−→ R(Z)

µ

x µ′x

Rnφ−−−−→ R

The ring R(Z) is a localization of the tensor product R⊗Rn Rn[Z] and Proposition6.4 applies. Thus, for (1), φ satisfies LFd−1 if and only if φ′ satisfies LFd−1.Similarly, for (2), φ satisfies LF1 if and only if φ′ satisfies LF1.

Since the inclusion map ψ taking R(Z) to its completion is faithfully flat, weobtain equivalences:

φ satisfies LFk ⇐⇒ φ′ satisfies LFk ⇐⇒ ψφ′ satisfies LFk.Since primary independence is equivalent to LFd−1 by (5.6) and residual alge-

braic independence is equivalent to LF1 by (5.4), statements (1) and (2) follow.For (3), if the τi are idealwise independence over R(Z), the morphism ψφ′ is

weakly flat. Thus φ′ is weakly flat and the statement follows by (6.1). 6.6.10 Corollary 20.65. For k a countable field, let Z be an uncountable set of inde-

terminates over k and let x, y be additional indeterminates. Let R := k(Z)[x, y](x,y).Then R is an uncountable excellent normal local domain and, for m a positive in-teger or m = ∞, there exist m primarily independent elements (and hence alsoresidually algebraically and idealwise independent elements) over R.

Proof. Apply (3.9), (4.4) and (6.9).

20.9. Passing to the Henselization6.7

In this section we investigate idealwise independence, residual algebraic inde-pendence, and primary independence as we pass from R to the Henselization Rh

of R. In particular, we show in Proposition (7.5) that for a single element τ ∈mRthe notions of idealwise independence and residual algebraic independence coincideif R = Rh. This implies that for every excellent normal local Henselian domainof dimension 2 all three concepts coincide for an element τ ∈ mR; that is, τ isidealwise independent ⇐⇒ τ is residually algebraically independent ⇐⇒ τ isprimarily independent.

We use the commutative square of (6.4) and obtain the following result forHenselizations:

6.7.1 Proposition 20.66. Let φ : (A,m) → (B,n) be an injective local morphismof normal local Noetherian domains, and let φh : Ah −→ Bh denote the inducedmorphism of the Henselizations. Then:

(1): φ satisfies LFk ⇐⇒ φh satisfies LFk, for each k with 1 ≤ k ≤dim(B). Thus, in particular, φ satisfies PDE ⇐⇒ φh satisfies PDE.

(2): (Descent) If φh is height-one preserving (respectively weakly flat), thenφ is height-one preserving (respectively weakly flat).

Using shorthand and diagrams, we show (7.1) schematically:

φ is LFk ⇐⇒ φh is LFk ; φ is PDE ⇐⇒ φh is PDE

↓ ↓φ ht-1 pres ⇐= φh ht-1 pres ; φ w.f. ⇐= φh w.f. .

20.9. PASSING TO THE HENSELIZATION 157

ALERT Not sure that the diagram above looks right.Compare to that of original paper before using it.

Proof. of (7.1) Consider the commutative diagram:

Ahφh

−−−−→ Bh

µ

x ν

xA

φ−−−−→ B

where µ and ν are the faithfully flat canonical injectionsN2[72, (43.8), page 182].

Since φ is injective and A is normal, φh is injective. By (6.4.2), (5.4) and (6.4.3)we need only show “⇒” in (1).

Let Q′ ∈ Spec(Bh) with ht(Q′) ≤ k. Put Q = Q′ ∩ B, P ′ = Q′ ∩ Ah, andP = P ′ ∩A. We consider the localized diagram:

AhP ′φh

Q′−−−−→ BhQ′

µP ′x νQ′

xAP

φQ−−−−→ BQ

The faithful flatness of ν implies ht(Q) ≤ k.In order to show that φhQ′ : AhP ′ −→ BhQ′ is faithfully flat, we apply (5.5.2). First

note that P ′ is a minimal prime divisor of PAh and that (Ah/PAh)P ′ = (Ah/P ′)P ′

is a fieldN2[72, (43.20)]. Thus

φhQ′ : (Ah/PAh)P ′ −→ (Bh/PBh)Q′

is faithfully flat and it remains to show that

PAhP ′ ⊗AhP ′ B

hQ′ ∼= PBhQ′ .

This can be seen as follows:

PAhP ′ ⊗AhP ′ B

hQ′ ∼= (P ⊗AP A

hP ′)⊗Ah

P ′ BhQ′ by flatness of µ

∼= P ⊗AP BhQ′

∼= (P ⊗AP BQ)⊗BQ BhQ′

∼= PBQ ⊗BQ BhQ′ by flatness of φQ

∼= PBhQ′ by flatness of ν.

6.7.2 Corollary 20.67. Let (R,m) and τimi=1 be as in the setting of (2.1), wherem is either a positive integer or m =∞ and dim(R) = d ≥ 2. Then:

(1): τimi=1 is primarily independent over R ⇐⇒ τimi=1 is primarilyindependent over Rh.

(2): τimi=1 is residually algebraically independent over R ⇐⇒ τimi=1

is residually algebraically independent over Rh.(3): (Descent) If τimi=1 is idealwise independent over Rh then τimi=1 is

idealwise independent over R.


Proof. For (1) and (2) it suffices to show the equivalence for every positiveinteger n ≤ m. Note that the local rings Rn = R[τ1, . . . , τn](m,τ1,...,τn) and Rn =Rh[τ1, . . . , τn](m,τ1,...,τn) have the same Henselization which we denote Rhn. AlsoRn ⊆ Rn. By (5.6) and (7.1) :

τ1, . . . , τn are primarily (respectively residually algebraically) independent overR ⇔

Rn −→ R satisfies LFd−1 (respectively LF1) ⇔Rhn −→ R = Rh satisfies LFd−1 (respectively LF1) ⇔Rn −→ R satisfies LFd−1 (respectively LF1).The third statement on idealwise independence follows from (6.4.3) by consid-

ering

Rnφ′

−−−−→ R

µ

x ∥∥∥Rn

φ−−−−→ R .

6.7.3 Remark 20.68. The examples given in (4.7) and (4.9) show the converse topart (3) of (7.2), fails, weak flatness need not lift to the Henselization. With thenotation of (7.1), if φ is weakly flat, then for every P ∈ Spec(A) of height one withPB 6= B there exists by (2.10), Q ∈ Spec(B) of height one such that P = Q ∩ A.In the Henselization Ah of A, the ideal PAh is a finite intersection of height-oneprime ideals P ′

i of AhN2[72, (43.20)]. Only one of the P ′

i is contained in Q. Thus asin (4.7) and (4.9), one of the minimal prime divisors P ′

i may fail the condition forweak flatness.

Let R be an excellent normal local domain and let K, respectively Kh, denotethe fraction fields of R, respectively Rh. Let L be an intermediate field with K ⊆L ⊆ Kh. It is shown in

R3.7[85] that the intersection ring T = L∩R is an excellent local

normal domain with Henselization T h = Rh. Excellent, Henselian, local, normaldomains are algebraically closed in their completion and we obtain:

6.7.4 Corollary 20.69. Let (R,m) and τimi=1 be as in the setting of (2.1), wherem denotes a positive integer or m = ∞. Suppose that T is a local Noetheriandomain dominating and algebraic over R and dominated by R with R = T . Then:

(1): τimi=1 is primarily independent over R ⇐⇒ τimi=1 is primarilyindependent over T .

(2): τimi=1 is residually algebraically independent over R ⇐⇒ τimi=1

is residually algebraically independent over T .(3): If τimi=1 is idealwise independent over T , then τimi=1 is idealwise

independent over R.

Proof. As mentioned above, R and T have a common Henselization and thestatement follows by (7.2).

We have seen in (4.4) that if τ ∈ mR is residually algebraically independentover R, then τ is idealwise independent over R. In Proposition 7.5 we show that ifR is Henselian or, more generally, if height-one prime ideals of R do not split in thecompletion of R, then idealwise independence and residual algebraic independence

20.10. SUMMARY DIAGRAM FOR THE INDEPENDENCE CONCEPTS 159

are equivalent for a single element τ in R. There is an example inAHW[6] of a normal

local domain R which is not Henselian but for each prime ideal P of R of height-one,the domain R/P is Henselian.

6.7.5 Proposition 20.70. Let (R,m) and τ ∈ m be as in the setting of (2.1). Sup-pose R has the property that for each P ∈ Spec(R) with ht(P ) = 1, the ideal PR isprime. Then τ is residually algebraically independent over R ⇐⇒ τ is idealwiseindependent over R.

In particular, if R is Henselian or if R/P is Henselian for each height-oneprime P of R, then τ is residually algebraically independent over R ⇐⇒ τ isidealwise independent over R.

Proof. By (4.4) it is enough to show τ idealwise independent =⇒ τ isresidually algebraically independent. Let P ∈ Spec(R) such that ht(P ) = 1 andP∩R 6= 0. Then ht(P∩R) = 1 and (P∩R)R1 is a prime ideal of R1 = R[τ ] of height1. Idealwise independence of τ implies that (P ∩R)R1 = (P ∩R)R1R ∩R1. Since(P ∩R)R is nonzero and prime, we have P = (P ∩R)R and P ∩R1 = (P ∩R)R1.Therefore ht(P∩R1) = 1 and Theorem 4.3.2 implies that τ is residually algebraicallyindependent over R.

For the last statement, suppose that P is a height-one prime of R such thatR/P is Henselian. Then the integral closure of the domain R/P in its fraction fieldis again local, in fact an excellent normal local domain and so analytically normal.But this implies that the extended ideal PR is prime, because of the behavior ofcompletions of finite integral extensions

N2[72, (17.7), (17.8)].

Apparently (7.5) cannot be extended to more than one algebraically indepen-dent element τ ∈mR, because even when R is Henselian, the localized polynomialring R[τ ](m,τ) fails to be Henselian.

6.7.6 Corollary 20.71. If R is an excellent Henselian normal local domain of di-mension 2, then τ is idealwise independent over R ⇐⇒ τ is residually algebraicallyindependent over R ⇐⇒ τ is primarily independent over R.

Proof. This follows from (7.5) and (4.4.3).

20.10. Summary diagram for the independence concepts6.8

With the notation of (2.1) for R,m, Rn, τ1 . . . , τn, let d = dim(R), L the quo-tient field of Rn, p ∈ Spec(Rn) such that dim(Rn/p) ≤ d− 1, P ∈ Spec(Rn) withht(P ) = 1, P ∈ Spec(R) with ht(P ) = 1, Rh = the Henselization of R in R, Ta local Noetherian domain dominating and algebraic over R and dominated by Rwith R = T , z an indeterminate over the quotient field of R and Z a possibly un-countable set of set of indeterminates over the quotient field of R. Then we have theimplications shown below. We use the abbreviations “prim. ind.”, “res. ind.” and“idw. ind” for “primarily independent”, “residually independent” and “idealwiseindependent”.

Note 20.72. Rn → R is always height-one preserving by (2.7).


(5.6.2)

(5.6.1)

(4.3)

(4.3)

(2.16.1)

(6.9.1)

(7.4.1)

(6.9.2)

(7.4.2)

(6.5.1)

(6.5.2)

(6.9.3)

(5.3)

(5.4)

(2.8b)

(2.12)

(2.8a)

(3.1)

(4.4.1)

(4.3)

(4.3)

(2.2)

(4.4.2) (4.4.2)

Rn → R LFd−1(5.3) τi prim. ind. /R (3.1) τi prim. ind. /R(Z) (3.1)

∀p,pR is mR-primary τi prim. ind./T (3.1)

Rn → R LF1(5.3) τi res. ind. /R (4.1) τi res. ind. /R(Z) (4.1)

Rn → R PDE (2.8b) ht(P ∩Ri) ≤ 1, ∀P∀i τi res. ind. /T (4.1)

ht(P ∩Rn) ≤ 1, ∀P τi res. ind. /R (4.1) τi res. ind. /R[z] (4.1)

Rn → R wf. (2.8a) τi idw. ind. /R(2.2) τi idw. ind. /R[z](m,z)(4.1)

PR ∩Rn = P, ∀P R ∩ L = Rn τi idw. ind. /R(Z) (4.1)

Independence concepts and results

CHAPTER 21

Intermediate rings between a local domain and its

completion May 17 2009

intsecLet (R,m) be an excellent normal local domain with field of fractions K and

m-adic completion R. In this chapter we consider the structure of an intermediatering A between R and R of the form A := K(τ1, τ2, . . . τs) ∩ R, where s ∈ N andτ1, τ2, . . . , τs ∈ m are certain algebraically independent elements over K. Thisconstruction follows a tradition begun by Nagata in the 1950’s. The intermediateintersection rings provide interesting examples of Noetherian and non-Noetherian,excellent and non-excellent rings. If the intersection ring A can be expressed as adirected union of localized polynomial extension rings of R the computation of Ais easier. We say τ1, τ2, . . . , τs are “limit-intersecting” if A is such a directed union.Two stronger forms of the limit-intersecting condition are useful for constructingexamples and for determining if A is Noetherian or excellent. We give criteriafor τ1, τ2, . . . , τs to have these properties. We include several concrete examplesinspired by the construction.

Also in this chapter the concepts of height-one preserving and weakly flat arediscussed and analyzed for completions: The height-one preserving property holdswithin completions, but weakly flat does not hold for a completion extension asshown by the example below.

Let S → T be an extension of Krull domains. We say that T is a height-onepreserving extension of S if for every height-one prime ideal P of S with PT 6= Tthere exists a height-one prime ideal Q of T with PT ⊆ Q.) We showed height-onepreserving property holds within completions.


the limit-intersecting and residual limit-intersecting properties.More on locally flat in height k.

??? alertThis is fromnoehom[39, (2.10), (2.14)]???

Proposition 21.1. Let φ : S → T be an extension of Krull domains and letF denote the fraction field of S.

(1) Suppose PT 6= T for every height-one prime ideal P of S. Then S → Tis weakly flat ⇐⇒ S = F ∩ T .

(2) If S → T is weakly flat, then φ is height-one preserving and, moreover, forevery height-one prime ideal P of S with PT 6= T , there is a height-oneprime ideal Q of T with Q ∩ S = P .

Remark 21.2. The height-one preserving condition does not imply weakly flat.To see this, consider a domain (D,m), as in (

7.3.321.5), such that dim(C⊗DQ(D)) = 0,

161

162 21. INTERMEDIATE RINGS I

where (C, n), τ , and S are as in the statement of (7.3.321.5), and so the local inclusion

morphism ϕ : S −→ C is height-one preserving. (For example, take C = k[[x, y]]and D = k[[x]][y](x,y).) There exists a height-one prime ideal P of S such thatP ∩D = 0; then PC 6= C. Since ϕ is height-one preserving, there exists a height-one prime ideal Q of C such that PC ⊆ Q. Also dim(C ⊗D Q(D)) = 0 impliesQ∩D 6= 0. We have P ⊆ Q∩S and P ∩D = 0. It follows that P is strictly smallerthan Q ∩ S, so Q ∩ S has height greater than one and so the extension ϕ is notweakly flat.

???alertSW put the entire inter here, except for what was removed and put inChapter

constr5 ???

21.1. Background

A fruitful source of examples of Noetherian local integral domains D are do-mains of the form D := Q(S) ∩ (S/a), for certain intermediate local rings (S,n)between R and R. Here S denotes the n-adic completion of S, Q(S) is the fractionfield of S, and a is an ideal of S such that the associated primes of a are in thegeneric formal fiber of S. Using this construction, examples D can be producedcontaining a coefficient field k such that D has finite transcendence degree overk, but nevertheless D is non-excellent and sometimes even non-Noetherian. Sincein our setting S/a ∼= R, we often use a simpler expression for the intersection D,namely D := L ∩ R, where L is an intermediate field between K and the fractionfield of R.

The following diagram displays the containments among the rings we are dis-cussing:

K⊆−−−−→ Q(S) L

⊆−−−−→ Q(R)x x x xR

⊆−−−−→ S⊆−−−−→ D := L ∩ (S/a) = L ∩ R ⊆−−−−→ R ∼= S/a.

A description of D as a directed union of localized polynomial rings over R is oftenused in the construction of the examples mentioned above, such as that of

N2[72,

(E7.1), p.210]. If such a realization of D as a directed union exists, it is easierto compute than the description of D as an intersection. In our setting there is anatural sequence of nested localized polynomial rings Bn over R having a directedunion B which is contained in D and has the same fraction field as D. The classicalmethod for demonstrating that the intersection D = L∩ R is Noetherian has beento show that B is a Noetherian domain having completion R, and therefore thatB = L ∩ R = D.

In this chapter we continue an investigation begun in Chapteridwisec20 but we modify

the focus and approach. In Chapteridwisec20 we found non-trivial examples of ideals a

such that the construction described above fails to produce a new ring; that is, whereD := Q(S) ∩ S/a = S, or using the expression D := L ∩ R for the intersection,the case where D is a localized polynomial ring over R. Our primary goal here isto obtain interesting Noetherian rings, but we expand our working setting to Krulldomains because such an intersection domain D may be a birational extension of S

21.2. OUTLINE 163

which is not Noetherian. Another modification in this chapter is that we considercompletions with respect to a principal ideal; this is because in most examples ofnew Noetherian domains produced using the D = Q(S) ∩ (S/a) construction, theideal a is extended from a completion of R with respect to a principal ideal. Finally,we analyze the intersection D := Q(S)∩ (S/a) (or L∩ R) more systematically herethan in Chapter

idwisec20; we focus on conditions in order that the intersection be a

directed union of localized polynomial rings over R.Suppose τ1, . . . , τs ∈ m are algebraically independent over K. Then A :=

K(τ1, . . . , τs)∩ R is a quasilocal Krull domain that dominates 1 R and is dominatedby R. Thus A, as a subring of R and an extension ring of R, is a special type ofintermediate ring between R and R with fraction field K(τ1, . . . , τs). Indeed, if adenotes the kernel of the canonical map to R from the completion A of A, thenA = Q(A)∩(A/a) has the form described above. The ring A birationally dominates2 the localized polynomial ring B0 = R[τ1, . . . , τs](m,τ1,...,τs). In the present chapterwe explore the nature of the birational domination of A over B0.

Many of the concepts from our earlier work are useful in this study. In Chapteridwisec20, the elements τ1, . . . , τs ∈ m are defined to be idealwise independent over R ifA = B0. Here, with the assumption that each τi is in the completion of R withrespect to a principal ideal (and the τi are algebraically independent over K), weinvestigate conditions in order that A can be realized as a directed union of localizedpolynomial rings over R; that is, A = B, where B := lim−→n∈N

Bn, and, for each n,

Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn) ⊆ Bn+1,

where τ1n, . . . , τsn ∈ m are series formed by the endpieces of the τi (see (4.2.36.1) ).

Essentially what we require in the present chapter is that the conditions of Chapteridwisec20 be satisfied off the proper closed subset of Spec(R) defined by a principal ideal.This leads to the analysis of “limit-intersecting” independence properties for ele-ments τ1, . . . , τs ∈ m which are algebraically independent over K; these propertiesare analogs to types of “idealwise independence” over R defined in Chapter

idwisec20.

As we show in Section7.621.6, these modified independence conditions enable us to

produce concrete examples illustrating the concepts.

21.2. Outline

The motivating example (formerly Section 7.2 in this part) has now been movedto Chapter

constr5, Section

4.24.3 . (We may still have some items to justify, so we leave

in what was not moved.) This example is now given in Theorem4.2.14.12, where there

is a description of the rings A and B for particular elements σ and τ of a powerseries ring in two variables over a field. In Section

7.321.3, we give some background

material from Chapteridwisec20, including some definitions, terminology, and related

results. Section7.421.4 contains a description of the intermediate rings Bn, whereas

Section7.521.5 gives the new definitions associated with the elements τi of m and

their basic properties. In Section7.621.6 we display concrete examples of idealwise

independent elements in the sense of Chapteridwisec20.

1That is, the maximal ideal of A intersects R in m.2That is, A dominates B0 and A is contained in the fraction field of B0.


21.3. Background material7.3

We review the main definitions and relevant results from Chapteridwisec20. The

flatness conditions ( (7.3.121.3), (

7.3.221.4) and (

7.3.421.6) ) are used in the limit-intersecting

independence definitions of Section7.521.5.

7.3.1 Definition 21.3. Let φ : S −→ T be an injective morphism of commutativerings and let k ∈ N be an integer with 1 ≤ k ≤ d = dim(T ) where d is an integeror d = ∞. Then φ is called locally flat in height k, abbreviated LFk, if, for everyprime ideal Q of T with ht(Q) ≤ k, the induced morphism on the localizationsφQ : SQ∩S −→ TQ is faithfully flat.

7.3.2 Definition 21.4. Let S → T be an extension of Krull domains. We say thatT is a height-one preserving extension of S if for every height-one prime ideal P ofS with PT 6= T there exists a height-one prime ideal Q of T with PT ⊆ Q.

The height-one preserving property is crucial for our work, and so it is fortunatethat it holds in the situations we consider. In particular the following result, whichextends the result (

6.2.720.12) from Chapter

idwisec20 by eliminating a Noetherian hypothesis,

shows that the height-one preserving property holds within completions.

7.3.3 Proposition 21.5. Suppose (C, n) is a complete normal Noetherian local do-main that dominates a quasilocal Krull domain (D,m) Assume the injection D −→C is height-one preserving, and suppose τ ∈ n is algebraically independent over thefraction field L of D. Let S = D[τ ](m,τ). Then the local inclusion morphismϕ : S −→ C is height-one preserving.

Proof. Let P be a height-one prime ideal of S.Case (i): If ht(P ∩ D) = 1, then P = (P ∩ D)S. Since D −→ C is height-one preserving, (P ∩ D)C ⊆ Q, for some height-one prime ideal Q of C. ThenPC = (P ∩D)SC ⊆ Q as desired.Case (ii): Suppose P ∩ D = (0). Let U denote the multiplicative set of nonzeroelements of D. Let t be an indeterminate over D and let S1 = D[t](m,t). Considerthe following commutative diagram where the map from S1 to S is the D-algebraisomorphism taking t to τ and λ is the natural extension to C[[t]].

U−1S1⊆−−−−→ U−1C[t](bn,t)

∪x ∪

xD

⊆−−−−→ S1 = D[t](m,t)⊆−−−−→ C[t](bn,t)

⊆−−−−→ C[[t]]

=

y ∼=y λ

yD

⊆−−−−→ S = D[τ ](m,τ)ϕ−−−−→ C.

Under the above isomorphism of S with S1, P correponds to a height-one prime

ideal P0 of S1 such that P0 ∩D = (0). Thus P0 is contracted from the localizationU−1S1. Since U−1S1 is a localization of the polynomial ring L[t], it is a principalideal domain. Hence P0 is contained in a proper principal ideal of U−1S1. ThereforeP0 is contained in a proper principal ideal of U−1C[t](bn,t), and hence in a height-one prime ideal of C[t](bn,t). Now C[t](bn,t) −→ C[[t]] is faithfully flat because C is

21.4. INTERSECTIONS AND DIRECTED UNIONS 165

Noetherian; thus P0C[[t]] is contained in a height-one prime ideal of C[[t]]. Since Cis catenary and ker(λ) = (t − τ) is principal, P0 + ker(λ) = P + ker(λ) has heighttwo in C[[t]]. It follows that PC is contained in a height-one prime of C .

Next we review the concept of weak flatness defined in Chapteridwisec20:

7.3.4 Definition 21.6. Let S → T be an extension of Krull domains. We say thatT is weakly flat over S if every height-one prime ideal P of S with PT 6= T satisfiesPT ∩ S = P .

7.3.5 Proposition 21.7. (See (6.2.1020.15) and (

6.2.1420.19) of Chapter

idwisec20.) Let φ : S → T

be an extension of Krull domains and let F denote the fraction field of S.

(1) Suppose PT 6= T for every height-one prime ideal P of S. Then S → Tis weakly flat ⇐⇒ S = F ∩ T .

(2) If S → T is weakly flat, then φ is height-one preserving and, moreover, forevery height-one prime ideal P of S with PT 6= T , there is a height-oneprime ideal Q of T with Q ∩ S = P .

7.3.6 Remark 21.8. The height-one preserving condition does not imply weakly flat.To see this, consider a domain (D,m), as in (

7.3.321.5Check??), such that dim(C ⊗D

Q(D)) = 0, where (C, n), τ , and S are as in the statement of (7.3.321.5Check??), and so

the local inclusion morphism ϕ : S −→ C is height-one preserving. (For example,take C = k[[x, y]] and D = k[[x]][y](x,y).) There exists a height-one prime ideal P ofS such that P ∩D = 0; then PC 6= C. Since ϕ is height-one preserving, there existsa height-one prime ideal Q of C such that PC ⊆ Q. Also dim(C ⊗D Q(D)) = 0implies Q∩D 6= 0. We have P ⊆ Q∩S and P ∩D = 0. It follows that P is strictlysmaller than Q∩ S, so Q∩ S has height greater than one and so the extension ϕ isnot weakly flat.

21.4. Intersections and directed unions7.4

In general the intersection of a normal Noetherian domain with a subfield ofits field of fractions is a Krull domain, but is not Noetherian. The Krull domainB in the motivating example of Section

4.24.3 and Chapter

motiv7 (in the case where

B 6= A) illustrates that a directed union of normal Noetherian domains may be anon-Noetherian Krull domain. Thus, in order to apply an iterative procedure inSection

7.521.5, we consider a quasilocal Krull domain (T,n) which is not assumed

to be Noetherian, but is assumed to have a Noetherian completion. To distinguishfrom the earlier Noetherian hypothesis on R, we let T denote the base domain.

As we mention in the introduction, completions with respect to principal idealsare used in our constructions.

7.4.1 Setting and Notation 21.9. Let (T,n) be a quasilocal Krull domain withfraction field F . Assume there exists a nonzero element y ∈ n such that the y-adic completion (T, (y)) := (T ∗,n∗) of T is an analytically normal Noetherian localdomain. It then follows that the n-adic completion T of T is also a normal localdomain, since the n-adic completion of T is the same as the n∗-adic completionof T ∗. Since T ∗ is Noetherian, if F ∗ denotes the field of fractions of T ∗, then


T ∗ = T ∩ F ∗. Therefore F ∩ T ∗ = F ∩ T . Let d denote the dimension of theNoetherian domain T ∗. It follows that d is also the dimension of T . 3

(1) Assume that T = F ∩ T ∗ = F ∩ T , or equivalently by (7.3.521.7.1), that T ∗ and

T are weakly flat over T .(2) Let Ty := T [1/y], the localization of T at the powers of y, and similarly, let

T ∗y := T ∗[1/y]. The domains Ty and T ∗

y are of dimension d− 1.(3) Let τ1, . . . , τs ∈ n∗ be algebraically independent over F .(4) For each i with 1 ≤ i ≤ s, we have an expansion τi := Σ∞

j=1cijyj where

cij ∈ T .(5) For each n ∈ N and each i, 1 ≤ i ≤ s, we define the nth-endpiece of τi with

respect to y as in (4.2.36.1), so that

τin := Σ∞j=n+1cijy

j−n, τin = yτi,n+1 + ciny.

(6) For each n ∈ N, we define Bn := T [τ1n, . . . , τsn](n,τ1n,...,τsn). In view of (5),we have Bn ⊆ Bn+1 and Bn+1 dominates Bn for each n. We define

B := lim−→n∈N

Bn =∞⋃n=1

Bn, and A := F (τ1, . . . , τs) ∩ T .

Thus, B and A are quasilocal domains and A birationally dominates B. We areespecially interested in conditions which imply that B = A.

(7) Let A∗ denote the y-adic completion (A, (y)) of A and B∗ the y-adiccompletion of B.

7.4.2 Remark 21.10. The motivating example of Section4.24.3 and Chapter

motiv7 as given

in Theorem4.2.117.2 with T := B 6= A (from the notation of (

4.7.n7.1) and Example

4.7.137.3

continued in Example4.7.13c8.11) shows that T −→ T ∗

y can satisfy the other conditionsof (

7.4.121.9) but not satisfy the assumption (

7.4.121.9.1); that is, such an extension is always

height-one preserving (by (7.3.321.5 Check??) ) but not in general weakly flat.

ALERT This independence is done in Proposition4.5.226.3

7.4.3 Proposition 21.11. The definitions of B and Bn are independent of repre-sentations for τ1, . . . , τs as power series in y with coefficients in T .


τi =∞∑j=1

aijyj and ωi =

∞∑j=1

bijyj,

where each aij , bij ∈ T . We define the nth-endpieces τin and ωin as in (2.3):

τin =∞∑

j=n+1

aijyj−n and ωin =

∞∑j=n+1

bijyj−n.

Then we have

τi = Σ∞j=1aijy

j = Σnj=1aijyj + ynτin = Σ∞

j=1bijyj = Σnj=1bijy

j + ynωin = ωi.

3If T is Noetherian, then dim(T ) = d. However, without the hypothesis that T is Noetherian,it is unclear whether T has dimension d.

21.4. INTERSECTIONS AND DIRECTED UNIONS 167


ynτin− ynωin = Σnj=1bijyj −Σnj=1aijy

j, and so τin−ωin =Σnj=1(bij − aij)yj

yn.

Since Σnj=1(bij − aij)yj ∈ T is divisible by yn in T ∗ and T = F ∩T ∗, it follows thatyn divides Σnj=1(bij − aij)yj in T . Therefore τin − ωin ∈ T . It follows that Bn andB = ∪∞n=1Bn are independent of the representation of the τi.

7.4.4 Theorem 21.12. Assume the setting and notation of (7.4.121.9). Then the inter-

mediate rings Bn, B and A have the following properties:

(1) yA = yT ∗ ∩ A and yB = yA ∩ B = yT ∗ ∩ B. More generally, for everyt ∈ N, we have ytA = ytT ∗ ∩A and ytB = ytA ∩B = ytT ∗ ∩B.

(2) T/ytT = B/ytB = A/ytA = T ∗/ytT ∗, for each positive integer t.(3) Every ideal of T,B or A that contains y is finitely generated by elements

of T . In particular, the maximal ideal n of T is finitely generated, andthe maximal ideals of B and A are nB and nA.

(4) For every n ∈ N: yB ∩Bn = (y, τ1n, . . . , τsn)Bn, an ideal of Bn of heights+ 1.

(5) If P ∈ Spec(A) is minimal over yA and Q = P ∩ B, W = P ∩ T , thenTW ⊆ BQ = AP , and all three localizations are DVRs.

(6) For every n ∈ N, B[1/y] is a localization of Bn, i.e., for each n ∈ N, thereexists a multiplicatively closed subset Sn of Bn such that B[1/y] = S−1

n Bn.(7) B = B[1/y]∩Bq1 ∩ · · · ∩Bqr , where q1, . . . ,qr are the prime ideals of B

minimal over yB.

Proof. Let K := F (τ1, . . . τs), the field of fractions of A and B. Then A =T ∗ ∩K implies yA ⊆ yT ∗ ∩A. Let g ∈ yT ∗ ∩A ⊆ yT ∗ ∩K. Then g/y ∈ T ∗ ∩K =A =⇒ g ∈ yA. Since B =

⋃∞n=1Bn, we have yB =

⋃∞n=1 yBn. It is clear that

yB ⊆ yA ∩ B ⊆ yT ∗ ∩ B. We next show yT ∗ ∩ B = yB. Let g ∈ yT ∗ ∩ B.Then there is an n ∈ N with g ∈ Bn and, multiplying g by a unit of Bn ifnecessary, we may assume that g ∈ T [τ1n, . . . , τsn]. Write g = r0 + g0 where g0 ∈(τ1n, . . . , τsn)T [τ1n, . . . , τsn] and r0 ∈ T . Substituting τjn = yτjn+1 + cjny ∈ yT ∗

from (7.4.121.9.5) yields that g0 ∈ yT ∗ and so r0 ∈ yT ∗ ∩ T = yT . Since by (

7.4.121.9.5),

(τ1n, . . . , τsn)Bn ⊆ yBn+1, it follows that g ∈ yB. Now yB = yT ∗ ∩ B impliesy2B = y(yT ∗ ∩ B) = y2T ∗ ∩ yB = y2T ∗ ∩ B. Similarly ytB = ytT ∗ ∩ B for everyt ∈ N.

Since ytT ∗ ∩ T = ytT , T/ytT = T ∗/ytT ∗, and T/(ytT ) → B/(ytB) →A/(ytA) → T ∗/ytT ∗ , the assertion in (2) follows.

Since T ∗ is Noetherian, the assertions of (3) follow from (2).For (4), let f ∈ yB ∩Bn. After multiplication by a unit of Bn, we may assume

that f ∈ T [τ1n, . . . , τsn], and hence f is of the form

f =∑

(i)∈Ns

a(i)τi11n . . . τ

issn

with a(i) ∈ T . Since τjn ∈ yB, we see that a(0) ∈ yB ∩ T ⊆ yT ∗ ∩ T , and wecan write a(0) = yb for some element b ∈ T ∗. This implies that b ∈ F ∩ T ∗ = T ;the last equality uses (

7.4.121.9.1). Therefore a(0) ∈ yT and f ∈ (y, τ1n, . . . , τsn)Bn.

Furthermore if g ∈ (y, τ1n, . . . , τsn)Bn, then τin ⊆ yB ∩Bn, so g ∈ yB ∩Bn.


For (5), since T ∗ and hence A is Krull, P has height one and AP is a DVR.Also AP has the same fraction field as BQ. By (2), W is a minimal prime of yT .Since T is a Krull domain, TW is a DVR and the maximal ideal of TW is generatedby u ∈ T . Thus by (2) the maximal ideal of BQ is generated by u and so BQ is aDVR dominated by AP . Therefore they must be the same DVR.

Items (6) follows from (7.4.121.9.5).

For (7), suppose β ∈ B[1/y] ∩ Bq1 ∩ · · · ∩ Bqr . Now Bq1 ∩ · · · ∩ Bqr = (B −(∪qi))−1B. There exist t ∈ N, a, b, c ∈ B with c /∈ q1 ∪ · · · ∪ qr such that β =a/yt = b/c. We may assume that either t = 0 (and we are done) or that t > 0and a /∈ yB. Since yB = yA ∩ B, it follows that q1, . . . ,qr are the contractionsto B of the minimal primes p1, . . . ,pr of yA in A. Since A is a Krull domain,A = A[1/y] ∩ Ap1 ∩ · · · ∩ Apn . Thus β ∈ A, and a = ytβ ∈ yA ∩ B = yB, acontradiction. Thus t = 0 and β = a ∈ B.

7.4.5 Theorem 21.13. With the setting and notation of (7.4.121.9), the intermediate

rings A and B have the following properties:

(1) A and B are quasilocal Krull domains.(2) B ⊆ A, with A dominating B.(3) A∗ = B∗ = T ∗.(4) If B is Noetherian, then B = A.

Moreover, if T is a unique factorization domain (UFD) and y is a prime elementof T , then B is a UFD.

Proof. As noted in the proof of (7.4.421.12), A is a Krull domain. By (

7.4.421.12.6),

B[1/y] is a localization of B0. Since B0 is a Krull domain, it follows that B[1/y]is a Krull domain. By (

7.4.421.12.7), B is the intersection of B[1/y] and the DVR’s

Bq1 , . . . , Bqr . Therefore B is a Krull domain. Items (2) and (3) are immediatefrom (

7.4.421.12). If B is Noetherian, then B∗ is faithfully flat over B, and hence

B = F (τ1, . . . , τs) ∩ B∗ = A. For the last statement, if T is a UFD, so is thelocalized polynomial ring B0. By (

7.4.421.12.6), B[1/y] = S−1

0 B0[1/y], which impliesthat B[1/y] is also a UFD. By (

7.4.421.12.2), y is a prime element of B; hence it follows

fromSam[88, (6.3), page 21] that B is a UFD.

21.5. Limit-intersecting elements7.5

As we state in the introduction, we are interested in the structure of L ∩ R,for intermediate fields L between the fraction fields of R and R. This is difficult todetermine in general. We show in Theorem

7.5.521.18 that each of the limit-intersecting

properties of (7.5.121.14) implies L ∩ R is a directed union of localized polynomial ring

extensions of R. These limit-intersecting properties are related to the idealwiseindependence concepts defined in Chapter

idwisec20.

7.5.1 Definition 21.14. Let (T,n) be a quasilocal Krull domain with fraction fieldF , let 0 6= y ∈ n be such that the y-adic completion (T, (y)) := (T ∗,n∗) of T is ananalytically normal Noetherian local domain of dimension d. Assume that T ∗ andT are weakly flat over T . Let τ1, . . . , τs ∈ n∗ be algebraically independent over F(as in (

7.4.121.9)).

21.5. LIMIT-INTERSECTING ELEMENTS 169

(1) The elements τ1, . . . , τs are said to be limit-intersecting in y over T pro-vided the inclusion morphism

B0 := T [τ1, . . . , τs](n,τ1,...,τs) −→ T ∗y is weakly flat (

7.5.121.14.1).

(2) The elements τ1 . . . , τs are said to be residually limit-intersecting in y overT provided the inclusion morphism

B0 := T [τ1, . . . , τs](n,τ1,...,τs) −→ T ∗y is LF1 (

7.5.121.14.2).

(3) The elements τ1 . . . , τs are said to be primarily limit-intersecting in y overT provided the inclusion morphism

B0 := T [τ1, . . . , τs](n,τ1,...,τs) −→ T ∗y is flat, or LFd−1 (

7.5.121.14.3).

Since T ∗y and Ty have dimension d−1, the condition LFd−1 is equivalent to primarily

limit-intersecting, that is, to the flatness of the map B0 −→ T ∗y .

7.5.2 Remarks 21.15. (1) The terms “residually” and “primarily” come from Chap-ter

idwisec20. We justify this terminology in (

7.5.721.20) and (

7.5.821.21). It is clear that pri-

marily limit-intersecting implies residually limt-intersecting and residually limit-intersecting implies limit-intersecting.

(2) Since Ty is faithfully flat over T ∗y , the statements obtained by replacing T ∗

y

by Ty give equivalent definitions to those of ((7.5.121.14) (see (

6.6.120.56) and (

6.6.320.58) of

Chapteridwisec20).

(3) We remark that

B −→ T ∗y is weakly flat ⇐⇒ B −→ T ∗ is weakly flat .

To see this, observe that by (7.4.421.12.2), every height-one prime of B containing y is

the contraction of a height-one prime of T ∗. If p is a height-one prime of B withy 6∈ p, then pT ∗ ∩B = p if and only if pT ∗

y ∩B = p.(4) Since by (

7.4.421.12.6), By is a localization S−1

0 B0 of B0, and since the canonicalmaps B0 −→ T ∗

y and B −→ T ∗y factor through the localization at the powers of y,

the elements τ1, . . . , τs are limit-intersecting in y over T if and only if the canonicalmap

S−10 B0 = By −→ T ∗

y

is weakly flat. In view of (7.5.321.16) below, we also have τ1, . . . , τs are residually (resp.

primarily) limit-intersecting in y over T if and only if the canonical map

S−10 B0 = By −→ T ∗

y

is LF1 (resp. LFd−1 or equivalently flat).(5) If d = 2, then obviously LF1 = LFd−1. Hence in this case primarily limit-

intersecting is equivalent to residually limit-intersecting.(6) Since T −→ Bn is faithfully flat for every n, it follows

B[10, Chap.1, Sec.2.3,

Prop.2, p.14] that T −→ B is always faithfully flat. Thus if residually limit-intersecting elements exist over T , then T −→ T ∗

y must be LF1. If primarilylimit-intersecting elements exist over T , then T −→ T ∗

y must be flat.(7) Items (

7.3.621.8) and (

7.4.221.10)

???? WHAT IS INTENDED HERE???show that in some situations T ∗ contains no limit-intersecting elements. Indeed,

if T is complete with respect to some nonzero ideal I, and y is outside every minimal


prime over I, then every algebraically independent τ =∑aiy

i ∈ T ∗ fails to be limit-intersecting in y. To see this, choose an element x ∈ I, x outside every minimalprime ideal of yT ; define σ :=

∑aix

i ∈ T . Then τ − σ ∈ (x − y)T ∗ ∩ T [τ ]. Thusa minimal prime over x− y in T ∗ intersects T [τ ] in an ideal of height greater thanone, because it contains x− y and τ − σ.

7.5.3 Proposition 21.16. Assume the notation and setting of (7.4.121.9) and let k be a

positive integer with 1 ≤ k ≤ d− 1. Then the following are equivalent:(1) The canonical injection φ : B0 := T [τ1, . . . , τs](m,τ1,...,τs) −→ T ∗

y is LFk.(1′) The canonical injection φ1 : B0 := T [τ1, . . . , τs](m,τ1,...,τs) −→ Ty is LFk.(2) The canonical injection φ′ : U0 := T [τ1, . . . , τs] −→ T ∗

y is LFk.(2′) The canonical injection φ′1 : U0 := T [τ1, . . . , τs] −→ Ty is LFk.(3) The canonical injection θ : Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn) −→ T ∗

y isLFk.

(3′) The canonical injection θ1 : Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn) −→ Ty isLFk.

(4) The canonical injection ψ : B −→ T ∗y is LFk.

(4′) The canonical injection ψ : B −→ Ty is LFk.Moreover, these statements are also all equivalent to LFk of the correspondingcanonical injections obtained by replacing B0, U0 and B by B0[1/y], U0[1/y] andB[1/y].

Proof. We have:

U0loc.−−−−→ B0

φ−−−−→ T ∗y

f.f.−−−−→ Ty.

The injection φ′1 : U0 −→ Ty factors as φ′ : U0 −→ T ∗y followed by the faithfully flat

injection T ∗y −→ Ty. Therefore φ′ is LFk if and only if φ′1 is LFk. The injection φ′

factors through the localization U0 −→ B0 and so φ is LFk if and only if φ′ is LFk.Now set Un := T [τ1n, . . . , τsn] for each n > 1 and U :=

⋃∞n=0 Un. For each

positive integer i, τi = ynτin +∑n

i=0 aiyi. Thus Un ⊆ U0[1/y], and U0[1/y] =⋃

Un[1/y] = U [1/y]. Moreover, for each n, Bn is a localization of Un, and hence Bis a localization of U .

We have:B[1/y] −→ T ∗

y is LFk ⇐⇒ U [1/y] −→ T ∗y is LFk ⇐⇒ U0[1/y] −→ T ∗

y is LFk⇐⇒ Bn[1/y] −→ T ∗

y is LFk ⇐⇒ B0[1/y] −→ T ∗y is LFk .

Thusψ : B −→ T ∗

y is LFk ⇐⇒ U −→ T ∗y is LFk ⇐⇒ φ′ : U0 −→ T ∗

y is LFk⇐⇒ θ : Bn −→ T ∗

y is LFk ⇐⇒ φ : B0 −→ T ∗y is LFk .

7.5.4 Remarks 21.17. (1) If (T,n) is a one-dimensional quasilocal Krull domain,then T is a rank-one discrete valuation domain (DVR). Hence T ∗ is also a DVRand T ∗

y is flat over U0 = T [τ1, . . . , τs]. Therefore, in this case, τ1, . . . , τs are primarilylimit-intersecting in y over T if and only if τ1, . . . , τs are algebraically independentover F .

(2) Let τ1, . . . , τs ∈ k[[y]] be transcendental over k(y), where k is a field. Thenτ1, . . . , τs are primarily limit-intersecting in y over k[y](y) by (1) above. In Chapter


idwisec20 (

6.3.320.24), we show that if x1, . . . , xm are additional indeterminates over k(y), then

τ1, . . . , τs are primarily limit-intersecting in y over k[x1, . . . , xm, y](x1,...,xm,y).(3) With the notation of (

7.4.121.9), if B is Noetherian, then τ1, . . . , τs are primarily

limit-intersecting in y over T . For B Noetherian implies T ∗ as the (y)-adic com-pletion of B is flat over B. Hence T ∗

y is also flat over B, and it follows from (7.5.321.16)

that τ1, . . . , τs are primarily limit-intersecting in y over T .(4) By the equivalence of (1) and (2) of (

7.5.321.16), we see that τ1, . . . , τs are

primarily limit-intersecting in y over T if and only if the endpiece power seriesτ1n, . . . , τsn are primarily limit-intersecting in y over T .

7.5.5 Theorem 21.18. With the setting and notation of (7.4.121.9), the following are

equivalent:(1) The elements τ1, . . . , τs are limit-intersecting in y over T .(2) The intermediate rings A and B are equal.(3) B −→ T ∗

y is weakly flat.(4) B −→ T ∗ is weakly flat.

Proof. (1)⇒(2): Since A and B are Krull domains with the same field offractions and B ⊆ A it is enough to show that every height-one prime ideal p of Bis the contraction of a (height-one) prime ideal of A. By Theorem

7.4.421.12.3 CHECK

THIS!!!, each height-one prime of B containing yB is the contraction of a height-oneprime of A.

Let p be a height-one prime of B which does not contain yB. Considerthe prime ideal q = T [τ1, . . . , τs] ∩ p. Since B[1/y] is a localization of the ringT [τ1, . . . , τs], we see that Bp = T [τ1, . . . , τs]q and so q has height one in T [τ1, . . . , τs].The limit-intersecting hypothesis implies qT ∗ ∩ T [τ1, . . . , τs] = q and there is aheight-one prime ideal w of T ∗ with w ∩ T [τ1, . . . , τs] = q. This implies thatw ∩B = p and thus also (w ∩A) ∩B = p. Hence every height-one prime ideal ofB is the contraction of a prime ideal of A. Since A is birational over B, this primeideal of A can be chosen to have height one.

(3)⇐⇒ (4): This is shown in (7.5.221.15.3).

(2)⇒(4): If B = A = F ∩ T ∗, then by (7.3.521.7) every height-one prime ideal of B

is the contraction of a height-one prime ideal of T ∗.(4)⇒(1): If B → T ∗ is weakly flat so is the localization By → T ∗

y . Since By =S−1

0 B0y for a suitable multiplicative subset S0 ⊆ B0y the embedding B0y → T ∗y is

weakly flat. Now (1) holds by (7.5.221.15.4).

7.5.6 Remarks 21.19. (1) If an injective morphism of Krull domains is weakly flat,then it is height-one preserving (

7.3.521.7.2). Thus any of the equivalent conditions of

(7.5.521.18) imply that B −→ T ∗ is height-one preserving.

(2) In (7.5.521.18) if B is Noetherian, then by (

7.4.521.13.4), A = B and all the conclu-

sions of (7.5.521.18) hold.

(3) In Chapternoeslocsec22, (

7.4.421.12), we give an example of a three-dimensional regular

local domain R dominating Q[x, y, z](x,y,z) and having completion Q[[x, y, z]], suchthat there exists an element τ in the (y)-adic completion of R that is residuallylimit-intersecting in y over R but fails to be primarily limit-intersecting in y overR. In particular, the rings A and B constructed using τ are equal, yet A and Bare not Noetherian. We also show in Chapter

noeslocsec22, (

8.2.1222.12) that if R is a Noetherian

semilocal domain, then τ1, . . . , τs ∈ yR∗ are primarily limit-intersecting in y overR if and only if B is Noetherian. If this holds, we also have B = A.


We now give criteria for elements to be residually limit-intersecting or primar-ily limit-intersecting similar to those in Chapter

idwisec20 for elements to be residually

algebraically independent or primarily independent.

7.5.7 Proposition 21.20. With the setting and notation of (7.4.121.9) and s = 1, the

following are equivalent:

(1) The element τ = τ1 is residually limit-intersecting in y over T .(2) If P is a height-one prime ideal of T such that y /∈ P and P ∩T 6= 0, then

ht(P ∩ T [τ ](n,τ)) = 1.(3) For every height-one prime ideal P of T such that y /∈ P and for ev-

ery minimal prime divisor P of P T in T , the image τ of τ in T /P isalgebraically independent over the fraction field of T/P .

(4) B −→ T ∗y is LF1 and height-one preserving

Proof. For (1)⇒ (2), suppose (2) fails; that is, there exists a prime ideal P ofT of height one such that y /∈ P , P ∩ T 6= 0, but ht(P ∩ T [τ ]) ≥ 2. Let Q := P Ty.Then Q := Q∩T [τ ](n,τ) has height greater than or equal to 2. But by the definitionof residually limit-intersecting in (

7.5.121.14), the injective morphism T [τ ](n,τ) −→ Ty

is LF1 and so by (7.3.121.3), (T [τ ](n,τ))Q −→ (Ty) bQ is faithfully flat, a contradiction to

ht(Q) > ht(P ) = ht(Q).For (2)⇒ (1), the argument of (1)⇒ (2) can be reversed since (T [τ ](n,τ))Q −→

(Ty) bQ is faithfully flat.

For (3)⇒ (2), again suppose (2) fails; that is, there exists a prime ideal P of Tof height one such that y /∈ P , P ∩T 6= 0, but ht(P ∩T [τ ]) ≥ 2. Now ht(P ∩T ) = 1,since LF1 holds for T → T . Thus, with P = P ∩ T , we have PT [τ ] < P ∩ T [τ ];that is, there exists f(τ) ∈ (P ∩ T [τ ]) − PT [τ ], or equivalently there is a nonzeropolynomial f(x) ∈ (T/(P ∩ T ) )[x] so that f(τ ) = 0 in T [τ ]/(P ∩ T [τ ]), where τdenotes the image of τ in T /P . This means that τ is algebraic over the fractionfield of T/(P ∩ T ), a contradiction to (3).

For (2)⇒ (3), let P be a height-one prime of R such that P ∩T = P 6= 0. Sinceht(P ∩ T [τ ]) = 1, P ∩ T [τ ] = PT [τ ] and T [τ ]/(PT [τ ]) canonically embeds in T /P .Thus the image of τ in T [τ ]/PT [τ ] is algebraically independent over T/P .

For (1) ⇐⇒ (4), we see by (7.5.321.16) that (1) is equivalent to the embedding

ψ : B −→ T ∗y being LF1. Now (

7.5.221.15.1) and (

7.3.521.7.2) imply that when ψ is LF1, it

is also height-one preserving.

7.5.8 Theorem 21.21. Assume the setting and notation of (7.4.121.9) and in addition

that (R,m) := (T,n) is excellent. The following are equivalent:

(1) The elements τ1, . . . , τs are primarily limit-intersecting in y over R.(2) For every prime ideal P of B0 := R[τ1, . . . , τs](n,τ1,...,τs) with y /∈ PR and

dim(B0/P ) ≤ s, the extension PR is primary for the maximal ideal of R.

Proof. For (1)⇒(2), let P ∈ Spec(B0) be such that y /∈ PR and dim(B0/P ) ≤s. Suppose that PR is not mR-primary. Then there exists a minimal prime divisorQ of PR such that y 6∈ Q. It follows that ht(Q) ≤ d − 1, where d = dim(R). Put


Q := Q ∩B0; now B0 −→ Ry is LFd−1 and so the morphism

φ bQ : (B0)Q −→ (Ry) bQ bRy

is faithfully flat. Hence by going-down, as inM1[66, Theorem 4, page 33], ht(Q) ≤

d−1. But P ⊆ Q and B0 is catenary, so d−1 ≥ ht(Q) ≥ ht(P ) ≥ d, a contradiction.For (2)⇒(1), let P ∈ Spec(R) with ht(P ) ≤ d − 1. Put P = P ∩ B0 and

p = P ∩R = P ∩R. We show that the induced morphism

φ bP : (B0)P −→ R bP

is faithfully flat. ByM[68, (1) ⇐⇒ (3) of Theorem 22.3] we have to verify two

conditions:

(a): The morphism φ bP : (B0/pB0)P −→ (R/pR) bP is faithfully flat.(b): p(B0)P ⊗(B0)P

R bP ∼= pR bP

Proof of (a): We observe that the ring (B0/pB0)P is a localization of thepolynomial ring (Rp/pRp)[τ1, . . . , τn]. Hence the ring (B0/pB0)pB0 is regular andso is the ring (R/pR) bP , since R is excellent. In particular, the ring (R/pR) bP isCohen-Macaulay, and

M[68, Theorem 23.1] applies. Therefore we need only show the

following dimension formula:

dim(R/pR) bP = dim(B0/pB0)P + dim(R/P R) bP .

Since PR is contained in P and ht(P ) ≤ d − 1, our hypothesis implies thatdim(B0/P ) > s. (If dim(B0/P ) ≤ s, then PR is mR-primary.)

Claim 21.22. ht(P ) in B0 is equal to ht(PR) in R; if W ∈ Spec(R) is aminimal prime divisor of PR, then ht(W ) = ht(P ).

Proof of Claim: Let S := R[t1, . . . , ts](m,t1,...,ts), where t1, . . . , ts are indetermi-nates over R, and consider the commutative diagram:

S := R[t1, . . . , ts](m,t1,...,ts) −−−−→ S

α

y λ

yR −−−−→ B0 := R[τ1, . . . , τs](m,τ1,...,τs) −−−−→ R,

where λ is the surjection with kernel (t1 − τ1, . . . , ts − τs), and α is the restriction,which is an isomorphism. Let Q ∈ Spec(S) correspond to P ∈ Spec(B0) (that is,Q := α−1(P ) and let V = λ−1(W ). Then V is minimal over (Q, ti−τi) in Ss. Wehave that ht(Q) = ht(P ) < d, y /∈ Q and dim(S/Q) > s. Let h = d−ht(P ); that is,ht(P ) = ht(Q) = d−h and dim(S/Q) = s+h. Now choose s1, . . . , sh ∈ S such thatI = (Q, s1, . . . , sh)S has height d in S. Now dim(B0/(P, α(s1), . . . , α(sh))B0)) = s.

Thus (P, α(s1), . . . , α(sh))R is primary for the maximal ideal of R by the hypothesis.Thus J = λ−1(P, α(s1), . . . , α(sh)) = (Q, s1, . . . , sh, ti − τi) is primary for themaximal ideal of S. Therefore ht(J) = s + d. But (V , s1, . . . , sh)S ⊇ JS, and soht(V ) ≥ s+ d− h. Also ht(V ) ≤ ht(Q)+ s = d− h+ s. That is, ht(V ) = s+ d− h.Now ht(W ) = d− h = ht(P ), so ht(P ) = ht(PR).


We proceed with the proof of (7.5.821.21) as follows: Let W ∈ Spec(R) be a minimal

prime divisor of PR contained in P . Then:

dim(R/P R) bP = dim(R bP )− ht(PR bP )

= dim(R bP )− ht(W )

= dim(R bP )− ht(P (B0)P )

= dim(R bP )− ht(pR bP )− (ht(P (B0)P )− ht(p(B0)P ))

= dim( (R/pR) bP )− dim( (B0/pB0)P ).

Now (a) is proved. Proof of (b): Since Rp −→ (R− p)−1(B0) is a flat extension we have that

p(B0)P ∼= pRp ⊗Rp (B0)P .

Therefore

p(B0)P ⊗(B0)PR bP ∼= (pRp ⊗Rp (B0)P )⊗(B0)P

R bP ∼= pRp ⊗Rp R bP ∼= pR bPwhere the last isomorphism is implied by the flatness of the canonical morphismRp −→ R bP . Thus (b) holds and the theorem is proved.

7.5.9 Question 21.23. It would be interesting to know if a similar statement tothat given in (

7.5.821.21) also holds without the hypothesis that T = R is an excellent

normal Noetherian domain, i.e., if T is a quasilocal Krull domain as in (7.4.121.9) does

condition (1) in (7.5.821.21) imply condition (2)?

We have the following transitive property of limit-intersecting elements.

7.5.10 Proposition 21.24. Assume the setting and notation of (7.4.121.9). Also assume

that s > 1 and for all j ∈ 1, . . . s, set A(j) := F (τ1, . . . , τj) ∩ T . Then thefollowing statements are equivalent:

(1) τ1, . . . , τs are limit-intersecting, respectively, residually limit-intersecting,respectively, primarily limit-intersecting in y over T

(2) ∀j ∈ 1, . . . , s, the elements τ1, . . . , τj are limit-intersecting, respectively,residually limit-intersecting, respectively, primarily limit-intersecting in yover T and the elements τj+1, . . . , τs are limit-intersecting, respectively,residually limit-intersecting, respectively, primarily limit-intersecting in yover A(j).

(3) There exists a j ∈ 1, . . . , s, such that the elements τ1, . . . , τj are limit-intersecting, respectively, residually limit-intersecting, respectively, pri-marily limit-intersecting in y over T and the elements τj+1, . . . , τs arelimit-intersecting, respectively, residually limit-intersecting, respectively,primarily limit-intersecting in y over A(j).

Proof. Set B(j) :=⋃∞n=1 T [τ1n, . . . , τjn](n,τ1n,...,τjn). It is clear that (2) =⇒

(3).For (3) =⇒ (1), items (

7.5.521.18) and (

7.5.221.15.1) imply that A(j) = B(j) under

each of the conditions on τ1, . . . , τj . The definitions of τj+1, . . . , τs being limit-intersecting, respectively, residually limit-intersecting, respectively, primarily limit-intersecting in y over A(j) together with (

7.5.221.15.4) imply the equivalence of the

21.6. SOME EXAMPLES 175

stated flatness properties for each of the morphisms

ϕ1 :A(j)[τj+1, . . . , τs](−) −→ A(j)∗y = T ∗y

ϕ2 :(A(j)[τj+1, . . . , τs](−))y −→ T ∗y

ϕ3 :(B(j)[τj+1, . . . , τs](−))y −→ T ∗y

ϕ4 :(T [τ1, . . . , τs](−))y −→ T ∗y

ϕ5 :T [τ1, . . . , τs](n,τ1,...,τs) −→ T ∗y .

Thusψ : B −→ T ∗

y is LFk ⇐⇒ U −→ T ∗y is LFk ⇐⇒ φ′ : U0 −→ T ∗

y is LFk⇐⇒ θ : Bn −→ T ∗

y is LFk ⇐⇒ φ : B0 −→ T ∗y is LFk .

The respective flatness properties for ϕ5 are equivalent to the conditions thatτ1, . . . , τs be limit-intersecting, respectively, residually limit-intersecting, respec-tively, primarily limit-intersecting in y over T . Thus (3) =⇒ (1).

For (1) =⇒ (2), we go backwards: The statement of (1) for τ1, . . . , τs is equiva-lent to the respective flatness property for ϕ5. This is equivalent to ϕ4 and thus ϕ3

having the respective flatness property. By (7.5.221.15.4), B(j)[τj+1, . . . , τs](−) −→ T ∗

y

has the appropriate flatness property. Also B(j) −→ B(j)[τj+1, . . . , τs](−) is flat,and so B(j) −→ T ∗

y has the appropriate flatness property. Thus τ1, . . . , τj arelimit-intersecting, respectively, residually limit-intersecting, respectively, primarilylimit-intersecting in y over T . Therefore A(j) = B(j), and so A(j) −→ T ∗

y hasthe appropriate flatness property. It follows that τj+1, . . . , τs are limit-intersecting,respectively, residually limit-intersecting, respectively, primarily limit-intersectingin y over A(j).

21.6. Some examples7.6

Let R = Q[x, y](x,y), the localized polynomial ring in two variables x and y overthe field Q of rational numbers. Then R = Q[[x, y]], the formal power series ringin x and y, is the m = (x, y)R-adic completion of R. In Chapter

idwisec20, an element

τ ∈ m = (x, y)R is defined to be residually algebraically independent over R if τ isalgebraically independent over the fraction field of R and for each height-one primeP of R such that P ∩R 6= (0), the image of τ in R/P is algebraically independentover the fraction field of R/(P ∩ R). It is shown in Theorem


idwisec20,

that if τ is residually algebraically independent over R and L is the fraction fieldof R[τ ], then L ∩ R is the localized polynomial ring R[τ ](m,τ).

In this section we present several examples of residually algebraically indepen-dent elements.

ALERT Note that this is in section 6 of the paper,

where the motivating example is given

7.6.1 Theorem 21.25. Let σ ∈ xQ[[x]] and ρ ∈ yQ[[y]] be such that the followingtwo conditions are satisfied:

(i) σ is algebraically independent over Q(x) and ρ is algebraically independentover Q(y).

(ii) trdegQQ(y, ∂nρ∂yn n∈N) > trdegQQ(x, ∂nσ

∂xn n∈N).


Then τ := σ + ρ is residually algebraically independent over Q[x, y](x,y).

Before proving Theorem7.6.121.25, we establish the existence of elements σ and

ρ satisfying properties (i) and (ii) of Theorem7.6.121.25. Let σ = ex − 1 ∈ Q[[x]]

and choose for ρ a hypertranscendental element in Q[[y]]. Recall that a powerseries ρ =

∑∞i=0 biy

i ∈ Q[[y]] is called hypertranscendental over Q[y] if the set ofall partial derivatives ∂nρ

∂yn n∈N is infinite and algebraically independent over Q(y).(Two examples of hypertranscendental elements are the Gamma function and theRiemann Zeta function.4) Thus σ, ρ satisfy the conditions of Theorem

7.6.121.25.

Alternatively, let σ = ex − 1 and ρ = e(ey−1) − 1. The conditions of Theorem

6.1 follow fromAx[9].

In either case, Theorem7.6.121.25 implies that τ := σ+ τ is residually algebraically

independent, and we have the following corollary.

7.6.2 Corollary 21.26. There exists an explicitly defined element τ ∈ (x, y)Q[[x, y]]such that τ is residually algebraically independent over Q[x, y](x,y). Therefore thelocalized polynomial ring Q[x, y, τ ](x,y,τ) is the intersection Q(x, y, τ) ∩ R.

Proof. of7.6.121.25 To show that the element τ = σ+ρ is residually algebraically

independent over R = Q[x, y](x,y), we introduce the following intermediate ring:

D := Q(x, σ) ∩Q[[x]].

Then D is an excellent discrete valuation domain with completion D = Q[[x]],and D has transcendence degree 2 over Q. There is a convenient way to de-scribe D as a directed union of polynomial rings in two variables over Q: Setσ :=

∑aix

i, where ai ∈ Q. Then the nth-endpiece for σ, defined as in (2.3),satisfies σn = x(σn+1 + an+1) and D can be obtained as

D = lim−→n→∞

Q[x, σn](x,σn) =∞⋃n=1

Q[x, σn](x,σn).

The displayed statement follows by (7.5.421.17.1): Every element of Q[[x]] which is

algebraically independent over Q(x) is also primarily limit-intersecting over thediscrete valuation domain Q[x](x).

Since σ ∈ Dx, the maximal ideal of D is (x). The structure morphism

Q[x, σn](x,σn) −→ Q[x, σn+1](x,σn+1)

is defined by the relation σn 7→ x(σn+1 + an+1).The ring T := D[y](x,y) is between R and its completion R and has completion

T = R:

R = Q[x, y](x,y) −→ T = D[y](x,y) −→ R = T = Q[[x, y]]

4The exponential function is, of course, far from being hypertranscendental.


R = Q[[x, y]]

T := D[y](x,y)

R := Q[x, y](x,y)

D := Q(x, σ) ∩Q[[x]]

= ∪Q[x, σn](x,σn)


To show that τ := σ+ρ is residually algebraically independent over R, let Q bea height-one prime ideal of R and assume that P := Q ∩ R 6= 0. Let W := Q ∩ T .It is easy to see for P = (x) or P = (y) that the image τ of τ in ¯

R = Q[[x, y]]/Qremains algebraically independent over R = Q[x, y](x,y)/P . We show:

7.6.3 Proposition 21.27. Let P ∈ Spec(R) and Q ∈ Spec(R) be height-one primesas in the paragraph above with P 6= (x) and P 6= (y). Then τ is transcendental overT := T/W , and the set σ, ρ is algebraically independent over R. In particularτ = σ + ρ is residually algebraically independent over R.

Proof. Consider the commutative diagram:

Q[[y]] ¯R = R/Q Q[[x]]

ψy→ ψx→

T = T/W

φy→ φx→Q[y](y) R = R/P Q[x](x)

All morphisms in the diagram are injective and we obtain:(a) The ring R is algebraic over the rings Q[x](x) and Q[y](y) since trdegQ(R) =

1.(b) The ring ¯

R is finite over both rings Q[[x]] and Q[[y]].

To complete the proof of (7.6.321.27) we prove the following claim:

7.6.4 Claim 21.28. trdegQ(T ) = 2, and thus trdegR(T ) = 1.

Proof. of (7.6.421.28) Let W0 = W ∩ Q[x, y, σ]. Since D[1/x] is a localization

of Q[x, σ](x,σ), we see that T [1/x] is a localization of Q[x, y, σ]/W0. Now W0 hasheight one because x /∈W0. This shows that trdegQ(T ) = 2.

Proof. of (7.6.321.27), continued We have seen that the element σ is algebraically

independent over R. Now σ ∈ T , and trdeg[T : Q] = 1+ trdeg[D : Q] = 3, whereas


trdeg[R : Q] = 2. Also trdeg[T : Q] ≤ 2, and trdeg[R : Q] ≤ 1. Thus to show that τis transcendental over T is equivalent to showing that the set σ, ρ is algebraicallyindependent over R. In order to show this we make use of the differential propertiesof the functions σ and ρ. We first pass to the embeddings of the fraction fields:

Q((y)) Q( ¯R) := Q(Q[[x, y]])

Q(T ) := Q(y, σ, x)

Q(y) Q(R) := Q(y, x)

We have Q(R) = Q(y, x) and Q(T ) = Q(y, σ, x) where x is algebraic over Q(y),and y and σ are algebraically independent over Q. Let d, respectively d, denotethe partial derivative map ∂

∂y on Q((y)), respectively on Q(y). Note that d is

the restriction of d to Q(y). Since all the horizontal field extensions are separablealgebraic, d and d extend uniquely to derivations d1 and d1 of Q( ¯

R) := Q(Q[[x, y]]),respectively Q(R) := Q(x, y). Again d1 is the restriction of d1 to Q(x, y). Supposethat the height-one prime ideal P in R = Q[x, y](x,y) is generated by the primeelement p(x)

p(x) :=m∑i=0

ai(y)xi ∈ Q[x, y], where ai(y) ∈ Q[y].

Then p(x) = 0. We assign the notation p′(x):

p′(x) :=∂p

∂x=

m∑i=1

iai(y)xi−1 6= 0,

because x is separable over Q[y]. Also, since 0 = d1(p(x)) ),

d1(p(x)) =∂p(x)∂y

=m∑i=0

[∂ai(y)∂y

xi + ai(y)i(x)i−1d1(x)] =⇒

−m∑i=0

∂ai(y)∂y

xi = d1(x)m∑i=0

iai(y)(x)i−1 = d1(x)p′(x) = d1(x)p′(x).

Thus, we have shown that p′(x)d1(x) ∈ ¯R.

Next we show

7.6.5 Claim 21.29. For every element λ ∈ ¯R we have that p′(x)d1(λ) ∈ ¯

R.

Proof. of (7.6.521.29) Let q(x, y) ∈ Q[[x, y]] be a prime element generating Q.

Since x and y are not contained in P , the element q(x, y) is regular in x (in thesense of Zariski-Samuel

ZSII[106, p.145]). Thus by

ZSII[106, Corollary 1, p.145] the element

q(x, y) can be written as:

q(x, y) = ε(x, y)(xn + bn−1(y)xn−1 + . . .+ b0(y)),


for some unit ε(x, y) ∈ Q[[x, y]], where each bi(y) ∈ Q[[y]]. Now Q is also gener-ated by ε−1q, and thus ¯

R = Q[[x, y]]/Q is a finite free Q[[y]]-module with basis1, x, . . . , xn−1. Thus every element λ ∈ ¯

R can be written as:

λ = cn−1(y)xn−1 + . . .+ c1(y)x+ c0(y), where ci ∈ Q[[y]].

This implies:

d1(λ) = d1(x)((n− 1)cn−1(y)xn−2 + . . .+ c1(y)) +n−1∑i=0

d1(ci(y))xi.

Now the sum expression on the right is in ¯R. But also, by the earlier argument,

p′(x)d1(x) ∈ ¯R and so p′(x)d1(λ) ∈ ¯

R.

Note For convenience we drop the bars on x, y, σ, ρ. For the remainder of thisproof, x, y are considered in ¯

R. Thus we rewrite the last result as: p′(x)d1(λ) ∈ ¯R.

7.6.6 Claim 21.30. d1(σ) = d1(x)∂σ∂x and for all n > 1, d n1 (σ) is a linear combina-

tion of ∂iσ∂xi over Q(x, y) = Q(R), where 1 ≤ i ≤ n. (Note that d1 : Q( ¯

R) −→ Q( ¯R)

and that its restriction d1|Q(R) : Q(R) 7→ Q(R) is a derivation of Q(R).)

Proof. of (7.6.621.30) For all m ∈ N we have that

σ =m∑i=1

aixi + xm+1λ where λ =

∞∑i=m+1

aixi−(m+1) ∈ ¯

R and ai ∈ Q.

Therefore

p′(x)d1(σ) = p′(x)d1(x)m∑i=1

iaixi−1 + p′(x)d1(x)xmλ+ xm+1p′(x)d1(λ).

By Claim7.6.521.29

p′(x)d1(σ)− p′(x)d1(x)σ ∈ (xm) ¯R for all m ∈ N.

Since we are in a domain, it follows that d1(σ) = d1(x)∂σ∂x , as desired. The secondstatment of (

7.6.621.30) follows by induction.

Proof. Completion of proof of Proposition7.6.321.27. The fieldQ(T , ∂nσ

∂xn n∈N) =Q(y, σ, x, ∂nσ

∂xn n∈N) is closed under d1 and has the same transcendence degreeover Q as the field Q(x, ∂nσ

∂xn n∈N). Now d1 extends to the algebraic closure ofQ(y, σ, x, ∂nσ

∂xn n∈N) uniquely. If τ is algebraic over Q(T ), then the set ∂nτ∂yn n∈N

is contained in the algebraic closure of the field Q(T , ∂nσ∂xn n∈N). But this is im-

possible, since the transcendence degree of Q(y, ∂nσ∂xn n∈N) is too large.

An alternative way of saying this is as follows: The fraction field of T , Q(T ) =Q(y, σ, x) is generated by y, σ, and x, which are all mapped into Q(T ) under d1.This shows that the fieldQ(T ) is closed under the derivation d1. Moreover, since therationals are contained in Q(T ), this derivation extends uniquely to the algebraicclosure. If ρ is algebraic over Q(T ) all its partial derivatives are algebraic overthe same field, a contradiction to our assumption that ∂nρ

∂yn is an algebraically

independent set of elements over Q. Now Q((y)) → Q( ¯R) and d1 is the partial


derivative : Q((y)) → Q((y)). This shows that ρ is transcendental over Q(T ) andhence τ is transcendental over R.

This completes the proof of Proposition7.6.321.27 and Theorem

7.6.121.25.

7.6.7 Example 21.31. The element τ = σ+ρ is residually algebraically independentover R = Q[x, y](x,y). Thus by Theorem

6.4.420.37 of Chapter idwisec, we have

Q(x, y, τ) ∩ R = Q[x, y, τ ](x,y,τ).

Since dim(R) = 2, by Theorem6.5.620.52 of Chapter

idwisec20, the element τ is also

primarily independent over R = Q[x, y](x,y) in the sense of Definition6.3.120.22 of

Chapteridwisec20, that is, for every prime ideal P of S = R[τ1, . . . , τn](m,τ1,...,τn) such

that dim(S/P ) ≤ n, the ideal PR is mR-primary.

7.6.8 Example 21.32. For S := Q[x, y, z](x,y,z), the construction of (7.6.321.27) yields an

example of a height-one prime ideal P of S = Q[[x, y, z]] in the generic formal fiberof S such that

Q(S) ∩ (S/P ) = S.

Proof. Let P := (z − τ) ⊆ Q[[x, y, z]], where τ is the element of Theo-rem

7.6.121.25. Then Q(x, y, z)∩S/P can be identified with the intersection Q(x, y, τ)∩

Q[[x, y]] of (6.1). Therefore

Q(x, y, z) ∩ (S/P ) = S = Q[x, y, z](x,y,z).

Since every prime ideal maximal in the generic formal fiber of a polynomialring in one variable over a two-dimensional ring has height 2, the prime ideal P isnot maximal in the generic formal fiber of S = Q[x, y, z](x,y,z).

Example7.6.821.32 demonstrates that the strong connection between the maximal

ideals of the generic formal fiber of a localized polynomial ring and certain bira-tional extensions of this localized polynomial ring does not extend to prime idealsnonmaximal in the generic formal fiber of this ring. (See

HRS[33] for more details.)

7.6.9 Example 21.33. Again let S = Q[x, y, z](x,y,z). With a slight modification ofExample

7.6.821.32, we exhibit a prime ideal P in the generic formal fiber of S which

does correspond to a nontrivial birational extension; that is, the intersection ring

A := Q(S) ∩ S/Pis a spot over S.

Proof. Let τ be the element from Theorem 6.1. Let P = (z−xτ) ⊆ Q[[x, y, z]].Since τ is transcendental over Q(x, y, z), the prime ideal P is in the generic formalfiber of S. The ring S can be identified with a subring of S/P ∼= Q[[x, y]] byconsidering S = Q[x, y, xτ ](x,y,xτ). By reasoning similar to that of Example

7.6.821.32,

Q(S) ∩Q[[x, y]] = Q(x, y, τ) ∩Q[[x, y]] = Q[x, y, τ ](x,y,τ).

The ring Q[x, y, τ ](x,y,τ) is then the essentially finitely generated birational exten-sion of S defined as S[z/x](x,y,z/x).


Example7.6.921.33 is of interest in connection with

HRS[33], where it is shown that

if the prime ideal P of Q[[x, y, z]] is maximal in the generic formal fiber of S =Q[x, y, z](x,y,z), then the intersection ring Q(x, y, z) ∩ Q[[x, y, z]]/P is well under-stood; whereas the last two examples show if P is not maximal in the generic formalfiber, then the intersection ring can be almost anything.

ALERT Note that many times there should be a ref7.– in place of 5.3ALERT: also for 5.4.1 or etc.

7.6.10 Example 21.34. Let σ ∈ xQ[[x]] and ρ ∈ yQ[[y]] be as in Theorem7.6.121.25. If

D := Q(x, σ)∩Q[[x]] =⋃∞n=1 Q[x, σn](x,σn) and T := D[y](x,y), so T is regular local

with completion T = Q[[x, y]], then the element ρ is primarily limit-intersecting iny over T .

Proof. We show that the morphism φy : T [ρ] −→ Q[[x, y]]y is LF1; that is,the induced map φ bP : T [ρ] bP∩T [ρ] −→ Q[[x, y]] bP is flat for every height-one prime

ideal P of Q[[x, y]] with y /∈ P . It is equivalent to show for every height-oneprime P of Q[[x, y]] that P ∩ T [ρ] has height ≤ 1. If P = (x), the statement isimmediate, since ρ is algebraically independent over Q(y). Next we consider thecase P ∩ Q[x, y, σ] = (0). Since Q(x, y, σ) = Q(x, y, σn) for every positive integern, P ∩Q[x, y, σ] = (0) if and only if P ∩Q[x, y, σn] = (0). Moreover, if this is true,then since the fraction field of T [ρ] has transcendence degree one over Q(x, y, σ),then P ∩T [ρ] has height ≤ 1. The remaining case is where P := P ∩Q[x, y, σ] 6= (0)and xy /∈ P . By Proposition 6.3, ρ is transcendental over T = T/(P ∩ T ), and thisis equivalent to ht(P ∩ T [τ ]) = 1. (For an alternative proof see ((

8.3.522.17) of Chapter

noeslocsec22)

Still referring to ρ, σ, σn as in (7.6.121.25 and (

7.6.1021.34) and using that σ is primarily

limit-intersecting in y over T , we have:

A := Q(T )(ρ) ∩Q[[x, y]] = lim−→T [ρn](x,y,ρn) = lim−→Q[x, y, σn, ρn](x,y,σn,ρn)

where the endpieces ρn are defined as in (2.3); viz., ρ :=∑∞n=1 biy

i and ρn =∑∞i=n+1 biy

i−n. The philosophy here is that sufficient “independence” of the alge-braically independent elements σ and ρ allows us to explicitly describe the inter-section ring A.

The previous examples have been over localized polynomial rings, where we arefree to exchange variables. The next example shows, over a different regular localdomain, that an element in the completion with respect to one regular parameterx may be residually limit-intersecting with respect to x whereas the correspondingelement in the completion with respect to another regular parameter y may betranscendental but fail to be residually limit-intersecting.

7.6.11 Example 21.35. There exists a regular local ring R with R = Q[[x, y]] suchthat σ = ex− 1 is residually limit-intersecting in x over R, whereas γ = ey − 1 failsto be limit-intersecting in y over R.

Proof. Let ωii∈I be a transcendence basis of Q[[x]] over Q(x) such that:

exnn∈N ⊆ ωii∈I .


Let D be the discrete valuation ring:

D = Q(x, ωii∈I,ωi 6=ex) ∩Q[[x]].

Obviously, Q[[x]] has transcendence degree 1 over D. The set ex is a transcen-dence basis of Q[[x]] overD. Let R = D[y](x,y). By (

7.5.421.17.1), the element σ = ex−1

is residually limit-intersecting in x over D. Moreover, by (8.3.322.15) of Chapter

noeslocsec22,

σ is also residually limit-intersecting over R := D[y](x,y). However, the elementγ = ey − 1 is not residually limit-intersecting in y over R. To see this, consider theheight-one prime ideal P := (y− x2)Q[[x, y]]. The prime ideal W := P ∩R[τ ](x,y,τ)contains the element γ − ex2 − 1 = ey − ex2

. Therefore W has height greater thanone and γ is not residually limit-intersecting in y over R.

Note that the intersection ring Q(R)(τ) ∩ Q[[x, y]] is a regular local ring withcompletion Q[[x, y]] by Valabrega

V[96].

21.7. Corrigendum to “Intermediate rings ...”,7.7.7

CHAPTER 22

Noetherian rings between a semilocal domain and

its completion May 17, 2009noeslocsec


We specify conditions in order that excellence be preserved in our general con-struction and use the construction to obtain a non-Noetherian ring A of the formK(τ) ∩R∗ which is a directed union of localized polynomial rings over R.

???alertThere is more to the introduction. We go through the history of con-structing examples with this technique. We talk about the papers idealwise andinter, and explain how the limit intersecting property arose in the inter paper. ???

In this chapter, we use the notation of Chapterappcom6. Let (R,m) be a Noetherian

semilocal domain with field of fractions K and m-adic completion (R, m), wherem denotes the Jacobson radical of R.

Let t1, . . . , tn be indeterminates over R and let (S,n) denote the localized poly-nomial ring R[t1, . . . , tn](m,t1,...,tn). Suppose that a is an ideal in the n-adic com-pletion S of S such that the associated primes of a are in the generic formal fiber ofS, i.e., each associated prime of a intersects S in (0). Then the fraction field Q(S)of S canonically embeds in the fraction field of S/a. We define the integral domainA := Q(S) ∩ (S/a). It is clear that A is a local birational extension domain of S.

The intersection domain A is the focus of our study; a wide variety of localdomains A which contain a coefficient field k and which have finite transcendencedegree over k can be obtained in this way. In particular this technique has providednormal local domains A ⊇ k, with tr.degk(A) <∞, such that A is Noetherian butnot excellent, and has provided other domains A which are non-Noetherian.

In our use of the basic construction, we consider elements τ1, . . . , τs of thecompletion R of R which are algebraically independent over K. We study theintersection ring A := K(τ1, . . . , τs)∩ R. The ring A may also be viewed as Q(S)∩(S/a), where S = R[t1, . . . , ts](m,t1,...,ts) is a localized polynomial ring over R, S isthe completion of S with respect to its maximal ideal (m, t1, . . . , ts)S and a is theprime ideal a := (t1−τ1, . . . , ts−τs)S. The ring A is the set of all fractions f/g ∈ R,where f = f(τ1, . . . , τs) and g = g(τ1, . . . , τs) ∈ R[τ1, . . . , τs](m,τ1,...,τs). In order tounderstand the intersection ring A, we investigate the arithmetic properties of theelements of the localized polynomial ring R[τ1, . . . , τs](m,τ1,...,τs) inside the completelocal ring R.

In Chapterintsec21 we introduce several “limit-intersecting” conditions in order to

analyze the birational domination of A over B0. Associated with the representationof the τi as power series in y there is a directed set Bn∞n=1 of localized polynomial

183

184 22. NOETHERIAN BETWEEN SEMILOCAL

rings in s variables over R such that Bn+1 birationally dominates Bn. The localring B =

⋃∞n=1Bn is birationally dominated by A. A main goal of Chapter

intsec21 and

this chapter is to analyze the structure of the rings B and A.In this chapter we show:(1) If B is Noetherian, then B = A (Proposition

8.2.222.2.3).

(2) If B is Noetherian, then the τi are “primarily” limit-intersecting, whichis the strongest limit-intersecting condition (cf. Definition

8.2.522.5, Remark

8.2.622.6.3).

(3) If τ1, . . . , τs ∈ yR∗ are primarily limit-intersecting, then B is Noetherianand B = A (Theorem

8.2.1222.12). 1

(4) The condition B = A does not imply that B is Noetherian even if s = 1.(In Section

8.423.1 we present a specific example where B is not Noetherian

but B = A.)Partial results are obtained in Section

8.322.3 on descent of excellence from R∗ to A.

In Section8.222.2 2 we consider a Noetherian semilocal domain R with Jacobson

radical m and field of fractions K. Let y be a nonzero element of m and let R∗ bethe (y)-adic completion of R. We define the concept of “ primarily limit-intersectingin y over R” for an element τ ∈ yR∗ which is algebraically independent over Kand, along with the results (1) and (2) listed above, we prove in Theorem

8.2.1222.12 that

τ1, . . . , τs ∈ yR∗ algebraically independent over K are primarily limit-intersectingin y over R if and only if A := K(τ1, . . . , τs)∩R∗ is simultaneously Noetherian anda directed union of localized polynomial rings in s variables over R.

In Section8.322.3 we describe a construction for intermediate domains C between

R and B. In Theorem8.3.222.14 we give equivalent conditions for the Noetherian

property to pass from B to C. We observe in (8.3.322.15) that if k is a field and

x1, . . . , xr, y are indeterminates over k, then elements τ1, . . . , τs ∈ yk[[y]] whichare algebraically independent over k(y) are primarily limit-intersecting in y overk[x1, . . . , xr, y](x1,...,xr,y). We use this in (

8.3.422.16) to discuss two classical examples.

In particular, Theorem8.3.222.14 provides a relatively easy new proof that the pseudo-

geometric non-excellent domain constructed inR1[81] is Noetherian. In Theorem

8.3.722.19 we give equivalent conditions for excellence to pass from B to C.

We present in Section8.423.1 an example of a specific excellent regular local ring

(R,m) and an element τ ∈ (R,m) which is residually limit-intersecting over R,but fails to be primarily limit-intersecting over R. This means that, although theintersection ring A = K(τ) ∩ R is well understood in the sense that we know whatis in A, A fails to be Noetherian.

22.2. Preserving Noetherian under directed unions of localizedpolynomial extensions

8.28.2.1 Setting 22.1. Let R be a Noetherian semilocal domain of dimension d > 0

with Jacobson radical m and fraction field K. Let y ∈ m be a nonzero element,let R∗ = (R, (y)) be the (y)-adic completion of R and let R∗

y = R∗[1/y]. Supposeτ1, . . . , τs ∈ yR∗ are regular elements 2 of R∗ that are algebraically independent

1We originally proved this result in the case where R is excellent and we have only one τi,i.e., where s = 1. A result of Heitmann in

H1[54] enables us to consider the situation where there

are finitely many τi and R is assumed only to be semilocal and Noetherian.2We say an element of a ring is a regular element if it is not a zero divisor.

22.2. PRESERVING NOETHERIAN 185

over K. We consider the localized polynomial ring

B0 := R[τ1, . . . , τs](m,τ1,...,τs),

where “localization at (m, τ1, . . . , τs)” denotes localization at the complement ofthe union of the finitely many maximal ideals of the polynomial ring R[τ1, . . . , τs]which are the contraction of a maximal ideal of R∗ (or, equivalently, which contain(τ1, . . . , τs) and lie over a maximal ideal of R). It is clear that B0 is a Noetheriansemilocal domain with dim(B0) = dim(R) + s such that R∗ dominates B0 and B0

dominates R.For every γ ∈ R∗ and every n > 0, we define the nth-endpiece γn with respect

to y of γ to be

8.2.1.18.2.1.1 (noeslocsec22.2.1.1) γn :=

∞∑j=n+1

cjyj−n, where γ :=

∞∑j=1

cjyj and each cj ∈ R.

In particular, we represent each of the τi by a power series expansion in y; we usethese representations to obtain for each positive integer n the nth-endpieces τinand corresponding nth-polynomial ring Un and nth-localized polynomial ring Bn.In more detail, for 1 ≤ i ≤ n and n ∈ N,

if τi :=∞∑j=1

aijyj , with aij ∈ R, then τin :=

∞∑j=n+1

aijyj−n,

Un := R[τ1n, . . . , τsn] and Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn).

Here the “localization of R[τ1n, . . . , τsn] at (m, τ1n, . . . , τsn)” is interpreted as forB0. We have a birational inclusion of polynomial rings Un ⊂ Un+1 and the localizedpolynomial ring Bn+1 birationally dominates Bn. We define

8.2.1.28.2.1.2 (noeslocsec22.2.1.2) B := ∪∞n=0Bn = lim−→Bn and A := K(τ1, . . . , τs) ∩R∗.

It is readily seen that A birationally dominates B. We say that the τi have goodlimit-intersecting behavior if B = A.

We observe the following properties of (y)-adic completions and an implicationof this concerning good limit-intersecting behavior.

8.2.2 Proposition 22.2. Assume the notation and setting of (8.2.122.1), and let B∗ and

A∗ denote the (y)-adic completions of B and A. Then(1) ykB = ykA ∩B for each positive integer k.(2) For each positive integer k, R/ykR = B/ykB = A/ykA = R∗/ykR∗ : thus

B∗ = A∗ = R∗.(3) If B is Noetherian, then R∗ is faithfully flat over B and B = A.

Proof. We have R ⊂ B ⊆ A ⊆ R∗. Since R is Noetherian, R∗ is faithfullyflat over R and hence ykR∗ ∩ R = ykR for each positive integer k. Furthermore,A = R∗ ∩K(τ1, . . . , τs) implies ykA = ykR∗ ∩A. It is clear that ykB ⊆ ykR∗ ∩B,thus for (1) and (2) it suffices to show ykR∗∩B ⊆ ykB. Moreover, if yR∗∩B = yB,it follows that ykR∗ ∩ B = ykR∗ ∩ yB = y(yk−1R∗ ∩ B), and by induction we seethat ykR∗ ∩B = ykB. Thus we show yR∗ ∩B ⊆ yB.

Let g ∈ yR∗ ∩ B. Then there is a positive integer n with g ∈ Bn and bymultiplying g by a unit of Bn if necessary, we may assume that g ∈ R[τ1n, . . . , τsn].Write g = r0 + g0 where g0 ∈ (τ1n, . . . , τsn)Bn and r0 ∈ R. From the definitionof τin, we have τin = yτin+1 + ainy, for each i with 1 ≤ i ≤ s. Thus r0 ∈


yR∗ ∩ R = yR, τinBn ⊆ yBn+1 and g ∈ yB. This completes the proof of (1)and (2). If B is Noetherian, then B∗ = R∗ is faithfully flat over B and henceB = K(τ1, . . . , τs) ∩R∗ = A.

With the notation and setting of (8.2.122.1), the representation of the τi as power

series in y with coefficients in R is, in general, not unique. However, as we observein (

8.2.322.3), the rings B and Bn are uniquely determined by the τi.

8.2.3 Proposition 22.3. Assume the notation and setting of (8.2.122.1). Then B and

the Bn are independent of the representation of the τi as power series in y withcoefficients in R.


τi =∞∑j=1

aijyj and ωi =

∞∑j=1

bijyj,

where each aij , bij ∈ R. We define the nth-endpieces τin and ωin as in (8.2.1.1noeslocsec22.2.1.1):

τin =∞∑

j=n+1

aijyj−n and ωin =

∞∑j=n+1

bijyj−n.

Then we have

τi = Σ∞j=1aijy

j = Σnj=1aijyj + ynτin = Σ∞

j=1bijyj = Σnj=1biJy

j + ynωin = ωi.


ynτin− ynωin = Σnj=1bijyj −Σnj=1aijy

j, and so τin−ωin =Σnj=1(bij − aij)yj

yn.

Since Σnj=1(bij − aij)yj ∈ R is divisible by yn in R∗ and R = K ∩ R∗, it followsthat yn divides Σnj=1(bij − aij)yj in R. Therefore τin −ωin ∈ R. It follows that Bnand B = ∪∞n=1Bn are independent of the representation of the τi.

The following elementary result on flatness is useful in considering the rings Bnand their directed union B.

8.2.4 Proposition 22.4. Assume the notation and setting of (8.2.122.1). Then the fol-

lowing are equivalent:(1) The canonical injection φ : B0 := R[τ1, . . . , τs](m,τ1,...,τs) −→ R∗

y is flat.(1′) The canonical injection φ1 : B0 := R[τ1, . . . , τs](m,τ1,...,τs) −→ Ry is flat.(2) The canonical injection φ′ : U0 := R[τ1, . . . , τs] −→ R∗

y is flat.(2′) The canonical injection φ′1 : U0 := R[τ1, . . . , τs] −→ Ry is flat.(3) The canonical injection θ : Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn) −→ R∗

y isflat.

(3′) The canonical injection θ1 : Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn) −→ Ry isflat.

(4) The canonical injection ψ : B −→ R∗y is flat.

Proof. We have:

U0loc.−−−−→ B0

φ−−−−→ R∗y

f.f.−−−−→ Ry.

The injection φ1 : B0 −→ Ry factors as φ : B0 −→ R∗y followed by the faithfully

flat injection R∗y −→ Ry. Therefore φ is flat if and only if φ1 is flat. Similarly the


injection φ′ : U0 −→ R∗y is flat if and only if φ′1 is flat, and θ : Bn −→ R∗

y is flat ifand only if θ1 is flat. The injection φ′ factors through the localization U0 −→ B0

and so φ is flat if and only if φ′ is flat.Now set Un := R[τ1n, . . . , τsn] for each n > 1 and U :=

⋃∞n=0 Un. For each

positive integer i, τi = ynτin +∑n

i=0 aiyi. Thus Un ⊆ U0[1/y], and U0[1/y] =⋃

Un[1/y] = U [1/y]. Moreover, for each n, Bn is a localization of Un, and hence Bis a localization of U .

We have:

B[1/y] −→ R∗y is flat ⇐⇒ U [1/y] −→ R∗

y is flat ⇐⇒ U0[1/y] −→ R∗y is flat

⇐⇒ Bn[1/y] −→ R∗y is flat ⇐⇒ B0[1/y] −→ R∗

y is flat.

Thus

ψ : B −→ R∗y is flat ⇐⇒ U −→ R∗

y is flat ⇐⇒ φ′ : U0 −→ R∗y is flat

⇐⇒ θ : Bn −→ R∗y is flat ⇐⇒ φ : B0 −→ R∗

y is flat.

In Chapter

intsec21 we study conditions on the τi which we call “limit-intersecting

properties” that are related to the condition B = A. In particular, we define therethe property of primarily limit-intersecting for τ1, . . . , τs, which is useful here.

8.2.5 Definition 22.5. Elements τ1, . . . , τs ∈ yR∗ are primarily limit-intersectingin y over R if τ1, . . . , τs are algebraically independent over K and the canonicalinjection φ : B0 := R[τ1, . . . , τs](m,τ1,...,τs) −→ R∗

y is flat. In view of (8.2.422.4), it is

equivalent to assume any one of φ1, φ′, or φ′1 as defined in (8.2.422.4) is flat.

8.2.6 Remarks 22.6. (1) With notation as in (8.2.522.5), if τ1, . . . , τs are primarily limit-

intersecting in y over R, then the flatness of R∗y over U0 = R[τ1, . . . , τs] implies that

every nonzero element of U0 is a regular element of R∗.(2) If R is a one-dimensional Noetherian semilocal domain, then R∗ is also

one-dimensional Noetherian and semilocal. Therefore in this case τ1, . . . , τs areprimarily limit-intersecting in y over R if and only if τ1, . . . , τs are algebraicallyindependent over K and each minimal prime of R∗ lies over (0) in U0.

(3) With the notation of (8.2.122.1.2), if B is Noetherian, then τ1, . . . , τs are pri-

marily limit-intersecting in y over R. For then R∗ is flat over B by (8.2.222.2.3), and

so is R∗y. It follows from (

8.2.422.4) that τ1, . . . , τs are primarily limit-intersecting in y

over R.

We show in (8.2.722.7) that primarily limit-intersecting elements behave well under

certain residue class formations.

8.2.7 Proposition 22.7. Let R be a Noetherian semilocal domain with Jacobsonradical m, let y be a nonzero element in m and let R∗ be the y-adic completionof R. Suppose τ1, . . . , τs ∈ yR∗ are primarily limit-intersecting in y over R andthat Q is a prime ideal of R with y /∈ Q. Let ¯ denote image in R∗

y/QR∗y. Then

τ1, . . . , τs are primarily limit-intersecting in y over R/Q.

Proof. Since the (y)-adic completion (R/Q)∗ of R/Q is canonically isomor-phic to R∗/QR∗, we see that τ1, . . . , τs are regular elements of (R/Q)∗. We showτ1, . . . , τs are algebraically independent over R/Q. 3 Since R[τ1, . . . , τs] −→ R∗

y is

3We say that elements are algebraically independent over a domain if they are algebraicallyindependent over its fraction field.


flat and QR∗y 6= R∗

y, we have QR[τ1, . . . , τs] = QR∗y ∩R[τ1, . . . , τs]. Thus

R[τ1, . . . , τs]/(QR∗y ∩R[τ1, . . . , τs]) ∼= (R/Q)[τ1, . . . , τs]

is a polynomial ring in s variables τ1, . . . , τs over R/Q. Therefore τ1, . . . , τs arealgebraically independent over R/Q.

We show flatness of the map:

φ : (R/Q)[τ1, . . . , τs](m,τ1,...,τs) −→ R∗y/QR

∗y = (R/Q)∗y.

Let P be a prime ideal of R∗/QR∗ with y /∈ P . The ideal P lifts to a prime idealP of R∗ with y /∈ P and QR∗ ⊆ P . By assumption the map

φP : R[τ1, . . . , τs](m,τ1,...,τs) −→ R∗P

is flat. The map on the residue class rings:

φP : (R/Q)[τ1, . . . , τs](m,τ1,...,τs) −→ (R∗/QR∗)Pis obtained from φP by tensoring with (R/Q)[τ1, . . . , τs](m,τ1,...,τs) over the ringR[τ1, . . . , τs](m,τ1,...,τs). Hence φ is flat.

8.2.8 Theorem 22.8. Assume the notation and setting of (8.2.122.1). Also assume that

s = 1 and that τ = τ10 ∈ yR∗ is primarily limit-intersecting in y over R. Then Bis Noetherian and B = R∗ ∩K(τ).

We use the same proof as inH1[54, Theorem 1.4] and prove first the following

lemma.

8.2.9 Lemma 22.9. With notation as in Theorem8.2.822.8, if P is a nonzero prime ideal

of B such that P ∩ R = (0), then there exists f ∈ P , u ∈ R and a positive integerN such that P = (fB :B uyN ).

Proof. The localization D := (R − 0)−1B of B at the nonzero elementsof R is also a localization of the polynomial ring U0 := R[τ10]. Hence PD is a aprincipal maximal ideal of D and there exists a polynomial f ∈ R[τ10] such thatPD = fD.

We use the fact that B is the directed union of the localized polynomial ringsBn = R[τ1n](m,τ1n), where Un := R[τ1n] and U = ∪∞n=0Un. Hence P is generated byits contraction to U . Let Pn = P ∩Un. Since DPD = (U0)P0 and U0 is Noetherian,there exists u ∈ R such that P0 = (fU0 :U0 u). Also for g ∈ U there exists apositive integer b(g), depending on g, such that yb(g)g ∈ U0. Hence for g ∈ P ∩ Uwe have uyb(g)g ∈ fU0.

The Artin-Rees LemmaN2[72, (3.7)] applied to the ideals yR∗ and fR∗ of the

Noetherian ring R∗ implies the existence of a positive integer N such that form ≥ N we have

fR∗ ∩ (yR∗)m = (yR∗)m−N ((fR∗ ∩ (yR∗)N ) = (ym−N )R∗(fR∗ ∩ yNR∗).

We may assume that b(g) ≥ N .Suppose g ∈ P ∩ U . Then uyb(g)g ∈ fU0 ⊆ fU , so

uyb(g)g ∈ fR∗ ∩ yb(g)R∗ = yb(g)−NR∗(fR∗ ∩ yNR∗).

Since y is not a zero-divisor in R∗, it follows that uyNg ∈ fR∗ ∩ yNR∗. ThusuyNg = ft, where t ∈ R∗. Since we also have uyb(g)g ∈ fU , it follows thatyb(g)−Nft ∈ fU , and therefore yb(g)−N t ∈ U , as f is not a zero-divisor in R∗.


Therefore yb(g)−N t ∈ yb(g)−NR∗ ∩B = yb(g)−NB by (8.2.222.2.1) and so t ∈ B. Hence

for every g ∈ P ∩U we have g ∈ (fB :B uyN). It follows that P = (fB :B uyN). As in

H1[54, Lemma 1.5], we have:

8.2.10 Lemma 22.10. With notation as in Theorem8.2.822.8, if each prime ideal P of B

such that P ∩R 6= (0) is finitely generated, then B is Noetherian.

Proof. By a Theorem of CohenN2[72, (3.4)], it suffices to show each P ∈

Spec(B). such that P ∩ R = (0) is finitely generated. Let P be a nonzero primeideal of B such that P ∩R = (0). Since the localization of B at the nonzero elementsofR is also a localization of the polynomial ring U0 := K[τ10], every prime ideal ofBproperly containing P has a nonzero intersection with R. Therefore the hypothesisimplies that B/P is Noetherian. By (

8.2.922.9), there exist u ∈ R and f ∈ P such

that P = (fB :B uyN). Since uyN is a nonzero element of R, every prime idealof B containing uyN is finitely generated, so B/uyNB is Noetherian. ThereforeB/(P ∩ uyNB) is Noetherian

N2[72, (3.16)]. Since uyn /∈ P and P is prime, we have

uyNB∩P = uyNP . Therefore B/uyNP is Noetherian. We have uyNP ⊆ fB ⊆ P .Hence B/fB, as a homomorphic image of B/uyNP , is Noetherian, and P/fB isfinitely generated. It follows that P is finitely generated.

Proof. (of Theorem8.2.822.8) Suppose B is not Noetherian and let Q ∈ Spec(R)

be maximal with respect to being the contraction to R of a non-finitely generatedprime ideal of B. Since R/yR = B/yB = R∗/yR∗ by (

8.2.222.2), we have y /∈ Q.

Since B = ∪∞n=0Bn and QBn is prime, we have QB is prime in B. Proposition8.2.722.7 implies that the image τ of τ in B/QB is primarily limit-intersecting in y overR/Q. But Lemma

8.2.1022.10 then implies that B/QB is Noetherian. This contradicts

the existence of a non-finitely generated prime ideal of B lying over Q in R. Weconclude that B is Noetherian. By (

8.2.222.2.3), we have B = R∗ ∩K(τ).

8.2.11 Remark 22.11. The proof of Theorem8.2.822.8 is essentially due to Ray Heitmann.

In his paperH1[54] Heitmann defines simple PS-extensions. For a regular element x

in a ring R and a formal power series in x transcendental over R, a simple PS-extension of R for x is an infinite direct union of simple transcendental extensionsof R. If R is Noetherian and T is a simple PS-extension of R, Heitmann provesin

H1[54, Theorem 1.4] that a certain monomorphism condition is equivalent to T

being Noetherian. Heitmann’s monomorphism condition insures that the elementf in the proof of Lemma

8.2.922.9 is a regular element in R∗. In our definition of

primarily limit-intersecting the regularity of f in R∗ follows from the flatness ofthe embedding B −→ R∗

y. Proposition8.2.722.7 implies that if τ is primarily limit

intersecting, then the ring U = lim−→R[τn] is a simple PS-extension satisfying themonomorphism condition of Heitmann. In view of Remark

8.2.622.6.3, Heitmann’s

monomorphism condition on the PS-extension determined by τ is equivalent toτ being primarily limit-intersecting. The concept of primarily limit-intersectinghowever extends to more than one element τ .

8.2.12 Theorem 22.12. Let (R,m) be a Noetherian semilocal domain and assume thenotation and setting of (2.1). Then τ1, . . . , τs ∈ yR∗ are primarily limit-intersectingin y over R if and only if B is Noetherian. If this holds, we also have B = A.

Proof. If B is Noetherian, then τ1, . . . , τs are primarily limit-intersectingin y over R by (

8.2.622.6.3). Assume, conversely, that τ1, . . . , τs are primarily limit-

intersecting in y over R. It follows that τ1 is primarily limit-intersecting in y over


R. By Theorem8.2.822.8, B(1), the directed union ring constructed with respect to τ1

in (8.2.122.1) is Noetherian and is equal to R∗ ∩ K(τ1). It also follows that τ2, . . . , τs

are primarily limit-intersecting in y over B(1) (cf. Proposition7.5.1021.24 of Chapter

intsec21). Hence a simple induction argument implies that B is Noetherian. By (

8.2.222.2.3),

B = A.

That is, we have established the following implications:

τi pr.lim.int. ⇐⇒ B Noetherian =⇒ A = B .

22.3. Applications and a classical construction of Noetherian rings8.3

We assume the following setting in this section.

8.3.1 Setting 22.13. Let (R,m) be a excellent local domain with fraction field Kand suppose that the s elements τ1, . . . , τs ∈ yR∗ are primarily limit-intersectingover R in y (Definition

8.2.522.5). Define B as in (

8.2.122.1.2). Assume that B is Noetherian

and so equal to A by (8.2.222.2.3); that is,

B := lim−→R[τ1n, . . . , τsn](m,τ1n,...,τsn) = A := K(τ1, . . . , τs) ∩R∗.

Let t1, . . . , ts, v be variables over R, let t denote the s-tuple (t1, . . . , ts), and letf(t) := f(t1, . . . , ts) be a nonzero polynomial in (m, t)R[t]. Also let τ denote thes-tuple (τ1, . . . , τs), let ρ := f(τ ) denote f(τ1, . . . , τs), let S := (R[t, v]/(f(t) −v))(m,t,v) and E := (R[f(t)])(m,f(t)). We have the following commutative diagramof R-algebra injections:

8.3.1.18.3.1.1 (noeslocsec22.3.1.1)

R[v](m,v)φ−−−−→ S := (R[t, v]/(f(t)− v))(m,t,v)yµ1

yν1E := R[f(t)](m,f(t))

φ−−−−→ R[t](m,t)yµ2

yν2R[ρ](m,ρ) = R[f(τ)](m,ρ)

φ−−−−→ B0 := R[τ ](m,τ)ψ−−−−→ R∗

y

where µ1 and ν1 are the R-algebra isomorphisms defined by µ1(v) = f(t), re-spectively ν1(ti) = ti and ν1(v) = f(t), and µ2 and ν2 take t → τ . We use theidentifications established by the µi and νi throughout the following and we identifythe three maps φ. With the same assumptions as above we put

ρn := (f(τ )− f(n∑j=1

a1jyj , . . . ,

n∑j=1

asjyj))/yn

and let

8.3.1.28.3.1.2 (noeslocsec22.3.1.2) C := lim−→R[ρn](m,ρn) = ∪∞n=1R[ρn](m,ρn) and D := K(ρ) ∩R∗.

Note that C[1/y] is a localization of E.Now define J to be the ideal of S such that

JSy =⋂P ∈ Spec(Sy)| (Sy)P is not flat over the ring R[ρ](m,ρ);

22.3. APPLICATIONS AND A CLASSICAL CONSTRUCTION OF NOETHERIAN RINGS 191

ByM[68, (24.3)] J describes the nonflat locus of Sy over the ring R[ρ](m,ρ), that is,

P ∈ Spec(Sy) |P ⊃ J = P ∈ Spec(Sy) | (Sy)P is not flat over R[ρ](m,ρ).Since Sy is an integral domain, the ideal J 6= 0.

We relate in (8.3.222.14) the Noetherian property of C to the ideal J .

8.3.2 Theorem 22.14. Assume the setting of (8.3.122.13). Then the following are equiv-

alent:(a) The element ρ := f(τ ) is primarily limit-intersecting in y over R.(b) JR∗

y = R∗y.

(c) The ring C is Noetherian.(d) The ring C is Noetherian and C = D.

(e) The composition map E := (R[f(t)])(m,f(t))φ→ S → Sy (or E → By) is

flat.

Proof. (a)=⇒(b) Let Q ∈ Spec(R∗) with y /∈ Q. Consider the morphisms:

R[f(τ)](m,f(τ))φ−−−−→ R[τ ](m,τ)

ψ−−−−→ R∗y

as in (8.3.1.1noeslocsec22.3.1.1). After localization we obtain:

R[f(τ )](φψ)−1(Q)φQ−−−−→ R[τ ]ψ−1(Q)

ψQ−−−−→ R∗Q.

Since ψQ φQ and ψQ are faithfully flat, the morphism φQ is also faithfully flat andthus J 6⊆ φ−1(Q). Hence for every prime ideal Q of R∗ with y /∈ Q we have thatJR∗ 6⊆ Q. Therefore JR∗

y = R∗y.

(b)=⇒(a) Consider again the morphisms:

R[f(τ )](φψ)−1(Q)φQ−−−−→ R[τ ]ψ−1(Q)

ψQ−−−−→ R∗Q,

where Q is a prime ideal in R∗ with y 6∈ Q. By assumption (b) the morphism φQis flat and so is the morphism ψQ, since the elements τ := τ1, . . . , τs are primarilylimit-intersecting in y over R. Thus the composition map ψQ φQ is flat and f(τ)is primarily limit-intersecting over R.

The equivalence of conditions (a) and (c) is established in (8.2.1222.12), and of (c)

and (d) in (8.2.222.2). The equivalence of (b) and (e) follows from the definition of

flatness.

???alert The following Proposition is referenced in Remark11.3.38.5: ???

We use the following result in applications of Theorem8.3.222.14.

8.3.3 Proposition 22.15. If k is a field and if τ1, . . . , τs ∈ yk[[y]] are algebraicallyindependent over k(y), then τ1, . . . , τs are primarily limit-intersecting in y overR = k[x1, . . . , xr, y](x1,...,xr,y).

Proof. By (8.2.622.6.2), τ1, . . . , τs ∈ yk[[y]] are primarily limit-intersecting in y

over k[y](y). Thus the map ψ : U0 := R[τ1, . . . , τs] −→ k[[y]]y = k((y)) is flat. Weshow that α : U0⊗k k[x1, . . . , xr] −→ R∗

y is flat, where R∗ is the (y)-adic completionof R. The flatness of ψ implies flatness of

ψx : U0 ⊗k k[x1, . . . , xr] −→ k((y))⊗k k[x1, . . . , xr].

The (y)-adic completion of k[[y]] ⊗k k[x1, . . . , xr] is k[x1, . . . , xr][[y]]. Let D =k[x1, . . . , xr][[y]](x1,...,xr,y). Then R∗ is the (y)-adic completion of D. The map α


factors as β ψx, where β is the canonical map of k[[y]][x1, . . . , xr] to the (y)-adiccompletion of D. Therefore α is flat.

8.3.4 Remark 22.16. With the setting of (8.3.122.13), Theorem

8.3.222.14 gives a necessary

and sufficient condition for the element f(τ1, . . . , τs) = ρ to be primarily limit-intersecting in y over R and thus for the ring C to be Noetherian. This constructionapplies to several classical examples:

(1) For the example inN2[72, Example 7, page 209], let k be a field of characteristic

different from two, let x and y be indeterminates over k, let R := k[x, y](x,y). Alsolet τ :=

∑∞i=1 aiy

i ∈ k[[y]] be transcendental over k(y). Let t and v be newindeterminates over k and let f(t) := (t + x)2. Let ρ := f(τ). Theorem

8.3.222.14

applies since

φ0 : k[x, y, v](x,y,v) −→ S := k[x, y, v, t](x,y,v,t)/((t+ x)2 − v)is flat. The flatness of φ0 follows, for example, from

M1[66, Corollary 2, page 152]

because φ0 is obtained by localization of a map where the coefficients of the poly-nomial (t+ x)2 − v in the indeterminate t generate the unit ideal of k[x, y, v]. 4

(2) For the example inR1[81], let k be a field of characteristic zero, let x, y, z

be indeterminates over k, let τ1 :=∑∞

i=1 aiyi ∈ k[[y]] and τ2 :=

∑∞i=1 biy

i ∈ k[[y]]be power series such that y, τ1, τ2 are algebraically independent over k. Let R :=k[x, y, z](x,y,z). Let t1, t2, v be indeterminates and let f(t1, t2) := (t1 + x)(t2 + z).Theorem

8.3.222.14 applies since the morphism

φ0 : k[x, y, z, v](x,y,z,v) −→ S := (k[x, y, z, t1, t2, v])(x,y,z,t1,t2,v)/((t1 +x)(t2 +z)−v)is flat. The flatness of φ0 follows, for example, from

M1[66, Corollary 2, page 152]

because φ0 is obtained by localization of a map where the coefficients of the poly-nomial (t1 + x)(t2 + z) − v in the indeterminates t1, t2, generate the unit ideal ofk[x, y, z, v]. 5

8.3.5 Remark 22.17. If k is a field of characteristic zero, x1, . . . , xr and y are indeter-minates over k and τ1, . . . , τs ∈ k[[y]] are algebraically independent over k(y), thenwith R := k[x1, . . . , xr, y](x1,...,xr,y) as in (

8.3.322.15), the rings A and B of (

8.2.1.2noeslocsec22.2.1.2)

are equal; furthermore

B = A := k(x1, . . . , xr, y, τ1, . . . , τs) ∩ k[[x1, . . . , xr, y]] = V [x1, . . . , xr](y,x1,...,xr),

where V := k[[y]] ∩ k(y, τ1, . . . , τs) is an excellent DVR of characteristic zeroG[30,

Chap. IV],R1[81, Folgerung 3, page 108]. To show the equality, note that D :=

V [x1, . . . , xr](y,x1,...,xr) is birationally dominated by B = A and D and B have thesame completion k[[x1, . . . , xr, y]]. Therefore D = B, since they are both equalto the intersection of k[[x1, . . . , xr, y]] with the field Q(D) = Q(B). Hence B isexcellent in this case.

By (8.2.622.6) and (

8.3.422.16), τ1, . . . , τs are primarily limit intersecting in y over R.

4To match the notation inN2[72] to our notation in (

8.3.122.13), note that k[x, y](x,y) in Nagata

corresponds to R in (8.3.122.13), x in Nagata corresponds to y in (

8.3.122.13), w corresponds to τ1, s = 1,

z corresponds to ρ, and the R in Nagata corresponds to C in (8.3.122.13).

5To match the notation inR1[81] to our notation in (

8.3.122.13), note that k[X,Y, Z](X,Y,Z) in

R1[81]

corresponds to R in (8.3.122.13), X, τ, σ, ω0 in

R1[81] correspond to y, τ1, τ2, ρ in (

8.3.122.13), (Y + τ)(Z +σ)

inR1[81] corresponds to f(τ1, τ2) = (x + τ1)(z + τ2) in (

8.3.122.13), and the ring R of

R1[81] corresponds

to C in (8.3.122.13).

22.3. APPLICATIONS AND A CLASSICAL CONSTRUCTION OF NOETHERIAN RINGS 193

With the notation of (8.3.122.13), we examine conditions under which B is excellent

implies C is excellent.

8.3.6 Notation 22.18. Assume the setting of (8.3.122.13). Also assume that B is excel-

lent. Consider the local injective morphism

R[ρ](m,ρ)φ−−−−→ S := R[τ1, . . . , τs](m,τ1,...,τs).

Let φy : R[ρ](m,ρ) −→ Sy denote the composition of φ followed by the canonicalmap of S to Sy. Let H be the ideal of S such that HSy describes the nonregularlocus of the map φy, that is,

HSy =⋂P ∈ Spec(Sy)|(φy)P : (R[ρ](m,ρ))φ−1

y (P ) −→ (Sy)P is not regular .If P ∈ Spec(Sy) and HSy ⊆ P then (φy)P is not a regular map by

G[30, IV, 6.8.7,

page 153]. For an explicit description of the ideal H , seeE[24].

8.3.7 Theorem 22.19. Assume the setting and notation of (8.3.122.13) and (

8.3.622.18). The

following are equivalent:(a) The ring C is excellent.(b) HR∗

y = R∗y.

(c) φy is a regular morphism.

Proof. That conditions (b) and (c) are equivalent is clear from the definitionof a regular morphism.

(a)=⇒(b) We have embeddings:

CΦ−−−−→ B

Ψ−−−−→ R∗ Γ−−−−→ R.

Since C,B and R∗ are all excellent with the same completion R,M[68, Theorem

32.1] implies Φ is regular. Let Q ∈ Spec(R∗) with y /∈ Q. After localization themorphism

C(ΨΦ)−1(Q)ΦQ−−−−→ BΨ−1(Q)

remains regular. Since y /∈ Q the morphism ΦQ can be identified with

R[v](ψφ)−1(Q)@ > φQ >> R[t1, . . . , ts, v]/(f(t)− v))ψ−1(Q) := Sψ−1(Q)

showing that φQ is regular. Hence H 6⊆ ψ−1(Q). It follows that HR∗y = R∗

y.(b)=⇒(a) By Theorem

8.3.222.14 the ring C is Noetherian with y-adic completion

R∗. Therefore the completion of C with respect to the powers of its maximal idealis R. Therefore C is formally equidimensional. Hence by

M[68, , Theorem 31.6,

page 251], C is universally catenary. To show C is excellent, it remains to showthat C is a G-ring. Consider the morphisms

CΦ−−−−→ B and C

ΓΨΦ−−−−→ R.

In view ofM[68, Theorem 32.2] it suffices to show that Φ is a regular morphism. Let

P ∈ Spec(C). If y ∈ P then R∗ ∩ Q(C) = C implies yR∗ ∩ Q(C) = yC. ThereforeC/yC = R∗/yR∗ = R/yR. Since R is excellent, the ring R⊗C k(P ) is geometricallyregular over k(P ) = (C/PC)P . If y /∈ P we show that for every finite field extensionL of k(P ) the ring B ⊗C L is regular. Let W be a prime ideal in B ⊗C L whichcontracts to a prime ideal W in B. Then W lies over P in C. Let Q ∈ Spec(R∗)be a prime ideal with Ψ−1(Q) = W . Since Φ−1(W ) = (Ψ Φ)−1(Q) = P , we knowthat y /∈ Q and thus H 6⊆ Q implying that H 6⊆W . Therefore the map

R[v](ψφ)−1(Q)@ > φQ >> R[t1, . . . , ts, v]/(f(t)− v))ψ−1(Q)


is regular. Since y /∈ Q we have the following identifications:

R[v](ψφ)−1(Q)φQ−−−−→ R[t1, . . . , ts, v]/(f(t)− v))ψ−1(Q)∥∥∥ ∥∥∥

CPΦQ−−−−→ BW

which imply that ΦQ is regular. Hence the ringBW⊗CPL is regular. But (B⊗CL)Wis a localization of BW ⊗CP L. Therefore B ⊗C L is regular. Thus Φ is a regularmorphism and C is excellent.

8.3.8 Remark 22.20. (1) For the example inN2[72, Example 7, page 209] mentioned

in (8.3.422.16.1), the map from R[ρ](m,ρ) → Sy may be identified as the inclusion map

E := k[x, y, (t+ x)2](x,y,(t+x)2)φy−−−−→ k[x, y, t](x,y,t)[1/y] = Sy.

Let Q = (x, t)Sy and let P = Q ∩E = (x, (t+ x)2)E. Then PSy is Q-primary andproperly contained in Q. Therefore the map φy is not regular.

(2) For the example inR1[81] mentioned in (

8.3.422.16.2), the map from R[ρ](m,ρ) →

Sy may be identified as the inclusion map

k[x, y, z, (t1 + x)(t2 + z)]nφy−−−−→ k[x, y, z, t1, t2]x,y,z,t1,t2 [1/y] = Sy,

where n = (x, y, z, (t1 + x)(t2 + z)k[x, y, z, (t1 + x)(t2 + z)]. Let E := k[x, y, z, (t1 +x)(t2 + z)](x,y,z,(t1+x)(t2+z) and let P be defined as P := (x, z, (t1 + x)(t2 + z))E.Then Sy/PSy = k(y)[t1, t2]/(t1t2) is a nonregular ring, so φy is not a regularmorphism.

CHAPTER 23

A non-Noetherian limit-intersecting example

nonnoeli23.1. An example where B = A is non-Noetherian8.4

In Chapterintsec21 we study several limit-intersecting concepts for elements over

a normal Noetherian local domain R, including the primarily limit-intersectingproperty and the weaker property “residually limit-intersecting”. We show in (

8.2.622.6.)

of the present paper that if the ring B from the setting of (8.2.122.1) is Noetherian then

the elements τ1 . . . , τs ∈ yR∗ are primarily limit-intersecting in y over R. In thischapter we present a specific example of an excellent regular local domain (R,m)of dimension three with m = (x, y, z)R and R = Q[[x, y, z]] such that there existsan element τ ∈ yR∗, where R∗ is the (y)-adic completion of R, with τ residuallylimit-intersecting and not primarily limit-intersecting in y over R. In this exampleB = A, but B is non-Noetherian.

We recall some definitions and results from Chapterintsec21.

8.4.1 Definition 23.1. Let φ : A −→ B be an injective morphism of Krull domains.Then φ is called LF1 (locally flat in height 1) if for every height-one prime idealQ ∈ Spec(B) the induced morphism on the localization

φQ : Aφ−1(Q) −→ BQ

is faithfully flat.

Using LF1-morphisms, we define:

8.4.2 Definition 23.2. Let (R,m) be a local analytically normal Noetherian domainand let y ∈ m be a nonzero element. Let R∗ = (R, (y)) denote the (y)-adiccompletion of R and suppose that τ ∈ yR∗ is algebraically independent over R.Then τ is residually limit-intersecting in y over R if the canonical embedding

φ : R[τ ](m,τ) −→ R∗y or, equivalently, as in (2.4) R[τ ](m,τ) −→ Ry

is an LF1-morphism.

We list a few properties of residually limit-intersecting elements from Chapterintsec21 (items (5.5) and (5.3)) :

8.4.3 Proposition 23.3. Let (R,m) be a local Noetherian analytically normal do-main and let R denote the m-adic completion of R. Let y ∈m be a nonzero elementand let R∗ denote the (y)-adic completion of R. For an element τ ∈ yR∗ which isalgebraically independent over R, the following conditions are equivalent:

(1) The element τ is residually limit-intersecting in y over R.(2) Let P ∗ be a height-one prime ideal of R∗ such that y /∈ P ∗ and P ∗ ∩R 6=

(0). Then ht(P ∗ ∩R[τ ](m,τ)) = 1.

195

196 23. A NON-NOETHERIAN LIMIT-INTERSECTING EXAMPLE

(3) If P is a height-one prime ideal in R such that y /∈ P and P ∩ R 6= (0),then ht(P ∩R[τ ](m,τ)) = 1.

(4) For every height-one prime ideal P of R such that y /∈ P and for everyminimal prime divisor P ∗ of PR∗ (resp. every minimal prime divisorP of PR), the image τ of τ in R∗/P ∗ (resp. in R/P ) is algebraicallyindependent over R/P .

With the hypothesis and notation of Proposition8.4.323.3, we define as in (

8.2.122.1):

τ = τ0 :=∞∑i=1

aiyi with ai ∈ R

τn :=∞∑

i=n+1

aiyi−n

B := lim−→R[τn](m,τn) = ∪∞n=1R[τn](m,τn)

A := K(τ) ∩R∗.

We use the following results:(a) In a Krull domain setting, if τ is residually limit-intersecting in y over R

then A = B (from Chapterintsec21, (5.5)).

(b) If B is Noetherian, then by (8.2.222.2.3), A = B, and by (

8.2.622.6.3), τ is primarily

limit-intersecting over R.We now present the specific example:

8.4.4 Theorem 23.4. There exist an excellent regular local three-dimensional do-main (R,m) contained in Q[[x, y, z]], a power series ring in the indeterminatesx, y, z over Q, the rational numbers, with m = (x, y, z)R, and an element τ in the(y)-adic completion R∗ of R such that(8.4.423.4.1) τ is residually limit-intersecting in y over R.

(8.4.423.4.2) τ is not primarily limit-intersecting in y over R.

In particular, the rings A and B constructed using τ and the process of (8.2.122.1) are

equal, yet A and B fail to be Noetherian.

Proof. We use the following elements of Q[[x, y, z]]:

γ := ex − 1 ∈ xQ[[x]], δ := ex2 − 1 ∈ xQ[[x]],

σ := γ + zδ ∈ Q[z](z)[[x]] and τ := ey − 1 ∈ yQ[[y]].

For each n, we define the endpieces γn, δn, σn and τn as in (8.2.1.1noeslocsec22.2.1.1), considering

γ, δ, σ as series in x and τ as a series in y. Thus, for example,

γ =∞∑i=1

aixi; γn =

∞∑i=n+1

aixi−n, and xnγn +

n∑i=1

aixi = γ.

(Here the ai := 1/i!. ) The δn, σn satisfy similar relations. Therefore for eachpositive integer n,

(1) Q[x, γn+1, δn+1](x,γn+1,δn+1) birationally dominates Q[x, γn, δn](x,γn,δn),(2) Q[x, γn, δn](x,γn,δn) birationally dominates Q[x, γ, δ](x,γ,δ), and(3) Q[x, z, σn+1](x,z,σn+1) birationally dominates Q[x, z, σn](x,z,σn).

23.1. AN EXAMPLE WHERE B = A IS NON-NOETHERIAN 197

For our proof of Theorem8.4.423.4 we first establish that certain subrings of Q[[x, y, z]]

can be expressed as directed unions:

8.4.5 Claim 23.5. For V := Q(x, γ, δ) ∩ Q[[x]] and D := Q(x, z, σ) ∩ Q[z](z)[[x]],the equalities (*1)-(*5) of the diagram below hold. Furthermore the ring V [z](x,z) isexcellent and the canonical local embedding ψ : D −→ V [z](x,z) is a direct limit of themaps ψn : Q[x, z, σn](x,z,σn) −→ Q[x, z, γn, δn](x,z,γn,δn), where ψ(σn) = γn + zδn.

Q[[x, y, z]]

Q[[x, y]] A := Q(x, y, z, σ, τ) ∩D[[y]]

B := ∪D[y, τn](x,y,z,τn)

R := D[y](x,y,z)V [y](x,y)

(∗5) = ∪Q[x, y, z, σn](x,y,z,σn)

D := Q(x, z, σ) ∩Q[z](z)[[x]](∗4) = ∪Q[x, z, σn](x,z,σn)

V [y, z](x,y,z)

V [z](x,z)(∗2) = Q(x, z, γ, δ) ∩Q[[x, z]]

(∗3) = ∪Q[x, z, γn, δn](x,z,γn,δn)

S := Q[z, x, γ, δ](x,z,γ,δ)

F [σ](x,z,σ)

F := Q[x, z](x,z)

V := Q(x, γ, δ) ∩Q[[x]]

(∗1) = ∪Q[x, γn, δn](x,γn,δn)

Q[x, γ, δ](x,γ,δ)


Proof. (of Claim8.4.523.5 ) By Remark

8.2.622.6.2, the elements γ and δ are primarily

limit-intersecting in x over Q[x](x) and thus by (8.2.1222.12), we have (*1):

V := Q(x, γ, δ) ∩Q[[x]] = lim−→(Q[x, γn, δn](x,γn,δn)) =⋃

Q[x, γn, δn](x,γn,δn).

By Remark8.3.522.17, V is an excellent discrete valuation domain and (*2) holds:

V [z](x,z) = Q(z, x, γ, δ) ∩Q[[z, x]].

Also V [z](x,z) is excellent.


Item (*3) is clear from (*1), and so V [z](x,z) is a directed union of the four-dimensional regular local domains given.

For (*4), for each positive integer n, the morphism

Q[x, z, σn](x,z,σn) −→ Q[x, z, γn, δn](x,z,γn,δn)

is faithfully flat. Thus the induced morphism on the direct limits:

ψn : lim−→Q[x, z, σn](x,z,σn) −→ lim−→Q[x, z, γn, δn](x,z,γn,δn)

is also faithfully flat. Now it was shown earlier that V = lim−→Q[x, γn, δn](x,γn,δn)

and thus V [z](x,z) = lim−→Q[x, z, γn, δn](x,z,γn,δn) is faithfully flat over the limitlim−→Q[x, z, σn](x,z,σn). Since V [z](x,z) is Noetherian, it follows that lim−→Q[x, z, σn](x,z,σn)

is Noetherian. Therefore by (8.2.222.2.3), ψ is a direct limit of ψn and

D = lim−→Q[x, z, σn](x,z,σn) =⋃

Q[x, z, σn](x,z,σn).

Now item (*5) follows:

R := D[y](x, y, z) = lim−→Q[x, y, z, σn](x,y,z,σn) =⋃

Q[x, y, z, σn](x,y,z,σn).

8.4.6 Claim 23.6. The ring D := Q(x, z, σ) ∩ Q[z](z)[[x]] is excellent and R :=D[y](x,y,z) is a three-dimensional excellent regular local domain with maximal idealm = (x, y, z)R and m-adic completion R = Q[[x, y, x]].

Proof. (of Claim8.4.623.6) By Valabrega

V[96, Proposition 3], D := Q(x, z, σ) ∩

Q[z](z)[[x]] is a two-dimensional regular local domain and the completion D of Dwith respect to the powers of its maximal ideal is canonically isomorphic to Q[[x, z]].

We observe that with an appropriate change of notation, Theorem8.3.722.19 applies.

Let F = Q[x, z](x,z) and let F ∗ denote the (x)-adic completion of F . Consider thelocal injective morphism

F [σ](x,z,σ)φ−−−−→ F [γ, δ](x,z,γ,δ) := S.

Let φx : F [σ](x,z,σ) −→ Sx denote the composition of φ followed by the canonicalmap of S to Sx. We have the setting of (

8.3.122.13) and (

8.3.722.19) where F plays the role

of R and V [z](x,z) plays the role of B.By Theorem

8.3.722.19, to show D is excellent, it suffices to show that φx is a

regular morphism. The map φx may be identified as the inclusion map

Q[z, x, t1 + zt2](z,x,t1+zt2)φx−−−−→ Q[z, x, t1, t2](z,x,t1,t2)[1/x]yµ yν

Q[z, x, γ + zδ](z,x,γ+zδ)φx−−−−→ Q[z, x, γ, δ](z,x,γ,δ)[1/x]

where µ and ν are the isomorphisms mapping t1 → γ and t2 → δ. Since Q[z, x, t1, t2] =Q[z, x, t1+zt2][t2] is isomorphic to a polynomial ring in one variable over its subringQ[z, x, t1 + zt2], φx is a regular morphism, so by Theorem 3.7, D is excellent. Thiscompletes the proof of (

8.4.623.6).

Proof. An alternate proof that D is excellent, again using Theorem8.3.722.19, is

as follows: The canonical embedding:

F [σ](x,z,σ)φ−−−−→ S := F [γ, δ](x,z,γ,δ)1654


may be rewritten as

F [σ](x,z,σ) −−−−→ S′ := F [σ, u, v](x,z,σ,u,v)/(σ − (u+ zv))

by means of an isomorphism identifying S and S′. Let f = σ− (u+ zv). The idealH of S′ which describes the nonsmooth locus of S′ over E[σ](x,z,σ) is given by theJacobian H = (∂f∂u ,

∂f∂v , f) = (1, z, f)S′ (see, for example,

E[24]). Again by Theorem

8.3.722.19, D is excellent.

8.4.7 Claim 23.7. The element τ := ey − 1 is in the (y)-adic completion R∗ ofR := D[y](x,y,z), but τ is not primarily limit-intersecting in y over R and the ringB (constructed using τ and the process in (2.1) ) is not Noetherian.

Proof. (of Claim8.4.723.7) Consider the height-two prime ideal P := (z, y − x)R

of R. Now y 6∈ P , so P Ry is a height-two prime ideal of Ry. Then the ideal Q :=P ∩R[τ ](m,τ) contains the element σ − τ . Thus ht(Q) = 3 and the canonical mapR[τ ](m,τ) −→ Ry is not flat. Thus by (

8.2.522.5), τ is not primarily limit-intersecting

in y over R. By (8.2.622.6.3), B is not Noetherian.

For the completion of the proof of Theorem8.4.423.4, it remains to show that τ is

residually limit-intersecting in y over R. We first establish the following claim.

8.4.8 Claim 23.8. Let P be a height-one prime ideal of R = Q[[x, y, z]], and supposeP ∩Q[[x, y]] 6= (0). Then the prime ideal P0 := P ∩R is extended from Q[x, y] ⊆ R.

Proof. We may assume P0 is distinct from (0), xR and yR since these areobviously extended.

Since R is faithfully flat over R, P0 has height one. Similarly P1 := P ∩V [y, z](x,y,z) has height at most one since R is also the completion of V [y, z](x,y,z).We also have P1∩R = P0, so P1 is nonzero and hence has height one. Since, for everyn ∈ N, σn ∈ Q[x, y, z, σ][1/x], the ring R[1/x] is a localization of Q[x, y, z, σ][1/x].Thus P ∩Q[x, y, z, σ] has height one and contains an element f which generates P0.Similarly, for every n ∈ N, γn and δn are in Q[x, y, z, γ, δ][1/x], which implies thering V [y, z](x,y,z)[1/x] is a localization of Q[x, y, z, γ, δ][1/x]. Thus P ∩Q[x, y, z, γ, δ]has height one and contains a generator g for P1. Let h ∈ Q[[x, y]] be a generatorof P ∩Q[[x, y]]. Then hR = P . The following diagram illustrates this situation:


P ⊂ Q[[x, y, z]]

P0 := P ∩R

f ∈ P ∩Q[x, y, z, σ]

P1 := P ∩ V [y, z](x,y,z)

g ∈ P ∩Q[x, y, z, γ, δ]h ∈ P ∩Q[[x, y]]

P ∩Q[x, y, γn]

P ∩Q[x, y]

Picture for proof of (8.4.823.8)

Subclaim 1: g ∈ P ∩Q[x, y, γ, δ].

Proof. (of Subclaim 1) : Write g = g0 + g1z + · · · + grzr, where the gi ∈

Q[x, y, γ, δ]. Since g ∈ P , we have g = h(x, y)φ(x, y, z), for some φ(x, y, z) ∈Q[[x, y, z]]. Since g is irreducible and P1 6= zV [y, z](x,y,z), we have g0 6= (0).

Setting z = 0, we have g0 = g(0) = h(x, y)φ(x, y, 0) ∈ Q[[x, y]]. Thus g0 ∈hQ[[x, y]] ∩Q[x, y, γ, δ] 6= (0).

Therefore gQ[x, y, γ, δ, z] = g0Q[x, y, γ, δ] is extended from Q[x, y, γ, δ], g =g0 ∈ Q[x, y, γ, δ],

Subclaim 2: If we express f as f = f0 +f1z+ · · ·+frzr, where the fi ∈ Q[x, y, γ, δ],

then f0 ∈ Q[x, y, γ].

Proof. (of Subclaim 2) Since f is an element of Q[x, y, σ, z], we can write fas a polynomial

f =∑

aijziσj =

∑aijz

i(γ + zδ)j, where aij ∈ Q[x, y].

Setting z = 0, we have f0 = f(0) =∑a0j(γ)j ∈ Q[x, y, γ].

Proof. Completion of proof of Claim8.4.823.8. Since f ∈ P ∩ Q[x, y, z, γ, δ], f =

dg, for some d ∈ Q[x, y, γ, δ, z]. Regarding d as a polynomial in z with coefficientsin Q[x, y, γ, δ] and setting z = 0, gives f0 = f(0) = d(0)g ∈ Q[x, y, γ, δ]. Thus f0 isa multiple of g. Hence g ∈ Q[x, y, γ], by Subclaim 2.

Now again using that f = dg and setting z = 1, we have d(1)g = f(1) ∈Q[x, y, γ + δ]. This says that f(1) is a multiple of the polynomial g ∈ Q[x, y, γ].Since γ and δ are algebraically independent over Q[x, y], this implies f(1) has degree0 in γ + δ and g has degree 0 in γ. Therefore g ∈ Q[x, y], d ∈ Q, 0 6= f ∈ Q[x, y],and P0 = fR is extended from Q[x, y].


By (4.3), τ is residually limit-intersecting provided we show the following:

8.4.9 Claim 23.9. Suppose P is a height-one prime ideal of R with y /∈ P and ht(P ∩R) = 1. Then the image τ of τ in R/P is algebraically independent over R/(P ∩R).

Proof. (of Claim8.4.923.9) Let P0 := P∩R and let π : Q[[x, y, z]] −→ Q[[x, y, z]]/P ;

we use ¯ to denote the image under π. If P = xR, then we have the commutativediagram:

R/P0 −−−−→ (R/P0)[τ ] −−−−→ Q[[x, y, z]]/P

∼=x ∼=

x ∼=x

Q[y, z](y,z) −−−−→ Q[y, z](y,z)[τ ] −−−−→ Q[[y, z]] .Since τ is transcendental over Q[y, z], the result follows in this case.

For the other height-one primes P of R, we distinguish two cases:case 1: P ∩Q[[x, y]] = (0).Let P1 := V [y, z](x,y,z) ∩ P . We have the following commutative diagram of localinjective morphisms:

R/P0 −−−−→ V [y, z](x,y,z)/P1 −−−−→ Q[[x, y, z]]/Px xV [y](x,y) −−−−→ Q[[x, y]] ,

where V [y, z](x,y,z)/P1 is algebraic over V [y](x,y). Since τ ∈ Q[[x, y]] is transcenden-tal over V [y](x,y), its image τ in Q[[x, y, z]]/P is transcendental over V [y, z](x,y,z)/P1

and thus is transcendental over R/P0.case 2: P ∩Q[[x, y]] 6= (0).In this case, by Claim

8.4.823.8, the height-one prime P0 := P ∩ R is extended from a

prime ideal in Q[x, y]. Let p be a prime element of Q[x, y] such that (p) = P∩Q[x, y].We have the inclusions:

G := Q[x, y](x,y)/(p) → R/P0 → R/P ,

where R/P0 = lim−→Q[x, y, z, σn](x,y,z,σn)/(p) has transcendence degree ≤ 1 overG[z](x,y,z). It suffices to show that σ and τ are algebraically independent over thefraction field Q(x, y, z) of G[z].

Let G be the integral closure of G in its fraction field and let H := Gn be alocalization of G at a maximal ideal n such that the completion H ofH is dominatedby the integral closure (R/P )′ of R/P .

Now H [z] has transcendence degree at least one over Q[z]. Also since P ∩Q[x, y] 6= 0, the transcendence degree of Q[x, y] and so of H is at most one over Q.Thus H [z] has transcendence degree exactly one over Q(z). There exists an elementt ∈ H that is transcendental over Q[z] and is such that t generates the maximalideal of the DVR H . Then H is algebraic over Q[t] and H may be regarded asa subring of C[[t]], where C is the complex numbers. In order to show that σand τ are algebraically independent over G[z], it suffices to show that σ and τ arealgebraically independent over H [z] and thus it suffices to show that these elementsare algebraically independent over Q(t, z). Thus it suffices to show that σ and τare algebraically independent over C(t, z).


We have the setup shown in the following diagram:

C(z)[[t]] (R/P )′

C[[t]] C(z)[t] H [z]

H := Gn Q[t, z] G[z](x,y,z)

Q[t] G := Q[x, y](x,y) Q[z]

ByAx[9], if x, x2, y ∈ tC[[t]] are linearly independent over Q, then:

trdegC(t)(C(t)(x, x2, y, ex, ex2, ey)) ≥ 3.

Since x, x2 and y are in H , these elements are algebraic over Q(t). Therefore ifx, x2, y are linearly independent over Q, then the exponential functions ex, ex

2, ey

are algebraically independent over Q(t) and hence σ and τ are algebraically inde-pendent over G(z).

We observe that if x, x2, y ∈ tH are linearly dependent over Q, then there exista, b, c ∈ Q such that

lp = ax+ bx2 + cy in Q[x, y].where l ∈ Q[x, y]. Since (p) 6= (x), we have c 6= 0. Hence we may assume c = 1 andlp = y− ax− bx2 with a, b ∈ Q. Since y− ax− bx2 is irreducible in Q[x, y], we mayassume l = 1. Also a and b cannot both be 0 since y 6∈ P . Thus if x, x2, y ∈ tH arelinearly dependent over Q, then we may assume

p = y − ax− bx2 for some a, b ∈ Q not both 0.

It remains to show that σ and τ are algebraically independent over G(z) pro-vided that p = y − ax− bx2, that is

y = ax+ bx2, for a, b ∈ Q, not both 0.

Suppose h ∈ G[z][u, v], where u, v are indeterminates and that h(σ, τ ) = 0. Thisimplies

h(ex + zex2, eax+bx

2) = 0.

We have eax = (ex)a and ebx2

are algebraic over G(z, ex, ex2) since a and b are

rational. By substituting z = 0 we obtain an equation over G:

h(ex, eax+bx2) = 0

which implies that b = 0 since x and x2 are linearly independent over Q. Now theonly case to consider is the case where p = y+ ax. The equation we obtain then is:

h(ex + zex2, eax) = 0

which implies that h must be the zero polynomial, since ex2

is transcendental overthe algebraic closure of the fraction field of G[z, ex]. This completes the proof ofClaim

8.4.923.9 and thus the proof of Theorem

8.4.423.4.


8.4.10 Remark 23.10. With notation as in Theorem8.4.423.4, let u, v be indeterminates

over Q[[x, y, z]]. Then the height-one prime ideal Q = (u − τ) in Q[[x, y, z, u]] isin the generic formal fiber of the excellent regular local ring R[u](x,y,z,u) and theintersection domain

K(u) ∩Q[[x, y, z, u]]/Q ∼= K(τ) ∩ R ,

where K is the fraction field of R, fails to be Noetherian. In a similar fashionthis intersection ring K(τ) ∩ R may be identified with the following ring: LetU = (u − τ, v − σ) be the height-two prime ideal in Q[[x, y, z, u, v]] which is in thegeneric formal fiber of the polynomial ring Q[x, y, z, u, v](x,y,z,u,v). Then we have:

Q(x, y, z, u, v) ∩ ((Q[[x, y, z, u, v]])/U) ∼= K(τ) ∩ R ,

and as shown in Theorem8.4.423.4, this ring is not Noetherian. We do not know an

example of a height-one prime ideal W in the generic formal fiber of a polynomialring T for which the intersection ring A = Q(T ) ∩ (T /W ) fails to be Noetherian.In Chapter intsec we present an example of such an intersection ring A whosecompletion is not equal to T , however in this example the ring A is still Noetherian.

??? alertExample 6.8 of Int is the example mentioned above. ???

CHAPTER 24

Approximating discrete valuation rings by regular

local rings (appdvrsec), May 17, 2009 12appdvrsec

24.1. Introduction

Let k be a field of characteristic zero and let (V,n) be a discrete rank-onevaluation domain containing k and having residue field V/n ∼= k. Assume that thefield of fractions L of V has finite transcendence degree s over k. For every positiveinteger d ≤ s, we prove in Theorem

12.1.224.3 that V can be realized as a directed union

of regular local k-subalgebras of V of dimension d.Suppose (R,m) is a Noetherian local domain with field of fractions F and (V,n)

is a valuation domain that birationally dominates R. In the classical case, the ringR is essentially finitely generated over a field. The concept of local uniformizationof R along V , in algebraic terms, asserts the existence of a regular local domainextension S ofR such that S is essentially finitely generated overR and is dominatedby V .

If R is a regular local ring and P is a prime ideal of R, embedded local uni-formization of R along V asserts the existence of a regular local domain extensionS of R such that S is essentially finitely generated over R and is dominated by V ,and has the property that there exists a prime ideal Q of S with Q ∩ R = P suchthat the residue class ring S/Q is a regular local ring.

If V/n is algebraic over R/m, then classical methods give a representationof V as a directed union of regular local rings all having dimension equal to thedimension of R

Abhy[2, Lemma 12]. We prove that certain rank-one discrete valuation

rings (DVRs) can be represented as a directed union of regular local domains ofdimension d for every positive integer d less than or equal to the dimension of R.We use a technique inspired by Nagata

N1[70] and examined in Chapters

fex4,

constr5,

appcom6,

motiv7,

noeslocsec22,

noehomsec8 for the construction of new Noetherian domains.The classical approach for obtaining embedded local uniformization, introduced

by Zariski in the 1940’sZ1[103], uses local quadratic transforms of R along V .

12.1.0 Definitions 24.1. Let (R,m) be a Noetherian local ring birationally domi-nated by a valuation ring (V,n).

(1) The first local quadratic transform of (R,m) along (V,n) is the ring R1 =R[m/a]m1 , where a ∈m is such that mV = aV and m1 := n ∩R[m/a].

(2) The ring R1 is also called the dilatation of R by the ideal m along VN2[72, page 141].

(3) More generally, if I ⊆ m is a nonzero ideal of R, the dilatation of R byI along V is the ring R[I/a]m1 , where a ∈ m is such that IV = aV andm1 = n ∩R[I/a].

205

206 24. APPROXIMATING DVRS

(4) For each positive integer i, we define the (i+1)st quadratic transform Ri+1

of R along V inductively: Ri+1 is the first local quadratic transform ofRi along V .

12.1.1 Remarks 24.2. Let (R,m) be a regular local ring.(1) It is well known that a local quadratic transform R1 of R is again a regular

local ringN2[72, 38.1], and it is straightforward to check that R1 is uniquely

determined by R, V and the ideal IN2[72, page 141].

(2) With the notation of Definition12.1.024.1.4, we have the following relationship

between iterated local quadratic transforms:

Ri+j = (Ri)j for all i, j ≥ 0.

(3) If R is pseudogeometric of dimension 2 and V is a valuation domain thatbirationally dominates R, a classical result of Zariski and Abhyankar as-serts that V = ∪∞n=1Rn

Abhy[2, Lemma 12].

(4) Suppose R has dimension d ≥ 3. For certain valuation rings V that bira-tionally dominate R, the union ∪∞n=1Rn of the local quadratic transformsof R along V is strictly smaller than V

Shan[90, 4.13]. In many cases Shan-

non inShan[90, (4.5), page 308] proves that V is a directed union of iterated

monoidal transforms of R, where a monoidal transform of R is a dilatationof R by a prime ideal P for which R/P is again regular.

(5) If (V,n) is a DVR that birationally dominates R, then V = ∪∞n=1Rn,where Rn is the nth quadratic transform of R along V

Abhy[2, page 336],

Z2[104,

pages 27-28]. In the case where V is essentially finitely generated over R,we have Rn = V for some n, and thus Rn+i = Rn for all i ≥ 0.

(6) Suppose R ⊆ S ⊆ V , where S is regular local ring birationally domi-nating R and V is a valuation domain birationally dominating S. Usingmonoidal transforms S. Cutkosky has shown that there exists an iteratedlocal monoidal transform T of S along V such that T is an iterated localmonoidal transform of R in

C1[19] and

C2[20].

A proof of item (5) goes as follows ( cf. also Example4.1.1r6.6): a nonzero element

η of V has the form η = b/c, where b, c ∈ R. If (b, c)V = V , then b/c ∈ Vimplies cV = V . Since V dominates R, it follows that cR = R, so b/c ∈ R inthis case. If η = b/c, with b, c ∈ R and (b, c)V = nk, we prove by induction thatη ∈ Rk. We have already done the case where k = 0. Assume for every regular localdomain (S,p) birationally dominated by V , and every nonzero element β/γ ∈ Vwith β, γ ∈ S, (β, γ)V = nj and 0 ≤ j < k we have β/γ ∈ Sj , where Sj is thej-th iterated local quadratic transform of S along V . Suppose β, γ ∈ S, β/γ ∈ Vwith (β, γ)V = nk. Let S1 = S[p/a]p1 , where a ∈ p is such that pV = aV andp1 := n ∩ S[p/a]; that is, S1 is the first local quadratic transform of S along V .Then β1 := β/a and γ1 := γ/a are in S1. Thus a ∈ n implies (β1, γ1)V = nj where0 ≤ j < k, so by induction

β/γ = β1/γ1 ∈ (S1)j = Sj+1 ⊆ Sk,which completes the proof.

24.2. Expressing a DVR as a directed union of regular local rings

We prove the following theorem:

24.3. PROOF OF THEOREM12.1.224.3 207

12.1.2 Theorem 24.3. Let k be a field of characteristic zero and let (V,n) be a DVRcontaining k with V/n = k. Assume that the fraction field L of V has finitetranscendence degree s over k. Then for every integer d with 1 ≤ d ≤ s, thereexists a nested family C(α)

n : n ∈ N, α ∈ Γ of d-dimensional regular local k-subalgebras of V such that V is the directed union of the C(α)

n and V dominateseach C(α)

n .Moreover, if the field L is finitely generated over k, then V is a countable union⋃∞

n=1 Cn, where, for each n ∈ N,

(1) Cn is a d-dimensional regular local k-subalgebra of V ,(2) Cn has field of fractions L,(3) Cn+1 dominates Cn and(4) V dominates Cn.

We have the following corollary to Theorem12.1.224.3.

12.1.3 Corollary 24.4. Let k be a field of characteristic zero and let (R,m) be alocal domain essentially of finite type over k with coefficient field k = R/m andfield of fractions L. Let (V,n) be a DVR birationally dominating R with V/n = k.For every integer d with 1 ≤ d ≤ s, s = trdegk(L), there exists a sequence of d-dimensional regular local k-subalgebras Cn of V such that V = ∪∞n=1Cn, and foreach n ∈ N, Cn+1 dominates Cn and V dominates Cn. Moreover Cn dominates Rfor all sufficiently large n.

12.1.4 Discussion 24.5. (1) If L/k is finitely generated of transcendence degree s,then the fact that V is a directed union of s-dimensional regular local rings followsfrom classical theorems of Zariski. The local uniformization theorem of ZariskiZ1[103] implies the existence of a regular local domain (R,m) containing the field ksuch that V birationally dominates R. Since k is a coefficient field for V , we have

• k → V V/n ∼= k; thus k is relatively algebraically closed in L.• R/m = k (because V dominates R).• Every iterated local quadratic transform of R along V has dimension s.

Now by Remark12.1.124.2.4, V is a directed union of s-dimensional RLRs.

(2) If d = 1, the main theorem is trivially true by taking each Cn = V . Thus ifL/k is finitely generated of transcendence degree s = 2, then the theorem is sayingnothing new.

(3) If s > 2, then the classical local uniformization theorem says nothing aboutexpressing V as a directed union of d-dimensional RLRs, where 2 ≤ d ≤ s − 1. If(S,p) is a Noetherian local domain containing k and birationally dominated by Vwith dim(S) = d < s, then S does not satisfy the dimension formula. It followsthat S is not essentially finitely generated over k

M[68, page 119].

24.3. Proof of Theorem12.1.224.3

12.1.55

We use the following remark and notation in the proof of Theorem12.1.224.3.

12.2.0 Remark 24.6. With the notation of Theorem12.1.224.3, let y ∈ n be such that

yV = n. Then the n-adic completion V of V is k[[y]], and we have

k[y](y) ⊆ V ⊆ k[[y]].


Then V = L ∩ k[[y]], for example since V → k[[y]] is flat. Since the transcendencedegree of L over k(y) is s−1, there are s−1 elements σ1, . . . , σs−d, τ1, . . . , τd−1 ∈ yVsuch that L is algebraic over F := k(y, σ1, . . . , σs−d, τ1, . . . , τd−1).

12.2.05 Notation 24.7. Continuing with the terminology of (12.2.024.6), we set

K := k(y, σ1, . . . , σs−d) and R := V ∩K.Then R is a DVR and the (y)-adic completion of R is R∗ = k[[y]]. We also haveB0 := R[τ1, . . . , τd−1](y,τ1,...,τd−1) is a d-dimensional regular local ring and V0 := V ∩F is a DVR that birationally dominates B0 and has y-adic completion V0 = k[[y]].The following diagram displays these domains:

k⊆−−−−→ K

⊆−−−−→ F F⊆−−−−→ L := Q(V )

‖x ∪|

x ∪|x ∪|

x ∪|x

k⊆−−−−→ R := V ∩K ⊆−−−−→ B0

⊆−−−−→ V0 := V ∩ F ⊆−−−−→ V

Let R∗ = (R, (y)) be the (y)-adic completion of R. Then τ1, . . . , τs ∈ yR∗

are regular elements of R∗ that are algebraically independent over K. As in No-tation

4.2.36.1 we represent each of the τi by a power series expansion in y; we use

these representations to obtain for each positive integer n the nth-endpieces τinand corresponding nth-localized polynomial ring Bn. For 1 ≤ i ≤ s, and τi :=∑∞j=1 rijy

j , where the rij ∈ R, we set, for each n ∈ N,

(12.2.0524.7.1)

τin :=∞∑

j=n+1

rijyj−n, Bn := R[τ1n, . . . , τsn](m,τ1n,...,τsn)

B :=∞⋃n=0

Bn = lim−→Bn and A := K(τ1, . . . , τs) ∩R∗.

Recall that A birationally dominates B. Also by Proposition4.5.226.3, the definition of

Bn is independent of the representations of the τi.Since V0 = F ∩ k[[y]], each τin ∈ V0 and Bn = R[τ1n, . . . , τd−1,n](y,τ1n,...,τd−1,n)

is, for each n ∈ N, the first quadratic transform of Bn−1 along V0.

In the proof of Theorem12.1.224.3, we make use of Theorem


noehomsec8 that

we summarize as Theorem12.1.624.8.

12.1.6 Theorem 24.8. Let (R,m) be a Noetherian local domain with field of fractionsK, let y ∈ m be a nonzero element, and let τ1, . . . , τs be elements of the (y)-adiccompletion R∗ of R that are algebraically independent over R. Then the followingconditions are equivalent:

(1) The ring R∗[1/y] is flat over B0 = R[τ1, . . . , τs](m,τ1,...,τs).(2) The ring R∗[1/y] is flat over R[τ1, . . . , τs].(3) The directed union B := ∪∞n=0Bn is Noetherian.

Moreover, if these equivalent conditions are satisfied, then the domain B is equalto A := K(τ1, . . . , τs) ∩R∗, and thus both are Noetherian.

We also use Proposition12.2.124.9 in the proof of Theorem

12.1.224.3:

24.3. PROOF OF THEOREM12.1.224.3 209

12.2.1 Proposition 24.9. With the notation of (12.2.0524.7), for each positive integer n, let

Bhn denote the Henselization of Bn. Then⋃∞n=1B

hn = V h0 = V h.

Proof. Since R∗y is a field, it is flat as an R[τ1, . . . , τd−1]-module. By Theo-

rem12.1.624.8, V0 =

⋃∞n=1Bn. An alternate way to justify this description of V0 is to

use Remark12.1.124.2.4 where the ring R is B0, and V is V0, and each Rn = Bn. We

haveV0 −→

algV −→ k[[y]],

where V0 and V are DVRs of characteristic zero having completion k[[y]]. SinceV0 and V are excellent, their Henselizations V h0 and V h are the set of elements ofk[[y]] algebraic over V0 or V

N2[72, (44.3)]. Thus V h0 = V h and V is a directed union

of etale extensions of V0

R3.7[85, (1.3)].

The ring C :=⋃Bhn is Henselian and contains V0, so V h0 = V h ⊆ C. Moreover,

the inclusion map V → C =⋃Bhn extends to a map V h σ→ C =

⋃Bhn . On the other

hand, the maps Bn → V extend to maps: Bhn → V h yielding a map ρ : C → V h

with σρ = 1C , and ρσ = 1V h . Thus⋃∞n=1B

hn = V h.

Proof of Theorem12.1.224.3 in the case L is finitely generated over k.

Proof. Since L is algebraic over F , it follows that L is finite algebraic overF . Since

⋃∞n=1B

hn = V h, we have

⋃∞n=1Q(Bhn) = Q(V h) and L ⊆ Q(V h). Since

L/F is finite algebraic, L ⊆ Q(Bhn) for all sufficiently large n. By relabeling, wemay assume L ⊆ Q(Bhn) for all n. Let Cn := Bhn ∩ L. Since Bn is a regular localring, Cn is a regular local ring with Chn = Bhn.

R3.7[85, (1.3)].

We observe that for every n, Cn+1 dominates Cn and V dominates Cn. Also⋃∞n=1 Cn = V . Indeed, since Bn+1 dominates Bn, we have Bhn+1 dominates Bhn and

hence Cn+1 = Bhn+1∩L dominates Cn = Bhn∩L. Since Cn = Bhn∩L ⊆ V h∩L = V ,it follows that V dominates Cn and V0 ⊆

⋃∞n=1 Cn ⊆ V . Since V birationally

dominates⋃∞n=1 Cn, it suffices to show that

⋃∞n=1 Cn is a DVR.

But by the same argument as before,⋃∞n=1 C

hn = (

⋃∞n=1 Cn)

h = V h. This showsthat

⋃∞n=1 Cn is a DVR, and therefore

⋃∞n=1 Cn = V . Thus in the case where L/k

is finitely generated we have completed the proof of Theorem12.1.224.3 including the

“moreover” statement.

12.2.2 Remark 24.10. An alternate approach to the definition of Cn is as follows.Since V is a directed union of etale extensions of V0 and Q(V ) = L is finite algebraicover Q(V0) = F , V is etale over V0 and therefore V = V0[θ] = V0[X ]/(f(X)),where f(X) is a monic polynomial such that f(θ) = 0 and f ′(θ) is a unit of V .Let B′

n denote the integral closure of Bn in L and let Cn = (B′n)(n∩B′

n). Since∪∞n=1Bn = V0, it follows that ∪∞n=1Cn = V . Moreover, for all sufficiently large n,f(X) ∈ Bn[X ] and f ′(θ) is a unit of Cn. Therefore Cn is a regular local ring forall sufficiently large n

N2[72, (38.6)]. As we note in (

12.2.324.11) below, this allows us to

deduce a version of Theorem12.1.224.3 also in the case where k has characteristic p > 0

provided the field F can be chosen so that L/F is separable.

Proof of Theorem12.2.124.9 in the case L is not finitely generated over k.

Proof. If L is not finitely generated over k, we choose a nested family of fieldsLα, with α ∈ Γ, such that


(1) F ⊆ Lα, for all α.(2) Lα is finite algebraic over F .(3) ∪α∈ΓLα = L.

The rings Vα = Lα ∩ V are DVRs with ∪α∈ΓVα = V and V hα = V h, since V0 ⊆ Vα,for each α ∈ Γ.

As above, ∪∞n=1Bhn = V h, ∪∞n=1Q(Bhn) = Q(V h) and L ⊆ Q(V h). Thus we see

that for each α ∈ Γ, there is an nα ∈ N such that Lα ⊆ Q(Bhn) for all n ≥ nα.Put C(α)

n = Lα ∩ Bhn for each n ≥ nα. Then Vα = ∪∞n=nαC

(α)n and Vα bira-

tionally dominates C(α)n . Hence

V = →α∈Γ,n≥nα

⋃C(α)n .

This completes the proof of Theorem12.1.224.3.

12.2.3 Remark 24.11. If the characteristic of k is p > 0 then the Henselization V h0 ofV0 = F ∩ k[[y]] may not equal the Henselization V h of V = L ∩ k[[y]], because thealgebraic field extension L/F may not be separable. But in the case where L/F isseparable algebraic, the fact that the DVRs V and V0 have the same completionimplies that V is a directed union of etale extensions of V0 (see, for example,

AH[4,

Theorem 2.7]). Therefore in the case where L/F is separable algebraic, V is adirected union of regular local rings of dimension d.

Thus for a local domain (R,m) essentially of finite type over a field k of char-acteristic p > 0, a result analogous to Corollary 1.3 is true provided there existsa subfield F of L such that F is purely transcendental over k, L/F is separablealgebraic, and F contains a generator for the maximal ideal of V .

In characteristic p > 0, with V excellent and the extension separable, the ringV0 need not be excellent (see for example

HRS[33, (3.3) and (3.4)]).

24.4. More general valuation rings as directed unions of RLRs12.4

A useful method for constructing rank-one valuation rings is to make use ofgeneralized power series rings as in

ZSII[106, page 101].

12.4.1 Definition 24.12. Let k be a field and let e0 < e1 < · · · be real numbers suchthat limn→∞ en =∞. For a variable t and elements ai ∈ k, consider the generalizedpower series expansion

z(t) := a0te0 + a1t

e1 + · · ·+ anten + · · ·

The generalized power series ring kt is the set of all generalized power seriesexpansions z(t) with the usual addition and multiplication.

12.4.2 Remarks 24.13. With the notation of Definition12.4.124.12, we have:

(1) The generalized power series ring kt is a field.(2) The field kt admits a valuation v of rank one defined by setting v(z(t))

to be the order of the generalized power series z(t). Thus v(z(t)) = e0 ifa0 is a nonzero element of k.

(3) The valuation ring V of v is the set of generalized power series of non-negative order together with zero. The value group of v is the additivegroup of real numbers.

24.4. MORE GENERAL VALUATION RINGS AS DIRECTED UNIONS OF RLRS 211

(4) If x1, . . . , xr are variables over k, then every k-algebra isomorphism ofthe polynomial ring k[x1, . . . , xr] into kt determines a valuation ring ofrank one on the field k(x1, . . . , xr). Moreover, every such valuation ringhas residue field k.

(5) Thus if z1(t), . . . , zr(t) ∈ kt are algebraically independent over k, thenthe k-algebra isomorphism defined by mapping xi 7→ zi(t) determines avaluation on the field k(x1, . . . , xr) of rank one. MacLane and Schillingprove in

MS[65] a result that implies for a field k of characteristic zero the

existence of a vaulation on k(x1, . . . , xr) of rank one with any preassignedvalue group of rational rank less than r. In particular, if r ≥ 2, then everyadditive subgroup of the group of rational numbers is the value group ofa suitable valuation on the field of rational functions in r variables over k.

(6) As a specific example, let k be a field of characteristic zero and considerthe k-algebra isomorphism of the polynomial ring k[x, y] into kt definedby mappping x 7→ t and y 7→ ∑∞

n=1 te1+···+en , where ei = 1/i for each

positive integer i. The result of MacLane and SchillingMS[65] mentioned

above implies that the value group of the valuation ring V defined by thisembedding is the group of all rational numbers.

CHAPTER 25

Intermediate rings between a local domain and its

completion II (intIIsec), 13, May 17 2009intIIsec

In this chapter we present results connecting flatness of extension rings to theNoetherian property for certain intermediate rings between an excellent normallocal domain and its completion. We consider conditions for these rings to haveCohen-Macaulay formal fibers. We also present several examples illustrating theseresults.


In the previous chaptersintsec21,

noeslocsec22,

noehomsec8, the authors have been analyzing and further

developing a technique for building new Noetherian (and non-Noetherian) com-mutative rings using completions, intersections, and homomorphic images. Thistechnique goes back at least to Akizuki

A[7]. It is used by Nagata in his famous

example of a normal Noetherian local domain for which the completion is not adomain

N1[70] or

N2[72, Example 7, pages 209-211]. The technique is also used by

RotthausR1[81],

R2[82], Ogoma

O1[76], Brodmann-Rotthaus

BR1[11],

BR2[12], and Heitmann

H4[57].

Let R be a commutative Noetherian integral domain with fraction field K, let ybe a nonzero nonunit of R, and let R∗ be the y-adic completion of R (the elements ofR∗ may be regarded as power series in y with coefficients in R). In Chapter

intsec21, we

consider a form of the construction involving “intermediate” rings between R andits y-adic completion: Let τ1, . . . , τs be elements of yR∗ algebraically independentover K, let S := R[τ1, . . . , τs] and define C := L ∩ R∗, where L (contained in thetotal ring of fractions of R∗) is the field of fractions of S. It follows from Chapternoeslocsec22 that C = L ∩R∗ is both Noetherian and a localization of a subring of S[1/y] ifand only if the extension

13.1.1.013.1.1.0 (intIIsec25.1.1) S → R∗[1/y]

is flat. (See Chapternoeslocsec22, Theorem

8.2.1222.12.)

In Chapternoehomsec8, we use a generalization of the intermediate rings construction

involving homomorphic images1, namely A := K ∩ (R∗/I), for certain ideals I ofR∗, where K is the field of fractions of R. For such ideals I, we show in Chapternoehomsec8, Theorem

11.3.28.3 that A = K ∩ (R∗/I) is both Noetherian and a localization of a

subring of R[1/y] if and only if R → (R∗/I)y is flat.

1As we point out in Chapternoehomsec8, the intermediate rings construction (without homomophic

images) is a special case of the homomorphic image form.

213

214 25. INTERMEDIATE OVER LOCAL DOMAINS II

In this chapter, we present test criteria, involving the heights of certain primeideals, for the extension in (

13.1.1.0intIIsec25.1.1) to be flat. 2 We give in Theorem

13.2.125.1 a

characterization of flatness in a more general related context under the conditionthat certain fibers are Cohen-Macaulay and other fibers are regular. We use this toshow in Corollary

13.2.425.4 that, in the intermediate rings version of the construction,

if C = L ∩ R∗ is both Noetherian and a localization of a subring of S[1/y], thenthe extension C = L ∩R∗ → R∗[1/y] always has Cohen-Macaulay fibers.

In contrast with this, we present an example in Section13.425.4, using the more

general homomorphic image form of the construction, where C = K ∩ (R∗/I) isNoetherian and is a localization of a subring of R[1/y], so the extension C →R∗[1/y] is flat, but fails to have Cohen-Macaulay fibers.

Rings with Cohen-Macaulay fibers are of interest in local cohomology (see forexample

BS[13]). In some instances the condition that the base ring R is a homo-

morphic image of a regular ring can be weakened to R is universally catenary withCohen-Macaulay formal fibers. Most notably Faltings’ annihilator theorem holdsfor finitely generated modules over such rings. A related application to universallycatenary local rings with Cohen-Macaulay formal fibers can be found in the theoryof generalized Cohen-Macaulay modules. These are finitely generated modules Mof dimension n > 0 for which the local cohomology modules Hi

m(M) are finitelygenerated in all possible cases, that is, for all i < n.

In Section13.325.3 we observe that every Noetherian local ring containing an ex-

cellent local subring R and having the same completion as R has Cohen-Macaulayformal fibers; this applies to the intermediate rings construction, but not to thehomomorphic images construction. In Remark

13.3.425.12, we discuss connections with

a famous example of Ogoma.We present in Section

13.425.4 non-Noetherian integral domains B ( A arising

from the intermediate rings construction. We also construct, using the homomor-phic images construction, a two-dimensional Noetherian local domain C that is ahomomorphic image of B and has the property that its generic formal fiber is notCohen-Macaulay.

Our conventions and terminology are as inM[68]. In particular, if f : A→ B is

a ring homomorphism and J is an ideal of B, then the ideal f−1(J) of A is denotedA ∩ J .

25.2. Flatness criteria13.2

We recall Theorem13.2.1p11.2.

13.2.1 Theorem 25.1. Assume that(i) R→ T is flat with Cohen-Macaulay fibers,(ii) R→ S is flat with regular fibers.

Then the following statements are equivalent:(1) S → T is flat with Cohen-Macaulay fibers.(2) S → T is flat.(3) For each prime ideal W of T , we have ht(W ) ≥ ht(W ∩ S).

2We discovered these results while making corrections to Theorem 7.5.8 of Chapterintsec21; the

theorem there was originally stated incorrectly, as is explained in Chapterintsec21. Theorem

13.2.225.2 of

this chapter is our restatement of that theorem.

25.2. FLATNESS CRITERIA 215

(4) For each prime ideal W of T such that W is minimal over nT , we haveht(W ) ≥ ht(n).

check restatement ????

The following theorem is a restatement of Chapterintsec21, Theorem

7.5.821.21:

13.2.2 Theorem 25.2. Let (R,m) be a normal excellent local domain and y ∈ m.Suppose that R∗ is the y-adic completion of R, that R is the m-adic completion ofR, and that τ1, . . . , τs ∈ yR∗ are algebraically independent over the fraction field ofR. Then the following statements are equivalent:

(1) S := R[τ1, . . . , τs](m,τ1,...,τs) → R[1/y] is flat.(2) Suppose P is a prime ideal of S and Q is a prime ideal of R minimal over

PR. If y /∈ Q, then ht(Q) = ht(P ).(3) If Q is a prime ideal of R with y /∈ Q, then ht(Q) ≥ ht(Q ∩ S).

Moreover, if any of (1)-(3) hold, then S → R[1/y] has Cohen-Macaulay fibers. Thecorresponding statements with R replaced by R∗ also hold.

Proof. (1)⇒ (2): Let P be a prime ideal of S and let Q be a prime ideal ofR that is minimal over PR and is such that y /∈ Q. The assumption of (1) impliesflatness of the map:

ϕ bQ : S bQ∩S −→ R bQ.

Furthermore the Going Down TheoremM[68, Theorem 9.5] implies that Q ∩ S = P

and byM[68, Theorem 15.1], ht(Q) = ht(P ).

(2) ⇒ (3): Let Q be a prime ideal of R with y /∈ Q. Set Q := Q ∩ S and letW be a prime ideal of R which is minimal over QR and is contained in Q. Thenht(Q) = ht(W ) by (2) since y /∈ W and therefore ht(Q) ≥ ht(Q).

(3)⇒ (1): Let Q be a prime ideal of R with y /∈ Q. Then for every prime idealW contained in Q, we also have y /∈ W , and by (3), ht(W ) ≥ ht(W ∩S). Therefore,by Theorem 2.1, ϕ bQ : S bQ∩S −→ R bQ is flat with Cohen-Macaulay fibers.

The corresponding statements, with R replaced by R∗, follow from Chapterintsec21,

Prop.7.5.321.16 or Chapter

noeslocsec22, Prop.

8.2.422.4.

As a corollary to Theorem13.2.225.2, we have the following result:

13.2.3 Corollary 25.3. With the notation of Theorem13.2.225.2, assume that R[1/y] is

flat over S. Let P ∈ SpecS with ht(P ) ≥ dim(R). Then

(1) For every Q ∈ Spec R minimal over PR we have y ∈ Q.(2) Some power of y is in PR.

Proof. Clearly items (1) and (2) are equivalent. To prove these hold, supposethat y /∈ Q. By Theorem

13.2.225.2.2, ht(P ) = ht(Q). Since dim(R) = dim(R), we have

ht(Q) ≥ dim(R). But then ht(Q) = dim(R) and Q is the maximal ideal of R. Thiscontradicts the assumption that y /∈ Q. We conclude that y ∈ Q.

Theorem13.2.225.2, together with results from Chapter

noeslocsec22 and Chapter

noehomsec8, gives the

following corollary.

13.2.4 Corollary 25.4. Let the notation be as Theorem13.2.225.2 and let L denote the

fraction field of S. Let C = L ∩R∗. Consider the following conditions:


(1) C is Noetherian and is a localization of a subring of S[1/y].(2) S → R∗[1/y] is flat.(3) S → R∗[1/y] is flat with Cohen-Macaulay fibers.(4) For every Q∗ ∈ Spec(R∗) with y 6∈ Q∗, we have ht(Q∗) ≥ ht(Q∗ ∩ S).(5) C is Noetherian.(6) C → R∗ is flat.(7) C → R∗[1/y] is flat.(8) C → R∗[1/y] is flat with Cohen-Macaulay fibers.

Conditions (1)-(4) are equivalent, conditions (5)-(8) are equivalent and (1)-(4) im-ply (5)-(8).

Proof. Item (1) is equivalent to (2) by Chapternoehomsec8 (

11.3.28.3), (2) is equivalent to

(3) and (7) is equivalent to (8) by Theorem13.2.125.1, and (2) is equivalent to (4) by

Theorem13.2.225.2.

It is obvious that (1) implies (5). By Chapternoeslocsec22, (

8.2.222.2), R∗ is the y-adic

completion of C, and so (5) is equivalent to (6). By Chapternoeslocsec22, (

8.2.222.2) and Chapter

noehomsec8, part (1) of (

11.3.18.2), (6) is equivalent to (7).

13.2.5 Remarks 25.5. (i) With the notation of Corollary13.2.425.4, if dimC = 2, it follows

that condition (7) of Corollary13.2.425.4 holds. For C is normal since R∗ is normal. Thus

if Q∗ ∈ SpecR∗ with y 6∈ Q∗, then CQ∗∩C is either a DVR or a field. The mapC → R∗

Q∗ factors as C → CQ∗∩C → R∗Q∗ . Since R∗

Q∗ is a torsionfree and hence flatCQ∗∩C -module, it follows that C → R∗

Q∗ is flat. Therefore C → R∗[1/y] is flat andC is Noetherian.(ii) There exist examples where dimC = 2 and conditions (5)-(8) of Corollary

13.2.425.4

hold, but yet conditions (1)-(4) fail to hold given in Chapterconstr5, Theorem

4.2.14.12.

13.2.6 Question 25.6. With the notation of Corollary13.2.425.4, suppose for every prime

ideal Q∗ of R∗ with y 6∈ Q∗ that ht(Q∗) ≥ ht(Q∗ ∩ C). Does it follow that R∗ isflat over C or, equivalently, that C is Noetherian?

Theorem13.2.225.2 also extends to give equivalences for the locally flat in height k

property of Chapterintsec21. This property is defined as follows:

Definition 25.7. Let φ : S −→ T be an injective morphism of commutativerings and let k ∈ N be an integer with 1 ≤ k ≤ d = dim(T ) where d is an integeror d = ∞. Then φ is called locally flat in height k, abbreviated LFk, if, for everyprime ideal Q of T with ht(Q) ≤ k, the induced morphism on the localizationsφQ : SQ∩S −→ TQ is faithfully flat.

13.2.7 Theorem 25.8. Let (R,m) be a normal excellent local domain and y ∈ m.Suppose that R∗ is the y-adic completion of R, that R is the m-adic completion ofR, and that τ1, . . . , τs ∈ yR∗ are algebraically independent over the fraction field ofR. Then the following statements are equivalent:

(1) S := R[τ1, . . . , τs](m,τ1,...,τs) → R[1/y] is LFk.(2) If P is a prime ideal of S and Q is a prime ideal of R minimal over PR

and if, moreover, y /∈ Q and ht(Q) ≤ k, then ht(Q) = ht(P ).(3) If Q is a prime ideal of R with y /∈ Q and ht(Q) ≤ k, then ht(Q) ≥

ht(Q ∩ S).

25.3. COHEN-MACAULAY FORMAL FIBERS AND OGOMA’S EXAMPLE 217

Proof. (1) ⇒ (2): Let P be a prime ideal of S and let Q be a prime idealof R that is minimal over PR with y /∈ Q and ht(Q) ≤ k. The assumption of (1)implies flatness for the map:

ϕ bQ : S bQ∩S −→ R bQ,

and we continue as in Theorem13.2.225.2.

(2) ⇒ (3): Let Q be a prime ideal of R with y /∈ Q and ht(Q) ≤ k. SetQ := Q∩S and let W be a prime ideal of R which is minimal over QR, and so thatW ⊆ Q. Then ht(Q) = ht(W ) by (2) since y /∈ W and therefore ht(Q) ≥ ht(Q).

(3) ⇒ (1): Let Q be a prime ideal of R with y /∈ Q and ht(Q) ≤ k. Thenfor every prime ideal W contained in Q, we also have y /∈ W and by (3) ht(W ) ≥ht(W ∩ S). To complete the proof it suffices to show that ϕ bQ : S bQ∩S −→ R bQ isflat, and this is a consequence of Theorem

13.2.125.1.

25.3. Cohen-Macaulay formal fibers and Ogoma’s example13.3

Our purpose in this chapter is to show that if R is excellent, then every Noe-therian example C obtained via the intermediate rings construction has Cohen-Macaulay formal fibers. We observe in Remark

13.3.425.12 that this implies the non-

Noetherian property of a certain intersection domain B that has Ogoma’s exampleas a homomorphic image.

The following is an analogue ofM[68, Theorem 32.1(ii)]. The distinction is that

we are considering regular fibers rather than geometrically regular fibers.

13.3.1 Proposition 25.9. Suppose R, S, and T are Noetherian commutative ringsand suppose we have maps R → S and S → T and the composite map R → T .Assume

(i) R→ T is flat with regular fibers,(ii) S → T is faithfully flat.

Then R→ S is flat with regular fibers.

As an immediate consequence of Theorem13.2.125.1 and Proposition

13.3.125.9, we have

the following implication concerning Cohen-Macaulay formal fibers.

13.3.2 Corollary 25.10. Every Noetherian local ring B containing an excellent localsubring R and having the same completion as R has Cohen-Macaulay formal fibers.

13.3.3 Remark 25.11. (Cohen-Macaulay formal fibers) Corollary13.3.225.10 implies that

every Noetherian local ring S that has as its completion S the formal power se-ries ring k[[x1, . . . , xd]] and that contains the polynomial ring k[x1, . . . , xd] hasCohen-Macaulay formal fibers. In connection with Cohen-Macaulay formal fibers,Luchezar Avramov pointed out to us that every homomorphic image of a regularlocal ring has formal fibers that are complete intersections and therefore Cohen-Macaulay

G[30, (3.6.4), page 118]. Also every homomorphic image of a Cohen-

Macaulay local ring has formal fibers that are Cohen-MacaulayM[68, page 181]. It

is interesting that while regular local rings need not have regular formal fibers, theymust have Cohen-Macaulay formal fibers.


13.3.4 Remark 25.12. (Ogoma’s example) Corollary13.3.225.10 sheds light on Ogoma’s

famous exampleO1[76] of a pseudo-geometric local domain of dimension three whose

generic formal fiber is not locally equidimensional.Ogoma’s construction begins with a countable field k of infinite but countable

transcendence degree over the field Q of rational numbers. Let x, y, z, w be variablesover k, and let S = k[x, y, z, w](x,y,z,w) be the localized polynomial ring. By aclever enumeration of the prime elements in S, Ogoma constructs three powerseries g, h, ` ∈ S = k[[x, y, z, w]] that satisfy the following conditions:

(a) g, h, ` are algebraically independent over k(x, y, z, w) = Q(S).(b) g, h, ` are part of a regular system of parameters for S = k[[x, y, z, w]](c) If P = (g, h, `)S, then P ∩ S = (0), i.e., P is in the generic formal fiber of

S.(d) If I = (gh, g`)S and R = Q(S)∩(S/I), then R is a pseudo-geometric local

domain 3 with completion R = S/I.(e) It is then obvious that the completion R = S/I of R has a minimal prime

gS/I of dimension 3 and a minimal prime (h, `)S/I of dimension 2. ThusR fails to be formally equidimensional. Therefore R is not universallycatenary

M[68, Theorem 31.7] and provides a counterexample to the cate-

nary chain condition.Since R is not universally catenary, R is not a homomorphic image of a regular

local ring. There exists a quasilocal integral domain B that dominates S, hascompletion S = k[[x, y, z, w]], and contains an ideal J such that R = B/J . If Bwere Noetherian, then B would be a regular local ring and R = B/J would beuniversally catenary. Thus B is necessarily non-Noetherian.

Theorem13.2.125.1 provides a different way to understand that the ring B is non-

Noetherian. To see this, we need more details about the construction of B. Thering B is defined as a nested union of rings:

Let λ1 = gh and λ2 = g` and define:

B =∞⋃n=1

S[λ1n, λ2n](x,y,z,w,λ1n,λ2n) ⊆ k[[x, y, z, w]]

where the λin are endpieces of the λi. The construction is done in such a way thatthe λ’s are in every completion of S with respect to a nonzero principal ideal. Byconstruction of the power series g, h, ` for every nonzero element f ∈ S the ringB/fB is essentially of finite type over the field k. This implies that the maximalideal of B is generated by x, y, z, w and that the completion with respect to themaximal ideal of B is the formal power series ring S = k[[x, y, z, w]]. Let K =k(x, y, z, w), then K ⊗S B is a localization of the polynomial ring in two variablesK[λ1, λ2]. Recall that I = (λ1, λ2)S and P = (g, h, `)S. Let J = I ∩ B. SinceP ∩ S = (0) we see that J = P ∩B is a prime ideal with J(K ⊗S B) a localizationof the prime ideal (λ1, λ2)K[λ1, λ2]. Thus J(K ⊗S S) = (λ1, λ2)(K ⊗S S) and P isin the formal fiber of B/J . Since (S/I) bP is not Cohen-Macaulay, Corollary

13.3.225.10

implies that B is not Noetherian.

3OgomaO1[76, page 158] actually constructs R as a directed union of birational extensions of

S. He proves that R is Noetherian and that bR = bS/I. It follows that R = Q(S)∩(bS/I). Heitmannobserves in

H4[57] that R is already normal.

25.4. RELATED EXAMPLES NOT HAVING COHEN-MACAULAY FIBERS 219

There is another intermediate ring between S and its completion k[[x, y, z, w]]that carries information about R. This is the intersection ring:

C = k(x, y, z, w, λ1, λ2) ∩ k[[x, y, z, w]].

It is shown inHRS[33, Claim 4.3] that the maximal ideal of C is generated by x, y, z, w,

and is shown inHRS[33, Claim 4.4] that C is non-Noetherian.

25.4. Related examples not having Cohen-Macaulay fibers13.4

In this section we adapt the two forms of the basic construction technique de-scribed in the introduction to obtain three rings A, B and C which we describe indetail. The setting is somewhat similar to that of Ogoma’s example. It is simplerin the sense that it is fairly easy to see that the ring C, which corresponds to R inOgoma’s example, is Noetherian. Also C is a birational extension of a polynomialring in 3 variables over a field. On the other hand this setting seems more compli-cated, since for A and B (which are the two obvious choices of intermediate rings)the ring B maps surjectively onto C, while A does not.

13.4.1 Setting and Notation 25.13. Let k be a field, and x, y, z variables over k.Let τ1, τ2 ∈ xk[[x]] be formal power series in x which are algebraically independentover k(x). Suppose that

τi =∞∑n=1

ainxn, with ain ∈ k, for i = 1, 2.

The intersection ring V := k(x, τ1, τ2) ∩ k[[x]] is a discrete valuation domain whichis a nested union of localized polynomial rings in 3 variables over k:V =

⋃∞n=1 k[x, τ1n, τ2n](x,τ1n,τ2n), where τ1n, τ2n are the endpieces:

τin =∞∑j=n

aijxj−n+1, for all n ∈ N and i = 1, 2.

We now define a 3-dimensional regular local ring D such that: (i) D is a local-ization of a nested union of polynomial rings in 5 variables, (ii) D has maximalideal (x, y, z)D and completion R = k[[x, y, z]], and (iii) D dominates the localizedpolynomial ring R := k[x, y, z](x,y,z):

13.4.1.113.4.1.1 (intIIsec25.4.1.1) D := V [y, z](x,y,z) = U(x,y,z)D∩U , where U :=

∞⋃n=1

k[x, y, z, τ1n, τ2n].

Moreover, D = k(x, y, z, τ1, τ2) ∩ R (see Chapternoeslocsec22, (

8.3.522.17)).

We consider the following elements of R:

s := y + τ1, t := z + τ2, ρ := s2 = (y+ τ1)2 and σ := st = (y + τ1)(z + τ2).

The elements s and t are algebraically independent over k(x, y, z) as are also theelements ρ and σ. The endpieces of ρ and σ are given as

ρn :=1xn

((y + τ1)2 − (y +n∑j=1

a1jxj)2)

σn :=1xn

((y + τ1)(z + τ2)− (y +n∑j=1

a1jxj)(z +

n∑j=1

a2jxj).


Obviously, the ideal I := (ρ, σ)R has height 1 and is the product of two primeideals I = P1P2 where P1 := sR and P2 := (s, t)R. Observe that P1 and P2 are theassociated prime ideals of I, and that P1 and P2 are in the generic formal fiber ofR.

We now define rings A and C as follows:

13.4.1.213.4.1.2 (intIIsec25.4.1.2) A := Q(R)(ρ, σ) ∩ k[[x, y, z]], C := Q(R) ∩ (k[[x, y, z]]/I).

In analogy with the rings D and U of (13.4.1.1intIIsec25.4.1.1), we have rings B ⊆ D and W ⊆ U

defined as follows:13.4.1.313.4.1.3 (

intIIsec25.4.1.3)

B :=∞⋃n=1

R[ρn, σn](x,y,z,ρn,σn) = W(x,y,z)B∩W , where W :=∞⋃n=1

k[x, y, z, ρn, σn].

It is clear that A, B and C are quasilocal domains and B ⊆ A with A birationallydominating B. Moreover (x, y, z)B is the maximal ideal of B.

We show in Propositions13.4.225.14 and

13.4.325.15 that A and B are non-Noetherian and

that B ( A. In Proposition13.4.525.17, we show that C is a Noetherian local domain

with completion C = R/I such that C has a non-Cohen-Macaulay formal fiber.

13.4.2 Proposition 25.14. With the notation of (13.4.125.13), the quasilocal integral do-

mains B ⊆ A both have completion R with respect to the powers of their maximalideals. Also:

(1) We have P1 ∩B = P2 ∩B,(2) B is factorial,(3) ht(P1 ∩B) > ht(P1) = 1,(4) B fails to have Cohen-Macaulay formal fibers, and(5) B is non-Noetherian.

Proof. It follows from Chapternoeslocsec22, Propositon

8.2.222.2 that R is the completion

of both A and B.For (1), it suffices to show P1 ∩W = P2 ∩W . It is clear that P1∩W ⊆ P2∩W .

Let v ∈ P2 ∩W . Then there is an integer n ∈ N such that xnv ∈ k[x, y, z, ρ, σ].Thus

xnv =∑

bijρiσj , where bij ∈ k[x, y, z], for all i, j ∈ N.

Since P2 ∩ k[x, y, z] = (0) and since ρ, σ ∈ P2 we have that b00 = 0. This impliesthat v ∈ P1. Thus (1) holds.

For (2), since B/xB = R/xR, the ideal xB = q is a principal prime ideal in B.Since B is dominated by R, we have ∩∞n=1q

n = (0). Hence Bq is a DVR. Moreover,by construction, Bx is a localization of (B0)x, where B0 := R[ρ, σ](x,y,z,ρ,σ), and(B0)x is factorial. Therefore B = Bx ∩Bq is factorial

Sam[88, (6.3), page 21].

For (3), we have Bx is a localization of the ring (B0)x and the ideal J = (ρ, σ)B0

is a prime ideal of height 2. LetQ = P1∩B; then x /∈ Q and BQ = (B0)J . ThereforehtQ = 2. Since P1 = sR has height one, this proves (3).

Item (3) implies (5), since ht(P1 ∩ B) > htP1 implies that B → R fails tosatisfy the going down property, so R is not flat over B and B is not Noetherian.

For (4), as we saw above, Qk[[x, y, z]]P2 = (ρ, σ)P2 = IP2 . Thus RP2/IRP2 isa formal fiber of B. Since k[[x, y, z]]/I = k[[x, s, t]]/(s2, st), we see that P2/I =(s, t)R/(s2, st)R is an embedded associated prime of the ring k[[x, y, z]]/I. Hence


(k[[x, y, z]]/I)P2 is not Cohen-Macaulay and the embedding B −→ k[[x, y, z]] failsto have Cohen-Macaulay formal fibers. This also implies that B is non-Noetherianby Corollary

13.2.425.4.

13.4.3 Proposition 25.15. With the notation of (13.4.125.13) we have:

(1) A is a quasilocal Krull domain with maximal ideal (x, y, z)A and comple-tion R,

(2) P1 ∩A ( P2 ∩A, so B ( A,(3) A is non-Noetherian.

Proof. For item (1), it follows from Chapternoeslocsec22,Propositon

8.2.222.2 that (x, y, z)A

is the maximal ideal of A. It remains to observe that A is a Krull domain, and thisis clear since by definition A is the intersection of a field with the Krull domain R.

For item (2), let Qi := Pi ∩A, for i = 1, 2. Observe that

σ2/ρ = (z + τ2)2 = t2 ∈ (Q2 −B)−Q1.

For item (3), assume A is Noetherian. Then A is a regular local ring and theembedding A −→ R = k[[x, y, z]] is flat. In particular, A is factorial and the idealP := sR ∩ A = P1 ∩ A is a prime ideal of height one in A. Thus P is principal.We have that ρ = s2 ∈ P and σ2 = ρ(σ2/ρ), therefore st = σ ∈ P . Let v be agenerator of P . Then v = sa where a is a unit in D ⊆ k[[x, y, z]]. We write:

13.4.3.113.4.3.1 (intIIsec254.3.1) v = sa = h(ρ, σ)/g(ρ, σ), where h(ρ, σ), g(ρ, σ) ∈ k[x, y, z][ρ, σ].

Now a ∈ D = U(x,y,z)D∩U , so a = g1/g2, where g1, g2 ∈ k[x, y, z, τ1n, τ2n], forsome n ∈ N, and g2 as a power series in k[[x, y, z]] has nonzero constant term.There exists m ∈ N such that xmg1 := f1 and xmg2 := f2 are in the poly-nomial ring k[x, y, z, τ1, τ2] = k[x, y, z][s, t]. We regard f2(s, t) as a polynomialin s and t with coefficients in k[x, y, z]. We have f2k[[x, y, z]] = xmk[[x, y, z]] =xmk[[x, s, t]]. Therefore f2 6∈ (s, t)k[[x, s, t]]. It follows that the constant term off2(s, t) ∈ k[x, y, z][s, t] is a nonzero element of k[x, y, z]. Since we have

13.4.3.213.4.3.2 (intIIsec25.4.3.2) a =

xmg1xmg2

=f1f2,

and a is a unit of D, the constant term of f1(s, t) ∈ k[x, y, z][s, t] is also nonzero.The equations (

13.4.3.1intIIsec254.3.1) and (

13.4.3.2intIIsec25.4.3.2) together yield

13.4.3.313.4.3.3 (intIIsec25.4.3.3) sf1(s, t)h(s2, st) = f2(s, t)g(s2, st).

The term of lowest total degree in s and t on the left hand side of (13.4.3.3intIIsec25.4.3.3) has

odd degree, while the term of lowest total degree in s and t on the right hand sidehas even degree, a contradiction. Therefore the assumption that A is Noetherianleads to a contradiction. We conclude that A is not Noetherian.

13.4.4 Remarks 25.16. (i) Although A is not Noetherian, the proof of Proposition13.4.325.15 does not rule out the possibility that A is factorial. The proof does show thatif A is factorial, then ht(P1 ∩A) > ht(P1). It would be interesting to know whetherthe non-flat morphism A → A = R has the property that ht(Q ∩ A) ≤ ht(Q), foreach Q ∈ Spec R. It would also be interesting to know the dimension of A.(ii) We observe the close connection of the integral domains A ⊆ D of (4.1). Theextension of fraction fields Q(A) ⊆ Q(D) has degree two and A = Q(A)∩D, yet Ais non-Noetherian while D is Noetherian.


13.4.5 Proposition 25.17. With the notation of (13.4.125.13), C is a two-dimensional

Noetherian local domain having completion R/I and the generic formal fiber ofC is not Cohen-Macaulay.

Proof. By Chapternoehomsec8 (

11.2.46.17), the completion of C is R/I. Hence if C is

Noetherian, then dim(C) = dim(R/I) = 2.To show that C is Noetherian, according to Chapter

noehomsec8 Theorem

11.3.28.3, it suffices

to show that the canonical morphism:

R = k[x, y, z](x,y,z)ϕ−−−−→ (R/I)x = (k[[x, y, z]]/I)x = (k[[x, s, t]]/(s2, st)k[[x, s, t]])x

is flat. It suffices to show for every prime ideal Q of R with x /∈ Q that the morphism

ϕ bQ : R −→ R bQ/IR bQ = (R/I) bQ

is flat. We may assume I = P1P2 ⊆ Q.If Q = P2 = (s, t)R then ϕ bQ is flat since P2 ∩R = (0).

If Q 6= P2, then P2R bQ = R bQ, because htP2 = 2. Hence IR bQ = P1R bQ = sR bQ.Thus we need to show

ϕ bQ : R −→ R bQ/sR bQ = (R/sR) bQ

is flat. To see that ϕ bQ is flat, we observe that, since R ⊆ D bQ∩D ⊆ R bQ and sR∩R =(0), the morphism ϕ bQ factors through a homomorphic image of D = V [y, z](x,y,z).That is, ϕ bQ is the composition of the following maps:

Rγ−−−−→ (D/sD)D∩ bQ

ψ bQ−−−−→ (R/sR) bQ.

Since D is Noetherian, the map ψ bQ is faithfully flat. Thus it remains to show that

γ is flat. Since x /∈ Q, the ring (D/sD)D∩ bQ is a localization of (D/sD)[1/x]. Thusit is a localization of the polynomial ring:

k[x, y, z, τ1, τ2]/sk[x, y, z, τ1, τ2] = k[x, y, z, s, t]/sk[x, y, z, s, t],

which is clearly flat over R. Thus C is Noetherian.Now P2/I = p is an embedded associated prime of (0) of C so Cp is not

Cohen-Macaulay. Since p ∩ C = (0) the generic formal fiber of C is not Cohen-Macaulay.

13.4.6 Proposition 25.18. The canonical morphism B −→ R/I factors through C.We have B/Q ∼= C, where Q = I ∩B = sR ∩B. On the other hand, the canonicalmorphism A −→ R/I fails to factor through C.

Proof. We have canonical maps B → R/I and C → R/I. We define a mapφ : B C such that the following diagram commutes:

13.4.6.113.4.6.1 (intIIsec25.4.6.1)

B −−−−→ R

φ

y yC −−−−→ R/I.

We write C as a nested union as is done in Chapternoehomsec8 (

11.3.28.3):

C =∞⋃n=1

R[ρn, σn](x,y,z,ρn,σn)


where ρn, σn are the nth frontpieces of ρ and σ:

ρn =1xn

(y +n∑j=1

a1jxj)2 and σn =

1xn

(y +n∑j=1

a1jxj)(z +

n∑j=1

a2jxj).

Then

B =∞⋃n=1

R[ρn, σn](x,y,z,ρn,σn) and C =∞⋃n=1

R[ρn, σn](x,y,z,ρn,σn).

It is clear that for each n ∈ N there is a surjection

R[ρn, σn](x,y,z,ρn,σn) −→ R[ρn, σn](x,y,z,ρn,σn)

which maps ρn 7→ ρn and σn 7→ σn and which extends to a surjective homomor-phism φ on the directed unions such that diagram (

13.4.6.1intIIsec25.4.6.1) commutes. This shows

that C ∼= B/Q is a homomorphic image of B.In order to see the canonical map ζ : A −→ R/I fails to factor through C, we

note that I ∩D = (ρ, σ)D and so ζ factors through D:

Aγ−−−−→ D/(ρ, σ)D δ−−−−→ R/I

where δ is injective. The map γ sends the element σ2/ρ = t2 to the residue classof t2 = (z + τ2)2 in D/(ρ, σ)D. This element is algebraically independent over R,which shows that the ring A/ker(γ) is transcendental over R. Since C is a birationalextension of R, the morphism A −→ R/I fails to factor through C.

CHAPTER 26

Building Noetherian and non-Noetherian integral

domains using power series

nonnoesecWe present several examples where this flatness condition holds, and other

examples where it fails to hold. In the first case this produces a Noetherian integraldomain and in the second case a non-Noetherian domain.

???alertWe took material from here and put it in the introduction to the memoir???

26.1. Noetherian results14.4.5

Suppose R is a Noetherian local domain and let the notation be as in (4.4.15.3).

We prove in Chapternoeslocsec22:

Theorem8.2.1222.12: The canonical map R[τ1, . . . , τs] → R∗[1/a] is flat if and only if

A is Noetherian and is a nested union of localized polynomial rings in s variablesover R as defined in Section

4.5.46.9.

???alertActually the statement says “primarily limit intersecting”, not flat. Wemay want to refer to chapter

noehomsec8 or to

4.5.6?? ???

In Chapternoehomsec8, we prove an analogous result in the more general setting of (

4.4.25.4):

Theorem11.1.18.1: The canonical map R → (R∗/I)[1/a] is flat if and only if A is

simultaneously Noetherian and a localization of a subring of R[1/a].The proof of the more general result in the setting of (

4.4.25.4) is actually shorter

and more direct than the earlier proof in Chapternoeslocsec22. In Chapters

noehomsec8 and

intIIsec25 examples

are constructed of the form (4.4.25.4) that cannot be realized by means of (

4.4.15.3).

26.2. Flatness, approximations14.5

In relation to Theorem11.1.18.1, we display, in the following diagram, the inclusions

that are always flat:

Q(R) Q(U) Q(B) Q(A) ⊆−−−−→ Q(R∗/I)

flat

x flat

x flat

x flat

x flat

xR[1/a] U [1/a] flat−−−−→ B[1/a]

⊆−−−−→ A[1/a]⊆−−−−→ (R∗/I)[1/a]

flat

x flat

x flat

x flat

x flat

xR

⊆−−−−→ U = ∪Un flat−−−−→ B⊆−−−−→ A

⊆−−−−→ R∗/I.

225

226 26. NOETHERIAN, NON-NOETHERIAN DOMAINS JULY 27, 2008

On the other hand, (i) if B ( A, then B → A is not flat, and (ii) the inclusionA ⊆ R∗/I is flat if and only if A is Noetherian.

??? alertOld message from Christel: Here is something I thought we may wantto consider for another paper. I don’t think it contains exciting news, but I feelit may be a good idea to compare all the methods we know of for constructingNoetherian local rings of finite transcendence degree over a field. If you prefer wecan (also) include this in our expository paper. I don’t know if I’m going to typemy transparencies this time. If I do I’ll send you a copy. I thought I would alsoinclude some of the stuff from below in my talk. I hope everything is fine and thatMary Ann is doing ok. Best wishes, Christel ???

26.3. Explicit constructions14.6

Using the flatness results described in Section14.526.2, we formulate two methods

for the construction of explicit examples. The first of these methods uses theconstruction technique of (

4.4.15.3) while the second uses (

4.4.25.4).

14.6.1 Method 26.1. Let k be a field, let a, y1, . . . , yn be variables over k, let Rbe the localized polynomial ring R := k[a, y1, . . . , yn](a,y1,...,yn) and let m denotethe maximal ideal of R. Let τ1, . . . , τs ∈ ak[[a]] be formal power series that arealgebraically independent over k(a) and let D0 := R[τ1, . . . , τs](m,τ1,...,τs) be theassociated localized polynomial ring over the field k in n+ s+ 1 variables. Observethat D0 is contained in the a-adic completion R∗ of R. It is readily seen that themap D0 → R∗[1/a] is flat and that D := Q(D0) ∩ R∗ is the localized polynomialring V [y1, . . . , yn](a,y1,...,yn), where V := k(a, τ1, . . . , τs)∩ k[[a]] is a DVR. Thus theconstruction method of (

4.4.15.3) gives a Noetherian limit-intersecting domain D.

We now investigate the construction of examples inside D. Let f1, . . . , fr beelements of the maximal ideal of D0 that are algebraically independent over Q(R),and let B0 := R[f1, . . . , fr](m,f1,...,fr) be the associated localized polynomial ring.The inclusion map B0 → D0 is an injective local R-algebra homomorphism.

Let A := Q(B0) ∩R∗ and let B be the associated nested union domain. ThenB is Noetherian and B = A if and only if the map B0 → R∗[1/a] is flat. Thismap factors as B0 → D0 → D0[1/a]→ R∗[1/a] and the map D0 → R∗[1/a] is flat.Therefore B is Noetherian (and so also B = A) if B0 → D0[1/a] is flat.

Since B0 and D0 are localized polynomial rings over a field, the nonflat locusof the inclusion map B0 → D0 is closed

M[68, Theorem 24.3] and is defined by the

ideal J := ∩P ∈ SpecD0 : B0 → (D0)P is not flat .We have established the following theorem.

14.6.2 Theorem 26.2. With the notation above, we have(1) If JD0[1/a] = D0[1/a], then B is Noetherian.(2) B is Noetherian if and only if JD[1/a] = D[1/a] if and only if JR∗[1/a] =

R∗[1/a].

Using again the factorization of B0 → R∗[1/a] through D0[1/a], Theorem7.5.521.18

of Chapterintsec21 implies the following result.

14.6.3 Theorem 26.3. With the notation above, if ht(JD0[1/a]) > 1, then B = A.

14.6.4 Remarks 26.4. With the notation of this section,

26.3. EXPLICIT CONSTRUCTIONS 227

(1) If B is Noetherian, then B is a regular local ring.(2) Example

4.3.14.8 of Nagata may be described by taking n = s = r = 1, y1 =

y, τ1 = τ , and f1 = f . Then R = k[a, y](a,y), D0 = k[a, y, τ ](a,y,τ),f = (y + τ)2, and B0 = k[a, y, f ](a,y,f). The Noetherian property of B isimplied by the flatness property of the map B0 → D0[1/a]. In this case,D0 is actually a free B0-module with < 1, y + τ > as a free basis.

(3) Example4.3.34.10 of Rotthaus may be described by taking n = s = 2, and

r = 1. Then R = k[a, y1, y2](a,y1,y2), D0 = R[τ1, τ2](m,τ1,τ2), f1 = (y1 +τ1)(y2 + τ2) and B0 = R[f1](m,f1). Again the Noetherian property of B isimplied by the flatness property of the map B0 → D0[1/a].

(4) The following example is given in ChapterintIIsec25, Section

13.425.4. Let n = s =

r = 2, let f1 = (y1 + τ1)2 and f2 = (y1 + τ1)(y2 + τ2). It is shown inChapter

intIIsec25 for this example that B ( A and that both A and B are

non-Noetherian.

14.6.5 Method 26.5. Let k be a field, let a, y1, . . . , yn be variables over k, let Rbe the localized polynomial ring R := k[a, y1, . . . , yn](a,y1,...,yn) and let m denotethe maximal ideal of R. Let τ1, . . . , τs ∈ ak[[a]] be formal power series that arealgebraically independent over k(a) and let D0 := R[τ1, . . . , τs](m,τ1,...,τs) be theassociated localized polynomial ring over the field k.

Assume that I is an ideal of D0 such that P ∩R = (0) for each P ∈ Ass(D0/I).Let A := Q(R) ∩ (R∗/IR∗); then the a-adic completion of A is R∗/IR∗. Usingthe frontpiece approximations of a generating set for I, it is shown in Chapternoehomsec8, Section

11.2?? that there exists a quasilocal integral domain B = ∪∞n=1Bn ⊆ A

birationally dominating R such that the a-adic completion of B is R∗/IR∗. ByChapter

noehomsec8, Theorem

11.3.28.3, R→ (R∗/IR∗)[1/a] is flat if and only if B is Noetherian.

Since the map D0 → R∗[1/a] is flat, the map D0/I → (R∗/IR∗)[1/a] is flat.Also the map R → (R∗/IR∗)[1/a] factors as R → (D0/I)[1/a] → (R∗/I)[1/a].Thus R→ (R∗/IR∗)[1/a] is flat if R→ (D0/I)[1/a] is flat. Since R and D0/I := Tare essentially finitely generated over a field, the nonflat locus of the inclusion mapR→ T is closed

M[68, Theorem 24.3] and is defined by the ideal J := ∩P ∈ SpecT :

R→ TP is not flat .This yields the following theorem.

14.6.6 Theorem 26.6. With the notation above, if J(D0/I)[1/a] = (D0/I)[1/a], thenB is Noetherian

14.6.7 Remark 26.7. Let the notation be as in part(4) of Remark14.6.426.4, so that

f1 = (y1 + τ1)2 and f2 = (y1 + τ1)(y2 + τ2), and let I := (f1, f2)R∗. In ChapterintIIsec25, Section

13.425.4 it is shown that C := Q(R) ∩ (R∗/I) is Noetherian and limit-

intersecting. Indeed, it is shown in Proposition13.4.525.17 that C is a two-dimensional

Noetherian local domain for which the generic formal fiber is not Cohen-Macaulay.

CHAPTER 27

What properties extend to an ideal-adic

completion? (wprop,25) May 17, 2009

25Many examples contained in earlier chapters are related to the question:

25.011 Question 27.1. For an ideal a of a Noetherian ring R, what properties of R,or of an ideal of R, extend to the a-adic completion R∗ of R ?

The a-adic completionR∗ ofR is always Noetherian. In the case where (R,m) isNoetherian local and a = m, so that R is the m-adic completion, among propertiesof R that are well-known to extend to R are

• R is a regular local ring• R is Gorenstein.• R is Cohen-Macaulay.• R has dimension n.• R is a complete intersection, that is R ∼= S/I, where S is a regular local

ring and I is generated by a regular sequence.Even in the case where (R,m) is local and a = m, there are a number of

variations to Question25.01127.1. We elaborate in Questions

25.00127.2.

25.001 Questions 27.2. Let (R,m) be a Noetherian local ring.

(1) If there is something special about R, is R also special in that way?(2) What properties do R and its completion R have in common?(3) What properties may R or R have but the other not have?(4) For an ideal I of R, what properties does I share with its extension IR ?

Examples of properties that extend to the completion as in Question25.00127.2.1 are

the five itemized statements listed below Question25.01127.1. Indeed, the first four of

these statements hold for R if and only if they hold for R, and thus are relevantto Question

25.00127.2.2. With regard to Question

25.00127.2.3, observe that Example

4.3.04.6 is

a classical example of a one-dimensional Noetherian local domain R such that itscompletion R is not an integral domain. Thus the property of being an integraldomain may be lost in extending to the completion.

If R is a one-dimensional Noetherian local domain, then its integral closureis a finitely generated R-module if and only if its completion R has no nonzeronilpotents, cf.

N2[72, (32.2) and Ex. 1, page 122] and

Katz[59, Corollary 5]. Thus the

examples of Akizuki and Schmidt mentioned in Remark4.3.014.7 are one-dimensional

Noetherian local integral domains for which the completion has nonzero nilpotents.That is, the property of having no nonzero nilpotents is not in general preservedunder completion.

A one-dimensional normal Noetherian local domain is a DVR and its comple-tion is then also a DVR, and so an integral domain. However, Example

4.3.14.8 of

229

230 27. WHAT PROPERTIES

Nagata provides a two-dimensional normal Noetherian local domain for which thecompletion is not an integral domain.

The catenary property of Noetherian rings is considered in Chaptercatsec15. Every

Noetherian ring of dimension less than or equal to two is catenary. There exist,however, examples of three-dimensional Noetherian local domains (R,m) that arenot catenary

ZSII[106, page 327]. It follows from Cohen’s structure theorems that a

complete Noetherian local ring is a homomorphic image of a regular local ring andhence is universally catenary cf. Remark

3.4353.20 . Thus, in relation to Question

25.00127.2.1,

the property of not being catenary does not extend to the completion.There exists a two-dimensional catenary Noetherian local domain (R,m) that

is not universally catenary even though R has geometrically normal formal fibers,cf. Example

11.4.59.8. Thus, in this example, R is universally catenary, whereas R is not

universally catenary.With regard to Question

25.00127.2.3, Example

11.4.59.8 mentioned above provides a Noe-

therian local domain R that is not a homomorphic image of a regular local ringwhereas the completion of R is a homomorphic image of a regular local ring. Aspecific question raised by Luchezar Avramov is:

25.002 Question 27.3. (Avramov) If R is a complete intersection, does it follow thatR is a homomorphic image of a regular local ring ?

??? Question: (for us) If R as in Question25.00227.3 is a homomorphic image of a

regular local ring, does it then follow that R is a c.i.???We recall in Definition

25.0127.4 a specific case of properties Ri and Si introduced

by SerreM[68, page 183].

25.01 Definition 27.4. A Noetherian ring A is said to satisfy Serre’s condition S1

if for each non-minimal prime P of A, the maximal ideal PAP of the local ring APcontains a regular element.


intIIsec25 demonstrates the existence of a two-dimensional

Noetherian local domain R such that its generic formal fiber is not Cohen-Macaulay.The completion of R is isomorphic to k[[s, t, x]]/(s2, st), where s, t and x are in-determinates over the field k. Since R is an integral domain, R satisfies Serre’scondition S1. However, in this example R does not satisfy S1.

27.1. Associated graded rings of ideals under completion25.2

Let (R,m) be a Noetherian local ring, and let R denote the m-adic completionof R. For an ideal I of R, let I = IR and let

G(I) =∞⊕n=0

In

In+1and G(I) =

∞⊕n=0

In

In+1

denote the associated graded rings of I in R and of I in R respectively.

25.21 Proposition 27.5. With notation as above, let M and M denote the gradedmaximal ideals of G(I) and G(I) respectively. We have dimG(I) = dimG(I), andthe length of a maximal regular sequence in M is equal to the length of a maximalregular sequence in M. Thus G(I) is Cohen-Macaulay if and only if G(I) is Cohen-Macaulay.

27.2. THE Sk CONDITION AND COMPLETION 231

Proof. The canonical map: G(I)→ G(I) is faithfully flat. Let x∗1, x∗2, . . . , x

∗n

be homogeneous elements of G(I) that form a regular sequence in M of maximallength. Since the map G(I) → G(I) is faithfully flat, the images of x∗1, . . . , x∗nare a regular sequence in M. Let xi be a preimage in R of x∗i . Now pass toG(I/(x1, x2, . . . , xn); thus we can assume that G(I) has depth 0. Thus M is anassociated prime of G(I). Then Matsumura 23.2 gives that M is an associatedprime of G(I) cM.

———————-

27.2. The Sk condition and completion

Use the example at the end of inter II:—————–Ray Heitmann (description) Loepp (description) Loepp-Rotthaus (description)xxxxxxxxxxxxxxx

Question 27.6. (Questions For Us)(1) In the example is P the unique prime of A that does not extend to an

integrally closed ideal of A?(2) Related to the previous item can this be done in such a way that all primes

of A of height less than d− 1 extend to integrally closed ideals of A?(3) If for every height-one prime P of A, the extension PA is a radical ideal,

does this property hold for all prime ideals of A? Does the condition onheight-one primes imply every prime ideal of A extends to an integrallyclosed ideal of A?

Remark 27.7. (Comments For Us)(1) Let S and T be Noetherian rings and assume that T is a flat S-algebra.

Let P ∈ SpecS. If for each Q ∈ Spec T with Q ∩ S = P , the map S → TQ isregular, then PT = rad(PT ), so PT is integrally closed. All the prime divisors ofPT contract in S to P and the fact that S → TQ is regular for each Q such thatQ ∩ S = P implies PT = rad(PT ).

(2) Related to the example given in Theorem 2.3, we have the following com-mutative diagram:

R∗[1/x] = B∗[1/x]

R := k[x, y, z](x,y,z) ⊆ S := R[f, g] T := R[α, β]

ψ

δ:=ψϕ

ϕ

Let γ : B → B∗[1/x]. Let Q ∈ SpecB∗ with x 6∈ Q and let q := Q∩T . ByM[68,

Theorem 32.1], if ϕq is regular, then δQ is regular. It follows that γQB → B∗[1/x].(3) If the height-two prime p of B contains an element that is a regular param-

eter for B∗, then it follows that pB∗ is integrally closed.(4) For p = (f, y2 − z3)B, it looks like pB∗ may be integrally closed. If we

consider ((y − α)2, y2 − z3)k[[x, y, z]], and then go modulo y2 − z3 by mappingy → t3, z → t2, then pB maps to the principal ideal (t3 − α)2k[[x, t2, t3]].

(5) Related to the example given in Theorem 2.3, the Jacobian ideal of theextension ϕ : S := k[x, y, z, f, g] → T := k[x, y, z, α, β]. is the ideal of T generated

232 27. WHAT PROPERTIES

by the determinant of the matrix

J :=

(∂f∂α

∂g∂α

∂f∂β

∂g∂β

).

Since the characteristic of the field k is zero, this ideal is (y − α)(z − β)T . LetQ ∈ SpecR∗ with x 6∈ Q and let q = Q ∩ T . By

insideps[45, (2.7)], if (y − α)(z − β) 6∈ q,

then ϕq is regular and so by item (2) γQ : B → B∗Q is regular. Notice that

fg = (y − α)2(z − β)2 ∈ S ⊆ B. We also have B∗/xB∗ = B/xB = R/xR.Let p ∈ SpecB. If x ∈ p, then pB is prime and hence integrally closed. If

x 6∈ p and fg 6∈ p, then for all Q ∈ SpecB∗ lying over p, the map γQ is regular.Hence pB∗ = rad(pB∗), so in particular pB∗ is integrally closed.

It is unclear as to whether P is the unique prime of B with the property thatPB∗ is not integrally closed.

CHAPTER 28

Integral closure under extension to the completion

(intclsec), May 10, 2010 18intclsec

Using the insider construction described in Chapterinsidecon13, we present in this

chapter an example of a 3-dimensional regular local domain (A,n) having a height-two prime ideal P with the property that the extension PA of P to the n-adiccompletion A of A is not integrally closed. The ring A in the example is a 3-dimensional regular local domains that is a nested union of 5-dimensional regularlocal domains.

More generally, we use this same technique to establish the existence for eachinteger d ≥ 3 and each integer h with 2 ≤ h ≤ d − 1 of a d-dimensional regularlocal domain (A,n) having a prime ideal P of height h with the property that theextension PA is not integrally closed. A regular local domain having a prime idealwith this property is necessarily not excellent.

We discuss in Section18.228.1 conditions in order that integrally closed ideals of a

ring R extend to integrally closed ideals of R′, where R′ is an R-algebra. In par-ticular, we consider under what conditions integrally closed ideals of a Noetherianlocal ring A extend to integrally closed ideals of the completion A of A.

28.1. Integral closure under ring extension18.2

For properties of integral closure of ideals, rings and modules we refer toSH[95].

In particular, we use the following definition.

18.2.0 Definitions 28.1. Let I be an ideal of a ring R.(1) An element r ∈ R is integral over I if there exists a monic polynomial

f(x) = xn +∑n

i=1 aixn−i such that f(r) = 0 and such that ai ∈ Ii for

each i with 1 ≤ i ≤ n.(2) If J is an ideal contained in I and JIn−1 = In, then J is said to be a

reduction of I.(3) The integral closure I of I is the set of elements of R integral over I.(4) If I = I, then I is said to be integrally closed.(5) The ideal I is said to be normal if In is integrally closed for every n ≥ 1.

18.2.01 Remarks 28.2. We record the following facts about an ideal I of a ring R.(1) An element r ∈ R is integral over I if and only if I is a reduction of

the ideal L = (I, r)R. To see this equivalence, observe that for a monicpolynomial f(x) as in Definition

18.2.028.1.1, we have

f(r) = 0 ⇐⇒ rn = −n∑i=1

airn−i ∈ ILn−1 ⇐⇒ Ln = ILn−1.

233


(2) It is well known that I is an integrally closed idealSH[95, Corollary 1.3.1].

(3) An ideal is integrally closed if and only if it is not a reduction of a properlybigger ideal.

(4) A prime ideal is always integrally closed. More generally, a radical idealis always integrally closed.

(5) Let a, b be elements in a Noetherian ring R and let I := (a2, b2)R. Theelement ab is integral over I. If a, b are a regular sequence, then ab 6∈ Iand the ideal I is not integrally closed. More generally, if h ≥ 2 anda1, . . . , ah are a regular sequence in R and I := (ah1 , . . . , a

hh)R, then I is

not integrally closed.

Our work in this chapter was motivated by the following questions:

18.2.02 Questions 28.3.(1) Craig Huneke: “Does there exist an analytically unramified Noetherian lo-

cal ring (A,n) that has an integrally closed ideal I for which the extensionIA to the n-adic completion A is not integrally closed?”

(2) Sam Huckaba: “If there is such an example, can the ideal of the examplebe chosen to be a normal ideal?”

Related to Question18.2.0228.3.1, we construct in Example

18.3.128.7 a 3-dimensional reg-

ular local domain A having a height-two prime ideal I = P = (f, g)A such thatIA is not integrally closed. Thus the answer to Question

18.2.0228.3.1 is “yes”. This

example also shows that the answer to Question18.2.0228.3.2 is again “yes”. Since f, g

form a regular sequence and A is Cohen-Macaulay, the powers Pn of P have noembedded associated primes and therefore are P -primary

M1[66, (16.F), p. 112],

M[68,

Ex. 17.4, p. 139]. Since the powers of the maximal ideal of a regular local domainare integrally closed, the powers of P are integrally closed. Thus the Rees alge-bra A[Pt] = A[ft, gt] is a normal domain while the Rees algebra A[ft, gt] is notintegrally closed.

Remarks 28.4. Without the assumption that A is analytically unramified,there exist examples even in dimension one where an integrally closed ideal of Afails to extend to an integrally closed ideal in A. If A is reduced but analyticallyramified, then the zero ideal of A is integrally closed, but its extension to A is notintegrally closed.

As we remark in Chapterintro1, examples exhibiting this behavior have been known

for a long time. An example in characteristic zero of a one-dimensional Noetherianlocal domain that is analytically ramified is given by Akizuki in his 1935 paper

A[7].

An example in positive characteristic is given by F.K. SchmidtSc[87, pp. 445-447].

Another example due to Nagata is given inN2[72, Example 3, pp. 205-207]. (See also

N2[72, (32.2), p. 114].)

Let R be a commutative ring and let R′ be an R-algebra. In Remark18.2.128.6 we

list cases where extensions to R′ of integrally closed ideals of R are again integrallyclosed. In this connection we use the following definition.

18.2.10 Definition 28.5. The R-algebra R′ is said to be quasi-normal if R′ is flat overR and the following condition (NR,R′) holds: If C is any R-algebra and D is aC-algebra in which C is integrally closed, then also C ⊗R R′ is integrally closed inD ⊗R R′.

28.1. INTEGRAL CLOSURE UNDER RING EXTENSION 235

18.2.1 Remarks 28.6. Let R be a commutative ring and let R′ be an R-algebra.

(1) ByL[62, Lemma 2.4], if R′ satisfies (NR,R′) and I is an integrally closed

ideal of R, then IR′ is integrally closed in R′.(2) Assume that R and R′ are Noetherian rings and that R′ is a flat R-

algebra. Let I be an integrally closed ideal of R. The flatness of R′ overR implies every P ′ ∈ Ass(R′/IR′) contracts in R to some P ∈ Ass(R/I)M[68, Theorem 23.2]. Since a regular map is quasi-normal, if the mapR → R′

P ′ is regular for each P ′ ∈ Ass(R′/IR′), then IR′ is integrallyclosed.

(3) Principal ideals of an integrally closed domain are integrally closed.(4) In general, integral closedness of ideals is a local condition. If R′ is an R-

algebra that is locally normal in the sense that for every prime ideal P ′ ofR′, the local ring R′

P ′ is an integrally closed domain, then the extensionto R′ of every principal ideal of R is integrally closed by item (3). Inparticular, if (A,n) is an analytically normal Noetherian local domain,then every principal ideal of A extends to an integrally closed ideal of A.

(5) Let (A,n) be a Noetherian local ring and let A be the n-adic completionof A. Since A/q ∼= A/qA for every n-primary ideal q of A, the n-primaryideals of A are in one-to-one inclusion preserving correspondence with then-primary ideals of A. It follows that an n-primary ideal I of A is areduction of a properly larger ideal of A if and only if IA is a reductionof a properly larger ideal of A. Therefore an n-primary ideal I of A isintegrally closed if and only if IA is integrally closed.

(6) If R is an integrally closed domain, then for every ideal I and element x ofR we have xI = xI. If (A,n) is analytically normal and also a UFD, thenevery height-one prime ideal of A extends to an integrally closed idealof A. In particular if A is a regular local domain, then PA is integrallyclosed for every height-one prime P of A. If (A,n) is a 2-dimensional localUFD, then every nonprincipal integrally closed ideal of A has the form xI,where I is an n-primary integrally closed ideal and x ∈ A. In particular,this is the case if (A,n) is a 2-dimensional regular local domain. In viewof item (5), it follows that every integrally closed ideal of A extends to anintegrally closed ideal of A in the case where A is a 2-dimensional regularlocal domain.

(7) If (A,n) is an excellent local ring, then the map A → A is quasi-normalby

G[30, (7.4.6) and (6.14.5)], and in this case every integrally closed ideal

of A extends to an integrally closed ideal of A.(8) Let (A,n) be a Noetherian local domain and let Ah denote the Henseliza-

tion of A. Every integrally closed ideal of A extends to an integrally closedideal of Ah. This follows because Ah is a filtered direct limit of etale A-algebras

L[62, (iii), (i), (vii) and (ix), pp. 800- 801]. Since the map from

A to its completion A factors through Ah, every integrally closed ideal ofA extends to an integrally closed ideal of A if and only every integrallyclosed ideal of Ah extends to an integrally closed ideal of A.


28.2. Extension to the completion18.3

We use results from Chaptersnoehomsec8,

flatholds9 and

insidecon13 In the construction of the following

example.

18.3.1 Construction of Example 28.7. Let k be a field of characteristic zero andlet x, y and z be indeterminates over k. Let R := k[x, y, z](x,y,z) and let R∗ be the(x)-adic completion of R. Thus R∗ = k[y, z](y,z)[[x]], the formal power series ringin x over the 2-dimensional regular local ring k[y, z](y,z).

Let α and β be elements of xk[[x]] that are algebraically independent over k(x).Set

f = (y − α)2, g = (z − β)2, and A = k(x, y, z, f, g) ∩ R∗.

Let T := R[α, β] and S := R[f, g]. Notice that S and T are both polynomialrings in 2 variables over R and that S is a subring of T . Indeed, with u := y − αand v := z − β, we have T = R[u, v] and S = R[u2, v2]. Let

Aout := R∗ ∩ k(x, y, z, α, β).

By Corollary11.4.119.4, the map T → R∗[1/x] is flat and Aout is a 3-dimensional regular

local domain that is a directed union of 5-dimensional regular local domains. AlsoAout has (x)-adic completion R∗

By Theorem16.3.213.2, the ring A is Noetherian and equal to its approximation

domain B provided that the map S → T [1/x] is flat. In our situation, the ring Tis a free S-module with 1, y − α, z − β, (y − α)(z − β) as a free module basis.Therefore the map S → T [1/x] is flat. The following commutative diagram whereall the labeled maps are the natural inclusions displays this situation:

(28.1)

B = A = R∗ ∩ Q(S)γ1−−−−→ Aout = R∗ ∩Q(T )

γ2−−−−→ R∗ = A∗

δ1

x δ2

x ψ

xS = R[, f, g]

ϕ−−−−→ T = R[α, β] T

In order to better understand the structure of A, we recall some of the detailsof the approximation domain B associated to A.

18.3.2 Approximation Technique 28.8. With k, x, y, z, f, g, R and R∗ as in Con-struction

18.3.128.7, we have

f = y2 +∞∑j=1

bjxj , g = z2 +

∞∑j=1

cjxj ,

where bj,∈ k[y] and cj ∈ k[z]. The rth endpieces for f and g are the sequencesfr∞r=1, gr∞r=1 of elements in R∗ defined for each r ≥ 1 by:

fr :=∞∑j=r

bjxj

xrand gr :=

∞∑j=r

cjxj

xr.

For each integer r ≥ 1, the ring Br is the polynomial ring k[x, y, z, fr, gr]localized at the maximal ideal generated by (x, y, z, fr − br, gr − cr), and the ap-proximation domain B =

⋃∞r=1Br.

28.2. EXTENSION TO THE COMPLETION 237

18.3.3 Theorem 28.9. Let A be the ring constructed in (18.3.128.7) and let P = (f, g)A,

where f and g are as in (18.3.128.7). Then

(1) The ring A = B is a 3-dimensional regular local domain that is a nestedunion of 5-dimensional regular local domains.

(2) The ideal P is a height-two prime ideal of A.(3) The (xA)-adic completion A∗ of A is R∗ = k[y, z](y,z)[[x] and PA∗ is not

integrally closed.(4) The completion A of A is R = k[[x, y, z]] and PA is not integrally closed.

Proof. Notice that the polynomial ring T = R[α, β] = R[y − α, z − β] is afree module of rank 4 over the polynomial subring R[f, g] since f = (y − α)2 andg = (z − β)2. Hence the extension

R[f, g] → R[α, β][1/x]

is flat. Thus item (1) follows from Theorem16.3.213.2.

For item (2), it suffices to observe that P has height two and that, for eachpositive integer r, Pr := (f, g)Br is a prime ideal of Br. We have f = xf1 + y2 andg = xg1 + z2. It is clear that (f, g)k[x, y, z, f, g] is a height-two prime ideal. SinceB1 is a localized polynomial ring over k in the variables x, y, z, f1 − b1, g1 − c1, wesee that

P1B1[1/x] = (xf1 + y2, xg1 + z2)B1[1/x]

is a height-two prime ideal of B1[1/x]. Indeed, setting f = g = 0 is equivalent to set-ting f1 = −y2/x and g1 = −z2/x. Therefore the residue class ring (B1/P1)[1/x] isisomorphic to a localization of the integral domain k[x, y, z][1/x]. Since B1 is Cohen-Macaulay and f, g form a regular sequence, and since (x, f, g)B1 = (x, y2, z2)B1 isan ideal of height three, we see that x is in no associated prime of (f, g)B1 (see, forexample

M[68, Theorem 17.6]). Therefore P1 = (f, g)B1 is a height-two prime ideal.

For r > 1, there exist elements ur ∈ k[x, y] and vr ∈ k[x, z] such that f =xrfr + urx + y2 and g = xrgr + vrx + z2. An argument similar to that givenabove shows that Pr = (f, g)Br is a height-two prime of Br. Therefore (f, g)B is aheight-two prime of B = A.

For items 3 and 4, R∗ = B∗ = A∗ by Construction18.3.128.7 and it follows that A =

k[[x, y, z]]. To see that PA∗ = (f, g)A∗ and PA = (f, g)A are not integrally closed,observe that ξ := (y − α)(z − β) is integral over PA∗ and PA since ξ2 = fg ∈ P 2.On the other hand, y − α and z − β are nonassociate prime elements in the localunique factorization domains A∗ and A. An easy computation shows that ξ 6∈ PA.Since PA∗ ⊆ PA, this completes the proof.

In Example18.3.428.10, we observe that this same technique may be used to construct

a d-dimensional regular local domain (A,n) having a prime ideal P of height h withthe property that the extension PA is not integrally closed for each integer d ≥ 3and each integer h with 2 ≤ h ≤ d− 1

18.3.4 Example 28.10. Let k be a field of characteristic zero and for an integer n ≥ 2let x, y1, . . . , yn be indeterminates over k. Let R denote the d := n+ 1 dimensionalregular local ring obtained by localizing the polynomial ring k[x, y1, . . . , yn] at themaximal ideal generated by (x, y1, . . . , yn). Let h be an integer with 2 ≤ h ≤ nand let τ1, . . . , τh ∈ xk[[x]] be algebraically independent over k(x). For each i with


1 ≤ i ≤ h, define fi = (yi − τi)h, and set ui = yi − τi. Consider the rings

S := R[f1, . . . , fh] and T := R[τ1, . . . , τh] = R[u1, . . . , uh].

Notice that S and T are polynomial rings in h variables over R and that T is afinite free integral extension of S. Let R∗ denote the (x)-adic completion of R, anddefine

A := R∗ ∩ Q(S) and Aout := R∗ ∩ Q(T ).The following commutative diagram where the labeled maps are the natural

inclusions displays this situation:

(28.2)

B = A = R∗ ∩ Q(S)γ1−−−−→ Aout = R∗ ∩Q(T )

γ2−−−−→ R∗ = A∗

δ1

x δ2

x ψ

xS = R[, f1, . . . , fh]

ϕ−−−−→ T = R[τ1, . . . , τh] T

Since the map S → T is flat, Theorem16.3.213.2 implies that A is a d-dimensional

regular local ring and is equal to its approximation domain B. Let P := (f1, . . . , fh)A.An argument similar to that given in Theorem

18.3.328.9 shows that P is a prime ideal of

A of height h. We have yi−τi ∈ A∗. Let ξ =∏hi=1(yi−τi). Then ξh = f1 · · · fh ∈ P h

implies ξ is integral over PA∗. Since y1−τ1, . . . , yh−τh is a regular sequence in A∗,we see that ξ 6∈ PA∗. Therefore the extended ideals PA∗ and PA are not integrallyclosed

28.3. Comments and Questions

In connection with Theorem18.3.328.9 it is natural to ask the following question.

18.4.1 Question 28.11. For P and A as in Theorem18.3.328.9, is P the only prime of A

that does not extend to an integrally closed ideal of A?

Com Comments 28.12. In relation to Example18.3.128.7 and to Question

18.4.128.11, con-

sider the following commutative diagram, where the labeled maps are the naturalinclusions:

(28.3)

B = A = R∗ ∩ Q(S)γ1−−−−→ Aout = R∗ ∩Q(T )

γ2−−−−→ R∗ = A∗

δ1

x δ2

x ψ

xS = R[, f, g]

ϕ−−−−→ T = R[α, β] T

Let γ = γ2 · γ1. Referring to the diagram above, we observe the following:(1) Theorem

16.3.213.2 implies that A[1/x] is a localization of S. Furthermore, by


noehomsec8, Aout is excellent. Notice, however, that A

is not excellent since there exists a prime ideal P of A such that PA isnot integrally closed. The excellence of Aout implies that if Q∗ ∈ SpecA∗

and x /∈ Q∗, then the map ψQ∗ : T → A∗Q∗ is regular

G[30, (7.8.3 v)].

(2) Let Q∗ ∈ SpecA∗ be such that x /∈ Q∗ and let q′ = Q∗ ∩ T . ByM[68,

Theorem 32.1] and Item 1 above, if ϕq′ : S → Tq′ is regular, then γQ∗ :A→ A∗

Q∗ is regular.

28.3. COMMENTS AND QUESTIONS 239

(3) Let I be an ideal of A. Since A′ and A∗ are excellent and both havecompletion A, Remark

18.2.128.6.3 shows that the ideals IA′, IA∗ and IA are

either all integrally closed or all fail to be integrally closed.(4) The Jacobian ideal of ϕ : S := k[x, y, z, f, g] → T := k[x, y, z, α, β] is the

ideal of T generated by the determinant of the matrix

J :=

( ∂f∂α

∂g∂α

∂f∂β

∂g∂β

).

Since the characteristic of the field k is zero, this ideal is (y−α)(z− β)T .

In Proposition18.4.328.13, we relate the behavior of integrally closed ideals in the

extension ϕ : S → T to the behavior of integrally closed ideals in the extensionγ : A→ A∗.

18.4.3 Proposition 28.13. With the setting of Theorem18.3.328.9 and Comment

Com28.12.2,

let I be an integrally closed ideal of A such that x 6∈ Q for each Q ∈ Ass(A/I). LetJ = I ∩ S. If JT is integrally closed (resp. a radical ideal) then IA∗ is integrallyclosed (resp. a radical ideal).

Proof. Since the map A → A∗ is flat, x is not in any associated prime ofIA∗. Therefore IA∗ is contracted from A∗[1/x] and it suffices to show IA∗[1/x] isintegrally closed (resp. a radical ideal). Our hypothesis implies I = IA[1/x] ∩ A.By Comment

Com28.12.1, A[1/x] is a localization of S. Thus every ideal of A[1/x] is

the extension of its contraction to S. It follows that IA[1/x] = JA[1/x]. ThusIA∗[1/x] = JA∗[1/x].

Also by CommentCom28.12.1, the map T → A∗[1/x] is regular. If JT is integrally

closed, then Remark18.2.128.6.7 implies that JA∗[1/x] is integrally closed. If JT is a

radical ideal, then the regularity of the map T → A∗[1/x] implies the JA∗[1/x] isa radical ideal.

Proposition 28.14. With the setting of Theorem18.3.328.9 and Comment

Com28.12,

let Q ∈ SpecA be such that QA (or equivalently QA∗) is not integrally closed. Then

(1) Q has height two and x 6∈ Q.(2) There exists a minimal prime Q∗ of QA∗ such that with q′ = Q∗ ∩ T , the

map ϕq′ : S → Tq′ is not regular.(3) Q contains f = (y − α)2 or g = (z − β)2.(4) Q contains no element that is a regular parameter of A.

Proof. By Remark18.2.128.6.6, the height of Q is two. Since A∗/xA∗ = A/xA =

R/xR, we see that x 6∈ Q. This proves item 1.By Remark

18.2.128.6.7, there exists a minimal prime Q∗ of QA∗ such that γQ∗ :

A→ A∗Q∗ is not regular. Thus item 2 follows from Comment

Com28.12.2.

For item 3, let Q∗ and q′ be as in item 2. Since γQ∗ is not regular it is notessentially smooth

G[30, 6.8.1]. By

insideps[45, (2.7)], (y−α)(z−β) ∈ q′. Hence f = (y−α)2

or g = (z − β)2 is in q′ and thus in Q. This proves item 3.Suppose w ∈ Q is a regular parameter for A. Then A/wA and A∗/wA∗ are

two-dimensional regular local domains. By Remark18.2.128.6.6, QA∗/wA∗ is integrally

closed, but this implies that QA∗ is integrally closed, which contradicts our hy-pothesis that QA∗ is not integrally closed. This proves item 4.



Com28.12, let Q ∈

SpecA with x /∈ Q and let q = Q ∩ S. If QA∗ is integrally closed, does it followthat qT is integrally closed?


Com28.12, if a

prime ideal Q of A contains f or g, but not both, and does not contain a regularparameter of A, does it follow that QA∗ is integrally closed ?

In Example18.3.128.7, the three-dimensional regular local domain A contains height-

one prime ideals P such that A/P A is not reduced. This motivates us to ask:

Question 28.17. Let (A,n) be a three-dimensional regular local domain andlet A denote the n-adic completion of A. If for each height-one prime P of A, theextension PA is a radical ideal, i.e., the ring A/P A is reduced, does it follow thatPA is integrally closed for each P ∈ SpecA?

Remark 28.18. A problem analogous to that considered here in the sense thatit also deals with the behavior of ideals under extension to completion is addressedby Loepp and Rotthaus in

LR[64]. They construct nonexcellent Noetherian local

domains to demonstrate that tight closure need not commute with completion.

CHAPTER 29

Generic formal fibers of mixed power

series/polynomial rings May 17 2009 (weiersec), 19

weiersecLet K be a field, m and n positive integers, and X = x1, . . . , xn, and

Y = y1, . . . , ym sets of independent variables over K. Let A be the localizedpolynomial ring K[X ](X). We prove that every prime ideal P in A = K[[X ]] thatis maximal with respect to P ∩ A = (0) has height n − 1. We consider the mixedpower series/polynomial rings B := K[[X ]] [Y ](X,Y ) and C := K[Y ](Y )[[X ]]. Foreach prime ideal P of B = C that is maximal with respect to either P ∩ B = (0)or P ∩ C = (0), we prove that P has height n+m− 2. We also prove each primeideal P of K[[X,Y ]] that is maximal with respect to P ∩K[[X ]] = (0) is of heighteither m or n+m− 2.

29.1. Introduction and Background19.1

Let (R,m) be a Noetherian local integral domain and let R denote the m-adiccompletion of R. The generic formal fiber ring of R is the localization (R\ (0))−1R

of R. The formal fibers of R are the fibers of the morphism Spec R→ SpecR; for aprime ideal P of R, the formal fiber over P is Spec( (RP /PRP )⊗R R ). The formalfibers encode important information about the structure of R. For example, thelocal ring R is excellent provided it is universally catenary and has geometricallyregular formal fibers

G[30, (7.8.3), page 214].

LetR → S be an injective homomorphism of commutative rings. IfR is an inte-gral domain, the generic fiber ring of the map R → S is the localization (R\(0))−1Sof S. In this section we study generic fiber rings for “mixed” polynomial and powerseries rings over a field. More precisely, for K a field, m and n positive integers,and X = x1, . . . , xn and Y = y1, . . . , ym sets of variables over K, we considerthe local rings A := K[X ](X), B := K[[X ]] [Y ](X,Y ) and C := K[Y ](Y )[[X ]], as wellas their completions A = K[[X ]] and B = C = K[[X,Y ]]. Notice that there is acanonical inclusion map B → C.

We have the following local embeddings.

A := K[X ](X) → A := K[[X ]], A → B = C = K[[X,Y ]] and

B := K[[X ]] [Y ](X,Y ) → C := K[Y ](Y )[[X ]] → B = C = K[[X ]] [[Y ]].

Matsumura proves inM1.5[67] that the generic formal fiber ring of A has dimension

n− 1 = dimA − 1, and the generic formal fiber rings of B and C have dimensionn+m− 2 = dimB − 2 = dimC − 2. However he does not address the question ofwhether all maximal ideals of the generic formal fiber rings for A, B and C have the

241

242 29. GENERIC FORMAL FIBERS

same height. If the field K is countable, it follows fromHRS[33, Prop. 4.10, page 36]

that all maximal ideals of the generic formal fiber ring of A have the same height.In answer to a question raised by Matsumura in

M1.5[67], Rotthaus in

R3.5[84] estab-

lishes the following result. Let n be a positive integer. Then there exist excellentregular local rings R such that dimR = n and such that the generic formal fiberring of R has dimension t, where the value of t may be taken to be any integerbetween 0 and dimR − 1. It is also shown in

R3.5[84, Corollary 3.2] that there exists

an excellent regular local domain having the property that its generic formal fiberring contains maximal ideals of different heights.

Let T be a complete Noetherian local ring and let C be a finite set of incom-parable prime ideals of T . Charters and Loepp in

CL[16] (see also

Lo[63, Theorem

17]) determine necessary and sufficient conditions for T to be the completion of aNoetherian local domain T such that the generic formal fiber of T has as maximalelements precisely the prime ideals in C. If T is of characteristic zero, Chartersand Loepp give necessary and sufficient conditions to obtain such a domain T thatis excellent. The finite set C may be chosen to contain prime ideals of differentheights. This provides many examples where the generic formal fiber ring containsmaximal ideals of different heights.

Our main results may be summarized as follows.

gffres Theorem 29.1. With the above notation, we prove that all maximal ideals ofthe generic formal fiber rings of A, B and C have the same height. In particular,we prove:

(1) If P is a prime ideal of A maximal with respect to P ∩ A = (0), thenht(P ) = n− 1.

(2) If P is a prime ideal of B maximal with respect to P ∩ B = (0), thenht(P ) = n+m− 2.

(3) If P is a prime ideal of C maximal with respect to P ∩ C = (0), thenht(P ) = n+m− 2.

(4) In addition, there are at most two possible values for the height of a max-imal ideal of the generic fiber ring (A \ (0))−1C of the inclusion mapA → C.

(a) If n ≥ 2 and P is a prime ideal of C maximal with respect toP ∩ A = (0), then either htP = n+m− 2 or htP = m.

(b) If n = 1, then all maximal ideals of the generic fiber ring(A \ (0))−1C have height m.

We were motivated to consider generic fiber rings for the embeddings displayedabove because of questions related to Chapters

ppssec30 and

tgfsec31 and ultimately because

of the following question posed by Melvin Hochster and Yongwei Yao.

Hochster Question 29.2. Let R be a complete local domain. Can one describe or some-how classify the local maps of R to a complete local domain S such that U−1S isa field, where U = R \ (0), i.e., such that the generic fiber of R → S is trivial?

Hochster and Yao remark that if, for example, R is equal characteristic zero,one obtains such extensions by starting with

19.1.2.119.1.2.1 (weiersec29.1.2.1) R = K[[x1, ..., xn]] → T = L[[x1, ..., xn, y1, ..., ym]]→ T/P = S,

29.2. VARIATIONS ON A THEME OF WEIERSTRASS 243

where K is a subfield of L, the xi, yj are formal indeterminates, and P is a primeideal of T maximal with respect to being disjoint from the image of R \ 0. Ofcourse, such prime ideals P correspond to the maximal ideals of the generic fiber(R \ (0))−1T .

In Theorem19.7.229.20, we answer Question

Hochster29.2 in the special case where the ex-

tension arises from the embedding in Sequence19.1.2.1weiersec29.1.2.1 with the field L = K. We

prove in this case that the dimension of the extension ring S must be either 2 or n.In Chapters

ppssec30 and

tgfsec31 we study extensions of integral domains R

ϕ→ S such

that, for every nonzero Q ∈ SpecS, we have Q∩R 6= (0). Such extensions are calledtrivial generic fiber extensions or TGF extensions. One obtains such an extensionby considering a composition R → T → T/P = S, where T is an extension ring ofR and P ∈ SpecT is maximal with respect to P ∩R = (0). Thus the generic fiberring and so also Theorem

gffres29.1 give information regarding TGF extensions in the

case where the smaller ring is a mixed polynomial/power series ring.In addition, Theorem

gffres29.1 is useful in the study of Sequence

19.1.2.1weiersec29.1.2.1, because

the map in Sequence19.1.2.1weiersec29.1.2.1 factors through:

R = K[[x1, . . . , xn]] → K[[x1, . . . , xn]] [y1, . . . , ym] → T = L[[x1, . . . , xn, y1, . . . , yn]].

Sectionweiersec29.2 contains implications of Weierstrass’ Preparation Theorem to the

prime ideals of power series rings. We first prove a technical proposition regarding achange of variables that provides a “nice” generating set for a given prime ideal P ofa power series ring; then in Theorem

19.2.529.7 we prove that, in certain circumstances, a

larger prime ideal can be found with the same contraction as P to a certain subring.In Sections

wimpB29.3 and

wimpC29.4, we prove parts 2 and 3 of Theorem

gffres29.1 stated above.

In Section19.529.5 we use a result of Valabrega for the two-dimensional case. We then

apply this result in SectionwimpA29.6 to prove part 1 of Theorem

gffres29.1, and in Section

gffps29.7 we prove part 4.

29.2. Variations on a theme of Weierstrass19.2

We apply the Weierstrass Preparation TheoremZSII[106, Theorem 5, page 139,

and Corollary 1, page 145] to examine the structure of a given prime ideal P in thepower series ring A = K[[X ]], where X = x1, . . . , xn is a set of n variables overthe field K. Here A = K[X ](X) is the localized polynomial ring in these variables.Our procedure is to make a change of variables that yields a regular sequence in Pof a nice form.

We recall the statement of the Weierstrass Preparation Theorem.

19.2.1 Theorem 29.3. (Weierstrass) Let (R,m) be a complete local ring, let f ∈R[[x]] be a formal power series and let f denote the image of f in (R/m)[[x]].Assume that f 6= 0 and that ord f = s > 0. There exists a unique ordered pair(u, F ) such that u is a unit in R[[x]] and F ∈ R[x] is a distinguished monic poly-nomial of degree s such that f = uF . Here F = xs + as−1x

s−1 + · · ·+ a0 ∈ R[x] isdistinguished if ai ∈m for 0 ≤ i ≤ s− 1.

19.2.2 Corollary 29.4. The ideal fR[[x]] is extended from R[x] and R[[x]]/(f) is afree R-module of rank s. Every g ∈ R[[x]] is of the form g = qf+r, where q ∈ R[[x]]and r ∈ R[x] is a polynomial with deg r ≤ s− 1.


19.2.3 Notation 29.5. By a change of variables, we mean a finite sequence of ‘poly-nomial’ change of variables of the type described below, where X = x1, . . . , xn isa set of n variables over the field K. For example, with ei, fi ∈ N, consider

x1 7→ x1 + xe1n = z1, x2 7→ x2 + xe2n = z2, . . . ,

xn−1 7→ xn−1 + xen−1n = zn−1, xn 7→ xn = zn,

followed by:

z1 7→ z1 = t1, z2 7→ z2 + zf21 = t2, . . . ,

zn−1 7→ zn−1 + zfn−11 = tn−1, xn 7→ zn + zfn

1 = tn.

Thus a change of variables defines an automorphism of A that restricts to an auto-morphism of A.

We also consider a change of variables for subrings of A and A. For example, ifA1 = K[x2, . . . , xn] ⊆ A and S = K[[x2, . . . , xn]] ⊆ A, then by a change of variablesinside A1 and S, we mean a finite sequence of automorphisms of A and A of thetype described above on x2, . . . , xn that leave the variable x1 fixed. In this case weobtain an automorphism of A that restricts to an automorphism on each of S, Aand A1.

19.2.4 Proposition 29.6. Let A := K[[X ]] = K[[x1, . . . , xn]] and let P ∈ Spec Awith x1 6∈ P and htP = r, where 1 ≤ r ≤ n − 1. There exists a change ofvariables x1 7→ z1 := x1 (x1 is fixed), x2 7→ z2, . . . , xn 7→ zn and a regular sequencef1, . . . , fr ∈ P so that, upon setting Z1 = z1, . . . , zn−r, Z2 = zn−r+1, . . . , znand Z = Z1 ∪ Z2, we have

f1 ∈ K[[Z1]] [zn−r+1, . . . , zn−1] [zn] is monic as a polynomial in zn

f2 ∈ K[[Z1]] [zn−r+1, . . . , zn−2] [zn−1] is monic as a polynomial in zn−1, etc...

fr ∈ K[[Z1]] [zn−r+1] is monic as a polynomial in zn−r+1.

In addition:(1) P is a minimal prime of the ideal (f1, . . . , fr)A.(2) The (Z2)-adic completion of K[[Z1]] [Z2](Z) is identical to the (f1, . . . , fr)-

adic completion and both equal A = K[[X ]] = K[[Z]].(3) If P1 := P ∩K[[Z1]] [Z2](Z), then P1A = P , that is, P is extended from

K[[Z1]] [Z2](Z).(4) The ring extension:

K[[Z1]] → K[[Z1]] [Z2](Z)/P1∼= K[[Z]]/P

is finite (and integral).

Proof. Since A is a unique factorization domain, there exists a nonzero primeelement f in P . The power series f is therefore not a multiple of x1, and so f mustcontain a monomial term xi22 . . . xinn with a nonzero coefficient in K. This nonzerocoefficient in K may be assumed to be 1. There exists an automorphism σ : A→ Adefined by the change of variables:

x1 7→ x1 x2 7→ t2 := x2+xe2n . . . xn−1 7→ tn−1 := xn−1+xen−1n xn 7→ xn,


with e2, . . . , en−1 ∈ N chosen suitably so that f written as a power series in thevariables x1, t2, . . . , tn−1, xn contains a term anx

snn , where sn is a positive integer,

and an ∈ K is nonzero. We assume that the integer sn is minimal among allintegers i such that a term axin occurs in f with a nonzero coefficient a ∈ K; wefurther assume that the coefficient an = 1. By Weierstrass we have that:

f = mε,

where m ∈ K[[x1, t2, . . . , tn−1]] [xn] is a monic polynomial in xn of degree sn andε is a unit in A. Since f ∈ P is a prime element, m ∈ P is also a prime element.Using Weierstrass again, every element g ∈ P can be written as:

g = mh+ q,

where h ∈ K[[x1, t2, . . . , tn−1, xn]] = A and q ∈ K[[x1, t2, . . . , tn−1]] [xn] is a poly-nomial in xn of degree less than sn. Note that

K[[x1, t2, . . . , tn−1]] → K[[x1, t2, . . . , tn−1]] [xn]/(m)

is an integral (finite) extension. Thus the ring K[[x1, t2, . . . , tn−1]] [xn]/(m) iscomplete. Moreover, the two ideals (x1, t2, . . . , tn−1,m) = (x1, t2, . . . , tn−1, x

snn )

and (x1, t2, . . . , tn−1, xn) of B0 := K[[x1, t2, . . . , tn−1]] [xn] have the same radical.Therefore A is the (m)-adic and the (xn)-adic completion of B0 and P is extendedfrom B0.

This implies the statement for r = 1, with f1 = m, zn = xn, z1 = x1, z2 =t2, . . . , zn−1 = tn−1, Z1 = x1, t2, . . . , tn−1 and Z2 = zn = xn. In particular,when r = 1, P is minimal over mA, so P = mA.

For r > 1 we continue by induction on r. Let P0 := P ∩K[[x1, t2, . . . , tn−1]].Sincem /∈ K[[x1, t2, . . . , tn−1]] and P is extended fromB0 := K[[x1, t2, . . . , tn−1]] [xn],then P ∩B0 has height r and htP0 = r− 1. Since x1 /∈ P , we have x1 /∈ P0, and bythe induction hypothesis there is a change of variables t2 7→ z2, . . . , tn−1 7→ zn−1 ofK[[x1, t2, . . . , tn−1]] and elements f2, . . . , fr ∈ P0 so that:

f2 ∈ K[[x1, z2 . . . , zn−r]] [zn−r+1, . . . , zn−2] [zn−1] is monic in zn−1

f3 ∈ K[[x1, z2 . . . , zn−r]] [zn−r+1, . . . , zn−3] [zn−2] is monic in zn−2, etc...

fr ∈ K[[x1, z2, . . . , zn−r]] [zn−r+1] is monic in zn−r+1,

and f2, . . . , fr satisfy the assertions of Proposition19.2.429.6 for P0.

It follows that m, f2, . . . , fr is a regular sequence of length r and that P is aminimal prime of the ideal (m, f2, . . . , fr)A. Set zn = xn. We now prove that mmay be replaced by a polynomial f1 ∈ K[[x1, z2, . . . , zn−r]] [zn−r+1, . . . , zn]. Write

m =sn∑i=0

aizn,

where the ai ∈ K[[x1, z2, . . . , zn−1]]. For each i < sn, apply Weierstrass to ai andf2 in order to obtain:

ai = f2hi + qi,


where hi is a power series inK[[x1, z2, . . . , zn−1]] and qi ∈ K[[x1, z2, . . . , zn−2]] [zn−1]is a polynomial in zn−1. With qsn = 1 = asn , we define

m1 =sn∑i=0

qizin.

Now (m1, f2, . . . , fr)A = (m, f2, . . . , fr)A and we may replace m by m1 which is apolynomial in zn−1 and zn. To continue, for each i < sn, write:

qi =∑j,k

bijzjn−1 with bij ∈ K[[x1, z2, . . . , zn−2]].

For each bij , we apply Weierstrass to bij and f3 to obtain:

bij = f3hij + qij ,

where qij ∈ K[[x1, z2, . . . , zn−3]] [zn−2]. Set

m2 =∑i,j

qijzjn−1z

in ∈ K[[x1, z2, . . . , zn−3]] [zn−2, zn−1, zn]

with qsn0 = 1. It follows that (m2, f2, . . . , fr)A = (m, f2, . . . , fr)A. Continuingthis process by applying Weierstrass to the coefficients of zkn−2z

jn−1z

in and f4, we

establish the existence of a polynomial f1 ∈ K[[Z1]] [zn−r+1, . . . , zn] that is monicin zn so that (f1, f2, . . . , fr)A = (m, f2, . . . , fr)A. Therefore P is a minimal primeof (f1, . . . , fr)A.

The extensionK[[Z1]] −→ K[[Z1]] [Z2]/(f1, . . . , fr)

is integral and finite. Thus the ring K[[Z1]] [Z2]/(f1, . . . , fr) is complete. This im-plies A = K[[x1, z2, . . . , zn]] is the (f1, . . . , fr)-adic (and the (Z2)-adic) completionof K[[Z1]] [Z2](Z) and that P is extended from K[[Z1]] [Z2](Z). This completes theproof of Proposition

19.2.429.6.

The following theorem is the technical heart of this section.

19.2.5 Theorem 29.7. Let K be a field and let y and X = x1, . . . , xn be variablesover K. Assume that V is a discrete valuation domain with completion V = K[[y]]and that K[y] ⊆ V ⊆ K[[y]]. Also assume that the field K((y)) = K[[y]] [1/y] hasuncountable transcendence degree over the quotient field Q(V ) of V . Set R0 :=V [[X ]] and R = R0 = K[[y,X ]]. Let P ∈ SpecR be such that:

(1) ”(i)” P ⊆ (X)R (so y /∈ P ), and(2) ”(ii)” dim(R/P ) > 2.

Then there is a prime ideal Q ∈ SpecR such that(1) P ⊂ Q ⊂ XR,(2) dim(R/Q) = 2, and(3) P ∩R0 = Q ∩R0.

In particular, P ∩K[[X ]] = Q ∩K[[X ]].

Proof. Assume that P has height r. Since dim(R/P ) > 2, we have 0 ≤ r <n − 1. If r > 0, then there exist a transformation x1 7→ z1, . . . , xn 7→ zn andelements f1, . . . , fr ∈ P , by Proposition

19.2.429.6, so that the variable y is fixed, and

f1 ∈ K[[y, z1, . . . , zn−r]] [zn−r+1, . . . , zn] is monic in zn,


f2 ∈ K[[y, z1, . . . , zn−r]] [zn−r+1, . . . , zn−1] is monic in zn−1 etc,...

fr ∈ K[[y, z1, . . . , zn−r]] [zn−r+1] is monic in zn−r+1,

and the assertions of Proposition19.2.429.6 are satisfied. In particular, P is a minimal

prime of (f1, . . . , fr)R. Let Z1 = z1, . . . , zn−r and Z2 = zn−r+1, . . . , zn−1, zn.By Proposition

19.2.429.6, if D := K[[y, Z1]] [Z2](Z) and P1 := P ∩D, then P1R = P .

The following diagram shows these rings and ideals.

R = K[[y,X ]] = K[[y, Z1, Z2]]

(X)R

D = K[[y, Z1]] [Z2](Z)

P = P1R

P1 = P ∩D

Note that f1, . . . , fr ∈ P1. Let g1, . . . , gs ∈ P1 be other generators such thatP1 = (f1, . . . , fr, g1, . . . , gs)D. Then P = P1R = (f1, . . . , fr, g1, . . . , gs)R. For each(i) := (i1, . . . , in) ∈ Nn and j, k with 1 ≤ j ≤ r, 1 ≤ k ≤ s, let aj,(i), bk,(i) denotethe coefficients in K[[y]] of the fj , gk, so that

fj =∑

(i)∈Nn

aj,(i)zi11 . . . zinn , gk =

∑(i)∈Nn

bk,(i)zi11 . . . zinn ∈ K[[y]] [[Z]].

Define

∆ :=

aj,(i), bk,(i) ⊆ K[[y]], for r > 0∅, for r = 0.

A key observation here is that in either case the set ∆ is countable.To continue the proof, we consider S := Q(V (∆))∩K[[y]], a discrete valuation

domain, and its field of quotients L := Q(V (∆)). Since ∆ is a countable set, thefield K((y)) is (still) of uncountable transcendence degree over L. Let γ2, . . . , γn−rbe elements of K[[y]] that are algebraically independent over L. We defineT := L(γ2, . . . , γn−r) ∩K[[y]] and E := Q(T ) = L(γ2, . . . , γn−r).

The diagram below shows the prime ideals P and P1 and the containmentsbetween the relevant rings.


R = K[[y, Z]]P = (fj , gk)R

D := K[[y, Z1]] [Z2](Z)

P1 = (fj, gk)DQ(K[[y]]) = K[[y]] [1/y] = K((y))

K[[y]]E := Q(T ) = L(γ2, . . . , γn−r)

T := L(γ2, . . . , γn−r) ∩K[[y]]

S := Q(V (∆)) ∩K[[y]]

L := Q(S) = Q(V (∆))

V

Q(V )

K[y]

Let P2 := P ∩ S[[Z1]] [Z2](Z). Since f1, . . . , fr, g1, . . . , gs ∈ S[[Z1]] [Z2](Z), wehave P2R = P . Since P ⊆ (x1, . . . , xn)R = (Z)R, there is a prime ideal P in L[[Z]]that is minimal over P2L[[Z]]. Since L[[Z]] is flat over S[[Z]], P ∩S[[Z]] = P2S[[Z]].Note that L[[X ]] = L[[Z]] is the (f1, . . . , fr)-adic (and the (Z2)-adic) completion ofL[[Z1]][Z2](Z). In particular,

L[[Z1]] [Z2]/(f1, . . . , fr) = L[[Z1]] [[Z2]]/(f1, . . . , fr)

and this also holds with the field L replaced by its extension field E.Since L[[Z]]/P is a homomorphic image of L[[Z]]/(f1, . . . , fr), it follows that

L[[Z]]/P is integral (and finite) over L[[Z1]]. This yields the commutative diagram:

19.2.5.019.2.5.0 (19.2.529.7.0)

E[[Z1]]−→E[[Z1]] [[Z2]]/PE[[Z]]

↑ ↑L[[Z1]]−→ L[[Z1]] [[Z2]]/P

with injective integral (finite) horizontal maps. Recall that E is the subfield ofK((y)) obtained by adjoining γ2, . . . , γn−r to the field L. Thus the vertical mapsin Diagram

19.2.529.7.0 are faithfully flat.


Let q := (z2 − γ2z1, . . . , zn−r − γn−rz1)E[[Z1]] ∈ Spec(E[[Z1]]) and let W be aminimal prime of the ideal (P ,q)E[[Z]]. Since

f1, . . . , fr, z2 − γ2z1, . . . , zn−r − γn−rz1is a regular sequence in T [[Z]] the prime ideal W := W ∩T [[Z]] has height n−1. LetQ be a minimal prime of WK((y))[[Z]] and let Q := Q∩R. Then W = Q∩T [[Z]],P ⊂ Q ⊂ ZR = XR, and pictorially we have:

(W ) ⊆ Q ⊂ K((y))[[Z]]

R := K[[y, Z]]

P = (fj, gk)R ⊆ Q ⊂ R

D := K[[y, Z1]] [Z2](Z)

P1 = (fj, gk)D ⊂ D

(P ,q) ⊆ W ⊂ E[[Z]]

(P2) ⊆ P ⊂ L[[Z]]

P2 = (fj , gk) ⊂ S[[Z1]] [Z2](Z)

W ⊂ T [[Z]]

S[[Z]]L[[Z1]] [Z2](Z)

q ⊂ E[[Z1]]

L[[Z1]]

Notice that q is a prime ideal of height n − r − 1. Also, since K((y))[[Z]] isflat over K[[y, Z]] = R, we have htQ = n− 1 and dim(R/Q) = 2. We clearly haveP2 ⊆W ∩ S[[Z1]] [Z2](Z).

19.2.6 Claim 29.8. q ∩ L[[Z1]] = (0).

To show this we argue as inM1.5[67]: Suppose that

h =∑m∈N

Hm ∈ q ∩ L[[z1, . . . , zn−r]],

where Hm ∈ L[z1, . . . , zn−r] is a homogeneous polynomial of degree m:

Hm =∑

|(i)|=mc(i)z

i11 . . . z

in−r

n−r ,

where (i) := (i1, . . . , in−r) ∈ Nn−r, |(i)| := i1 + · · · + in−r and c(i) ∈ L. Considerthe E-algebra homomorphism π : E[[Z1]] → E[[z1]] defined by π(z1) = z1 andπ(zi) = γiz1 for 2 ≤ i ≤ n− r. Then kerπ = q, and for each m ∈ N:

π(Hm) = π(∑

|(i)|=mc(i)z

i11 . . . z

in−r

n−r ) =∑

|(i)|=mc(i)γ

i22 . . . γ

in−r

n−r zm1

andπ(h) =

∑m∈N

π(Hm) =∑m∈N

∑|(i)|=m

c(i)γi22 . . . γ

in−r

n−r zm1 .


Since h ∈ q, π(h) = 0. Since π(h) is a power series in E[[z1]], each of its coefficientsis zero, that is, for each m ∈ N,∑

|(i)|=mc(i)γ

i22 . . . γ

in−r

n−r = 0.

Since the γi are algebraically independent over L, each c(i) = 0. Therefore h = 0,and so q ∩ L[[Z1]] = (0). This proves Claim

19.2.629.8.

Using the commutativity of Diagram19.2.529.7.0 and that the horizonal maps of this

diagram are integral extensions, we deduce that (W∩E[[Z1]]) = q, and q∩L[[Z1]] =(0) implies W∩L[[Z1]] = (0). We conclude thatQ∩S[[Z]] = P∩S[[Z]] and thereforeQ ∩R0 = P ∩R0.

We record the following corollary.

19.2.7 Corollary 29.9. Let K be a field and let R = K[[y,X ]], where X = x1, . . . , xnand y are independent variables over K. Assume P ∈ SpecR is such that:

(i) P ⊆ (x1, . . . , xn)R and(ii) dim(R/P ) > 2.

Then there is a prime ideal Q ∈ SpecR so that(1) P ⊂ Q ⊂ (x1, . . . , xn)R,(2) dim(R/Q) = 2, and(3) P ∩K[y](y)[[X ]] = Q ∩K[y](y)[[X ]].

In particular, P ∩K[[x1, . . . , xn]] = Q ∩K[[x1, . . . , xn]].

Proof. With notation as in Theorem19.2.529.7, let V = K[y](y).

29.3. Weierstrass implications for the ring B = K[[X ]] [Y ](X,Y )wimpB

As beforeK denotes a field, n andm are positive integers, andX = x1, . . . , xnand Y = y1, . . . , ym denote sets of variables over K. Let B := K[[X ]] [Y ](X,Y ) =K[[x1, . . . , xn]] [y1, . . . , ym](x1,...,xn,y1,...,ym). The completion of B is B = K[[X,Y ]].

19.3.1 Theorem 29.10. With the notation as above, every ideal Q of B = K[[X,Y ]]maximal with the property that Q ∩B = (0) is a prime ideal of height n+m− 2.

Proof. Suppose first that Q is such an ideal. Then clearly Q is prime. Mat-sumura shows in

M1.5[67, Theorem 3] that the dimension of the generic formal fiber of

B is at most n+m− 2. Therefore htQ ≤ n+m− 2.Now suppose P ∈ Spec B is an arbitrary prime ideal of height r < n +m − 2

with P ∩ B = (0). We construct a prime Q ∈ Spec B with P ⊂ Q, Q ∩ B = (0),and htQ = n+m− 2. This will show that all prime ideals maximal in the genericfiber have height n+m− 2.

For the construction of Q we consider first the case where P 6⊆ XB. Thenthere exists a prime element f ∈ P that contains a term θ := yi11 · · · yimm , wherethe ij’s are nonnegative integers and at least one of the ij is positive. Notice thatm ≥ 2 for otherwise with y = y1 we have f ∈ P contains a term yi. By Weierstrassit follows that f = gε, where g ∈ K[[X ]] [y] is a nonzero monic polynomial in y andε is a unit of B. But g ∈ P and g ∈ B implies P ∩B 6= (0), a contradiction to ourassumption that P ∩B = (0).

29.3. WEIERSTRASS IMPLICATIONS FOR THE RING B = K[[X]] [Y ](X,Y ) 251

For convenience we now assume that the last exponent im appearing in θ aboveis positive. We apply a change of variables: ym → tm := ym and, for 1 ≤ ` < m, lety` → t` := y` + tm

e` , where the e` are chosen so that f , expressed in the variablest1, . . . , tm, contains a term tqm, for some positive integer q. This change of variablesinduces an automorphism of B. By Weierstrass f = g1h, where h is a unit in Band g1 ∈ K[[X, t1, . . . , tm−1]] [tm] is monic in tm. Set P1 = P ∩K[[X, t1, . . . , tm−1]].If P1 ⊆ XK[[X, t1, . . . , tm−1]], we stop the procedure and take s = m− 1 in whatfollows. If P1 6⊆ XK[[X, t1, . . . , tm−1]], then there exists a prime element f ∈ P1

that contains a term t1j1 · · · tm−1

jm−1 , where the jk’s are nonnegative integers andat least one of the jk is positive. We then repeat the procedure using the primeideal P1. That is, we replace t1, . . . tm−1 with a change of variables so that a primeelement of P1 contains a term monic in some one of the new variables. Aftera suitable finite iteration of changes of variables, we obtain an automorphism ofB that restricts to an automorphism of B and maps y1, . . . , ym 7→ z1, . . . , zm.Moreover, there exist a positive integer s ≤ m − 1 and elements g1, . . . gm−s ∈ Psuch that

g1 ∈ K[[X, z1, . . . , zm−1]] [zm] is monic in zmg2 ∈ K[[X, z1, . . . , zm−2]] [zm−1] is monic in zm−1, etc

...

gm−s ∈ K[[X, z1, . . . , zs]] [zs+1] is monic in zs+1,

and such that, for Rs := K[[X, z1, . . . , zs]] and Ps := P ∩Rs, we have Ps ⊆ XRs.As in the proof of Proposition

19.2.429.6, we replace the regular sequence g1, . . . .gm−s

by a regular sequence f1, . . . , fm−s so that:

f1 ∈ Rs[zs+1, . . . , zm] is monic in zmf2 ∈ Rs[zs+1, . . . , zm−1] is monic in zm−1, etc

...

fm−s ∈ Rs[zs+1] is monic in zs+1.

and (g1, . . . .gm−s)B = (f1, . . . , fm−s)B.Let G := K[[X, z1, . . . , zs]] [zs+1, . . . , zm] = Rs[zs+1, . . . , zm]. By Proposition

19.2.429.6, P is extended from G. Let q := P ∩ G and extend f1, . . . , fm−s to a gen-erating system of q, say, q = (f1, . . . , fm−s, h1, . . . , ht)G. For integers k, ` with1 ≤ k ≤ m − s and 1 ≤ ` ≤ t, express the fk and h` in G as power series inB = K[[z1]][[z2, . . . , zm]] [[X ]] with coefficients in K[[z1]]:

fk =∑

ak(i)(j)zi22 . . . zimm xj11 . . . xjnn and h` =

∑b`(i)(j)z

i22 . . . zimm xj11 . . . xjnn ,

where ak(i)(j), b`(i)(j) ∈ K[[z1]], (i) = (i2, . . . , im) and (j) = (j1, . . . , jn). The set∆ = ak(i)(j), b`(i)(j) is countable. We define V := K(z1,∆) ∩K[[z1]]. Then V isa discrete valuation domain with completion K[[z1]] and K((z1)) has uncountabletranscendence degree over Q(V ). Let Vs := V [[X, z2, . . . , zs]] ⊆ Rs. Notice thatRs = Vs, the completion of Vs. Also f1, . . . , fm−s ∈ Vs[zs+1, . . . , zm] ⊆ G and(f1, . . . , fm−s)G ∩Rs = (0). Furthermore the extension

Vs := V [[X, z2, . . . , zs]] → Vs[zs+1, . . . , zm]/(f1, . . . , fm−s)


is finite. Set P0 := P ∩ Vs. Then P0 ⊆ XRs ∩ Vs = XVs.Consider the commutative diagram:

31.1.131.1.1 (29.1)

Rs := K[[X, z1, . . . , zs]]−→Rs[[zs+1, . . . , zm]]/(f1, . . . , fm−s)

↑ ↑Vs := V [[X, z2, . . . , zs]]−→Vs[zs+1, . . . , zm]/(f1, . . . , fm−s) .

The horizontal maps are injective and finite and the vertical maps are completions.The prime ideal q := PRs[[zs+1, . . . , zm]]/(f1, . . . , fm−s) lies over Ps in Rs.

By assumption Ps ⊆ (X)Rs and by Theorem19.2.529.7 there is a prime ideal Qs of Rs

such that Ps ⊆ Qs ⊆ (X)Rs, Qs ∩ Vs = Ps ∩ Vs = P0, and dim(Rs/Qs) = 2.There is a prime ideal Q in Rs[[zs+1, . . . , zm]]/(f1, . . . , fm−s) lying over Qs withq ⊆ Q by the “going-up theorem”

M[68, Theorem 9.4]. Let Q be the preimage in

B = K[[X, z1, . . . , zm]] of Q. We show the rings and ideals of Theorem19.3.129.10 below.

B = K[[X,Y ]] = K[[X, z1, . . . , zm]] = Rs[[zs+1, . . . , zm]](q, Qs)B ⊆ Q

P * XB

G := Rs[zs+1, . . . , zm]q := P ∩G

q = (fi, hj)G

fi /∈ Rs := K[[X, z1, . . . , zs]]Ps ⊆ Qs ⊂ Rs

Ps := P ∩Rs ⊆ XRs

V = K[[z1]]Vs := V [[X, z2, . . . , zs]]P0 := P ∩ Vs

V := K(z1,∆) ∩K[[z1]]

Then Q has height n + s− 2 +m− s = n+m − 2. Moreover, it follows fromDiagram

31.1.129.1 that Q and P have the same contraction to Vs[zs+1, . . . , zm]. This

implies that Q ∩B = (0) and completes the proof in the case where P 6⊆ XB.In the case where P ⊆ XB, let h1, . . . , ht ∈ B be a finite set of generators of

P , and as above, let b`(i)(j) ∈ K[[z1]] be the coefficients of the h`’s. Consider thecountable set ∆ = b`(i)(j) and the valuation domain V := K(z1,∆)∩K[[z1]]. SetP0 := P ∩V [[X, z2, . . . , zm]]. By Theorem

19.2.529.7, there exists a prime ideal Q of B =

K[[X, z1, . . . , zm]] of height n+m−2 such that P ⊂ Q and Q∩V [[X, z2, . . . , zm]] =P ∩ V [[X, z2, . . . , zm]] = P0. Therefore Q ∩ B = (0). This completes the proof ofTheorem

19.3.129.10.

29.4. WEIERSTRASS IMPLICATIONS FOR THE RING C = K[Y ](Y )[[X]] 253

29.4. Weierstrass implications for the ring C = K[Y ](Y )[[X ]]wimpC

As beforeK denotes a field, n andm are positive integers, andX = x1, . . . , xnand Y = y1, . . . , ym denote sets of variables over K. Consider the ring C =K[y1, . . . , ym](y1,...,ym)[[x1, . . . , xn]] = K[Y ](Y )[[X ]]. Then the completion of C isC = K[[Y,X ]].

19.4.1 Theorem 29.11. With notation as above, let Q ∈ Spec C be maximal with theproperty that Q ∩ C = (0). Then htQ = n+m− 2.

Proof. Let B = K[[X ]] [Y ](X,Y ) ⊂ C. If P ∈ Spec C = Spec B and P ∩ C =(0), then P ∩B = (0), so htP ≤ n+m− 2 by Theorem

19.3.129.10. Consider a nonzero

prime P ∈ Spec C with P ∩ C = (0) and htP = r < n +m − 2. If P ⊆ XC thenTheorem

19.2.529.7 implies the existence of Q ∈ Spec C with htQ = n+m− 2 such that

P ⊂ Q and Q ∩ C = (0).Assume that P is not contained in XC and consider the ideal J := (P,X)C.

Since C is complete in the XC-adic topology,R3[83, Lemma 2] implies that if J is

primary for the maximal ideal of C, then P is extended from C. Since we areassuming P ∩ C = (0), J is not primary for the maximal ideal of C and we havehtJ = n + s < n + m, where 0 < s < m. Let W ∈ Spec C be a minimal primeof J such that htW = n + s. Let W0 = W ∩ K[[Y ]]. Then W = (W0, X)C andW0 is a prime ideal of K[[Y ]] with htW0 = s. By Proposition

19.2.429.6 applied to

K[[Y ]] and the prime ideal W0 ∈ SpecK[[Y ]], there exists a change of variablesY 7→ Z with y1 7→ z1, . . . , ym 7→ zm and elements f1, . . . , fs ∈ W0 so that withZ1 = z1, . . . , zm−s, we have

f1 ∈ K[[Z1]] [zm−s+1, . . . , zm] is monic in zmf2 ∈ K[[Z1]] [zm−s+1, . . . , zm−1] is monic in zm−1, etc

...

fs ∈ K[[Z1]] [zm−s+1] is monic in zm−s+1.

Now z1, . . . , zm−s, f1, . . . , fs is a regular sequence in K[[Z]] = K[[Y ]]. Let T =tm−s+1, . . . , tm be a set of additional variables and consider the map:

ϕ : K[[Z1, T ]] −→ K[[z1, . . . , zm]]

defined by zi 7→ zi for all 1 ≤ i ≤ m − s and tm−i+1 7→ fi for all 1 ≤ i ≤ s. Theembedding ϕ is finite (and free) and so is the extension to power series rings in X :

ρ : K[[Z1, T ]] [[X ]] −→ K[[z1, . . . , zm]] [[X ]] = C.

The contraction ρ−1(W ) ∈ SpecK[[Z1, T,X ]] of the prime ideal W of C has heightn+ s, since htW = n+ s. Moreover ρ−1(W ) contains (T,X)K[[Z1, T,X ]], a primeideal of height n + s. Therefore ρ−1(W ) = (T,X)K[[Z1, T,X ]]. By construction,P ⊆W which yields that ρ−1(P ) ⊆ (T,X)K[[Z1, T,X ]].

To complete the proof we construct a suitable base ring related to C. Considerthe expressions for the fi’s as power series in z2, . . . , zm with coefficients in K[[z1]]:

fj =∑

aj(i)zi22 . . . zimm ,


where (i) := (i2, . . . , im), 1 ≤ j ≤ s, aj(i) ∈ K[[z1]]. Also consider a finite generatingsystem g1, . . . , gq for P and expressions for the gk, where 1 ≤ k ≤ q, as power seriesin z2, . . . , zm, x1, . . . , xn with coefficients in K[[z1]]:

gk =∑

bk(i)(`)zi22 . . . zimm x`11 . . . x`nn ,

where (i) := (i2, . . . , im), (`) := (`1, . . . , `n), and bk(i)(`) ∈ K[[z1]]. We take thesubset ∆ = aj(i), bk(i)(`) of K[[z1]] and consider the discrete valuation domain:

V := K(z1,∆) ∩K[[z1]].

Since V is countably generated over K(z1), the field K((z1)) has uncountable tran-scendence degree over Q(V ) = K(z1,∆). Moreover, by construction the ideal P isextended from V [[z2, . . . , zm]] [[X ]]. Consider the embedding:

ψ : V [[z2, . . . , zm−s, T ]] −→ V [[z2, . . . , zm]],

which is the restriction of ϕ above, so that zi 7→ zi for all 2 ≤ i ≤ m − s andtm−i+1 7→ fi for all i with 1 ≤ i ≤ s.

Let σ be the extension of ψ to the power series rings:

σ : V [[z2, . . . , zm−s, T ]] [[X ]] −→ V [[z2, . . . , zm]] [[X ]]with σ(xi) = xi for all i with 1 ≤ i ≤ n.

Notice that ρ defined above is the completion σ of the map σ, that is, theextension of σ to the completions. Consider the commutative diagram: 4.1.0

19.4.1e19.4.1e (weiersec29.4.1.0)

K[[Z1, T ]] [[X ]] bσ=ρ−−−−→ K[[Z]] [[X ]] = Cx xV [[z2, . . . , zm−s, T ]] [[X ]] σ−−−−→ V [[z2, . . . , zm]] [[X ]]

where σ = ρ is a finite map.Recall that ρ−1(W ) = (T,X)K[[Z1, T,X ]], and so ρ−1(P ) ⊆ (T,X)K[[Z1, T,X ]]

by Diagram19.4.1eweiersec29.4.1.0. By Theorem

19.2.529.7, there exists a prime ideal Q0 of the ring

K[[Z1, T,X ]] such that ρ−1(P ) ⊆ Q0, htQ0 = n+m− 2, and

Q0 ∩ V [[z2, . . . , zm−s, T ]] [[X ]] = ρ−1(P ) ∩ V [[z2, . . . , zm−s, T ]] [[X ]].

By the “going-up theorem”M[68, Theorem 9.4], there is a prime ideal Q ∈ Spec C

that lies over Q0 and contains P . Moreover, Q also has height n + m − 2. Thecommutativity of Diagram

19.4.1eweiersec29.4.1.0 implies that

P1 := P ∩ V [[z2, . . . , zm−s, T ]] [[X ]] ⊆ Q1 := Q ∩ V [[z2, . . . , zm−s, T ]] [[X ]].

Consider the finite homomorphism:

λ : V [[z2, . . . , zm−s]] [T ](Z1,T )[[X ]] −→ V [[z2, . . . , zm−s]] [zm−s+1, . . . , zm](Z)[[X ]]

(determined by ti 7→ fi for 1 ≤ i ≤ m) and the commutative diagram:

V [[z2, . . . , zm−s]] [[T ]] [[X ]] σ−−−−→ V [[z2, . . . , zm]] [[X ]]x xV [[z2, . . . , zm−s]] [T ](Z1,T )[[X ]] λ−−−−→ V [[z2, . . . , zm−s]] [zm−s+1, . . . , zm](Z)[[X ]].

29.5. SUBRINGS OF THE POWER SERIES RING K[[z, t]] 255

Since Q ∩ V [[z2, . . . , zm−s, T ]] [[X ]] = P ∩ V [[z2, . . . , zm−s, T ]] [[X ]] and since λis a finite map we conclude that

Q1 ∩ V [[z2, . . . , zm−s]] [zm−s+1, . . . , zm](Z)[[X ]]

= P1 ∩ V [[z2, . . . , zm]] [zm−s+1, . . . , zm](Z)[[X ]].

Since C ⊆ V [[z2, . . . , zm−s]] [zm−s+1, . . . , zm](Z)[[X ]], we obtain that Q ∩ C =P ∩ C = (0). This completes the proof of Theorem

19.4.129.11.

19.4.2 Remark 29.12. With B and C as in SectionswimpB29.3 and

wimpC29.4, we have

B = K[[X ]] [Y ](X,Y ) → K[Y ](Y )[[X ]] = C and B = K[[X,Y ]] = C.

Thus for P ∈ SpecK[[X,Y ]], if P ∩ C = (0), then P ∩ B = (0). By Theorems3.1 and 4.1, each prime of K[[X,Y ]] maximal in the generic formal fiber of B or Chas height n+m− 2. Therefore each P ∈ SpecK[[X,Y ]] maximal with respect toP ∩ C = (0) is also maximal with respect to P ∩B = (0). However, if n+m ≥ 3,the generic fiber of B → C is nonzero (see Chapter

ppssec30, so there exist primes of

K[[X,Y ]] maximal in the generic formal fiber of B that are not in the generic formalfiber of C.

29.5. Subrings of the power series ring K[[z, t]]19.5

In this section we establish properties of certain subrings of the power seriesring K[[z, t]] that will be useful in considering the generic formal fiber of localizedpolynomial rings over the field K.

19.5.1 Notation 29.13. Let K be a field and let z and t be independent variablesover K. Consider countably many power series:

αi(z) =∞∑j=0

aijzj ∈ K[[z]]

with coefficients aik ∈ K. Let s be a positive integer and let ω1, . . . , ωs ∈ K[[z, t]]be power series in z and t, say:

ωi =∞∑j=0

βijtj , where βij(z) =

∞∑k=0

bijkzk ∈ K[[z]] and bijk ∈ K,

for each i with 1 ≤ i ≤ s. Consider the subfield K(z, αi, βij) of K((z)) and thediscrete rank-one valuation domain

V := K(z, αi, βij) ∩K[[z]].

The completion of V is V = K[[z]]. Assume that ω1, . . . , ωr are algebraicallyindependent over Q(V )(t) and that the elements ωr+1, . . . , ωs are algebraic overthe field Q(V )(t, ωiri=1). Notice that the set αi ∪ βij is countable, and thatalso the set of coefficients of the αi and βij

∆ := aij ∪ bijkis a countable subset of the field K. Let K0 denote the prime subfield of K andlet F denote the algebraic closure in K of the field K0(∆). The field F is count-able and the power series αi(z) and βij(z) are in F [[z]]. Consider the subfield


F (z, αi, βij) of F ((z)) and the discrete rank-one valuation domain

V0 := F (z, αi, βij) ∩ F [[z]].

The completion of V0 is V0 = F [[z]]. Since Q(V0)(t) ⊆ Q(V )(t), the elementsω1, . . . , ωr are algebraically independent over the field Q(V0)(t).

Consider the subfield E0 := Q(V0)(t, ω1, . . . , ωr) of Q(V0[[t]]) and the subfieldE := Q(V )(t, ω1, . . . , ωr) of Q(V [[t]]). A result of Valabrega

V[96] implies that the

integral domains:

19.5.1.119.5.1.1 (weiersec29.5.1.1) D0 := E0 ∩ V0[[t]] and D := E ∩ V [[t]]

are two-dimensional regular local rings with completions D0 = F [[z, t]] and D =K[[z, t]], respectively. Moreover, Q(D0) = E0 is a countable field.

19.5.2 Proposition 29.14. Let D0 be as defined in (19.5.129.13). Then there exists a power

series γ ∈ zF [[z]] such that the prime ideal (t − γ)F [[z, t]] ∩ D0 = (0), i.e., (t −γ)F [[z, t]] is in the generic formal fiber of D0.

Proof. Since D0 is countable there are only countably many prime ideals inD0 and since D0 is Noetherian there are only countably many prime ideals in D0 =F [[z, t]] that lie over a nonzero prime of D0. There are uncountably many primesin F [[z, t]], which are generated by elements of the form t− σ for some σ ∈ zF [[z]].Thus there must exist an element γ ∈ zF [[z]] with (t− γ)F [[z, t]] ∩D0 = (0).

For ωi = ωi(t) =∑∞

j=0 βijtj as in (

19.5.129.13) and γ an element of zK[[z]], let ωi(γ)

denote the following power series in K[[z]]:

ωi(γ) :=∞∑j=0

βijγj ∈ K[[z]].

19.5.3 Proposition 29.15. Let D be as defined in (19.5.1.1weiersec29.5.1.1). For an element γ ∈

zK[[z]] the following conditions are equivalent:(1) ”(i)” (t− γ)K[[z, t]] ∩D = (0).(2) ”(ii)” The elements γ, ω1(γ), . . . , ωr(γ) are algebraically independent overQ(V ).

Proof. (i)⇒ (ii): Assume by way of contradiction that the set γ, ω1(γ), . . . , ωr(γ)is algebraically dependent over Q(V ) and let d(k) ∈ V be finitely many elementssuch that ∑

(k)

d(k)ω1(γ)k1 . . . ωr(γ)krγkr+1 = 0

is a nontrivial equation of algebraic dependence for γ, ω1(γ), . . . , ωr(γ), where each(k) = (k1 . . . , kr, kr+1) is an (r + 1)-tuple of nonnegative integers. It follows that∑

(k)

d(k)ωk11 . . . ωkr

r tkr+1 ∈ (t− γ)K[[z, t]] ∩D = (0).

Since ω1, . . . , ωr are algebraically independent over Q(V )(t), we have d(k) = 0 forall (k), a contradiction. This completes the proof that (i)⇒ (ii).

(ii) ⇒ (i): If (t− γ)K[[z, t]] ∩D 6= (0), then there exists a nonzero element

τ =∑(k)

d(k)ωk11 . . . ωkr

r tkr+1 ∈ (t− γ)K[[z, t]] ∩ V [t, ω1, . . . , ωr].

29.6. WEIERSTRASS IMPLICATIONS FOR THE LOCALIZED POLYNOMIAL RING A 257

But this implies that

τ(γ) =∑(k)

d(k)ω1(γ)k1 . . . ωr(γ)krγkr+1 = 0.

Since γ, ω1(γ), . . . , ωr(γ) are algebraically independent over Q(V ), it follows thatall the coefficients d(k) = 0, a contradiction to the assumption that τ is nonzero.

Let γ ∈ zF [[z]] be as in Proposition19.5.229.14 with (t−γ)F [[z, t]]∩D0 = (0). Then:

19.5.4 Proposition 29.16. With notation as above, we also have (t−γ)K[[z, t]]∩D =(0), that is, (t− γ)K[[z, t]] is in the generic formal fiber of D.

Proof. Let tii∈I be a transcendence basis ofK over F and let L := F (tii∈I).Then K is algebraic over L. Let αi, βij ⊂ F [[z]] be as in (5.1) and define

V1 = L(z, αi, βij) ∩ L[[z]] and D1 = Q(V1)(t, ω1, . . . , ωr) ∩ L[[z, t]].

Then V1 is a discrete rank-one valuation domain with completion L[[z]] and D1

is a two-dimensional regular local domain with completion D1 = L[[z, t]]. Notethat Q(V ) and Q(D) are algebraic over Q(V1) and Q(D1), respectively. Since(t− γ)K[[z, t]] ∩ L[[z, t]] = (t− γ)L[[z, t]], it suffices to prove that (t− γ)L[[z, t]] ∩D1 = (0). By Proposition

19.5.329.15, it suffices to show that γ, ω1(γ), . . . , ωr(γ) are

algebraically independent over Q(V1). The commutative diagram

F [[z]]tialgebraically ind.−−−−−−−−−−−−−→ L[[z]]x x

Q(V0)transcendence basis ti−−−−−−−−−−−−−−−→ Q(V1)

implies that the set γ, ω1(γ), . . . , ωr(γ) ∪ ti is algebraically independent overQ(V0). Therefore γ, ω1(γ), . . . , ωr(γ) is algebraically independent over Q(V1),which completes the proof of Proposition

19.5.429.16.

19.5.5 Remark 29.17. We remark that with ωr+1, . . . , ωs algebraic overQ(V )(ω1, . . . , ωr)as in (5.1), if we define

D := Q(V )(t, ω1, . . . , ωs) ∩ V [[t]],

then again by ValabregaV[96], D is a two-dimensional regular local domain with

completion K[[z, t]]. Moreover, Q(D) is algebraic over Q(D) and (t− γ)K[[z, t]] ∩D = (0) implies that (t− γ)K[[z, t]] ∩ D = (0).

29.6. Weierstrass implications for the localized polynomial ring AwimpA

Let n be a positive integer, let X = x1, . . . , xn be a set of n variables overa field K, and let A := K[x1, . . . , xn](x1,...,xn) = K[X ](X) denote the localizedpolynomial ring in these n variables over K. Then the completion of A is A =K[[X ]].

19.6.1 Theorem 29.18. For the localized polynomial ring A = K[X ](X) defined above,if Q is an ideal of A maximal with respect to Q ∩A = (0), then Q is a prime idealof height n− 1.


Proof. Again it is clear that Q as described in the statement is a prime ideal.Also the assertion holds for n = 1. Thus we assume n ≥ 2. By Proposition

19.5.429.16,

there exists a nonzero prime p in K[[x1, x2]] such that p ∩K[x1, x2](x1,x2) = (0).It follows that pA ∩A = (0). Thus the generic formal fiber of A is nonzero.

Let P ∈ Spec A be a nonzero prime ideal with P ∩A = (0) and htP = r < n−1.We construct Q ∈ Spec A of height n − 1 with P ⊆ Q and Q ∩ A = (0). ByProposition

19.2.429.6, there exists a change of variables x1 7→ z1, . . . , xn 7→ zn and

polynomials

f1 ∈ K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn] monic in znf2 ∈ K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn−1] monic in zn−1, etc

...

fr ∈ K[[z1, . . . , zn−r]] [zn−r+1] monic in zn−r+1,

so that P is a minimal prime of (f1, . . . , fr)A and P is extended from

R := K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn].

Let P0 := P ∩R and extend f1, . . . , fr to a system of generators of P0, say:

P0 = (f1, . . . , fr, g1, . . . , gs)R.

Using an argument similar to that in the proof of Theorem19.2.529.7, write

fj =∑

(i)∈Nn−1

aj,(i)zi22 . . . zinn and gk =

∑(i)∈Nn−1

bk,(i)zi22 . . . zinn ,

where aj,(i), bk,(i) ∈ K[[z1]]. Let

V0 := K(z1, aj,(i), bk,(i)) ∩K[[z1]].

Then V0 is a discrete rank-one valuation domain with completion K[[z1]], andK((z1)) has uncountable transcendence degree over the field of fractions Q(V0)of V0. Let γ3, . . . , γn−r ∈ K[[z1]] be algebraically independent over Q(V0) anddefine

q := (z3 − γ3z2, z4 − γ4z2, . . . , zn−r − γn−rz2)K[[z1, . . . , zn−r]].

We see that q∩ V0[[z2, . . . , zn−r]] = (0) by an argument similar to that inM1.5[67] and

in Claim19.2.629.8. Let R1 := V0[[z2, . . . , zn−r]] [zn−r+1, . . . , zn], let P1 := P ∩ R1 and

consider the commutative diagram:

K[[z1, . . . , zn−r]]−→R/P0

↑ ↑V0[[z2, . . . , zn−r]]−→R1/P1

The horizontal maps are injective finite integral extensions. Let W be a minimalprime of (q, P )A. Then htW = n− 2 and q ∩ V0[[z2, . . . , zn−r]] = (0) implies thatW ∩ R1 = P1. We have found a prime ideal W ∈ Spec A such that htW = n − 2,W ∩A = (0) and P ⊆W . Since f1, . . . , fr ∈ W and since A = K[[z1, . . . , zn]] is the(f1, . . . , fr)-adic completion of K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn], the prime ideal Wis extended from K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn].

29.6. WEIERSTRASS IMPLICATIONS FOR THE LOCALIZED POLYNOMIAL RING A 259

We claim that W is actually extended from K[[z1, z2]] [z3, . . . , zn]. To see thislet g ∈ W ∩K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn] and write:

g =∑(i)

a(i)zin−r+1n−r+1 . . . z

inn ∈ K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn],

where the sum is over all (i) = (in−r, . . . , in) and a(i) ∈ K[[z1, . . . , zn−r]]. For alla(i) by Weierstrass we can write

a(i) = (zn−r − γn−rz2)h(i) + q(i),

where h(i) ∈ K[[z1, . . . zn−r]] and q(i) ∈ K[[z1, . . . , zn−r−1]]. If n− r > 3, we write

q(i) = (zn−r−1 − γn−r−1z2)h′(i) + q′(i),

where h′(i) ∈ K[[z1, . . . zn−r−1]] and q′(i) ∈ K[[z1, . . . , zn−r−2]]. In this way wereplace a generating set for W in K[[z1, . . . , zn−r]] [zn−r+1, . . . , zn] by a generatingset for W in K[[z1, z2]] [z3, . . . , zn].

In particular, we can replace the elements f1, . . . , fr by elements:

h1 ∈ K[[z1, z2]] [z3, . . . , zn] monic in znh2 ∈ K[[z1, z2]] [z3, . . . , zn−1] monic in zn−1, etc

...

hr ∈ K[[z1, z2]] [z3 . . . , zn−r+1] monic in zn−r+1

and set hr+1 = z3−γ3z2, . . . , hn−2 = zn−r−γn−rz2, and then extend to a generatingset h1, . . . , hn+s−2 for

W0 = W ∩K[[z1, z2]] [z3, . . . , zn]

such that W0A = W . Consider the coefficients in K[[z1]] of the hj:

hj =∑(i)

cj(i)zi22 . . . zinn

with cj(i) ∈ K[[z1]]. The set cj(i) is countable. Define

V := Q(V0)(cj(i)) ∩K[[z1]]

Then V is a rank-one discrete valuation domain that is countably generated overK[z1](z1) and W is extended from V [[z2]] [z3, . . . , zn].

We may also write each hi as a polynomial in z3, . . . , zn with coefficients inV [[z2]]:

h =∑

ω(i)zi33 . . . zinn

with ω(i) ∈ V [[z2]] ⊆ K[[z1, z2]]. By the result of ValabragaV[96], the integral

domainD := Q(V )(z2, ω(i)) ∩K[[z1, z2]]

is a two-dimensional regular local domain with completion D = K[[z1, z2]]. LetW1 := W ∩D[z3, . . . , zn]. Then W1A = W . We have shown in Section

19.529.5 that

there exists a prime element q ∈ K[[z1, z2]] with qK[[z1, z2]] ∩D = (0). Considerthe finite extension

D −→ D[z3, . . . , zn]/W1.


Let Q ∈ Spec A be a minimal prime of (q,W )A. Since htW = n − 2 and q 6∈ W ,htQ = n− 1. Moreover, P ⊆W implies P ⊆ Q. We claim that

Q ∩D[z3, . . . , zn] = W1 and therefore Q ∩A = (0).

To see this consider the commutative diagram:

K[[z1, z2]]−→ K[[z1, . . . , zn]]/W

↑ ↑D −→ D[z3, . . . , zn]/W1 ,

which has injective finite horizontal maps. Since qK[[z1, z2]] ∩D = (0), it followsthat Q ∩D[z3, . . . , zn] = W1. This completes the proof of Theorem

19.6.129.18.

29.7. Generic fibers of power series ring extensionsgffps

In this section we apply the Weierstrass machinery from Section19.229.2 to the

generic fiber rings of power series extensions.

19.7.1 Theorem 29.19. Let n ≥ 2 be an integer and let y, x1, . . . , xn be variablesover the field K. Let X = x1, . . . , xn. Consider the formal power series ringR1 = K[[X ]] and the extension R1 → R1[[y]] = R. Let U = R1 \ (0). ForP ∈ SpecR such that P ∩ U = ∅ we have:

(1) If P 6⊆ XR, then dimR/P = n and P is maximal in the generic fiberU−1R.

(2) If P ⊆ XR, then there exists Q ∈ SpecR such that P ⊆ Q, dimR/Q = 2and Q is maximal in the generic fiber U−1R.

If n > 2 for each prime ideal Q maximal in the generic fiber U−1R, we have

dimR/Q =

n and R1 → R/Q is finite, or2 and Q ⊂ XR.

Proof. Let P ∈ SpecR be such that P ∩U = ∅ or equivalently P ∩R1 = (0).Then R1 embeds in R/P . If dim(R/P ) ≤ 1, then the maximal ideal of R1 generatesan ideal primary for the maximal ideal of R/P . By

M[68, Theorem 8.4] R/P is finite

over R1, and so dimR1 = dim(R/P ), a contradiction. Thus dim(R/P ) ≥ 2.

???alertMatsumura Thm 8.4 should be in the tools and be referred to there.

If P 6⊆ XR, then there exists a prime element f ∈ P that contains a term ys forsome positive integer s. By Weierstrass, it follows that f = gε, where g ∈ K[[X ]] [y]is a nonzero monic polynomial in y and ε is a unit of R. We have fR = gR ⊆ P isa prime ideal and R1 → R/gR is a finite integral extension. Since P ∩ R1 = (0),we must have gR = P .

If P ⊆ XR and dim(R/P ) > 2, then Theorem19.2.529.7 implies there exists Q ∈

SpecR such that dim(R/Q) = 2, P ⊂ Q ⊂ XR and P ∩ R1 = (0) = Q ∩ R1, andso P is not maximal in the generic fiber. Thus Q ∈ SpecR maximal in the genericfiber of R1 → R implies that the dimension of dim(R/Q) is 2, or equivalently thathtQ = n− 1.

19.7.2 Theorem 29.20. Let n and m be positive integers, and let X = x1, . . . , xnand Y = y1, . . . , ym be sets of independent variables over the field K. Considerthe formal power series rings R1 = K[[X ]] and R = K[[X,Y ]] and the extension

29.7. GENERIC FIBERS OF POWER SERIES RING EXTENSIONS 261

R1 → R1[[Y ]] = R. Let U = R1 \ (0). Let Q ∈ SpecR be maximal with respect toQ ∩ U = ∅. If n = 1, then dimR/Q = 1 and R1 → R/Q is finite.

If n ≥ 2, there are two possibilities(1) R1 → R/Q is finite, in which case dimR/Q = dimR1 = n, or(2) dimR/Q = 2.

Proof. First assume n = 1, and let x = x1. Since Q is maximal with respectto Q ∩ U = ∅, for each P ∈ SpecR with Q ( P we have P ∩ U is nonempty andtherefore x ∈ P . It follows that dimR/Q = 1, for otherwise,

Q =⋂P | P ∈ SpecR and Q ( P ,

which implies x ∈ Q. ByM[68, Theorem 8.4], R1 → R/Q is finite.

It remains to consider the case where n ≥ 2. We proceed by induction onm. Theorem

19.7.129.19 yields the assertion for m = 1. Suppose Q ∈ SpecR is max-

imal with respect to Q ∩ U = ∅. As in the proof of Theorem19.7.129.19, we have

dimR/Q ≥ 2. If Q ⊆ (X, y1, . . . , ym−1)R, then by Theorem19.2.529.7 with R0 =

K[ym](ym)[[X, y1, . . . , ym−1]], there exists Q′ ∈ SpecR with Q ⊆ Q′, dimR/Q′ = 2,and Q ∩ R0 = Q′ ∩ R0. Since R1 ⊆ R0, we have Q′ ∩ U = ∅. Since Q is maximalwith respect to Q ∩ U = ∅, we have Q = Q′, so dimR/Q = 2.

Otherwise, if Q 6⊆ (X, y1, . . . , ym−1)R, then there exists a prime element f ∈ Qthat contains a term ysm for some positive integer s. Let R2 = K[[X, y1, . . . , ym−1]].By Weierstrass, it follows that f = gε, where g ∈ R2[ym] is a nonzero monicpolynomial in ym and ε is a unit of R. We have fR = gR ⊆ Q is a prime ideal andR2 → R/gR is a finite integral extension. Thus R2/(Q∩R2) → R/Q is an integralextension. It follows that Q ∩ R2 is maximal in R2 with respect to being disjointfrom U . By induction dimR2/(Q∩R2) is either n or 2. Since R/Q is integral overR2/(Q ∩R2), dimR/Q is either n or 2.

19.7.3 Remark 29.21. In the notation of Theoremgffres29.1, Theorem

19.7.229.20 proves the

second part of the theorem, since dimR = n + m. Thus if n = 1, htQ = m. Ifn ≥ 2, the two cases are (i) htQ = m and (ii) htQ = n+m− 2, as in (a) and (b)of Theorem

gffres29.1, part 4.

Using the TGF terminology discussed in the introduction, we have the followingcorollary to Theorem

19.7.229.20.

19.7.4 Corollary 29.22. With the notation of Theorem19.7.229.20, assume P ∈ SpecR

is such that R1 → R/P =: S is a TGF extension. Then dimS = dimR1 = n ordimS = 2.

CHAPTER 30

Mixed polynomial/power series rings and relations

among their spectra (ppssec), 20, May 17, 2009

ppssecTakehiko Yasuda in

yasuda[102] gives additional information on the TGF property.

In particular, he shows inyasuda[102, Theorem 2.7] that

C[x, y][[z]] → C[x, x−1, y][[z]]

is not TGF, where C is the field of complex numbers.

30.1. Introduction and Background20.1

In this section we study the nested mixed polynomial/power series rings:

20.1.0.120.1.0.1 (ppssec30.1.0.1) A := k[x, y] → B := k[[y]] [x] → C := k[x] [[y]] → E := k[x, 1/x] [[y]],

where k is a field and x and y are indeterminates over k. In Sequence20.1.0.2ppssec30.1.0.2 the

maps are all flat. Also we consider

20.1.0.220.1.0.2 (ppssec30.1.0.2) C → D1 := k[x] [[y/x]] → · · · → Dn := k[x] [[y/xn]] → · · · → E.

With regard to Sequence20.1.0.2ppssec30.1.0.2, for n a positive integer, the map C → Dn is not

flat, but Dn → E is a localization followed by an adic completion of a Noetherianring and therefore is flat. We discuss the spectra of these rings and consider themaps induced on the spectra by the inclusion maps on the rings. For example,we determine whether there exist nonzero primes of one of the larger rings thatintersect a smaller ring in zero. We were led to consider these rings by questionsthat came up in two contexts.

The first motivation is from the introduction to the paperAJL[8] by Alonzo-Tarrio,

Jeremias-Lopez and Lipman: If a map between Noetherian formal schemes canbe factored as a closed immersion followed by an open one, can this map alsobe factored as an open immersion followed by a closed one? This is not true ingeneral. As mentioned in

AJL[8], Brian Conrad observed that a counterexample can

be constructed for every triple (R, x, p), where(1): R is an adic domain, that is, R is a Noetherian domain that is separated

and complete with respect to the powers of a proper ideal I.(2): x is a nonzero element of R such that the completion of R[1/x] with re-

spect to the powers of IR[1/x], denoted S := Rx, is an integral domain.(3): p is a nonzero prime ideal of S that intersects R in (0).

The compositionR→ S → S/p determines a map on formal spectra Spf (S/p)→Spf (S)→ Spf (R) that is a closed immersion followed by an open one. To see this,recall that a surjection such as S → S/p of adic rings gives rise to a closed immer-sion Spf (S/p) → Spf (S) while a localization, such as that of R with respect to

263

264 30. MIXED POLYNOMIAL/POWER SERIES

the powers of x, followed by the completion of R[1/x] with respect to the powersof IR[1/x] to obtain S gives rise to an open immersion Spf (S) → Spf (R)

G[30,

(10.2.2)].The map Spf (S/p) → Spf (R) cannot be factored, however, as an open im-

mersion followed by a closed one. This is because a closed immersion into Spf (R)corresponds to a surjective map of adic rings R → R/J , where J is an ideal of RG[30, page 441]. Thus if the immersion Spf (S/p)→ Spf (R) factored as an open im-mersion followed by a closed one, we would have R-algebra homomorphisms fromR → R/J → S/p, where Spf (S/p) → Spf (R/J) is an open immersion. Sincep ∩ R = (0), we must have J = (0). This implies Spf (S/p) → Spf (R) is an openimmersion, that is, the composite map Spf (S/p)→ Spf (S)→ Spf (R), is an openimmersion. But also Spf (S) → Spf (R) is an open immersion. It follows thatSpf (S/p) → Spf (S) is both open and closed. Since S is an integral domain thisimplies Spf (S/p) ∼= Spf (S). This is a contradiction since p is nonzero.

An example of such a triple (R, x, p) is described inAJL[8]: For w, x, y, z indeter-

minates over a field k, set R := k[w, x, z] [[y]], S := k[w, x, 1/x, z] [[y]]. Notice thatR is complete with respect to yR and S is complete with respect to yS. An indirectproof is given in

AJL[8] that there exist nonzero primes p of S for which p ∩ R = (0).

In Proposition20.4.530.24 below we give a direct proof of this fact.

A second motivation is from a question raised by Mel Hochster: “Can onedescribe or somehow classify the local maps R → S of complete local domains Rand S such that every nonzero prime ideal of S has nonzero intersection with R?”In Chapter

tgfsec31 we study this question and define:

20.1.1 Definition 30.1. For R and S integral domains with R a subring of S we saythat S is a trivial generic fiber extension of R, or a TGF extension of R, if everynonzero prime ideal of S has nonzero intersection with R.

In some correspondence to Lipman regarding closed and open immersions, Con-rad asked: “Is there a nonzero prime ideal of E := k[x, 1/x] [[y]] that intersectsC = k[x] [[y]] in zero?” If there were such a prime ideal, then C := k[x] [[y]] →E := k[x, 1/x] [[y]] would be a simpler counterexample to the assertion that a closedimmersion followed by an open one also has a factorization as an open immersionfollowed by a closed one. In the terminology of Definition

20.1.130.1, one can ask:

20.1.2 Question 30.2. Let x and y be indeterminates over a field k. Is C := k[x] [[y]] →E := k[x, 1/x] [[y]] a TGF extension?

We show in Proposition20.2.630.10.2 below that the answer to Question

20.1.230.2 is “yes”.

This is part of our analysis of the prime spectra of A, B, C, Dn and E, and themaps induced on these spectra by the inclusion maps on the rings.

The following example is a local map of the type described in Hochster’s ques-tion.

20.1.3 Example 30.3. Let x and y be indeterminates over a field k and consider theextension R := k[[x, y]] → S := k[[x]] [[y/x]]. To see this extension is TGF, itsuffices to show P ∩ R 6= (0) for each P ∈ SpecS with htP = 1. This is clear ifx ∈ P , while if x 6∈ P , then k[[x]] ∩ P = (0), and so k[[x]] → R/(P ∩ R) → S/Pand S/P is finite over k[[x]]. Therefore dimR/(P ∩R) = 1, and so P ∩R 6= (0).

20.1.4 Remark 30.4. (1) The extension k[[x, y]] → k[[x, y/x]] is, up to isomorphism,the same as the extension k[[x, xy]] → k[[x, y]].

30.2. TRIVIAL GENERIC FIBER (TGF) EXTENSIONS AND PRIME SPECTRA 265

(2) We show in Chaptertgfsec31 that the extension R := k[[x, y, xz]] → S :=

k[[x, y, z]] is not TGF.

30.2. Trivial generic fiber (TGF) extensions and prime spectra20.2

The following two propositions about trivial generic fiber extensions from Chap-ter

tgfsec31 are useful and straightforward to check.

20.2.1 Proposition 30.5. Let R → S be an injective map where R and S are integraldomains. Then the following are equivalent:

(1): S is a TGF extension of R.(2): Every nonzero element of S has a nonzero multiple in R.(3): For U = R \ 0, U−1S is a field.

20.2.2 Proposition 30.6. Let R → S and S → T be injective maps where R, S andT are integral domains.

(1): If R → S and S → T are TGF extensions, then so is R → T . Equiv-alently if R → T is not TGF, then at least one of the extensions R → Sor S → T is not TGF.

(2): If R → T is TGF, then S → T is TGF.(3): If the map Spec T → SpecS is surjective, then R → T is TGF impliesR → S is TGF.

More information about TGF extensions is in Chaptersweiersec29 and

tgfsec31.

20.2.3 Remark 30.7. Let R be a commutative ring and let R[[y]] denote the formalpower series ring in the variable y over R. Then

(1): Each maximal ideal of R[[y]] is of the form (m, y)R[[y]] where m is amaximal ideal of R. Thus y is in every maximal ideal of R[[y]].

(2): If R is Noetherian with dimR[[y]] = n and x1, . . . , xm are independentindeterminates over R[[y]], then y is in every height n+m maximal idealof the polynomial ring R[[y]] [x1, . . . , xm].

Proof. Item (1) follows fromN2[72, Theorem 15.1]. For item (2), let m be a

maximal ideal of R[[y]] [x1, . . . , xm] with ht(m) = n + m. ByKap[58, Theorem 39],

ht(m∩R[[y]]) = n ; thus m∩R[[y]] is maximal in R[[y]], and so, by item (1), y ∈m.

20.2.4 Proposition 30.8. Let n be a positive integer, let R be an n-dimensional Noe-therian domain, let y be an indeterminate over R, and let q be a prime ideal ofheight n in the power series ring R[[y]]. If y 6∈ q, then q is contained in a uniquemaximal ideal of R[[y]].

Proof. The assertion is clear if q is maximal. Otherwise S := R[[y]]/q hasdimension one. Moreover, S is complete with respect to the yS-adic topology

M[68,

Theorem 8.7] and every maximal ideal of S is a minimal prime of the principal idealyS. Hence S is a complete semilocal ring. Since S is also an integral domain, itmust be local by

M[68, Theorem 8.15]. Therefore q is contained in a unique maximal

ideal of R[[y]].

In Section20.330.3 we use the following corollary to Proposition

20.2.430.8.


20.2.5 Corollary 30.9. Let R be a one-dimensional Noetherian domain and let q bea height-one prime ideal of the power series ring R[[y]]. If q 6= yR[[y]], then q iscontained in a unique maximal ideal of R[[y]].

20.2.6 Proposition 30.10. Consider the nested mixed polynomial/power series rings

A := k[x, y] → B := k[[y]] [x] → C := k[x] [[y]]→ D1 := k[x] [[y/x]] → D2 := k[x] [[y/x2]] → · · ·→ Dn := k[x] [[y/xn]] → · · · → E := k[x, 1/x] [[y]],

where k is a field and x and y are indeterminates over k. Then(1): If S ∈ B,C,D1, D2, · · · , Dn, · · · , E, then A → S is not TGF.(2): If R,S ⊂ B,C,D1, D2, · · · , Dn · · · , E are such that R ⊆ S, thenR → S is TGF.

(3): Each of the proper associated maps on spectra fails to be surjective.

Proof. For item (1), let σ(y) ∈ yk[[y]] be such that σ(y) and y are alge-braically independent over k. Then (x − σ(y))S ∩ A = (0), and so A → S is notTGF.

For item (2), observe that every maximal ideal of C, Dn or E is of height twowith residue field finite algebraic over k. To show R → S is TGF, it suffices toshow q ∩ R 6= (0) for each height-one prime ideal q of S. This is clear if y ∈ q. Ify 6∈ q, then k[[y]] ∩ q = (0), and so k[[y]] → R/(q ∩ R) → S/q are injections. ByCorollary

20.2.530.9, S/q is a one-dimensional local domain. Since the residue field of

S/q is finite algebraic over k, it follows that S/q is finite over k[[y]]. Therefore S/qis integral over R/(q ∩R). Hence dim(R/(q ∩R) = 1 and so q ∩R 6= (0).

For item (3), observe that xDn is a prime ideal of Dn and x is a unit of E.Thus SpecE → SpecDn is not surjective. Now, considering C = D0 and n > 0, wehave xDn ∩Dn−1 = (x, y/xn−1)Dn−1. Therefore xDn−1 is not in the image of themap SpecDn → SpecDn−1. The map from SpecC → SpecB is not onto, because(1+xy) is a prime ideal of B, but 1+xy is a unit in C. Similarly SpecB → SpecAis not onto, because (1 + y) is a prime ideal of A, but 1 + y is a unit in B. Thiscompletes the proof.

20.2.7 Question and Remarks 30.11. Which of the Spec maps of Proposition20.2.630.10

are one-to-one and which are finite-to-one?(1): For S ∈ B,C,D1, D2, · · · , Dn, · · · , E, the generic fiber ring of the

map A → S has infinitely many prime ideals and has dimension one.Every height-two maximal ideal of S contracts in A to a maximal ideal.Every maximal ideal of S containing y has height two. Also yS ∩A = yAand the map SpecS/yS → SpecA/yA is one-to-one.

(2): Suppose R → S is as in Proposition20.2.630.10.2. Each height-two prime

of S contracts in R to a height-two maximal ideal of R. Each height-oneprime of R is the contraction of at most finitely many prime ideals of Sand all of these prime ideals have height one. If R → S is flat, which istrue if S ∈ B,C,E, then “going-down” holds for R → S, and so, for Pa height-one prime of S, we have ht(P ∩R) ≤ 1.

(3): As mentioned inHW[53, Remark 1.5], C/P is Henselian for every nonzero

prime ideal P of C other than yC.

30.3. SPECTRA FOR TWO-DIMENSIONAL MIXED POLYNOMIAL/POWER SERIES RINGS267

30.3. Spectra for two-dimensional mixed polynomial/power series rings20.3

Let k be a field and let x and y be indeterminates over k. We consider theprime spectra, as partially ordered sets, of the mixed polynomial/power series ringsA, B, C, D1, D2, · · · , Dn, · · · and E as given in Sequences

20.1.0.1ppssec30.1.0.1 and

20.1.0.2ppssec30.1.0.2 of

the introduction.Even for k a countable field there are at least two non-order-isomorphic partially

ordered sets that can be the prime spectrum of the polynomial ring A := k[x, y].Let Q be the field of rational numbers, let F be a field contained in the algebraicclosure of a finite field and let Z denote the ring of integers. Then, by

rW1[99] and

rW2[100], Spec Q[x, y] 6∼= SpecF [x, y] ∼= Spec Z[y].

The prime spectra of the rings B, C, D1, D2, · · · , Dn, · · · , and E of Sequencess20.1.0.1ppssec30.1.0.1 and

20.1.0.2ppssec30.1.0.2 are simpler since they involve power series in y. Remark

20.2.330.7.2

implies that y is in every maximal ideal of height two of each of these rings.The partially ordered set SpecB = Spec k[[y]] [x] is similar to a prime ideal

space studied inHW[53] and

Shah[91]. The difference from

HW[53] is that here k[[y]] is un-

countable, even if k is countable. It follows that SpecB is also uncountable. As apartially ordered set, SpecB can be described uniquely up to isomorphism by theaxioms of

Shah[91] (similar to the CHP axioms of

HW[53]), since k[[y]] is Henselian and

has cardinality at least equal to c, the cardinality of the real numbers R.The following theorem characterizes U := SpecB as a Henselian affine partially

ordered set (where the “≤” relation is “set containment”):

20.3.1 Theorem 30.12.HW[53, Theorem 2.7]

Shah[91, Theorem 2.4] Let B = k[[y]] [x] be

as in Sequence20.1.0.1ppssec30.1.0.1, where k is a field, the cardinality of the set of maximal

ideals of k[x] is α and the cardinality of k[[y]] is β. Then the partially orderedset U := SpecB is called Henselian affine of type (β, α) and is characterized as apartially ordered set by the following axioms:

(1): |U | = β.(2): U has a unique minimal element.(3): dim(U) = 2 and | height-two elements of U | = α.(4): There exists a unique special height-one element u ∈ U such that u is

contained in every height-two element of U .(5): Every nonspecial height-one element of U is in at most one height-two

element.(6): Every height-two element t ∈ U contains cardinality β many height-one

elements that are only contained in t. If t1, t2 ∈ U are distinct height-two elements, then the special element from (4) is the unique height-oneelement less than both.

(7): There are cardinality β many height-one elements that are maximal.

20.3.2 Remark 30.13. (1) The axioms of Theorem20.3.130.12 are redundant. We feel this

redundancy helps in understanding the relationships between the prime ideals.(2) The theorem applies to the spectrum of B by defining the unique minimal

element to be the ideal (0) of B and the special height-one element to be the primeideal yB. Every height-two maximal ideal m of B has nonzero intersection withk[[y]]. Thus m/yB is principal and so m = (y, f(x)), for some monic irreduciblepolynomial f(x) of k[x]. Consider f(x) + ay | a ∈ k[[y]]. This set has cardinalityβ and each f(x) + ay is contained in a nonempty finite set of height-one primes


contained in m. If p is a height-one prime contained in m with p 6= yB, thenp∩ k[[y]] = (0), and so pk((y))[x] is generated by a monic polynomial in k((y))[x].But for a, b ∈ k[[y]] with a 6= b, we have (f(x) + ay, f(x) + by)k((y))[x] = k((y))[x].Therefore no height-one prime contained in m contains both f(x)+ay and f(x)+by.Since B is Noetherian and |B| = β is an infinite cardinal, we conclude that thecardinality of the set of height-one prime ideals contained in m is β. Examples ofheight-one maximals are (1 + xyf(x, y) ), for various f(x, y) ∈ k[[y]] [x]. The set ofheight-one maximal ideals of B also has cardinality β.

(3) These axioms characterize SpecB in the sense that every two partiallyordered sets satisfying these axioms are order-isomorphic.

The picture of SpecB is shown below:

β (y) β β · · ·

• • • · · ·(# bullets = α)

(0)

Spec k[[y]] [x]

In the diagram, β is the cardinality of k[[y]], and α is the cardinality of theset of maximal ideals of k[x] (and also the cardinality of the set of maximal idealsof k[[y]] [x] ); the boxed β means there are cardinality β height-one primes in thatposition with respect to the partial ordering.

Next we consider SpecR[[y]], for R a Noetherian one-dimensional domain.Then SpecR[[y]] has the following picture by Theorem

3.4?? below:

(y) κi κj · · ·

• • • · · ·(# bullets = α)

(0)

Spec(R[[y]])

30.3. SPECTRA FOR TWO-DIMENSIONAL MIXED POLYNOMIAL/POWER SERIES RINGS269

Here α is the cardinality of the set of maximal ideals of R (and also the cardi-nality of the set of maximal ideals of R[[y]] by Remarks

20.2.330.7.1 ); the boxed κi (one

for each maximal ideal of R) means that there are cardinality κi prime ideals in thatposition, where each κi is uncountable. If R satisfies certain cardinality conditionsdescribed in Remark 20.1.0.

20.3.330.14.3, for example, if R = k[x], for k a countable field,

then each κi equals the cardinality of R[[y]]. By Remark20.3.330.14.2 below, each κi is

at least γℵ0 , where γ is the cardinality of R/m and m is the maximal ideal of Rsuch that (m, y) is the maximal ideal of R[[y]] above the κi height-one primes inthe picture above.

20.3.3 Remark 30.14. Let ℵ0 denote the cardinality of the set of natural numbers.Suppose that T is a commutative ring of cardinality δ, that m is a maximal idealof T and that γ is the cardinality of T/m. Then

(1) The cardinality of T [[y]] is δℵ0 , because the elements of T [[y]] are in one-to-one correspondence with ℵ0-tuples having entries in T . If T is Noetherian, thenT [[y]] is Noetherian, and so every prime ideal of T [[y]] is finitely generated. Sincethe cardinality of the finite subsets of T [[y]] is δℵ0 , it follows that T [[y]] has at mostδℵ0 prime ideals.

(2) If T is Noetherian, there are at least γℵ0 distinct height-one prime ideals(other than (y)T [[y]] ) of T [[y]] contained in (m, y)T [[y]]. To see this, choose a setC = ci | i ∈ I of elements of T so that ci + m | i ∈ I gives the distinct cosetrepresentatives for T/m. Thus there are γ elements of C, and for ci, cj ∈ C withci 6= cj , we have ci − cj /∈m. Now also let a ∈m, a 6= 0. Consider the set

G = a+∑n∈N

dnyn | dn ∈ C ∀n ∈ N.

Each of the elements of G is in (m, y)T [[y]] \ yT [[y]] and hence is contained in aheight-one prime contained in (m, y)T [[y]] distinct from yT [[y]].

Moreover, |G| = |C|ℵ0 = γℵ0 . Let P be a height-one prime ideal of T [[y]]contained in (m, y)T [[y]] but such that y /∈ P . If two distinct elements of G, sayf = a +

∑n∈N dny

n and g = a +∑

n∈N enyn, with the dn, en ∈ C, are both in P ,

then so is their difference; that is

f − g =∑n∈N

dnyn −

∑n∈N

enyn =

∑n∈N

(dn − en)yn ∈ P.

Now let t be the smallest power of y so that dt 6= et. Then (f − g)/yt ∈ P , since Pis prime and y /∈ P , but the constant term, dt − et /∈m, which contradicts the factthat P ⊆ (m, y)T [[y]]. Thus there must be at least |C|ℵ0 = γℵ0 distinct height-oneprimes contained in (m, y)T [[y]].

(3′) A new claim for this paper:

Claim 30.15. If T is Noetherian, then γℵ0 = δℵ0 and

(3) Using (1) and (2), if T is Noetherian, then there are exactly γℵ0 = δℵ0 dis-tinct height-one prime ideals (other than yT [[y]] ) of T [[y]] contained in (m, y)T [[y]].

Proof. First suppose that γ is infinite. We note that, for each i with i ∈N, that the following sequence with inclusion for the first map and the naturalsurjection for the second map is exact:

0→mi−1/mi → R/mi →mi−2/mi−1 → 0.


Since all of the terms are finite-dimensional vector spaces over R/m it followsthat γ = |R/m| = |R/m2| = . . . . Now R embeds into

(3) Using (1) and (2), if T is Noetherian, and if γℵ0 = δℵ0 , then there areexactly γℵ0 = δℵ0 distinct height-one prime ideals (other than yT [[y]] ) of T [[y]]contained in (m, y)T [[y]]. This is the case, for example, if T is countable, sayT = k[x] where k is a countable field, for then |T | = ℵ0 and for every γ with1 < γ ≤ ℵ0, γℵ0 = ℵℵ0

0 = c, the cardinality of the real numbers.

20.3.4 Theorem 30.16. Suppose that R is a one-dimensional Noetherian domain withcardinality δ := |R| = and that the cardinality of the set of maximal ideals of R is α(α can be finite). Let U = SpecR[[y]], where y is an indeterminate over R. Then(a) U as a partially ordered set (where the “≤” relation is “set containment”)satisfies the following axioms:

(1): |U | ≤ δℵ0 .(2): U has a unique minimal element, namely (0).(3): dim(U) = 2 and | height-two elements of U | = α.(4): There exists a unique special height-one element u ∈ U (namely u =

(y)) such that u is contained in every height-two element of U .(5): Every nonspecial height-one element of U is in exactly one height-two

element.(6): Every height-two element t ∈ U contains uncountably many height-

one elements that are contained only in t. (The number of height-oneelements contained only in t is at least γℵ0 , where γ is the cardinality ofthe residue field of the corresponding maximal ideal of R.) If t1, t2 ∈ Uare distinct height-two elements, then the special element from (4) is theunique height-one element less than both.

(7): There are no height-one maximal elements in U . Every maximal ele-ment has height two.

(b) If R is countable, or more generally if δ = |R| satisfies the condition of Remarks20.3.330.14.3, that is, γℵ0 = δℵ0 , for every γ that occurs as |R/m|, where m is a maximalideal of R, then U = SpecR[[y]] satisfies (1)-(7), with the stronger axioms (1′) and(6′):

(1′): |U | = δℵ0 . (For R countable, this is c, the cardinality of the realnumbers.)

(6′): Every height-two element t ∈ U contains δℵ0 (uncountably many)height-one elements that are contained only in t.

(c) With the additional hypotheses of (b), U is characterized as a partially orderedset by the axioms given in (a) and (b). Every partially ordered set satisfying theaxioms (1)-(7) in (a) and (b) is order-isomorphic to every other such partiallyordered set.

Proof. In part(a), item (1) is from Remark20.3.330.14.1. Item (2) and the first

part of (3) are clear. The second part of (3) follows immediately from Remark20.2.330.7.1.

For items (4) and (5), suppose that P is a height-one prime of R[[y]]. If P =yR[[y]], then P is contained in each maximal ideal of R[[y]] by Remark

20.2.330.7.1, and

30.4. HIGHER DIMENSIONAL MIXED POWER SERIES/POLYNOMIAL RINGS 271

so yR[[y]] is the special element. If y /∈ P , then, by Corollary20.2.530.9, P is contained

in a unique maximal ideal of R[[y]].For item (6) and items (1′) and (6′) of part (b) use Remarks

20.3.330.14.2 and

20.3.330.14.3.

For item (c), all partially ordered sets satisfying the axioms of Theorem20.3.130.12

are order-isomorphic, and the partially ordered set U of the present theorem satisfiesthe same axioms as in Theorem

20.3.130.12 except axiom (7) that involves height-one

maximals. Since U has no height-one maximals, an order-isomorphism betweentwo partially ordered sets as in item (c) can be deduced by adding on height-onemaximals and then deleting them.

Corollary 30.17. In the terminology of Sequences20.1.0.1ppssec30.1.0.1 and

20.1.0.2ppssec30.1.0.2 of

the introduction, we have SpecC ∼= SpecDn∼= SpecE, but SpecB SpecC.

Proof. The rings C,Dn, and E are all formal power series rings in one variableover a one-dimensional Noetherian domain R, where R is either k[x] or k[x, 1/x].Thus the domain R satisfies the hypotheses of Theorem

20.3.430.16 with the cardinality

conditions of parts (b) and (c). If k is finite, then |R| = |k[x]| = ℵ0 and α, thenumber of maximal ideals of R, is also ℵ0; in this case |R/m| = γ is finite for eachmaximal ideal m of R and δ = |R| = γ · ℵ0 = α, and so γℵ0 = δℵ0 . On the otherhand, if k is infinite, then |k| = |k[x]| = |R| = α, and |k| = |R/m| = γ is the samefor every maximal ideal m of R. Hence also in this case δ = |R| = γ · ℵ0 = α, andso γℵ0 = δℵ0 .

Also the number of maximal ideals is the same for C,Dn, and E, because in eachcase, it is the same as the number of maximal ideals of R which is |k[x]| = |k| · ℵ0.

Thus in the picture of R[[y]] shown above, for R[[y]] = C,Dn or E, the κi are allequal to |k|ℵ0 and α = |k| · ℵ0, and so the spectra are isomorphic. The spectrum ofB is not isomorphic to that of C, however, because B contains height-one maximalideals, such as that generated by 1 + xy, whereas C has no height-one maximalideals.

20.3.5 Remark 30.18. As mentioned at the beginning of this section, it is shown inrW1[99] and

rW2[100] that Spec Q[x, y] 6∼= SpecF [x, y] ∼= Spec Z[y], where F is a field

contained in the algebraic closure of a finite field. Corollary3.6?? shows that the

spectra of power series extensions in y behave differently in that Spec Z[[y]] ∼=Spec Q[x] [[y]] ∼= SpecF [x] [[y]].

20.3.6 Corollary 30.19. If Z is the ring of integers, Q is the rational numbers, F isa field contained in the algebraic closure of a finite field, and R is the real numbers,then

Spec Z[[y]] ∼= Spec Q[x] [[y]] ∼= SpecF [x] [[y]] 6∼= Spec R[x] [[y]].

Proof. The rings Z,Q[x] and F [x] are all countable with countably infinitelymany maximal ideals. Thus if R = Z,Q[x] or F [x], then R satisfies the hypothesesof Theorem

20.3.430.16 with the cardinality conditions of parts (b) and (c). On the

other hand, R[x] has uncountably many maximal ideals; thus R[x] [[y]] also hasuncountably many maximal ideals.

30.4. Higher dimensional mixed power series/polynomial rings20.4

In analogy to Sequence (20.1.0.1ppssec30.1.0.1), we display several embeddings involving

three variables.


20.4.020.4.0 (ppssec30.4.0.1)

k[x, y, z]α→ k[[z]] [x, y]

β→ k[x] [[z]] [y]

γ→ k[x, y] [[z]]

δ→ k[x] [[y, z]],

k[[z]] [x, y]ε→ k[[y, z]] [x]

ζ→ k[x] [[y, z]]

η→ k[[x, y, z]],

where k is a field and x, y and z are indeterminates over k.

20.4.1 Remark 30.20. (1): By Proposition20.2.630.10.2 every nonzero prime ideal

of C = k[x] [[y]] has nonzero intersection with B = k[[y]] [x]. In three ormore variables, however, the analogous statements fail. We show belowthat the maps α, β, γ, δ, ε, ζ, η in Sequence

20.4.0ppssec30.4.0.1 fail to be TGF. Thus,

by Proposition20.2.230.6.2, no proper inclusion in (20.4.0) is TGF. The dimen-

sions of the generic fiber rings of the maps in the diagram are either oneor two.

(2): For those rings in (20.4.0) of form R = S[[z]] (ending in a power seriesvariable) where S is a ring, such as R = k[x, y][[z]], we have some informa-tion concerning the prime spectra. By Proposition

20.2.430.8 every height-two

prime ideal not containing z is contained in a unique maximal ideal. ByN2[72, Theorem 15.1] the maximal ideals of S[[z]] are of the form (m, z)S[[z]],where m is a maximal ideal of S, and thus the maximal ideals of S[[z]] arein one-to-one correspondence with the maximal ideals of S. As in section20.330.3, using Remarks

20.2.330.7, we see that maximal ideals of Spec k[[z]] [x, y]

can have height two or three, that (z) is contained in every height-threeprime ideal, and that every height-two prime ideal not containing (z) iscontained in a unique maximal ideal.

(3): It follows by arguments analogous to that in Proposition20.2.630.10.1, that

α, δ, ε are not TGF. For α, let σ(z) ∈ zk[[z]] be transcendental over k(z);then (x − σ)k[[z]] [x, y] ∩ k[x, y, z] = (0). For δ and ε: let σ(y) ∈ k[[y]] betranscendental over k(y); then (x − σ)k[x] [[z, y]] ∩ k[x] [[z]] [y] = (0), and(x− σ)k[[y, z]] [x] ∩ k[[z]] [x, y] = (0).

(4): By the main theorem of Chapterweiersec29, η is not TGF and the dimension

of the generic fiber ring of η is one.

In order to show in Proposition20.4.330.22 below that the map β is not TGF, we

first observe:

20.4.2 Proposition 30.21. The element σ =∑∞

n=1(xz)n! ∈ k[x] [[z]] is transcenden-

tal over k[[z]] [x].

Proof. Consider an expression

Z := a`σ` + a`−1σ

`−1 + · · ·+ a1σ + a0,

where the ai ∈ k[[z]] [x] and a` 6= 0. Let m be an integer greater than ` + 1 andgreater than degx ai for each i such that 0 ≤ i ≤ ` and ai 6= 0. Regard each aiσi asa power series in x with coefficients in k[[z]].

For each i with 0 ≤ i ≤ `, we have i(m!) < (m + 1)!. It follows that thecoefficient of xi(m!) in σi is nonzero, and the coefficient of xj in σi is zero for every jwith i(m!) < j < (m+1)!. Thus if ai 6= 0 and j = i(m!)+degx ai, then the coefficientof xj in aiσ

i is nonzero, while for j such that i(m!) + degx ai < j < (m + 1)!, the


coefficient of xj in aiσi is zero. By our choice of m, for each i such that 0 ≤ i < `and ai 6= 0, we have

(m+ 1)! > `(m!) + degx a` ≥ i(m!) +m! > i(m!) + degx ai.

Thus in Z, regarded as a power series in x with coefficients in k[[z]], the coefficientof xj is nonzero for j = `(m!) + degx a`. Therefore Z 6= 0. We conclude that σ istranscendental over k[[z]] [x].

20.4.3 Proposition 30.22. k[[z]] [x, y]β→ k[x] [[z]] [y] is not TGF.

Proof. Fix an element σ ∈ k[x] [[z]] that is transcendental over k[[z]] [x]. Wedefine π : k[x] [[z]] [y]→ k[x] [[z]] to be the identity map on k[x] [[z]] and π(y) = σz.Let q = kerπ. Then y − σz ∈ q. If h ∈ q ∩ (k[[z]] [x, y]), then

h =s∑j=0

t∑i=0

(∑`∈N

aij`z`)xiyj , for some s, t ∈ N and aij` ∈ k, and so

0 = π(h) =s∑j=0

t∑i=0

(∑`∈N

aij`z`)xi(σz)j =

s∑j=0

t∑i=0

(∑`∈N

aij`z`+j)xiσj .

Since σ is transcendental over k[[z]] [x], we have that x and σ are algebraicallyindependent over k((z)). Thus each of the aij` = 0. Therefore q∩(k[[z]][x, y]) = (0),and so the embedding β is not TGF.

20.4.4 Proposition 30.23. k[[y, z]] [x]ζ→ k[x] [[y, z]] and k[x] [[z]] [y]

γ→ k[x, y] [[z]]

are not TGF.

Proof. For ζ, let t = xy and let σ ∈ k[[t]] be algebraically independent overk(t). Define π : k[x] [[y, z]]→ k[x] [[y]] as follows. For

f :=∞∑`=0

∑m+n=`

fmn(x)ymzn ∈ k[x] [[y, z]],

where fmn(x) ∈ k[x], define

π(f) :=∞∑`=0

∑m+n=`

fmn(x)ym(σy)n ∈ k[x] [[y]].

In particular, π(z) = σy. Let p := kerπ. Then z − σy ∈ p, and so p 6= (0). Leth ∈ p ∩ k[[y, z]] [x]. We show h = 0. Now h is a polynomial with coefficients ink[[y, z]], and we define g ∈ k[[y, z]] [t], by, if ai(y, z) ∈ k[[y, z]] and

h :=r∑i=0

ai(y, z)xi, then set g := yrh =r∑i=0

(∞∑`=0

∑m+n=`

bimnymzn)ti.

The coefficients of g are in k[[y, z]], since yrxi = yr−iti. Thus

0 = π(g) =r∑i=0

(∞∑`=0

∑m+n=`

bimnym(σy)n)ti =

r∑i=0

(∞∑`=0

∑m+n=`

bimnσny`)ti

=∞∑`=0

(∑

m+n=`

(r∑i=0

bimnti)σn)y`.


Now t and y are analytically independent over k, and so the coefficient of each y`

(in k[[t]]) is 0; since σ and t are algebraically independent over k, the coefficient ofeach σn is 0. It follows that each bimn = 0, that g = 0 and hence that h = 0. Thusthe extension ζ is not TGF.

To see that γ is not TGF, we switch variables in the proof for ζ, so thatt = yz. Again choose σ ∈ k[[t]] to be algebraically independent over k(t). Defineψ : k[x, y] [[z]] → k[y] [[z]] by ψ(x) = σz and ψ is the identity on k[y] [[z]]. Thenψ can be extended to π : k[y] [[x, z]] → k[y] [[z]], which is similar to the π inthe proof above. As above, set p := kerπ; then p ∩ k[[x, z]] [y] = (0). Thusp ∩ k[x] [[z]] [y] = (0) and γ is not TGF.

20.4.5 Proposition 30.24. Let k be a field and let x and t be indeterminates over k.Then σ =

∑∞n=1 t

n! is algebraically independent over k[[x, xt]].

Proof. Let ` be a positive integer and consider an expression

γ := γ`σ` + · · ·+ γiσ

i + · · ·+ γ1σ, where γi :=∞∑j=0

fij(x)(xt)j ∈ k[[x, xt]],

that is, each fij(x) ∈ k[[x]] and 1 ≤ i ≤ `. Assume that γ` 6= 0. Let a` bethe smallest j such that f`j(x) 6= 0, and let m` be the order of fà`

(x), that is,fà`

(x) = xm`g`(x), where g`(0) 6= 0. Let n be a positive integer such that

n ≥ 2 + max`,m`, a`.Since ` < n, for each i with 1 ≤ i ≤ `, we have

eqeq (30.1) σi = σi1(t) + citi(n!) + t(n+1)!τi(t),

where ci is a nonzero element of k, σi1(t) is a polynomial in k[t] of degree at most(i− 1)n! + (n− 1)! and τi(t) ∈ k[[t]].

20.4.6 Claim 30.25. The coefficient of t`(n!)+a` in σ`γ` = σ`(∑∞

j=a`f`j(x)(xt)j) as a

power series in k[[x]] has order m` + a`, and hence, in particular, is nonzero.

Proof. By the choice of n, (n+ 1)! > `(n!) + a`. Hence by the expression forσ` given in Equation

eq30.1, we see that all of the terms in σ`γ` of the form bt`(n!)+a` ,

for some b ∈ k[[x]], appear in the product

(σ`1(t) + c`t`(n!))(

`(n!)+a`∑j=a`

f`j(x)(xt)j).

One of the terms of the form bt`(n!)+a` in this product is

c`t`(n!)fà`

(x)(xt)a` = (c`xm`+a`g`(x))t`(n!)+a` = (c`xm`+a`g`(0) + . . .)t`(n!)+a` .

Since c`g`(0) is a nonzero element of k, c`xm`+a`g`(x) ∈ k[[x]] has orderm`+a`. Theother terms in the product σ`γ` that have the form bt`(n!)+a` , for some b ∈ k[[x]],are in the product

(σ`1(t))(`(n!)+a`∑j=a`

f`j(x)(xt)j) =`(n!)+a`∑j=a`

f`j(x)(xt)jσ`1(t).

Since degt σ`1 ≤ (`− 1)n! + (n− 1)! and since, for each j with f`j(x) 6= 0, we havedegt f`j(x)(xt)

j = j, we see that each term in f`j(xt)jσ`1(t) has degree in t less


than or equal to j + (`− 1)n! + (n− 1)!. Thus each nonzero term in this productof the form bt`(n!)+a` has

j ≥ `(n!) + a` − (`− 1)(n!)− (n− 1)! = a` + (n− 1)!(n− 1) > m` + a`,

by choice of n. Moreover, for j such that f`j(x) 6= 0, the order in x of f`j(x)(xt)j

is bigger than or equal to j. This completes the proof of Claim20.4.630.25.

20.4.7 Claim 30.26. For i < `, the coefficient of t`(n!)+a` in σiγi as a power series ink[[x]] is either zero or has order greater than m` + a`.

Proof. As in the proof of Claim20.4.630.25, all of the terms in σiγi of the form

bt`(n!)+a` , for some b ∈ k[[x]], appear in the product

(σi1 + citi(n!))(

`(n!)+a`∑j=0

fij(x)(xt)j) =`(n!)+a`∑j=0

fij(x)(xt)j(σi1 + citi(n!)).

Since degt(σi1 + citi(n!)) = i(n!), each term in fij(x)(xt)j(σi1 + cit

i(n!)) has degreein t at most j + i(n!). Thus each term in this product of the form bt`(n!)+a` , forsome nonzero b ∈ k[[x]], has

j ≥ `(n!) + a` − i(n!) ≥ n! + a` > m` + a`.

Thus ordx b ≥ j > m` + a`. This completes the proof of Claim20.4.730.26. Hence

γ 6∈ k[[x, xz]] and so Proposition20.4.530.24 is proved.

20.4.8 Question and Remarks 30.27. (1): As we show in Proposition20.2.630.10,

the embeddings from Equation 1 involving two-dimensional mixed powerseries/polynomial rings over a field k with inverted elements are TGF. Inthe paper

ppsgilmer[49] we ask whether the same is true in the three-dimensional

case. For example, is θ below TGF?

k[x, y] [[z]]θ→ k[x, y, 1/x] [[z]]

The answer for this example is shown to be “no” inyasuda[102]. Dumitrescu

establishes the answer is “no” in more generality, cf. Theorem20.4.930.28 below.

(2): For the four dimensional case, as observed in the introduction, it fol-lows from

HR[32, Theorem 1.12, p. 364] that the extension k[x, y, u] [[z]] →

k[x, y, u, 1/x] [[z]] is not TGF. We provide in Proposition20.4.930.28 a direct

proof of this fact.

20.4.9 Theorem 30.28.D[21, Corollary 4] Let D be a integral domain and x, y, z in-

determinates over D. Then the extension B[[z]] → Bx[[z]] is not TGF for everysubring B of D[[x, y]] that contains D[x, y].

???alert we want to make what follows into an exercise with hints ?? Fork a field and x, y, u and z indeterminates over k, the extension k[x, y, u] [[z]] →k[x, y, u, 1/x] [[z]] is not TGF.

Proof. Let t = z/x and let σ ∈ k[[t]] be algebraically independent overk[[x, z]]. (By Proposition

20.4.530.24, we may take σ =

∑∞r=1 t

r! . )Consider

π : k[[x, y, u]] [1/x] [[z]]→ k[[x, u]] [1/x] [[z]]


defined by mapping∞∑i=0

ai(x, y, u, 1/x)zi 7→∞∑i=0

ai(x, σu, u, 1/x)zi,

where ai(x, y, u, 1/x) ∈ k[[x, y, u]][1/x]. Let p = kerπ. Then y − σu ∈ p. We showthat p ∩ k[[x, y, u, z]] = (0), and so also p ∩ k[x, y, u] [[z]] = (0). Let

f :=∞∑`=0

(∑i+j=`

dijuiyj) ∈ k[[x, y, u, z]],

where dij ∈ k[[x, z]]. If f ∈ p, then

0 = π(f) =∞∑`=0

(∑i+j=`

dijuiσjuj) =

∞∑`=0

(∑i+j=`

dijσj)u`.

This is a power series in u, and so, for each `,∑i+j=` dijσ

j = 0. Since σ isalgebraically independent over k[[x, z]], each dij = 0. Thus f = 0. This completesthe proof of Proposition

20.4.930.28.

CHAPTER 31

Extensions of local domains with trivial generic

fiber May 17 2009(tgfsec), 21

tgfsec31.1. Introduction

Our work in this section originates with the following question asked by MelvinHochster and Yongwei Yao.

Question 31.1.Hochster29.2 Let R be a complete local domain. Can one describe or

somehow classify the local maps of R to a complete local domain S such that U−1Sis a field, where U = R \ (0), i.e., such that the generic fiber of R → S is trivial?

In connection with QuestionHochster29.2, Hochster and Yao remark: if R is equal

characteristic zero, then every extension as in QuestionHochster29.2 may be obtained by

starting with

R0 := K[[x1, ..., xn]] → T := L[[x1, ..., xn, y1, .., ym]],

where K is a subfield of L and the xi, yj are formal indeterminates. Let P be aprime ideal of T maximal with respect to being disjoint from the image of R0 \0.Then the composite map R0 → T → T/P =: S is an extension of this type. To seein equicharacteristic zero that it is possible to reduce the original inclusion R → Sto R0 → T → T/P =: S, one can argue as follows: By Cohen’s structure theoremsCo[18],

N2[72, (31.6)], a complete local domain R is a finite integral extension of a

complete regular local domain R0. If R has the same characteristic as its residuefield, then R0 is a formal power series ring over a field. The generic fiber of R → Sis trivial if and only if the generic fiber of R0 → S is trivial.

21tgf Definition 31.2. If R → S is an injective map of integral domains, we saythat S is a trivial generic fiber extension, TGF extension, of R if each nonzeroideal of S has a nonzero intersection with R, or equivalently, if each nonzero elementof S has a nonzero multiple in R. Since ideals of S maximal with respect to notmeeting the multiplicative system of nonzero elements of R are prime ideals, S isa TGF extension of R if and only if P ∩R 6= (0) for each nonzero prime ideal P ofS. Another condition equivalent to S is a TGF extension of R is that U−1S is afield, where U = R \ (0).

Let (R,m) → (S,n) be an injective local homomorphism of complete localdomains, so that n∩R = m. We say that S is a TGF-complete extension of Rif S is a TGF extension of R.

We discuss several topics and questions related to QuestionHochster29.2.

Previous work on these questions appears in articles of Matsumura, Rotthaus,Abhyankar, Moh.

R3.5[84],

Abhy1[3]

AM[5]. In particular, if R and S contain a field and are both

complete regular local rings of dimension two, then the map from R to S factors asa composition of maps R := k[[x, xy]] → S := k[[x, y]], by

AM[5, Section 3].

277

278 31. TGF EXTENSIONS

31.2. General remarks about TGF extensions21.2

We record in Propositiontgftrans31.3 several basic facts about TGF extensions. We

omit the proofs since they are straightforward.

tgftrans Proposition 31.3. Let R → S and S → T be injective maps where R, S andT are integral domains.

(1) If R → S and S → T are TGF extensions, then so is the composite mapR → T . Equivalently if the composite map R → T is not TGF, then atleast one of the extensions R → S or S → T is not TGF.

(2) If R → T is TGF, then S → T is TGF.(3) If the map SpecT → SpecS is surjective and R → T is TGF, then R → S

is TGF.

We consider in Proposition21.3.1.131.4 the relatively easy case where the base ring

has dimension one.

21.3.1.1 Proposition 31.4. Let (R,m) be a complete one-dimensional local domain.Assume that (S,n) is a TGF-complete extension of R. Then

(1) dim(S) = 1 and mS is n-primary.(2) If [S/n : R/m] <∞, then S is a finite integral extension of R.

Thus, if R → S is a TGF-extension with finite residue extension and dimS ≥2, then dimR ≥ 2.

Proof. By Krull’s Principal Ideal TheoremM[68, Theorem 13.5], n is the union

of the height-one primes of S. If dimS > 1, then S has infinitely many height-oneprimes. Each nonzero element of n is contained in only finitely many of theseheight-one primes. If dimS > 1, then the intersection of the height-one primes ofS is zero. Since dimR = 1, every nonzero prime of S contains m. Thus dimS = 1and mS is n-primary. Moreover, if [S/n : R/m] < ∞, then S is finite over R byM[68, Theorem 8.4].

21.2.3 Remarks 31.5. (1) Notice that there exist TGF-complete extensions of R thathave an arbitrarily large extension of residue field. For example, if k is a subfieldof a field F and x is an indeterminate over F , then R := k[[x]] ⊆ S := F [[x]] is aTGF-complete extension.

(2) Let (R,m) → (T,q) be an injective local homomorphism of complete localdomains. For P ∈ SpecT , S := T/P is a TGF-complete extension of R if and onlyif P is an ideal of T maximal with respect to the property that P ∩R = (0).

21.2.4 Remarks 31.6. Let X = x1 . . . , xn, Y = y1 . . . , ym and Z = z1 . . . , zrbe algebraically independent finite sets of indeterminates over a field k, where n ≥ 2,m, r ≥ 1. Set R := k[[X ]] and let P be a prime ideal of k[[X,Y, Z]] that is maximalwith respect to P ∩R = (0). Then we have the inclusions:

R := k[[X ]]σ→ S := k[[X,Y ]]/(P ∩ k[[X,Y ]])

τ→ T := k[[X,Y, Z]]/P.

By Remark21.2.331.5.2, τ · σ is a TGF extension. By Proposition

tgftrans31.3.3, S → T is

TGF.(1) If the map SpecT → SpecS is surjective, then σ : R → S is TGF by

Propositiontgftrans31.3.2.

(2) If R → T is finite, then R → S is also finite, and so σ : R → S is TGF.

31.3. TGF-COMPLETE EXTENSIONS WITH FINITE RESIDUE FIELD EXTENSION 279

(3) If R → T is not finite, then dimT = 2 byweier[48, Theorem 7.2].

(4) If P ∩ k[[X,Y ]] = 0, then S = R[[Y ]] and R → S is not TGF. (We showin Example

rel2.431.18 that this can occur.)

21.2.4.5 Remarks and Question 31.7. (1) With notation as in Remarks21.2.431.6 and

with Y = y, a singleton set, it is always true that ht(P ∩ R[[y]]) ≤ n − 1. (SeeTheorem


weiersec29.) Moreover, if ht(P ∩R[[y]]) = n− 1, then R → S is

TGF. Thus if n = 2 and P ∩R[[y]] 6= 0, then R → S is TGF.(2) With notation as in (1) and n = 3, it can happen that P ∩ k[[X, y]] 6= (0)

and R → R[[y]]/(P ∩ R[[y]]) is not a TGF extension. To construct an example ofsuch a prime ideal P, we proceed as follows: Since dimk[[X,y]]=4, there exists aprime ideal Q of k[[X, y]] with htQ = 2 and Q ∩ k[[X ]] = (0), [11, Theorem 7.1].Let p ⊂ Q be a prime ideal with htp = 1. Since p ( Q and Q ∩ k[[X ]] = (0),the extension k[[X ]] → k[[X, y]]/p is not a TGF extension. In particular, it is notfinite. Let P ∈ Spec k[[X, y, Z]] be maximal with respect to P ∩ k[[X, y]] = p. ByCorollary

21.3.4431.12 below, dim k[[X, y, Z]]/P = 2. Hence P is maximal in the generic

fiber over k[[X ]].(3) If (R,m) → (S,n) is a TGF-complete extension with S/n finite algebraic

over R/m, can the transcendence degree of S over R be finite but nonzero?(4) If (R,m) → (S,n) is a TGF-complete extension as in (3) with R equichar-

acteristic and dimR ≥ 2, then by Corollarygendim2th31.11 below it follows that either S is

a finite integral extension of R or dimS = 2.

21.2.66 Proposition 31.8. Let A → B be a TGF extension, where B is a Noetherianintegral domain. For each Q ∈ SpecB, we have htQ ≤ ht(Q ∩ A). In particular,dimA ≥ dimB.

Proof. If htQ = 1, it is clear that htQ ≤ ht(Q ∩ A) since Q ∩ A 6= (0). LethtQ = n ≥ 2, and assume by induction that htQ′ ≤ ht(Q′∩A) for eachQ′ ∈ SpecBwith htQ′ ≤ n− 1. Since B is Noetherian,

(0) =⋂Q′ |Q′ ⊂ Q and htQ′ = n− 1.

Hence there exists Q′ ⊂ Q with htQ′ = n − 1 and Q′ ∩ A ( Q ∩ A. We haven− 1 ≤ ht(Q′ ∩A) < ht(Q ∩A), so ht(Q ∩A) ≥ n.

31.3. TGF-complete extensions with finite residue field extension

set3 Setting 31.9. Let n ≥ 2 be an integer, let X = x1, . . . , xn be a set ofindependent variables over the field k and let R = k[[X ]] be the formal power seriesring in n variables over the field k.

dim2th Theorem 31.10. Let R = k[[X ]] be as in Settingset331.9. Assume that R → S

is a TGF-complete extension, where (S,n) is a complete Noetherian local domainand S/n is finite algebraic over k. Then either dimS = n and S is a finite integralextension of R or dimS = 2.

Proof. It is clear that if S is a finite integral extension of R, then dimS = n.Assume S is not a finite integral extension of R. Let b1, . . . , bm ∈ n be such thatn = (b1, . . . , bm)S, and let Y = y1, . . . , ym be a set of independent variables over


R. Since S is complete the R-algebra homomorphism ϕ : T := R[[Y ]] → S suchthat ϕ(yi) = bi for each i with 1 ≤ i ≤ m is well defined. Let Q = kerϕ. We have

R → T/Q → S.

ByM[68, Theorem 8.4], S is a finite module over T/Q. Hence dimS = dim(T/Q)

and the map SpecS → Spec T/Q is surjective, so by Propositiontgftrans31.3(3) R → T/Q

is TGF. Byweier[48, Corollary 7.4], dim(T/Q) = 2, so dimS = 2.

gendim2th Corollary 31.11. Let (A,m) and (S,n) be complete equicharacteristic localdomains with dimA = n ≥ 2 and suppose that A → S is a local injective homo-morphism and that the residue field S/n is finite algebraic over the residue fieldA/m := k. If A → S is a TGF-complete extension, then either dimS = n and Sis a finite integral extension of A or dimS = 2.

Proof. ByM[68, Theorem 29.4(3)],A is a finite integral extension ofR = k[[X ]],

whereX is as in Settingset331.9. We haveR → A → S. By Proposition

tgftrans31.3(1), R → S

is TGF. By Theoremdim2th31.10, either dimS = n and S is a finite integral extension of

A or dimS = 2. For example, if R = k[[x1, . . . , x4]] and S = k[[y1, y2, y3]], then every k-algebra

embedding R → S fails to be TGF.

21.3.44 Corollary 31.12. Let R = k[[X ]] be as in Settingset331.9. Let Y = y1, . . . , ym

be a set of m independent variables over R and let S = R[[Y ]]. If P ∈ SpecR issuch that dimR/P ≥ 2 and Q ∈ SpecS is maximal with respect to Q∩R = P , theneither

(i) dimS/Q = 2, or(ii) R/P → S/Q is a finite integral extension (and so dimR/P = dimS/Q).

Proof. Let A := R/P → S/Q =: B, and apply Corollarygendim2th31.11.

gen General Example 31.13. It is known that, for each positive integer n, thepower series ring R = k[[x1, . . . , xn]] in n variables over a field k can be embeddedinto a power series ring in two variables over k. The construction is based on thefact that the power series ring k[[z]] in the single variable z contains an infiniteset of algebraically independent elements over k. Let fi∞i=1 ⊂ k[[z]] with f1 6= 0and fi∞i=2 algebraically independent over k(f1). Let (S := k[[z, w]],n := (z, w))be the formal power series ring in the two variables z, w. Fix a positive integer nand consider the subring Rn := k[[f1w, . . . , fnw]] of S with maximal ideal mn =(f1w, . . . , fnw). Let x1, . . . , xn be new indeterminates over k and define a k-algebrahomomorphism ϕ : k[[x1, . . . , xn]]→ Rn by setting ϕ(xi) = fiw for i = 1, . . . , n.

genclaim Claim 31.14. (cf.ZSII[106, pp. 219-220]) ϕ is an isomorphism.

Proof. Suppose g =∑∞

m=0 gm, where gm is a form of degreem in k[x1, . . . , xn].Then

ϕ(g) =∞∑m=0

ϕ(gm) and ϕ(gm) = gm(f1w, . . . , fnw) = wmgm(f1, . . . , fn),

where gm(f1, . . . , fn) ∈ k[[z]]. If ϕ(g) = 0, then gm(f1, . . . , fn) = 0 for each m.Thus

0 = gm(f1, . . . , fn) =∑

i1+···+in=m

ai1,...,infi11 · · · f inn ,

31.3. TGF-COMPLETE EXTENSIONS WITH FINITE RESIDUE FIELD EXTENSION 281

where the ai1,...,in ∈ k and the ij are nonnegative integers. Our hypothesis on thefj implies that each of the ai1,...,in = 0, and so gm = 0 for each m.

genex Proposition 31.15. With notation as in Examplegen31.13, for each integer n ≥

2, the extension (Rn,mn) → (S,n) is nonfinite TGF-complete with trivial residueextension. Moreover ht(P ∩Rn) ≥ n− 1, for each nonzero prime P ∈ SpecS.

Proof. We have k = Rn/mn = S/n, so the residue field of S is a trivialextension of that of Rn. Since mnS is not n-primary, S is not finite over Rn. IfP ∩Rn = mn, then ht(P ∩Rn) = n ≥ n−1. Since dimS = 2, if mn is not containedin P , then htP = 1, S/P is a one-dimensional local domain, and mn(S/P ) isprimary for the maximal ideal n/P of S/P . It follows that Rn/(P ∩ Rn) → S/Pis a finite integral extension

M[68, Theorem 8.4]. Therefore dimRn/(P ∩ Rn) = 1.

Since Rn is catenary and dimRn = n, ht(P ∩Rn) = n− 1. Corollary 31.16. Let X and R = k[[X ]] be as in Setting

set331.9. Then there

exists an infinite properly ascending chain of two-dimensional TGF-complete ex-tensions R =: S0 → S1 → S2 → · · · such that each Si has the same residue fieldas R and Si+1 is a nonfinite TGF-complete extension of Si for each i.

Proof. Examplegen31.13 and Proposition

genex31.15 imply that R can be identified

with a proper subring of the power series ring in two variables so that k[[y1, y2]] is aTGF-complete extension of R and the extension is not finite. Now Example

gen31.13

and Propositiongenex31.15 can be applied again, to k[[y1, y2]], and so on.

21.3.4 Example 31.17. A particular case of Examplegen31.13.

For R := k[[x, y]], the extension ring S := k[[x, y/x]] has infinite transcendenceover R (see

Shel[93]). The method used in

Shel[93] to prove that S has infinite transcen-

dence degree over R is by constructing power series in y/x with ‘special large gaps’.Since k[[x]] is contained in R, it follows that S is a TGF-complete extension of R.To show this, it suffices to show P∩R 6= (0) for each P ∈ SpecS with htP = 1. Thisis clear if x ∈ P , while if x 6∈ P , then k[[x]]∩P = (0), so k[[x]] → R/(P ∩R) → S/Pand S/P is finite over k[[x]]. Therefore dimR/(P ∩R) = 1, so P ∩R 6= (0).

Notice that the extension k[[x, y]] → k[[x, y/x]] is, up to isomorphism, the sameas the extension k[[x, xy]] → k[[x, y]].

In Examplerel2.431.18 we show the situation of Remark

21.2.431.6.4 does occur.

rel2.4 Example 31.18. Let k, X = x1, x2, Y = y, Z = z and R = k[[x1, x2]]be as in Remarks

21.2.431.6. Let f1, f2 ∈ k[[z]] be algebraically independent over k. Let

P denote the ideal of k[[x1, x2, y, z]] generated by (x1 − f1y, x2 − f2y), Then P isthe kernel of the k-algebra homomorphism θ : k[[x1, x2, y, z]] → k[[y, z]] obtainedby defining θ(x1) = f1y, θ(x2) = f2y, θ(y) = y and θ(z) = z. In the notation ofRemark

21.2.431.6,

T = k[[x1, x2, y, z]]/P ∼= k[[y, z]].Let ϕ := θ|R and τ := θ|R[[y]]. The proof of Claim

genclaim31.14 shows that ϕ and τ are

embeddings. Hence P ∩R[[y]] = (0). By Propositiongenex31.15, ϕ and τ are TGF. We

have

Rσ→ S =

R[[y]]P ∩R[[y]]

= R[[y]]τ→ R[[y, z]]

P∼= k[[y, z]],

where σ : R → S is the inclusion map. Also τ · σ = ϕ is TGF as in Remark21.2.431.6.4.

Since yS ∩R = (0), σ : R → S is not TGF.


21.3.6 Questions 31.19. (1) If ϕ : R → S is a TGF-complete nonfinite extensionwith finite residue field extension, is it always true that ϕ can be extended to aTGF-complete nonfinite extension R[[y]] → S?

(2) Suppose that R → S is a TGF-complete extension and y is an indetermi-nate over S. It is natural to ask: Does R[[y]] → S[[y]] have the TGF property?Computing with elements, one may ask if for s ∈ S \R, does y+ s have a multiplein R[[y]]? There is a t ∈ S with ts ∈ R, but is there a t′ ∈ S with both t′t andt′ts ∈ R?

(3) A related question is whether the given R → S is extendable to an injectivelocal homomorphism ϕ : R[[y]] → S. For example, with k a field, k[[x1]][y](x1,y) →k[y][[x1]](x1,y) is TGF. Can we extend to k[[x1]][y][[x2]](x1,x2,y) → k[y][[x1]](x1,y),say by x2 →

∑∞n=0(yx)

n, which is still local injective?

We show in Proposition21.3.1031.20 that the answer to (

21.3.631.19.2) is ‘no’ if the answer

to (21.3.631.19.3) is “yes”, that is, the given R → S is extendable to an injective local

homomorphism R[[y]] → S. In Example21.3.831.21 we present an example where this

occurs.

21.3.10 Proposition 31.20. Let R → S be a TGF-complete extension and let y be anindeterminate over S. If R → S is extendable to an injective local homomorphismϕ : R[[y]] → S, then R[[y]] → S[[y]] is not TGF.

Proof. Let a := ϕ(y) and consider the ideal Q = (y− a)S[[y]]. The canonicalmap S[[y]]→ S[[y]]/Q = S extends ϕ. Thus Q∩R[[y]] = (0) and R[[y]] → S[[y]] isnot TGF.

21.3.8 Example 31.21. Let R := Rn = k[[f1w, . . . , fnw]] → S := k[[z, w]] be asin Example

gen31.13 with n ≥ 2. Define the extension ϕ : R[[y]] → S by setting

ϕ(y) = fn+1w ∈ S. By Propositiongenex31.15, ϕ : R[[y]] → S is TGF-complete. Thus

by Proposition21.3.1031.20, R[[y]] → S[[y]] is not TGF.

tgfnf Remark and Questions 31.22. Let (R,m) → (S,n) be a TGF-complete ex-tension. Assume that [S/n : R/m] < ∞ and that S is not finite over R. By

M[68,

Theorem 8.4], mS is not n-primary. Thus dimS > ht(mS). Therefore dimS > 1,so by Proposition

21.3.1.131.4, dimR > 1.

(1) If (R,m) is equicharacteristic, then by Corollarygendim2th31.11, dimS = 2. Is it

true in general that dimS = 2?(2) Is it possible to have dimS − ht(mS) > 1?

Examples 31.23. (1) Let R := k[[x, xy, z]] → S := k[[x, y, z]]. We showthis is not a TGF extension. By (

21.3.431.17), ϕ : k[[x, xy]] → k[[x, y]] is TGF-

complete. By Proposition21.3.1031.20, it suffices to extend ϕ to an injective

local homomorphism of k[[x, xy, z]] to k[[x, y]]. Let f ∈ k[[x]] be suchthat x and f are algebraically independent over k, so (1, x, f) is not asolution to any nonzero homogeneous form over k. As in (

dim2th31.10) and

(gen31.13), the extension of ϕ obtained by mapping z → fy is an injective

local homomorphism.(2) The extension R = k[[x, xy, xz]] → S = k[[x, y, z]] is not a TGF-complete

extension, since R = k[[x, xy, xz]] → k[[x, xy, z]] → S = k[[x, y, z]] is acomposition of two extensions that are not TGF by part (1). Now applyProposition

tgftrans31.3.

31.4. THE CASE OF TRANSCENDENTAL RESIDUE EXTENSION 283

31.4. The case of transcendental residue extension21.4

In this section we address but do not fully resolve the following question.

21.2.5 Question 31.24. If (S,n) is a TGF-complete extension of (R,m) and if S/nis transcendental over R/m does it follow that dimS ≤ 1?

In Proposition21.2.631.25 we prove every complete local domain of positive dimen-

sion has a one-dimensional TGF-complete extension.

21.2.6 Proposition 31.25. Let (R,m) be a local domain of positive dimension.(1) There exists a one-dimensional complete local domain (S,n) that is a TGF

extension of R.(2) If R is complete, there exists a one-dimensional TGF-complete extension

of R.

Proof. It is well known that there exists a discrete rank-one valuation domain(S,n) that dominates R (see, for example,

C[17]). The n-adic completion S of S

is a one-dimensional local ring that dominates R and each minimal prime pi ofS intersects S in zero, so S/pi is a one-dimensional complete local domain thatdominates R. Moreover, if (S,n) is a one-dimensional local domain that dominatesa local domain (R,m) of positive dimension, then it is obvious that S is a TGFextension of R, so if R and S are also complete, then S is a TGF-complete extensionof R.

set4 Setting 31.26. Let n ≥ 2 be an integer, let X = x1, . . . , xn be a set ofindependent variables over the field k and let R = k[[X ]] be the formal power seriesring in n variables over the field k. Let z, w, t, v be independent variables over R.

dim1 Proposition 31.27. Let notation be as in Settingset431.26.

(1) There exists a TGF embedding θ : k[[z, w]]→ k(t)[[v]] defined by θ(z) = tvand θ(w) = v.

(2) Moreover, the composition ψ = θ ϕ of θ with ϕ : R → k[[z, w]] given inGeneral Example

gen31.13 is also TGF.

Proof. Suppose f ∈ ker θ. Write f =∑∞

n=0 fn(z, w), where fn is a homoge-neous form of degree n with coefficients in k. We have

0 = θ(f) =∞∑n=0

fn(tv, v) =∞∑n=0

vnfn(t, 1) .

This implies fn(t, 1) = 0 for each n. Since t is algebraically independent over k, wehave fn(z, w) = 0 for each n. Thus f = 0 and θ is an embedding. Since θ is a localhomomorphism and dim k(t)[[v]] = 1, it is clear that θ is TGF.

For the second part, we use Propositiontgftrans31.3 together with the observation in

General Examplegen31.13 that ϕ is a TGF embedding.

As a consequence of Propositiondim131.27, we prove:

genfib Corollary 31.28. Let R = k[[X ]] be as above and let A = k(t)[[X ]]. Thereexists a prime ideal P ∈ SpecA in the generic fiber over R with htP = n − 1. Inparticular, the inclusion map R = k[[X ]] → A = k(t)[[X ]] is not TGF.


Proof. Define ϕ : R→ k[[z, w]] := S, by

ϕ(x1) = z, ϕ(x2) = h2(w)z, . . . , ϕ(xn) = hn(w)z,

where h2(w), . . . , hn(w) ∈ k[[w]] are algebraically independent over k. Also defineθ : S → k(t)[[v]] := B by θ(z) = tv and θ(w) = v. Consider the following diagram

R = k[[X ]] ⊂−−−−→ A = k(t)[[X ]]

ϕ

y Ψ

yS = k[[z, w]] θ−−−−→ B = k(t)[[v]],

where Ψ : A→ B is the identity map on k(t) and is defined by

Ψ(x1) = tv, Ψ(x2) = h2(v)tv, . . . , Ψ(xn) = hn(v)tv.

Notice that Ψ|R = ψ = θϕ. Therefore the diagram is commutative. Let P = kerΨ.Since Ψ is surjective, htP = n − 1. Commutativity of the diagram implies thatP ∩R = (0).

Discussion 31.29. Let us describe generators for the prime ideal P = kerΨgiven in Corollary

genfib31.28. Under the map Ψ, x1 7→ tv, and so x1

t 7→ v. Since alsox2 7→ h2(v)tv, . . . , xn 7→ hn(v)tv, we see that

(x2 − h2(x1

t)x1, x3 − h3(

x1

t)x1, . . . , xn − hn(x1

t)x1)A ⊆ P

(that is, Ψ(x2 − h2(x1/t)x1) = h2(v)tv − h2(v)tv = 0 etc.) Since the ideal on theleft-hand-side is a prime ideal of height n − 1, the inclusion is an equality. Thuswe have generators for the prime ideal P = kerΨ resulting from the definitions ofϕ and θ given in the corollary.

On the other hand, in Corollarygenfib31.28 if we change the definition of θ and we

define θ′ : k[[z, w]]→ k(t)[[v]] by θ′(z) = v and θ′(w) = tv (but we keep ϕ as above),then ψ′ defined by ψ′|R = θ′ · ϕ maps x1 → v, x2 → h2(tv)v, . . . , xn → hn(tv)v.In this case

(x2 − h2(tx1)x1, x3 − h3(tx1)x1, . . . , xn − hn(tx1)x1)A ⊆ kerΨ′ = P ′.

Again the ideal on the left-hand-side is a prime ideal of height n − 1, so we haveequality. This yields a different prime ideal P ′.

In this case one can also see directly for

P ′ = (x2 − h2(tx1)x1, x3 − h3(tx1)x1, . . . , xn − hn(tx1)x1)A

that P ′ ∩ R = (0). We have Ψ : A → A/P ′ = k(t)[[v]]. Suppose f ∈ R ∩ P ′. Wewrite f =

∑∞`=0 f`(x1, . . . , xn), where f` ∈ k[x1, . . . , xn] is a homogeneous form of

degree `. We have

0 = Ψ′(f) =∞∑`=0

f`(v, h2(tv)v, . . . , hn(tv)v) =∞∑`=0

v`f`(1, h2(tv), . . . , hn(tv)).

This implies f`(1, h2(tv), . . . , hn(tv)) = 0 for each `. Since h2, . . . , hn are alge-braically independent over k, each of the homogeneous forms f`(x1, . . . , xn) = 0.Hence f = 0.

Question 31.30. With notation as in Corollarygenfib31.28, does every prime ideal

of the ring A maximal in the generic fiber over R have height n− 1?

31.4. THE CASE OF TRANSCENDENTAL RESIDUE EXTENSION 285

21.4.8 Theorem 31.31. Let (A,m) → (B,n) be an extension of two-dimensional reg-ular local domains. Assume that B dominates A and that B/n as a field extensionof A/m is not algebraic. Then A → B is not TGF.

Proof. Since dimA = dimB, the assumption that B/n is transcendental overA/m implies that B is not algebraic overA

M[68, Theorem 15.5]. If mB is n-primary,

then B is faithfully flat over AM[68, Theorem 23.1], and

HR[32, Theorem 1.12] implies

that A → B is not TGF in this case.If mB is principal, then mB = xB for some x ∈m since B is local. It follows

that m/x ⊂ B. Localizing A[m/x] at the prime ideal n ∩ A[m/x] gives a localquadratic transform (A1,m1) of A. If dimA1 = 1, then A1 → B is not TGFbecause only finitely many prime ideals of B can contract to the maximal idealof A1. Hence A → B is not TGF if dimA1 = 1. If dimA1 = 2, then (A1,m1)is a 2-dimensional regular local domain dominated by (B,n) and the field A1/m1

is finite algebraic over A/m, and so B/n is transcendental over A1/m1. Thus wecan repeat the above analysis: If m1B is n-primary, then as above A → B is notTGF. If m1B is principal, we obtain a local quadratic transform (A2,m2) of A1. Ifthis process does not end after finitely many steps, we have a union V = ∪∞n=1Anof an infinite sequence An of quadratic transforms of a 2-dimensional regular localdomains. Then V is a valuation domain of rank at most 2 contained in B, andso at most finitely many of the height-one primes of B have a nonzero intersectionwith V . Therefore V → B is not TGF and hence also A → B is not TGF.

Thus by possibly replacing A by an iterated local quadratic transform An ofA, we may assume that mB is neither n-primary nor principal. Let m = (x, y)A.There exist f, g, h ∈ B such that x = gf, y = hf and g, h is a regular sequence inB. Hence (g, h)B is n-primary. Let f = fe11 · · · fer

r , where f1B, . . . frB are distinctheight-one prime ideals and the ei are positive integers. Then f1B, . . . , frB areprecisely the height-one primes of B that contain m.

Let t ∈ B be such that the image to t in B/n is transcendental over A/m.Modifying t if necessary by an element of n we may assume that t is transcendentalover A. We have n ∩ A[t] = m[t]. Let A(t) = A[t]m[t]. Notice that A(t) is a2-dimensional regular local domain with maximal ideal mA(t) that is dominatedby (B,n). We have

A → A[t] → A(t) → B.

Let P = (xt−y)A(t). Then P ∩A = (0). We have PB = (gft−hf)B = f(gt−h)B.Also gt − h is a nonunit of B. Let Q be a minimal prime of (gt − h)B. ThenQ 6∈ f1B, . . . , frB. Hence mA(t) 6⊆ Q. Therefore Q ∩A(t) has height one. SinceP ⊆ (gt− h)B ⊆ Q, we have Q∩A(t) = P . Thus Q∩A = (0). This completes theproof.

We have the following immediate corollary to Theorem21.4.831.31.

21.4.9 Corollary 31.32. Let x, y, z, w, t be indeterminates over the field k and let

ϕ : R = k[[x, y]] → S := k(t)[[z, w]]

be an injective local k-algebra homomorphism. Then ϕ(R) → S is not TGF.

In relation to Question21.2.531.24, Example

tgfdim2rf31.33 is a TGF extension A → B that

is not complete for which the residue field of B is transcendental over that of Aand dimB = 2.


tgfdim2rf Example 31.33. Let A = k[x, y, z, w](x,y,z,w), where k is a field and xw =yz. Thus A is a 3-dimensional normal local domain with maximal ideal m :=(x, y, z, w)A and residue field A/m = k. Notice that C := A[y/x] = k[y/x =w/z, x, z] is a polynomial ring in 3 variables over k. Thus B := C(x,z) is a 2-dimensional regular local domain with maximal ideal n = (x, z)B. Notice that(B,n) birationally dominates (A,m). Hence (A,m) → (B,n) is a TGF extension.Also B = k(y/x)[x, z](x,z), so k(y/x) is a coefficient field for B. The image t of y/xin B/n is transcendental over k and B/n = k(t). The completion of A is the normallocal domain A = k[[x, y, z, w]], where xw = yz. By a form of Zariski’s subspacetheorem

Abhy1[3, (10.6)], A is dominated by B = k(t)[[x, z]]. Thus we have ϕ : A → B,

where ϕ(x) = x, ϕ(z) = z, ϕ(y/x) = t = ϕ(w/z) and so also ϕ(y) = tx, ϕ(w) = tz,ϕ(xw) = xtz = ϕ(yz).

In Exampletgfdim2rf31.33, A → B is not a TGF-complete extension. Equivalently, the

inclusion mapR := k[[x, z, tx, tz]] → k(t)[[x, z]] := S

is not a TGF-extension.

CHAPTER 32

Multi-ideal-adic Completions of Noetherian Rings

(multsec), 22 Jan. 18, 2009

multsecIn this chapter we consider a variation of the usual I-adic completion of a

Noetherian ring R. Instead of successive powers of a fixed ideal I, we use a multi-adic filtration formed from a more general descending sequence In∞n=0 of ideals.We develop the mechanics of a multi-adic completion R∗ of R. With additionalhypotheses on the ideals of the filtration, we show that R∗ is Noetherian. In thecase where R is local, we prove that R∗ is excellent, or Henselian or universallycatenary if R has the stated property.

32.1. Introductionmultintro

Let R be a commutative ring with identity. A filtration on R is a decreasingsequence In∞n=0 of ideals of R. Associated to a filtration there is a well-definedcompletion R∗ = lim←−

n

R/In, and a canonical homomorphism ψ : R → R∗ Nor[75,

Chapter 9]. If⋂∞n=0 In = (0), then ψ is injective and R may be regarded as a

subring of R∗ Nor[75, page 401]. In the terminology of Northcott, a filtration In∞n=0

is said to be multiplicative if I0 = R and InIm ⊆ In+m, for all m ≥ 0, n ≥ 0Nor[75,

page 408]. A well-known example of a multiplicative filtration on R is the I-adicfiltration In∞n=0, where I is a fixed ideal of R.

In this chapter we consider filtrations of ideals of R that are not multiplicative,and examine the completions associated to these filtrations. We assume the ringR is Noetherian. Instead of successive powers of a fixed ideal I, we use a filtrationformed from a more general descending sequence In∞n=0 of ideals. We require that,for each n > 0, the nth ideal In is contained in the nth power of the Jacobson radicalof R, and that Ink ⊆ Ikn for all k, n ≥ 0. We call the associated completion a multi-adic completion, and denote it by R∗. The basics of the multi-adic constructionand the relationship between this completion and certain ideal-adic completions areconsidered in Section

multmech32.2. In Section

multNoeth32.3, we prove, for In as described above,

that R∗ is Noetherian. Let (R,m) be a Noetherian local ring. If R is excellent,Henselian or universally catenary, we prove in Section

22.432.15 that R∗ has the same

property.The process of passing to completion gives an analytic flavor to algebra. Often

we view completions in terms of power series, or in terms of coherent sequences asin

AM[5, pages 103-104]. Sometimes results are established by demonstrating for each

n that they hold at the nth stage in the inverse limit.Multi-adic completions are interesting from another point of view. Many coun-

terexamples in commutative algebra can be considered as subrings of R∗/J , whereR∗ is a multi-adic completion of a localized polynomial ring R over a countable

287

28832. MULTI-IDEAL-ADIC COMPLETIONS OF NOETHERIAN RINGS (MULTSEC), 22 JAN. 18, 2009

ground field and J is an ideal of R∗. In particular, certain counterexamples ofBrodmann and Rotthaus, Heitmann, Nishimura, Ogoma, Rotthaus and Westoncan be interpreted in this way, cf.

BR1[11],

BR2[12],

H4[57],

Ni[73],

Ni2[74],

O1[76],

O2[77],

R2[82],

R3[83],

W[98]. For most of these examples, a particular enumeration, p1, p2, . . . , of count-ably many non-associate prime elements is chosen and the ideals In are definedto be In := (p1p2 . . . pn)n. The Noetherian property in these examples is a trivialconsequence of the fact that every ideal of R∗ that contains one of the ideals In, ora power of In, is extended from R. In general, an advantage of R∗ over the In-adiccompletion Rn is that an ideal of R∗ is more likely to be extended from R than isan ideal of Rn.

32.2. Basic mechanics for the multi-adic constructionmultmech

22.1.05 Setting 32.1. Let R be a Noetherian ring with Jacobson radical J , and let Ndenote the set of positive integers. For each n ∈ N, let Qn be an ideal of R. Assumethat the sequence Qn is descending, that is Qn+1 ⊆ Qn, and that Qn ⊆ J n, foreach n ∈ N. Also assume, for each pair of integers k, n ∈ N, that Qnk ⊆ Qkn.

Let F = Qkk≥0 be the filtration

R = Q0 ⊇ Q1 ⊇ · · · ⊇ Qk ⊇ Qk+1 ⊇ · · ·of R and define

(22.1.0532.1.1) R∗ := lim←−

k

R/Qk

to be the completion of R with respect to F .Let R := lim←−

k

R/J k denote the completion of R with respect to the powers of

the Jacobson radical J of R, and for each n ∈ N, let

(22.1.0532.1.2) Rn := lim←−

k

R/Qkn

denote the completion of R with respect to the powers of Qn.

22.1r Remark 32.2. Assume notation as in Setting22.1.0532.1. For each fixed n ∈ N, we

haveR∗ = lim←−

k

R/Qk = lim←−k

R/Qnk,

where k ∈ N varies. This holds because the limit of a subsequence is the same asthe limit of the original sequence.

We establish in Proposition22.1.232.3 canonical inclusion relations among R and the

completions defined in (22.1.0532.1.1) and (

22.1.0532.1.2).

22.1.2 Proposition 32.3. Let the notation be as in Setting22.1.0532.1. For each n ∈ N, we

have canonical inclusions

R ⊆ R∗ ⊆ Rn ⊆ Rn−1 ⊆ · · · ⊆ R1 ⊆ R.Proof. The inclusion R ⊆ R∗ is clear since the intersection of the ideals Qk is

zero. For the inclusion R∗ ⊆ Rn, by Remark22.132.13, R∗ = lim←−

k

R/Qnk. Notice that

Qnk ⊆ Qkn ⊆ Qkn−1 ⊆ · · · ⊆ J k.

32.2. BASIC MECHANICS FOR THE MULTI-ADIC CONSTRUCTION 289

To complete the proof of Proposition22.1.232.3, we state and prove a general result

about completions with respect to ideal filtrations (see alsoNor[75, Section 9.5]). We

define the respective completions using coherent sequences as inAM[5, pages 103-104].

22.1.5 Lemma 32.4. Let R be a Noetherian ring with Jacobson radical J and letHkk∈N, Ikk∈N and Lkk∈N be descending sequences of ideals of R such that,for each k ∈ N, we have inclusions

Lk ⊆ Ik ⊆ Hk ⊆ J k.We denote the families of natural surjections arising from these inclusions as:

δk : R/Lk → R/Ik, λk : R/Ik → R/Hk and θk : R/Hk → R/J k,and the completions with respect to these families as:

RL := lim←−k

R/Lk, RI := lim←−k

R/Ik RH := lim←−k

R/Hk and R := lim←−k

R/J k.

Then(1) These families of surjections induce canonical injective maps ∆, Λ and Θ

among the completions as shown in the diagram below.(2) For each positive integer k we have a commutative diagram as displayed

below, where the vertical maps are the natural surjections.

R/Lkδk−−−−→ R/Ik

λk−−−−→ R/Hkθk−−−−→ R/J kx x x x

RL∆−−−−→ RI

Λ−−−−→ RHΘ−−−−→ R .

(3) The composition Λ · ∆ is the canonical map induced by the natural sur-jections λk · δk : R/Lk → R/Hk. Similarly, the other compositions inthe bottom row are the canonical maps induced by the appropriate naturalsurjections.

Proof. In each case there is a unique homomorphism of the completions. Forexample, the family of homomorphisms δkk∈N induces a unique homomorphism

(32.1) RL∆−−−−→ RI .

To define ∆, let x = (xk)k∈N ∈ RL, where each xk ∈ R/Lk. Then δk(xk) ∈ R/Ikand we define ∆(x) := (δk(xk))k∈N ∈ RI .

To show the maps on the completions are injective, consider for example themap ∆. Suppose x = (xk)k∈N ∈ lim←−

k

R/Lk with ∆(x) = 0. Then δk(xk) = 0 in

R/Ik, that is, xk ∈ IkR/Lk, for every k ∈ N. For v ∈ N, consider the followingcommutative diagram:

(32.2)

R/Lkδk−−−−→ R/Ik

βk,kv

x αk,kv

xR/Lkv

δkv−−−−→ R/Ikv

where βk,kv and αk,kv are the canonical surjections associated with the inverselimits. We have xkv ∈ IkvR/Lkv. Therefore

xk = βk,kv(xkv) ∈ Ikv(R/Lk) ⊆ J kv(R/Lk),


for every v ∈ N. Since J (R/Lk) is contained in the Jacobson radical of R/Lk andR/Lk is Noetherian, we have ⋂

v∈N

J kv(R/Lk) = (0).

Therefore xk = 0 for each k ∈ N, so ∆ is injective. The remaining assertions areclear.

22.3l Lemma 32.5. With R∗ and Rn as in Setting22.1.0532.1, we have

R∗ =⋂n∈N

Rn.

Proof. The inclusion “⊆” is shown in Proposition22.1.232.3. For the reverse in-

clusion, fix positive integers n and k, and let L` = Qnk`, I` = Q`nk and H` = Q`nfor each ` ∈ N. Then L` ⊆ I` ⊆ H` ⊆ J `, as in Lemma

22.1.532.4 and

RL := lim←− R/Qnk` = R∗, RI := lim←− R/Q`nk = Rnk, RH := lim←− R/Q`n = Rn.

(Also, as before, R := lim←− R/J `.) We define ϕn, ϕnk, ϕnk,n, θ and ϕ to be the

canonical injective homomorphisms given by Lemma22.1.532.4 among the rings displayed

in the following diagram.

R Rnθ

R∗ Rnk

(22.3l32.5.1)

ϕ ϕnk,nϕn

ϕnk

By Lemma22.1.532.4, Diagram

22.3l32.5.1 is commutative.

Let y ∈ ⋂n∈N Rn. We show that there is an element ξ ∈ R∗ such that ϕ(ξ) = y.This is sufficient to ensure that y ∈ R∗, since the maps θt are injective and Diagram22.3l32.5.1 is commutative.

First, we define ξ: For each t ∈ N, we have

y = (y1,t, y2,t, . . . , ) ∈ lim←−R/Q`t = Rt,

where y1,t ∈ R/Qt, y2,t ∈ R/Q2t and y2,t + Qt/Q

2t = y1,t in R/Qt, · · · and so

forth, is a coherent sequence as inAM[5, pp. 103-104]. Now take zt ∈ R so that

zt + Qt = y1,t. Thus y − zt ∈ QtRt. For positive integers s and t with s ≥ t, wehave Qs ⊆ Qt. Therefore zt − zs ∈ QtRt ∩R = QtR. Thus ξ := (zt)t∈N ∈ R∗. Wehave y−zt ∈ QtRt ⊆ J tR, for all t ∈ N. Hence ϕ(ξ) = y. This completes the proofof Lemma

22.3l32.5.

The following special case of Setting22.1.0532.1 is used by Brodmann, Rotthaus,

Ogoma, Heitmann, Weston and Nishimura for the construction of numerous exam-ples.

22.1.0 Setting 32.6. Let R be a Noetherian ring with Jacobson radical J . For eachi ∈ N, let pi ∈ J be a non-zero-divisor (that is a regular element) on R.

For each n ∈ N, let qn = (p1 · · · pn)n. Let F0 = (qk)k≥0 be the filtration

R ⊇ (q1) ⊇ · · · ⊇ (qk) ⊇ (qk+1) ⊇ · · ·

32.3. PRESERVING NOETHERIAN UNDER MULTI-ADIC COMPLETION 291

of R and define R∗ := lim←−k

R/(qk) to be the completion of R with respect to F0.

22.2.7 Remark 32.7. In Setting22.1.032.6, assume further thatR = K[x1, . . . , xn](x1,...,xn),

the localized polynomial ring over a countable field K, and that p1, p2, . . . is anenumeration of all the prime elements (up to associates) in R. As in

22.1.032.6, let

R∗ := lim←−n

R/(qn), where each qn = (p1 · · · pn)n. The ring R∗ is often useful for the

construction of Noetherian local rings with a bad locus (regular, CM, normal). Inparticular, the authors listed in the paragraph preceding Setting

22.1.032.6 make use of

special subrings of this multi-adic completion R∗ for their counterexamples. Thefirst such example was constructed by Rotthaus in

R2[82]. In this paper, a regular

local Nagata ring A is obtained that contains a prime element ω so that the singularlocus of the quotient ring A/(ω) is not closed. This ring A is situated between thelocalized polynomial ring R and its ∗-completion R∗; thus, in general R∗ is biggerthan R. In the Rotthaus example, the singular locus of (A/(ω))∗ is defined by aheight one prime ideal Q that intersects A/(ω) in (0). Since all ideals Q+ (pn) areextended from A/(ω), the singular locus of A/(ω) is not closed.

32.3. Preserving Noetherian under multi-adic completionmultNoethr*noeth Theorem 32.8. Let the notation be as in Setting

22.1.0532.1. Then the ring R∗ de-

fined in (22.1.0532.1.1) is Noetherian.

Proof. It suffices to show each ideal I of R∗ is finitely generated. Since R isNoetherian, there exist f1, . . . , fs ∈ I such that IR = (f1, . . . , fs)R. Since Rn → R

is faithfully flat, IRn = IR ∩ Rn = (f1, . . . , fs)Rn, for each n ∈ N.Let f ∈ I ⊆ R∗. Then f ∈ IR1, and so

f =s∑i=1

bi0fi,

where bi0 ∈ R1. Consider R as “Q0”, and so bi0 ∈ Q0R1. Since R1/Q1R1∼= R/Q1,

for all i with 1 ≤ i ≤ s, we have bi0 = ai0 + ci1, where ai0 ∈ R = Q0R andci1 ∈ Q1R1. Then

f =s∑i=1

ai0fi +s∑i=1

ci1fi.

Notice that

d1 :=s∑i=1

ci1fi ∈ (Q1I)R1 ∩R∗ ⊆ R2.

By the faithful flatness of the extension R2 → R1, we see d1 ∈ (Q1I)R2, andtherefore there exist bi1 ∈ Q1R2 with

d1 =s∑i=1

bi1fi.

As before, using that R2/Q2R2∼= R/Q2, we can write bi1 = ai1+ ci2, where ai1 ∈ R

and ci2 ∈ Q2R2. This implies that ai1 ∈ Q1R2 ∩R = Q1. We have:

f =s∑i=1

(ai0 + ai1)fi +s∑i=1

ci2fi.


Now set

d2 :=s∑i=1

ci2fi.

Then d2 ∈ (Q2I)R2 ∩R∗ ⊆ R3 and, since the extension R3 → R2 is faithfully flat,we have d2 ∈ (Q2I)R3. We repeat the process. By a simple induction argument,

f =s∑i=1

(ai0 + ai1 + ai2 + . . .)fi,

where aij ∈ Qj and ai0 + ai1 + ai2 + . . . ∈ R∗. Thus f ∈ (f1, . . . , fs)R∗. Hence I isfinitely generated and R∗ is Noetherian.

ff Corollary 32.9. With notation as in Setting22.1.0532.1, the maps R → R∗, R∗ →

Rn and R∗ → R are faithfully flat.

We use Proposition22.232.10 in the next section on preserving excellence.

22.2 Proposition 32.10. Assume notation as in Setting22.1.0532.1, and let the ring R∗

be defined as in (22.1.0532.1.1). If M is a finitely generated R∗-module, then

M ∼= lim←−k

(M/QkM),

that is, M is ∗-complete.

Proof. If F = (R∗)n is a finitely generated free R∗-module, then one can seedirectly that

F ∼= lim←−k

F/QkF,

and so F is ∗-complete.Let M be a finitely generated R∗-module. Consider an exact sequence:

0 −→ N −→ F −→M −→ 0,

where F is a finitely generated free R∗-module. This induces an exact sequence:

0 −→ N −→ F ∗ −→M∗ −→ 0,

where N is the completion ofN with respect to the induced filtration QkF∩Nk≥0,cf.

AM[5, (10.3)].This gives a commutative diagram:

0 −−−−→ N −−−−→ F −−−−→ M −−−−→ 0y ∼=y γ

y0 −−−−→ N −−−−→ F ∗ −−−−→ M∗ −−−−→ 0

where γ is the canonical map γ : M −→ M∗. The diagram shows that γ issurjective. We have

∞⋂k=1

(QkM) ⊆∞⋂k=1

JkM = (0),

where the last equality is byAM[5, (10.19)]. Therefore γ is also injective.

32.3. PRESERVING NOETHERIAN UNDER MULTI-ADIC COMPLETION 293

22.22 Remark 32.11. Let the notation be as in Setting22.1.0532.1, and let B be a finite

R∗-algebra. Let Bn ∼= B ⊗R∗ Rn denote the Qn-adic completion of B. By Propo-sition

22.1.232.3, and Corollary

ff32.9, we have a sequence of inclusions:

B → · · · → Bn+1 → Bn → . . . → B1 → B,

where B denotes the completion of B with respect to JB. Let J0 denote theJacobson radical of B. Since every maximal ideal of B lies over a maximal ideal ofR∗, we have JB ⊆ J0.

22.3 Theorem 32.12. With the notation of Setting22.1.0532.1, let B be a finite R∗-algebra

and let Bn ∼= B ⊗R∗ Rn denote the Qn-adic completion of B. Let I be an ideal ofB, let I := I ∩B, and let In := I ∩ Bn, for each n ∈ N. If I = InB, for all n, thenI = IB.

Proof. By replacing B by B/I, we may assume that (0) = I = I ∩ B. Toprove the theorem, it suffices to show that I = 0.

For each n ∈ N, we define ideals cn of Bn and an of B:

cn := In +QnBn, an := cn ∩B.Since B/QnB = Bn/QnBn, the ideals of B containing Qn are in one-to-oneinclusion-preserving correspondence with the ideals of Bn containing QnBn, and so

(22.332.12.1) anBn = cn, an+1Bn = an+1Bn+1Bn = cn+1Bn.

Since B is faithfully flat over Bn and I is extended,

(22.332.12.2) In+1Bn = (In+1B) ∩ Bn = I ∩ Bn = In.

Thus, for all n ∈ N, we have, using (22.332.12.1), (

22.332.12.2) and Qn+1Bn ⊆ QnBn:

anBn = cn = In +QnBn = In+1Bn+QnBn = cn+1Bn+QnBn = an+1Bn+QnBn.

Since Bn is faithfully flat over B, the equation above implies that

(22.332.12.3) an+1 +QnB = (an+1Bn +QnBn) ∩B = anBn ∩B = an.

Thus also

(22.332.12.4) anB ⊆ an+1B +QnB ⊆ In+1B +QnB = I +QnB.

Now Qn ⊆ J nB and J ⊆ J0, and so using (22.332.12.4)⋂

n∈N

(anB) ⊆⋂n∈N

(I +QnB) ⊆⋂n∈N

(I + J nB) = I .

Since I ∩B = (0), we have

0 = I ∩B ⊇ (⋂nnN

(anB)) ∩B ⊇⋂n∈N

((anB) ∩B) =⋂n∈N

an,

where the last equality is because B is faithfully flat over B. Thus⋂n∈N an = (0).

Claim. I = (0).


Proof of Claim. Suppose I 6= 0. Then there exists d ∈ N so that I * J d0 B. Byhypothesis, I = IdB, and so IdB * J d0 B. Since B is faithfully flat over Bd, wehave Id * J d0 Bd. By (

22.332.12.1),

adBd = cd = Id +QdBd * J d0 Bd,and so there exists an element yd ∈ ad with yd /∈ J d0 .

By (22.332.12.3), ad+1 +QdB = ad. Hence there exists yd+1 ∈ ad+1 and qd ∈ QdB

so that yd+1 + qd = yd. Recursively we construct sequences of elements yn ∈ an andqn ∈ QnB, for each n ≥ d, so that yn+1 + qn = yn.

The sequence ξ = (yn + QnB) ∈ lim←−n

B/QnB = B corresponds to a nonzero

element y ∈ B so that, for every n ≥ d, we have y = yn + gn, for some elementgn ∈ QnB. This shows that y ∈ an, for all n ≥ d, and therefore

⋂n∈N an 6= 0, a

contradiction. Thus I = (0).

32.4. Preserving excellence or Henselian under multi-adic completionmulthens

The first four results of this section concern preservation of excellence.

22.1 Theorem 32.13. Assume notation as in Setting22.1.0532.1, and let the ring R∗ be

defined as in (22.1.0532.1.1). If (R,m) is an excellent local ring, then R∗ is excellent.

The following result is critical to the proof of Theorem22.132.13.

matgr Lemma 32.14.M[68, Theorem 32.5, page 259] Let A be a semilocal Noetherian

ring. Assume that (B)Q is a regular local ring, for every local domain (B,n) thatis a localization of a finite A-algebra and for every prime ideal Q of the n-adiccompletion B such that Q ∩ B = (0). Then A is a G-ring, that is, A → Ap isregular for every prime ideal p of A; thus all of the formal fibers of all the localrings of A are geometrically regular.

We use Proposition22.432.15 in the proof of Theorem

22.132.13.

22.4 Proposition 32.15. Let (R,m) be a Noetherian local ring with geometricallyregular formal fibers. Then R∗ has geometrically regular formal fibers.

Proof. Let B be a domain that is a finite R∗-algebra and let P ∈ Sing(B),that is, BP is not a regular local ring. To prove that R∗ has geometrically regularformal fibers, by Lemma

matgr32.14, it suffices to prove that P ∩B 6= (0).

The Noetherian complete local ring R has the property J-2 in the sense ofMatsumura, that is, for every finite R-algebra, such as B, the subset Reg(Spec(B)),of primes where the localization of B is regular is an open subset, c.f.

M1[66, pp. 246–

249]. Thus there is a reduced ideal I in B so that

Sing(B) = V(I).

If I = (0), then B is a reduced ring and, for all minimal primes Q of B, thelocalization BQ is a field, contradicting Q ∈ Sing(B). Thus I 6= (0). For all n ∈ N:

Bn ∼= Rn ⊗R∗ B

is a finite Rn-algebra. Since byR1[81] Rn has geometrically regular formal fibers so

has Bn. This implies that I is extended from Bn for all n ∈ N. By Theorem22.332.12,

32.4. PRESERVING EXCELLENCE OR HENSELIAN UNDER MULTI-ADIC COMPLETION295

I is extended from B, so I = IB, where 0 6= I := I ∩ B. Since I ⊆ P , we have(0) 6= I ⊆ P ∩B.

Proof of Theorem22.132.13 It remains to show that R∗ is universally catenary. We

have injective local homomorphisms R → R∗ → R, and R∗ is Noetherian withR∗ = R. Proposition

ucint32.16 below implies that R∗ is universally catenary.

ucint Proposition 32.16. Let (A,m) be a Noetherian local universally catenary ringand let (B,n) be a Noetherian local subring of the m-adic completion A of A withA ⊆ B ⊆ A and B = A, where B is the n-adic completion of B. Then B isuniversally catenary.

Proof. ByM[68, Theorem 31.7], it suffices to show for P ∈ Spec(B) that A/P A

is equidimensional. We may assume that P ∩ A = (0), and hence we may assumethat A is a domain.

Let Q and W in Spec(A) be minimal primes over PA.

Claim: dim(A/Q) = dim(A/W ).

Proof of Claim: Since B is Noetherian, the canonical morphisms BP −→ AQ andBP −→ AW are flat. By

M[68, Theorem 15.1],

dim(AQ) = dim(BP ) + dim(AQ/PAQ), dim(AW ) = dim(BP ) + dim(AW /PAW ).

Since Q and W are minimal over PA, it follows that:

dim(AQ) = dim(AW ) = dim(BP ).

Let q ⊆ Q and w ⊆W be minimal primes of A so that:

dim(AQ) = dim(AQ/qAQ) and dim(AW ) = dim(AW /wAW ).

Since we have reduced to the case where A is a universally catenary domain, itscompletion A is equidimensional and therefore:

dim(A/q) = dim(A/w).

Since a complete local ring is catenaryM[68, Theorem 29.4], we have:

dim(A/q) = dim(AQ/qAQ) + dim(A/Q),

dim(A/w) = dim(AW /wAW ) + dim(A/W ).

Since dim(A/q) = dim(A/w) and dim(AQ) = dim(AW ), it follows that

dim(A/Q) = dim(A/W ).

This completes the proof of Propositionucint32.16.

Remark 32.17. Let R be a universally catenary Noetherian local ring. Propo-sition

ucint32.16 implies that every Noetherian local subring B of R with R ⊆ B and

B = R is universally catenary. Hence, for each ideal I of R, the I-adic completionof R is universally catenary. Also R∗ as in Setting

22.1.0532.1 is universally catenary.

Propositionucint32.16 also implies that the Henselization of R is universally catenary.

Seydi shows that the I-adic completions of universally catenary rings are univer-sally catenary in

S[89]. Proposition

ucint32.16 establishes this result for a larger class of

rings.


Proposition 32.18. With notation as in Setting22.1.0532.1, let (R,m, k) be a Noe-

therian local ring. If R is Henselian, then R∗ is Henselian.

Proof. Assume that R is Henselian. It is well known that every ideal-adiccompletion of R is Henselian, cf.

R2[82, p.6]. Thus Rn is Henselian for all n ∈ N. Let

n denote the nilradical of R. Then n ∩R∗ is the nilradical of R∗, and to prove R∗

is Henselian, it suffices to prove that R′ := R∗/(n ∩R∗) is HenselianN2[72, (43.15)].

To prove R′ is Henselian, byR2[82, Prop. 3, page 76], it suffices to show:

If f ∈ R′[x] is a monic polynomial and its image f ∈ k[x] has a simple root,then f has a root in R′.

Let f ∈ R′[x] be a monic polynomial such that f ∈ k[x] has a simple root.Since Rn/(n ∩ Rn) is Henselian, for each n ∈ N, there exists αn ∈ Rn/(n ∩ Rn)with f(αn) = 0. Since f is monic and R/(n ∩ R) is reduced, f has only finitelymany roots in R/(n ∩ R). Thus there is an α so that α = αn, for infinitely manyn ∈ N. By Lemma

22.332.12, R∗ =

⋂n∈N Rn. Hence

R′ = R∗/(n ∩R∗) =⋂n∈N

Rn/(n ∩ Rn,

and so there exists α ∈ R′ such that f(α) = 0.

CHAPTER 33

Outline/Plan May 17 2009

(1) Introduction and History.(a) Goals(b) Discussion of ideal-adic completions as power series, or more pre-

cisely as homomorphic images of formal power series rings.(c) Mention of previous contributors to this area, and some history

and some indication of the construction.

???alertNeed to put in Akizuki here ???

(d) Discussion of universality.(e) Summary of what we do in this memoir, including questions we

partially answer and would like to know more about.

???alert Need to put this in. ???alert ???

(2) Basic tools(3) The construction. (include notation, pictures and some examples)

(0) Easy examples and Motivating examples (intsec21.1.1) - (

intsec21.1.10.) From

sectionintsec21: Motivating example; In one case dim a = 3 , the max ideal is

2-generated, A is a 2-dim rlr. an example in Section4.24.3 has 2-dim A rlr,

B not Noeth; see also7.6.1021.34. Section

13.425.4 has A = B, not Noetherian.

(i) The basic construction as an intersection A = R∗ ∩ F . (Chapterbuildsec18) ( Chapter

noeslocsec22 gives the construction for semilocal. Chapter 9? gives it

for any domain.)(ii) The homomorphic image form.(iii) The insider construction.

??? Other examples:Notes Chapter

noesloc?? section 4 (ref 8) (

8.4.323.3 uses

7.5.521.18;

8.3.422.16 uses

7.5.321.16

8.3.422.16 discusses Nagata and Rotthaus example in context. ??? alertWe

should put this in.???

(iv) Revisit Nagata’s example of a normal domain that is analyticallyreducible and Christel’s example of a regular local ring having geometri-cally reduced formal fibers that are not geometrically regular.

(4) Flatness(i) flatness of maps of polynomial rings.Notes The corrigendum

7.7.721.7 has flatness/

(ii) flatness in the construction.(iii) weakly flat and height-one preserving extensions.Notes (

8.2.422.4) has flatness of various extensions.

297

298 33. OUTLINE/PLAN MAY 17 2009

(iv) idealwise independence, residual algebraic indpendence, and pri-mary independence (relation to flatness).

Notes (8.2.722.7) has primary limit intersection for R/Q.

(v) the limit-intersecting and residual limit-intersecting properties.(i) Chapter

idwisec20 has height-one preserving, weakly flat, PDE, locally

flat of height k (LFk)(ii) Chapter

intsec21 has more on the items in (i) and also has residually

limit intersecting, properties in 7.5.5, 7.5.3, used in 8.4.3 example.(iii) Chapter

noeslocsec22 revisits limit intersecting, def of primarily limit in-

tersecting, 8.27 a result for primarily lim. for R/Q; 8.3.2: flat ⇐⇒ pr.lim.

(5) PROOFS, First results such as equivalent conditions for the intersectiondomain to be the associated directed union: Chapter

noeslocsec22 and other places.

(Chapternoeslocsec22 shows B Noeth ⇐⇒ B = A,A Noeth. ⇐⇒ primarily

limit-intersecting.)(6) Further examples.

(i) What properties extend to the completion? (Lisbon paper). Inte-gral closure

(ii) catenary, universally catenary(iii) geometrically normal formal fibers(iv) Non-Noetherian examples.

(7) Prime ideals and fibers.(i) prime ideal pictures.(ii) fibers.

(8) Applications.(i) Approximating a DVR with regular local rings.(ii) The Weierstraus construction paper.(iii) The Hochster-Yao question about trivial generic fiber extensions.(iv) When B excellent =⇒ C excellent for R ⊆ B ⊆ B from Chapter

noeslocsec22

(9) Miscellaneous(i) The idealwise construction in Chapter

idwisec20 is of a different nature

to the construction in Chapterbuildsec18. It shows the intersection construction

can sometimes be small or large Idealwise considerations : Some flatnessis discussed in this.

(ii)Inter considerations(iii) What domains lie between a Noetherian domain and its quotient

field? (e.g. from Chapternoehomsec8

CHAPTER 34

AMS Monograph Series Sample

This is an unnumbered first-level section head

This is an example of an unnumbered first-level heading.

This is a Special Section Head

This is an example of a special section head1

34.1. This is a numbered first-level section head

This is an example of a numbered first-level heading.

34.1.1. This is a numbered second-level section head. This is an exam-ple of a numbered second-level heading.

This is an unnumbered second-level section head. This is an exampleof an unnumbered second-level heading.

34.1.1.1. This is a numbered third-level section head. This is an example of anumbered third-level heading.

This is an unnumbered third-level section head. This is an example of an un-numbered third-level heading.

Lemma 34.1. Let f, g ∈ A(X) and let E, F be cozero sets in X.

(1) If f is E-regular and F ⊆ E, then f is F -regular.(2) If f is E-regular and F -regular, then f is E ∪ F -regular.(3) If f(x) ≥ c > 0 for all x ∈ E, then f is E-regular.

The following is an example of a proof.

Proof. Set j(ν) = max(I\a(ν))− 1. Then we have

∑i/∈a(ν)

ti ∼ tj(ν)+1 =j(ν)∏j=0

(tj+1/tj).

1Here is an example of a footnote. Notice that this footnote text is running on so that it canstand as an example of how a footnote with separate paragraphs should be written.

And here is the beginning of the second paragraph.

299

300 34. AMS MONOGRAPH SERIES SAMPLE

Hence we have∏ν

( ∑i/∈a(ν)

ti

)|a(ν−1)|−|a(ν)|∼∏ν

j(ν)∏j=0

(tj+1/tj)|a(ν−1)|−|a(ν)|

=∏j≥0

(tj+1/tj)P

j(ν)≥j(|a(ν−1)|−|a(ν)|).(34.1)

By definition, we have a(ν(j)) ⊃ c(j). Hence, |c(j)| = n − j implies (5.4). Ifc(j) /∈ a, a(ν(j))c(j) and hence we have (5.5).

This is an example of an ‘extract’. The magnetization M0 of theIsing model is related to the local state probability P (a) : M0 =P (1)− P (−1). The equivalences are shown in Table

eqtable1.

Table 1eqtable

−∞ +∞f+(x, k) e

√−1kx + s12(k)e−√−1kx s11(k)e

√−1kx

f−(x, k) s22(k)e−√−1kx e−

√−1kx + s21(k)e√−1kx

Definition 34.2. This is an example of a ‘definition’ element. For f ∈ A(X),we define

(34.2) Z(f) = E ∈ Z[X ] : f is Ec-regular.Remark 34.3. This is an example of a ‘remark’ element. For f ∈ A(X), we

define

(34.3) Z(f) = E ∈ Z[X ] : f is Ec-regular.Example 34.4. This is an example of an ‘example’ element. For f ∈ A(X),

we define

(34.4) Z(f) = E ∈ Z[X ] : f is Ec-regular.Exercise 34.5. This is an example of the xca environment. This environment

is used for exercises which occur within a section.

Some extra text before the xcb head. The xcb environment is used for exercisesthat occur at the end of a chapter. Here it contains an example of a numbered list.

Exercises(1) First item. In the case where in G there is a sequence of subgroups

G = G0, G1, G2, . . . , Gk = e

such that each is an invariant subgroup of Gi.(2) Second item. Its action on an arbitrary element X = λαXα has the form

eq:actioneq:action (34.5) [eαXα, X ] = eαλβ [XαXβ ] = eαcγαβλβXγ ,

(a) First subitem.−2ψ2(e) = cδαγc

γβδe

αeβ.

34.2. SOME MORE LIST TYPES 301

Figure 1. This is an example of a figure caption with text. firstfig

Figure 2 otherfig

(b) Second subitem.(i) First subsubitem. In the case where in G there is a sequence of

subgroups

G = G0, G1, G2, . . . , Gk = e

such that each subgroupGi+1 is an invariant subgroup ofGi and eachquotient group Gi+1/Gi is abelian, the group G is called solvable.

(ii) Second subsubitem.(c) Third subitem.

(3) Third item.

Theorem 34.6. This is an example of a theorem.

Theorem 34.7 (Marcus Theorem). This is an example of a theorem with aparenthetical note in the heading.

34.2. Some more list types

This is an example of a bulleted list.• Jg of dimension 3g − 3;• E2

g = Pryms of double covers of C = with normalization of C hyper-elliptic of genus g − 1 of dimension 2g;• E2

1,g−1 = Pryms of double covers of C = HP 1 with H hyperelliptic of

genus g − 2 of dimension 2g − 1;• P2

t,g−t for 2 ≤ t ≤ g/2 = Pryms of double covers of C = C′C′′ with

g(C′) = t− 1 and g(C′′) = g − t− 1 of dimension 3g − 4.This is an example of a ‘description’ list.

Zero case: ρ(Φ) = 0.Rational case: ρ(Φ) 6= 0 and ρ(Φ) is contained in a line through 0 with

rational slope.Irrational case: ρ(Φ) 6= 0 and ρ(Φ) is contained in a line through 0 with

irrational slope.

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Index

Absorbing barrier, 4Adjoint partial differential operator, 20A-harmonic function, 16, 182

A∗-harmonic function, 182

Boundary condition, 20, 22Dirichlet, 15

Neumann, 16Boundary value problem

the first, 16the second, 16the third, 16

Bounded set, 19

Diffusion

coefficient, 1equation, 3, 23

Dirichlet

boundary condition, 15boundary value problem, 16

Elliptic

boundary value problem, 14, 158partial differential equation, 14

partial differential operator, 19

Fick’s law, 1Flux, 1

Formally adjoint partial differentialoperator, 20

Fundamental solutionconceptional explanation, 12

general definition, 23temporally homogeneous case, 64, 112

Genuine solution, 196Green function, 156Green’s formula, 21

Harnack theoremsfirst theorem, 185inequality, 186

lemma, 186second theorem, 187

third theorem, 187

Helmhotz decomposition, 214Hilbert-Schmidt expansion theorem, 120

Initial-boundary value problem, 22Initial condition, 22Invariant measure (for the fundamental

solution), 167

Maximum principlefor A-harmonic functions, 183for parabolic differential equations, 65strong, 83

Neumannboundary condition, 16boundary value problem, 16function, 179

One-parameter semigroup, 113

Parabolic initial-boundary value problem,22

Partial differential equationof elliptic type, 14of parabolic type, 22

Positive definite kernel, 121

Reflecting barrier, 4Regular (set), 19Removable isolated singularity, 191Robin problem, 16

Semigroup property (of fundamentalsolution), 64, 113

Separation of variables, 131Solenoidal (vector field), 209Strong maximum principle, 83Symmetry (of fundamental solution), 64,

112

Temporally homogeneous, 111

Vector field with potential, 209

Weak solutionof elliptic equations, 195of parabolic equation, 196

307

308 INDEX

associated with a boundary condition,204

ConstructionHomomorphic ImageInclusionInsider

Excellence,

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