Simple Extensions with the Minimum Degree Relations of Integral Domains (Lecture Notes in Pure and Applied Mathematics) German

Chapman & Hall/CRC Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2007 by Taylor & Francis Group, LLC Chapman & Hall/CRC is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number-10: 1-58488-851-2 (Softcover) International Standard Book Number-13: 978-1-58488-851-2 (Softcover) This book contains information obtained from authentic and highly regarded sources. Reprinted material is quoted with permission, and sources are indicated. A wide variety of references are listed. Reasonable efforts have been made to publish reliable data and information, but the author and the publisher cannot assume responsibility for the validity of all materials or for the consequences of their use. No part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www. copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC) 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data Oda, Susumu, 1950Simple extensions with the minimum degree relations of integral domains / Susumu Oda, Ken-ichi Yoshida. p. cm. -- (Lecture notes in pure and applied mathematics) Includes bibliographical references and index. ISBN-13: 978-1-58488-851-2 (alk. paper) ISBN-10: 1-58488-851-2 (alk. paper) 1. Ring extensions (Algebra) 2. Noetherian rings. 3. Integral domains. I. Yoshida, Ken-ichi, 1948- II. Title. III. Series. QA247.S556 2007 512’.4--dc22 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

2006052720

To the authors’ families, especially to Masako Oda and Sayaka Oda

Preface

This monograph consists of eleven chapters. We study several properties about simple ring-extensions of Noetherian integral domains. The simplest one is a polynomial extension. Many properties of polynomial rings are easily seen. So we may restrict ourselves to algebraic extensions. But arbitrary algebraic simple extensions are not always so simple and are rather complicated. For this reason, we treat simple extensions satisfying a certain condition. Let R be a Noetherian domain and let K be its quotient field. Let α be an element in some algebraic field-extension L of K . The investigation of a simple extension R[α] is equivalent to studying the natural exact sequence: 0 −→ Ker(π ) −→ R[X ] −→ R[α] −→ 0 where π : R[X ] → R[α] denotes the natural R-algebra homomorphism sending X to α and R[X ] is a polynomial ring. Many properties of R[α] are represented by the Ker(π ). We say that R[α] is anti-integral over R if Ker(π) = I[α] ϕα (X )R[X ], where ϕα (X ) = X d + η1 X d−1 + · · · + ηd denotes d the minimal polynomial of α over K and I[α] denotes the ideal i=1 (R : R ηi ) in R. Put J[α] := I[α] (1, η1 , . . ., ηd ). Then J[α] is an ideal of R. If J[α] ⊆ p for every prime ideal p of grade one, we say that α is a super-primitive element over R and that R[α] is a super-primitive extension of R. Let R denote the integral closure of R in K . If grade(I[α] + C(R/R)) > 1, where C(R/R) := R : R R, then we say that α is an ultra-primitive element over R and that R[α] is an ultra-primitive extension of R. We can show the implications: “ultra-primitive ⇒ super-primitive ⇒ anti-integral.” These extensions have the good tool I[α] . We also imply that an algebraic element α over a Noetherian Krull domain R is an ultra-primitive element. Thus, ultra-primitiveness (super-primitiveness, anti-integrality) is not so rare and not so special. So we study anti-integral, super-primitive, or ultra-primitive extensions, of R. We divide the inclusion R → R[α] into R → R[α] ∩ R[α −1 ] → R[α]. The subring R[α] ∩ R[α −1 ] seems a somewhat interesting subring and it is completely determined if α is anti-integral over R. Note that R[α] ∩ R[α −1 ] and R[α] are birational. We study mainly anti-integral, super-primitive, or ultra-primitive extensions, and we are especially interested in studying the flatness, the integrality, and the unramifiedness of R[α] over R by use of I[α] . As will be seen in this monograph,

simple extensions of a Noetherian domain R are not so simple even if they are birationally equal to R. This monograph owes its existence to the generous efforts of our colleagues and friends, Professors M. Kanemitsu, J. Sato, T. Sugatani, and others, and is based on their joint works with the authors. The authors are deeply grateful to them.

Susumu Oda and Ken-ichi Yoshida June 30, 2006

Comments

The sources of each section are in the following references: Chapter 1 §1 [OY1] [K], §2 [OY1] [Ri], §3 [OY1] [Y], §4 [S2] [L] [KY5] §5 [Mc] [S1] [KY3] [OY3], §6 [KY1] Chapter 2 §1 [OSaY1] [Sh], §2 [OSaY1], §3 [OSaY1], §4 [OSaY2] [KY6] [SaOY2], §5 [OSaY1] [OY17] Chapter 3 §1 [OSuY2] [KY6] [KSaY2] [OY19], §2 [OY6], §3 [TOY], §4 [BOY1] [SaOY3] [OYaY] [OY20] Chapter 4 §1 [S2], §2 [OSaY3], §3 [KSuY] [OnSuY1] [U], §4 [KY1] Chapter 5 §1 [OSaY1], §2 [KY1], §3 [OnSuY2] Chapter 6 §1 [KYJ] [KY2], §2 [KSaY1] Chapter 7 §1 [OSuY1], §2 [YSaO], §3 [OY4] [OY17] [OY18], §4 [KOY], §5 [OY9] Chapter 8 §1 [OY14], §2 [OY11] [KSaY3], §3 [SO], §4 [OY13], §5 [OY16] Chapter 9 §1 [S2], §2 [SaOY1], §3 [OY12], §4 [OY10], §5 [OY15] [OY18] Chapter 10 §1 [YOSa1], §2 [YOSa1], §3 [YOSa2] Chapter 11 §1 [OY8], §2 [OKY], §3 [SY1], §4 [OY3]

Convention

Alphabets in brackets refer to the references. All rings considered in this monograph are assumed to be commutative and have an identity, and our general references for unexplained technical terms are [M1], [M2], and [N]. In particular, throughout this monograph, R denotes an integral domain, K denotes its quotient field, L denotes an algebraic extension field of K , and U (R) or R x denotes the set of all units in R unless otherwise specified.

Contents Chapter 1

Birational Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1 The Ring R[α] ∩ R[α −1 ] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Anti-Integral Extension and Flat Simple Extensions . . . . . . . . . . . . . . . . . 5 1.3 The Ring R(Iα ) and the Anti-Integrality of α . . . . . . . . . . . . . . . . . . . . . . 12 1.4 Strictly Closedness and Integral Extensions . . . . . . . . . . . . . . . . . . . . . . . . 14 1.5 Upper-Prime, Upper-Primary, or Upper-Quasi-Primary Ideals . . . . . . . 15 1.6 Some Subsets of Spec(R) in the Birational Case . . . . . . . . . . . . . . . . . . . 20 Chapter 2 2.1 2.2 2.3 2.4 2.5

Simple Extensions of High Degree . . . . . . . . . . . . . . . . . . . . . . . . . 23

Sharma Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Anti-Integral Elements and Super-Primitive Elements . . . . . . . . . . . . . . 27 Integrality and Flatness of Anti-Integral Extensions . . . . . . . . . . . . . . . . 33 Anti-Integrality of α and α −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Vanishing Points and Blowing-Up Points . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Chapter 3

Subrings of Anti-Integral Extensions . . . . . . . . . . . . . . . . . . . . . . . 49

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R . . . . . . . . . . . . . . . 49 3.2 The Integral Closedness of the Ring R[α] ∩ R[α −1 ] (I) . . . . . . . . . . . . . 65 3.3 The Integral Closedness of the Ring R[α] ∩ R[α −1 ] (II) . . . . . . . . . . . . 71 3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α) . . . . . . . . . . . . . . . . . 80 Chapter 4

Denominator Ideals and Excellent Elements . . . . . . . . . . . . . . . . 97

4.1 Denominator Ideals and Flatness (I) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 4.2 Excellent Elements of Anti-Integral Extensions . . . . . . . . . . . . . . . . . . . . 99 4.3 Flatness and LCM-Stableness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 4.4 Some Subsets of Spec(R) in the High Degree Case . . . . . . . . . . . . . . . . 108

Chapter 5

Unramified Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.1 Unramifiedness and Etaleness of Super-Primitive Extensions . . . . . . . 111 5.2 Differential Modules of Anti-Integral Extensions . . . . . . . . . . . . . . . . . . 114 5.3 Kernels of Derivations on Simple Extensions . . . . . . . . . . . . . . . . . . . . . 120 Chapter 6 6.1 6.2

The Unit-Groups of Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 125

The Unit-Groups of Anti-Integral Extensions . . . . . . . . . . . . . . . . . . . . . 126 Invertible Elements of Super-Primitive Ring Extensions . . . . . . . . . . . 129

Chapter 7

Exclusive Extensions of Noetherian Domains . . . . . . . . . . . . . . 135

7.1 Subring R[α] ∩ K of Anti-Integral Extensions . . . . . . . . . . . . . . . . . . . . 136 7.2 Exclusive Extensions and Integral Extensions . . . . . . . . . . . . . . . . . . . . . 143 7.3 An Exclusive Extension Generated by a Super-Primitive Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 7.4 Finite Generation of an Intersection R[α] ∩ K over R . . . . . . . . . . . . . 155 7.5 Pure Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Chapter 8

Ultra-Primitive Extensions and Their Generators . . . . . . . . . . . 167

8.1 Super-Primitive Elements and Ultra-Primitive Elements . . . . . . . . . . . 168 8.2 Comparisons of Subrings of Type R[aα] ∩ R[(aα)−1 ] . . . . . . . . . . . . . 175 8.3 Subrings of Type R[H α] ∩ R[(H α)−1 ] . . . . . . . . . . . . . . . . . . . . . . . . . . 183 8.4 A Linear Generator of an Ultra-Primitive Extension R[α] . . . . . . . . . . 189 8.5 Two Generators of Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Chapter 9 9.1 9.2 9.3 9.4 9.5

Flatness and Contractions of Ideals . . . . . . . . . . . . . . . . . . . . . . . 201

Flatness of a Birational Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 Flatness of a Non-Birational Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 Anti-Integral Elements and Coefficients of its Minimal Polynomial. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .209 Denominator Ideals and Flatness (II) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 Contractions of Principal Ideals and Denominator Ideals . . . . . . . . . . . 224

Chapter 10

Anti-Integral Ideals and Super-Primitive Polynomials . . . . . 233

10.1 Anti-Integral Ideals and Super-Primitive Ideals . . . . . . . . . . . . . . . . . . 234 10.2 Super-Primitive Polynomials and Sharma Polynomials . . . . . . . . . . . 239 10.3 Anti-Integral, Super-Primitive, or Flat Polynomials . . . . . . . . . . . . . . 243

Chapter 11

Semi Anti-Integral and Pseudo-Simple Extensions . . . . . . . . 249

11.1 Anti-Integral Extensions of Polynomial Rings . . . . . . . . . . . . . . . . . . . 249 11.2 Subrings of R[α] Associated with Ideals of R . . . . . . . . . . . . . . . . . . . 253 11.3 Semi Anti-Integral Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 11.4 Pseudo-Simple Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

Chapter 1 Birational Simple Extensions

It is known that if α is a nonzero element in the quotient field K of an integral domain R, then Rα := R[α] ∩ R[α −1 ] is integral over R (cf. [K, Ex.4, page 12]). The ring Rα seems rather interesting to us, but we do not recall seeing any other results in the literature about such rings except in a few (e.g., [MR]). Let R be a Noetherian domain, K be its quotient field, and α be a nonzero element in K , and put Rα = R[α] ∩ R[α −1 ]. We call α anti-integral over R if Rα = R. In this chapter, we investigate a simple extension R[α] of R with α anti-integral over R, which is called an anti-integral extension of R. First recall the following fact: Let A be an over-ring of R (i.e., R ⊆ A ⊆ K ), which is of finite type over R. Assume that A is quasi-finite over R and R is normal. If the contraction map Spec(A) −→ Spec(R) is surjective, we have R = A (cf. [Ra]). This is known as a special case of the Zariski Main Theorem (ZMT). If R is not integral closed, we can find a counterexample. Even if Spec( A) −→ Spec(R) is surjective, A is not necessarily integral over R (in particular, R = A). But we can show that if A is R[α1 , . . . , αn ], each αi anti-integral over R, then the surjectivity of Spec(A) −→ Spec(R) yields A = R. This result is our first motive for writing this chapter (see Section 1.2). Throughout this chapter, we fix the notation as follows (unless otherwise specified): Let R denote a Noetherian domain, K the quotient field of R, and R the integral closure of R in K . For α ∈ K , Iα denotes an ideal R : R α := {α ∈ R | aα ∈ R} of R, which is called a denominator ideal.

1

2

Birational Simple Extensions

1.1

The Ring R[α] ∩ R[α −1 ]

We begin with the following definition. Definition 1.1.1

For a nonzero element α of K , let Rα := R[α] ∩ R[α −1 ]

in K . The following lemma is seen in [K, Ex.4], but we prove this for the sake of completeness. Lemma 1.1.2 R.

Let α is a nonzero element in K . Then Rα is integral over

Proof We know that the integral closure R of R can be expressed as ∩Vλ , where Vλ ’s are discrete valuation over-rings containing R. Since Vλ is a valuation ring, we have that α ∈ Vλ or α −1 ∈ Vλ . So Rα = R[α] ∩ R[α −1 ] ⊆ Vλ for each Vλ . Hence Rα ⊆ R. The following elementary lemma is useful in the sequel. Lemma 1.1.3 Let α be a non-zero element in K . Then the following statements are equivalent: (1) α is integral over R; (2) Rα = R[α]; (3) α ∈ Rα. Proof (2) ⇔ (3) is obvious. (2) ⇔ (1) follows from Lemma 1.1.2. (1) ⇒ (2): Let α be integral over R. Then α n + a1 α n−1 + · · · + an = 0

(ai ∈ R)

So α = −(a1 + a2 (α −1 ) + · · · + an (α −1 )n−1 ) ∈ R[α −1 ], and hence R[α] ⊆ R[α −1 ]. Thus R[α] ⊆ R[α] ∩ R[α −1 ] = Rα. Definition 1.1.4 Assume that R = Rα for a non-zero α ∈ K . Then we say that α is an anti-integral element or that α is anti-integral over R. When α is

1.1 The Ring R[α] ∩ R[α −1 ]

3

anti-integral over R, the simple extension R[α] is said to be an anti-integral extension of R. The property of being anti-integral is local-global as is seen in the next lemma. Lemma 1.1.5 equivalent:

For a nonzero element α in K , the following statements are

(1) Rα = R; (2) R p α = R p for each p ∈ Spec(R); (3) R p α = R p for each p ∈ Dp1 (R) := { p ∈ Spec(R) | depth(R p ) = 1}. Proof

(1) ⇒ (2): Take β ∈ (Rα) p . Then β = an α n + · · · + a0 = bm (α −1 )m + · · · + b0

with ai , b j ∈ R p . Choose a common denominator d ∈ R of ai ’s and b j ’s such that d ∈ p. Then dβ ∈ Rα = R, and hence β ∈ R p . (2) ⇒ (3) is obvious. (3) ⇒ (1): Suppose that β ∈ R for some β ∈ Rα. Then Iβ = R. Let p be a prime divisor of Iβ . Then depth(R p ) = 1 (cf. [Y]). Hence β ∈ Rα p = R p , so that Iβ ⊆ p, which is a contradiction. Remark 1.1.6 (1) If α ∈ K is anti-integral over R, α is also anti-integral over R p for each p ∈ Spec(R). (2) For a nonzero element α ∈ K , α is anti-integral and integral over R if and only if α ∈ R.

Theorem 1.1.7 equivalent:

For a nonzero element α ∈ K , the following statements are

(1) R[α] ∩ R = R (i.e., R is integrally closed in R[α]); (2) For any nonzero element β ∈ R[α], Rβ := R[β] ∩ R[β −1 ] = R (i.e., β ∈ R[α] is anti-integral over R). Proof (2) ⇒ (1): Suppose that R[α] ∩ R contains R properly. Then there exists β ∈ R[α] ∩ R with β ∈ R. Being β ∈ R implies that Rα = R[β] contains R properly by Lemma 1.1.3. Thus Rα contains R properly. (1) ⇒ (2): Suppose that R[α] ∩ R = R. Since R ⊆ Rα ⊆ R ∩ R[α] = R, we

4

Birational Simple Extensions

have Rα = R. For any nonzero element β ∈ R[α], we have R ⊆ R[β] ∩ R ⊆ R[α] ∩ R = R. So R[β] ∩ R = R and Rβ = R. Remark 1.1.8 For each nonzero element α ∈ K , we have Rα[α] ∩ Rα[α −1 ] = Rα, i.e., α is anti-integral over Rα. In other words, R[α] is an anti-integral extension of Rα. In fact, this is implied as follows: Rα ⊆ Rα[α] ∩ Rα[α −1 ] = (R[α] ∩ R[α −1 ])[α] ∩ (R[α] ∩ R[α −1 ])[α −1 ] ⊆ R[α] ∩ R[α −1 ] = Rα. Lemma 1.1.9 Let α be an anti-integral element in K over R. Let β = uα + a (u, a ∈ R) with u a unit in R. Then β is an anti-integral element over R. Proof It is easy to see that uα is an anti-integral element over R. So we may assume that u = 1 and β = α + a (a ∈ R). Let ϕ

0 −→ I −→ R[X ] −→ R[β] −→ 0 be an exact sequence, where R[X ] is a polynomial ring and  ϕ(X ) = β. By [M1], it suffices to show that I has a linear basis, that is, I = (ci X − di )R[X ] with β = di /ci (ci , di ∈ R). Take f (X ) ∈ I ⊆ R[X ]. Putting Y = X − a, let g(Y ) =  f (X ). Since f (β) = g(α)  = 0 as α is anti-integral overR, we have g(Y ) = (ai Y − bi )h i (Y ) = (ai (X − a) − bi )h i (X − a) = (ai − X − (bi + ai a))h i (X − a), and (bi + ai a)/ai = bi /ai + a = α + a = β. Thus I has a linear basis, which means that β is anti-integral over R. Proposition 1.1.10 For a nonzero element α ∈ K , if Rα contains R properly then there exists r ∈ R such that r α ∈ Rα \ R. Proof

Let β be an element in Rα \ R. Then (∗)

β = an α − n + · · · + a0 = bm (α −1 )m + · · · + b0

with ai , b j ∈ R. Changing β and the expression (∗), we may assume that n is the smallest among such n’s. If n = 1, then there is nothing to prove. Suppose n > 1. We can assume that a0 = 0. Thus β/α = an α n−1 + · · · + a1 = bm (α −1 )m+1 + · · · + b0 (α −1 ) ∈ Rα By the choice of n, we have β/α ∈ R, and hence we conclude that β = r α with r ∈ R.

1.2 Anti-Integral Extension and Flat Simple Extensions

5

Corollary 1.1.11 For a nonzero element α ∈ K , let Iα = q1 ∩ . . . ∩ qt be a √ primary decomposition. If qi ∈ Ass R (R/R) for each i, then α is anti-integral over R. Proof Suppose that Rα contains R properly. Then there exists r ∈ R such that r α ∈ Rα \ R by Proposition 1.1.10, since Ir α = Iα : R r = (q1 : r ) ∩ . . . ∩ (qt : r ). Since r α is integral over R (Lemma 1.1.2), there exists i such √ that qi ∈ Ass R (R/R). Remark 1.1.12 Let α ∈ K and let S be a multiplicatively closed subset of R. Then S −1 (Rα) = S −1 R[α] ∩ S −1 R[α −1 ]. If an over-ring A is an anti-integral extension of R, S −1 A is an anti-integral extension of S −1 R, and a polynomial extension A[X ] is an anti-integral extension of R[X ].

1.2

Anti-Integral Extension and Flat Simple Extensions

To begin with, we give the following definition. Definition 1.2.1 Let A := R[α], α a nonzero element in K . For p ∈ Spec(R), we denote A p = R p [α]. We say that an extension A/R is a blowing-up at p if the following two conditions are satisfied: (i) p A p ∩ R p = p R p , (ii) A p / p A p is isomorphic to a polynomial ring (R p / p R p )[T ]. Lemma 1.2.2 Let (R, m) be a local domain and put A := R[α], α ∈ K (α = 0). Assume that m A ∩ R = m. In this case we have a canonical injection R/m −→ A/m A = (R/m)[α]. Then (1) A/R is a blowing-up at m if and only if α is transcendental over R/m (where α denotes the residue class of α in A/m A). (2) Assume that R contains an infinite field k. If A/R is not a blowing-up at m, then for some µ ∈ k, 1/(α − µ) is integral over R and i is integral and j is a localization ( flat) in the following sequence: i

j

R −→ Rα − µ = R[1/(α − µ)] −→ R[α] = R[α − µ, 1/(α − µ)]

6

Birational Simple Extensions

Proof (1) is obvious. (2) By the preceding (1), α is algebraic over R/m. So we have (∗∗) α k (α  + λ1 α −1 + · · · + λ ) = 0

(λi ∈ R/m, λ = 0)

If k > 0, we can find µ = 0 in the infinite field k such that µ + λ1 µ−1 + · · · + λ = 0. So putting β = α − µ, from (∗∗) we have an algebraic relation: n

β + λ1 β

n−1

+ · · · + λn = 0 (λn = 0, λi ∈ R/m)

Let ai denote the representative of λi in R. Then we have β n + a1 β n−1 + · · · + an ∈ m R[β] Since a n = λn = 0, we have an ∈ m. By the expression of an element in m R[β], we get b β  + · · · + b0 = 0 with bi ∈ R and b0 = a n = λn = 0. Hence b0 is a unit in R. Multiplying 1/b0 , c β  + · · · + c1 β + 1 = 0 which means that β −1 is integral over R. By Lemma 1.1.3, Rβ = Rβ −1  = j

i

R[β −1 ] and R −→ Rβ −1  = R[β −1 ] −→ R[β, β −1 ], where i is integral and j is a localization (hence flat). Definition 1.2.3 Let R −→ A be a ring extension and p ∈ Spec(R). We say that p is contracted from A if p A ∩ R = p. It is easy to see that p ∈ Spec(R) is contracted from A if and only if P ∩ R = p for some P ∈ Spec(A) if and only if p A p ∩ R p = P R p . If A = R[α] is a blowing-up at p ∈ Spec(R), then p is contracted from A. Lemma 1.2.4 α ∈ K . Then

Let (R, m) be a local domain and put A = R[α] with a nonzero

(1) m is not contracted from A if and only if m A = A; i

(2) if m is not contracted from A, then α −1 is integral over R and R −→ j

Rα = R[α −1 ] −→ R[α] = R[α, α −1 ], where i is integral and j is flat. Proof (1) is obvious. (2) Suppose that m A = A. Then 1 = a0 + a1 α + · · · + an α n

(ai ∈ m)

1.2 Anti-Integral Extension and Flat Simple Extensions

7

Since 1 − a0 is a unit in R, α −1 is integral over R. Hence noting that α −1 ∈ Rα ⊆ R[α], we have a sequence j

i

R −→ Rα = R[α −1 ] −→ R[α] = R[α, α −1 ] where i is integral and j is a localization. Proposition 1.2.5 Let (R, m) be a local domain and let α be a nonzero element in K which is anti-integral over R. Assume that R[α] contains R properly. Then j

m R[α] = R[α] if and only if R = Rα = R[α −1 ] −→ R[α], where j is flat. Proof

(⇒): By Lemma 1.2.4, α −1 is integral over R and Rα = Rα −1  = j

R[α −1 ] −→ R[α], where j is flat. Since α is anti-integral, we have Rα = R. (⇐): By assumption, α ∈ R. Since α −1 ∈ R, we have α −1 ∈ m, which implies that m R[α] = R[α]. The ring-extension R[α] of R is characterized by an ideal Iα as is seen in the following proposition. Theorem 1.2.6 Let α ∈ K be anti-integral over R and let p be a prime ideal of R. Assume that R contains an infinite field k. Then (1) Iα ⊆ p if and only if R p = R p [α]; (2) Iα ⊆ p but α Iα ⊆ p if and only if R p = R p [α], and p is not contracted from R[α]; (3) Iα + α Iα ⊆ p if and only if R[α] is blowing-up at p over R. Proof (1): Iα ⊆ p ⇔ α ∈ R p ⇔ R p [α] = R p . (2) (⇒): Since α Iα ⊆ p, there exists b ∈ α Iα but b ∈ p. Hence α = b/a for some a ∈ Iα ⊆ p. So aα = b ∈ p R p [α] and hence p R[α] ∩ R contains p properly. i

j

(⇐): By Lemma 1.2.4, R p −→ R p [α −1 ] −→ R p [α −1 ], where i is integral and j is flat, and α −1 is integral over R p . Since α is anti-integral over R p , we have R p = (Rα) p = (Rα −1 ) p = R p [α −1 ]. Hence α −1 ∈ R p , which implies that α Iα = Iα−1 ⊆ p. Since p R p [α] = R p [α], we have R p [α] contains R p properly, and hence Iα ⊆ p. (3) (⇒): Since R p [α]/ p R p [α] = (R p / p R p )[α], we must show that α is transcendental over R p / p R p . Suppose the contrary. We have by Lemma 1.2.2 that β −1 is integral over R p with β = α − µ for some µ ∈ k, and β is anti-integral over R by Lemma 1.1.9. So β −1 ∈ R P . Hence β Iβ = Iβ −1 ⊆ p. But it is easy to

8

Birational Simple Extensions

see that Iα + α Iα = Iα (R + α R) = Iβ (R + β R) = Iβ + β Iβ , which yields that Iβ + β Iβ ⊆ p, a contradiction. (⇐): Suppose that Iα + α Iα ⊆ p. If Iα ⊆ p, then R p = R p [α] and R[α] is not a blowing-up at p by (1). If Iα ⊆ p and α Iα ⊆ p, then p R p [α] ∩ R p = p R p by (2). Thus R[α] is not a blowing-up at p by Definition 1.2.1. Therefore in any case R[α] is not a blowing-up at p. The following result is shown in [Ri]: Lemma 1.2.7 Let A be an integral domain (not necessarily Noetherian), L its quotient field, and α = b/a ∈ L (a,b ∈ A,ab = 0). If A[α] is flat over A and A is integrally closed in L, the ideal a(1, α)A = (a,b)A is invertible. Proof Since A ⊆ A[α] is flat, we have that (b : Aa)A[α] = A[α], i.e., there exist h j ∈ (b : A a) such that h n a n /y n +h n−1 a n−1 /y n−1 +· · ·+h 1 a/b+h 0 = 1, for some nonnegative integer n. Multiplying by bn /a n , we get that b/a is integral over A[1/s], where s = h 0 − 1, and so b/a ∈ A[1/s] since A[1/s] is integrally closed. Therefore b/a = h/(h 0 − 1)t for some h ∈ A and some nonnegative integer t, and so ha/b + ch 0 = 1 for some c ∈ A, i.e., (h/b)a + (ch 0 /b)b = 1, where h/b and ch 0 /b ∈ (a, b)−1 . Theorem 1.2.8

Let α ∈ K be a nonzero element. Then

(1) If Iα + α Iα = R then R[α] is flat over R. (2) Assume that α = b/a (a, b ∈ R) is anti-integral over R. Then the following statements are equivalent: (2.i) R[α] is flat over R; (2.ii) Iα + α Iα = R; (2.iii) (a,b)R is an invertible ideal of R. Proof (1) Since Iα (R + α R) = R, a fractional ideal R + α R is invertible. Since R + α R  a(R + α R) = (a, b)R is invertible, so R[α] is flat (cf. [Ri]). (2) (2.iii) ⇒ (2.i) follows from [Ri]. (2.ii) ⇒ (2.iii) is shown above. (2.i) ⇒ (2.ii): Note first that p ⊇ Iα + α Iα if and only if R[α] is a blowing-up at p over R by Theorem 1.2.6. Since R[α] is flat over R, R[α] is not a blowing-up at any p ∈ Spec(R) over R. Thus Iα + α Iα = R. Proposition 1.2.9 Assume that α ∈ K is an anti-integral element over R. Then R[α] is flat over R if and only if R[α −1 ] is flat over R.

1.2 Anti-Integral Extension and Flat Simple Extensions

9

Proof By Theorem 1.2.8, R[α] is flat over R ⇔ Iα + Iα−1 = Iα + α Iα = R ⇔ Iα + Iα−1 = (α −1 )Iα−1 + Iα−1 = R (Note that α Iα = Iα−1 ) ⇔ R[α −1 ] is flat over R. Lemma 1.2.10 Let α ∈ K (α = 0) and P ∈ Spec(Rα). Put p = P ∩ R. Then either p R[α] = R[α] or p R[α −1 ] = R[α −1 ] holds. Proof Suppose that p R[α] = R[α]. Then we have (Rα) p = R p [α −1 ] by Lemma 1.2.4 since p R p [α] = R p [α]. Hence (Rα) p ⊃ P(Rα) p ⊇ p(Rα) p = −

p R p [α −1 ] and consequently we have p R p [α −1 ] = R p [α −1 ]. Thus p R[α −1 ] = R[α −1 ]. The next result is well known (cf. [Ri]). Lemma 1.2.11 Let A be an over-ring of R. Then A is flat over R if and only if A P = R P∩R for each P ∈ Spec(A). Proposition 1.2.12 Let α be a nonzero element in K . Assume that both R[α] and R[α −1 ] are flat over R. Then Rα = R, i.e., α is anti-integral over R. Proof Take P ∈ Spec(Rα) and put p = P ∩ R. Then by Lemma 1.2.10, either p R[α] = R[α] or p R[α −1 ] = R[α −1 ] holds. We may assume p R[α] = R[α] (by symmetry). Then there exists Q ∈ Spec(R[α]) such that Q ∩ R = p. Since R p [α] is flat over R p , we have R p = (R p [α]) Q ⊇ R p [α] ⊇ R p . Thus R p [α] = R p and hence (Rα) p = Rα P = R p , which implies that Rα is flat over R by Lemma 1.2.11. So Rα = R by [Ri] since Rα is integral over R. Corollary 1.2.13 Let α = b/a ∈ K (a, b ∈ R, ab = 0). Then if both R[α] and R[α −1 ] are flat over R, (a, b)R is an invertible ideal of R. Proof The element α is anti-integral over R by Proposition 1.2.12. The conclusion follows from Theorem 1.2.8. Corollary 1.2.14 conditions:

Let α ∈ K be a nonzero element. Consider the following

10

Birational Simple Extensions

(1) R[α] is an anti-integral extension of R; (2) R[α] is flat over R; (3) R[α −1 ] is flat over R. Then two of these conditions imply the other. Proof If (1) holds, (2) ⇔ (3) follows from Proposition 1.2.9. (2) + (3) implies (1) by Proposition 1.2.12. We recall that: Let A be an over-ring of R, which is of finite type over R. Assume that R is integrally closed in K and that A is quasi-finite over R (cf. [Ra]). Then the contraction map ϕ : Spec(A) −→ Spec(R) is an open immersion, and if furthermore ϕ is surjective, then A = R. This is known as the Zariski Main Theorem (ZMT) (cf.[Ra]). We extend this in a certain sense (see Theorem 1.2.18). We recall the following definitions. Definition 1.2.15 ([R,(1.1.11)], [M2]). If T is an integral domain, then T satisfies the altitude formula or the dimension formula in case, for all finitely generated integral domain B over T and for all P ∈ Spec(B), ht(P) + Tr.degT /(P∩T ) B/P = ht(P ∩ T ) + Tr.degT B Proposition 1.2.16 Let α be a nonzero element in K , which is anti-integral over R. Assume that R contains an infinite field k and that R satisfies the altitude formula. If the contraction map ϕ : Spec(R[α]) −→ Spec(R) is such that Im(ϕ) ⊇ Dp1 (R), then R[α] = R. Proof Take p ∈ Spec(R). If Iα ⊆ p, then R[α] ⊆ R p , and so R p [α] = R p . Dp1 (R).√Since Suppose that Iα ⊆ p and that p is a prime divisor of Iα . Then p ∈ √ Im(ϕ) ⊇ Dp1 (R), we have α Iα ⊆ p by Theorem 1.2.6. Thus Iα ⊇ α Iα . Note here that Iα = Iα−µ for any µ ∈ k, that R[α] −µ], and that α −µ is  = R[α anti-integral over R by Lemma 1.1.9. We have Iα−µ ⊇ (α − µ)Iα−µ for any µ ∈ k. Now let q be in Ht1 (R) and suppose q ⊇ Iα . Then R[α] is not a blowingup at q over R. [Indeed, suppose that Rq [α]/q Rq [α]  (Rq /q Rq )[T ], where T is an indeterminate. Then q Rq [α] is a prime ideal and ht(q Rq [α]) = ht(q Rq ) + Tr.deg R R[α] − Tr.degk(q) Rq [α]/q Rq [α] = 1 + 0 − 1 = 0 (k(q) := Rq /q Rq ) by the altitude formula (cf. Definition 1.2.15 in Section 23), which implies that q Rq [α] = (0), a contradiction.] Hence β −1 is integral over R for β = α − µ (some µ ∈ k) by Lemma 1.2.2. Since β is anti-integral over R by Lemma 1.1.9,

1.2 Anti-Integral Extension and Flat Simple Extensions

11

  β −1 ∈ Rq . Thus β Iβ = Iβ −1 ⊆ q, which contradicts with q ⊇ Iβ ⊇ β Iβ . Therefore ht(Iα ) > 1. But since R satisfies the altitude formula, the element α is integral over R. [Indeed, putting p = P ∩ R for arbitrary P ∈ Ht1 (R), we have ht( p) = 1 by the altitude formula. Thus Iα ⊆ p and hence α ∈ R p ⊆ R p , which yields that α ∈ R.] Therefore Rα = R[α] by Lemma 1.1.3. So R[α] = Rα = R. As a corollary to the proof of Proposition 1.2.16, we have the following: Corollary 1.2.17 Let α be an anti-integral element in K . Assume that R contains an infinite field k and that R[α] is quasi-finite over R. If the contraction map ϕ : Spec(R[α]) −→ Spec(R) satisfies Im(ϕ) ⊇ Dp1 (R), then R[α] = R. Proof √In the same way as in the proof of Proposition 1.2.16, we can show that √ Iα ⊇ α Iα . Since R[α] is quasi-finite over R, R[α] is not a blowing-up at any p ∈ Spec(R) with p R p [α] = R p [α]. Hence by √Theorem √ 1.2.6, Iα +α Iα ⊆ p for any p ∈ Spec(R) with p R p [α] = R p [α]. So Iα ⊇ α Iα implies that Iα ⊆ p for any p ∈ Spec(R) with p R p [α] = R p [α]. Thus α ∈ R p for any p ∈ Spec(R) with p R p [α] = R p [α]. But since Im(ϕ)  ⊇ Dp 1 (R), we have p R[α] = R[α] for any p ∈ Dp1 (R), and hence R = R p ⊇ R p [α] ⊇ R[α] ( p ∈ Dp1 (R)), which implies that R = R[α]. Theorem 1.2.18 Assume that R contains an infinite field k and that R satisfies the altitude formula. Let A = R[α1 , . . . , αn ] with αi ∈ K anti-integral over R. If the contraction map ϕ : Spec(A) −→ Spec(R) satisfies Im(ϕ) ⊇ Dp1 (R), then A = R. Proof

From the commutative diagram: Spec(A)   ϕ Spec(R)

−→ ===

Spec(R[α  i ])  ϕ i Spec(R)

we have Im(ϕi ) ⊇ Im ϕ ⊇ Dp1 (R). Thus αi ∈ R for each i by Proposition 1.2.16. Hence A = R. Corollary 1.2.19 Under the same assumption as in Theorem 1.2.18, The following statements are equivalent: (1) ϕ is a surjection; (2) A is integral over R; (3) A = R.

12

Birational Simple Extensions

Corollary 1.2.20 Let A = R[α1 , . . . , αn ] with nonzero element αi ∈ K anti-integral over R. Assume that R contains an infinite field k and that A is quasi-finite over R. If the contraction map ϕ : Spec(A) −→ Spec(R) satisfies Im(ϕ) ⊇ Dp1 (R), then A = R. Remark 1.2.21 Let A and R be the same as in Theorem 1.2.18. If A is integral over R, ϕ : Spec(A) −→ Spec(R) is obviously surjective. But the surjectivity of ϕ does not necessarily ensure the integrality of A over R. For example: A = k[T, 1/(T − 1)] ⊇ k[T ] ⊇ R = k[T (T − 1), T 2 (T − 1)], where k is a field. Theorem 1.2.18 asserts that if A is anti-integral over R, the surjectivity of ϕ ensures the integrality of A over R.

1.3

The Ring R(Iα ) and the Anti-Integrality of α

To begin with, we introduce the following over-ring of R. Definition 1.3.1

Let J be a fractional ideal of R. Define R(J ) := {α ∈ K |α J ⊆ J }

It is easy to see that R(J ) is an over-ring of R, which is integral over R, and R(J p ) = R(J ) p for any p ∈ Spec(R). Proposition 1.3.2 Let J be an integral ideal of R. Then the following statements are equivalent: (1) R(J −1 ) = R (where J −1 := R : K J = Hom R (J, R) ); (2) J p is a principal ideal of R p for each p ∈ Dp1 (R). Proof (2) ⇒ (1): Since R = ∩R p ( p ∈ Dp1 (R)), it suffices to show that R(J −1 ) ⊆ R p for any p ∈ Dp1 (R). As J p = α R p for some α ∈ R, we have J p−1 = α −1 R p . Hence R p ⊆ R(J p−1 ) = R(α −1 R p ) ⊆ R p . The containment R(J −1 ) ⊆ R(J p−1 ) is obvious. Thus R(J −1 ) ⊆ R(J p−1 ) ⊆ R p , as was wanted. (1) ⇒ (2): We may assume that (R, m) is a local domain and that m ∈ Dp1 (R). If R is a discrete valuation ring, R is a principal ideal domain. Hence J is a principal ideal of R. Suppose that R is not a discrete valuation ring (DVR) of rank one. Then m : K m = R. If there exists φ ∈ J −1 = Hom R (J, R) with

1.3 The Ring R(Iα ) and the Anti-Integrality of α

13

φ

φ(J ) = R, an exact sequence: J −→ R −→ 0 splits. Since rankJ = 1, we have J  R. So we suppose that φ(J ) = R for any φ ∈ J −1 . In this case, φ(J ) ⊆ m. We have αφ(J ) ⊆ αm ⊆ m ⊆ R for any α ∈ m : K m, and hence α ∈ R(J −1 ). Thus m : K m ⊆ R(J −1 ). But since R(J −1 ) ⊇ m :k m⊃ R, we − have R = R(J −1 )⊃ R, which is a contradiction. −

Proposition 1.3.3 Let J be a divisorial integral ideal of R. If R(J −1 ) = R, then there exists a nonzero element α ∈ K such that J = Iα . Proof We first show the following claim: If p ∈ Dp1 (R) contains J , then p is a prime divisor of J . In fact, since R(J −1 ) = R, it follows that J p = a R p for some a ∈ J by Proposition 1.3.2. As p R p is a prime divisor of a principal ideal a R p = J p (cf. √ [Y]), p is a prime divisor of J . Let J = q1 ∩ · · · ∩ qn ( qi = pi ) be a primary  decomposition. Then pi ∈ Dp1 (R) because J is divisorial. Put S = R \ pi . Then A := S −1 R is semi-local and each maximal ideal of A is of depth 1. Since J A is invertible by the above claim and since A is semi-local, J A is a principal ideal. So there exists a ∈ J such that J A = a A. Hence we have a primary decomposition:  a R = q1 ∩ · · · ∩ qn ∩ Q 1 ∩ · · · ∩ Q  ( Q i = Pi , Pi = p1 , . . . , pn ) Suppose that Pi ⊇ J for some i. Then Pi ∈ Dp1 (R) implies that Pi is a prime divisor of J by the first claim. But since the prime divisors of J are p1 , . . . , pn , this is absurd. Hence J ⊆ Pi for all i = 1, . . . , m. Next we will show that Pi ⊆ p j for any i, j. Suppose that Pi ⊆ p j . Then J R p j = a R p j , and a ∈ Pi implies J R p j ⊆ Pi R p j . Thus J ⊆ Pi , a contradiction. Hence we have Q 1 ∩ · · · ∩ Q m ⊆ p1 , . . . , pn So there exists b ∈ Q 1 ∩ · · · ∩ Q m , b ∈ p1 , . . . , pn . We have (a R : R b R) = q1 ∩ · · · ∩ qn = J . Putting α = b/a ∈ K , we have Iα = a R : R b R = J .

Theorem 1.3.4 Let α be a nonzero element in K . If R(Iα ) = R, then Rα = R, i.e., α is anti-integral over R. Proof Claim 1: For each p ∈ Dp1 (R), either (Iα ) p = R p or (Iα−1 ) p = R p holds. Indeed, let p ∈ Dp1 (R). If Iα ⊆ p, then (Iα ) p = R p . Suppose that Iα ⊆ p and Iα−1 ⊆ p. Then (Iα ) p = a R p for some a ∈ Iα by Proposition 1.3.2. So αa R p = α(Iα ) p = (Iα−1 ) p ⊆ p R p . Thus αa = r for some r ∈ p R p

14

Birational Simple Extensions

and α = r/a. So a R p = (Iα ) p = (a R p : R p r R p ), which asserts that a, r is an R p -regular sequence. But depth(R p ) = 1, a contradiction. Therefore either (Iα ) p = R p or (Iα−1 ) p = R p holds. Claim 2: If R(I α) = R, then Iα + Iα−1 ⊆ p for each p ∈ Dp1 (R). In fact, suppose that Iα + Iα−1 ⊆ p with p ∈ Dp1 (R). Then both (Iα ) p and (Iα−1 ) p are principal ideals in R p by Proposition 1.3.2. But by Claim 1, either (Iα ) p = R p or (Iα−1 ) p = R p holds. This contradicts with (Iα ) p +(Iα−1 ) p ⊆ p R p . Thus for p ∈ Dp1 (R), Iα + Iα−1 ⊆ p by Claim 2. Hence α ∈ R p or α −1 ∈ R p . Therefore Rα = R[α] ∩ R[α −1 ] ⊆ R p for all p ∈ Dp1 (R). So Rα = R by Lemma 1.1.5. Corollary 1.3.5 Let J be an invertible ideal contained in R. Then there exists an anti-integral element α of K such that J = Iα . Proof Since R(J −1 ) = R by Proposition 1.3.2, J = Iα for some nonzero element α in K by Proposition 1.3.2. Hence Rα = R by Theorem 1.3.4.

Remark 1.3.6 The converse of Theorem 1.3.4 does not necessarily hold as can be seen in the following example: Let (R, m) be a Noetherian local domain such that Ass R (R/R) = {m}, ht(m) > 1. Take α ∈ K such that Iα = q1 ∩ . . . ∩ qt , where qi is a primary ideal √ with ht(qi ) = 1 for each i. Then qi ∈ Ass R (R/R) implies that Rα = R by Corollary 1.1.11. But since depth(R) = 1, Iα cannot be a principal ideal, and hence R(Iα ) = R (cf. Proposition 1.3.2).

1.4

Strictly Closedness and Integral Extensions

We consider integral extensions. Let R be a Noetherian domain with quotient field K . Assume that the integral closure R of R in K is a finite R-module. We recall the following definition introduced by Lipman ([L]). Definition 1.4.1 defined as

Let A be an over-ring of R and ∗A R be the subring of A ∗ AR

:= {α ∈ A | α ⊗ 1 = 1 ⊗ α ∈ A ⊗ R A}

We say that R is strictly closed in A if ∗A R = R.

1.5 Upper-Prime, Upper-Primary, or Upper-Quasi-Primary Ideals

15

An intermediate ring A between R and R is called a birational-integral extension of R. For the strictly closedness of a birational-integral extension of R, the following proposition is important. Proposition 1.4.2 ([KY5, Proposition 2]) Let A be a birational-integral extension of R. If R is strictly closed in A, then Iα ∩ Iβ ⊆ Iαβ for any elements α, β ∈ A. We show the following proposition, which is a characterization of birationalintegral extensions in terms of denominator ideals. Note that Iα−1 = α Iα . Theorem 1.4.3 Let A = R[α] with α ∈ K . Suppose that R is strictly closed in A. Then α is integral if and only if Iα−1 ⊆ Iα . Proof (⇒): By the determinant trick, this implication is clear. n (⇐): Let C be the conductor of A over R. Then it is easy to see that C = i=1 Iα i , for some n. Since R is strictly closed in A by assumption, we have Iα ∩ Iβ ⊆ Iαβ for any elements by Proposition 1.4.2. Hence it holds that Iα ⊆ Iαi for all i, and Iα ⊆ C. Thus C = Iα . Since the conductor C is an ideal of R, Iα is an ideal in A, and α Iα ⊆ Iα .

1.5

Upper-Prime, Upper-Primary, or Upper-Quasi-Primary Ideals

Let R be a Noetherian domain and K its quotient field. Let α be a nonzero element in K . We call α anti-integral over R if R = R[α] ∩ R[α −1 ] and, in this case, a simple birational-extension R[α] is called an anti-integral extension of R (see Section 1). For α ∈ K , Iα denotes an ideal R : R α = {a ∈ R|aα ∈ R} of R. A. Mirbagheri and L. J. Ratliff, Jr. proved that if R[α] is an anti-integral extension of R, then Ker(π) is generated by linearly related polynomials of α (that is, a X − b, α = b/a ), where π : R[X ] −→ R[α] is the natural R-algebra homomorphism. Thus if p ⊇ Iα + α Iα for p ∈ Spec(R), then it follows that R[α]/ p R[α] ∼ = (R/ p)[T ], where T is an indeterminate (Theorem 1.2.6). Anti-integral extensions were studied before. We list some known results (Theorem 1.2.8), which will be used frequently: Let α ∈ K be a nonzero element. Then (1) If Iα + α Iα = R, then R[α] is flat over R. (2) Assume that α = b/a (a, b ∈ R) is anti-integral over R. Then the following statements are equivalent:

16

Birational Simple Extensions (2.i) R[α] is flat over R; (2.ii) Iα + α Iα = R; (2.iii) (a, b)R is an invertible ideal of R.

If for p ∈ Spec(R), p ⊇ Iα + α Iα , then R p [α] = (R[α]) p is flat over R p by (1). Let S be a birational extension of R. Definition 1.5.1 (cf. [Mc]) Let q be an ideal of R. We say that q is an upperprime ideal of R with respect to S if aα ∈ q S for a ∈ R, α ∈ S implies a ∈ q or α ∈ q S. The following results are proved in [S1]. Remark 1.5.2 (1) Let I be an ideal of R such that I S = S. If I is an upper-prime ideal of R with respect to S, then I S ∩ R = I and I is a prime ideal of R. (2) Let p ∈ Spec(R). Then p is an upper-prime ideal of R with respect to S if and only if pS = S or pS p ∩ S = pS. (3) Let p ∈ Spec(R). Then p is an upper-prime ideal of R with respect to S if and only if either pS = S or pS = S, and P ∩ R = p for any prime divisor P of the ideal pS. Now, we give the following theorem. Theorem 1.5.3 Let R be a Noetherian domain and K its quotient field. Let α be an anti-integral element in K . Put S = R[α], Iα := {a ∈ R | aα ∈ R} and Jα := Iα + α Iα . For p ∈ Spec(R), the following statements hold: (i) assume that p ⊇ Jα . Then pS ∈ Spec(S) and pS ∩ R = p; Consequently p is an upper-prime ideal of R with respect to S. Moreover, S/ pS = (R/ p)[T ] holds, where T is an indeterminate. (ii) assume that Jα ⊆ p and p + Jα = R; (ii.a) if p ⊇ Iα and p ⊇ α Iα , then pS p = S p . Also, the converse is valid. Hence it follows that pS ∩ R ⊃ p and p is not an upper-prime ideal − of R with respect to S; (ii.b) if α Iα , then p is not an upper-prime ideal of R with respect to S. Thus there exists some primary components Q i of pS such that pS = ( pS p ∩ S) ∩ Q i ∩ · · · ∩ Q l

1.5 Upper-Prime, Upper-Primary, or Upper-Quasi-Primary Ideals and



17

Qi ∩ R ⊃ p −

(iii) assume that p + Jα = R. Then p is an upper-prime ideal of R with respect to S. In particular, it follows that pS ∈ Spec(S), pS ∩ R = p and S/ pS = R/ p. Proof (i) Let π : R[X ] −→ R[α] be the canonical ring homomorphism. Then Ker(π ) is generated by linear related polynomials of α. And Jα is the ideal of R generated by the coefficients of the elements of Ker(π ). Since p ⊇ Jα by assumption, we have p R[X ] ⊆ Ker(π). Hence R[α]/ p R[α] ∼ = (R/ p)[T ]. Consequently pS is a prime ideal of S and pS ∩ R = p. Thus p is an upperprime ideal of R with respect to S. (ii.a) There exist two elements a and b of R such that a ∈ Iα ⊆ p and b = aα ∈ p by assumption. Then b = a(b/a) ∈ pS p . Thus we have pS p = S p . Hence p is not an upper-prime ideal of R with respect to S. Conversely, assume that pS p = S p . Since (i) does not hold in this case, it holds that p ⊇ Jα . Suppose that p ⊇ Iα , we would have that S p = R p and so pS p = S p . This contradicts the assumption. Hence p ⊇ Iα . Since p ⊇ Jα , it follows that p ⊇ α Iα . (ii.b) By assumption, there exists a, b ∈ R such that α = b/a with a ∈ Iα \ p. Then aα = b ∈ α Iα ⊆ p ⊆ pS. Suppose that p is an upper-prime ideal of R with respect to S. Since a ∈ p and Iα ⊆ p, it holds that α ∈ pS. Hence S/ pS = R/ p. Since p + Jα = R, there is a maximal ideal m of R such that p + Jα ⊆ m. It is obvious that α ∈ pS ⊆ m S and so S/m S = R/m. But, since Jα ⊆ m, we have S/m S = R/m[T ] by (ii,a) (replacing p by m), which is a contradiction. Thus p is not an upper-prime ideal of R with respect to S. There exist some primary components Q i such that pS = ( pS p ∩ S) ∩ Q 1 ∩ · · · ∩ Q l where

√

Q i ∩ R ⊃ p. −

(iii) Suppose that p is not an upper-prime ideal of R with respect to S. Then there exist some primary components Q i such that a primary decomposition is pS = ( pS p ∩ S) ∩ Q 1 ∩ · · · ∩ Q l where

√

Q i ∩ R (= qi ) ⊃ p. Since p + Jα = R and qi ⊃ p, we have that −

−

qi = q ⊇ Jα . Since Sq is a flat extension of /Rq , it holds that p Rq is an upperprime ideal of R with respect to S. There exists some element a ∈ Iα \ p. Since b = aα ∈ α Iα ⊆ p ⊆ pS, it follows that α ∈ pS. Hence S/ pS = R/ p. Thus pS is a prime ideal of S and pS ∩ R = p.

18

Birational Simple Extensions

Remark 1.5.4 If p ⊇ Jα and p ⊇ α Iα , then we have p ⊇ Iα . Therefore there exists an element a ∈ Iα \ p. Put b = aα. Then b ∈ α Iα ⊆ p. This is the case (iii) in Theorem 1.5.3. So another case is that p ⊇ Iα and p ⊇ α Iα . In this case, it is not known whether p is an upper-prime ideal of R with respect to S or not. In the case (ii), if pS = S, it follows that pS = q S and S/ pS = (R/q)[T ], where q = pS ∩ R and T is an indeterminate. It is not known whether q is prime or √ (iii), the minimal prime divisors of p + Jα appear √ not. In the case among Q 1 ∩ R, . . . , Q l ∩ R. Example 1.5.5 Let k be a field and R = k[x, y] be a polynomial ring. Put α = y/x, p = x R and S = R[y/x]. Then R ⊆ S and y ∈ x S ∩ R. Hence x S ∩ R⊃ p and pS p = S p . This is the case that p ⊇ Iα and p ⊇ Jα . =

Example 1.5.6 R, k, S, α are the same as in Example 1.5.5. Put p = y R. Then we have that p ⊇ Jα , p ⊇ Iα , p ⊇ α Iα , pS p ∩ R = p and pS = p. But p is not an upper-prime ideal of R with respect to S. It holds that pS = (y/x)S ∩ x S, (y/x)S ∩ R = y R and x S = (x, y)R. The same results hold for p = (x + y)R. Example 1.5.7 Let k be a field and R = k[x, y, z] be a polynomial ring in three variables over k. Put α = y/x and p = z R. Then pS is a prime ideal of S and pS ∩ R = p. Hence p is an upper-prime ideal of R with respect to S. This example is applied to the case that p ⊇ Iα , p ⊇ α Iα . Consequently it follows that p ⊇ Jα . For a primary decomposition of pS, the following result holds. Proposition 1.5.8 Let p ∈ Spec(R). Assume that pS p = S p and p is not an upper-prime ideal of R with respect to S. Then pS has a primary decomposition: pS = ( pS p ∩ S) ∩ Q 1 ∩ · · · ∩ Q l √ such that qi ⊇ p + Jα , where qi = Q i ∩ R. Proof Since p ⊆ pS ⊆ Q i , it is clear that p ⊆ qi . Assume that qi ⊇ Jα . We have that Sqi /Rqi is a flat extension by Theorem 1.2.8 and pSqi = Sqi . It follows that Ass R (Sqi / pSqi ) = { p Rqi }, but qi Rqi ∈ Ass R (Sqi / pSqi ). This is a contradiction. Let R be a Noetherian domain and K its quotient field and let S ⊇ R be a birational extension of integral domains. For α ∈ K , Iα denotes an ideal R : R α = {a ∈ R|aα ∈ R} of R. Put Jα = Iα + α Iα .

1.5 Upper-Prime, Upper-Primary, or Upper-Quasi-Primary Ideals

19

Definition 1.5.9 Let q be an ideal of R. We say that q is an upper-primary ideal√of R with respect to S if aα ∈ q S for a ∈ R, α ∈ S implies a ∈ q or α ∈ q S, and that q is an upper-quasi-primary ideal of R with respect to S if √ aα ∈ q S for a ∈ R, α ∈ S implies a ∈ q or α ∈ q S. Example 1.5.10 Let Z denote the ring of rational integers. Put R = Z and S = Z [1/2]. Then every primary ideal of R is an upper-primary ideal of R with respect to S. Moreover, the upper-primary ideal of R with respect to S ideals of R coincide with the upper-quasi-primary ideals of R with respect to S. Also, if q is an ideal of R such that q S = S, then q is an upper-primary ideal of R with respect to S if and only if q is a primary ideal. Then principal ideal 2a R such that a is not 2n (n is any nonnegative integer ) is a an upper-primary ideal of R with respect to S ideal of R, but not a primary ideal. Remark 1.5.11 (1) Let q be a prime ideal of R. If q is an upper-prime ideal of R with respect to S, then q is an upper-primary ideal of R with respect to S. (2) Let q be an ideal of R such that q S = S. If q is an upper-primary ideal of R with respect to S, then q S ∩ R = q and q is a primary ideal of R. (3) Let q be an ideal of R and let q S = Q 1 ∩ · · · ∩ Q n ∩ T1 ∩ · · · ∩ Tm denote an irredundant primary decomposition of the ideal q S, where the Q i are the isolated primary components and the T j are the embedded primary components. Then q is an upper-primary ideal of R with respect to S if and only if Q i ∩ R = q for all Q i . Let α be an anti-integral element in K and put S := R[α]. We consider the upper-primary ideals of R with respect to S. Example 1.5.12 Let k be a field and R = k[x, y] be the polynomial ring in two indeterminates over k. Put α = y/x, S = R[α] and q = (x, y)2 R = (x 2 , x y, y 2 )R. Then we have that q S = (x 2 , x 2 α, x 2 α 2 )S = x 2 S and q S is a primary ideal of S. Since x 2 S∩ R = q, it follows that q is an upper-primary ideal of R with respect to S by Remark 1.5.11. Put q  = (x 2 , y 2 )R. Since q  S = x 2 S and q  S ∩ R = q = q  , we see that q  is not an upper-primary ideal of R with respect to S but both q and q  are primary ideals belonging to (x, y)R.

20

Birational Simple Extensions

The following are the properties of a primary ideal which is neither an upperprimary ideal of R with respect to S nor an upper-quasi-primary ideal of R with respect to S. Proposition 1.5.13 Let S = R[α] and α be an anti-integral element in K . Let p be a prime ideal of R. Assume that α = b/a(a, b ∈ R, a ∈ p, b ∈ p) and p + Jα = R. If q is a p-primary ideal, then q is neither an upper-primary ideal of R with respect to S nor an upper-quasi-primary ideal of R with respect to S. Proof Suppose that q is p-primary. Since q is a an upper-primary ideal of R n n n with respect to √ S, there exists a positive integer n such that b = a α ∈ q ⊆ q S. n Hence α ∈ q S. Now, since p + Jα = R, there exists a maximal ideal m of R such that p + Jα ⊆ m. Thus S/m S ∼ ], where√ T is an indeterminate. = (R/m)[T√ Therefore m S is a prime ideal of S. Since α n ∈ q S ⊆ pS ⊆ m S, it follows that α ∈ m S. Hence S/m S = R/m, it contradicts. The later part of Proposition 1.5.13 is clear from the above proof.

1.6

Some Subsets of Spec(R) in the Birational Case

Let R be a Noetherian integral domain with quotient field K , and let α be a nonzero element of K . Write A1 := R[α] (=: A), A2 := R[α −1 ], Iα := {r ∈ R | r α ∈ R}, and Jα = Iα + α Iα . We recall that α is an anti-integral element or that α is anti-integral over R if A1 ∩ A2 = R. When α is anti-integral over R, R[α] is said to be an anti-integral extension of R. Let ϕ Ai /R : Spec(Ai ) −→ Spec(R) be maps (i = 1, 2) which satisfy the condition ϕ Ai /R (P) = P ∩ R for any P ∈ Spec(Ai ). For any ideal N of R, we define V (N ) = {P ∈ Spec(R) | P ⊇ N } We write D(N ) to denote the set Spec(R) \ V (N ). Let  A/R := { p ∈ Spec(R)| p A = A} and  N := { p ∈ Spec(R)| p + N = R}. Theorem 1.6.1 hold:

Assume that α is anti-integral over R. The following results

(1)  A/R = V (Iα ) ∩  Jα ; (2) Im(ϕ A/R ) = D(Iα ) ∪ V (Jα ), where ϕ = ϕ A/R : Spec(A) → Spec(R) is a map such that ϕ A/R (P) = P ∩ R for any P ∈ Spec(A);

1.6 Some Subsets of Spec(R) in the Birational Case

21

(3)  A1 /R ∩  A2 /R = ∅; (4) Im(ϕ A1 /R ) ∪ Im(ϕ A2 /R ) = Spec(R). Proof (2) Suppose that p ∈ D(Iα ) ∪ V (Jα ). When p ∈ V (Jα ), we claim that p ∈ Im(ϕ). Indeed, we have that A/ p A ∼ = (R/ p)[T ] by Theorem 1.2.6, where T is an indeterminate. Thus p A is a prime ideal of A and p A ∩ R = p. Hence we have p ∈ Im(ϕ). In the case that p ∈ D(Iα ), it follows that A p = R p . Write p R p ∩ A = P. Then P ∈ Spec(A) and P ∩ R = p. So it follows that p ∈ Im(ϕ). Hence we have proved that D(Iα ) ∪ V (Jα ) ⊆ Im(ϕ). To prove the opposite inclusion, assume that p ∈ D(Iα ) ∪ V (Jα ). We claim that p ∈ Im(ϕ). Suppose that p ∈ Im(ϕ). Then there exists a prime ideal P of A such that P ∩ R = p, since p ⊇ Iα .

Chapter 2 Simple Extensions of High Degree

Let R be a Noetherian integral domain and R[X ] a polynomial ring. Let α be an element of an algebraic field extension L of the quotient field K of R and let π : R[X ] −→ R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with
d deg ϕα (X ) = d (R : R ηi ). For and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Let I[α] := i=1 f (X ) ∈ R[X ], let c( f (X )) denote the ideal generated by the coefficients of f (X ). Let J[α] := I[α] c(ϕα (X )), which is an ideal of R and contains I[α] . The element α is called an anti-integral element of degree d over R if Ker(π ) = I[α] ϕα (X )R[X ]. When α is an anti-integral element over R, R[α] is called an anti-integral extension of R. In the case K (α) = K , an anti-integral element α is the same as an anti-integral element (i.e., R = R[α] ∩ R[α −1 ]) defined in Definition 1.1.4. The element α is called a super-primitive element of degree d over R if J[α] ⊆ p for all primes p of depth 1, i.e., p ∈ Dp1 (R). For p ∈ Spec(R), k( p) denotes the residue field R p / p R p and rankk( p) R[α]⊗ R k( p) denotes the dimension as a vector space over k( p). In this chapter, we are mainly interested in characterizing the flatness and the integrality of an antiintegral extension R[α] of R. Among others we obtain the following results: R[α] is flat over R if and only if rankk( p) R[α] ⊗ R k( p) ≤ d for all p ∈ Spec(R). Concerning a super-primitive element, we obtain that if R is a Krull domain and α is an algebraic element over R, then α is a super-primitive element. We also obtain that a super-primitive element is an anti-integral element. More precisely, α is super-primitive over R if and only if α is anti-integral over R and R[α] p is flat over R p for any prime ideal p of depth 1.

Notations and Conventions Throughout this chapter, we use the following notations unless otherwise specified: R: a Noetherian integral domain, K = K (R): the quotient field of R,

23

24

Simple Extensions of High Degree L: an algebraic field extension of K , α: a nonzero element of L , d = [K (α) : K ], and ϕα (X ) = X d + η1 X d−1 + · · · + ηd , the minimal polynomial of α over K .

Let π : R[X ] −→ R[α] be an R-algebra homomorphism defined by X −→ α and let E (α) := Ker(π). Then E (α) is a prime ideal of R[X ] with E (α) ∩ R = (0). By definition, E
(α) = {ψ(X ) ∈ R[X ]|ψ(α) = 0}. d (R : R ηi ), which is an ideal of R. For f (X ) ∈ K [X ], Let I[α] := i=1 c( f (X )) := the ideal generated by all coefficients of f (X ), that is, c( f (X )) is the content ideal of f (X ). Let J[α] := I[α] c(ϕα (X )), which is an ideal of R and contains I[α] and let J˜ [α] := I[α] (1, η1 , . . ., ηd−1 ). We also use the standard notations as follows: k( p) :=the residue field R p / p R p for p ∈ Spec(R), Dp1 (R) := { p ∈ Spec(R) | depth(R p ) = 1}, and Ht1 (R) := { p ∈ Spec(R) | ht( p) = 1}.

2.1

Sharma Polynomials

We begin with the following definition. Definition 2.1.1 Let f (X ) = a0 X n + a1 X n−1 + · · · + an be a polynomial in R[X ]. We say that f (X ) is a Sharma polynomial in R[X ] if there does not exist t ∈ R with t ∈ a0 R such that tai ∈ a0 R for 1 ≤ i ≤ n. We give an equivalent condition for a polynomial to be a Sharma polynomial in the following proposition. Theorem 2.1.2 Let f (X ) be a polynomial in R[X ]. Then f (X ) is a Sharma polynomial if and only if c( f (X )) ⊆ p for any p ∈ Dp1 (R). Proof Let f (X ) = a0 X n + · · · + an (ai ∈ R). (⇒) Suppose that c( f (X )) ⊆ p for some p ∈ Dp1 (R). Then a0 ∈ p, and there exists t ∈ a0 R such that p = (a0 R : R t). In this case, ai ∈ p implies that ai t ∈ a0 R (1 ≤ i ≤ n), which asserts that f (X ) is not a Sharma polynomial.

2.1 Sharma Polynomials

25

(⇐) Suppose that f (X ) is not a Sharma polynomial. Then there exists t ∈ R such that t ∈ a0 R, tai ∈ a0 R (1 ≤ i ≤ n). Since there exists p ∈ Dp1 (R) such that (a0 R : R t) ⊆ p, we have ai ∈ (a0 R : R t) ⊆ p (1 ≤ i ≤ n) and obviously a0 ∈ p. So c( f (X )) = (a0 , . . ., an ) ⊆ p, a contradiction. Proposition 2.1.3 Let R[X ] be a polynomial ring over R and let P be a nonzero prime ideal in R[X ] such that P ∩ R = (0). Let f (X ) ∈ P be a polynomial of least positive degree. Then P = f (X )R[X ] if f (X ) is a Sharma polynomial. Proof Assume P = f (X )R[X ] and put f (X ) = a0 X d + · · · + ad . Suppose there exists t ∈ (a0 ) such that tai ∈ (a0 ) for 1 ≤ i ≤ d. We will arrive at a contradiction. Let tai = bi a0 with bi ∈ R (i = 1, . . ., d). Then a0 (t X d + b1 X d−1 + · · · + bd ) = t f (X ) ∈ P, which implies that t X d + b1 X d−1 + · · · + bd−1 X + bd ∈ P since P is prime and P ∩ R = (0). Thus, since the degree of f (X ) is d, there exists c ∈ R such that c f (X ) = t X d + b1 X d−1 + · · · + bd . Hence t = ca0 ∈ (a0 ), which contradicts the assumption that t ∈ (a0 ).

The converse statement of the above proposition holds (see [Sh]). Corollary 2.1.4 Let R be a factorial domain and P be a nonzero prime ideal in a polynomial ring R[X ] such that P ∩ R = (0). Then there exists an irreducible polynomial f (X ) with least positive degree among elements in P such that P = f (X )R[X ]. Proof Choose f (X ) ∈ P as in Proposition 2.1.3 such that the greatest common divisor of the coefficients a0 , . . ., ad of f (X ) is 1. Let there exist t ∈ a0 R such that tai ∈ a0 R for all i ≥ 1. Let tai = bi a0 , i = 1, . . ., d. This implies that t · GCD(a1 , . . ., ad ) = a0 · GCD(b1 , . . ., bd ). Hence a0 divides t since GCD(a0 , a1 , . . ., ad ) = 1, which is a contradiction to the assumption that t ∈ a0 R. Corollary 2.1.5 Let a, b be two nonzero elements of R. Then (a X + b)R[X ] is a prime ideal in R[X ] if and only if a, b is an R-sequence. Proof If (a X + b)R[X ] = P is a prime ideal, the proof is immediate by Proposition 2.1.3. Conversely, let a, b be an R-sequence. It is clear that a X + b generates a prime ideal Q in K [X ]. Let P = Q ∩ R[X ]. Then clearly P is

26

Simple Extensions of High Degree

a prime ideal in R[X ] containing a X + b such that P ∩ R = (0). The rest is immediate. Corollary 2.1.6 Let I be a nonzero ideal in R[X ] such that I ∩ R = (0). If there exists a polynomial f (X ) ∈ I such that f (X ) is of least positive degree in I and that c( f (X )) = R, then I is generated by the polynomial f (X ). Proof Let f (X ) = a0 X d + · · · + ad . Let t ∈ R be such that tai ∈ (a0 ) for all 1 ≤ i ≤ d, then clearly t ∈ (a0 ) because the content c( f (X )) = R. Now, the result follows as in Proposition 2.1.3. Corollary 2.1.7 Let R be a regular ring. Any prime ideal P of R[X ] such that P ∩ R = (0) is invertible in R[X ]. Proof The result will be proved by showing that P(R[X ]) M is principal for every maximal ideal M of R[X ]. Take any maximal ideal M in R[X ], then M ∩ R = Q is a prime ideal in R. We shall show that P R Q [X ] is principal. If Q = (0), then R Q = K , the field of fractions of R. In this case, we can see the result trivially. Let Q = (0). Then R Q is a factorial domain. Therefore the result follows from Corollary 2.1.4. Thus P is invertible in R[X ].

Theorem 2.1.8

The following statements are equivalent:

(i) E (α) is a principal ideal of R[X ]; (ii) I[α] is a principal ideal of R; (iii) there exists a Sharma polynomial in E (α) of degree d. If one of the above conditions holds, then E (α) is generated by a Sharma polynomial. Proof (iii) ⇒ (i): Let f (X ) be a Sharma polynomial in E (α) of degree d. Since deg ϕα (X ) = d, this Sharma polynomial has the least degree. Thus E (α) is principal by Proposition 2.1.3. (i) ⇒ (ii): Let E (α) = f (X )R[X ]. Then f (X )R[X ] ⊇ I[α] ϕα (X )R[X ]. Note that E (α) ⊗ R K = f (X )K [X ] = ϕα (X )K [X ] and hence that deg f (X ) = deg ϕα (X ) = d. Take a ∈ I[α] . Then aϕα (X ) = b f (X ). Let f (X ) = a0 X d + · · · + ad with ai ∈ R. Then a = ba0 , so that I[α] ⊇ a0 R for some b ∈ R. Since ba0 ηi = aηi = bai (1 ≤ i ≤ d), we have a0 ηi = ai ∈ R. Hence we have a0 ∈ I[α] , which implies that I[α] = a0 R.

2.2 Anti-Integral Elements and Super-Primitive Elements

27

(ii) ⇒ (iii): Let I[α] = b R. Then I[α] ϕα (X )R[X ] = bϕα (X )R[X ] ⊆ E (α) and bηi ∈ R (1 ≤ i ≤ d). Suppose that there exists t ∈ b R with tbηi ∈ b R (1 ≤ i ≤ d). Then tηi ∈ R and hence t ∈ I[α] = b R, a contradiction. Thus we conclude that bϕα (X ) ∈ R[X ] is a Sharma polynomial of degree d.

2.2

Anti-Integral Elements and Super-Primitive Elements

We extend the anti-integrality developed in Chapter 1 to the high degree case. Definition 2.2.1 (i) α ∈ L is called an anti-integral element of degree d over R if E (α) = I[α] ϕα (X )R[X ]. When α is an anti-integral element over R, we say that R[α] is an anti-integral extension of R. (ii) α ∈ L is called a super-primitive element of degree d over R if J[α] ⊆ p for all p ∈ Dp1 (R). When α is a super-primitive element, we say that R[α] is a super-primitive extension of R. Remark 2.2.2 (i) In Chapter 1, we studied the anti-integrality in the birational case, which is defined as follows: An element α ∈ K is called anti-integral over R if R = R[α] ∩ R[α −1 ] (:= R α ). We knew that α is anti-integral over R in this sense if and only if E (α) has a linear basis, that is,  E (α) = (ci X − di )R[X ] with di /ci = α (Proof of Lemma 1.1.9). The last condition is equivalent to E (α) = I[α] ϕα (X )R[X ], where ϕα (X ) = X − α. So α ∈ K is anti-integral over R in this sense if and only if α is an anti-integral element of degree 1 over R in the sense of Definition 2.2.1, that is, the anti-integrality defined in Definition 2.2.1 is equivalent to the one defined in Definition 1.1.4 in the case of degree 1 (i.e., birational). (ii) It is immediate that α ∈ L is a super-primitive element of degree d over R if and only if α is a super-primitive element of degree d over R p for any p ∈ Spec(R). Thus R[α] is a super-primitive extension of R if and only if R[α] p is a super-primitive extension of R p for all p ∈ Spec(R), where R[α] p denotes the localization S −1 R[α] with S = R \ p, that is, R[α] p = R[α] p . Proposition 2.2.3 Let α be an element in an algebraic field extension L of K with d = [K (α) : K ]. Let π : R[X ] −→ R[α] be the natural R-homomorphism. If Ker(π ) is generated by polynomials of degree d, then Ker(π)

28

Simple Extensions of High Degree

= I ϕα (X )R[X ] for some divisorial ideal I of R. Moreover I coincides with I[α] . Proof Assume that f i (X ) (1 ≤ i ≤ s) generate Ker(π ). Then f i (X ) = ai ϕα (X ) with some ai ∈ R. It is easy to see that ai η j ∈ R for all i, j. Hence
d we have ai ∈ i=1 Iηi = I[α] . Put I = (a1 , . . ., as )R. Then I ⊆ I[α] . We see that I ϕα (X )R[X ] ⊇ ( f 1 (X ), . . ., f s (X ))R[X ] = Ker(π ) ⊇ I[α] ϕα (X )R[X ]. Accordingly it follows that I = I[α] and Ker(π ) = I ϕα (X )R[X ]. Lemma 2.2.4 Let f (X ) be an element of a polynomial ring R[X ] and let p ∈ Spec(R). Then p ⊇ c( f (X )) if and only if R p [X ]/ f (X )R p [X ] is not flat over R p . Proof The implication (⇐) follows from [M1,(20.F)]. (⇒) Since c( f (X )) ⊆ p, it follows that p R[X ] contains f (X ) and hence that Q = p R[X ]/ f (X )R[X ] is a prime ideal of B := R[X ]/ f (X )R[X ]. Suppose that B p = R p [X ]/ f (X )R p [X ] is flat over R p . Then B Q is obtained from B p by localizing at Q B p . So depth(B Q ) ≥ depth(B p ), and so depth(B Q ) ≥ depth(R p ). It is easy to see that depth(B p B ) = depth(B Q ) and B p B = R[X ] p R[X ] / f (X )R[X ] p R[X ] . Since R is an integral domain, it follows that depth(B p B ) = depth(R[X ] p R[X ] ) − 1 = depth(R p ) − 1, which is a contradiction. What will be considered in this chapter is based on the following result. Theorem 2.2.5 Assume that α is an anti-integral element of degree d over R. Then for p ∈ Spec(R), the following are equivalent: (i) rankk( p) R[α] ⊗ R k( p) ≤ d; (ii) rankk( p) R[α] ⊗ R k( p) < ∞; (iii) R[α] ⊗ R k( p) is not isomorphic to a polynomial ring k( p)[T ]; (iv) J[α] ⊆ p; (v) p R[X ] ⊇ E (α) ; (vi) R[α] p is flat over R p . Proof Since α is an anti-integral element over R, we have E (α) = I[α] ϕα (X ) R[X ]. (iv) ⇒ (vi): Since R p = (J[α] ) p = (I[α] ) p c(ϕα (X )) p , the ideal (I[α] ) p is a principal ideal b R p for some b ∈ I[α] . So (E (α) ) p = bϕα (X )R p [X ]. We have R[α] p  R p [X ]/(E (α) ) p = R p [X ]/bϕα (X )R p [X ]. Thus R[α] p is flat over R p by Lemma 2.2.4 because R p = (J[α] ) p = c(bϕα (X )) p .

2.2 Anti-Integral Elements and Super-Primitive Elements

29

(iv) ⇒ (i): By the same argument as above, we have R[α] p  R p [X ]/(E (α) ) p = R p [X ]/bϕα (X )R p [X ]. Since R p = (J[α] ) p = c(bϕα (X )) p , there exists i (0 ≤ i ≤ d) such that bηi ∈ p R p [X ]. We take i minimal among such ones. Then bϕα (X ) = bX d + bη1 X d−1 + · · · + bηd ≡ bηi X d−i + · · · + bηd ≡ 0 (mod p R p [X ]), which means that rankk( p) R[α] ⊗ R k( p) ≤ d − i ≤ d. (i) ⇒ (ii) is trivial. (ii) ⇒ (iv): Note that R[α] p / p R[α] p  R p [X ]/( p R[X ] + E (α) ) p . The ideal ( p R[X ] + E (α) ) p contains an element f (X ) ∈ R[X ] such that c( f (X )) p = R p since rankk( p) R[α] ⊗ R k( p) < ∞. Indeed, if not, we conclude that R[α] ⊗ R k( p)  k( p)[T ], a polynomial ring, a contradiction. We may assume that f (X ) ∈ E (α) . So (E (α) ) p = I[α] ϕα (X )R p [X ] yields that (J[α] ) p = (I[α] ) p c(ϕα (X )) p = R p . (vi) ⇒ (iv): Suppose that J[α] ⊆ p. Localizing at p, we may assume that R is a local ring (R, m). Consider the exact sequence: 0 → E (α) → R[X ] → R[α] → 0. Then E (α) is flat over R because R[X ] and R[α] are flat over R. The isomorphism E (α) = I[α] ϕα (X )R[X ]  I[α] R[X ] yields that I[α] R[X ] is flat over R[X ] and hence I[α] is flat over R. Since R is local, I[α] = b R for some b ∈ I[α] . So J[α] = bc(ϕα (X )) and E (α) = bϕα (X )R[X ]. Thus c(bϕα (X )) ⊆ m, and hence R[α] is not flat over R by Lemma 2.2.4. (iv) ⇒ (v): Since J[α] = I[α] c(ϕα (X )) ⊆ p, there exists a ∈ I[α] such that ac(ϕα (X )) = c(aϕα (X )) ⊆ p. Thus aϕα (X ) ∈ p R[X ] and therefore E (α) ⊆ p R[X ]. (v) ⇒ (iv): Since E (α) = I[α] ϕα (X )R[X ], there exists a ∈ I[α] such that c(aϕα (X )) ⊆ p. So J[α] = I[α] c(ϕα (X )) ⊆ p. (v) ⇒ (iii): There exists f (X ) ∈ E (α) with f (X ) ∈ p R[X ]. So R[α]/ p R[α] = (R/ p)[α  ], where α  denotes the residue class of α in R[α]/ p R[α], and f (α  ) = 0. Thus α  is algebraic over R/ p. (iii) ⇒ (v): Suppose that E (α) ⊆ p R[X ]. Then R[α]/ p R[α] = (R[X ]/E (α) )/ ( p R[X ]/E (α) ) = R[X ]/ p R[X ] = (R/ p)[X ], which is a polynomial ring over R/ p. Recall here Definition 1.2.1 for convenience: Let A be an extension of R and let p ∈ Spec(R). We say that A is a blowingup at p or p is a blowing-up point of A/R if the following two conditions are satisfied: (i) p A p ∩ R p = p R p (equivalently p A ∩ R = p); (ii) A p / p A p is isomorphic to a polynomial ring (R p / p R p )[T ].

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Simple Extensions of High Degree

Making use of the above definition, we get the following corollary to Theorem 2.2.5. Corollary 2.2.6 When α is an anti-integral element over R, the blowing-up locus { p ∈ Spec(R) | p is a blowing-up point of R[α]} is given by V (J[α] ), and is the same as { p ∈ Spec(R) | R[α] p is not flat over R p }. Proof

This follows from Theorem 2.2.5 and Lemma 2.2.4.

The next proposition gives rise to the relation between Sharma polynomials and the ideal E (α) . Theorem 2.2.7 (a) R[α] is not a blowing-up at any point in Dp1 (R) if and only if E (α) contains a Sharma polynomial. (b) R[α] is not a blowing-up at any point in Spec(R) if and only if there exists a polynomial f (X ) in E (α) such that c( f (X )) = R. Proof (a) Take g0 (X ) ∈ E (α) \ (0). If g0 (X ) is a Sharma polynomial, then we are done. Suppose that g0 (X ) is not a Sharma polynomial. Let { p1 , . . ., pt } be the set of all members in Dp1 (R) satisfying c(g0 (X )) ⊆ pi . Such pi exists by Theorem 2.1.2. Since E (α) ⊆ p R[X ] for any p ∈ Dp1 (R), there are gi (X ) ∈ E (α) such that c(gi (X )) ⊆ pi (1 ≤ i ≤ t). Put N (0) := deg(g N (i) :=  0 (X )) and gi (X )X N (i) . Then N (i − 1) + deg(gi (X )) + 1 inductively. Let f (X ) := c( f (X )) = c(g0 (X )) + · · · + c(gt (X )). By the choice of pi , there does not exist p ∈ Dp1 (R) such that c( f (X )) ⊆ p. Whence f (X ) is a Sharma polynomial. Assume that E (α) contains a Sharma polynomial. Then E (α) ⊆ p R[X ] for any p ∈ Dp1 (R) by Theorem 2.1.2. So a blowing-up does not occur for R[α] on Dp1 (R) over R. (b) Let E (α) = ( f 1 (X ), . . ., f n (X ))R[X ]. Take p ∈ Spec(R). Then E (α) ⊆ p R[X ]. So there exists i such that c( f i (X )) ⊆ = 0 and N (i) = p. Put NN(0) f i (X )X (i) . Then c( f (X )) = N (i − 1) + deg( f i (X )) + 1, and let f (X ) = c( f 1 (X )) + · · · + c( f n (X )) = R. The converse is obvious. By the following theorem, we see that a super-primitive element is an antiintegral element. Theorem 2.2.8

The following statements are equivalent:

(i) α is a super-primitive element of degree d; (ii) α is an anti-integral element of degree d over R and R p [α] is flat over R p for all p ∈ Dp1 (R);

2.2 Anti-Integral Elements and Super-Primitive Elements

31

(iii) α is an anti-integral element of degree d over R and p R[X ] ⊇ E (α) for all p ∈ Dp1 (R); (iv) α is an anti-integral element of degree d over R and there exists a Sharma polynomial in E (α) ; (v) J[α] −1 = R, where J[α] −1 := R : K J[α] . Proof (i) ⇒ (ii): It is clear that I[α] ϕα (X )R[X ] ⊆ E (α) , and hence I[α] R[X ] ⊆ ϕα (X )−1 E (α) . Put J = ϕα (X )−1 E (α) . Let I[α] R[X ] = Q 1 ∩ · · · ∩ Q n be √ an irredundant primary decomposition of the ideal I[α] R[X ] and let Pi = Q i (1 ≤ i ≤ n). Assume that Q (resp. P) represents some Q i (resp. Pi ). Since I[α] is a divisorial ideal of R, I[α] R[X ] is a divisorial ideal of R[X ], and hence depth(R[X ] P ) = 1. Put p = P ∩ R. As p ⊇ I[α] , we see that p = (0). Thus we have P = p R[X ] and depth(R p ) = 1. Since α is a super-primitive element, J[α] ⊆ p by definition. Therefore there exists an element a ∈ I[α] such that (E (α) ) p = aϕα (X )R p [X ] by Theorem 2.1.8. Hence we have J p = a R p [X ] ⊆ I[α] R p [X ] ⊆ Q R p [X ]. Thus we get J ⊆ R[X ] ∩ Q R p [X ] = Q, that is, J ⊆ I[α] R[X ] because Q (resp. P, p) is every Q i (resp. Pi , pi := Pi ∩ R) for 1 ≤ i ≤ n. This implies that α is an anti-integral element. Hence the assertion follows from Theorem 2.2.5. (ii) ⇔ (iii) ⇔ (iv): These are immediate from Theorem 2.2.5 and Theorem 2.2.7. (iv) ⇒ (i): Since α is an anti-integral element, E (α) = I[α] ϕα (X )R[X ]. By Theorem 2.2.7 (a), we have E (α) ⊆ p R[X ] for all p ∈ Dp1 (R). Hence there exists an element a( p) ∈ I[α] such that f (X ) = a( p)ϕα (X ) and c( f (X )) ⊆ p. Thus J[α] ⊆ p for any p ∈ Dp1 (R). Therefore α is a super-primitive element over R. (i) ⇔ (v): Assume that J[α] ⊆ p for any p ∈ Dp1 (R). Then (J[α] −1 ) p = (R : K J[α] ) p = (R p : K (J[α] ) p ) = R p : K R p =
R p for any Dp1 (R). Since J[α] −1
p ∈ −1 is a divisorial ideal of R, we have R = R p = (J[α] ) p ⊇ J[α] −1 , where p ranges over the prime ideals of depth one. So it follows that R = J[α] −1 . Conversely, suppose that R = J[α] −1 and J[α] ⊆ p for some p ∈ Dp1 (R). Then J[α] −1 ⊇ p −1 and hence R = (J[α] −1 )−1 ⊆ ( p −1 )−1 = p, a contradiction. By the following result, we see that a super-primitive element is not so special. Theorem 2.2.9 Assume that R is a Krull domain. Then any element α which is algebraic over R is a super-primitive element over R. Proof Since R is a Krull domain, Dp1 (R) = Ht1 (R). Take p ∈ Ht1 (R). Then R p is a DVR. Let v denote the valuation corresponding to R p . Let ϕα (X ) =

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Simple Extensions of High Degree

X d + η1 X d−1 + · · · + ηd be the minimal polynomial of α. Put η0 = 1. Then there exists j such that v(η j ) ≤ v(ηi ) for all i. Thus ηi /η j = ai /b ∈ R p , where b ∈ R \ p, ai ∈ R. In particular, a j = b ∈ p. Hence ϕα (X ) = η j (a0 /η j )X d + · · · + η j (ad /η j )ηd Note that I[α] R p = Iη j R p by the choice of η j . Hence f (X ) := (b/η j )ϕα (X ) = a0 X d + · · · + ad ∈ Iη j ϕα (X )R p [X ] = I[α] ϕα (X )R p [X ]. Since a j = b ∈ p, we have c( f (X )) ⊆ p R p . Thus J[α] ⊆ p, which means that α is a super-primitive element over R. Once we find a super-primitive element, we can get many such elements. Indeed we obtain the following. Proposition 2.2.10 Assume that α is a super-primitive element of degree d over R. Then for any unit u of R and any element b ∈ R, β = uα + b is a super-primitive element of degree d over R. Proof We may assume that u = 1. It is clear that ϕβ (X ) = ϕα (X − b) because K (β) = K (α), d = deg ϕα (X − b), and ϕα (X − b) is monic in K [X ]. We see that I[α] ⊆ I[β] and c(ϕα (X )) = c(ϕα (X − b)) = c(ϕβ (X )). Since (J[α] ) p = (I[α] ) p c(ϕα (X )) p = R p for any p ∈ Dp1 (R) by Theorem 2.2.8, we have R p = (J[α] ) p ⊆ (J[β] ) p and hence (J[β] ) p = R p for any p ∈ Dp1 (R). Thus we conclude that β is a super-primitive element of degree d over R by Theorem 2.2.8. Proposition 2.2.11 Assume that R is a local ring containing an infinite field k and that J[α] = R. Then there exists an element λ ∈ k which satisfies the following three conditions: (a) 1/(α − λ) belongs to R[α]; (b) 1/(α − λ) is a super-primitive element of degree d over R; (c) 1/(α − λ) is integral over R. Proof Since R is local, there exists an element λ in k such that I[α] ϕα (X + λ) contains a degree d polynomial g(X ) in R[X ] of which constant term is 1. Put β = α − λ. Then g(β) = 0. Put h(X ) = X d g(1/ X ) ∈ R[X ]. Then h(β −1 ) = (β −1 )d g(β) = 0. So β −1 is integral over R. Since [K (α) : K ] = [K (β) : K ] = d, we conclude that ϕβ −1 (X ) = h(X ) ∈ R[X ]. Thus I[β −1 ] = R and hence J[β −1 ] = I[β −1 ] c(ϕβ −1 (X )) = R. In particular, β −1 is a super-primitive element of degree d over R by Theorem 2.2.8.

2.3 Integrality and Flatness of Anti-Integral Extensions

2.3

33

Integrality and Flatness of Anti-Integral Extensions

For p ∈ Spec(R), k( p) denotes the residue field R p / p R p and rankk( p) R[α]⊗ R k( p) denotes the dimension as a vector space over k( p). We are interested in characterizing the flatness and the integrality of an anti-integral extension R[α] of R. Among others, we obtain the following: R[α] is integral over R if and only if rankk( p) R[α] ⊗ R k( p) = d for all p ∈ Spec(R). Thus if an anti-integral extension R[α] is integral over R, then R[α] is flat over R (Theorem 2.2.5). The following result asserts that the integrality of an extension of R is determined by localizing at the prime ideals in Dp1 (R). Proposition 2.3.1 Let A be an integral domain containing R. Then A is integral over R if and only if A p (:= A ⊗ R R p ) is integral over R p for any p ∈ Dp1 (R). Proof The implication (⇒) is trivial. Now consider the converse. We assume that A p is integral over R p for any p ∈ Dp1 (R). It suffices to show that α is integral over R. Let R be the integral closure of R in K . Then R is a Krull domain [M1, p.144]. It suffices to show that α is integral over R. Let R˜ be the integral closure of R in K (A) and let C = R˜ : R˜ α, a denominator ideal ˜ Then K ( R) ˜ = K (A) and C is a divisorial ideal of R. ˜ There exists P ∈ of R. ˜ = Ht1 ( R) ˜ such that C ⊆ P. Since R/R ˜ Dp1 ( R) is integral and R is integrally closed in K , the Going-Down Theorem, holds between R˜ and R. Thus P ∩ R ∈ Ht1 (R) = Dp1 (R). In particular, P ∩ R is a divisorial ideal of R. So R˜ : R α = C ∩ R ⊆ P ∩ R ∈ Dp1 (R). By [F,(4.6)], (P ∩ R) ∩ R is a divisorial ideal of R. Hence R˜ : R α = (C ∩ R) ∩ R ⊆ (P ∩ R) ∩ R ∈ Dp1 (R). Put p = (P ∩ R) ∩ R. Then we have p ∈ Dp1 (R) and R˜ : R α ⊆ p, which is a contradiction. The integrality of anti-integral extensions is characterized as follows: Theorem 2.3.2 Assume that α is an anti-integral element of degree d over R. Then the following statements are equivalent: (i) R[α] is integral over R; (ii) ϕα (X ) ∈ R[X ]; (iii) I[α] = R;

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(iv) rankk(q) R[α] ⊗ R k(q) = d for any q ∈ Dp1 (R); (v) rankk(q) R[α] ⊗ R k(q) = d for any q ∈ Spec(R). Proof Since α is anti-integral, E (α) = I[α] ϕα (X )R[X ]. So the equivalence of (i), (ii), and (iii) are immediate because R[X ]/E (α)  R[α], and implications (ii) ⇒ (v) ⇒ (iv) are obvious. (iv) ⇒ (ii): Suppose that I[α] ⊆ p for some p ∈ Dp1 (R). Since J[α] = I[α] c(ϕα (X )) ⊆ p by Theorem 2.2.5, (I[α] ) p is an invertible ideal of R p and hence (I[α] ) p is a principal ideal b R p of R p for some b. Whence we have (E (α) ) p = (I[α] ) p ϕα (X )R p [X ] = (bϕα (X ))R p [X ]. Since I[α] ⊆ p, bϕα (X ) ∈ R p [X ] is not monic. Hence either R[α] ⊗ R k( p)  k( p)[T ], a polynomial ring or rankk( p) R[α] ⊗ R k( p) < d holds, a contradiction. By the above theorem, we see that the obstruction of integrality of anti-integral extensions is given by I[α] . Namely, we obtain the following. Corollary 2.3.3 Assume that α is an anti-integral element over R. Then V (I[α] ) = { p ∈ Spec(R)|R[α] p is not integral over R p }. Proof It is easy to see that the integrality is a local-global property. So our conclusion follows from Theorem 2.3.2. Remark 2.3.4 Let R be a Noetherian normal domain and let α be an element in a field L containing R. If α is integral over R, then R[α] is flat and superprimitive over R. Indeed, when ϕα (X ) ∈ K [X ] denotes the minimal polynomial of α over R, it is known that α is integral over R if and only if ϕα (X ) belongs to R[X ] ([M2, (9.2)]). Since R is normal, p ∈ Dp1 (R) ⇒ ht( p) = 1 ⇒ R p is a DVR. As R[α] is a finite R-module, R[α] p is free over R p for any p ∈ Dp1 (R). By Theorem 2.2.9, α is a super-primitive element over R. Moreover R[α] is flat over R by Theorems 2.2.5 and 2.3.2 because R[α] is a super-primitive, integral, and flat extension of R. Summing up the results in the preceding argument, we obtain the following: Assume that α is an anti-integral element of degree d. Let p be a prime ideal of R. Then (1) R[α] p is flat over R p if and only if rankk( p) R[α] ⊗ R k( p) ≤ d; (2) R[α] p is integral over R p if and only if rankk( p) R[α] ⊗ R k( p) = d. In particular, we conclude:

2.3 Integrality and Flatness of Anti-Integral Extensions

35

Corollary 2.3.5 Assume that α is an anti-integral element of degree d. If R[α] is integral over R, then R[α] is flat over R. In view of Theorem 2.2.7, we extend Theorem 2.2.5 to the following. Theorem 2.3.6 Assume that α is an anti-integral element of degree d over R. Then the following are equivalent: (i) R[α] is flat over R; (ii) J[α] = R; (iii) rankk( p) R[α] ⊗ R k( p) < ∞ for any p ∈ Spec(R); (iv) rankk( p) R[α] ⊗ R k( p) ≤ d for any p ∈ Spec(R); (v) R[α] is not a blowing-up at any point in Spec(R); (vi) R[α] is quasi-finite over R; (vii) E (α) contains a polynomial f (X ) with c( f (X )) = R. Proof

The proof follows from Theorems 2.2.5 and 2.2.7(b).

Remark 2.3.7 Let A be an over-ring of R (i.e., R ⊆ A and K (A) = K ). If A is integral and flat over R on Dp1 (R), then A = R. Indeed, it is known that
R = p∈Dp (R) R p . For p ∈ Dp1 (R), A p is integral and flat over R p by the 1 assumption. So A p is a free R p -module of rank 1. Thus A p = R p and hence
R = p∈Dp (R) R p ⊇ A, which means that R = A. 1

Relating to this remark, we have the following. Theorem 2.3.8 Let α be an algebraic element over R. If R[α] is integral and flat at any point in Dp1 (R), then R[α] is a free R-module and α is a superprimitive element over R. In particular, if α is both anti-integral and integral over R, then R[α] is a free R-module R ⊕ Rα ⊕ · · · ⊕ Rα d−1 . Proof First, we shall show that I[α] = R. Suppose that I[α] = R. Then there exists p ∈ Dp1 (R) such that I[α] ⊆ p because I[α] is a divisorial ideal of R. Since R[α] p is integral over R p by assumption, R[α] p is a flat extension of R p . As R[α] p is flat over R p , R[α] p is a free R p -module of rank d. Here we want to show that R[α] p = R p + R p α + · · · + R p α d−1 . To this end, we have only to show that 1 , α  , . . ., α  d−1 ∈ R[α] p / p R[α] p are linearly independent over k( p), where α  denotes the residue class of α in R[α] p / p R[α] p . [Suppose the

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Simple Extensions of High Degree

contrary. Then R[α] p / p R[α] p = k( p)[α  ] = k( p) + k( p)α  + · · · + k( p)α  s for some s < d. But R[α] p is a free R p -module of rank d, which asserts that rankk( p) R[α] p / p R[α] p = d, a contradiction.] Thus we have shown that R[α] p = R p + R p α + · · · + R p α d−1 . So we have a relation: α d = λ0 + λ1 α + · · · + λd−1 α d−1 (λi ∈ R p ). Since the minimal monic polynomial ϕα (X ) of α is unique, we have ϕα (X ) = X d − λd−1 X d−1 − · · · − λ0 . So I[α] ⊆ p, a contradiction. Thus ϕα (X ) ∈ R[X ], which implies that E (α) = ϕα (X )R[X ] and R[α] is free R-module. Since c(ϕα (X )) = R, we conclude that J[α] = R, which asserts that α is a super-primitive element over R by Theorem 2.2.8. Now we consider a certain over-ring of R, which is seen in Chapter 1. Let J be a fractional ideal of R. Recall R(J ) := J : K J, which is an over-ring of R. Lemma 2.3.9 Let J be a divisorial ideal of R. Then R(J ) = R if and only if R(J −1 ) = R. Proof Since J is divisorial, we have (J −1 )−1 = J. So it suffices to prove one of the implications. Assume that R(J ) = R. The implication R(J −1 ) ⊇ R is obvious. Take λ ∈ R(J −1 ). Then λJ −1 ⊆ J −1 . Thus R : K λJ −1 ⊇ R : K J −1 = (J −1 )−1 = J. On the other hand, we have R : λJ −1 = λ−1 R : K J −1 = λ−1 (R : K J −1 ) = λ−1 (J −1 )−1 = λ−1 J. Thus λ−1 J ⊇ J, which shows that J ⊇ λJ, and hence λ ∈ R(J ) = R. By these arguments, we extend Theorem 2.2.8 as follows: Theorem 2.3.10

The following conditions are equivalent:

(i) α is a super-primitive element over R; (ii) for each p ∈ Dp1 (R), there exists f (X ) ∈ E (α) with (E (α) ) p = f (X )R p [X ]; (iii) for each p ∈ Dp1 (R), there exists a ∈ I[α] with (I[α] ) p = a R p ; (iv) R(I[α] ) = R. Proof Denote the degree of α by d. (i) ⇒ (ii) : Since J[α] = I[α] c(ϕα (X )) ⊆ p for any p ∈ Dp1 (R), there exists a ∈ I[α] with f (X ) := aϕα (X ) ∈ p R[X ]. Note that (E (α) ) K ∩ R p [X ] = (E (α) ) p and f (X ) ∈ (E (α) ) p . By Theorem 2.1.2, f (X ) is a Sharma polynomial of degree d in R p [X ]. So (E (α) ) p = f (X )R p [X ]. (ii) ⇒ (iii): Suppose that (E (α) ) p = f (X )R p [X ]. Then deg f (X ) = d. Let a be the leading coefficient of f (X ). Then ϕα (X ) = (1/a) f (X ) by the uniqueness

2.4 Anti-Integrality of α and α −1

37

of the minimal polynomial of α. So f (X ) = aϕα (X )R[X ], and hence a ∈ I[α] . Since (E (α) ) p = f (X )R p [X ], (I[α] ) p = a R p . (iii) ⇔ (iv): We know that R(I[α] ) = R if and only if R(I[α] −1 ) = R by Lemma 2.3.9. So apply the result of Proposition 1.3.2 and we conclude that (iii) and (iv) are equivalent. (iii) ⇒ (i): Since (I[α] ) p is a principal ideal of R p for any p ∈ Dp1 (R), there exists f (X ) ∈ E (α) such that deg f (X ) = d and (E (α) ) p = f (X )R p [X ]. Since f (X ) is a Sharma polynomial in R p [X ] by Theorem 2.1.2 and depth(R p ) = 1, c( f (X )) ⊆ p. Thus J[α] ⊆ p for any p ∈ Dp1 (R) and hence α is a superprimitive element over R by definition.

2.4

Anti-Integrality of α and α −1

One objective of this section is to show that the anti-integrality is also characterized by the set Dp1 (R). In fact, we prove the following statements are equivalent: (1) α is an anti-integral element over R; (2) α is an anti-integral element over R p for all p ∈ Dp1 (R). The other objective is to study several conditions equivalent to the statement that α is an anti-integral element of degree d over R by use of the ring R[α] ∩ R[α −1 ]. Put R α := R[α] ∩ R[α −1 ]. We begin with the following theorem, which characterizes anti-integrality. Theorem 2.4.1

The following statements are equivalent:

(1) α is an anti-integral element of degree d over R; (2) the ideal I[α] ηd of R is generated by the set {g(0) | g(X ) ∈ E (α) }. Proof (1) ⇒ (2): Let J be the ideal of R generated by the set {g(0) | g(X ) ∈ E (α) }. Since I[α] ϕα (X ) ⊆ E (α) and the constant term of I[α] ϕα (X ) is I[α] ηd , it follows that I[α] ηd ⊆ J . Conversely take a ∈ J , and let an α n + an−1 α n−1 + · · · + a1 α + a = 0 be a relation, where ai ∈ R and an = 0. Put f (X ) = an X n + · · · + a1 X + a. element of degree d over R, we Then f (X ) ∈ E (α) . Since α is an anti-integral have E (α) = I[α] ϕα (X )R[X ]. Hence f (X ) = h i (X )gi (X ) for some h i (X ) ∈

38

Simple Extensions of High Degree

 I[α] ϕα (X ) and gi (X ) ∈ R[X ]. Thus a = f (0) = h i (0)gi (0) ∈ I[α] ηd , as desired. (2) ⇒ (1): Let 0 = f (X ) ∈ E (α) and write f (X ) = an X n + · · · + a1 X + a. Since [K (α) : K ] = d, we have n ≥ d. By the assumption that a ∈ J = I[α] ηd , it follows that a = bηd for some b ∈ I[α] . Put g(X ) = bX d + (bη1 )X d−1 + · · · + (bηd ). Note that g(X ) ∈ E (α) . As f (0) = g(0), we get f (X ) − g(X ) = X (h(X )) ∈ E (α) for some h(X ) ∈ R[X ]. Since R[α] is an integral domain, E (α) is a prime ideal of R[X ], and hence h(X ) ∈ E (α) . Since deg h(X ) ≤ n − 1, we can prove f (X ) ∈ I[α] ϕα R[X ] by induction. Therefore α is an anti-integral element of degree d over R. Remark 2.4.2 A similar result to Theorem 2.4.1 is valid when R is a Noetherian reduced ring and α is algebraic over R and is not a zero-divisor in R[α]. Under this preparation, we obtain the following result mentioned before. Theorem 2.4.3

The following statements are equivalent to each other:

(1) α is an anti-integral element of degree d over R; (2) α is an anti-integral element of degree d over R p for all p ∈ Dp1 (R). Proof

(1) ⇒ (2): By the assumption, we have the following exact sequence: 0 → I[α] ϕα (X )R[X ] → R[X ] → R[α] → 0

Take p ∈ Dp1 (R). Tensoring − ⊗ R R p , we have an exact sequence: 0 → I[α] ϕα (X )R p [X ] → R p [X ] → R p [α] → 0 This exact sequence implies that α is an anti-integral element of degree d over Rp. (2) ⇒ (1): Consider the following canonical exact sequence: 0 → E (α) → R[X ] → R[α] → 0 Let J denote the ideal generated by the set {g(0) | g(X ) ∈ E (α) }. We need to show that I[α] ηd = J by (2). Since I[α] ϕα (X ) ⊆ E (α) , we have I[α] ηd ⊆ J . We shall show the converse inclusion. Since α is an anti-integral element of degree d over R p by assumption, we conclude that (I[α] ηd ) p = J p for all p ∈ Dp1 (R) by (2). Thus we have J ⊆ J p = (I[α] ηd ) p . Let q ∈ Spec(R) be a prime divisor of I[α] ηd . Since I[α] ηd is a
divisorial ideal, we see that q ∈ Dp1 (R) ([Y, Proposition 5.6]). Hence J ⊆ p∈Dp (R) (I[α] ηd ) p = I[α] ηd 1

([Y, Proposition 5.6]), which completes the proof.

2.4 Anti-Integrality of α and α −1

39

The rest of this section, we investigate several conditions for α to be an antiintegral element over R by use of the properties of the ring R α = R[α] ∩ R[α −1 ]. Lemma 2.4.4

The following assertions are valid:

(1) if {a ∈ R | aα ∈ R α } = I[α] , then α is an anti-integral element of degree d over R; (2) if α −1 is an anti-integral element of degree d over R, then {a ∈ R | aα ∈ R α } = I[α] . Proof (1) Take f (X ) ∈ R[X ] satisfying f (α) = 0, i.e., f (X ) ∈ E (α) . We must show that f (X ) ∈ I[α] ϕα (X )R[X ] by induction on deg f (X ). Put deg f (X ) = n. If n = d, then f (X ) ∈ I[α] ϕα (X )R[X ]. Suppose that n > d. Put f (X ) = an X n + an−1 X n−1 + · · · + a0 . Then an α + an−1 ∈ R[α] ∩ R[α −1 ]. By induction hypothesis, we have an ∈ I[α] . Put g(X ) = f (X ) − an X n−d ϕα (X ) ∈ R[X ]. We see that g(α) = 0 and deg g(X ) ≤ n − 1. By induction hypothesis, we have that g(X ) ∈ I[α] ϕα (X )R[X ]. Hence f (X ) ∈ I[α] ϕα (X )R[X ]. Since I[α] R[X ] ⊆ E (α) is obvious, we have E (α) = I[α] ϕα (X )R[X ]. Therefore α is an anti-integral element of degree d over R. (2) To prove the opposite inclusion, take a ∈ S, then aα ∈ R α . Thus −aα = c0 + c1 α −1 + · · · + cn (α −1 )n where c0 , c1 , . . ., cn are elements of R. Hence cn (α −1 )n+1 + · · · + c0 α −1 + a = 0. Since α −1 is an anti-integral element of degree d  over R, there exist Fi (X ) ∈ R[X ] and G i (X ) ∈ R[X ] such that −1 f (X ) = F i (X )G i (X ) where deg Fi (X ) = d and Fi (α ) = 0. Therefore d a = f (0) = Fi (0)G i (0). Let Fi (X ) = ad X + · · · + a1 X + a0 . It follows d d−1 that ad (α −1 )d + · · · + a1 α −1 + a0 = 0. Hence + · · · + ad = 0.
da0 α + a1 α Since ηi = ai  /a0 , we see that a0 = Fi (0) ∈ i=1 Iηi = I[α] . Consequently, we get that a = Fi (0)G i (0) ∈ I[α] . Take any element a of I[α] . Put ai = aηi . Then we have that aα d + a1 α d−1 + · · · + ad = 0 (ai ∈ R). Hence aα = −(a1 + a2 α −1 + · · · + ad (α −1 )d−1 ) ∈ R α . Putting S = {a ∈ R | aα ∈ R α }, we see that I[α] ⊆ S. By use of the above lemma, we have the following result. Theorem 2.4.5 An element α is an anti-integral element of degree d over R if and only if so is α −1 . Proof Suppose that α −1 is anti-integral. By Lemma 2.4.4(2), it follows that {a ∈ R | aα ∈ R[α] ∩ R[α −1 ]} = I[α] . By Lemma 2.4.4(1), α is anti-integral. By this result, the converse is also valid.

40

Simple Extensions of High Degree

Another Proof of Theorem 2.4.5 Let πα : R[X ] → R[α], πα−1 : R[X −1 ] → R[α −1 ] and π : R[X, X −1 ] → R[α, α −1 ] be R-algebra homomorphism sending X to α. Then we have the following commutative diagram: 0

→

0

→

0

→

Ker(π) ∩ R[X −1 ] ∩ Ker(π) ∪ Ker(π) ∩ R[X ]

→ → →

R[X −1 ] ∩ R[X, X −1 ] ∪ R[X ]

→ → →

R[α −1 ] ∩ R[α, α −1 ] ∪ R[α]

→

0

→

0

→

0

It is easy to see that Ker(πα−1 ) = Ker(π ) ∩ R[X −1 ] and Ker(πα ) = Ker(π ) ∩ R[X ]. It is also easy to see that Ker(π ) = (Ker(π ) ∩ R[X ])R[X, X −1 ] = (Ker(π) ∩ R[X −1 ])R[X, X −1 ]. Now we claim: I[α−1 ] ϕα−1 (X −1 )R[X, X −1 ] = I[α] ϕα (X )R[X, X −1 ] [Indeed, we see that Iα−1 ϕα−1 (X −1 ) = ηd I[α] ηd−1 ϕα (X )X −d . Hence it follows that Iα−1 ϕα−1 (X −1 )R[X, X −1 ] = I[α] ϕα (X )X −d R[X, X −1 ] = I[α] ϕα (X )R[X, X −1 ].] Since α is anti-integral over R, we have Ker(πα ) = I[α] ϕα (X )R[X ] Thus =

Ker(πα−1 )R[X, X −1 ] (Ker(π ) ∩ R[X −1 ])R[X, X −1 ]

=

I[α] ϕα (X )R[X, X −1 ]

=

I[α−1 ] ϕα−1 (X −1 )R[X, X −1 ]

Hence

=

Ker(πα−1 ) Ker(πα−1 )R[X, X −1 ] ∩ R[X −1 ] I[α−1 ] ϕα−1 (X −1 )R[X, X −1 ] ∩ R[X −1 ]

=

I[α−1 ] ϕα−1 (X −1 )R[X −1 ]

=

Therefore we conclude that Ker(πα−1 ) = I[α−1 ] ϕα−1 (X −1 )R[X −1 ], which means that α −1 is anti-integral over R. Now we shall see that an analogous result of Theorem 2.4.5 holds for superprimitiveness.
d Note that I[α] = i=1 Iηi . Since α d + η1 α d−1 + · · · + ηd = 0, we have −1 1 + η1 (α ) + · · · + ηd (α −1 )d = 0, and hence (α −1 )d + ηd−1 ηd−1 (α −1 )d−1 +

2.4 Anti-Integrality of α and α −1

41

· · · + ηd−1 = 0. So ϕα−1 (X ) = X d + ηd−1 ηd−1 C d−1 + · · · + ηd−1 η1 X + ηd−1 . Hence
d−1 I[α−1 ] = ( i=1 Iηd−1 ηi ) ∩ Iηd−1 . Lemma 2.4.6

For an element α ∈ L , I[α−1 ] = ηd I[α] .

Proof Take ηd x ∈ ηd I[α] with x ∈ I[α] . Then ηd−1 · ηd x = x ∈ R and ηd−1 ηi · ηd x = xηi ∈ R. Thus ηd x ∈ I[α−1 ] . Conversely, take y ∈ I[α−1 ] =
d−1 Iηd−1 ηi ) ∩ Iηd−1 . Then ηd−1 y ∈ R. Put x := ηd−1 y. Then xηd = y ∈ R. We ( i=1 have that xηi = ηd−1 ηi ηd x = (ηd−1 ηi )y ∈ R. So we have x ∈ I[α] . The reverse implication is clear. Therefore I[α−1 ] = ηd I[α] . For an element α ∈ L, J[α] = J[α−1 ] .

Proposition 2.4.7 Proof J[α]

Theorem 2.4.8 so is α −1 .

=

I[α] (1, η1 , . . ., ηd )

=

I[α] + I[α] η1 + · · · + I[α] ηd

=

ηd−1 (ηd I[α] ) + ηd−1 η1 (ηd I[α] ) + · · · + (ηd I[α] )

=

ηd−1 I[α−1 ] + ηd−1 η1 I[α−1 ] + · · · + I[α−1 ]

=

I[α] (1, ηd−1 ηd−1 , . . ., ηd−1 η1 , ηd−1 )

=

J[α−1 ]

It holds that α ∈ L is super-primitive over R if and only if

Proof The element α is super-primitive over R if and only if J[α] ⊆ p for any p ∈ Dp1 (R). Since J[α] = J[α−1 ] by Proposition 2.4.7, J[α−1 ] ⊆ p for any p ∈ Dp1 (R), which means that α −1 is super-primitive over R. Corollary 2.4.9 Assume that α is anti-integral over R. Then R[α] is flat over R if and only if R[α −1 ] is flat over R. Proof The extension R[α] is flat over R if and only if J[α] = R by Theorem 2.3.6. Since J[α−1 ] = J[α] = R, R[α −1 ] is flat over R.

42

2.5

Simple Extensions of High Degree

Vanishing Points and Blowing-Up Points

Assume that α is an anti-integral element over R. For p ∈ Spec(R), rankk( p) R[α] ⊗ R k( p) < ∞ if and only if R[α] p is flat over R p by Theorem 2.2.5. So it may be natural to ask when rankk( p) R[α] ⊗ R k( p) is infinite or zero. Let α be an element which is algebraic over R. Recall that ϕα (X ) = X d + η1 X d−1 + · · · + ηd is the minimal polynomial of α over K , where d = [K (α) : K ] and that J[α] := I[α] c(ϕα (X )) = I[α] + I[α] η1 + · · · + I[α] ηd . Let J˜ [α] := I[α] (1, η1 , . . ., ηd−1 ), which is an ideal of R. Recall that E (α) = Ker(π ), where π : R[X ] → R[α] is an R-algebra homomorphism sending X to α. We use this notation throughout this section. Lemma 2.5.1 Assume that α is an anti-integral element over R. For q ∈ Spec(R), the following are equivalent: (i) q R[α]q = R[α]q ; (ii) q R[α] ∩ R ⊆ q; (iii) q ⊇ J˜ [α] and q ⊇ I[α] ηd . Proof (i) ⇒ (ii): Since q R[α] βi ∈ R[α]  q = R[α]q , there exist ai ∈ q,  and si ∈ R \ q such that 1 = ai βi /si . Put s = si . Then s = ai βi bi ∈ q R[α] ∩ R with s ∈ q, where sβi /si = bi ∈ R[α]. Thus q R[α] ∩ R ⊆ q. (ii) ⇒ (i): Take s ∈ q R[α] ∩ R with s ∈ q. Then s ∈ q R[α]q and s is invertible in R[α]q . Thus q R[α]q = R[α]q . (iii) ⇒ (ii): Take a ∈ I[α] with aηd ∈ q. Put f (X ) = aϕα (X ) and aηi = bi , a = b0 . Then f (X ) = b0 X d + b1 X d−1 + · · · + bd . Since f (α) = 0, we have b0 α d + b1 α d−1 + · · · + bd = 0. Noting that bd ∈ q, bd is a unit in R[α]q . Since b0 , . . ., bd−1 ∈ q, bd ∈ q R[α] ⊆ q R[α]q . Thus q R[α]q = R[α]q . (ii) ⇒ (iii): Since q R[α]q = R[α]q , 1 = b0 + b1 α + · · · + bn α n for some bi ∈ q Rq . Put f (x) = bn X n + · · · + b1 X + b0 − 1. Then f (α) = 0 and b0 − 1 is a unit in Rq . The kernel of Rq [X ] −→ R[α]q is (I[α] )q ϕα (X )Rq [X ]. So f (X ) ∈ (I[α] )q ϕα (X )Rq [X ] and c( f (X ))q = Rq . Thus it follows that (J[α] )q = (I[α] )q c(ϕα (X ))q = Rq , which means that R[α]q is flat over Rq by Theorem 2.2.5. Thus (I[α] )q ϕα (X )Rq [X ] is an invertible ideal of Rq [X ]. Hence (I[α] )q is a principal ideal of Rq . Let (I[α] )q = a Rq . We shall show that all of a, aη1 , . . ., aηd−1 belong to q Rq . Note that f (X ) ∈ aϕα (X )Rq [X ] because f (α) = 0. So there exists h(X ) ∈ Rq [X ] such that f (X ) = aϕα (X )h(X ).

2.5 Vanishing Points and Blowing-Up Points

43

We have −1 ≡ aϕα (X )h(X ) (mod q Rq [X ]). Thus aηi , a ∈ q Rq , for 1 ≤ i ≤ d − 1 and aηd ∈ q Rq . Therefore I[α] , I[α] η1 , . . ., I[α] ηd−1 ⊆ q and I[α] ηd ⊆ q.

Definition 2.5.2 Let A be an extension of R. We say that a prime ideal p of R is a vanishing point of R with respect to A if p A p = A p . Recall that A is a blowing-up at p or p is a blowing-up point of R with respect to A if the following two conditions are satisfied: (i) p A p ∩ R p = p R p ; (ii) A p / p A p is isomorphic to a polynomial ring (R p / p R p )[T ]. By Lemma 2.5.1, we obtain the following theorem. Theorem 2.5.3 Assume that α is an anti-integral element over R. Then the set of vanishing points of R with respect to A (i.e., {q ∈ Spec(R) | q R[α]q =
d−1 R[α]q }) is given by V ( J˜ [α] ηi ) \ V (I[α] ηd ) = i=0 V (I[α] ηi ) \ V (I[α] ηd ), where η0 = 1. Proposition 2.5.4 Assume that α is an anti-integral element of degree d over R. Consider the following conditions: (i) R[α] is flat over R; (ii) J[α] = R; (iii) If p R[α] p = R[α] p for p ∈ Spec(R), then p R[α] = R[α]. Then we have implications (i) ⇔ (ii) ⇒ (iii). If moreover R is a local ring and  J˜ [α] ⊇ I[α] ηd , then (i), (ii), and (iii) are equivalent to each other. Proof (i) ⇔ (ii) was proved in Theorem 2.3.6. (ii) ⇒ (iii): Take p ∈ Spec(R) and assume that p R[α] p = R[α] p . Then p ⊇ J˜ [α] = I[α] + I[α] η1 + · · · + I[α] ηd−1 and p ⊇ I[α] ηd by Lemma 2.5.1. Take a ∈ I[α] and put f (X ) = aϕα (X ) = a X d + aη1 X d−1 + · · · + aηd . Since f (α) = 0, we get aηd ∈ p R[α] and hence I[α] ηd ⊆ p R[α]. So J[α] = J˜ [α] + I[α] ηd ⊆ p R[α]. Since J[α]= R, we conclude that p R[α] = R[α]. We will show the last part. Since J˜ [α] ⊇ I[α] ηd , there exists q ∈ Spec(R) such that q ⊇ J˜ [α] but q ⊇ I[α] ηd . Thus q R[α]q = R[α]q and so q R[α] = R[α]. Let m denote the maximal ideal of R. Suppose that m ⊇ J[α] . Then we have R[α]/m R[α]  (R/m)[T ], a polynomial ring (cf. Theorem 2.2.5). Hence m R[α] = R[α]. But q ⊆ m implies that m R[α] = R[α], a contradiction. Thus we conclude that J[α] = R.

44 Remark 2.5.5

Simple Extensions of High Degree Let the notation be the same as in Proposition 2.5.4.

(1) When d = 1 (i.e., α is an anti-integral element of R), then (i), (ii), and (iii) of Proposition 2.5.4 are equivalent. (2) p R[α] ∩ R = p if and only if there exists P ∈ Spec(R[α]) such that P ∩ R = p. Remark 2.5.6 Let the notation be the same as in Lemma 2.5.1. If J˜ [α] ⊆ q, then q is either a vanishing point (i.e.,  I[α] ηd ⊆ q) or a blowing-up point (i.e.,  I[α] ηd ⊆ q). So if J[α] contains J˜ [α] properly, there exists a vanishing point. Thus Spec(R[α]) −→ Spec(R) is not surjective. Proposition 2.5.7 Assume that α is an anti-integral elementof degreed over R. Then Spec(R[α]) −→ Spec(R) is surjective if and only if J[α] = J˜ [α] .   Proof (⇒): Since J[α] ⊇ J˜ [α] , J˜ [α] . If J˜ [α] ⊆ q for some J[α] ⊇ q ∈ Spec(R), there exists Q ∈ Spec(R[α]) such that Q ∩ R = q because Spec(R[α]) −→ Spec(R) is surjective. So q R[α]q = R[α]q , which means that q is not a vanishing point.  Remark 2.5.6, q is a blowing-up point, that  Thus by is, q ⊇ J[α] . Therefore J[α] = J˜ [α] . (⇐): Suppose that Spec(R[α]) −→ Spec(R) is not surjective. There  exists q ∈ Spec(R) such that q R[α]q = R[α]q . So it follows that q ⊇ J˜ [α] =  J[α] ⊇ J[α] ⊇ I[α] ηd , a contradiction. Proposition 2.5.8 Let the notation be the same as in Proposition 2.5.7 and let p ∈ Spec(R) satisfy p R[α] p = R[α] p . If q ⊇ p R[α] ∩ R, then q is a blowing-up point of R with respect to R[α]. Proof Since p ∈ Spec(R) satisfies p R[α] p = R[α] p , we have p ⊇ J˜ [α] by Lemma 2.5.1. Thus ηd I[α] ⊆ α d I[α] + · · · + ηd−1 α I[α] ⊆ J˜ [α] R[α] ⊆ p R[α]. So q ⊇ p R[α] ∩ R ⊇ J˜ [α] + I[α] ηd = J[α] , which means that q is a blowing-up point. Remark 2.5.9 Let k be a field, a, b indeterminates, and R = k[a, b]. Let α be a root of an equation a X 2 + bX + a = 0. Then J[α] = (a, b)R and grade((a, b)R) = 2 so that α is a super-primitive element by Definition 2.2.1. In this case, J[α] = J˜ [α] = (a, b)R. Thus Spec(R[α]) −→ Spec(R) is surjective, but not flat. Hence the implication (iii) ⇒ (i) in Proposition 2.5.4 does not necessarily hold.

2.5 Vanishing Points and Blowing-Up Points

45

Theorem 2.5.10 Assume that α is an anti-integral element over R and let p ∈ Spec(R). If R[α] is not a blowing-up at q, then depth(R[α] Q ) = depth(Rq ) for Q ∈ Spec(R[α]) with Q ∩ R = q. Proof Since α is an anti-integral element over R and q is not a blowing-up point of R with respect to R[α], R[α]q is flat over Rq by Corollary 2.2.6. Since R[α] Q is obtained from R[α]q by localizing at Q R[α]q , R[α] Q is flat over Rq . So we have depth(Rq ) ≤ depth(R[α] Q ). As q is not a blowing-up point, there exists a ∈ I[α] such that aϕα (X )Rq [X ] = (E (α) )q . Put f (X ) := aϕα (X ). Since Q ∈ Spec(R[α]), there exists P ∈ Spec(R[X ]) such that P ⊇ E (α) and Q = P/E (α) . Then Q q = Pq /(E (α) )q = Pq / f (X )Rq [X ]. Since Q R[α] Q = P R[X ] P / f (X )R[X ] P , it follows that depth(R[α] Q ) = depth(R[X ] P ) −1. Now since P ∩ R = q, we have P ⊇ q R[X ]. Suppose that P = q R[X ]. Then q R[X ] = P ⊇ E (α) , which asserts that q is a blowing-up point. So we have P = q R[X ]. Since P Rq [X ]/q Rq [X ](⊆ k(P)[X ]) = 0, we have P Rq [X ] = q Rq [X ]+ g(X )Rq [X ] for some g(X ) ∈ R[X ]\q R[X ]. Hence depth(R[X ] P ) ≤ depth(R[X ]q R[X ] ) + 1. We obtain that depth(R[α] Q ) ≤ depth(Rq ) because depth(R[X ]q R[X ] ) = depth(Rq ). Thus depth(Rq ) = depth(R[α] Q ). Proposition 2.5.11 Assume that α is an anti-integral element over R. Then the following statements are equivalent: (i) I[α] R[α] = R[α]; (ii) ηd ∈ R[α], I[α] = Iηd , and Jηd = R. Proof If I[α] = R, then the implications (i) ⇔ (ii) are trivially valid because ηd ∈ R by the definition of I[α] . So we may assume that I[α] = R. −1 −1 R[α] = I[α] I[α] R[α] ⊆ R[α], we have (i) ⇒ (ii): Since η1 , . . ., ηd ∈ I[α] ηd ∈ R[α]. Put C := R[η1 , . . ., ηd ]. Then R[α] is a free C-module and R[α] is integral over C. So I[α] R[α] = R[α] yields that I[α] C = C. Thus C is flat over R. Since R[α] is flat over C, R[α] is flat over R, and hence J[α] = R. Now we shall show that I[α] = Iηd . It is easy to see that I[α] ⊆ Iηd . Take p ∈ Spec(R) with p ⊇ I[α] . Since I[α] R[α] = R[α], we have 1 = a0 + a1 α + · · · + an α n , where ai ∈ I[α] ⊆ p. Since 1 − a0 is a unit in R p , α −1 is integral over R. Since α is anti-integral over R, α −1 is also anti-integral over R. Thus ϕα−1 (X ) = X d + (ηd−1 ηd−1 )X d−1 + · · · + (ηd−1 η1 )X + ηd−1 ∈ R p [X ]. So there exists x ∈ R p such that ηd = 1/x, ηi = yi /x with some yi ∈ R p . Therefore we have (I[α] ) p = x R p = (Iηd ) p . Since p is arbitrary, we conclude that I[α] = Iηd . Next we shall show that Jηd = Iηd (1, ηd ) = R. Suppose that Iηd (1, ηd ) ⊆ p for some p ∈ Spec(R). Then I[α] ⊆ p. Thus by using the above notation, we have (Iηd (1, ηd )) p = x(1, 1/x)R p = R p , which contradicts the assumption Iηd (1, ηd ) ⊆ p. Therefore we conclude that Jηd = R.

46

Simple Extensions of High Degree

(ii) ⇒ (i) follows the implications: R[α] ⊇ I[α] R[α] ⊇ I[α] (1, ηd ) = Iηd (1, ηd ) = R  1, and hence R[α] ⊇ I[α] R[α] ⊇ R[α]. Corollary 2.5.12 Assume that α is an anti-integral element over R. If I[α] R[α] = R[α], then R[α] ∩ K = R[ηd ]. Proof Put C := R[η1 , . . ., ηd ]. Then by the proof of Proposition 2.5.11, we have C ⊆ R[α]. Since R[α] is a free C-module C + Cα + · · · + Cα d−1 , we conclude that R[α] ∩ K = C. Besides, as in the proof of Proposition 2.5.11, if I[α] ⊆ p, then there exists x in R p such that ηd = 1/x and ηi = yi /x ∈ R p [ηd ] for some yi ∈ R p . If I[α] ⊆ p, then C p = R p = R p [ηd ]. Therefore we have C = R[ηd ]. Theorem 2.5.13 Assume that α is an anti-integral element over R and that ηd ∈ R[α]. The following statements are equivalent: (i) I[α] R[α] = R[α]; (ii) I[α] + I[α−1 ] = R; (iii) either α or α −1 is integral over R p for each p ∈ Spec(R). Proof (i) ⇒ (ii): We have ηd ∈ R[α], I[α] = Iηd , and Jηd = R by Proposition 2.5.11. So we obtain R = Jηd = Iηd (1, ηd ) = I[α] (1, ηd ) = I[α] + ηd I[α] = I[α] + I[α−1 ] (cf. Lemma 2.4.6). (ii) ⇒ (i): We have R = I[α] + I[α−1 ] = I[α] + ηd I[α] = I[α] (1, ηd ). Thus R[α] = I[α] (1, ηd )R[α] = I[α] R[α] + I[α] ηd R[α] ⊆ I[α] R[α] because ηd ∈ R[α]. The converse inclusion is obvious. Thus I[α] R[α] = R[α]. (ii) ⇒ (iii): Take p ∈ Spec(R). Then I[α] ⊆ p or I[α−1 ] ⊆ p because I[α] + I[α−1 ] = R. If I[α] ⊆ p, we have η1 , . . ., ηd ∈ R p . Hence α d + η1 α d−1 + · · · + ηd = 0, which means that α is integral over R p . Next, we assume that I[α−1 ] ⊆ p. The ηd−1 η1 , . . ., ηd−1 ηd−1 , ηd−1 ∈ R p . Thus (α −1 )d +(ηd−1 ηd−1 )α d−1 + · · · + ηd−1 η1 α + ηd−1 = 0, which means that α −1 is integral over R p . (iii) ⇒ (ii): Suppose that there exists a prime ideal p of R such that I[α] + I[α−1 ] ⊆ p. Note that α or α −1 is integral over R p . Assume that α is integral over R p . Then α satisfies a monic relation of degree d over R p because α is anti-integral over R p . But since ϕα (X ) = X d + η1 X d−1 + · · · + ηd ∈ R p [X ], we have I[α] ⊆ p, a contradiction. Similarly we come to a contradiction when we assume α −1 is integral over R p .

2.5 Vanishing Points and Blowing-Up Points

47

Proposition 2.5.14 Assume element of degree d  that α is an anti-integral  over R. If ηd ∈ R[α], then I[α] R[α] ∩ R = I[α] + I[α−1 ] . Proof Take p ∈ Spec(R). Then I[α] R[α] ∩ R ⊆ p ⇔ I[α] R p [α] = R p [α] ⇔ (I[α] + I[α−1 ] ) p = R p ⇔ I[α] + I[α−1 ] ⊆ p. Thus we come to our conclusion by Theorem 2.5.13.

Chapter 3 Subrings of Anti-Integral Extensions

Let R be a Noetherian domain and R[X ] a polynomial ring. Let α be an element of an algebraic field extension L of the quotient field K of R and π : R[X ] → R[α] be the R-algebra homomorphism sending X to α. Put E (α) := Ker(π ), an ideal of R. Let ϕα (X ) be the monic minimal polynomial of α over K with deg ϕα (X ) = d and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd Then ηi (1 ≤ i ≤ d) are uniquely determined by α. Let Iηi := R : R ηi and d ηi := i=1 Iηi , the latter of which is called a generalized denominator ideal of α. We say that α is an anti-integral element if and only if E (α) = I[α] ϕα (X )R[X ]. For f (X ) ∈ R[X ], let c( f (X )) denote the ideal of R generated by the coefficients of f (X ). For an ideal J of R[X ], let c(J ) denote the ideal generated by the coefficients of the elements in J . If α is an anti-integral element, then c(E (α) ) = c(I[α] ϕα (X )R[X ]) = I[α] (1, η1 , . . ., ηd ). Put J[α] = I[α] (1, η1 , . . ., ηd ). If J[α] ⊆ p for all p ∈ Dp1 (R) := { p ∈ Spec(R)|depth(R p ) = 1}, then α is called a super-primitive element. It is known that a super-primitive element is an anti-integral element (cf. Theorem 2.2.8). By definition, the super-primitive is characterized by the set of Dp1 (R). Put Rα := R[α] ∩ R[α −1 ].

3.1

Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

By definition (cf. Definition 1.1.4), R[α]∩ R[α −1 ] = R if α is an anti-integral element of degree 1 over R. So we are interested in the case α is an algebraic element of degree d > 1. Therefore we come to study the following problem: Problem. Let R be a Noetherian integral domain, K denote the quotient field of R, and α be an algebraic element over R of degree d. What is R[α]∩ R[α −1 ]?

49

50

Subrings of Anti-Integral Extensions

In the first part of this section, our results are the following: Let α be an anti-integral element of degree d. (i) (ii) (iii) (iv)

We find a set of generators of Rα as an R-algebra. Rα is an R-algebra of finite type. For d = 1, 2, 3, Rα is completely determined. When α is anti-integral over R and I[α] is a radical ideal of R, Rα is integrally closed in R[α].

In the second part of this section, we investigate an R-module structure of Rα for an anti-integral element α of degree d over R. The R-module structure of Rα is completely determined. We start with the following definition. Definition 3.1.1 (α) := {a0 α

d−

Remark 3.1.2

Let + · · · + ad− | a0 X d + · · · + ad ∈ I[α] ϕα (X )R[X ], 0 ≤  ≤ d}

By definition, it is easy to see the following:

(1) (α) ⊆ Rα; (2) the R-module generated by (α) is finitely generated because I[α] is a finitely generated ideal of R. Lemma 3.1.3 Assume that α is an anti-integral element of degree d over R. If a0 α n + · · · + an = b0 (α −1 )m + · · · + bm ∈ Rα with ai , b j ∈ R, then b0 ∈ I[α] ηd . Proof Put f (X ) = a0 X n+m + · · · + an X m − bm X m − · · · − b0 ∈ R[X ]. Then f (α) = 0 and hence f (X ) ∈ Ker(π ) = I[α] ϕα (X )R[X ]. Thus f (0) = −b0 ∈ I[α] ηd . Theorem 3.1.4 Assume that α is an anti-integral element of degree d. Then Rα is an R-algebra generated by (α), i.e., Rα = R[(α)]. Proof The inclusion R[(α)] ⊆ Rα is obvious. We must show the converse inclusion. Take β ∈ Rα and put β = a0 α n + · · · + an = b0 (α −1 )m + · · · + bm with ai , b j ∈ R. We shall prove our assertion by induction on m. If m = 0, then obviously β ∈ R ⊆ R[(α)]. First, we will show the case m < d. By Lemma 3.1.3, we have b0 ∈ I[α] ηd , so that there exist c0 , . . ., cd

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

51

with cd = b0 such that c0 α d + c1 α d−1 + · · · + cd = 0. Since m < d, putting γ = cd (α −1 )m + · · · + cd−m = −(cd−m−1 α + · · · + c0 α d−m ), the element β − γ satisfies our induction hypothesis. Note that γ ∈ (α) and hence β ∈ R[(α)]. Second, we consider the case m ≥ d. There exist c0 , . . ., cd with cd = b0 such that c0 α d + c1 α d−1 + · · · + cd = 0. Put γ = cd (α −1 ) = −(c0 α d−1 + · · · + cd−1 ) ∈ (α). Then b0 (α −1 )m = b0 (/α)(α −1 )m−1 = cd (α −1 )(α −1 )m−1 = γ (α −1 )m−1 . Hence we have β = γ (α −1 )m−1 + b1 (α −1 )m−1 + · · · + bm = −(c0 α d−1 + · · · + cd−1 )(α −1 )m−1 + b1 (α −1 )m−1 + · · · + bm , which satisfies the induction hypothesis. Thus β ∈ R[(α)]. Theorem 3.1.5 Assume that α is anti-integral over R of degree d. Then Rα is an R-algebra of finite type. Proof

This follows from Remark 3.1.2(2) and Theorem 3.1.4.

Theorem 3.1.6

The following statements are equivalent:

(1) α is an anti-integral element of degree d over R; (2) {a ∈ R|aα ∈ Rα} = I[α] ; (3) each element of Rα is of the form h(α) such that h(X ) ∈ R[X ] with deg h(X ) ≤ d − 1. Proof By Lemma 2.4.4 and Theorem 2.4.5, we see that (1) is equivalent to (2). (1) ⇒ (3) will follow from Theorem 3.1.4 in the sequel. (3) ⇒ (1): Consider the exact sequence: 0 −→ E (α) −→ R[X ] −→ R[α] −→ 0 Take f (X ) ∈ E (α) . Suppose that deg f (X ) = n and n ≥ d. We shall prove that f (X ) ∈ I[α] ϕα (X )R[X ] by induction on n. In the case n = d, we can show that I[α] ϕα (X )R[X ] = { f (X ) ∈ E (α) | deg f (X ) = d} So (1) is valid. We will consider the case n > d. Put f (X ) = an X n + · · · + a1 X + a0 . Since f (α) = 0, we have that an α n−1 + · · · + a1 = −a0 α −1 ∈ R[α] ∩ R[α −1 ] = Rα So we see that there exist b0 , b1 , . . ., bd−1 ∈ R such that an α n−1 + · · · + a1 = bd−1 α d−1 + · · · + b0 by the hypothesis. Putting g(X ) = an X n−1 + · · · + (ad −

52

Subrings of Anti-Integral Extensions

bd−1 )X d−1 + · · · + (a1 − b0 ), we see that g(α) = 0 and deg g(X ) = n − 1. By the induction hypothesis, g(X ) ∈ I[α] ϕα (X )R[X ]. Also, since p(X ) = f (X ) − Xg(X ) ∈ E (α) and deg p(X ) ≤ n − 1, we have that p(X ) ∈ I[α] ϕα (X )R[X ] by induction. Hence f (X ) ∈ I[α] ϕα (X )R[X ]. Therefore we have proved that α is an anti-integral element of degree d over R. Theorems 2.4.5 and 3.1.6 yield the following result. Corollary 3.1.7

The following statements are equivalent:

(1) α is an anti-integral element of degree d over R; (2) α −1 is an anti-integral element of degree d over R; (3) {a ∈ R|aα ∈ Rα} = I[α] ; (4) {a ∈ R|a(α −1 ) ∈ Rα} = I[α−1 ] ; (5) each element of Rα is of the form h(α) such that h(X ) ∈ R[X ] with deg h(X ) ≤ d − 1; (6) each element of Rα is of the form h(α −1 ) such that h(X ) ∈ R[X ] with deg h(X ) ≤ d − 1. Now we require the following general result for later use. Theorem 3.1.8 over R.

Let α be an algebraic element over R. Then Rα is integral

Proof Let V be a valuation ring on the field K (α) which contains R. Then either α or α −1 ∈ V . The integral closure of R in K (α) is the intersection of the family of valuation rings on K (α) which contains R by [G,(19.8)]. Hence Rα is integral over R. Corollary 3.1.9

Assume that α is anti-integral over R of degree d.

(i) If d = 1 then Rα = R. (ii) If d = 2 then Rα = R[α I[α] ]. (iii) If d = 3 then Rα = R[α I[α] , α −1 η3 I[α] ]. Proof (i) The case d = 1 is shown in Definition 1.1.4 and Remark 2.2.2. (ii) For the case d = 2, aα 2 +bα+c = 0 with a, b, c ∈ R. So aα = −c(α −1 )−b and hence aα ∈ (α). Any element in (α) is the type aα with a ∈ R. Thus (α) ⊆ α I[α] .

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

53

(iii) The case d = 3: Since aα 3 + bα 2 + cα + d = 0 with a, b, c, d ∈ R, every element of (α) is either a type aα 2 + bα + c = −d(α −1 ) ∈ α −1 η3 I[α] or a type aα = −d(α −1 )2 − c(α −1 ) − b ∈ α I[α] . Lemma 3.1.10 Assume that α −1 is anti-integral over R. Let g(X ) = a0 X n + a1 X n−1 + · · · + an be a polynomial in R[X ] with g(α) = 0. Then a0 ∈ I[α] . Proof Put Y = 1/ X and put f (Y ) = (1/ X )n g(X ). Thenf (Y ) ∈ R[Y ] and f (α −1 ) = 0. Since α −1 is anti-integral over R, f (Y ) = f i (Y )gi (Y ) with f i (0)gi (0) and f i (0) ∈ deg f i (Y ) = d and f i (α −1 ) = 0. In this case, f (0) = I[α] . Indeed, put f i (Y ) = bd (Y )d + · · · + b0 with bi ∈ R =⇒ bd + · · · + b0 X d ∈ R[X ] =⇒ bd + · · · + b0 α d  = 0 =⇒ bd /b0 + (bd−1 /b0 )α + · · · + α d = 0 d Iηi = I[α] . Therefore the constant term of =⇒ ηi = bi /b0 =⇒ b0 ∈ i=1 (1/ X )n g(X ) is f (0) = a0 , which belongs to I[α] . Theorem 3.1.11 Assume that α is anti-integral over R and that I[α] is a radical ideal (i.e., I[α] = I[α] ) of R. Then Rα is integrally closed in R[α]. Proof Take β ∈ R[α] and write β = an α n + · · · + a0 with ai ∈ R. We have only to show that if β is integral over R, then β ∈ Rα. We prove this assertion by induction on n. The case n = 0 is obvious. Let β  + b1 β −1 + · · · + b = 0 be an integral dependence of β over Rα. Since Rα is integral over R by Theorem 3.1.8, we may assume that bi ∈ R for all i. Hence (an α n + · · · + a0 ) + b1 (an α n + · · · + a0 )−1 + · · · + b = 0. Note that α −1 is an anti-integral element of degree d over R by Theorem 2.4.5. By use of Lemma 3.1.10, we therefore obtain an ∈ I[α] . Since I[α] is a radical ideal, we have an ∈ I[α] . So there exist an = c0 , c1 , . . ., cd ∈ R such that (∗)

c0 α d + c1 α d−1 + · · · + cd = 0

If n ≥ d, then β = an α n + · · · + a0 = c0 α n + an−1 α d−1 + · · · + a0 = − (c1 α d−1 + · · · + cd )α n−d + an−1 α n−1 + · · · + a0 and hence β ∈ Rα by induction hypothesis. Next suppose that n ≤ d − 1. Multiply (*) by (α −1 )n−d and put γ := c0 α n + c1 α n−1 + · · · + cn = −(cn+1 (α −1 ) + · · · + cd (α −1 )d−n ). Then γ is integral over R and belongs to Rα. Considering the element β − γ , c0 = an yields β − γ ∈ Rα by induction hypothesis. Therefore we conclude that β ∈ Rα. Corollary 3.1.12 Assume that α is an anti-integral element of degree 1, i.e., α ∈ K . If Iα := (R : R α) is a radical ideal of R, then R is integrally closed in R[α].

54

Subrings of Anti-Integral Extensions

Proof Since the degree d = 1, we have Rα = R (see Remark 2.2.2) and I[α] = Iα . Our conclusion follows from Theorem 3.1.11. Proposition 3.1.13 (i) I[α] = I[α−a] for every a ∈ R. (ii) If α is anti-integral over R, then so is α − a for every a ∈ R. Proof (i) Put X = Y +a. Then ϕα−a (Y ) = (Y +a)d +η1 (Y +a)d−1 +· · ·+ηd = Y d + λ1 Y d−1 + · · · + λd . Note that λi ∈ R + Rη1 + · · · + Rηd for all i. Take x ∈ I[α] . Then xζi ∈ R for all I . Hence I[α] ⊆ I[α−a] . Since α = (α − a) + a, we have I[α−a] ⊆ I[α] , whence I[α] = I[α−a] . (ii) Let a ∈ R. Let τ : R[X ] −→ R[X ] be an R-algebra homomorphism sending X to X − a and πa : R[X ] −→ R[α − a] = R[α] be an R-algebra homomorphism sending X to α − a. Then πa = π · τ and K er (πa ) = τ −1 (Ker(π )) = τ −1 (I[α] ϕα (X )R[X ]) = I[α] ϕα (X + a)R[X ] = I[α] ϕα−a (X )R[X ] = I[α−a] ϕα−a (X )R[X ] Thus α − a is anti-integral over R. Corollary 3.1.14 If α is anti-integral over R and if I[α] is a radical ideal of R, then R[α] ∩ R[1/(α − a)] is integrally closed in R[α] for each a ∈ R. In particular, R[α] ∩ R[1/(α − a)] is the integral closure of R in R[α]. Proof

This follows from Theorem 3.1.11 and Proposition 3.1.13.

Proposition 3.1.15 Let a be an element in R. Assume that α is anti-integral over R and that I[α] is a radical ideal of R. Then Rα = R[α − a] ∩ R[1/(α − a)] = Rα − a Proof Since I[α] is a radical ideal, Rα is the integral closure of R in R[α] by Theorem 3.1.11. By Proposition 3.1.13, I[α] = I[α−a] , which is a radical ideal, and α is anti-integral over R. So R[α]∩ R[1/(α −a)] = R[α −a]∩ R[1/(α −a)]

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

55

is the integral closure of R in R[α − a] = R[α] by Corollary 3.1.14. Thus Rα = R[α − a] ∩ R[1/(α − a)]. Proposition 3.1.15 yields the following corollary. Corollary 3.1.16 Assume that α is super-primitive over R and that I[α] is a radical ideal of R. Then Rα ⊆ R[1/(α − a)] for every a ∈ R. Theorem 2.4.8 and Proposition 2.2.10 yield the following corollary. Proposition 3.1.17 If α is a super-primitive element over R, then so are both α − a and (α − a)−1 for every a ∈ R. Next we investigate an R-module structure of Rα when α is an anti-integral element of degree d over R. Let ϕi (X ) := X i + η1 X i−1 + η2 X i−2 + · · · + ηi ∈ K [X ] (1 ≤ i ≤ d − 1), where ϕα (X ) = X d + η1 X d−1 + · · · + ηd is the minimal polynomial of α over K , and put ζi := ϕi (α) and i := I[α] ζi (1 ≤ i ≤ d − 1), which is an R-module. Under this preparation, we have the following Structure Theorem of Rα. Theorem 3.1.18 Assume that α is an anti-integral element of degree d over

d−1 ∼ R. Then Rα = R ⊕  ⊕ · · · ⊕  R ⊕ ( = 1 d−1 i=1 I[α] ζi ), as R-modules,

where I[α] denotes the direct sum of d − 1 copies of I[α] . Proof We shall show that Rα = R + 1 + · · · + d−1 as R-modules. The inclusion R + 1 + · · · + d−1 ⊆ Rα is obvious. We must show the converse inclusion. Take β ∈ Rα and put β = a0 α n + · · · + an = b0 (α −1 )m + · · · + bm with ai , b j ∈ R. We prove our assertion by induction on m. If m = 0, then β = b0 ∈ R. First, we will show the case m < d. Put f (X ) = a0 X n+m + · · · + an X m − bm X m − · · · − b0 ∈ R[X ]. Then f (α) = 0 and hence f (X ) ∈ I[α] ϕα (X )R[X ]. Thus f (0) = −b0 ∈ I[α] ηd , so that there exist c0 , . . ., cd ∈ R with cd = b0 such that c0 α d + c1 α d−1 + · · · + cd = 0. Indeed, since b0 = aηd for some a ∈ I[α] , we may put c0 = a, ci = aηi (1 ≤ i ≤ d). Since m < d, putting γ = cd (α −1 )m + · · · + cd−m+1 = −(cd−m + cd−m−1 α + · · · + c0 α d−m ), we see that γ ∈ d−m and the element β − γ satisfies our induction hypothesis. Since γ ∈ d−m , it follows that β ∈ R + 1 + · · · + d−1 . Second, we consider the case m ≥ d. As the above statements, there exist c0 , . . ., cd ∈ R with cd = b0 such that c0 α d + c1 α d−1 + · · · + cd = 0. Put γ := cd (α −1 ) = −(c0 α d−1 + · · · + cd−1 ) ∈ d−1 . Then b0 (α −1 )m = γ (α −1 )m . Hence we have

56

Subrings of Anti-Integral Extensions

β = γ (α −1 )m−1 +· · ·+bm = −(c0 α d−1 +· · ·+cd−1 )(α −1 )m−1 +b1 (α −1 )m−1 + · · · bm = (b1 −cd−1 )(α −1 )m−1 +(b2 −cd−2 )(α −1 )m−2 +· · ·+bm , which satisfies the induction hypothesis. Thus β ∈ R +1 +· · ·+d−1 . Hence we have proved that Rα = R + 1 + · · · + d−1 . Next, we consider the R-homomorphism φi : I[α] → i defined by φi (a) = aϕi (α). This R-homomorphism is an Risomorphism, that is, I[α] ∼ = i as R-modules. Since (Rα) ⊗ R K = K (α) is a d-dimensional vector space over K , it follows that Rα

is an R-module of rank d. Hence Rα = R ⊕ 1 ⊕ · · · ⊕ d−1 ∼ = R ⊕ ( I[α] ) as R-modules. Combining Theorem 3.1.18 with Corollary 3.1.7, we have the following result. Proposition 3.1.19

The following statements are equivalent:

(1) α is an anti-integral element of degree d over R; (2) Rα = R ⊕ 1 ⊕ · · · ⊕ d−1 as R-modules. Proof By Theorem 3.1.6, (2) ⇒ (1) and the implication (1) ⇒ (2) follows Theorem 3.1.18. Proposition 3.1.20 [KY6] Assume that α is an anti-integral element of degree d over R. Then Rα is a flat R-module if and only if I[α] is an invertible ideal of R. Moreover, in this case, α is a super-primitive element of degree d over R. ∼ R ⊕ I[α] ⊕ · · · ⊕ I[α] by Theorem 3.1.18 because α Proof Note that Rα = is anti-integral over R. Thus Rα is flat over R if and only if I[α] is invertible. The last part follows from Theorems 2.2.8 and 2.1.8. Proposition 3.1.21 is a flat R-module.

If R is a Noetherian locally factorial domain, then Rα

Proof Since R is a Noetherian locally factorial domain, each divisorial ideal d is an invertible ideal ([F, 9.2]). Noting that I[α] = i=1 Iηi is a divisorial ideal, I[α] is an invertible ideal. Hence Rα is flat R-module. Proposition 3.1.22 Assume that α is an anti-integral element of degree d over R. Then Rα ∩ K = R. Proof [K].

The proof is easily seen by Theorem 3.1.18. The case d = 1 is seen in

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

57

In the rest of this section, our objective is to investigate the flatness of a ring extension R[α] ⊇ Rα. When α is an anti-integral element over Rα, we give an equivalent condition for an extension R[α] ⊇ Rα to be flat in terms of the ideal I[α] + I[α−1 ] . From this result, we give a property of an extension R[α] ⊇ R with respect to the flatness when α is a super-primitive element over R. Also, when α is an anti-integral element over R, we give an equivalent condition for the element α to be super-primitive over Rα using the ideal I[α] + I[α−1 ] . In what follows, we fix the following notation unless otherwise specified: IαRα

:=

{b ∈ Rα|bα ∈ Rα},

Iα

:=

R : R α R for α ∈ K and Jα := Iα + α Iα ,

ζi

:=

α i + η1 α i−1 + · · · + ηi = ϕi (α).

Theorem 3.1.23 Assume that α is an anti-integral element of degree d over R. Then IαRα = I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Proof From Theorem 3.1.19, we have Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Take an element β ∈ IαRα . Since Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 , we may put β = a + b1 ζ1 + · · · + bd−1 ζd−1 with a ∈ R and bi ∈ I[α] (1 ≤ i ≤ d − 1). It follows that aα ∈ Rα since βα ∈ Rα. Thus a ∈ Rα : R α = I[α] by Theorem 3.1.6. Therefore IαRα ⊆ I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . We shall show the converse inclusion. Take an element a ∈ I[α] . Since aηi ∈ R and aζ1 ∈ I[α] ζ1 , we have aα = a(ζ1 −η1 ) = aζ1 −aη1 ∈ Rα. Hence I[α] ⊆ IαRα . Furthermore for any a ∈ I[α] , we have α(aζd−i ) = aα(α d−i + η1 α d−i−1 + · · · + ηd−1 ) = −aα{ηd−i+1 α −1 + · · · + ηd (α −1 )i } = −a{ηd−i+1 + ηd−i+2 α −1 + · · · + ηd (α −1 )i−1 } ∈ Rα. Thus IαRα ⊇ I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Remark 3.1.24 Since Rα ⊆ Rα[α] ∩ Rα[α −1 ] ⊆ Rα, we see that α is an anti-integral element over Rα. Proposition 3.1.25 Assume that α is an anti-integral element of degree d over R. Then the following assertions hold. (1) if p is a prime ideal of R containing I[α] , then P := p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 is a prime ideal of Rα. (2) if p is a prime divisor of I[α] , then P is a prime ideal of Rα with depth 1. Proof (1) By Theorem 3.1.23, IαRα = I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 is an ideal of Rα. Thus (I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 )Rα ⊆ I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 .

58

Subrings of Anti-Integral Extensions

So P Rα = p Rα ⊕ (I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 )Rα ⊆ p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 = P. Therefore P is an ideal of Rα. Also, since Rα/P = R/ p is an integral domain, P is a prime ideal of Rα. (2) Since p is a prime divisor of I[α] , there exists an element a of R such that p = I[α] : R a. Now we shall show that P = IαRα : Rα a. Indeed, for any β ∈ P, we can write β = x + b1 ζ1 + · · · + bd−1 ζd−1 , where x ∈ p and bi ∈ I[α] (1 ≤ i ≤ d − 1). We have only to prove that x ∈ I[α] : R a = p. Since aβ ∈ IαRα , it follows that ax ∈ I[α] . Thus x ∈ I[α] : R a = p and so β ∈ P. Hence P = IαRα : Rα a. Since P is a prime divisor of IαRα and IαRα is a divisorial ideal, we conclude that P ∈ Dp1 (Rα) by [Y, Proposition 1.10].

Proposition 3.1.26 Assume that α is an anti-integral element of degree d over R. Then a prime ideal P of Rα satisfies one of the following three statements. Put p := P ∩ R. (1) The case that p ⊇ I[α] . Then Rα p = R p [α] and R p [α] is a free R p module of rank d. Furthermore P is a prime ideal of Rα with depth 1 if and only if p is a prime ideal of R with depth 1. (2) The case that p ⊇ I[α] but P ⊇ IαRα . Then R[α] ⊆ Rα P and Rα P = R[α] p , where p = P Rα P ∩ R[α]. (3) The case that P ⊇ IαRα . Then p ⊇ I[α] and P = p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . In particular, if P ∈ Ass Rα (R[α]/Rα), then P is a prime ideal of depth 1. Proof (1) Since p ⊇ I[α] , we see that (I[α] ) p = R p . So it is clear that Rα p = R p ⊕ R p ζ1 ⊕ · · · ⊕ R p ζd−1 . Thus Rα p is a free R p -module of rank d. Take a ∈ I[α] \ p. Then (aα)d +(aη1 )(aα)d−1 +· · · a d ηd = 0. Hence aα is integral over R and so aα is integral over R p . Since Rα is integral over R, we conclude that R p ⊆ R p [α] ⊆ R p [α −1 ]. Thus we have R p [α] = R p [α] ∩ R p [α −1 ] = Rα p . Since Rα p is flat over R p , it follows that depth(Rα p ) = depth(R p ). In particular, P ∈ Dp1 (Rα) if and only if p ∈ Dp1 (R). (2) Since P ⊇ IαRα and α ∈ Rα P , it follows that R[α] ⊆ Rα P . Hence we have Rα P = R[α] p . (3) Put P  = p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Then P  is a prime ideal of Rα by Proposition 3.1.25(1). It is obvious that P, P  ⊇ I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Thus P ∩ R = P  ∩ R = p and so P = P  . If P ∈ Ass Rα (R[α]/Rα), then P = Rα : Rα x for some x ∈ R[α] \ Rα. Since P is a divisorial ideal, we see that P is an ideal of depth 1 in Rα by [Y, Proposition 1.10]. Suppose that P ⊇ IαRα . In the same way as in (2), we can prove that R[α] ⊆

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

59

Rα P and hence Rα P = R[α] p , which contradicts the assumption that P ∈ Ass Rα (R[α]/Rα). Thus p ⊇ IαRα and so p ⊇ IαRα ∩ R = I[α] . We obtain the following theorem from the arguments above. Theorem 3.1.27 Assume that α is an anti-integral element of degree d over R. Then the following two conditions are equivalent to each other: (1) R[α] is a flat R-module; (2) Rα[α] = R[α] is a flat Rα-module. Proof (1) ⇒ (2): We must show that Rα[α] is flat over Rα. For this end, we have only to show that JαRα = Rα. Since JαRα = IαRα + α IαRα = Rα Rα IαRα + Iα−1 , it is enough to prove that IαRα + Iα−1 = Rα. Suppose that Rα IαRα +Iα−1 ⊆ P for some P ∈ Spec(Rα). By Proposition 3.1.25, we have that P = p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . On the other hand, since IαRα = I[α] ⊕ I[α] ζ1 ⊕ Rα · · · ⊕ I[α] ζd−1 by Theorem 3.1.23, we conclude that I[α] ⊆ p. Note that Iα−1 = −1 −1    I[α−1 ] ⊕I[α−1 ] ζ1 ⊕· · ·⊕I[α−1 ] ζd−1 ⊆ P, where ζi = −ηd ζd−1 +ηd ηd−i . Take an  element a ∈ I[α] . Then aηd ζd−i = a(−ζd−i +ηd−i ) = −aηd−1 ζd−i +aζd−i . Since aζd−i ∈ P, we get aηd−i ∈ P ∩ R = p for all i (1 ≤ i ≤ d −1). Also we see that I[α−1 ] = ηd I[α] ⊆ P ∩ R = p and hence that J[α] = I[α] (1, η1 , . . ., ηd ) ⊆ p. Since R[α] is a flat extension of R, we have J[α] = R by Theorem 2.3.6, a contradiction. Thus Rα[α] is a flat Rα-module. (2) ⇒ (1): By Remark 3.1.25, the element α is an anti-integral element over Rα. Since Rα[α] is a flat Rα-module, we see that Rα = JαRα = Rα IαRα + α IαRα = IαRα + Iα−1 from Theorem 1.2.8. Note that IαRα = I[α] ⊕ Rα  I[α] ζ1 ⊕· · ·⊕ I[α] ζd−1 and Iα−1 = I[α−1 ] ⊕ I[α−1 ] ζ1 ⊕· · ·⊕ I[α−1 ] ζd−1 , where ζi = Rα −1 −1 Rα −ηd ζd−1 + ηd ηd−i . It is easy to see that I[α−1 ] = ηd I[α] , α Iα = Iα−1 and Rα) that Rα[α]/Rα is a flat extension. We obtain that Rα = IαRα + Iα−1 ⊆ I[α] ⊕ I[α−1 ] ⊕ I[α] (η1 , . . ., ηd−1 ) ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 = I[α] (1, η1 , . . ., ηd ) ⊕ I[α] ζ1 ⊕ · · · I[α] ζd−1 . Since Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 , we see that R = I[α] (1, η1 , . . ., ηd ) = J[α] . From Theorem 2.3.6, it follows that R[α] is a flat extension of R. Lemma 3.1.28 Assume that α is an anti-integral element of degree d over R. Then P ∩ R =: p ∈ Dp1 (R) for every P ∈ Dp1 (Rα). Proof Let P ∈ Dp1 (Rα) and put P ∩ R = p. Then there exists an element β of K [α] such that P = Rα : Rα β by [Y, Proposition 1.10]. We have K [α] = K ⊕ K ζ1 ⊕ K ζ2 ⊕· · ·⊕ K ζd−1 and Rα = I[α] ⊕ I[α] ζ1 ⊕ I[α] ζ2 ⊕· · ·⊕ I[α] ζd−1 .

60

Subrings of Anti-Integral Extensions

We write β = λ0 +λ1 ζ1 +· · ·+λd−1 ζd−1 , where λi ∈K for all i (1 ≤ i ≤ d −1). d−1 (I[α] : R λi )). Since I[α] Thus we conclude that p = P ∩ R = (R : R λ0 ) ∩ ( i=1 is a divisorial ideal, we get p ∈ Dp1 (R) by [Y, Proposition 1.10]. Theorem 3.1.29 Assume that α is an anti-integral element of degree d over R. Then the following two conditions are equivalent: (1) α is a super-primitive element over Rα; (2) grade R (I[α] + I[α−1 ] ) > 1. Proof (1) ⇒ (2): Suppose that I[α] + I[α−1 ] ⊆ p for some p ∈ Dp1 (R). Since α is super-primitive over Rα, we have (I[α] ) p = a R p for some a ∈ R by Theorem 2.3.10. Thus p R p is a prime divisor of a R p . Thus p is a prime divisor of I[α] . Hence P = p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 ∈ Dp1 (Rα) by Proposition 3.1.25. Since p ⊇ I[α] , we have that P ⊇ I[α] ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 = IαRα . Rα  Similarly, we have P ⊇ Iα−1 = I[α−1 ] ⊕ I[α−1 ] ζ1 ⊕ · · · ⊕ I[α−1 ] ζd−1 because  p ⊇ I[α−1 ] , where ζi is the same as in the proof of Theorem 3.1.28. Thus we get P ⊇ I[α] + I[α−1 ] = JαRα , which contradicts the assumption that α is a superprimitive element over Rα. Hence we conclude that grade R (I[α] + I[α−1 ] ) > 1. (2) ⇒ (1): Suppose that α is not a super-primitive element over Rα. Then there exists a prime ideal P of Rα such that depth(Rα P ) = 1 and that Rα P ⊇ JαRα = IαRα + α IαRα = IαRα + Iα−1 . From Lemma 3.1.23, it follows Rα that Iα = I[α] ⊕ I[α] ζ1 ⊕ I[α] ζ2 ⊕ · · · ⊕ I[α] ζd−1 . So we have IαRα ∩ R = I[α] . Rα Similarly, we get Iα−1 ∩ R = I[α−1 ] . Thus we have P ∩ R ⊇ IαRα ∩ R = I[α] Rα and P ∩ R ⊇ Iα−1 ∩ R = I[α−1 ] . Hence P ∩ R ⊇ I[α] + I[α−1 ] . By Lemma 3.1.29, we see that P ∩ R ∈ Dp1 (R), which contradicts the assumption that grade R (I[α] + I[α−1 ] ) > 1. Consequently, α is a super-primitive element over Rα. For an ideal I , recall that R(I ) := I : K I , which is an over-ring of R. Definition 3.1.30

For α ∈ L, let

D(α) := R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1

(as R-modules)

where ζi := α i + η1 α i−1 + · · · + ηi for i (1 ≤ i ≤ d − 1). Remark 3.1.31

Rα, R(R), and D(α) are integral over R.

The proof of the following lemma is obtained from the proof of Theorem 3.1.18 without the assumption that α is anti-integral over R.

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

61

Lemma 3.1.32 Under the same situation as in Definition 3.1.30, D(α) is a subring of Rα and they are birational i.e., D(α) and Rα have the same quotient fields K (α). Proof When d = 1, D(α) = R, we are done. Assume that d = 2. Note ζ1 = α + η1 and hence ζ12 = ζ1 (α + η1 ) = αζ1 + η1 ζ1 = ζ2 − η2 + η1 ζ1 = η1 ζ1 − η2 (here ζ2 = 0). Assume that d = 3. Note ζ1 = α + η1 , ζ2 = αζ1 + η2 , and ζ3 = αζ2 + η3 . Thus ζ12 = ζ1 (α + η1 ) = αζ1 + η1 ζ1 = ζ2 − η2 + η1 ζ1 , ζ2 ζ1 = ζ2 (α + η1 ) = αζ2 + ζ2 η1 = ζ3 − η3 + η1 ζ2 = −η3 + η1 ζ2 , and ζ22 = ζ2 (αζ1 + η2 ) = (αζ2 )ζ1 + ζ2 η2 = (ζ3 − η3 )ζ1 + ζ2 η2 = −η3 ζ1 + η2 ζ2 because ζ3 = 0. Assume that d ≥ 3. Note first that η0 := 1, ζ0 := 1, ζd = 0 and ζi+1 = αζi + ηi+1 . Now we compute ζi ζ j as follows: ζi ζ j

=

ζi (αζ j−1 + η j )

=

αζi ζ j−1 + η j ζi

=

(ζi+1 − ηi+1 )ζ j−1 + η j ζi

=

ζi+1 ζ j−1 − ηi+1 ζ j−1 + η j ζi

=

ζi+2 ζi−2 − ηi+2 ζ j−2 + η j−1 ζi+1 − ηi+1 ζ j−1 + η j ζi

=

ζi+2 ζi−2 − (ηi+2 ζ j−2 + ηi+1 ζ j−1 ) + (η j ζi + η j−1 ζi+1 ) ········· ·········

(i) Repeat the above process, we have ζi ζ j = ζi+ j ζ0 −

j 

ηi+t ζ j−t +

t=1

that is, ζi ζ j = ζi+ j −

j 

η j−s ζi+s

s=0

ηi+t ζ j−t +

t=1

j−1 

j−1 

η j−s ζi+s

s=0

(ii) Put  := i + j − d. Then j ≤ i < d yields  < d − 1 and j −  ≥ 1. Thus continuing the above process yields ζi ζ j = ζd ζ −

j−  t=1



j−−1

ηi+t ζ j−t +

s=0

η j−s ζi+s

62

Subrings of Anti-Integral Extensions that is, ζi ζ j = −

d−i 

ηi+t ζ j−t +

t=1

d−i−1 

η j−s ζi+s

s=0

2 Thus we have I[α] ζi I[α] ζ j = I[α] ζi ζ j ∈ D(α) . Hence for β, γ ∈ D(α) , we have β + γ , βγ ∈ D(α) , which shows that D(α) is a subring of R[α]. Next, it is obvious that D(α) ⊆ Rα. Take ζ1 = α + η1 ∈ D(α) . Then α = ζ1 − η1 ∈ K (D(α) ). So we have K (α) = K (D(α) ) ⊆ K (Rα) = K (α).

Definition 3.1.33

For α ∈ L, define:

Ant(α) := {c ∈ R \ (0) | Rα[1/c] = D(α) [1/c]} ∪ {0} and Sup(α) := {c ∈ R \ (0) | R(I[α] )[1/c] = R[1/c]} ∪ {0} The set Ant(α) is called an obstruction of anti-integrality of α over R and the set Sup(α) is called an obstruction of super-primitiveness of α over R. Theorem 3.1.34

For α ∈ L,

(1) Ant(α) is a radical ideal of R which contains the ideal I[α] ; (2) for p ∈ Spec(R), Ant(α) ⊆ p if and only if α is an anti-integral element over R p ; (3) the following conditions are equivalent: (3.i) (3.ii) (3.iii) (3.iv)

α is anti-integral over R, Ant(α) = R, Ant(α) ⊆ p for all p ∈ Spec(R), Ant(α) ⊆ p for all p ∈ Dp1 (R).

Proof (1) First, take a, b ∈ Ant(α) and put c := a + b. Take x ∈ Rα. Then Rα[1/a] = D(α) [1/a] and Rα[1/b] = D(α) [1/b]. So there exists n Thus we have c2n x = (a + n ∈ N such that a n x ∈ D (α) and b x ∈ D(α) .  2n i j i j b) = i+ j=2n a b x = i+ j=2n,i≥n a b x + i+ j=2n, j≥n a i b j x ∈ D(α) + D(α) = D(α) , which implies that c2n x ∈ D(α) and hence x ∈ D(α) [1/c]. Therefore Ant(α) is an ideal of R. Second, since Rα[1/e] = Rα[1/en ] and D(α) [1/e] = D(α) [1/en ] for every n ∈ N. So Ant(α) is a radical ideal. Third, take a nonzero a ∈ I[α] . Then I[α] [1/a] = R[1/a]. Hence ϕα (X ) ∈ R[1/a][X ], that is, ηi ∈ R[1/a] for all i (1 ≤ i ≤ d). Since deg(ϕα (X )) = d, α

3.1 Extensions R[α] ∩ R[α −1 ] of Noetherian Domains R

63

is integral over R[1/a]. Thus α ∈ R[1/a][η1 , . . ., ηd−1 ] = D(α) [1/a]. So we have Rα[1/a] ⊆ R[α][1/a] ⊆ D(α) [1/a] ⊆ Rα[1/a] and hence Rα[1/a] = D(α) [1/a], which shows that a ∈ Ant(α). Therefore we have I[α] ⊆ Ant(α). (2) Let p ∈ Spec(R). Assume that Ant(α) ⊆ p. Then there exists a ∈ Ant(α)\ p such that Rα[1/a] = D(α) [1/a]. Then Rα p = (D(α) ) p . Take f (X ) ∈ Ker(π ). with n := deg( f (X ))(≥ d). Put f (X ) := a0 X n + a1 X n−1 + · · · + an ∈ R[X ]. Then a0 α n + a1 α n−1 + · · · + an = 0 and hence a0 α d−1 + a1 α d−2 + · · · + ad−1 = −(ad (1/α) + · · · + an (1/α)n−d ) ∈ Rα p = R p α = (D(α) ) p = R p ⊕ (I[α] ) p ζ1 ⊕ · · · ⊕ (I[α] ) p ζd−1 . Thus we have a0 ∈ (I[α] ) p . Consider f 2 (X ) := f (X ) − a0 ϕα (X )X n−d . Then deg( f 2 (X )) ≤ n − 1 and f 2 (X ) ∈ Ker(π ). By induction on n, we have (Ker(π )) p ⊆ I[α] ϕα (X )R p [X ]. Hence Ker(π) p = I[α] ϕα (X )R p [X ], which shows that α is anti-integral over R p . Conversely, assume that α is anti-integral over R p . Then we have R p α = (D(α) ) p by Theorem 3.1.18. Since (D(α) ) p is a finitely generated R p -module, so is R p α. Thus there exists a ∈ R \ (0) such that Rα[1/a] = D(α) [1/a]. (3)(3.iii) ⇔ (3.ii) is obvious. (3.iv) ⇔ (3.ii) follows from Theorem 2.4.3. (3.i) ⇔ (3.ii): Since α is anti-integral over R p for every p ∈ Spec(R), we have Ant(α) p = R p by (2). Thus Ant(α) = R. The converse implication is obvious.

Theorem 3.1.35

For α ∈ L,

(1) Sup(α) is a radical ideal of R which contains the ideal I[α] ; (2) for p ∈ Spec(R), Sup(α) ⊆ p if and only if α is a super-primitive element over R p ; (3) The following conditions are equivalent: (3.i) α is super-primitive over R, (3.ii) Sup(α) = R, (3.iii) Sup(α) ⊆ p for all p ∈ Spec(R), (3.iv) Sup(α) ⊆ p for all p ∈ Dp1 (R). Proof (1) Take a ∈ I[α] . Then R(I[α] )[1/a] = R(I[α] [1/a]) = R[1/a]. So a ∈ Sup(α) and hence I[α] ⊆ Sup(α). In the same manners as in the proof of Theorem 5(1), we can show that I[α] is a radical ideal of R. (2) By using Theorem 2.3.10 instead of Theorem 3.1.18, our conclusion is obtained in the same manners as in Theorem 3.1.34. (3) The equivalences (3.i) ⇔ (3.ii) ⇔ (3.iii) follow from the same argument as in Theorem 3.1.34, and the equivalence (3.i) ⇔ (3.iv) follows from Theorem 2.2.8.

64

Subrings of Anti-Integral Extensions From Theorems 3.1.34 and 3.1.35, we have the following result.

Theorem 3.1.36

Let α ∈ L. Then Ass R (R/Ant(α)) ⊆ Dp1 (R)

and Ass R (R/Sup(α)) ⊆ Dp1 (R) Remark 3.1.37 Sup(α) ⊆ Ant(α). In fact, if α is super-primitive over R then α is anti-integral over R by Theorem 2.2.8. Put In := {a ∈ R|a ∈ R is the leading coefficient of some polynomial f (X ) ∈ Ker(π ) of degree n}. Note that Ker(π ) contains no polynomial of degree less than d = [K (α) : K ]. If f (X ) ∈ Ker(π ), then X f (X ) ∈ Ker(π ) and deg(X f (X )) = deg( f (X )) + 1. Hence we have an ascending chain: Id ⊆ Id+1 ⊆ · · · ⊆ In ⊆ · · · Since R is a Noetherian domain, the above chain stops, that is, In = In+1 = · · · for some large n. Put I∞ := In . Proposition 3.1.38

Under the above notation,

(1) Id = I[α] ; (2) for p ∈ Spec(R), α is anti-integral over R p if and only if p ⊇ Id : R I∞ . Proof (1) Take a ∈ I[α] . Then there exists a polynomial f (X ) = a X d + a1 X d−1 + · · · + ad ∈ Ker(π ). Since F(α) = 0 and deg( f (X )) = d, we have f (X ) = aϕα (X ) ∈ R[X ]. Thus a ∈ I[α] . Conversely, let aϕα (X ) ∈ R[X ]. Then a ∈ Id and hence I[α] ⊆ Id . Therefore we have Id = I[α] . (2) (⇒): Assume that a ∈ L is anti-integral over R p . Then the kernel of π p := π ⊗ R R p : R p [X ] → R p [α] is generated by some polynomials of degree d := [K (α] K ]. Hence (I[α] ) p = (I ∞) p . Thus p ⊇ Id : R I∞ . (⇐): Assume that Id : R I∞ ⊆ p with p ∈ Spec(R). Since (Id ) p = (I∞ ) p , Ker(π p ) is generated by some polynomial of degree d. So α is anti-integral over R p . [Indeed, if ( f 1 (X ), . . ., f n (X ))R p [X ] = Ker(π p ) with deg( f i (X )) = d for all i, then f i (X ) = ai ϕi (X ) with ai ∈ R, where ϕi (X ) denotes the monic polynomial in K [X ]. Since deg(ϕi (X )) = d and ϕi (α) = 0, ϕ1 (X ) = · · · = ϕn (X ). Thus Ker(π p ) = (a1 , . . ., an )ϕ1 (X )R p [X ]. So we have (a1 , . . ., an )R p = I[α] R p .] Combining Theorem 3.1.34 and Proposition 3.1.38, we have the following result.

3.2 The Integral Closedness of the Ring R[α] ∩ R[α −1 ] (I) Corollary 3.1.39 Ant(α) =

3.2



65

Id : R I∞ .

The Integral Closedness of the Ring R[α] ∩ R[α −1 ] (I)

Consider the following statements: (I) {a ∈ R|aα is integral over R } = I[α] ; (II) Rα is the integral closure of R in R[α]; (III) {a ∈ R|aα ∈ Rα} = I[α] ; (IV) α is an anti-integral element of degree d over R. The following implication is shown in Theorem 3.1.6: (III) ⇐⇒ (IV) First we show the following implications: (II) + (I) ⇐⇒ (II) + (IV) (I) =⇒ (IV) Proposition 3.2.1 If {a ∈ R | aα is integral over R} = I[α] holds, then α is an anti-integral element of degree d over R. Proof Assume that aα ∈ Rα, where a ∈ R. Then aα is integral over R by Proposition 3.1.8 (cf. [K, p.12 Exercise 4]). Therefore a ∈ I[α] . Hence {a ∈ R | aα ∈ Rα} ⊆ I[α] . Next, take an element a ∈ I[α] . By the assumption, aα is integral. So aα ∈ Rα. Hence {a ∈ R | aα ∈ Rα} = I[α] . By Corollary 3.1.7 (1) ⇔ (3), α is an anti-integral element of degree d over R. Example 3.2.2 Let a be a nonzero and nonunit element of R and take an integer d ≥ 2. Assume that ϕα (X ) = X d + (1/a)X d−1 + · · · + (1/a)d is the minimal polynomial of an algebraic element α over K . Then aα is integral over R. It follows that (aα)d + (aα)d−1 + · · · + 1 = 0, but a ∈ I[α] = a d R. Remark 3.2.3 In the above example, Rα is not necessarily the integral closure of R in R[α] and J[α] := I[α] c(ϕα (X )) = R, where c(ϕα (X )) denotes the R-submodule of K generated by the coefficients of ϕα (X ). Indeed, since a

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super-primitive element is an anti-integral element of degree d over R (Theorem 2.2.8, α is an anti-integral element of degree d over R. Suppose that Rα is the integral closure R of R in R[α]. Since aα is integral over R, it is obvious that aα ∈ Rα. By Theorem 3.1.6, it follows that a ∈ I[α] . This contradicts to the fact that a ∈ I[α] = a d R. And so Rα⊂ R. This implies that aα ∈ R and − aα ∈ Rα. Remark 3.2.4 Assume that Rα is the integral closure of R in R[α]. Then {a ∈ R | aα is integral over R} = I[α] if and only if α is an anti-integral element of degree d over R. Our second objective of this section is to prove the implication (I) ⇐⇒ (II) + (IV) and to show that the implications (II) =⇒ (I), (II) =⇒ (IV), and (IV) =⇒ (I) do not hold by examples. Definition 3.2.5

Let N[α] := {a ∈ R|aα is integral over R}

Remark 3.2.6 (i) It is easy to see that N[α] is an ideal of R. (ii) N[α] ⊇ I[α] . Indeed, take a ∈ I[α] . Then putting ai = aηi ∈ R (1 ≤ i ≤ d), we have aα d + a1 α d−1 + · · · + ad = 0 and hence (aα)d + a1 (aα)d−1 + · · · + ad a d−1 = 0. Thus we have aα is integral over R, which means that a ∈ N[α] . The implication (IV) =⇒ (I) does not necessarily hold as will be shown in the following example. Example 3.2.7 Let k be a field and let u, v denote two indeterminates. Put R := k[u 2 , u 3 , v 2 , v 3 ] ⊆ k[u, v] and α := v 2 /u 2 . Then α is anti-integral over R, i.e., Rα = R. Indeed, take f ∈ Rα. Then f = g(u 2 , u 3 , v 2 , v 3 )/u 2 = h(u 2 , u 3 , v 2 , v 3 )/v 2m for some polynomials g(W, X, Y, Z ), h(W, X, Y, Z ) ∈ k[W, X, Y, Z ]. In this case, we have u 2 (h(u 2 , u 3 , v 2 , v 3 )/v 2m ) = g(u 2 , u 3 , v 2 , v 3 ) ∈ R. So h(u 2 , u 3 , v 2 , v 3 )/v 2m = h 0 (u 2 , u 3 ) + v 2 h 2 (u 2 , u 3 , v 2 , v 3 ) where h 0 (W, X ), h 2 (W, X, Y, Z ) ∈ k[W, X, Y, Z ]. The right hand side of this equality does not contain a linear term in v. By the symmetric argument, g(u 2 , u 3 , v 2 , v 3 )/u 2 = g0 (v 2 , v 3 ) + u 2 g2 (u 2 , u 3 , v 2 , v 3 )

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where g0 (Y, Z ), g2 (W, X, Y, Z ) ∈ k[W, X, Y, Z ], has no linear term in u. Thus f ∈ R. Moreover u 3 α = uv 2 ∈ k[u, v] \ R and the element uv 2 is integral over R, which means that u 3 ∈ N[α] but u 3 ∈ I[α] . Proposition 3.2.8 R[α].

If N[α] = I[α] , then Rα is the integral closure of R in

Proof Since Rα is integral over R, it is enough to prove the following: If β ∈ R[α] is integral over R, then β ∈ Rα. Let β = an α n + · · · + a1 α + a0 ∈ R[α] be integral over R. Then β has an integral dependence: β m + b1 β m−1 + · · · + bm = 0, (bi ∈ R). Hence (an α n + · · · + a1 α + a0 )m + b1 (an α n +· · ·+a1 α+a0 )m−1 +· · ·+bm = 0. Multiplying an(n−1)m and expanding it, we have (an α)nm +c1 (an α)nm−1 +· · ·+cnm = 0, so that an α is integral over R. By the assumption, we have an ∈ N[α] = I[α] . When n ≥ d, an ϕα (α) = 0 holds. So we may assume that n < d. By induction on n, we shall show that β ∈ Rα. The case n = 0 is trivial. Assume that n > 0. Put e0 = an and ei = an ηi , where ϕα (X ) = X d + η1 X d−1 + · · · + ηd is the minimal polynomial of α over K . Then γ := e0 α n + · · · + ed−n = −(ed−n+1 α −1 + · · · + ed (α −1 )d−n ) ∈ Rα. Now β − γ is integral over R, and so β − γ ∈ Rα by induction hypothesis. Thus β ∈ Rα. Proposition   3.2.9 N[α] = I[α] .

Assume that α is an anti-integral element over R. Then

  Proof By Remark 3.2.6(ii), we have I[α] ⊆ N[α] . So I[α] ⊆ N[α] . Take a ∈ N[α] . Since aα is integral over R, we have (aα)n +a1 (aα)n−1 +· · ·+an = 0. Since α is anti-integral over R, so is α −1 by Theorem 2.4.5. Then from Lemma 3.1.10, it follows that a n ∈ I[α] . As a corollary, we have the following result. Corollary 3.2.10 Assume that α is an anti-integral element over R. If I[α] is a radical ideal of R, then N[α] = I[α] .   Proof By Proposition 3.2.9, we have N[α] ⊆ N[α] = I[α] = I[α] . Hence N[α] = I[α] by Remark 3.2.6(ii) because N[α] is a radical ideal of R. The following example shows that even if Rα is the integral closure of R in R[α], α is not necessarily anti-integral and N[α] = I[α] does not always hold.

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Example 3.2.11 Let k be a field and let t be an indeterminate. Put R := k[t 2 , t 3 ] and α := t. Then R[α] = R[t] = k[t] and R[α −1 ] = R[1/t] = k[t, 1/t]. Thus Rα = k[t] is the integral closure of R in R[α] = k[t]. But α is not anti-integral because Rα = R. Moreover N[α] = R and I[α] = (t 2 , t 3 )R. Theorem 3.2.12

The following statements are equivalent:

(1) N[α] = I[α] : (2) α is anti-integral over R and Rα is the integral closure of R in R[α]. Proof (1) ⇒ (2) follows from Propositions 3.2.1 and 3.2.8. (2) ⇒ (1) follows from the equivalence: (II) + (I) ⇒ (II) + (IV). As a corollary, we have the following result. Corollary 3.2.13 Let α be an element in K and assume that α is anti-integral over R. If the equality R : R α = R : R α holds, then R is integrally closed in R[α]. Proof In the birational case, we have R : R α = Iα = I[α] and R : R α = N[α] . Hence the assumption implies that I[α] = N[α] . Thus R = Rα is integrally closed in R[α] by Theorem 3.2.12. Let A ⊆ B be two integral domains. We denote by c(B/A) the ideal {a ∈ A|a B ⊆ A} of A. We also say that the dimension formula holds between A and B if ht(P) = ht( p) + Tr.deg A B − Tr.degk( p) k(P) for every P ∈ Spec(B), where p = P ∩ A (cf. [M2, p.119]). It is clear that if B is a finite A-module, then the dimension formula (if it holds) implies that ht(P) = ht(P ∩ A) for each P ∈ Spec(B). Recall for α ∈ K , Iα denotes the ideal {a ∈ R|aα ∈ R} of R. Proposition 3.2.14

Assume the following conditions:

(i) the integral closure R of R in K is a finite R-module; (ii) the dimension formula holds between R and R; (iv) α1 , . . ., αn ∈ K are super-primitive over R. Then R is integrally closed in R[α1 , . . ., αn ] if and only if grade(Iα1 ∩ · · · ∩ Iαn + C(R/R)) > 1. Proof (⇒): Suppose that there exists p ∈ Dp1 (R) such that Iα1 ∩ · · · ∩ Iαn + C(R/R) ⊆ p. Then Iα1 ∩· · ·∩ Iαn ⊆ p and hence Iαi ⊆ p for some i. Since αi is

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a super-primitive element over R, Iαi is divisorially flat. So p is a prime divisor of Iαi . Thus there exists a ∈ R such that Iaαi = p. Suppose that ht( p) > 1. Then aαi ∈ R \ R by the dimension formula. [Indeed, P ∈ Ht1 (R) satisfies P ∩ R ∈ Ht1 (R) by the dimension formula. The  inclusion aαi p ⊆ R ⊆ R implies aαi ∈ R P∩R ⊆ R P and hence aαi ∈ P∈Ht1 (R) R P = R because R is a Krull domain.] Hence aαi ∈ R[αi ] ⊆ R[α1 , . . ., αn ], which contradicts the assumption that R is integrally closed in R[α1 , . . ., αn ]. Next suppose that ht( p) = 1. Since C(R/R) ⊆ p, p is a conductor ideal. So (aαi ) p ⊆ p and hence aαi ∈ R, a contradiction. (⇐): Suppose that β ∈ R[α1 , . . ., αn ] with β ∈ R \ R. Since β  there exists is expressed as bi1 ···in α1i1 · · · αnin , any element a ∈ Iα1 ∩ · · · ∩ I αn satisfies a N β ∈ R for a sufficiently large integer N . Thus Iα1 ∩ · · · ∩ Iαn ⊆ Iβ . Take a prime divisor p of Iβ . Then β ∈ R implies C(R/R) = R : R R ⊆ R : R β = Iβ ⊆ p. Thus C(R/R) ⊆ p and hence Iα1 ∩ · · · ∩ Iαn + C(R/R) ⊆ p. Since Iβ is divisorial ideal, depth(R p ) = 1, which contradicts the assumption grade(Iα1 ∩ · · · ∩ Iαn + C(R/R)) > 1. Theorem 3.2.15 Assume that α ∈ K is a super-primitive element over R and that the dimension formula holds between R and R. Then the following statements are equivalent: (1) R is integrally closed in R[α]; (2) putting Nα := {x ∈ R|xα ∈ R}, Iα = Nα holds; (3) none of the prime divisors p of Iα contains C(R/R); (4) grade(Iα + C(R/R)) > 1. Proof Since α is super-primitive over R, α is anti-integral over R by Theorem 2.2.8. Hence Rα = R. Thus (1) ⇔ (2) follows from Theorem 3.2.12. (1) ⇔ (4) follows from Proposition 3.2.14. (1) ⇒ (3): Let p be a prime divisor of I . Then there exists a ∈ R such that Iaα = p. Assume that C(R/R) ⊆ p. Then ht( p) = 1. [ Indeed, suppose that ht( p) > 1. Then the dimension formula and depth(R p ) = 1 imply aα ∈ R (cf. the proof of Proposition 3.2.14). Since aα ∈ R and aα ∈ R[α], this is a contradiction.] Now since C(R/R) ⊆ p, p is a minimal prime divisor of C(R/R). Hence (aα) p ⊆ p. [ In fact, suppose that (aα) p ⊆ p. Then (aα) p R p = R p and hence 1 = (aα) · y for some y ∈ R. So we have p R p = y R p , which means that R p is a DVR and hence p ⊇ C(R/R), a contradiction.] Thus aα ∈ R, aα ∈ R and aα ∈ R[α], which is a contradiction. Therefore p ⊇ C(R/R). (3) ⇒ (1): Suppose that there exists β ∈ R[α] such that β ∈ R \ R. Then we n with ai ∈ R. So for any a ∈ Iα , we can write β = a0 + a1 α + · · · + an α n have a β ∈ R. Thus we have Iα ⊆ Iβ . Let p be a prime divisor of Iβ .

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Since β ∈ R, p ∈ Ass R (R/R). Hence p ⊇ C(R/R). Thus p ⊇ Iα + C(R/R), which is a contradiction. Therefore R is integrally closed in R[α].  Theorem 3.2.16 Let L be an algebraic field extension of K , α ∈ L, and R denote the integral closure of R in K (α). Assume that the dimension formula  and that R  is a finite R-module. Then the following holds between R and R, statements are equivalent: (1) Rα is integrally closed in R[α]; (2) I[α] = N[α] ;  ∩ R) > 1; (3) grade(I[α] + C( R/Rα)  (4) none of the prime divisors p of I[α] contains C( R/Rα) ∩ R.  Proof (1) ⇒ (3): Take p ∈ Dp1 (R) such that I[α] + C( R/Rα) ∩ R ⊆ p. Since α is super-primitive over R, I[α] is divisorially flat. So p is a prime divisor of I[α] . It follows that I[α] = N[α] = Rα : R α from Theorem 3.2.12 and Definition 3.2.5. Hence there exists an element a ∈ R such that p = Rα : R  with ht(P) = 1. Then aα. Suppose now that ht( p) > 1. Take P ∈ Spec( R) ht(P ∩ R) = 1 by the dimension formula. Thus P ∩ R = p, which implies that  and hence aα ∈ R  P . So we have aα ∈ R  P = R.  Thus we (aα) p ⊆ Rα ⊆ R have aα ∈ R[α] and aα ∈ Rα. This contradicts the assumption that Rα is  integrally closed in R[α]. Therefore ht( p) = 1. In this case, C( R/Rα)∩ R⊆p  holds. Thus there exists a prime ideal P of Rα such that C( R/Rα) ⊆ P with P ∩ R = p. Since Rα is integral over R, we have ht(P) = 1. So  P is a minimal prime divisor of C( R/Rα). Hence aα P ⊆ P (cf. the proof of Theorem 3.2.15). Thus aα ∈ R[α] and aα ∈ Rα. This contradicts the assumption that Rα is integrally closed in R[α]. Therefore we conclude that  grade(I[α] + C( R/Rα) ∩ R) > 1. (3) ⇒ (1): Take a ∈ I[α] . Then putting aηi = ai , aα d + a1 α d−1 + · · · + ad = 0. So we have aα = −(a1 + a2 α −1 + · · · + ad (α −1 )d−1 ) ∈ Rα. Conversely, if a ∈ R and aα ∈ Rα, then aα = (b0 + b1 α −1 + · · · + bn (α −1 )n ). Thus aα n+1 + b0 α n + · · · + bn = 0. Since α is anti-integral over R, so is α −1 . Hence  a ∈ I [α] . Therefore Rα : R α = I[α] . Take β ∈ R[α] with β ∈ R \ Rα. Then i β = bi α for bi ∈ R. Since Rα : R α = I[α] , there exists a sufficiently large integer N such that any b ∈ I[α] satisfies b N β ∈ Rα. Let p be a prime divisor √  of Rα : R β. Then I[α] ⊆ Rα : R β ⊆ p and C( R/Rα) ∩ R = R(α) : R   ⊆ Rα : R β ⊆ p. So I[α] + C( R/Rα) ∩ R ⊆ p. We have only to show that R depth(R p ) = 1. Let ϕα (X ) = X d +η1 X d−1 +· · ·+ηd be the minimal polynomial of α over K . Put ζi = α d−i + η1 α d−i−1 + · · · + ηd−i . Then Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 as R-modules (cf. Proposition 3.2.19). Since L = Rα ⊗ R K , L = K + K ζ1 +· · ·+ K ζd−1 . Write β = λ0 +λ1 ζ1 +· · ·+λd−1 ζd−1 . Then xβ ∈ Rα if and only if xλ0 ∈ R, xλ1 ∈ I[α] , . . ., xλd−1 ∈ I[α] . Thus Rα : R β is a

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divisorial ideal (R : λ0 )∩(I[α] : λ1 )∩· · ·∩(I[α] : λd−1 ). Since p is a prime divisor  ∩ R) = 1, of this ideal, we have depth(R p ) = 1. Thus grade(I[α] + C( R/Rα) a contradiction. Therefore Rα is integrally closed in R[α]. (1) ⇔ (2) follows from Theorem 3.2.12, and (3) ⇔ (4) is obtained by the similar way to the proof (3) ⇔ (4) of Theorem 3.2.15. Proposition 3.2.17 Let k be a field contained in R and assume that α is an anti-integral element over R of degree d. If I[α] = N[α] , then Rα = R[α + λ] ∩ R[1/(α + λ)] for each λ ∈ k. Proof Note that N[α] = {a ∈ R|aα is integral over R } and that aα is integral over R if and only if a(α + λ) is integral over R. Hence N[α] = N[α+λ] . It is easy to see that I[α] = I[α+λ] . Thus we have I[α] = N[α] and I[α+λ] = N[α+λ] . So both Rα and R[α + λ] ∩ R[1/(α + λ)] are the integral closure of R in R[α] = R[α + λ], which means that Rα = R[α + λ] ∩ R[1/(α + λ)]. Let J be an ideal of R and let R  be a ring containing R. An element b ∈ R  is said to be integral over J if there exist elements a1 , . . ., an such that ai ∈ J i for each i and such that bn + a1 bn−1 + · · · + an = 0. When R  = R, the set of the elements integral over J is an ideal of R, which is called the integral closure of J and is denoted by J . Proposition 3.2.18 Assume that α is anti-integral over R of degree d. If Rα is integrally closed in R[α], then I[α] = I[α] . Proof Take x ∈ I[α] . Then x n + a1 x n−1 + · · · + an = 0 with ai ∈ (I[α] )i . Since I[α] α = Nα α is integral over R by Theorem 3.2.12 and is contained in R[α], ai α i ∈ R[α] is integral over R. Thus (xα)n + a1 α(xα)n−1 + (a2 α 2 )(xα)n−2 + · · · + an α n = 0, which means that xα belongs to R[α] and is integral over R. Hence x ∈ N[α] = I[α] by Theorem 3.2.12. Therefore we have I[α] = I[α] .

3.3

The Integral Closedness of the Ring R[α] ∩ R[α −1 ] (II)

We start with the following lemma. Lemma 3.3.1 Every nonzero principal ideal of R is integrally closed if and only if R is integrally closed in K .

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Proof Assume that a/b ∈ K (a, b ∈ R) is integral over R. Then there exists an integral dependence over R: (a/b)r + c1 (a/b)r −1 + · · · + cr = 0 We have

a r + bc1 a r −1 + · · · + br cr = 0

Since the principal ideal b R is integrally closed, it follows that a ∈ b R, and hence a/b ∈ R. Conversely consider a principal ideal b R and an integral dependence over b R: a r + c1 a r −1 + · · · + cr = 0 Then we can put ci = h i bi for some h i ∈ R by definition, we have: (a/b)r + h 1 (a/b)r −1 + · · · + h r = 0 Since R is integrally closed in K , it follows that a/b ∈ R, that is, a ∈ b R. So b R is integrally closed. The converse implication of Proposition 3.2.18 is not always valid as will be seen in the following example. Example 3.3.2 Let R be a polynomial ring k[a] over a field k, which is a normal domain. Let ϕα (X ) = X 2 +(1/a)X +(1/a 2 ). Then I[α] = a 2 R. By [KY, Theorem 1], Rα = R ⊕ I[α] ζ1 = R ⊕ I[α] (α + (1/a)) = R ⊕ (a 2 α + a)R, as R-modules. Note that aα ∈ R[α] and aα ∈ Rα. Since a 2 α 2 + aα + 1 = 0, we have (aα)2 = a 2 α 2 = −(aα + 1) and (aα)3 = aα(aα)2 = −aα(aα + 1) = −(aα)2 − aα = (aα + 1) − aα = 1. Hence aα is integral over Rα, which means that Rα is not integrally closed in R[α]. Since R is integrally closed in K , the ideal I[α] = a 2 R is integrally closed by Lemma 3.3.1. Lemma 3.3.3 Let I be an ideal of R. Then I is integrally closed if and only if I p is integrally closed for each p ∈ Ass R (R/I ). Proof (⇒): Take p ∈ Ass R (R/I ), let x ∈ R p be integral over I p . Then we have an integral dependence: x  + a1 x −1 + · · · + a = 0 with ai ∈ (I p )i (1 ≤ i ≤ ). We can find an element s ∈ R \ p such that sai ∈ I i (1 ≤ i ≤ ) and sx ∈ R. Hence we obtain (sx) + (sa1 )(sx)−1 + · · · + s  a = 0 with sai ∈ I i (1 ≤ i ≤ ). Since I is integrally closed, we have sx ∈ I . Thus x ∈ Ip. (⇐): Take x ∈ R which is integral over I . Then there is an integral dependence: x  + a1 x −1 + · · · + a = 0 with ai ∈ I i (1 ≤ i ≤ ). So we have

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a ∈ I i ⊆ (I p )i for each p ∈ Ass R (R/I ). Since I p is integrally closed, we have  x ∈ I p . Hence x ∈ p∈Ass R (R/I ) I p = I . Thus I is integrally closed. Let C(R/R) denote the conductor between R and R, that is, C(R/R) = {a ∈ R|a R ⊆ R}. Theorem 3.3.4 Assume that R is a finitely generated R-module and that the dimension formula holds between R and R. Assume that α is a super-primitive element of degree d over R. Then the following statements are equivalent to each other: (i) the ideal I[α] is integrally closed; (ii) grade(I[α] + C(R/R)) > 1. Proof (ii) ⇒ (i): Let p be a prime divisor of I[α] . Since I[α] is a divisorial ideal, we have p ∈ Dp1 (R). Since α is super-primitive over R, we have (I[α] ) p = a R p for some a ∈ I[α] . Thus the assumption grade(I[α] + C(R/R)) > 1 yields that C(R/R) ⊆ p. Hence R p = R p , which means that (I[α] ) p is integrally closed for each p ∈ Ass R (R/I[α] ) by Lemma 3.3.1. So we conclude that I[α] is integrally closed by Lemma 3.3.3. (i) ⇒ (ii): Note first that C(R/R) = 0 because R is a finitely generated R-module. Suppose that grade(I[α] + C(R/R)) ≤ 1. Then there exists a prime ideal p ∈ Dp1 (R) such that I[α] + C(R/R) ⊆ p. We now claim that p ∈ Ass R (R/R). [In fact, when ht( p) = 1, the inclusion C(R/R) ⊆ p implies that p is a minimal prime divisor of Ann R (R/R) and hence that p ∈ Ass R (R/R). Assume that ht( p) > 1. Since depth(R p ) =1, there exists β ∈ K such that Iβ = p. Since R is a Krull domain, R = P∈Ht1 (R) R P . Take P ∈ Ht1 (R) and put q := P ∩ R. Then by the dimension formula, we have ht(q) = 1. Sinceht( p) > 1, it follows that q ⊆ p = Iβ . Thus β ∈ Rq ⊆ R P and so β ∈ R P = R. Hence we have p ∈ Ass R (R/R). Therefore we have shown that p ∈ Ass R (R/R) in any case.] Since p ∈ Ass R (R/R) by the above claim, there exists γ ∈ R such that γ ∈ R and p = Iγ . Since α is super-primitive and p ∈ Dp1 (R), J[α] R p = R p and hence (I[α] ) p is invertible. So there exists a ∈ R such that (I[α] ) p = a R p . The implications a R p = (I[α] ) p ⊆ p R p = Iγ R p yield that aγ ∈ R p and γ ∈ R p . Since γ ∈ R p , there exists an integral dependence: γ n + b1 γ n−1 + · · · + bn = 0 with bi ∈ R p for some n. Hence we have (aγ )n +(ab1 )(aγ )n−1 +· · ·+a n bn = 0. Since I[α] is integrally closed, so is (I[α] ) p = a R p . Thus aγ ∈ a R p and hence γ ∈ R p , a contradiction. Therefore grade(I[α] + C(R/R)) > 1.

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Corollary 3.3.5 Assume that R is a finitely generated R-module and that the dimension formula holds between R and R. Assume that α is a super-primitive element of degree d over R. If I[α] is integrally closed, then R p is a DVR for each p ∈ Ass R (R/I[α] ). Proof Since I[α] is integrally closed, we have grade(I  [α] + C(R/R)) > 1 by Theorem 3.3.4. Take p ∈ Ass R (R/I[α] ). Since I[α] = 1≤i≤d Iηi is a divisorial ideal, the ideal p is divisorial prime ideal of R. Hence grade( p) ≤ 1. Thus C(R/R) ⊆ p, which means R p = R p . So we have p ∈ Ht1 (R) and R p is a Noetherian normal local domain of dimension 1, that is, a DVR. Recall that N[α] := {a ∈ R|aα is integral over R}. Proposition 3.3.6 Assume that aα is anti-integral over R for all a ∈ R. Then N[α] = {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R}. Proof Let a be an element in R such that aη1 , a 2 η2 , . . ., a d ηd ∈ R. Put b1 = aη1 , b2 = a 2 η2 , . . ., bd = a d ηd . Then (aα)d + b1 (aα)d−1 + · · · + bd = 0. Hence aα is integral over R, which means that a ∈ N[α] by the definition. Conversely, take a ∈ N[α] . Then aα is integral over R. Since aα is anti-integral over R by the assumption, the element aα satisfies a monic polynomial equation of degree d over R by Theorem 2.3.2, that is, ϕaα (X ) ∈ R[X ]. Since α d + η1 α d−1 + · · · + ηd = 0, we have (aα)d + aη1 (aα)d−1 + · · · + a d ηd = 0. Since deg ϕaα (X ) = d, comparing the coefficients of ϕaα (aα) with those of (aα)d + aη1 (aα)d−1 + · · · + a d ηd in K [α], we have aη1 , . . ., a d ηd ∈ R. Hence a ∈ {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R}. Combining Proposition 3.3.6 with Theorem 3.3.12, we have the following result: Proposition 3.3.7 Assume that aα is anti-integral over R for all a ∈ R. Rα is integrally closed if and only if I[α] = {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R}. Proposition 3.3.8 Assume that α is super-primitive over R. If I[α] is integrally closed, then aα is anti-integral over R for every nonzero a ∈ R. Proof Since α is super-primitive over R, (J[α] ) p = R p for every p ∈ Dp1 (R), which means that (I[α] ) p is invertible, that is, (I[α] ) p is a principal ideal of R p . So R p = R p by Lemma 3.3.1 because that I[α] is integrally closed (and hence (I[α] ) p is integrally closed). Since R p − R p for every p ∈ Dp1 (R), aα

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75

is super-primitive over R p for all p ∈ Dp1 (R) by Theorem 2.2.8. Hence aα is anti-integral over R p for all p ∈ Dp1 (R) by Theorem 2.2.8, which implies that aα is anti-integral over R by Theorem 2.4.3. Proposition 3.3.9 Assume that α is a super-primitive element of degree d over R. Then the following statements are equivalent: (i) I[α] = Iη1 and this ideal is integrally closed; (ii) Rα is integrally closed in R[α] and Iηi 1 ⊆ Iηi for all i (1 ≤ i ≤ d). Proof (ii) ⇒ (i): Since Rα is integrally closed in R[α], I[α] is integrally closed by Proposition 3.2.18 and N[α] = I[α] by Theorem 3.2.12. Take a ∈ Iη1 . Then aη1 , a 2 η2 , . . ., a d ηd ∈ R by the assumption. Thus aα satisfies a monic equation of degree d over R, that is, ϕaα (aα) = (aα)d + aη1 (aα)d−1 + · · · + a d ηd = 0. So aα is integral over R, which means that a ∈ N[α] = I[α] . Thus Iηi ⊆ I[α] . The inclusion I[α] ⊆ Iη1 is obvious by definition. (i) ⇒ (ii): Since Iη1 = I[α] ⊆ Iηi , we have Iηi 1 ⊆ Iηi . We must show that Rα is integrally closed in R[α]. For this end, we have only to prove that I[α] = N[α] by Theorem 3.2.12. Take a ∈ N[α] . Then aα is integral over R. Since I[α] is integrally closed and since α is super-primitive over R, aα is antiintegral over R by Proposition 3.3.8. Thus aα satisfies a monic polynomial equation of degree d over R by Theorem 2.3.2, that is, ϕaα (X ) ∈ R[X ]. Since α d + η1 α d−1 + · · · + ηd = 0 in K [α], we have (aα)d + aη1 (aα)d−1 + · · · + a d ηd = 0. So a i ∈ Iηi (1 ≤ i ≤ d). Since deg ϕaα (X ) = d, comparing the coefficients of ϕaα (aα) with those of (aα)d + aη1 (aα)d−1 + · · · + a d ηd , we have aη1 , . . ., a d ηd ∈ R. Hence a ∈ Iη1 = I[α] . Therefore we obtain I[α] = N[α] .

The following result determines the ideal N[α] . Theorem 3.3.10

N[α] = {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R}.

Proof Take a ∈ R such that aη1 , a 2 η2 , . . ., a d ηd ∈ R. Since (aα)d + aη1 (aα)d−1 + · · · + a d ηd = 0, aα is integral over R and hence is integral over R. Hence a ∈ N[α] . Conversely, take a ∈ N[α] . Then aα is integral over R by definition and hence is integral over R. Since R is a Krull domain, aα has a monic relation of degree d over R by Gauss’ lemma. The desired monic relation must be (aα)d +aη1 (aα)d−1 +· · ·+a d ηd = 0. Hence aη1 , a 2 η2 , . . ., a d ηd ∈ R.

Theorem 3.2.12 and Theorem 3.3.10 give rise to the following corollary.

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Corollary 3.3.11

Rα is integrally closed in R[α] if and only if I[α] = {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R}

Example 3.3.12 Even if Rα is integrally closed in R[α], the ideal I[α] is not always a radical ideal of R. For instance, let R be a polynomial ring k[a] over a field k, and let ϕα (X ) = X 2 + (1/a 2 )X + (1/a 2 ). Then I[α] = a 2 R. Since (a 2 α)2 + (a 2 α) + a 2 = 0, we have N[α] = a 2 R. So Rα is integrally closed in R[α] by Theorem 3.2.12. But I[α] is not a radical ideal of R. Let (V, m) be a DVR and let J be a nonzero ideal of V . Then m and J are principal ideals t V and t  V for some  ∈ N, respectively. Let v denote the valuation associated with V . In this case, we write v(J ) := . To obtain the next result, we recall the following: Assume that α is an anti-integral element of degree d over R. Let ϕα (X ) = X d +η1 X d−1 +· · ·+ηd be the minimal polynomial of α over K . Put ζi = α d−i + η1 α d−i−1 +· · ·+ηd−i (1 ≤ i ≤ d −1). Then Rα = R ⊕ I[α] ζ1 ⊕· · ·⊕ I[α] ζd−1 as R-modules (cf. Proposition 3.1.19). Lemma 3.3.13 Assume that R is a DVR with maximal ideal m and that Rα is integrally closed in R[α]. If η1 ∈ R, then I[α] = m, that is, v(I[α] ) = 1, where v( ) denotes the valuation associated with R. d Proof Put v(I[α] ) = e. Since I[α] = i=1 Iηi , I[α] ⊆ Iηi . So v(ηi ) + e ≥ 0 and hence v(ηi ) ≥ −e (1 ≤ i ≤ d − 1). Suppose that e ≥ 3. Take a ∈ R such that v(a) = e − 1. Then a ∈ I[α] . Consider the element (aζ1 )3 . Then (aζ1 )3 = a 3 (−(η1 η2 + η3 ) + (η12 − η1 )ζ1 + 2η1 ζ2 + ζ3 ) We have the following: (1) since v(a 3 η1 η2 ) ≥ 3v(a) + v(η2 ) ≥ 3e − 3 − e = 2e − 3 > 0, then a 3 η1 η2 ∈ R; (2) since v(a 3 η3 ) ≥ 2e − 3 > 0, then a 3 η3 ∈ R; Note that v(a 3 ) = 3e − 3 ≥ e. (3) since v(a 3 η2 ) ≥ 3e − 3 − e = 2e − 3 ≥ e, then a 3 η2 ∈ I[α] . Moreover we see that a 3 η1 , a 3 ∈ I[α] . Hence we have that (aζ1 )3 ∈ Rα but aζ1 ∈ Rα because Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Since aζ1 ∈ R[α], we conclude that Rα is not integrally closed in R[α], which is a contradiction.

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Therefore we have e ≤ 2, that is, v(I[α] ) = 1 or 2, and consequently v(η2 ), v(η3 ), v(η4 ) = 0, −1, −2, respectively. Next suppose that e = 2. Take a ∈ R such that v(a) = e − 1 = 1. We have (aζ1 )4

=

a 4 ((η2 (η2 − η12 ) − 2η1 η3 − η4 ) + (−2η1 η2 − η3 + η13 )ζ1 + (3η12 − η2 )ζ2 + 3η1 ζ3 + ζ4 )

Since v(a) = 1, η1 ∈ R, and v(η2 ), v(η3 ), v(η4 ) ≥ −2, we have v(a 4 η22 ) ≥ 4 + 2(−1) = 0, v(a 4 η2 η12 ) ≥ 4 − 2 = 2 > 0, v(a 4 η1 η3 ) ≥ 4 − 2 = 2 > 0, v(a 4 η4 ) ≥ 4 − 2 = 2 > 0. Thus a 4 (η2 (η2 − η12 ) − 2η1 η3 − η4 ) ∈ R. Next we have: v(a 4 η1 ) ≥ 4 − 2 = 2, v(a 4 η3 )4 − 2 = 2, v(a 4 η13 ) ≥ 4 > 2. Thus a 4 (−2η1 η2 − η3 + η13 ) ∈ I[α] . Similarly, we have a 4 (3η12 − η2 ) ∈ I[α] , a 4 η1 ∈ I[α] , and a 4 ∈ I[α] . Thus (aζ1 )4 ∈ Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 but aζ1 ∈ Rα, which means that Rα is not integrally closed in R[α], a contradiction. Therefore we conclude that e = 1, that is, I[α] = m. Lemma 3.3.14 Assume that R is a DVR with maximal ideal m. Let v( ) denote the valuation associated with R. Put v(I[α] ) = e and v(Iη1 ) = e − f . If I[α] ⊆ Iη1 = R and Rα is integrally closed in R[α], then f = 0 or 1. Proof Suppose that f ≥ 2. Then v(Iη1 ) = e − f ≥ 1 implies that e ≥ 3. Take a ∈ R such that v(a) = e − 1. Then a ∈ I[α] , aζ1 = aα + aη1 ∈ R[α] and aζ1 ∈ Rα. Consider the element (aζ1 )3 = a 3 ζ13 = a 3 (−(η1 η2 + η3 ) + (η12 − η2 )ζ1 + 2η1 ζ2 + ζ3 ) We have v(a 3 η1 η2 ) = 3v(a) + v(η1 ) + v(η2 ) ≥ 3e − (e − f ) − e = e + f − 3 ≥ 0 Thus a 3 η1 η2 ∈ R. We have v(a 3 η3 ) = 3v(a) + v(η3 ) ≥ 3(e − 1) − e = 2e − 3 ≥ 0.

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Thus a 3 η3 ∈ R. We have v(a 3 η12 ) = 3v(a) + 2v(η1 ) = 3(e − 1) − 2(e − f ) = e + 2 f − 3 ≥ e. v(a 3 η2 ) = 3v(a) + v(η2 ) ≥ 3(e − 1) − e = 2e − 3 ≥ e. Thus a 3 η12 , a 3 η2 ∈ I[α] . Similarly we have a 3 η1 , a 3 ∈ I[α] . Therefore (aζ1 )3 ∈ Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . But aζ1 ∈ Rα. So R[α] ∩ R[α −1 ] is not integrally closed in R[α], a contradiction. Lemma 3.3.15 Assume that R is a DVR with maximal ideal m. Let v( ) denote the valuation associated with R. Assume that v(I[α] ) ≥ 2. If Rα is integrally closed in R[α], then v(I[α] ) = v(Iη1 ). Proof Since v(I[α] ) ≥ 2, we have v(Iηi ) ≥ 2 − f ≥ 1, and hence η1 ∈ R by Lemma 3.3.14, whence v(Iη1 ) ≤ v(I[α] ) ≤ v(Iη1 ) + 1 by Lemma 3.3.14. Put g = v(Iη1 ) and suppose that v(I[α] ) = g + 1. Take a ∈ m such that v(a) = g. Consider the element (aζ1 )3 = a 3 ζ13 = a 3 (−(η1 η2 + η3 ) + (η12 − η2 )ζ1 + 2η1 ζ2 + ζ3 ) Then (aζ1 )3 + (a 2 η2 − (aη1 )2 )(aζ1 ) = a 3 (−(η1 η2 + η3 ) + 2η1 ζ2 + ζ3 ) Note here that v(a 2 η2 ) ≥ 2g − (g + 1) = g − 1 ≥ 0 and that v(aη1 ) = v(a) + v(η1 ) = g − g = 0. Thus a 2 η2 − (aη1 )2 ∈ R. Since v(a 3 η1 η2 ) ≥ 3g − g − (g + 1) = g − 1 ≥ 0 and v(a 3 η3 ) ≥ 3g − (g + 1) = 2g − 1 > 0, we have a 3 (−(η1 η2 + η3 )) ∈ R. Since v(a 3 η1 ) = 3g − g = 2g ≥ g + 1, we have a 3 η1 ∈ I[α] . Moreover since v(a 3 ) = 3g ≥ g + 1, we have a 3 ∈ I[α] . Hence (aζ1 )3 + (a 2 η2 − (aη1 )2 )(aζ1 ) ∈ Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . So aζ1 ∈ R[α] is integral over Rα but aζ1 ∈ Rα, which is a contradiction. Hence v(I[α] ) = g = v(Iη1 ). The following theorem gives a perfect form of Proposition 3.2.18. Theorem 3.3.16 Assume that R is a finitely generated R-module and that the dimension formula holds between R and R. Assume further that α is a super-primitive element of degree d over R. Then the following statements are equivalent:

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(i) Rα is integrally closed in R[α]; (ii) I[α] is integrally closed and either (I[α] ) p = p R p or (Iη1 ) p = (I[α] ) p holds for each p ∈ Ass R (R/I[α] ). Proof (ii) ⇒ (i): For this implication, by Propositions 3.3.7 and 3.3.8 we need to show that I[α] = {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R} We have only to show this equality by localizing at each prime divisor of I[α] . Since I[α] is integrally closed, R p is a DVR for each prime divisor p of I[α] by Corollary 3.3.5. By the assumption, (I[α] ) p = p R p or (Iη1 ) p = (I[α] ) p for each p ∈ Ass R (R/I[α] ). In the former case, the above equality holds because I[α] ⊆ {a ∈ R|aη1 , a 2 η2 , . . ., a d ηd ∈ R} is always valid. In the latter case, the condition a ∈ R, aη1 , . . ., a d ηd ∈ R implies aη1 ∈ R and hence a ∈ (Iη1 ) p = (I[α] ) p for each prime divisor p of I[α] . (i) ⇒ (ii) follows from Proposition 3.2.18. Since R p is a DVR for each prime divisor of I[α] by Corollary 3.3.5, the latter statement follows from Lemma 3.3.15. Corollary 3.3.17 Assume that R is a finitely generated R-module and that the dimension formula holds between R and R. Assume further that α is a super-primitive element of degree d over R and that η1 ∈ R. Then Rα is integrally closed in R[α] if and only if I[α] is integrally closed. Proof In this case, condition (ii) of Theorem 3.3.16 is that I[α] is integrally closed and (I[α] ) p = p R p for each p ∈ Ass R (R/I[α] ). Hence I[α] is a radical ideal. Thus Theorem 3.3.11 and Theorem 3.3.16 give rise to the equivalence. We end this section by showing that the matter mentioned in Theorem 3.3.16 is quite plain in the birational case. When α ∈ K , that is, R[α] is a birational extension of R, the following result holds. Proposition 3.3.18 Assume that α ∈ K is a super-primitive element. Then R is integrally closed in R[α] if and only if I[α] = Iα is integrally closed. Proof (⇐): We have only to show that N[α] ⊆ I[α] . Take a ∈ N[α] . Then aα is integral over Rα = R. Since α is super-primitive over R and since I[α] is integrally closed, aα is anti-integral over R by Proposition 3.3.8. Thus aα is integral and anti-integral over R, and hence we have aα ∈ R by Remark 2.2.8. So we conclude that a ∈ Iα = I[α] .

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(⇒): Since α is anti-integral over R, we have Rα = R by definition. Hence I[α] is integrally closed by Proposition 3.2.18.

3.4

Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α)

Lemma 3.4.1

Let a be an element of R. Then the following assertions hold:

(1) ϕα−a (X ) = ϕα (X + a); (2) I[α−a] = I[α] ; (3) if α is an anti-integral element of degree d over R, then so is α − a. Proof

(1) For some λ1 , . . ., λd ∈ K , we can write ϕα (X ) = (X − a)d + λ1 (X − a)d−1 + · · · + λd

Then ϕα−a (X ) = X d + λ1 X d−1 + · · · λd = ϕα (X +a). d (2) In the proof of the assertion (1), we get I[α−a] = i=1 (R : R λi ). Note that λi is an R-linear combination of 1, η1 , . . . , ηi and η is an R-linear i d dcombination (R : R λi ) = i=1 (R : R ηi ). of 1, λ1 , . . ., λi . Hence it is easy to see that i=1 Therefore I[α−a] = I[α] . (3) Let π  : R[X ] → R[α − a] = R[α] be the R-algebra homomorphism defined by π  (X ) = α − a. Take f (X ) ∈ R[X ]. Then we know the following implications: f (X ) ∈ Ker(π  )

⇔

f (X − a) ∈ Ker(π )

⇔

f (X − a) ∈ I[α] ϕα (X )R[X ]

⇔

f (X ) ∈ I[α] ϕα (X + a)R[X ]

⇔

f (X ) ∈ I[α−a] ϕα−a (X )R[X ]

Hence Ker(π  ) = I[α−a] ϕα−a (X )R[X ], which implies that α − a is an antiintegral element of degree d over R. Lemma 3.4.2 Assume that α is an anti-integral element of degree d over R. Let a be a nonzero element of R. Then α is also an anti-integral element of degree d over R[1/a]. Proof

Note that R[1/a] is flat over R. By tensoring the exact sequence: 0 → I[α] ϕα (X )R[X ] → R[X ] → R[α] → 0

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we have an exact sequence: 0 → I[α] ϕα (X )R[1/a][X ] → R[1/a][X ] → R[1/a][α] → 0 Besides, we know that I[α] R[1/a] =

d

(R[1/a] : R[1/a] ηi )

i=1

because R[1/a] is flat over R. Hence α is also an anti-integral element of degree d over R[1/a]. Lemma 3.4.3 Assume that α is a nonzero anti-integral element of degree d over R. Put A := R[α]. Then the following two assertions hold: (1) I[α−1 ] = I[α] ηd = I[α] ϕα (0); (2) I[α−1 ] = α A ∩ R. Proof (1) It is easily verified that ϕα−1 (X ) = X d + ηd−1 ηd−1 X d−1 + · · · + ηd−1 d−1 and I[α−1 ] = i=1 (R : R ηd−1 ηi ) ∩ (R : R ηd−1 ). Let a be an element of I[α−1 ] . Since a is in (R : R ηd−1 ), there exists an element b of R such that a = ηd b. We see that bηi = ηd−1 aηi ∈ R for i = 1, . . ., d − 1 and bηd = a ∈ R. Thus b is in I[α] . So a = bηd ∈ I[α] ηd . Hence we conclude that I[α−1 ] ⊆ I[α] ηd . Conversely, assume that a is in I[α] ηd . Then we can take an element b of I[α] such that a = bηd . Note that a is in R. Since aηd−1 ηi = bηi ∈ R for i = 1, . . ., d − 1 and aηd−1 = b ∈ R, we know that a is in I[α−1 ] . Hence I[α] ηd ⊆ I[α−1 ] . (2) By assertion (1), we have I[α] ηd = I[α−1 ] . So we have only to prove that α A ∩ R ⊇ I[α] ηd . Take a in I[α] . Then ϕα (α) = 0 implies that −α(aα d−1 + aη1 α d−2 +· · ·+aηd−1 ) = aηd . Hence aα d−1 +aη1 α d−2 +· · ·+aηd−1 belongs to A and aηd ∈ R. Hence aηd is in α A ∩ R, and so α A ∩ R ⊇ I[α] ηd = I[α−1 ] . Now suppose that a is in α A∩ R. Then we can write −a = α(a1 +a2 α+· · ·+an α n−1 ) for some a1 , . . . , an ∈ R and a positive integer n. Set f (X ) := an X n + an−1 X n−1 + · · · + a1 X + a. Then f (X ) is in Ker(π ). Therefore  there exist elements h i (X ) ∈ I[α] ϕα (X ) and g ∈ R[X ] such that f (X ) = h i (X )gi (X ).  i This shows that a = f (0) = h i (0)gi (0) ∈ I[α] ηd = I[α−1 ] because h i (0) ∈ I[α] ηd and gi (0) ∈ R. The following result is a generalization of Lemma 3.4.3. Proposition 3.4.4 Assume that α is an anti-integral element of degree d over R. Put A := R[α]. Let a be an element of R such that α − a is not zero. Then I[(α−a)−1 ] = I[α] ϕα (a) = (α − a)A ∩ R

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Proof By Lemma 3.4.1(3), we have that α − a is an anti-integral element of degree d over R. Hence Lemma 3.4.3(2) yields that I[(α−a)−1 ] = (α − a)A ∩ R because R[α − a] = R[α]. On the other hand, Lemma 3.4.3(1) implies that I[(α−a)−1 ] = I[α−a] ϕα−a (0). By Lemmas 3.4.3(1) and 3.4.1(2), we get I[α−a] = I[α] and ϕα−a (0) = ϕα (a). Thus I[(α−a)−1 ] = I[α] ϕα (a). Corollary 3.4.5 Under the same assumptions as in Proposition 3.4.4, α − a is a unit of A if and only if I[α] ϕα (a) = R. Proof We can easily see that α−a is a unit of A if and only if (α−a)A∩ R = R. Hence we reach the conclusion by Proposition 3.4.4. Lemma 3.4.6 Assume that R is a Noetherian domain and that α is a nonzero anti-integral element of degree d over R. Let a be a nonzero element of R. Then the following assertions hold: (1) α −1 is an anti-integral element of degree d over R; (2) α −1 is an anti-integral element of degree d over R[1/a]. Proof (1) is referred to Theorem 3.4.5. (2) is immediate from the assertion (1) and Lemma 3.4.2. Lemma 3.4.7 Assume that R is a Noetherian domain and that α is a nonzero anti-integral element of degree d over R. Let a be a nonzero element of R. Then the following conditions are equivalent to each other: (i) α −1 ∈ R[1/a][α]; (ii) α −1 is integral over R[1/a];  (iii) a ∈ I[α−1 ] . Proof

(i) ⇒ (ii): We can write α −1 = b0 + b1 α + · · · + bn α n

(1)

for some b1 , . . . , bn ∈ R[1/a]. Dividing both sides of equality (1) by α n , we get (α −1 )n+1 = b0 (α −1 )n + b1 (α −1 )n−1 + · · · + bn This shows that α −1 is integral over R[1/a]. (ii) ⇒ (iii): Assume that α −1 is integral over R[1/a]. Then there exists a monic polynomial f (X ) of R[1/a][X ] such that f (α −1 ) = 0. By Lemma 3.4.6(2), we

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see that α −1 is an anti-integral element of degree d over R[1/a]. Hence f (X ) is in I[α−1 ] ϕα−1 (X )R[1/a][X ]. Considering the leading coefficient of f (X ), we see that 1= c/a n for some c ∈ I[α−1 ] and a nonnegative integer n. If n > 0, then a is in I[α−1 ] . If n = 0, then c = 1 and I[α−1 ] = R. Again we have a ∈ I[α−1 ] .  (iii) ⇒ (i): Suppose that a ∈ Iα−1 ] . Then there exists a positive integer r such that a r ∈ I[α−1 ] . Write ϕα−1 (X ) := X d + µ1 X d−1 + · · · + µd , where µ1 , . . . , µd ∈ K . Thus, we have α −1 = −(µ1 + µ2 α + · · · + µd α d−1 ) Hence

a r α −1 = −(a r µ1 + a r µ2 α + · · · a r µd α d−1 )

and a r µ1 , a r µ2 , . . . , a r µd are in R. This means that α −1 is in R[1/a][α].

Proposition 3.4.8 Assume that R is a Noetherian domain and that α is a nonzero anti-integral element of degree d over R. Let a be an element  of R such that α−a is not zero. Then R[α −1 ] ⊆ R[(α−a)−1 ] if and only if a ∈ I[α] ϕα (0). Proof

(⇒): We can write α −1 = a0 + a1 (α − a)−1 + · · · + an (α − a)−1

(1)

for some a0 , a1 , . . ., an ∈ R. There exists a polynomial g(X ) ∈ R[X ] such that X n = (X + a)g(X ) + (−a)n . Then we have (α − a)n = αg(α − a) + (−a)n by substituting α − a for X . Multiplying both sides of equality (1) by (α − a)n , we know that α −1 (α − a)n = g(α − a) + (−a)n α −1 belongs to R[α − a] = R[α]. n −1 Hence (−a) in R[α], and so α −1 belongs to R[1/a][α]. Thus we have  α is  −1 that a ∈ I[α ] = I[α] ϕα (0) by Lemmas 3.4.6 and 3.4.7. (⇐): Lemma 3.4.7 asserts that α −1 is in R[1/a][α = R[1/a][α − a]. So there exist elements b0 , b1 , . . ., bn ∈ R such that (−a)n α −1 = b0 + b1 (α − a) + · · · + bn (α − a)n . Note that bn may be zero. Moreover, there exists a polynomial g(X ) ∈ R[X ] satisfying X n = (X + a)g(X ) + (−a)n with deg(g(X )) ≤ n − 1. Put β := α − a. Then we have (−a)n = β n − αg(β) and (−a)n α −1 = b0 + b1 β + · · · + bn β n

(2)

Substituting β n − αg(β) for (−a)n in the equality (2), we can write β n α −1 = a0 + a1 β + · · · + an β n for some a0 , a1 , . . ., an ∈ R. Hence α −1 = a0 β −n + a1 β−(n − 1)+· · ·+an is in R[β −1 ] = R[(α −a)−1 ]. This shows that R[α −1 ] ⊆ R[(α − a)−1 ]. Theorem 3.4.9 Assume that R is a Noetherian domain and that α is a nonzero anti-integral element of degree d over R. Let a be an element of R such that

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Subrings of Anti-Integral Extensions

α − a is not zero. Then R[α −1 ] = R[(α − a)−1 ] if and only if a ∈  I[α] ϕα (a).



I[α] ϕα (0) ∩

 Proof (⇒): Since R[α −1 ] ⊆ R[(α − a)−1 ], we see that a is in I[α] ϕα (0) by Proposition 3.4.8. On the other hand, we have R[(α − a)−1 ] ⊆ R[α −1 ]. By Lemma 3.4.1, we know that α − a is an anti-integral element of degree d over R and  that I[α−a] ϕα−a (0) = I[α] ϕα (a). Thus Proposition 3.4.8 yields that −a is in I[α−a] ϕα−a (0). So we have the assertion. −1 ⊆ R[(α − a)−1 ] because a is in (⇐):  By Proposition 3.4.8, we get R[α ] −1 I ϕ (0). Similarly, we have R[(α − a) ] ⊆ R[α −1 ] because −a is in   [α] α I[α] ϕα (a) = I[α−a] ϕα−a (0). So we complete the proof. Lemma 3.4.10 Assume that α is an anti-integral element of degree d over R. Then { p ∈ Spec(R) | α is integral over R p } = { p ∈ Spec(R) | I[α] ⊆ p} Proof Assume that I[α] ⊆ p. Then there exists an element a of I[α] such that a ∈ p. Hence aη1 , . . ., aηd are in R, and η1 , . . ., ηd are in R p . Therefore α is integral over R p because ϕα (α) = 0. Conversely, assume that α is integral over R p . Then there exists a monic polynomial f (X ) in R p [X ] such that f (α) = 0. This implies that f (X ) is in I[α] ϕ[α] ϕα (X )R p [X ] because α is an anti-integral element of degree d over R. Considering the leading coefficient of f (X ), we see that I[α] R p = R p . Hence I[α] ⊆ p. Proposition 3.4.11 Assume that R is a Noetherian domain and that α is a nonzero anti-integral element of degree d over R. Let a be an element of R such that α − a is not zero. Then the following two conditions are equivalent to each other:   (i) a ∈ I[α] ϕα (0) ∩ I[α] ϕα (a);   (ii) a ∈ I[α] ϕα (0) = I[α] ϕα (a). Proof (ii) ⇒ (i) is obvious. (i) ⇒ (ii): By Theorem 3.4.9, we see that R[α −1 ] = R[(α − a)−1 ]. Lemma −1 3.4.10 shows  obstruction ideal of integrality of the extension R[α ]  that the of R is I[α−1 ] = I[α] ϕα (0). Also the obstruction ideal of integrality of the    extension R[(α −a)−1 ] of R is I[(α−a)−1 ] = I[α] ϕα (a). Hence I[α] ϕα (0) =  I[α] ϕα (a). Recall that ϕα (X ) = X d + η1 X d−1 + · · · + ηd−1 X + ηd .

3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α)

85

Proposition 3.4.12 Let (R, m) be a Noetherian local domain and assume that α is an anti-integral element of degree d ≥ 2 over R. Assume that I[α] ηd−1 ⊆ m. Then for every a ∈ I[α] ηd ∩ m, the following hold: (1) R[α −1 ] = R[(α − a)−1 ]; (2) R[α] ∩ R[α −1 ] = R[α − a] ∩ R[(α − a)−1 ]. Proof (1) Note that α − a = 0 because d ≥ 2. Since a is in I[α] ηd , there exists an element b of I[α] with a = bηd . Set ai = bηi for i = 1, . . ., d −1 and a0 = b. Then bϕα (a) = a0 a d +a1 a d−1 +· · ·+ad−1 a+a = a(a0 a d−1 +· · ·+ad−1 +1). Put c := a0 a d−1 + · · · + ad−1 + 1. Then c is a unit of R because a and ad−1 are in m. Thus a = c−1 bϕα (a) is in I[α] ϕα (a). Hence a is in I[α] ηd = I[α] ϕα (0)∩ I[α] ϕα (a). Therefore Theorem 3.4.9 implies that R[α −1 ] = R[(α − a)−1 ]. (2) is clear from the assertion (1). Lemma 3.4.13

Let a be an element of R. If α( = 0) is integral over R, then R[α] ∩ R[α −1 ] = R[α − a] ∩ R[(α − a)−1 ]

Proof Since α is integral over R, there exist elements a1 , . . ., an of R such that α n + a1 α n−1 + · · · + an = 0. Hence α = −(a1 + a2 α −1 + · · · + an α −(n−1) ) belongs to R[α −1 ]. By using the fact that α − a is integral over R, we also get R[α − a] ∩ R[(α − a)−1 ] = R[α − a] = R[α]. We come to the conclusion.

Theorem 3.4.14 Assume that R is a Noetherian domain and that α is an anti-integral element of degree  d ≥ 2 over R. Assume that I[α] ηd−1 ⊆ I[α] . Then for every element a ∈ I[α] ∩ I[α] ηd , the following holds: R[α] ∩ R[α −1 ] = R[α − a] ∩ R[(α − a)−1 ] Proof Take p ∈ Spec(R). If I[α] ⊆ p, then α is integral over R p by Lemma 3.4.10. Thus we have R p [α] ∩ R p [α −1 ] = R p [α − a] ∩ R p [(α − a)−1 ] by Lemma 3.4.13. If I[α] ⊆ p, then we have R p [α] ∩ R p [α −1 ] = R p [α − a] ∩ R p [(α − a)−1 ] by Proposition 3.4.12(2) and the assumption I[α] ηd−1 ⊆ I[α] . Therefore R[α] ∩ R[α −1 ] = R[α − a] ∩ R[(α − a)−1 ]. We will generalize Theorem 3.4.9 as follows. Theorem 3.4.15 Assume that R is a Noetherian domain and that α is an anti-integral element of degree d over R. Let a and b be elements of R such

86

Subrings of Anti-Integral Extensions

that α − a  = 0 and α − b = 0. Then R[(α − a)−1 ] = R[(α − b)−1 ] if and only if a − b ∈ I[α] ϕα (a) = I[α] ϕα (b). Proof Put β := α −a. Then α − b = β − (b −a). By Lemma 3.4.1(3), we see that β is also an anti-integral element of degree d over R. Hence Theorem 3.4.9  asserts that R[β −1 ] = R[(β − (b − a))−1 ] if and only if b − a ∈ I[β] ϕβ (0) ∩  I[β] ϕβ (b − a). By Lemma 3.4.1, we see that ϕβ (X ) = ϕα (X + a) and I[β] = I [α]. Hence ϕβ (0) = ϕα (a) and ϕβ (b − a) = ϕ α (b). Similarly,  by using Proposition 3.4.11, we can easily verify that a − b ∈ I[α] ϕα (a) ∩ I[α] ϕα (b)   if and only if a − b ∈ I[α] ϕα (a) = I[α] ϕα (b). So we get the required conclusion. Next we show that the equality R[α − a] ∩ R[(α − a)−1 ] = R[α] ∩ R[α −1 ] holds for any a ∈ R with α − a = 0. Lemma 3.4.16

Let a be an element of R. Then the following assertions hold:

(1) ϕα−a (X ) = ϕα (X + a); (2) I[α−a] = I[α] ; (3) if α is an anti-integral element of degree d over R, then so is α − a. Proof

(1) For some λ1 , . . ., λd ∈ K , we can write ϕα (X ) = (X − a)d + λ1 (X − a)d−1 + · · · + λd

Then ϕα−a (X ) = X d + λ1 X d−1 + · · · λd = ϕα (X +a). d (2) In the proof of the assertion (1), we get I[α−a] = i=1 (R : R λi ). Note that λi is an R-linear combination of 1, η1 , . . . , ηi and η is an R-linear i d dcombination of 1, λ1 , . . ., λi . Hence it is easy to see that i=1 (R : R λi ) = i=1 (R : R ηi ). Therefore I[α−a] = I[α] . (3) Let π  : R[X ] → R[α − a] = R[α] be the R-algebra homomorphism defined by π  (X ) = α − a. Take f (X ) ∈ R[X ]. Then we know the following implications: f (X ) ∈ Ker(π  ) ⇔ f (X − a) ∈ Ker(π ) ⇔ f (X − a) ∈ I[α] ϕα (X )R[X ] ⇔ f (X ) ∈ I[α] ϕα (X +a)R[X ] ⇔ f (X ) ∈ I[α−a] ϕα−a (X )R[X ]. Hence Ker(π  ) = I[α−a] ϕα−a (X )R[X ], which implies that α − a is an antiintegral element of degree d over R. Let ζi := α i + η1 α i−1 + η2 α i−2 + · · · + ηi ∈ K [α] (1 ≤ i ≤ d − 1) and put i := I[α] ζi (1 ≤ i ≤ d −1), which is an R-module. When R is Noetherian, the following result is seen in Theorem 3.1.18. But the same result is valid for any integral domains and the same proof is effective. We show it for convenience sake.

3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α)

87

Lemma 3.4.17 Assume that α is an anti-integral element of d over

degree d−1 R. Then R[α] ∩ R[α −1 ] = R ⊕ 1 ⊕ · · · ⊕ d−1 = R ⊕ ( i=1 I[α] ζi ), as R-modules. Proof We shall show that R[α] ∩ R[α −1 ] = R + 1 + · · · + d−1 , as R-modules. The inclusion R + 1 + · · · + d−1 ⊆ R[α] ∩ R[α −1 ] is obvious. We must show the converse inclusion. Take β ∈ R[α] ∩ R[α −1 ] and put β = a0 α n +· · ·+an = b0 (α −1 )m +· · ·+bm with ai , b j ∈ R. We prove our assertion by induction on m. If m = 0, then β = b0 ∈ R. First we will show the case m < d. Put f (X ) = a0 X n+m + · · · + an X m − bm X m − · · · − b0 ∈ R[X ]. Then f (α) = 0 and hence f (X ) ∈ I[α] ϕα (X )R[X ]. Thus f (0) = −b0 ∈ I[α] ηd , so that there exist c0 , . . ., cd ∈ R with cd = b0 such that c0 α d + c1 α d−1 + · · · + cd = 0. Indeed, since b0 = aηd for some a ∈ I[α] , we may put c0 = a, ci = aηi (1 ≤ i ≤ d). Since m < d, putting γ = cd (α −1 )m + · · · + cd−m+1 = −(cd−m + cd−m−1 α + · · · + c0 α d−m ), we see that γ ∈ d−m and the element β − γ satisfies our induction hypothesis. Since γ ∈ d−m , it follows that β ∈ R + 1 + · · · + d−1 . Secondly we consider the case m ≥ d. As the above statements, there exists c0 , . . ., cd ∈ R with cd = b0 such that c0 α d + c1 α d−1 + · · · + cd = 0. Put γ := cd (α −1 ) = −(c0 α d−1 + · · · + cd−1 ) ∈ d−1 . Then b0 (α −1 )m = γ (α −1 )m . Hence we have β = γ (α −1 )m−1 + · · · + bm = −(c0 α d−1 +· · ·+cd−1 )(α −1 )m−1 +b1 (α −1 )m−1 +· · · bm = (b1 −cd−1 )(α −1 )m−1 + (b2 − cd−2 )(α −1 )m−2 + · · · + bm , which satisfies the induction hypothesis. Thus β ∈ R + 1 + · · · + d−1 . Hence we have proved that R[α] ∩ R[α −1 ] = R + 1 +· · ·+d−1 . Since (R[α]∩ R[α −1 ])⊗ R K = K (α) is a d-dimensional vector space over K , it follows that R[α] ∩ R[α −1 ] is an

R-module of rank d. Hence d−1 R[α] ∩ R[α −1 ] = R ⊕ 1 ⊕ · · · ⊕ d−1 = R ⊕ ( i=1 I[α] ζi ) as R-modules.

Let f : Z → R be a ring homomorphism sending n ∈ Z to n · 1 ∈ R, where Z denotes the ring of integers. Let F := Z/Ker( f ), which is a subdomain of R. Let a be an element of R such that α − a = 0. Then α − a is of degree d over R. Let ϕα−a (X ) := X d + η1 X d−1 + · · · + ηd be the monic minimal polynomial of α − a over K , where η1 , . . ., ηd ∈ K . Put ζi := (α − a)i + η1 (α − a)i−1 + η2 (α − a)i−2 + · · · + ηi ∈ K [α − a] (1 ≤ i ≤ d − 1). Then putting X := Y + a, we have ϕα (X ) = ϕα (Y + a) = (Y + a)d + η1 (Y + a)d−1 + · · · + ηd = Y d +η1 Y d−1 +· · ·+ηd = ϕα−a (Y ) (cf. Lemma 3.4.1(1)). Note that ϕα (a) = ηd . Put β := α − a. Then we have for each k = 1, . . ., d − 2,

 ζk+1

=

 βζk + ηk+1

ζk+1

=

αζk + ηk+1 ,

i.e., αζk = ζk+1 − ηk+1

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Subrings of Anti-Integral Extensions

Comparing the coefficients of ϕα (Y + a) = ϕα−a (Y ), we have





 d − 1 k−1 d − 2 k−2 d k ηk = a a a + η1 + η2 k−1 k−2 k

 d −k+1 a + ηk + · · · + ηk−1 1 for k = 1, . . ., d − 1. So we have ηk ∈ F[a] + F[a]η1 + · · · + F[a]ηd−1 for each k (1 ≤ k ≤ d − 1). Proposition 3.4.18

Under the notation as above, for each k (1 ≤ k ≤ d −1),

ζk − ζk ∈ F[a]ζk−1 + · · · + F[a]ζ1 + F[a] + F[a]η1 + · · · + F[a]ηd−1 Proof

We shall prove this equality by induction on k.

Case (I) k = 1: ζ1

=

β + η1

=

α + a + η1

= =

α + η1 + (η1 − a − η1 ) ζ1 + (η1 − a − η1 )

Hence ζ1 − ζ1 = η1 − a − η1 ∈ F[a] + F[a]η1 . Case (II) Assume that the equality holds for k(≥ 1). That is, there exist λi (a) ∈ F[a] (1 ≤ i ≤ k−1) and λ0 (a) ∈ F[a]+F[a]η1 + · · · + F[a]ηd−1 such that ζk = ζk + λk−1 (a)ζk−1 + · · · + λ1 (a)ζ1 + λ0 (a). Then  ζk+1

=

 βζk + ηk+1

=

(α − a)(ζk + λk−1 (a)ζk−1 + λk−2 (a)ζk−2 (a) + · · · + λ0 (a))

=

αζk + λk−1 (a)αζk−1 + λk−2 (a)αζk−2 + · · · + λ0 (a)α − aζk  − aλk−1 (a)ζk−1 − aλk−2 (a)ζk−2 − · · · − aλ0 (a) + ηk+1

=

(ζk+1 − ηk+1 ) + λk−1 (a)(ζk − ηk ) + · · · + λ1 (a)(ζ2 − η2 ) + λ0 (a)(ζ1 − η1 ) − aζk − aλk−1 (a)ζk−1 − · · · − aλ1 (a)ζ1  − aλ0 (a) + ηk+1

3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α) =

89

ζk+1 + (λk−1 (a) − a)ζk + (λk−2 (a) − aλk−1 (a))ζk−1 + · · · + (λ0 (a) − aλ1 (a))ζ1 − (ηk+1 + λk−1 (a)ηk + · · · + λ0 (a)η1  + aλ0 (a)) + ηk+1

 Thus ζk+1 − ζk+1 ∈ F[a]ζk + F[a]ζk−1 + · · · + F[a]ζ1 + F[a] + F[a]η1 + · · · + F[a]ηd−1 . Hence we are done.

Under these preparations, we have the following result. Theorem 3.4.19 Let R be an integral domain and let α be an anti-integral element of degree d over R. Then for any a ∈ R with α − a = 0, R[α] ∩ R[α −1 ] = R[α − a] ∩ R[(α − a)−1 ] Proof We see that α −a is an anti-integral element of degree d over R and that I[α] = I[α−a] by Lemma 3.4.1. Hence R[α]∩R[α −1 ] = R+I[α] ζ1 +· · ·+I[α] ζd−1  and R[α−a]∩ R[(α−a)−1 ] = R + I[α−a] ζ1 +· · ·+ I[α−a] ζd−1 by Lemma 3.4.17.   By Proposition 3.4.18, we have R + I[α−a] ζ1 + · · · + I[α−a] ζd−1 = R + I[α] ζ1 +  · · · + I[α] ζd−1 ⊆ R + I[α] ζ1 + · · · + I[α] ζd−1 , whence R[α − a] ∩ R[(α − a)−1 ] ⊆ R[α]∩R[α −1 ]. By symmetry, we have R[α]∩R[α −1 ] ⊆ R[α−a]∩R[(α−a)−1 ]. Therefore R[α] ∩ R[α −1 ] = R[α − a] ∩ R[(α − a)−1 ]. Example 3.4.20 Case (d = 3): ζ1 = ζ1 + 2a, ζ2 = ζ2 + aζ1 + a 2 . Case (d = 4): ζ1 = ζ1 + 3a, ζ2 = ζ2 + 2aζ1 + 3a 2 , ζ3 = ζ3 + aζ2 + a 2 ζ1 + a 3 . Now we show the more explicit calculation in the proof of Theorem 3.4.19. Let R be an integral domain with quotient field K and let R[X ] denote a polynomial ring over R. Let α be an element of an algebraic field extension of K , a an element of R, and ϕα−a (X ) the monic minimal polynomial of α − a over K . Then the degree of ϕα−a (X ) is unchanged for any a ∈ R. So let d be this constant value and write ϕα−a (X ) := X d + η1(a) X d−1 + · · · + ηd(a) where η1(a) , . . ., ηd(a) ∈ K . Let ζi(a) := (α − a)i + η1(a) (α − a)i−1 + η2(a) (α − a)i−2 + · · · + ηi(a) , ηi = ηi(0) , and ζi = ζi(0) , for i = 1, . . ., d − 1.

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Subrings of Anti-Integral Extensions

Let π : R[X ] → R[α] be the R-algebra homomorphism defined by π (X ) = α. Define I[α] :=

d

(R : R ηi )

i=1

where (R : R ηi ) := {c ∈ R|cηi ∈ R}. Then I[α] is an ideal of R. An element α is called an anti-integral element of degree d over R if Ker(π ) = I[α] ϕα (X )R[X ]. Let f : Z → R be a ring homomorphism sending n ∈ Z to n · 1 ∈ R, where Z denotes the ring of integers and F := Z/Ker( f ), which is a subdomain of R. Lemma 3.4.21 (1) ϕα−a (X ) = ϕα (X + a). (a) (2) K (ζ1 , . . ., ζd−1 ) = K (ζ1(a) , . . ., ζd−1 ) = K (α). Proof

(1) For some λ1 , . . ., λd ∈ K , we can write ϕα (X ) = (X − a)d + λ1 (X − a)d−1 + · · · + λd

since K [X ] = K [X −a]. Then ϕα−a (X ) = X d +λ1 X d−1 +· · ·+λd = ϕα (X +a). (2) It is clear that K (ζ1(a) ) = K (α). Lemma 3.4.22





 d − 1 k−1 d − 2 k−2 d k a a a + η1 + η2 k−1 k−2 k

 d −k+1 a + ηk + · · · + ηk−1 1

ηk(a) =

for k = 1, . . ., d. Proof

Compare the coefficients of ϕα (X +a) = ϕα−a (X ) (cf. Lemma 3.4.1(1)).

Lemma 3.4.23 ζk+1 (a) ζk+1

For k = 1, . . ., d − 1, = =

αζk + ηk+1 , (α −

a)ζk(a)

ζ 1 = α + η1

(a) + ηk+1 , and ζ1(a) = (α − a) + η1(a)

3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α) Proof

91

Trivial.

It is well-known that

   n+i n+i −1 n+i −1 = + i +1 i +1 i for i ≥ 0. The aim is to give the following explicit equality. Lemma 3.4.24





 d −k d −k+1 2 d −k+2 3 (a) ζ k = ζk + aζk−1 + a ζk−2 + a ζk−3 1 2 3





 d − 3 k−2 d − 2 k−1 d −1 k +··· + a ζ2 + a ζ1 + a k−2 k−1 k i. e., ζk(a)



 k−1  d −k +i −1 i d −1 k a ζk−i + = a i k i=0

for k = 1, . . ., d − 1. Proof  Let theright-hand side of the above equality,  Ak denote d−k+1 d−3i.e.,k−2Ak := d−k+2 3 2 aζ a a a ζ2 + ζk + d−k + ζ + ζ + · · · + k−1 k−2 k−3 1 3 k−2 d−1 k 2 d−2 k−1 (a) a ζ1 + k a . We shall prove ζk = Ak , by induction on k. For k−1   k = 1, A1 = ζ1 + d−1 a = ζ1 + (d − 1)a, and ζ1(a) = (α − a) + η1(a) = 1 d  (α −a)+{ 1 a +η1 } = α −a +da +η1 = α −a +da +(ζ1 −α) = ζ1 +(d −1)a. (a) Hence ζ1(a) = A1 . Assume that ζk(a) = Ak . We shall show that ζk+1 = Ak+1 . By use of Lemmas 3.4.22 and 3.4.23, and by induction hypothesis, we have the following equality: (a) ζk+1

= =

(a) (α − a)ζk(a) + ηk+1 



 d −k d −k+1 2 (α − a) ζk + aζk−1 + a ζk−2 + · · · 1 2



  d − 2 k−1 d −1 k (a) + a ζ1 + a + ηk+1 k−1 k

92

Subrings of Anti-Integral Extensions 

 d −k d −k+1 2 αζk + aαζk−1 + a αζk−2 + · · · 1 2





 d − 2 k−1 d −1 k d −k 2 + a αζ1 + a α − aζk − a ζk−1 k−1 k 1

   d −k+1 3 d −2 k d − 1 k+1 (a) − + ηk+1 a ζk−2 − · · · − a ζ1 − a 2 k−1 k



 d −k d −k+1 2 (ζk+1 − ηk+1 ) + a(ζk − ηk ) + a (ζk−1 − ηk−1 ) 1 2



 d − 2 k−1 d −1 k +··· + a (ζ2 − η2 ) + a (ζ1 − η1 ) − aζk k−1 k





 d −k 2 d −k+1 3 d −2 k − a ζk−1 − a ζk−2 − · · · − a ζ1 1 2 k−1

 d − 1 k+1 (a) − a + ηk+1 k





 d −k d −k+1 2 d − 2 k−1 ζk+1 + aζk + a ζk−1 + · · · + a ζ2 1 2 k−1





 d −1 k d −k 2 d − 3 k−1 + a ζ1 − aζk − a ζk−1 − · · · − a ζ2 k 1 k−2







 d −2 k d − 1 k+1 d −1 k d − 2 k−1 − a ζ1 − a − a η1 − a η2 k−1 k k k−1

  

 d −k d d −1 k k+1 −··· − aηk − ηk+1 + a + a η1 + · · · 1 k+1 k 



 d −k+1 2 d −k + a ηk−1 + aηk + ηk+1 2 1       d −k d −k+1 d −k ζk+1 + − 1 aζk + − a 2 ζk−1 1 2 1      d −1 d −2 d k +··· + − a ζ1 + k k−1 k+1

 d −1 − a k+1 k

=

=

=

=

=

=

3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α) 93





 d −k−1 d −k 2 d −2 k ζk+1 + aζk + a ζk−1 + · · · + a ζ1 1 2 k

 d − 1 k+1 + a k+1 Ak+1

as desired. Lemma 3.4.24 shows that for k = 1, . . ., d − 1,

Remark 3.4.25

ζk(a)

− ζk ∈ F[a]ζk−1 + · · · + F[a]ζ1 + F[a]

Corollary 3.4.26 As R-submodules of K [α], R + Rζ1 + · · · + Rζd−1 = (a) , for any a ∈ R. R + Rζ1(a) + · · · + Rζd−1 Let H be an ideal of R and let a ∈ R. Put (a) C H(a) := R + H ζ1(a) + · · · + H ζd−1

and C H := R + H ζ1 + · · · + H ζd−1 which are R-submodules of K [α]. Corollary 3.4.27

C H = C H(a) , for any a ∈ R.

Proof By Corollary 3.4.26, we have R + Rζ1 + · · · + Rζd−1 = R + Rζ1(a) + (a) (a) , and hence H + H ζ1 + · · · + H ζd−1 = H + H ζ1(a) + · · · + H ζd−1 . · · · + Rζd−1 (a) (a) Thus C H = R + H ζ1 + · · · + H ζd−1 = R + H ζ1 + · · · + H ζd−1 = C H(a) .

We close this section by showing: Let R be an integral domain and α an anti-integral element over R. Let β := (cα − k)/(aα − h) with a, h, c, k ∈ R and ak − hc ∈ R × . We prove that β is anti-integral over R and that the equality R[β] ∩ R[β −1 ] = R[α] ∩ R[α −1 ] holds. Theorem 3.4.28 Assume that α is an anti-integral element of degree d over R. Let β = (cα − k)/(aα − h) with a, h, c, k ∈ R and ak − hc ∈ R × . Then β is anti-integral over R and the equality R[β] ∩ R[β −1 ] = R[α] ∩ R[α −1 ] holds.

94

Subrings of Anti-Integral Extensions

Note that when α, a, h, c, k ∈ C, the field of complex numbers, β is called a linear fractional transformation of α. To begin with, we show the following lemma. Lemma 3.4.29 Assume that α is an anti-integral element of degree d over R. Then the following assertions hold: (1) if u ∈ R × , then uα is an anti-integral element over R; (2) if a ∈ R, then α − a is an anti-integral element over R; (3) α −1 is an anti-integral element over R. Proof (2) and (3) are seen in Lemma 3.4.16. (1) We have only to prove (⇒). Let πuα : R[X ] → R[uα] be a R-algebra homomorphism sending X to uα and ϕuα (X ) ∈ K [X ] be the monic minimal polynomial of uα over K . It is easy to see that deg(ϕuα (X )) = deg(ϕα (X )) = d. Since (uα)d + uη1 (uα)d−1 + · · · + u d ηd = u d ϕα (α) = 0, we have ϕuα (X ) = X d + uη1 X d−1 + · · · + u d ηd d d Since u is a unit in R, we have I[uα] = i=1 (R : R u i ηi ) = i=1 (R : R ηi ) = I[α] . Take f (X ) ∈ R[X ] such that f (uα) = 0 and put f (X ) = an X n + an−1 X n−1 + · · · + a0 with ai ∈ R. Then f (uα) = an u n α n + an−1 u n−1 α n−1 + · · · + a0 . Note that n ≥ d because d is the degree of minimal polynomial of uα. Set g(X ) := an u n X n + an−1 u n−1 X n−1 + · · · + a0 ∈ R[X ]. Then g(α) = f (uα) = 0. So we have g(X ) ∈ Ker(π ) = I[α] ϕα (X )R[X ] because α is anti-integral over R. Thus g(X ) = ϕα (X )h(X ) for some h(X ) ∈ I[α] R[X ]. Put h(X ) := b0 + b1 X + · · · + bn−d X n−d with b j ∈ I[α] and let m(X ) := b0 u −d + b1 u −d−1 X + · · · + bn−d X n−d ∈ R[X ]. Then m(X ) ⊆ I[α] R[X ] = I[uα] R[X ]. Then it is easy to see that g(X ) = ϕuα (u X )m(u X ). Put Y := u X . Then R[X ] = R[Y ] and I[α] R[X ] = I[uα] R[X ] = I[uα] R[Y ]. We see that f (Y ) = g(X ) = ϕuα (Y )m(Y ) ∈ R[Y ]. Therefore f (X ) = ϕuα (X )m(X ) ∈ R[X ] by replacing Y by X formally. Hence Ker(πuα ) ⊆ I[uα] ϕuα (X )R[X ] and the reverse inclusion is obvious. So we have Ker(πuα ) = I[uα] ϕuα (X )R[X ], which means that uα is an anti-integral element of degree d over R. Proof of Theorem 3.4.28. Considering Lemma 3.4.29 and Theorem 2.2.8, we have only to show that for each maximal ideal m of R, β or β −1 is anti-integral over Rm and Rm [β] ∩ Rm [β −1 ] = Rm [α] ∩ Rm [α −1 ]. Let m be a maximal ideal of R. Since ak − hc ∈ R × ⊆ (Rm )× , either (i) ak ∈ (Rm )× or (ii) hc ∈ (Rm )× holds.

3.4 Extensions of Type R[β] ∩ R[β −1 ] with β ∈ K (α)

95

Case (i): ak ∈ (Rm )× means that both a and k are units in Rm . In this case, we have k c 1 α− (hc − ak) cα − k c 2 a a a β= = + = h h aα − h a α− α− a c Put e := h/a, f := c/a, and g := (1/a 2 )(hc − ak). Then e, f ∈ Rm and g ∈ (Rm )× ; β = f + g/(α − e). Put γ := g/(α − e). Then β = γ + f . Since α is anti-integral over Rm , so is α − e; so is (α − e)−1 ; so is γ = g/(α − e); and so is β = γ + f by Lemma 3.4.28. Since γ is anti-integral over Rm , we have Rm [γ ] ∩ Rm [γ −1 ] = Rm [γ + f ] ∩ Rm [(γ + f )−1 ] = Rm [β] ∩ Rm [β −1 ] by Theorem 2.2.8. Moreover Rm [γ ] ∩ Rm [γ −1 ] = Rm [g/(α − e)] ∩ Rm [(α − e)/g] = Rm [1/(α − e)] ∩ Rm [α − e] = Rm [1/(α)] ∩ Rm [α], where the last equality follows from Theorem 2.2.8. Hence we have shown that β is antiintegral over Rm and that Rm [β] ∩ Rm [β −1 ] = Rm [α] ∩ Rm [α −1 ] holds. Case (ii): We have only to consider β −1 instead of β in the argument of Case (i).

Chapter 4 Denominator Ideals and Excellent Elements

Let R be a Noetherian domain and R[X ] a polynomial ring. Let α be an element of an algebraic field extension L of the quotient field K of R and let π : R[X ] → R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with deg ϕα (X ) = d and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd Then ηi ∈ K (1 ≤ i ≤ d) are uniquely determined by α. Let Iηi := R : R ηi and
d I[α] := i=1 Iηi , the latter of which is called a generalized denominator ideal of α. We say that α is an anti-integral element if Ker(π ) = I[α] ϕα (X )R[X ]. For f (X ) ∈ R[X ], let c( f (X )) denote the ideal of R generated by the coefficients of f (X ). For an ideal J of R[X ], let c(J ) denote the ideal generated by the coefficients of the elements in J . If α is an anti-integral element, then c(Ker(π )) = c(I[α] ϕα (X )R[X ]) = I[α] (1, η1 , . . ., ηd ). Put J[α] = I[α] (1, η1 , . . ., ηd ). Let J˜[α] := I[α] (1, η1 , . . ., ηd−1 ). If J[α] ⊆ p for all p ∈ Dp1 (R) := { p ∈ Spec(R)|depth(R p ) = 1}, then α is called a superprimitive element. It is known that a super-primitive element is an anti-integral element (cf. Theorem 2.2.8). It is known that any algebraic element over a Krull domain R is anti-integral over R (cf. Theorem 2.2.9). When α is an element in K , ϕα (X ) = X −α. So we have J[α] = I[α] (1, α) = Iα (1, α) = Iα +α Iα = Iα +Iα−1 , where Iα := R : R α, a denominator ideal of α ∈ K .

4.1

Denominator Ideals and Flatness (I)

For the flatness of anti-integral extension, denominator ideals play important roles. In fact we have the following proposition. Theorem 4.1.1 Assume that α ∈ L is an anti-integral element of degree d over R. If I[α] R[α] = R[α], then R[α] is flat over R.

97

98

Denominator Ideals and Excellent Elements

Proof SinceI[α] R[α] = R[α], there exist elements ai ∈ I[α] (0 ≤ i ≤ n) n ai α i . Take p ∈ Spec(R). Replacing R and R[α] by R p and such that 1 = i=0 R[α] p , we may assume that R is a local ring with maximal ideal m. If I[α] is not contained in m, then I[α] = R, and hence ϕα (X ) ∈ R[X ]. Thus is a R[α] n ai X i . free R-module in this case. If I[α] ⊆ m, then we put f (X ) = 1 − i=0 Then f (α) = 0. Since α is an anti-integral element over R, exist elements there n gi (X ), h i (X ) ∈ R[X ] (0 ≤ i ≤ n) such that f (X ) = i=0 gi (X )h i (X ) and gi (X ) ∈ I[α] ϕα (X ) for all i. Hence we have f (X ) ∈ m[X ]. However, considering a constant term of f (X ), we see that 1 − a0 ∈ m. Since a0 ∈ I[α] ⊆ m, this is a contradiction. Therefore we see J[α] = c(I[α] ϕα (X )) = R. Hence R[α] is flat over R by Theorem 2.3.6. Next we determine the nonflat locus for an integral extension. For this purpose, we prove the following lemma. Lemma 4.1.2 Assume that R[α] is a free R-module of rank d. Then we have R[α] = R + Rα + · · · + Rα d−1 . Proof Put B := R + Rα + · · · + Rα d−1 ⊆ R[α]. Then we have only to show that R[α] p = B p for all p ∈ Spec(R). Hence we may assume that R is a local ring with maximal ideal m. Since R[α] is free R-module of rank d, R[α]/m R[α] is a free R/m-module of rank d, and hence R[α]/m R[α] is a d-dimensional vector space over R/m. Note that R[α]/m R[α] = (R/m)[α] with α ∈ R[α]/m R[α]. Now suppose that 1, α, . . ., α s (s < d) are linearly dependent over R/m and 1, α, . . ., α s−1 are linearly independent over R/m. Then we have R[α]/m R[α] = R/m + (R/m)α + · · · + (R/m)α s−1 . This contradicts the fact that dim(R[α]/m R[α]) = d. Hence 1, α, . . ., α d−1 are linearly independent over R/m and we see that 1, α, . . ., α d are linearly dependent over R/m. Therefore we see that R[α] = R + Rα + · · · + Rα d−1 by Nakayama’s lemma, as desired. Hence we have the following: Proposition 4.1.3 Assume that R[α] is integral over R. Then V (I[α] ) = { p ∈ Spec(R) | R[α] p is not flat over R}. Proof Let [K (α) : K ] = d and consider p ∈ Spec(R) such that I[α] is not contained in p. Then ϕα (X ) is monic polynomial of degree d over R p . Since [K (α) : K ] = d, we see that R[α] p = R p [α] is a free R p -module of rank d, and R[α] p is a flat R p -module, then R[α] p = R p + R p α + · · · + R p α d−1 by Lemma 4.1.2. Thus there exists a monic relation of α of degree d over R p .

4.2 Excellent Elements of Anti-Integral Extensions

99

Therefore ϕα (X ) ∈ R p [X ], and I[α] is not contained in p. This completes the proof. Proposition 4.1.4 Assume that I[α] R[α] = R[α]. Then V (I[α] + I[α−1 ] ) ⊇ { p ∈ Spec(R)|R[α] p is not flat over R p }. Proof Let [K (α) : K ] = d and assume that p does not contain I[α] + I[α−1 ] for p ∈ Spec(R). Then p does not contain I[α] or I[α−1 ] . First, assume that p does not contain I[α] . Then ϕα (X ) ∈ R p [X ]. Hence R[α] p is a free R p -module. Next, assume that p does not contain I[α−1 ] and p ⊇ I[α] . Since p does not contain I[α−1 ] , by the definition of I[α−1 ] , we have that R p [α −1 ] is a free R p -module of rank d. Since I[α] R[α] = R[α] by the assumption, it holds that α −1 is integral over R p . Thus we have an integral relation (α −1 )n + (α −1 )n−1 + · · · + an = 0, ai ∈ R p (1 ≤ i ≤ n), and hence α −1 = −(a1 +a2 α +· · ·+an α n−1 ) ∈ R p [α]. Therefore R[α] p = R p [α] = R p [α, α −1 ]. Since R[α] p is flat over R p [α −1 ], we conclude that R[α] p is flat over R p . Remark 4.1.5 Under the above circumstances, we can see easily that if I[α] + I[α−1 ] = R and I[α] R[α] = R[α], then R[α] is flat over R.

4.2

Excellent Elements of Anti-Integral Extensions

Now we are interested in giving a characterization for an anti-integral extension R[α] of R to be flat. Let R[α] be an anti-integral extension of R and let γ be an element of R[α]. Then we consider the following conditions: (1) R[α] is flat over R; (2) R[α, γ −1 ] is flat over R. It is easy to see that (1) implies (2). Our question is what γ implies (2) ⇒ (1). For this, we introduce the following definition: An element γ in R[α] is said to be an excellent element in R[α] if there exist elements c0 , c1 , . . ., cn ∈ R such that γ = c0 + c1 α + · · · + cn α n and (c0 , c1 , . . ., cn )R = R. It is obvious that α itself is an excellent element in R[α]. Our main theorem in this section is the following:

100

Denominator Ideals and Excellent Elements

Let R[α] be an anti-integral extension of R and let γ ∈ R[α] be an excellent element. Then the following statements are equivalent: (i) R[α] is flat over R; (ii) R[α, γ −1 ] is flat over R. We now start with the following proposition. Proposition 4.2.1 Assume that α is anti-integral over R and let γ ∈ R[α] be a nonzero element. Then the following statements are equivalent: (1) R[α, γ −1 ] is flat over R;  (2) γ ∈ J[α] R[α]. Proof Put A = R[α, γ −1 ]. (2) ⇒ (1): Take P ∈ Spec(A), and let P = P ∩ R[α] andP ∩ R = p. It is obvious that γ ∈ P. Suppose that p ⊇ J[α] . Then P ⊇ J[α] R[α] γ , which is a contradiction. Hence we have p ⊇ J[α] . Since R p [α] is flat over R p (Corollary 2.2.6) and R[α] P is flat over R p [α], we see that R[α] P is flat over R p . Note that R[α] P = A P because γ ∈ P. Thus A P is flat over R p . (1) ⇒ (2): Let P be a minimal prime divisor of J[α] R[α]. We have only to show that γ ∈ P. Suppose the contrary, that is, suppose that γ ∈ P. Put p = P ∩ R. Then J[α] ⊆ p. Hence p is a blowing-up point (Theorem 2.2.5), that is, R p [α]/ p R p [α] is isomorphic to a polynomial ring R p / p R p [T ] = k( p)[T ]. Thus p R p [α] is a prime ideal of R p [α]. Since P R p [α] ⊇ p R p [α] ⊇ J[α] R[α] and the minimality of P, we have P R p [α] = p R p [α]. We also have (∗)

Tr.degk( p) k(P) = 1

Since γ ∈ P, it follows that A P = R[α, γ −1 ] P = R[α] P . The flatness of A over R implies that R[α] P = A P is flat over R p . Hence by [M1,(13.B)], we have dim A P

=

dim R p + dim A P / p A P

= =

dim R p + dim R[α] P / p R[α] P dim R p + dim R[α] P /P R[α] P

=

dim R p + dim k(P)

=

dim R p

Thus we have (∗∗)

dim A P = dim R p

On the other hand, we have Tr.deg R p R p [α] = 0 since α is algebraic over R. Note that R p [α] is an R p -algebra of finite type. So from [M1, (14.C) Theorem 23],

4.2 Excellent Elements of Anti-Integral Extensions

101

it follows that ht(P)

=

ht(P R p )

≤

ht( p R p ) + Tr.deg R p R p [α] − Tr.degk( p) k(P)

= =

ht( p) − Tr.degk( p) k(P) ht( p) − 1 (by (∗))

This implies ht(P) ≤ ht( p) − 1, which contradicts the equality (∗∗). Remark 4.2.2 Let f (X ) and g(X ) be elements in a polynomial ring R[X ]. If c( f (X )) = c(g(X )) = R, then c( f (X )g(X )) = R. Indeed, we may assume that R is a local domain (R, M). Let f (X ) = a0 X s + · · · + as and g(X ) = b0 X t + · · · + bt , where ai ∈ M (respectively b j ∈ M) for 0 ≤ i ≤ m − 1 (respectively for 0 ≤ j ≤ n − 1) and an ∈ M (respectively bm ∈ M). Then the coefficient of X n+m in f (X )g(X ) is not in M. An excellent element has the following property, that is, the set of excellent elements are closed under multiplication. Proposition 4.2.3 excellent element.

If β, γ ∈ R[α] are excellent elements, then βγ is also an

Proof Since β, γ are excellent elements, we can write β = a0 + a1 α + · · · + an α n and γ = b0 +b1 α+· · ·+bt α t such that (a0 , . . ., an )R = (b0 , . . ., bt )R = R for some ai , b j ∈ R. Put f (X ) = a0 + a1 X + · · · + an X n and g(X ) = b0 + b1 X + · · · + bt X t . Then c( f (X )) = c(g(X )) = R. Thus by Remark above, c( f (X )g(X )) = R, which means that βγ is an excellent element in R[α].

In view of Proposition 4.2.3, we have the following result. Proposition 4.2.4 Assume that α is an anti-integral element over R and that γ ∈ R[α] is an excellent element. Then the following statements are equivalent:  (1) γ ∈ J[α] R[α]; (2) J[α] = R. Proof The implication(2) ⇒ (1) is trivial. (1) ⇒ (2): Since γ ∈ J[α] R[α], we have γ n ∈ J[α] R[α] for some n > 0. Since γ is an excellent element in R[α], it follows that γ n is also an excellent

102

Denominator Ideals and Excellent Elements

element by Proposition 4.2.3, and so there exist ci ∈ R such that γ n = c0 + c1 α + · · · + ct α t and (c0 , c1 , . . ., ct )R = R. On the other hand, since γ n ∈ J[α] R[α], we see that γ n = a0 + a1 α + · · · + am α m for some ai ∈ J[α] . Put f (X ) = a0 + a1 X + · · · + am X m − (c0 + c1 X + · · · + ct X t ). Then f (α) = 0 and hence f (X ) ∈ Ker π = I[α] ϕα R[X ] (because α is anti-integral over R). Thus c( f (X )) ⊆ c(Ker π ) = c(I[α] ϕα ) = J[α] . Since a0 , . . ., am belong to J[α] , we conclude that c0 , . . ., ct ∈ J[α] , whence J[α] = R. Combining the above result with Proposition 4.2.1, we have the following result. Theorem 4.2.5 Assume that α is an anti-integral element over R and let γ ∈ R[α] be an excellent element. Then the following statements are equivalent to each other: (1) R[α, γ −1 ] is flat over R; (2) R[α] is flat over R. Proof (2) ⇒ (1) is trivial.  (1) ⇒ (2): Since R[α, γ −1 ] is flat over R, it follows that γ ∈ J[α] R[α] by Proposition 4.2.1. So we have J[α] = R by Proposition 4.2.4, which means that R[α] is flat over R by Theorem 2.3.6. Noting that α itself is an excellent element in R[α], we have the following Corollary. Corollary 4.2.6 Assume that α is an anti-integral element over R. The following statements are equivalent: (1) R[α] is flat over R; (2) R[α, α −1 ] is flat over R;  (3) α ∈ J[α] R[α]; (4) J[α] = R;  (5) every excellent element belongs to J[α] R[α]. Proof (1) ⇔ (2) follows from Theorem 4.2.5, (2) ⇔ (3) follows from Proposition 4.2.1, and (3) ⇔ (4) and (4) ⇔ (5) follow from Proposition 4.2.4.

Finally, we give a characterization of excellent elements.

4.3 Flatness and LCM-Stableness

103

Proposition 4.2.7 Assume that α is an anti-integral element over R and let γ ∈ R[α]. Then the following statements are equivalent: (1) γ is an excellent element; (2) (a0 , a1 , . . ., an )R + J[α] = R for each representation γ = a0 + a1 α + · · · + an α n of γ with ai ∈ R. Proof (1) ⇒ (2): Suppose that (a0 , a1 , . . ., an )R + J[α] = R for some representation γ = a0 + a1 α + · · · + an α n . Then there exists p ∈ Spec(R) such that (a0 , a1 , . . ., an )R + J[α] ⊆ p. Since γ is an excellent element in R[α], we have γ = c0 + c1 α + · · · + ct α t with (c0 , c1 , . . ., ct )R = R for some ci ∈ R. Put f (X ) = (c0 + c1 X + · · · + ct X t ) − (a0 + a1 X + · · · + an X n ). Then we see f (α) = 0, and hence c( f (X )) ⊆ J[α] ⊆ p. Since a0 , a1 , . . ., an ∈ p, it follows that c0 , c1 , . . ., ct ∈ p. This implies that p = R, a contradiction. γ = (2) ⇒ (1): Since (a0 , a1 , . . ., an )R + J[α] = R for any representation  a0 + a1 α + · · · + an α n of γ with ai ∈ R, we have 1 = ( ai xi ) + y for some xi ∈ R, y ∈ J[α] . Since J[α] = c(I[α] ϕα (X )) by definition, t there exist ci ∈ J[α] , f i (X ) ∈ I[α] ϕα (X ), and bi ∈ R such that y = i=1 ci bi , ci is one of the coefficients of f i (X ), and f i (α) = 0 foreach i. Thus we have t γ = a0 + a1 α + · · · + an α n = a0 + a1 α + · · · + an α n + i=1 bi α n+di f i (α) for a sufficiently large integer di > 0 and (a0 , a1 , . . ., an , b1 c1 , b2 c2 , . . ., bt ct ) = R. It follows that γ is an excellent element in R[α].

4.3

Flatness and LCM-Stableness

Let C be a ring extension of a ring D. We say that C is LCM-stable over D if (a D ∩ bD)C = aC ∩ bC for any a, b ∈ D. Let R be a Noetherian integral domain and α be a nonzero element of an algebraic field extension of the quotient field K of R. Let R α := R[α] ∩ R[α −1 ], let IαR α := {b ∈ R α |bα ∈ R α }, and JαR α := IαR α + α IαR α . In the birational case, F. Richman, H. Uda, J. Sato, and K. Yoshida showed that if R[α] is LCM-stable over R, then R[α] is flat over R ([R], [SY2], and [U]). But, in general, this is not the case [U] (cf. Example 4.3.14) ]. Lemma 4.3.1 Assume that α is an anti-integral element over R. Then R[α] is flat over R if and only if R[α] is flat over R α . Proof Since R α [α]∩ R α [α −1 ] ⊆ R α = R α , it follows that R α [α]∩ R α [α −1 ] = R α , and hence α is an anti-integral element over R α (cf. Definition 1.1.4).

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Denominator Ideals and Excellent Elements

(⇐) Suppose that R[α] is flat over R α . Since R α is integral over R, R α is not a blowing-up at any point in Spec(R). Then R[α] is flat over R by Theorem 2.3.6. (⇒) Suppose that R[α] is flat over R. Take P ∈ Spec(R[α]) and put p = P ∩ R. Since k(P)/k( p) is an algebraic field extension, R[α] is not a blowing-up at any point in Spec(R). Then Theorem 2.3.6 implies that R[α] is flat over R α .

Lemma 4.3.2 Let a, b ∈ R be an R-regular sequence. Then there exists P ∈ Dp1 (R[b/a]) such that P ∩ R is of grade > 1. Proof Put α = b/a. Since a, b is an R-regular sequence, we have that Iα = a R. Since Jα = (a, b)R is of grade > 1, α is super-primitive, and so α is anti-integral. Now R[α]/Jα R[α] ∼ = (R[X ]/Iα (X − α)R[X ])/Jα R[X ]) ∼ = ∼ R[X ]/Jα R[X ] = (R/Jα )[X ]. Thus, in particular, a R[α] = R[α]. Let P be a minimal prime divisor of a R[α]. It then follows that a, b ∈ p = P ∩ R. Hence depth(R p ) > 1. Using Lemma 4.3.2, we can give the following result. Proposition 4.3.3 If α ∈ K and Iα is an invertible ideal of R, then the following statements are equivalent: (1) for each P ∈ Dp1 (R[α]), P ∩ R belongs to Dp1 (R); (2) R[α] is flat over R. Proof We may assume that R is a local ring. Then Iα = a R for some a ∈ R because Iα is invertible. So there exists an element b ∈ R such that α = b/a. Assume that (1) holds. Then by Lemma 4.3.2, a, b cannot be an R-regular sequence. Since Iα = a R, we have (a, b)R = R. Now x, y ∈ R satisfy ax +by = 1. Then 1/a = x +α ∈ R[α], and so R[α] = R[1/a], which implies that R[α] is flat over R. Conversely, assume that (2) holds. Let P ∈ Dp1 (R[α]), and put p = P ∩ R. If a, b ∈ p is an R-regular sequence, then a R : b R = a R. Since R[α] is flat over R, we conclude that a R[α] : b R[α] = a R[α]. Thus a, b is an R[α]-regular sequence, which is a contradiction. Lemma 4.3.4 Assume that α is an anti-integral element over R. Then we  R α ⊇ J[α] . have that Jα Proof Take R ∈ Spec(R α ) and put p = P ∩ R. Now suppose that p ⊇ J[α] . Since R[α] p is flat over R p (Theorem 2.3.6) and R[α] p is anti-integral over

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R α R p , we see that R[α] p is flat over  R α p by Lemma 4.3.1. Hence P ⊇ Jα

(Theorem 2.3.6). Thus we have

R α

Jα

⊇ J[α] .

Proposition 4.3.5 Assume that α is a super-primitive element over R. Then α is also a super-primitive element over R α . Proof Take P ∈ Dp1 (R α ). Suppose that P ⊇ JαR α . Then it follows from Lemma 4.3.4 that p = P ∩ R ⊇ J[α] . So by Lemma 3.1.28, we see that p ∈ Dp1 (R). This contradicts that α is a super-primitive over R. Hence α is super-primitive over R α . Theorem 4.3.6 Assume that α is a super-primitive element over R. Then the following statements are equivalent: (1) IαR α is an invertible ideal of R α and P ∩ R ∈ Dp1 (R) for each P ∈ Dp1 (R[α]); (2) R[α] is a flat R-module. Proof (2) ⇒ (1) Suppose that R[α] is flat over R. By Lemma 4.3.1, R[α] is flat over R α , R[α] = R α [α] is anti-integral over R α by Proposition 4.3.5, it follows from Proposition 2.5.4 that R α = JαR α = IαR α (1, α). Hence IαR α is an invertible ideal of R α . Then by Proposition 4.3.3 and Lemma 3.1.28, we see that P ∩ R ∈ Dp1 (R) for each P ∈ Dp1 (R[α]). (1) ⇒ (2) First, suppose that IαR α R[α] = R[α]. Then R[α] is flat over R α (Theorem 4.1.1 and Proposition 4.3.5). Next, suppose that IαR α R[α] = R[α]. Let P be a prime divisor of IαR α R[α]. Put q = P ∩ R α , and localize at q. We have that IαR α R[α]q is a principal ideal of R[α]q . Hence P ∈ Dp1 (R[α]). Further, we have JαR α = IαR α + α IαR α ⊆ q. Then Proposition 4.3.5 implies that depth(R α q ) > 1. On the other hand, we see that Lemma 4.3.4 implies that p = q ∩ R ⊇ J[α] . However, we have that depth(R p ) > 1 since α is a superprimitive element over R, which is a contradiction. Therefore IαR α R[α] = R[α], and so R[α] is flat over R α . Hence by Lemma 4.3.1, R[α] is flat over R. Lemma 4.3.7 If R[α] is LCM-stable over R, then P ∩ R ∈ Dp1 (R) for each P ∈ Dp1 (R[α]). Proof Let P ∈ Dp1 (R[α]), and put p = P ∩ R. Suppose that depth(R p ) > 1. Then there exists an R p -regular sequence a, b of R p . Since R[α] is LCM-stable over R, we know that (a R p ∩b R p )R[α] p = a R[α] p ∩b R[α] p = (ab R p )R[α] p . Hence a, b is an R[α] p -regular sequence, a contradiction.

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Theorem 4.3.8 Assume that α is an anti-integral element over R. If IαR α is an invertible ideal of R α and R[α] is LCM-stable over R, then R[α] is flat over R. Proof First note that by Lemma 4.3.7, we have P ∩ R ∈ Dp1 (R) for each P ∈ Dp1 (R[α]). Since IαR α is an invertible ideal of R α , α is a super-primitive element over R α (Theorem 2.3.10). So we conclude that R[α] is flat over R by Theorem 4.3.6. Lemma 4.3.9 Let R ⊆ A be an extension of integral domains such that A is LCM-stable over R, and let P be a prime ideal of A of depth 1. Then P ∩ R is also of depth 1. Proof Put p = P ∩ R and assume that depth(R p ) > 1. Then there exists an R p -regular sequence a, b in p R p . We have a R p ∩ b R p = ab R p . Since A is LCM-stable over R, it follows that a A p ∩ b A p = ab A p , and hence a A P ∩ b A P = ab A P . This shows that a, b is an A P -regular sequence in P A P , which contradicts depth(A P ) = 1. Lemma 4.3.10 Let A := R[X ]/I , where I is a nonzero prime ideal of R[X ] with I ∩ R = (0), and let p be a prime ideal of R containing c(I ). Then p A is a prime ideal of A such that p A ∩ R = p and that ht( p A) ≤ ht( p) − 1. In particular, ht(c(I )) ≥ 2. Proof Since p contains c(I ), we have I ⊆ p R[X ], and hence p A = p R[X ]/I . Thus p A is a prime ideal satisfying p A∩ R = p. Recall that ht( p R[X ]) = ht( p) because R is Noetherian. Therefore ht( p A) = ht( p R[X ])/I ) ≤ ht( p R[X ]) − 1 = ht( p) − 1 Note that ht( p A) = 0, and if ht( p A) = 0, then p R[X ] = I , which contradicts I ∩ R = (0). Hence we have ht( p) ≥ 2, which implies that ht(c(I )) ≥ 2.

Lemma 4.3.11 Let the notation and the assumption be the same as in the previous lemma, and assume that A is LCM-stable over R. If ht( p) = 2, then depth(R p ) = 1. Proof Let P = p A. Then we have P ∩ R = p and ht(P) = 1 by Lemma 4.3.10. The assertion then follows from Lemma 4.3.9. Now we are in a position to show the following theorem.

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107

Theorem 4.3.12 Assume that α is a super-primitive element over R and that J[α] = R. If R[α] is LCM-stable over R, then ht(J[α] ) ≥ 3. Proof

Our conclusion is an immediate consequence of Lemma 4.3.11.

Concerning Lemma 4.3.11, we note the following. Proposition 4.3.13 Assume that R satisfies the S2 -condition, that is, prime ideals of depth 1 are of height 1, and α is an anti-integral element over R. Then α is super-primitive over R. Proof Let p be a prime ideal of R containing J[α] . Then we have ht( p) ≥ 2 by Lemma 4.3.10, and hence depth(R p ) ≥ 2 because of the S2 -condition. This shows that J[α] is not contained in any depth 1 prime ideal of R, and therefore α is super-primitive over R. We conclude this section by giving the following example, which shows that the converse statement of Theorem 4.3.12 does not hold in general. Example 4.3.14 Let B := k[s, t, u] be a polynomial ring in three variables over a field k and α be a root of the polynomial s X 2 + t X + u in a suitable extension field L of k(s, t, u). Let R := B[sα] and consider the extension R ⊆ A := R[α]. We shall show that ht(J[α] ) = grade(J[α] ) = 3 and that A is not LCM-stable over R. Note that R = B + sα B and R is a free B-module of rank 2. By making use of this fact, it is easy to check that I[α] = (s, sα + t)R. Hence we have J[α] = I[α] + α I[α] = (s, t, u, sα)R which is a maximal ideal of height 3. Since s, t, u is a B-regular sequence and R is a free B-module, s, t, u is an R-regular sequence. On the other hand, we have α ∈ (s, t, u)R and (sα)2 ∈ (s, t, u)R. Thus grade(J[α] ) = 3, as claimed. Recall that, in the case of a birational extension of integral domains, LCM-stability is equivalent to flatness (cf. [Ri]). Since A is not flat over R, we know that A is not LCM-stable over R. Remark 4.3.15 H. Uda [U] proved: If R is locally a GCD-domain, not necessarily Noetherian, then R[α] being LCM-stable over R is equivalent to grade(J[α] ) ≥ 3. Using this, H. Uda presented the following example [U]:

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Let R := k[s, t, u] be a polynomial ring in three variables over a field k. Then R[X ]/(s X 2 + t X + u) is LCM-stable over R, but not flat over R. Example 4.3.14 shows that Uda’s theorem above does not hold in general without the assumption that R is locally a GCD-domain, even if R is Noetherian.

4.4

Some Subsets of Spec(R) in the High Degree Case

Let S be an R-algebra of finite type. A prime ideal P ∈ Spec(S) is called unramified over R (or S/R unramified at P) if pS P = P S P , where p := P ∩ R, and k(P)/k( p) is a finite separable extension. A prime ideal p ∈ Spec(R) is called unramified in S if all P ∈ Spec(S) lying over p are unramified over R (or S/R is unramified at p) (cf. [AK]). We say that S is unramified over R if every P ∈ Spec(S) is unramified over R. We recall that S is unramified over R if and only if the differential module  R (S) = (0). Let R be a Noetherian domain with the quotient field K and L be a finite algebraic extension field of K such that [L : K ] = d. Let α ∈ L. Let L be the quotient field of R[α]. When R[α] is an unramified extension of R, we call α an unramified element over R. Let α ∈ L be an element unramified over R. For an ideal N of R, put V (N ) := { p ∈ Spec(R)|N ⊆ p}. Let  R[α]/R := { p ∈ Spec(R)| p R[α] = R[α]} and let J[α] := { p ∈ Spec(R)| p + J[α] = R}. Recall that J˜[α] := I[α] (1, η1 , . . ., ηd−1 )R. Proposition 4.4.1 Assume that α is anti-integral over R. J˜[α] = R if and only if the following two conditions hold: (1) R[α] is a flat R-module; (2) ϕ : Spec(R[α]) −→ Spec(R) is surjective. Proof (⇐) Suppose that J˜[α] = R. Then there exists a prime ideal p of R such that p ⊇ J˜[α] . Since ϕ is surjective, there exists a prime ideal P of R[α] such that P ∩ R = p. Thus J˜[α] R[α] ⊆ P. We claim that J˜[α] R[α] = R[α]. Let a be any element of I[α] . Since ϕα (α) = 0, we have that aα d + (aη1 )α d−1 + · · · + (aηd−1 )α = −aηd ∈ J˜[α] R[α]. Hence J˜[α] R[α] ⊇ J˜[α] R[α] + I[α] ηd ⊇ J˜[α] + I[α] ηd = J[α] . Since R[α] is R-flat, we have that J[α] = R by Theorem 2.2.5 or Proposition 2.5.4, and so J˜[α] R[α] ⊇ J[α] R[α] = R[α]. Therefore J˜[α] R[α] = R[α]. Hence P = R[α], which is a contradiction. Thus J˜[α] = R. (⇒) If J˜[α] = R, then J[α] = R. By Theorem 2.2.5 or Proposition 2.5.5, we have that R[α] is R-flat. Now, by the following Theorem 4.4.2(1), Im(ϕ) =

4.4 Some Subsets of Spec(R) in the High Degree Case

109

D( J˜[α] ) ∪ V (J[α] ), we will have that Im(ϕ) = Spec(R). Hence ϕ is surjective.

Theorem 4.4.2 Assume that α is an anti-integral element of degree d over R. Then the following results hold: (1) Im(ϕ) = D( J˜[α] ) ∪ V (I[α] ηd ) = D( J˜[α] ) ∪ V (J[α] ); (2)  R[α]/R = V ( J˜[α] ) ∩ I[α] ηd = V ( J˜[α] ) ∩ J[α] . Proof First, we shall prove (1). Let p ∈ D( J˜[α] ) and π : R[X ] → R[α] be the canonical map. Then R[α] p = R p [X ]/(Ker(π ))R p [X ] and R[α] p / p R[α] p ∼ = (R p / p R p )[X ]/(Ker(π ))(R p / p R p )[X ] where Ker(π) denotes the image of Ker(π ) in (R p / p R p )[X ]. Since p ⊇ J˜[α] ,  be a prime divisor of p R[α] p . Ker(π ) is not 0. Thus p R[α] p = R[α] p . Let P  ∩ R[α] = P. Then P  ∩ R p = p R p and so P ∩ R = p. Hence p ∈ Im(ϕ). Put P Next, let p ⊇ ηd I[α] . We recall that if p ⊇ J˜[α] then p ∈ Im(ϕ). So, we suppose that p ⊇ J˜[α] . Then p ⊇ J[α] . It follows that R[α]/ p R[α] ∼ = (R/ p)[T ], where T is an indeterminate. Since p R[X ] ⊇ Ker(π ), P = p R[α] ∈ Spec(R[α]), and P ∩ R = p, we have that p ∈ Im(ϕ). Thus we proved that D( J˜[α] ) ∪ V (J[α] ) ⊆ D( J˜[α] ) ∪ V (I[α] ηd ) ⊆ Im(ϕ). To prove the opposite inclusion, assume that p ∈ D( J˜[α] ) ∪ V (I[α] ηd ). We claim that p ∈ Im(ϕ). Suppose that p ∈ Im(ϕ). There exists a prime ideal P of R[α] such that P∩R = p. Since p ⊇ ηd I[α] , there exists some element a ∈ I[α] such that aηd ∈ p. Since p ⊇ J˜[α] a, aη1 , . . ., aηd−1 , it follows that P (aα d + aη1 α d−1 + · · · + aηd−1 α) = −aηd . So p −aηd . This is a contradiction. We have that p ∈ Im(ϕ). Hence we have that Im ϕ = D( J˜[α] ) ∪ V (I[α] ηd ). We claim that D( J˜[α] ) ∪ V (I[α] ηd ) = D( J˜[α] ) ∪ V (J[α] ). Clearly, it follows that D( J˜[α] ) ∪ V (I[α] ηd ) ⊇ D( J˜[α] ) ∪ V (J[α] ). To prove the opposite inclusion, assume that p ∈ D( J˜[α] ) ∪ V (J[α] ). Then p ⊇ J˜[α] and p ⊇ J[α] . Suppose that p ∈ V (I[α] ηd ) ∪ D( J˜[α] ). Then p ⊇ I[α] ηd + J˜[α] = J[α] .This is a contradiction. Therefore we have that p ∈ V (I[α] ηd ) ∪ D( J˜[α] ). Thus we complete the proof of the claim. (2) Let p ∈ V ( J˜[α] ) ∩ I[α] ηd . Suppose that p ∈  R[α]/R . Then there exists a prime ideal P of R[α] such that P ⊇ p R[α]. Put P ∩ R = q. Then we have that q ⊇ p and q ∈ Im(ϕ). Since a, aη1 , . . ., aηd−1 ∈ J˜[α] ⊆ p ⊆ P, for any a ∈ I[α] , we have that aα d + aη1 α d−1 + · · · + aηd−1 α = −aηd ∈ P ∩ R = q Consequently, we have that I[α] ηd ⊆ q. Since p ∈ I[α] ηd , it follows that q ⊇ I[α] ηd + p = R. This is a contradiction. Therefore we proved that V ( J˜[α] )∩

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I[α] ηd ⊆  R[α]/R . Conversely, we shall prove that  R[α]/R ⊆ V ( J˜[α] ) ∩ I[α] ηd . Let p ∈ V ( J˜[α] ) ∩ I[α] ηd . Then p ⊇ J˜[α] or p + I[α] ηd = R. If p ⊇ J˜[α] , then p ∈ Im(ϕ) by Theorem 4.4.2(1). Therefore there exists a prime ideal P of R[α] such that P ∩ R = p. Suppose that p ∈  R[α]/R . Then P ⊇ p R[α] = R[α]. This is a contradiction. Thus p ∈  R[α]/R . Next, if p + I[α] ηd = R, then there exists a prime ideal q of R such that p + I[α] ηd ⊆ q. Since q ∈ Im(ϕ) by Theorem 4.4.2(1), it follows that q R[α] = R[α]. Since p ⊆ q, we have that p R[α] = R[α]. So we get p ∈  R[α]/R . Hence we complete the proof of  R[α]/R = V ( J˜[α] ) ∩ J[α] . At last, we shall prove that V ( J˜[α] ) ∩ I[α] ηd = V ( J˜[α] ) ∩ J[α] . Clearly, we have that V ( J˜[α] ) ∩ I[α] ηd ⊆ V ( J˜[α] ) ∩ J[α] . So we shall prove that V ( J˜[α] ) ∩ J[α] ⊆ V ( J˜[α] ) ∩ I[α] ηd . Let p ∈ V ( J˜[α] ) ∩ J[α] . Suppose that p ∈ V ( J˜[α] ) ∩ I[α] ηd . Then p + I˜[α] = R. Also there exists a prime ideal q of R such that q ⊇ p + I[α] ηd . Since q ⊇ J[α] and p + J[α] = R, we have that q ⊇ p + J[α] = R. This is a contradiction. Hence p ∈ V ( J˜[α] ) ∩ I[α] ηd . This completes the proof.

Chapter 5 Unramified Extensions

Throughout this chapter, the following notations will be in force unless otherwise specified: R: a Noetherian integral domain, K := K (R): the quotient field of R, L: an algebraic field extension of K , α: a nonzero element of L , d = [K (α) : K ], ϕα (X ) = X d + η1 X d−1 + · · · + ηd , the minimal polynomial of α over K . d d I[α] := i=1 (R : R ηi ) = i=1 Iηi , which is an ideal of R. Ia := R : R a R for a ∈ K . It is clear that for a ∈ K , I[a] stands for Ia from definitions. J[α] := I[α] (1, η1 , . . ., ηd ), J˜[α] := I[α] (1, η1 , . . ., ηd−1 ). We also use the following standard notation: Dp1 (R) := { p ∈ Spec(R)|depth(R p ) = 1}. Let A be an R-algebra. Then  R (A) denotes the universal module of differentials of A over R.

5.1

Unramifiedness and Etaleness of Super-Primitive Extensions

Let S be an R-algebra of finite type. A prime ideal P ∈ Spec(S) is called unramified over R (or S/R unramified at P) if pS P = P S P , where p := P ∩ R, and k(P)/k( p) is a finite separable algebraic extension. A prime ideal

111

112

Unramified Extensions

p ∈ Spec(R) is called unramified in S if all P ∈ Spec(S) lying over p are unramified over R (or S/R is unramified at p) (cf. [AK]). It is known that  R (S) = 0 if and only if S is unramified over R. The following result can be proved by using [AK, V1(6.8)] but we give a direct proof. If α is super-primitive and integral over R, R[α] is finite, flat over R (cf. Theorems 2.3.2 and 2.3.6.). Proposition 5.1.1 Assume that α is an anti-integral element and is integral over R. Then R[α] is unramified over R if and only if R[α] p is unramified over R p for each p ∈ Dp1 (R). Proof Since R[α] is integral over R, ϕα (X ) ∈ R[X ] by Theorem 2.3.2. For a polynomial f , we denote the derivative of f by f  . Then ϕα (α) = dα d−1 + (d − 1)η1 α d−2 + · · · + ηd−1 and let p ∈ Spec(R). Then ϕα (α)R[α] ⊆ P for any P ∈ Spec(R[α]) with P ∩ R = p if and only if R[α] p is unramified over R p (cf. [AK, V1(6.12)]). Suppose that ϕα (α)R[α] = R[α]. Then there exists P ∈ Ht1 (R[α]) such that ϕα (α) ∈ P. Put p = P ∩ R. Then depth(R[α] P ) = 1 implies depth(R p ) = 1 because R[α] p is flat over R p . Thus R[α] p is unramified over R p by the assumption. Hence R[α] P is unramified over R p , which is a contradiction. So ϕα (α)R[α] = R[α], which means that R[α] is unramified over R. Remark 5.1.2 Let the notation be the same as in Proposition 5.1.1 and its proof. Let B = R[α, α −1 ]. Then for P ∈ Spec(B), B P is unramified over R R∩P if and only if P ⊇ ϕα (α)B. Indeed, let P ⊆ B be a prime ideal and put Q = P ∩ R[α] and p = P ∩ R. When B P /R p is ramified, R[α] Q /R p is ramified. So ϕα (α) ∈ Q ⊆ P. Conversely, if ϕα (α) ∈ P, then Q = P ∩ R[α] # ϕα (α). So B P = R[α] Q is ramified over R p . It is known that the purity of branch locus holds for a finite flat extension [AK]. The following is a result similar to this fact. Proposition 5.1.3 Assume that α is a super-primitive element and is flat over R and that R contains an infinite field k. Then R[α] is unramified over R if and only if R[α] p is unramified over R p for any p ∈ Dp1 (R). Proof We have only to consider the case that R is a local ring. So we may assume that (R, m) is a local ring. If R[α] is integral over R, we have shown this in Proposition 5.1.1. Assume that R[α] is not integral over R. Since J[α] = R by Theorem 2.3.6, replacing α by α − λ for some λ ∈ k, we may assume by

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113

Proposition 2.2.10 that α satisfies that (a) α −1 ∈ R[α]; (b) α −1 is a super-primitive element of degree d over R; (c) α −1 is integral over R. Hence we have R ⊆ R[α −1 ] ⊆ R[α, α −1 ] = R[α] Apply Remark 5.1.2 to B = R[α −1 ][(α −1 )−1 ] = R[α]. We conclude that for P ∈ Spec(R[α]), R[α] P is unramified over R P∩R if and only if P ⊇ ϕα −1 (α −1 )R[α]. In the same way as in the proof of Proposition 5.1.1, the assumption that R[α] p is unramified over R p for any p ∈ Dp1 (R) yields that R[α] is unramified over R. Let S denote an R-algebra of finite type. We say that S is etale over R if S is flat and unramified over R (cf. [AK]). As a consequence of Propositions 5.1.1 and 5.1.3, we obtain the following theorem. Theorem 5.1.4 Assume that α is a super-primitive element over R and that R contains an infinite field k. Then there exist p1 , . . ., pt ∈ Dp1 (R)(t may be 0) t such that the nonetale locus of R[α] over R is given by V (J[α] ) ∪ ( i=1 V ( pi )). Example 5.1.5 Let k be a field, a, b indeterminates, and R = k[a, b]. Let α be a root of an equation a X 2 + bX + a = 0. Then J[α] = (a, b)R. Assume that p ∈ Spec(R) and p ⊇ J[α] . When a ∈ p, (2α + b/a)R[α] p is the ramification locus. When a ∈ p and b ∈ p, (α + 1)R[α] p is the ramification locus. Definition 5.1.6

Let A be an extension of R with [K (A) : K ] = d. Define

(A) := {q ∈ Spec(R)|rankk(q) A ⊗ R k(q) = d} It is easy to see that when α is a super-primitive element of degree d over R, we have: (R[α]) ⊇ Dp1 (R) ⇔ R[α] is integral over R ⇒ R[α] is flat over R. When A is a finitely generated extension of R, define: Ur (A) := { p ∈ Spec(R)|A p is unramified over R p } which is an open set of Spec(R). Under these preparations, we finally obtain the following.

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Unramified Extensions

Theorem 5.1.7 Assume that [L : k] = d, and that α1 , . . ., αn ∈ L are super-primitive elements of degree d, and let A = R[α1 , . . ., αn ]. If (R[αi ]) ⊇ Dp1 (R) (1 ≤ i ≤ n) and Ur (R[α j ]) ⊇ Dp1 (R) for some j, then A is integral over R, and A p is etale over R p for any p ∈ (A). If (A) = Spec(R) in addition to the preceding assumptions, then A is integral and etale over R. Proof The assumption Dp1 (R) ⊆ (R[αi ]) implies that αi is integral over R and (R[αi ]) = Spec(R) by Theorem 2.3.2, and hence A is integral over R. Take p ∈ (A). Then p ∈ (R[α j ]) and R[α j ] is finite, flat over R as was shown in Theorem 2.2.5. Thus R[α j ] p is an R p -free module of rank d. Since Ur (R[α j ]) ⊇ Dp1 (R), R[α j ] is unramified over R by Proposition 5.1.1. Hence p R[α j ] p is a radical ideal. Noting that A is integral over R[α j ], we have p A p ∩ R[α j ] p = p R[α j ] p . Thus R[α j ] p / p R[α j ] p ⊆ A p / p A p . As both sides have the same dimension d as vector spaces over k( p), we have R[α j ] p / p R[α j ] p = A p / p A p , which means that A p = R[α j ] p + p A p . By Nakayama’s lemma, we get A p = R[α j ] p . Therefore A p is unramified and flat (i.e., etale) over R p for any p ∈ (A).

5.2

Differential Modules of Anti-Integral Extensions

In this section, we treat the module of differentials. The definition of such modules are seen in [M2] and [AK]. Theorem 5.2.1 Assume that α is an anti-integral element of degree d over R. Let ϕα (X ) = X d + η1 X d−1 + · · · + ηd be the monic minimal polynomial of α over K . Then the following conditions are equivalent: (1)  R (R[α]) = (0); (2) I[α] ϕα (α)R[α] = R[α]. Proof Put N = I[α] ϕα (X )R[X ]. Let 0 −→ N −→ R[X ] −→ R[X ]/N −→ 0 be an exact sequence. From this exact sequence, we get the following exact sequence: N /N 2 −→ R[α] ⊗ R[X ] R[X ]d X ∼ = R[α] −→  R (R[α]) −→ 0 where ρ : N /N 2 −→ R[α] ⊗ R[X ] R[X ]d X is a map such that ρ(x mod N 2 ) = d R[X ]/R x ⊗ 1 for x ∈ N . Note that R[α] R[X ]  R (R[X ]) ∼ = R[α]. Since N =

5.2 Differential Modules of Anti-Integral Extensions

115

I[α] ϕα (X )R[X ], it follows that Im(ρ) = I[α] ϕα (α)R[α]. Hence  R (R[α]) = (0) if and only if I[α] ϕα (α)R[α] = R[α]. Example 5.2.2 Let k be a field with characteristic = 2 and let R = k[X 2 , 1/ X 2 ] be a subring of k[X, 1/ X ]. Put α := X . Then K = k(X 2 ) is the quotient field of R and L = k(X ) is the quotient field of R[α]. Then ϕα (Y ) = Y 2 − X 2 ∈ R[Y ] is the monic minimal polynomial of α over K . Since ϕα (Y ) = 2Y, ϕα (X ) = 2X is a unit of R[α] = k[X, 1/ X ]. Hence R[α] is unramified over R (cf. [N]). In the rest of this section, we use the notation of Theorem 5.2.1 unless otherwise specified. Remark 5.2.3 Let α denote an anti-integral element of degree d over R. If R[α] is unramified over R, then J˜[α] R[α] = I[α] R[α] = R[α]. In fact, since I[α] ϕα (α)R[α] ⊆ I[α] R[α] and I[α] ⊆ J˜[α] , it follows that I[α] R[α] = J˜[α] R[α] = R[α]. Theorem 5.2.4 Assume that α is an anti-integral element over R. If  R (R[α]) = (0), then R[α] is a flat R-module. Proof Suppose that R[α] is not a flat R-module. Then there exists a prime ideal p of R such that p ⊇ c(I[α] ϕα (X )) by Theorem 2.2.5. Since p R[X ] ⊇ I[α] ϕα (X )R[X ] = N , we have that R[α]/ p R[α] ∼ = (R/ p)[X ]. Hence p R[α] ∈ Spec(R[α]). From an exact sequence 0 −→ p R[α] −→ R[α] −→ R[α]/ p R[α] −→0, we have the following exact sequence: p R[α]/ p 2 R[α] −→  R (R[α]) R[α] R[α]/ p R[α] −→  R/ p (R[α]/ p R[α]) −→ 0. But  R (R[α]) = (0), and so  R/ p (R[α]/ p R[α]) = (0). Also, since R[α]/ p R[α] ∼ = (R/ p)[X ], we have that  R/ p (R[α]/ p R[α]) = ((R/ p)[X ])d X = (0). This is a contradiction. Hence R[α] is a flat R-module. Remark 5.2.5 Let α be an anti-integral over R. Although R[α] is a flat Rmodule, R[α] is not necessarily unramified over R. For example, let R = k[X 2 ] be a subring of k[X ] where k denotes a field with characteristic = 2. Set R[α] = R[X ] = k[X ]. Since  R (R[α]) = R[α]/ X R[α] = (0), we have that R[α] is not an unramified extension over R. Proposition 5.2.6 Under the assumption in Theorem 5.2.1, it follows that V (Ann R ( R (R[α]))) ⊇ V (J[α] ), where Ann R ( R (R[α])) denotes the annihilator ideal of  R (R[α]). Proof Assume that p ∈ V (Ann R ( R (R[α]))). So there exists an element s ∈ Ann R ( R (R[α])) such that s ∈ p. Therefore s R (R[α]) = (0) and so

116

Unramified Extensions

 R p (R[α] p ) = (0). Note that R[α] p = R p [α]. Using Theorem 5.2.4, it follows that R[α] p is a flat R p -module. From this fact and Theorem 2.2.5, we obtain that p ⊇ J[α] . Therefore p ∈ V (J[α] ). Proposition 5.2.7 The element α is an anti-integral element of degree d over R if and only if α is an anti-integral element of degree d over R p for any p ∈ Spec(R). Proof Let 0 −→ N −→ R[X ] −→ R[α] −→ 0 be an exact sequence. (⇒) N is an ideal generated by some polynomials of degree d, where d = [L : K ]. Also, 0 −→ N p −→ R p [X ] −→ R p [α] −→ 0 is an exact sequence. Hence N p is also an ideal generated by some polynomials of degree d. Thus α is an anti-integral element of degree d over R p . (⇐) Put B = J[α] ϕα (X )R[X ]. Then B ⊆ N . And N = B if and only if α is anti-integral over R. By the assumption, we have that N p = B p for any p ∈ Spec(R). Hence we get N = B. Theorem 5.2.8 If α is an anti-integral element in K , then the following four statements are equivalent: (i) Iα R[α] = R[α]; (ii) R[α] is a flat R-module; (iii) J[α] R[α] = R[α]; (iv) R[α] is an unramified extension of R. Proof Using the fact that Iα R[α] = J[α] R[α] and Theorem 1.2.8, the equivalence of (i), (ii), and (iii) are already proved. Also, it can be proved that (iv) implies (ii) from Theorem 5.2.4. Now, we claim that (ii) implies (iv). Since d = 1, we have that J[α] ϕα (α)R[α] = J[α] R[α], where ϕα (X ) = X − α. Since R[α] is a flat R-module, it follows that J[α] = R. So we get that J[α] R[α] = R[α]. Therefore we have that Iα ϕα (α)R[α] = J[α] ϕα (α)R[α] = R[α]. Using Theorem 5.2.1, we see that  R (R[α]) = (0). Hence R[α] is an unramified extension of R. Remark 5.2.9 In Theorem 5.2.8, when α ∈ K is anti-integral over R, that is, a simple birational anti-integral extension is flat if and only if it is an unramified extension, but, in case of a nonbirational extension, flatness is not equivalent to unramifiedness (cf. Remark 5.2.5).

5.2 Differential Modules of Anti-Integral Extensions

117

Remark 5.2.10 Assume that α is anti-integral over R. Let ϕα (X ) = X d + η1 X d−1 + · · · + ηd be the monic minimal polynomial of α over K, where d = [L : K ] > 1. Then I[α] ϕα (α)R[α] ⊆ J˜[α] R[α], for I[α] ϕα (α)R[α] = I[α] (dα d−1 + (d − 1)η1 α d−2 + · · · + ηd−1 )R[α] ⊆ I[α] (1, η1 , . . ., ηd−1 )R[α] ⊆ J˜[α] R[α]. From Theorem 5.2.1, if  R (R[α]) = (0), then J˜[α] R[α] = R[α]. As an example of I[α] ϕα (α) = J˜[α] R[α], we give the following example. Put ϕα (X ) = X 2 − a ∈ R[X ] where a is not a unit of R. Then I[α] ϕα (α)R[α] = 2α R[α] = R[α] and J˜[α] R[α] = R[α] (we assume that 2 is not a unit of R[α]).

Proposition 5.2.11 Assume that R[α] is an unramified extension of R and let ϕ : Spec(R[α]) −→ Spec(R) be a restriction map. Then ϕ is surjective if and only if J˜[α] = R. Proof (⇐): When J˜[α] = R, R = J˜[α] ⊆ J[α] and hence α is a super-primitive element over R. From Theorem 4.3.2, we know that Im (ϕ) = D( J˜[α] )∪V (J[α] ). So Im(ϕ) = Spec(R) by assumption. Hence ϕ is surjective. (⇒): Suppose that J˜[α] = R. There exists a prime ideal p of R such that J˜[α] ⊆ p. Since ϕ is surjective, there exists a prime ideal P of R[α] such that P ∩ R = p. Thus J˜[α] R[α] ⊆ P. But R[α] is an unramified extension over R, J˜[α] R[α] = R[α] from Theorem 5.2.1 and Remark 5.2.3. Hence R[α] = P, a contradiction. Therefore J˜[α] = R. Remark 5.2.12 Assume that α is anti-integral over R. When ϕ is surjective, for an ideal N of R, N R[α] = R[α] if and only if N = R. Proposition 5.2.13 Let A be a ring extension of R and N be an ideal of R. √ Let N = q1 ∩ q2 ∩ . . . ∩ qn be a primary decomposition of N , where qi = pi for 1 ≤ i ≤ n. Then N A = A if and only if pi A = A for 1 ≤ i ≤ n. Proof (⇒) Since N A ⊆ pi A, we have pi A = A. (⇐) From pi A = A, we have that 1 =  j ai  j αi j (ai j ∈ pi , αi j ∈ A). Clearly, it can be assumed that ai j ∈ qi . Then 1 = i ( ai j αi j ) ∈ N A and so N A = A.

Theorem 5.2.14 Assume that α is an anti-integral element of degree d over R. R[α] is a flat R-module if and only if J˜[α] R[α] = R[α]. √ Proof Let J˜[α] = q1 ∩ q2 ∩ · · · ∩ qn ( qi = pi ) be a primary decomposition of J˜[α] . (⇒) From Theorem 4.3.2 (2),  R[α]/R = V ( J˜[α] ) ∩ ηd I[α] , where J[α] =

118

Unramified Extensions

J˜[α] + ηd I[α] . Since R[α] is R-flat, it follows that J[α] = R from Theorem 2.3.6. Since J˜[α] ⊆ pi , we get that pi ∈ V ( J˜[α] ) and pi ∈ ηd I[α] . Therefore pi ∈ V ( J˜[α] )∩ηd I[α] =  R[α]/R and so pi R[α] = R[α]. From Proposition 5.2.13, we have J˜[α] R[α] = R[α]. (⇐) Since J˜[α] R[α] = R[α], it follows that pi R[α] = R[α] from Proposition 5.2.13. So pi ∈  R[α]/R = V ( J˜[α] ) ∩ ηd I[α] . Since pi + ηd I[α] = R, we have that J[α] = ηd I[α] + J˜[α] = R. Hence R[α] is a flat R-module. We recall that α is called an unramified element over R if R[α] is an unramified extension of R. Theorem 5.2.15 Let α1 , α2 , . . ., αn ∈ L be unramified elements over R. Then R[α1 , α2 , . . ., αn ] is an unramified extension over R. Proof Clearly, it can be assumed that n = 2. Let B = R[α1 ], C = R[α2 ], and A = B[α2 ] = R[α1 , α2 ]. Then we have the following exact sequence [M1]:  R (B) ⊗ B A −→  R (A) −→  B (A) −→ 0 Since B and C are unramified extensions over R, we see that  R (B) = (0) and  R (C) = (0). So it suffices to show that  B (A) = (0). Since C is unramified over R, it follows that B ⊗ R C is unramified over B, that is,  B (B ⊗ R C) = (0). From the exact sequence 0 −→ N −→ B ⊗ R C −→ A −→ 0, and the fact that  B (A) =  B (B ⊗ R C)/ (N ) where (N ) denotes the submodule generated by {da | a ∈ N }, we get that  B (A) = (0). Thus we have that  R (A) = (0). The proof is complete. Remark 5.2.16 Assume that α is an element of the quotient field of R and that R[α] is an anti-integral extension of R. Then R[α] is a flat R-module if and only if for each p ∈ Spec(R), R[α] p = R p , or R[α] p = R p [1/a] for some element a ∈ R. Proof (⇐) It is trivial to show this implication. (⇒) In the case p ⊇ Iα , it follows that R[α] p = R p . In the remaining case, since J[α] = Iα + α Iα = R, we have that p ⊇ α Iα . And so there exists an element a of Iα such that aα ∈ p. Put b = aα. Then R[α] p = R p [α] = R p [1/a].

In the rest of this section, we examine the ramification locus of Rα over R when α is an anti-integral element of degree d > 1 over R.

5.2 Differential Modules of Anti-Integral Extensions

119

Proposition 5.2.17 Assume that α is an anti-integral element of degree d > 1 over R. If Rα is unramified over R, then I[α] = R and hence R[α] is integral over R. Proof Suppose that I[α] = R. Then there exists a prime ideal p of R such that p ⊇ I[α] . By Proposition 3.1.25 (1), P := p ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 is a prime ideal of B. By construction, P is a unique prime ideal lying over p and B/P = R/ p. Since B is unramified over R, p B is a radical ideal. Since B is integral over R, we have p B = P. Hence B = R + p B. Since B is a finite R-module by Theorem 3.1.18, localizing at p and using Nakayama’s lemma, we have B = R. But B = R because d > 1 by assumption. Thus I[α] = R. The last statement follows from Theorem 2.3.2. Remark 5.2.18 Assume that I[α] = R. Since α is anti-integral and integral over R, we have R[α] = Rα, which is a free R-module of rank d. Moreover R[α] is unramified over R if and only if ϕα (α) is an invertible element in R[α], where ϕα (X ) denotes the derivative of ϕα (X ). Proposition 5.2.19 Assume that α is an anti-integral element of degree d > 1 over R. Then for each prime ideal p of R, the following statements are equivalent: (i) Rα p is ramified over R p ; (ii) either I[α] ⊆ p holds or both I[α] ⊆ p and ϕα (α)Rα p = Rα p hold. Proof (i) ⇒ (ii): Suppose that I[α] ⊆ p and ϕα (α)Rα p = Rα p . Since I[α] ⊆ p, we have Rα p = R p [α]. Since ϕα (α)Rα p = Rα p , we have ϕα (α)R p [α] = R p . Thus Rα p is unramified over R p . (ii) ⇒ (i): Note that Rα p = R p [α] ∩ R p [α −1 ] and that α is anti-integral over R p of degree d. By Proposition 5.2.17, the unramifiedness of Rα p /R p yields I[α] ⊆ p. By Remark 5.2.18, we have ϕα (α)Rα p = Rα p . Finally we determine the ramification locus of Rα over R. Theorem 5.2.20 Assume that α is an anti-integral element of degree d > 1 d over R. Then the ramification locus of Rα over R is given by I[α] ϕα (α)Rα ∩  √ d R. Consequently, Ann R ( R (Rα)) = I[α] ϕα (α)Rα ∩ R. Proof By Theorem 3.1.6, we have Rα : R α = I[α] . Since ϕα (α) is red−1  garded as a polynomial of α of degree d − 1, it follows that I[α] ϕα (α) ⊆ Rα.

120

Unramified Extensions

d Let P be a prime ideal of Rα and put p = P ∩ R. Then P ⊇ I[α] ϕα (α)Rα d  if and only if Rα p is unramified over R p . Indeed, if P ⊇ I[α] ϕα (α)Rα, then d−1  d−1  Rα ⊇ I[α] ϕα (α)Rα, and so either P ⊇ I[α] or P ⊇ I[α] ϕα (α)Rα. If d−1  P ⊇ I[α] , then p ⊇ I[α] . If P ⊇ I[α] , then P ⊇ I[α] ϕα (α)Rα p implies that ϕα (α)Rα p ⊆ P Rα p and that ϕα (α)Rα p = Rα p . Hence Rα p is ramified over R p by Proposition 5.2.19. Conversely assume that Rα p is unramified over R p . Then p ⊇ I[α] and ϕα (α)Rα p = Rα p by Proposition 5.2.19. d So we have P ⊇ I[α] ϕα (α)Rα p .

5.3

Kernels of Derivations on Simple Extensions

Let Z denote the set of all integers and Q the field of the rational numbers. For an element α ∈ L, put A := R[α]. Let π : R[X ] → R[α] be the Ralgebra homomorphism of a polynomial ring R[X ] to R[α] sending X to α. Let E (α) := Ker(π ). We denote by  R (A) the module of differentials of A over R and by d A/R the universal R-derivation A →  R (A). Then D := Ker(d A/R ) is an R-subalgebra of A. If the prime ring k of A is equal to Z/ pZ with p a positive prime number, then we have aα p ∈ D, and hence A is a finite D-module. Therefore D is Noetherian by Eakin-Nagata theorem (cf. [M2]). In the case k = Z, D is not necessarily Noetherian. Let c(E (α) ) denote the ideal of R generated by the coefficients of the polynomials in E (α) . Then it is easy to see that E (α) ⊆ c(E (α) )R[X ] and c(E (α) )A ∩ R = c(E (α) ). It is  (α) := { f  (X ) | f (X ) ∈ E (α) }, where also easy to see that E (α) = (0). Let E  f (X ) denotes the derivative of f (X ) in the usual sense. Then it is easy to see  (α) ) is an ideal of A and that the map δ : A → A/π ( E  (α) ) defined that π ( E  (α) by δ(g(α)) = g (α) mod π (E ) is a well-defined R-derivation. In this case, there exists an A-module isomorphism ρ : A/π (E (α) ) →  R (A) such that ρδ = d A/R (cf. [AK]). Hence we have Ker(d A/R ) = Ker(δ). Note that, for an element f (X ) ∈ R[X ], there exists an element F(X ) ∈ R[X ] such that α F  (X ) = f (X ) and F(0) = 0. We then set 0 f (X )d X = F(α). Lemma 5.3.1

Let D := Ker(d A/R ). Then D = R +

α

 f (X )∈E (α)

0

f (X )d X .

α Proof Let f (X ) be an element of E (α) and let a = 0 f (X )d X . Then we have  (α) ) = 0, and hence a ∈ D. Conversely, let g(α) be an δ(a) = f (α) mod π( E element of D. Then we have g  (α) ∈ π (E (α) ) and hence we can find an element f (X ) of E (α) satisfying g  (α) = f  (α). Let h(X ) = g  (X ) − f  (X  )α and c = g(0)− f (0). Then h(X ) ∈ E (α) and c ∈ R. Since g(α)− f (α) = c+ 0 h(X )d X

5.3 Kernels of Derivations on Simple Extensions α

and f (α) = 0, we have g(α) = c +

0

121

h(X )d X , which completes the proof.

Lemma 5.3.2 Let a be a nonzero element of c(E (α) ). Then there exists a positive integer r such that aαr +i ∈ D + Dα + · · · + Dαr −1 for every i ≥ 0. Proof As is seen, there exists an element f (X ) = a0 X n + a1 X n−1 + · · · + an of E (α) such that a = a j for some 0 ≤ j ≤ n. Let s = n(n + 1)/2.  α We show that r = s + n + 1 − j satisfies the required condition. Let γi − 0 X i f (X )d X for each i ≥ 0. Then each γi is an element of D by Lemma 5.3.1. Moreover, we have 



 a0 α i+n+1   a1 α i+n   = Mi   ..   .  an α i+1

γi

 γ  i+1   ..  . γi+n

    

for every i ≥ 0, where 

1 i+n+1 α i+n+2

1 i+n α i+n+1

.. . αn i+2n+1

  Mi =   

Let

    m i = det   

··· ···

1 i+1 α i+2

.. .

···

.. .

αn i+2n

···

αn i+n+1

1 i+n+1 1 i+n+2

1 i+n 1 i+n+1

.. . 1 i+2n+1

···

     



···

1 i+1 1 i+2

.. .

···

.. .

1 i+2n

···

1 i+n+1

     

Then we can check that m i is a nonzero rational number (cf. [OnSuY2]) and det Mi = m i α s . Hence, putting L i be the cofactor matrix of Mi , we have  m i−1 L i

    

γi γi+1 .. . γi+n





     = αs     

a0 α i+n+1 a1 α i+n .. . an α i+1

     

122

Unramified Extensions

Note that every component of m i−1 L i belongs to Q + Qα + · · · + Qα s . Since each γk is an element of D and r ≥ s + 1, it follows that a j αr +i ∈ D + Dα + · · · + Dα s ⊆ D + Dα + · · · + Dαr −1 for every i ≥ 0. This completes the proof. Lemma 5.3.3

The element α is integral over D if and only if c(E (α) ) = R.

Proof If c(E (α) ) = R, then it follows from Lemma 5.3.2 that there exists a positive integer r such that αr ∈ D + Dα + · · · + Dαr −1 . Thus α is integral over D. Conversely, assume that α is integral over D. Then we have αr + c1 αr −1 + · · · + cr = 0

(1)

for some positive integer r and c1 , . . ., cr ∈ D. Suppose that c(E (α) ) = R. Let ψ : A → A/c(E (α) )A be the canonical homomorphism. Then we have A/c(E (α) )A ∼ = (R[X ]/E (α) )/(c(E (α) )R[X ]/E (α) ) ∼ = R[X ]/c(E (α) )R[X ] ∼ = (α) (R/c(E ))[X ], which yields that ψ(α) is algebraically independent over R/c(E (α) ). On the other hand, it follows by relation (1) that ψ(α)r + ψ(c1 )ψ(α)r −1 + · · · + ψ(cr ) = 0

(2)

Note that we have ψ(α) ∈ R/c(E (α) ) for every c ∈ D since D ⊆ R + c(E (α) )A by Lemma 5.3.1. Then relation (2) shows that ψ(c) is not algebraically independent over R/c(E (α) ), which is a contradiction. We are now ready to state and prove the following theorem. Theorem 5.3.4 Assume that R contains the rational number field Q and let A := R[α]. Let D := Ker(d A/R ). Then the following three conditions are equivalent: (1) D is finitely generated R-algebra; (2) D is Noetherian; (3) c(E (α) ) = R. Proof We have only to check that (3) implies (1) and that (2) implies (3). Suppose that c(E (α) ) = R. Then α is integral over D by Lemma 5.3.3, and so αr + c1 αr −1 + · · · + cr = 0 for some positive integer r and c1 , . . ., cr ∈ D. Put B := R[c1 , . . ., cr ]. Then A is a finite D-module. Thus D is finitely generated over R. Next, suppose that D

5.3 Kernels of Derivations on Simple Extensions

123

is Noetherian and let a ∈ c(E (α) ) be a nonzero element of A. Then there exists a positive integer r such that aαr +i ∈ M := D + Dα + · · · + Cαr −1 for every i ≥ 0 by Lemma 5.3.2. Since a is a nonzero element of A, this implies that D[αr ] ⊆ (1/a)M. Hence D[αr ] is a finite D-module because D is Noetherian and (1/a)M is a finite D-module. Thus αr is integral over D, and so is α over D. So we have c(E (α) ) = R by Lemma 5.3.3. Corollary 5.3.5 Assume that R contains Q. Let A := R[α]. If A is flat over R, then D := Ker(d A/R ) is finitely generated over R. Proof Since c(E (α) ) = (0), if A is flat over R, then we have c(E (α) ) = R (cf. [OR, Corollary 1.1.3]). So the assertion follows from Theorem 5.3.4. The converse of Corollary 5.3.5 seems not to hold in general (cf. [OnSuY2, Example 2.3]). However, the converse holds if the extension R ⊆ A = R[α] is anti-integral. In fact, when A := R[α] is anti-integral over R, A is flat over R if and only if c(E (α) ) = J[α] = R by Proposition 2.5.4. Hence we have the following theorem. Theorem 5.3.6 Assume that α ∈ L an anti-integral element of degree d over R. Let A := R[α] and let D := Ker(d A/R ). If R contains Q, then the following three conditions are equivalent: (1) D is finitely generated R-algebra; (2) D is Noetherian; (3) A is flat over R. Remark 5.3.7 Let R ⊆ A := R[α] be an anti-integral extension of degree d and assume that R contains Q. Then I[α] ϕα (X )R[X ] = Ker(π ), and hence, α putting βi := 0 X i ϕα (X )d X for each i ≥ 0, we have D = R + I[α] β0 + I[α] β1 + · · · by Lemma 5.3.1. Example 5.3.8 Let R be a polynomial ring k[x, y] over a field k := Q, and let α := y/x. Then ϕα (X ) = X − (y/x) and I[α] = x R. It is easy to see that the extension R ⊆ A := R[α] is anti-integral. Since  α y i+2 y βi = X i (X − )d X = − x (i + 1)(i + 2)x i+2 0

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we have I[α] βi = y i+2 /x i+1 R for every i ≥ 0. It follows from Remark 5.3.7 that D = k[x, y, y 2 /x, y 3 /x 2 , . . . ] which is not Noetherian. Note that J[α] = c(I[α] ϕα (X ))R = R, i.e., A is not flat over R by Theorem 2.3.6.

Chapter 6 The Unit-Groups of Extensions

Let R be a Noetherian integral domain with the quotient field K and L be an algebraic extension field of K such that the degree [L : K ] of L over K equals to d. Let α be a nonzero element of L. We assume that L = K (α); that is, L is the quotient field of R[α]. We denote the unit group of R by U (R); that is, U (R) := {a ∈ R | ab = ba = 1 for some element b ∈ R} In this chapter, we consider the relations among U (R[α]), U (R[α]) ∩ R, and the subgroup of U (R[α]) generated by the semigroup U (R[α]) ∩ R, in the case of an anti-integral extension R[α] of R. Let Iβ := {r ∈ R | rβ ∈ R} = R : R β for any β ∈ L and IβS := {r ∈ S | rβ ∈ S} = S : S β for a ring S such that R ⊆ S ⊆ L . Note that Iβ = IβR . The natural numbers, the ring of integers, and a polynomial ring over R are written as N, Z, and R[X ], respectively. When R[α] is a super-primitive and LCM-stable extension over R, we shall give a criterion whether β ∈ U (R[α]) or not for an element β of R[α]. Let P be a prime ideal of height 1 in R[α]. Set p := P ∩ R. Since R[α] is LCM-stable over R, it is known that p is a prime ideal of height 1 in R from [SY2]. It is easy to see that β ∈U (R[α]) if and only if β −1 ∈ R p [α] for all p ∈ Dp1 (R), where Dp1 (R) denotes the set of all prime ideals of R of depth 1. So we may assume that (R, m) is a local domain of depth 1. First, we shall give a criterion whether β ∈ U (R[α]) or not in the case that α is integral over R and R[α] is a free R-module of rank d. In short, we call this case an “integral case.” Next, in the general case, replacing α by a suitable element of R[α], we may assume that α satisfies that R[α −1 ] ⊆ R[α] = R[α], and moreover we get that (α −1 ) β ∈ R[α −1 ] and (α −1 ) β is integral over R for a suitable integer  > 0. So the general case is reduced to an integral case. Throughout this chapter, we use the following notation. Let R be a Noetherian integral domain and R[X ] a polynomial ring. Let α be an element of an algebraic field extension L of the quotient field K of R and π : R[X ] −→ R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with
d deg ϕα (X ) = d (R : R ηi ). For and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Let I[α] := i=1 f (X ) ∈ R[X ], let c( f (X )) denote the ideal generated by the coefficients of

125

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f (X ), that is, the content ideal of f (X ). Let J[α] := I[α] c(ϕα (X )), which is an ideal of R and contains I[α] . The element α is called an anti-integral element of degree d over R if Ker(π ) = I[α] ϕα (X )R[X ]. When α is an anti-integral element over R, R[α] is called an anti-integral extension of R. The element α is called a super-primitive element of degree d over R if J[α] ⊆ p for all primes p of depth 1. It is known that a super-primitive element is an anti-integral element (Theorem 2.2.8). (See Chapters 2 and 3 for details.)

6.1

The Unit-Groups of Anti-Integral Extensions

√ For α ∈ L, the set {a ∈√R| a R ⊇ I[α] and J[α] + a R = R} and U (R[α]) ∩ R are semigroups, where a R denotes a radical ideal of a R. When R[α] is an anti-integral extension of R and α ∈ K , we proved that √ U (R[α]) ∩ R = {a ∈ R| a R ⊇ Iα

and

J[α] + a R = R}

in [KYJ, Theorem 9]. We obtain the same result as above for any algebraic element α ∈ L. Theorem 6.1.1 Let α ∈ L √ be an anti-integral element of degree d over R. Then U (R[α]) ∩ R = {a ∈ R| a R ⊇ I[α] and J[α] + a R = R}. Proof (⇒) Take a ∈ U (R[α]) ∩ R and let p1 , . . ., pn be the minimal prime divisors of a R. Since α is anti-integral over R, we see that  R[α]/R = V ( J˜[α] ) ∩  J[α] by Theorem 4.3.2, where  R[α]/R := { p ∈ Spec(R)| p R[α] = R[α]} and  J[α] := { p ∈ Spec(R)| p + J[α] = R}. Since a R[α] = R[α], it follows ˜ that pi R[α] = R[α] and hence pi ∈  R[α]/R √ . Thus pi ⊇ J[α] ⊇ I[α] and pi + J√ a R = p1 ∩ · · · ∩ pn ⊇ [α] = R for all i, which means that √ I[α] and J[α] + a R = J[α] + ( p1 ∩ · · · ∩ pn ) = R. It is easy to see that J[α] + a R = R if and only if J[α] + a R = R.√ (⇐) Take a ∈ R satisfying a R ⊇ J[α] and J[α] + a R = R. Suppose that a ∈ U (R[α]) ∩ R. Then a R[α] = R[α]. Let P1 , . . ., P be the minimal prime divisors of √ i (1 ≤ i ≤ ) √a R[α]. Put pi = Pi ∩ R. Then pi R[α] = R[α] for some because a R[α] = R[α]. Put p := pi . We have I[α] ⊆ a R ⊆ p, but J[α] + a R ⊆ p = R, which is a contradiction. Thus a ∈ U (R[α]) ∩ R. We give the following theorem.

6.1 The Unit-Groups of Anti-Integral Extensions

127

Theorem 6.1.2 Assume that α is an  anti-integral element of degree d over R and R satisfies the altitude formula. If I[α] = p1 ∩ p2 ∩ · · · ∩ pn (I[α] = R, and pi ∈ Spec(R) for each i), then ht( pi ) = 1 for each i (1 ≤ i ≤ n). First, we prove the following lemma. Lemma 6.1.3 If α is an anti-integral element of degree d over R, then α is an anti-integral element of degree d over R p for each p ∈ Spec(R). Proof Consider an exact sequence: 0 −→ I −→ R[X ] −→ R[α] −→ 0. Since α is anti-integral over R, I is generated by some polynomials of degree d over R. From the exact sequence 0 −→ I p −→ R p [X ] −→ R p [α] −→ 0 we have that I p is generated by some polynomials of degree d over R p . Hence α is anti-integral of degree d over R p . [The proof of Theorem 6.1.2] Fix i and put p = pi . Since p is a minimal prime divisor of I[α] , it follows that (I[α] ) p is p R p -primary. Suppose that ht( p) > 1. Since R p satisfies the altitude formula, we have that ht((I[α] ) p ) > 1, noting
that (I[α] ) p = i (R p : R p ηi ). We recall that ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Also, α is integral over R p by Lemma 6.1.4. By Lemma 6.1.3, we have that α is an anti-integral element of degree d over R p . Therefore R p [α] is a free R p -module of rank d by Corollary 2.3.5 and Theorem 2.3.8. Since α is integral over R p , it follows that ηi ∈ R p for each i. Hence we have   (I[α] ) p = (Iηi ) p = (R p : R p ηi ) = R p i

i

which is a contradiction. Thus we conclude ht( p) = 1. Lemma 6.1.4 Assume that R satisfies the altitude formula and ht(I[α] ) > 1. If α is anti-integral of degree d over R, then it is integral over R.  denote the integral closure of R in L. Note that R  is a Krull domain Proof Let R  (cf. [M2]). Let P be any prime ideal of height 1 in R and put p := P ∩ R. Since R satisfies the altitude formula, it follows that p is also of height 1. Since ht(I[α] ) > 1, we have that p does not contain I[α] . Therefore ht(Iηi ) > 1 and so p does not contain Iηi for each i. Thus we have ηi ∈ R p and so ϕα (X ) ∈ R p [X ].  = ∩R P, Hence ϕα (α) = 0 yields that α is integral over R p . From the fact that R  we see that α ∈ R.

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Corollary 6.1.5 We assume that R satisfies the altitude formula (dimension formula) and that α is anti-integral of degree d over R. If ht(I[α] ) > 1, then R[α] is a free R-module of rank d, i.e., R[α] = R ⊕ Rα ⊕ · · · ⊕ Rα d−1 . Remark 6.1.6 In Corollary 6.1.5, if have U (R[α]) ∩ R = U (R).

√ a R ⊇ I[α] , then a ∈ U (R). So we

Theorem 6.1.7 Let α be a nonzero element of L. If R[α] is an√anti-integral extension of R and a flat R-module, then U (R[α]) ∩ R = {a ∈ R| a R ⊇ I[α] }. √ √ Proof Take a ∈ R such that a R ⊇ I[α] . Then a R+ J[α] ⊇ √ J[α] . Since R[α] is flat over R, we have that J[α] = R by Theorem 2.3.6 and so a R+ J[α] = R. Hence a R + J[α] = R . Thus we have that a ∈ U (R[α]) ∩ R by Theorem 6.1.1. The converse has been proved already. Remark 6.1.8 We assume that x ∈ R and that 1/x is anti-integral over R. Then the following conditions are equivalent: (1) a ∈ U (R[1/x]) ∩ R; √ (2) a R ⊇ x R; (3) There exists an element b ∈ R such that x n = ab for some natural number n. Let R ⊆ A be Noetherian integral domains. We consider the subgroup of U (A) generated by the semigroup U (A) ∩ R. Definition 6.1.9 We denote by S 0 the subgroup of U (A) generated by a subset S ⊆ U (A). Next we consider some examples. Example 6.1.10 (An example such that U (A)/U (R) is not a torsion-free group.) Let k be a field. Put R = k[X, Y, (Y / X )2 , (X/Y )2 ] and A = R[1/ X ], where X, Y are indeterminates. Then A = k[X, Y, 1/ X, 1/Y ]. We have that U (R) = U (k), U (A)/U (R) ∼ = Z ⊕ Z, and U (A) ∩ R/U (R) ∼ = N ⊕ N. Hence 0 (U (A) ∩ R) = U (A). Also, Y / X is a torsion-element in U (A)/U (R). Example 6.1.11 Let k be a field and X, Y are indeterminates. Put R = k[X, Y, X/Y ] and A = R[1/ X ]. Then A = k[X, Y, 1/ X, 1/Y ]

6.2 Invertible Elements of Super-Primitive Ring Extensions

129

and U (R) = U (k) U (A)/U (R) ∼ =Z⊕Z and

U (A) ∩ R/U (R) ∼ =N⊕N

Thus (U (A) ∩ R)0 = U (A). Remark 6.1.12 U (R[1/x]).

Let x ∈ R, x = 0. Then we have (U (R[1/x]) ∩ R)0 =

Theorem 6.1.13 We assume that R is a Noetherian local domain containing an infinite field k and p ∈ Spec(R) is of height 1 at most. Also, we assume that p is not a radical ideal of a principal ideal in R. Let α ∈ L be anti-integral √ over R. If Iα = p and J[α] = R, that is, R[α] is a flat R-module, then U (R[α]) ∩ R = U (R) and U (R[α]) = U (R). Proof Since R is local and J[α] = R, we may assume that a0 α d +a1 α d−1 +· · ·+ ad = 0 and ad = 1 where each ai ∈ R. Since α(a0 α d−1 +· · ·+ad−1 ) = −1, we have that α ∈ U (R[α]) and α is not contained √ in U (R). If a ∈ U (R[α])∩ R such that a is not contained in U (R), then a R⊇ I[α] and a R + J[α] = √R by √ Theorem 6.1.1. From the fact that a R ⊇ I[α] = p, we have p = a R. This contradicts the assumption. Thus U (R[α]) ∩ R = U (R).

6.2

Invertible Elements of Super-Primitive Ring Extensions

We start with the following lemma. Lemma 6.2.1 Let R be a ring and let R[α] = R + Rα + · · · + Rα d−1 be a ring extension of R. Then every element β ∈ R[α] has an integral relation of degree d over R, that is, there exists a monic polynomial f (X ) of degree d over R such that f (β) = 0. Proof

Writing βα i−1 = ai1 + ai2 α + · · · + aid α d−1 , where ai j ∈ R, we have      1 1 a11 a12 · · · a1d  α   a21 a22 · · · a2d   α       β . = . ..   ..  ..  ..   ..   . . .  d−1 d−1 ad1 ad2 · · · add α α

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The Unit-Groups of Extensions

Therefore we get that det(ai j − δi j β) = 0. By expansion of this determinant, we obtain that β has an integral relation of degree d over R. We recall the following definitions (Definition 2.2.1): (i) α ∈ L is called an anti-integral element of degree d over R if Ker(π) = I[α] ϕα (X )R[X ]. When α is an anti-integral element, we say that R[α] is an anti- integral extension of R. (ii) α ∈ L is called a super-primitive element of degree d over R if J[α] ⊆ p for all p ∈ Dp1 (R). When α is a super-primitive element, we say that R[α] is a super-primitive extension of R. Using the result of Lemmas 6.2.1, we have the following proposition. Proposition 6.2.2 Let R[α] be a free R-module of rank d, that is, R[α] = R+Rα+· · ·+Rα d−1 . Let β ∈ R[α] be an element satisfying β d +a1 β d−1 +· · ·+ ad = 0, where all ai ∈ R. Assume that [K (β) : K ] = d. Then β ∈ U (R[α]) if and only if ad ∈ U (R). Proof (⇐) Writing β(β d−1 + a1 β d−2 + · · · + ad−1 ) = −ad ∈ U (R), we have that β ∈ U (R[α]). (⇒) Assume that β ∈ U (R[α]). Since β −1 has a monic relation of degree d from Lemma 6.2.1, we conclude that J[β −1 ] = R and so β −1 is a superprimitive element. Since a super-primitive element is an anti-integral element (Theorem 2.2.8), we conclude that R[β −1 ] is integral over R by Theorem 2.3.2. So β −1 has a monic relation of degree d, which yields that ad ∈ U (R). Using the result of Proposition 6.2.2, we shall have a criterion whether β ∈ U (R[α]) or not, where β is an integral or anti-integral element of R[α]. The following theorem plays an important role in this section. Theorem 6.2.3 Assume that α is an integral and anti-integral element of degree  d over R. For β ∈ R[α], we assume that [K (β) : K ] = d. Write j βα i = d−1 j=0 ai+1 j+1 α where i = 0, 1, . . ., d − 1. Then β ∈ U (R[α]) if and only if det(ai j ) ∈ U (R). Proof By the assumption, it follows that R[α] = R[α] = R + Rα + · · · + Rα d−1 is a free R-module of rank d. Now, det(ai j − δi j β) = 0 is a monic relation of degree d with respect to β and the constant term of this equation of β is det(ai j ). From Proposition 6.2.2, we complete the proof.

6.2 Invertible Elements of Super-Primitive Ring Extensions

131

Theorem 6.2.4 Assume that α is a super-primitive element of degree d over R and grade (I[α] + ηd I[α] ) > 1. If β ∈ R α and [K (β) : K ] = d, then β has a monic relation of degree d over K . Therefore β is also a super-primitive element of degree d over R. Proof There exists a monic relation ϕβ (β) = 0, where ϕβ (X ) = X d + λ1 X d−1 + · · · + λd ∈ K [X ]. We claim that all λi ∈ R. Suppose that p ⊇ I[α] for p ∈ Dp1 (R). Then we have that R p [α] is a free R p -module of rank d from Theorem 2.2.5 and so β ∈ R p [α] = R p + R p α + · · · + R p α d−1 . Therefore β has a monic relation of degree d over R p . This monic relation equals to ϕβ (β) = 0. Hence λi ∈ R p . Next, we assume that p ⊇ I[α] for p ∈ Dp1 (R). Since grade (I[α] + ηd I[α] ) > 1, we obtain that p ⊇ ηd I[α] = I[α−1 ] . Hence β ∈ R p [α −1 ] = R p + R p (α −1 )+· · ·+ R p (α −1 )d−1
. λi ∈ R p is proved along the same lines as the case of p ⊇ I[α] . Hence λi ∈ p∈Dp1 (R) R p = R(1 ≤ i ≤ d). Consequently, we proved the claim. Hence I[β] = R and so β is a super-primitive element of degree d over R by definition. Assume that R ⊆ R[α −1 ] ⊆ R[α]. Then, for β ∈ R[α], there exists an integer  > 0 such that (α −1 ) β ∈ R[α −1 ]. Since α −1 ∈ R[α], it follows that α ∈ U (R[α]). Consequently, β ∈ U (R[α]) if and only if (α −1 ) β ∈ U (R[α]). Since R[α −1 ] ⊆ R[α], we have that R[α −1 ] is integral over R and so (α −1 ) β is an integral element over R. Hence we give a criterion whether (α −1 ) β ∈ U (R[α]) or not from Theorem 6.2.3. Now, we shall consider the general case. Assume that R contains an infinite field k and assume that α is a super-primitive element of degree d over R and R[α] is LCM-stable over R. For β ∈ R[α], β ∈ U (R[α]) if and only if β ∈ U (R[α] p ) for all p ∈ Dp1 (R). In case that I[α] ⊆ p, since a ring extension R[α] p /R p is integral, we give a criterion from Theorem 6.2.3. Next, in case that I[α] ⊆ p, since α is a super-primitive element, we see that grade(J[α] ) > 1. Replacing α by α − µ for some µ ∈ k, we may assume that R p ⊆ R p [α −1 ] ⊆ R p [α]. Hence we have reduced to the case that R p ⊆ R p [α −1 ] ⊆ R p [α]. Summing up the above statements, we obtain the following remark. Remark 6.2.5 Assume that R contains an infinite field k and R[α] is a superprimitive extension of R. Also, assume that R[α] is LCM-stable over R. Then, for β ∈ R[α], (1) β ∈ U (R[α]) if and only if β ∈ U (R[α] p ) for all p ∈ Dp1 (R). (2) If I[α] ⊆ p, then we give a criterion whether β ∈ U (R[α] p ) or not, using Theorem 6.2.3. If I[α] ⊆ p, then replacing α by a suitable element, we give a remark whether β ∈ U (R[α] p ) or not, using Theorem 6.2.3.

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The Unit-Groups of Extensions

Remark 6.2.6 If grade(Iα + ηd Iα ) > 1, then the later case (2) in Remark 6.2.5 is the following: We give a criterion whether α ∈ U (R p ) or not. Indeed, if p ∈ Dp1 (R), then p ⊇ Iα or p ⊇ ηd Iα . If p ⊇ Iα , then α is integral over R p . If p ⊇ Iα and p ⊇ ηd Iα , then R p ⊆ R p [α −1 ] ⊆ R p [α]. Now we shall consider the problem along the same lines as the preceding argument for a general ring extension. Let A be a ring extension of R. Assume that A is not necessarily a simple ring extension of R but L is an algebraic extension of K , where L is the quotient field of A. Theorem 6.2.7 Let A be a ring extension of R. Suppose that A is a finite R-module of rank n and A p is a flat R p -module for all p ∈ Dp1 (R). For β ∈ A, assume that [K (β) : K ] = n and ϕβ (X ) = X n + λ1 X n−1 + · · · + λn is the monic minimal polynomial of β over K . Then β ∈ U (A) if and only if λ1 , . . ., λn ∈ R, λn ∈ U (R). (Thus β is an anti-integral element in this case.) n−1 Proof (⇐) Writing β −1 = −λ−1 + λ1 β n−2 + · · · + λn−1 ) ∈ A, we have n (β that β ∈ U (A).  (⇒) A p = R p α1 + · · · + R p αn for all p ∈ Dp1 (R). Therefore βαi = j ai j α j , where ai j ∈ R p . Hence β has a monic relation of degree n and so ϕβ (β) = 0. Hence λi ∈ R p for all p ∈ Dp1 (R). Thus λi ∈ R. Since β −1 ∈ A, we get that λn ∈ U (R).

Especially, when R is a normal domain, we have the following result. Theorem 6.2.8 Let R be a Noetherian normal domain and let A be a ring extension of R such that [L : K ] = n. Also, let β ∈ A,β is integral over R, and [K (β) : K ] = n. Let ϕβ (X ) = X n + λ1 X n−1 + · · · + λn (λi ∈ K ) be the monic minimal polynomial of β over K . Then β ∈ U (A) if and only if λ1 , . . ., λn ∈ R and λn ∈ U (R). n−1 Proof (⇐) Writing β −1 = −λ−1 + λ1 β n−2 + · · · + λn−1 ) ∈ A, we see n (β that β ∈ U (A). (⇒) Let B be the integral closure of R in A. Then β ∈ B and so λ1 , . . ., λn ∈ R and λn ∈ U (R) from Theorem 6.2.7.

For β ∈ A, if A is integral over R, then we get a powerful criterion whether β ∈ U (A) or not by Theorem 6.2.3. Consequently, we come to a problem when a ring extension A/R is an integral extension. On this problem, we prove the following theorem in the

6.2 Invertible Elements of Super-Primitive Ring Extensions

133

case that L is a finite separable algebraic extension of K , where L is the quotient field of A. Theorem 6.2.9 Let A be an integral domain with the quotient field L and a ring extension of R. Assume that L/K is a finite separable algebraic extension. If A p is flat and integral over R p for all p ∈ Dp1 (R), then A is an integral extension over R. Proof Since L/K is a finite separable algebraic extension, L has only a finite number of subfields containing K . Case (1).There exists β ∈ A such that [K (β) : K ] = [L : K ] = n. (L is a simple field extension over K .) We are done in this case. Case (2). [K (γ ) : K ] < n for all γ ∈ A. In this case, there exists an element β ∈ A p such that [K (β) : K ] = n and [K (β + γ ) : K ] = n. Consequently, if β is integral over R for all β such that [K (β) : K ] = n, then A is integral over R. Let [K (β) : K ] = n and ϕβ (X ) = X n + η1 X n−1 + · · · + ηn (ηi ∈ K ) be the monic minimal polynomial of β over K . Since A p is a free R p -module of rank n for all p ∈ Dp1 (R), we get that A p = R p η1 + R p η2 + · · · + R p ηn . Therefore βηi = j ai j ηi where ai j ∈ R p and so ϕβ (X ) = det (δi j X − ai j ). Consequently, ηi ∈ R p for all p ∈ Dp1 (R). Hence ϕβ (X ) ∈ R[X ]. Thus we have that A is integral over R. Remark 6.2.10 If R is a normal domain, we get that A p is a flat R p -module for all p ∈ Dp1 (R) because R p is a DVR.

Chapter 7 Exclusive Extensions of Noetherian Domains

In this chapter, we use the following notations used before unless otherwise specified: R: a Noetherian integral domain, K := K (R): the quotient field of R, L: an algebraic field extension of K , α: a nonzero element of L , d = [K (α) : K ], ϕα (X ) = X d + η1 X d−1 + · · · + ηd , the minimal polynomial of α over K ,
d I[α] := i=1 (R : R ηi ), Ia := R : R a R for a ∈ K ,
d−1 (R : R ηi ). I˜[α] := i=1 It is clear that for a ∈ K , I[a] = Ia from definitions. J[α] := I[α] (1, η1 , . . ., ηd ), J˜[α] := I[α] (1, η1 , . . ., ηd−1 ). Note that I[α] , I˜[α] , J[α] , and J˜[α] are ideals of R. We also use the following standard notation: Dp1 (R) := { p ∈ Spec(R)|depth(R p ) = 1} Recall the following notation: Let π : R[X ] −→ R[α] be the R-algebra homomorphism sending X to α. The element α is called an anti-integral element of degree d over R if Ker(π ) = I[α] ϕα (X )R[X ]. When α is an anti-integral element over R, R[α] is called an anti-integral extension of R. (See Chapter 3 for details.) For f (X ) ∈ R[X ], let c( f (X )) denote the ideal generated by the coefficients of f (X ), that is, the

135

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Exclusive Extensions of Noetherian Domains

content ideal of f (X ). Let J[α] := I[α] c(ϕα (X )), which is an ideal of R and contains I[α] . The element α is called a super-primitive element of degree d over R if J[α] ⊆ p for all primes p of depth 1. It is known that a super-primitive element is an anti-integral element (see Theorem 2.2.8 for details).

7.1

Subring R[α] ∩ K of Anti-Integral Extensions

We examine the following problem: Problem. Let R be a Noetherian integral domain, K denote the quotient field of R, and α be a super-primitive element over R of degree d. What is a subring R[α] ∩ K ? Especially we intend to study when R[α] ∩ K = R holds and to investigate what is R[α] ∩ K if R[α] ∩ K = R. We start with the following result. Proposition 7.1.1 Assume that α ∈ L is anti-integral over R of degree d. Then  R[α] ∩ K ⊆ R p ( p ∈ Dp1 (R) with I[α] ⊆ p) Proof It suffices to show that R p ⊇ R p [α] ∩ K for any p ∈ Dp1 (R) with p ⊇ I[α] . Since (I[α] ) p = R p , R p [α] is integral over R p , i.e., α is integral over R p (cf. Theorem 2.3.2). Hence R p [α] is a finitely generated free R p -module and we have an expression: R p [α] = R p + R p α + · · · + R p α d−1 (a direct sum). We see that R p [α] ∩ K is contained in the first term R p on the right-hand side. So we have our conclusion. Lemma 7.1.2 Assume that α is anti-integral over R. For any β ∈ R[α] ∩ K , there exists an integer n > 0 such that Iβ ⊇ (I[α] )n . Proof Since β ∈ K , we have I[β] = Iβ := R : R β and so Iβ is a divisorial ideal of R. Then any prime divisor p of Iβ is of depth 1. To show Iβ ⊇ (I[α] )n for some n, we have only to show that I[α] ⊆ p for each prime divisor of Iβ . Suppose that p ⊇ I[α] . Then β ∈ R[α] ∩ K ⊆ R p by Proposition 7.1.1 and hence Iβ ⊆ p, which contradicts the assumption Iβ ⊆ p. Lemma 7.1.3 Assume that α is anti-integral over R. For any β ∈ R[α] ∩ K , there exists a ∈ R such that (β − a)I[β] ⊆ I[α] ηd .

7.1 Subring R[α] ∩ K of Anti-Integral Extensions

137

Proof Let β = a0 + a1 α + · · · + an α n with some ai ∈ R. Then (a0 − β) + a1 α + · · · + an α n = 0. For any b ∈ I[β] , we have b(a0 − β) + ba1 X + · · · + ban X n ∈ E (α) = I[α] ϕα (X )R[X ], where the last equality results from the antiintegrality of α. Considering the constant term, we have b(a0 − β) ∈ I[α] ηd . Thus (β − a0 )I[β] ⊆ I[α] ηd . The element a is a0 , which we are looking for.

Lemma 7.1.4

Assume that α is anti-integral over R and that R[α]∩ K = R.

(1) There exists β ∈ R[α] ∩ K \ R such that I[α] ⊆ I[β] . (2) If γ ∈ R[α] ∩ K \ {0}, then Iγ −a = Iγ for any a ∈ R. Proof The second assertion (2) is obvious by definition. To prove (1), take β ∈ R[α] ∩ K \ R. Then there exists an integer n > 0 such that (I[α] )n ⊆ I[β] = Iβ , i.e., (I[α] )n β ⊆ R, by Lemma 7.1.2. Take a minimal n having the above property. If aβ ∈ R for all a ∈ (I[α] )n−1 , then (I[α] )n−1 β ⊆ R, that is, (I[α] )n−1 ⊆ Iβ = I[β] , which contradicts the minimality of n. Hence there exists b ∈ (I[α] )n−1 such that bβ ∈ R and I[α] bβ ⊆ (I[α] )n β ⊆ R. By the choice of b, we have bβ ∈ R[α] ∩ K \ R. So we have only to take bβ as a required element β. Let J be a fractional ideal of R and let R(J ) := J : K J, which is an over-ring of R. Remark 7.1.5 It is known that α is anti-integral over R and R(I[α] ) = R if and only if α is super-primitive over R (see Theorem 2.3.10). Theorem 7.1.6 Assume that α is a super-primitive element over R and that ηd ∈ R. Then R[α] ∩ K = R. Proof Suppose that R[α]∩K = R. Then there exists an element β ∈ R[α]∩K with β ∈ R such that I[α] ⊆ I[β] by Lemma 7.1.4. Replace β by β − a for some a ∈ R, we can assume that β I[β] ⊆ I[α] ηd by Lemma 7.1.3. Recall that I[β] = I[β−a] for any a ∈ R by Lemma 7.1.4. Since ηd ∈ R, we have I[α] ηd ⊆ I[α] and so β I[α] ⊆ β I[β] ⊆ I[α] ηd ⊆ I[α] , which means that β ∈ R(I[α] ) = R, a contradiction. Therefore R[α] ∩ K = R. Example 7.1.7 Let k be a field and s, t denote indeterminates. Let R = k[s, t] be a polynomial ring. Let α be an element which has the minimal polynomial ϕα (X ) = X 2 + (t/s)X + 1. Then ηd = 1 ∈ R, I[α] = s R, and J[α] = (s, t)R.

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Hence α is a super-primitive element over R. So we have R[α] ∩ K = R by Theorem 7.1.6. Note that R[α] is not integral over R. A strong version of Theorem 7.1.6 is given by the following: Theorem 7.1.8

Assume that α is super-primitive over R. Then

R[α] ∩ K ⊆



R p ( p ∈ Dp1 (R) with p ⊇ Iηd )

p

Proof Take p ∈ Dp1 (R) with p ⊇ Iηd . Then ηd ∈ R p and R((I[α] ) p ) = R p by Remark 7.1.5. So R p [α] ∩ K = R p by Theorem 7.1.6. Thus R[α] ∩ K ⊆ R p
and therefore R[α] ∩ K ⊆ R p ( p ∈ Dp1 (R) with p ⊇ Iηd ). For a ∈R, we have ϕα (a) = a d + η1 a d−1 + · · · + ηd ∈ K . Let Mα denote the ideal a∈R Iϕα (a) of R, where Iϕα (a) = R : R ϕα (a). Theorem 7.1.9 Assume that α is a super-primitive element over R. If Mα ⊆ p for any p ∈ Dp1 (R), then R[α] ∩ K = R. Proof The assumption Mα ⊆ p for any p ∈ Dp1 (R) yields that there exists Iϕα (a) ⊆ p for each p ∈ Dp1 (R). Note that ϕα (a) is the constant term of ϕα (X + a) = ϕα−a (X ) and that α − a is super-primitive over R (cf. Proposition 2.2.10). Hence R[α] p ∩ K = R[α − a] p ∩ K ⊆ R p by Theorem 7.1.6. Thus R[α] ∩ K ⊆ R[α] p ∩ K ⊆ R p for all p ∈ Dp1 (R). So we have R[α] ∩ K ⊆
p∈Dp1 (R) R p = R. Now we are devoted to investigating what is R[α] ∩ K in the case it is not equal to R. In the next proposition, we remark that α is not necessarily assumed to be anti-integral over R. Proposition 7.1.10

If η1 , . . ., ηd−1 ∈ R, then R[α] ∩ K = R[ηd ].

Proof Since ηd = −(α d + η1 α d−1 + · · · + ηd−1 α) ∈ R[α] ∩ K , we have R[ηd ] ⊆ R[α] ∩ K . Note that ϕα (X ) is a minimal polynomial of α over K and ηd ∈ K . It follows that R[ηd ][α] = R[ηd ]+ R[ηd ]α +· · ·+ R[ηd ]α d−1 (a direct sum). Considering R[ηd ] ⊆ K [α] = K + K α + · · · + K α d−1 , we conclude that R[α] ∩ K ⊆ R[ηd ][α] ∩ K = R[ηd ]. Hence R[α] ∩ K = R[ηd ]. Example 7.1.11 Let R be the same as in Example 7.1.7. Let α be an element which has the minimal polynomial ϕα (X ) = X 2 + X + (1/s). Then I[α] =

7.1 Subring R[α] ∩ K of Anti-Integral Extensions

139

Mα = s R, J[α] = R, and R[α] ∩ K = R[1/s] by Proposition 7.1.10. Note here that α is not super-primitive over R. The example above is generalized as follows: Proposition 7.1.12 Assume that α is super-primitive over R. If I˜[α] +Iηd = R, then R[α] ∩ K = R[ I˜[α] ηd ]. Proof Put D = R[ I˜[α] ηd ]. Then it is obvious that R[α] ∩ K ⊇ D. Take P ∈ Spec(D) with depth(D P ) = 1, and put p = P ∩ R. Since I˜[α] + Iηd = R, it follows that p ⊇ Iηd or p ⊇ I˜[α] . If p ⊇ Iηd , then ηd ∈ R p , and hence R[α] ∩ K ⊆ R p [α] ∩ K = R p ⊆ D P by Theorem 7.1.6. If p ⊇ I˜[α] , then η1 , . . ., ηd−1 ∈ R p . So by Proposition 7.1.10, we have R[α]∩
K ⊆ R p [α]∩ K = R p [ηd ] ⊆ D P . In any case, we have R[α] ∩ K ⊆ P∈Dp1 (D) D P = D = R[ I˜[α] ηd ]. This completes the proof. The proof of Proposition 7.1.12 also shows the following. Corollary 7.1.13 Assume that α is super-primitive over R. If P ∩ R ⊇ Iηd + I˜[α] for every P ∈ Dp1 (R[ I˜[α] ηd ]), then R[α] ∩ K = R[ I˜[α] ηd ]. In the next two propositions, we show that α anti-integral (respectively super-primitive) over R is still anti-integral (respectively super-primitive) over a Noetherian intermediate ring B in the case B is flat over R. B Let B be an intermediate ring between R and K . Put IηBi = B : B ηi , I[α] = 
d B B B B B b∈B Iϕα (b) . If B is flat over R, then i=1 Iηi , J[α] = I[α] c(ϕα (X )) and Mα = B B IηBi = Iηi B, I[α] = I[α] B, J[α] = J[α] B and MαB = Mα B. Theorem 7.1.14 Assume that α is anti-integral over R, that R ⊆ B ⊆ K and that B is a Noetherian ring and is flat over R. Then α is anti-integral over B. Proof

Consider the following exact sequence: π

0 −→ Ker(π ) −→ R[X ] −→ R[α] −→ 0 Tensoring it with B over R, we have the exact sequence of B-modules: πB

0 −→ Ker(π) ⊗ R B −→ B[X ] −→ B[α] −→ 0

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Exclusive Extensions of Noetherian Domains

Since α is anti-integral over R, we have E (α) := Ker(π ) = I[α] ϕα (X )R[X ] (α) B (α) and E (α) ⊗ R B = I[α] ϕα (X ) B := Ker(π ). Since B is flat over R, E B = E R[X ] ⊗ R B = I[α] ϕα (X )B[X ]. So we have the exact sequence: πB

0 −→ Ker(π B )ϕα (X )B[X ] −→ B[X ] −→ B[α] −→ 0 This shows that E (α) B is generated by polynomials of degree d in B[X ]. Thus α is anti-integral over B by Proposition 2.2.3. Note that if R ⊆ B ⊆ K and B is a Noetherian ring and is flat over R, then B I[α] ϕα (X )B[X ] = I[α] ϕα (X )B[X ] = Ker(π B ), where π B : B[X ] −→ B[α] is the canonical B-algebra homomorphism sending X to α. Theorem 7.1.15 Assume that α is super-primitive over R, that R ⊆ B ⊆ K , and that B is a Noetherian ring and is flat over R. Then α is also super-primitive over B. Proof Since B is flat over R, we have IηBi = Iηi B for all i = 1, . . ., d. Thus we have: B J[α]

=

B I[α] c(ϕα (X ))

=

(

d 

IηBi )(1, η1 , . . ., ηd )

i=1

=

(

d 

Iηi B)(1, η1 , . . ., ηd )

i=1

=

(

d 

Iηi )(1, η1 , . . ., ηd )B

i=1

=

J[α] B

Since α is super-primitive over R, J[α] ⊆ p for every p ∈ Dp1 (R). Since B B = J[α] B ⊆ P for is flat over R, J[α] B ⊆ P for every P ∈ Dp1 (B). Thus J[α] every P ∈ Dp1 (B), which means that α is super-primitive over B. Corollary 7.1.16 Let R, α, and B be the same as in Theorem 7.1.15. Assume that ηd ∈ B. Then R[α] ∩ K ⊆ B. Proof Theorem 7.1.15 yields that α is super-primitive over B. Since ηd ∈ B, we have B[α]∩ K = B by Theorem 7.1.12, whence R[α]∩ K ⊆ B[α]∩ K = B.

7.1 Subring R[α] ∩ K of Anti-Integral Extensions

141

Corollary 7.1.17 Assume that α is super-primitive over R and that R[ηd ] is flat over R. Then R[α] ∩ K ⊆ R[ηd ]. Proof

Apply Corollary 7.1.16 to the case B = R[ηd ].

Proposition 7.1.18 Let B be an intermediate Noetherian ring between R and K and assume that α is super-primitive over R and that R[α] is flat over R. Then α is super-primitive over B and B[α] is flat over B. Proof Since R[α] is flat over R, it follows that J[α] = R by Theorem 2.3.6. B B So J[α] ⊇ J[α] B = B and hence J[α] = B. Thus α is super-primitive over B. B[α] is flat over B by Theorem 2.3.6. Proposition 7.1.19 Let B be an intermediate Noetherian ring between R and R[α] ∩ K . Assume that α is super-primitive over B and that MαB ⊆ p for all p ∈ Dp1 (B). Then R[α] ∩ K = B. Proof Since α is super-primitive over B and MαB ⊆ p for all p ∈ Dp1 (B), we have B ⊆ R[α] ∩ K ⊆ B[α] ∩ K = B by Theorem 7.1.9, and hence R[α] ∩ K = B. Corollary 7.1.20 Assume that α is super-primitive over R. Let a1 , . . ., an be elements in R such that (a1 , . . ., an )R ⊆ p for all p ∈ Dp1 (R). Let Bi := R[1/ai ] for 1 ≤ i ≤ n. Assume that MαBi ⊆ p for all p ∈ Dp1 (Bi ) and all i. Then R[α] ∩ K = R. Proof Since Bi (1 ≤ i ≤ n) is flat over R, α is super-primitive over Bi for all i by Theorem
n 7.1.15. So R[α] ∩ K = Bi by Proposition 7.1.19. Hence R[α] ∩ K ⊆
n
i=1 Bi . Since (a1 , . . ., an )R ⊆ p for all p ∈ Dp1 (R), we have p∈Dp1 (R) R p = R. Thus R ⊆ R[α]∩ K ⊆ R and hence R[α]∩ K = i=1 Bi ⊆ R. In the rest of this section, we consider the case of lower degree and some examples. We restrict ourselves to the case d = 2 and show some examples. Proposition 7.1.21 Let a, b be elements in R with a = 0 and let ϕα (X ) = X 2 + (1/a)X + (b/a). Then R[α] ∩ K = R.

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Exclusive Extensions of Noetherian Domains

Proof Since J[α] = a(1, 1/a, b/a)R = R, the element α is a super-primitive element over R. The constant term of ϕα+b (X ) belongs to R. Hence R[α]∩ K = R by Theorem 7.1.6. Proposition 7.1.22

Let ϕα (X ) = X 2 + η1 X + η2 with ηi ∈ K .

(1) Assume that η1 = b/a with a regular sequence a, b ∈ R and that Iη1 ⊆ Iη2 . Then R[α] ∩ K = R. (2) If R[α] ∩ K = R, then Iη1 ⊆ Iη2 . Proof (1) Since Iη1 = a R and Iη1 ⊆ Iη2 , η2 = c/a. Hence ϕα (X ) = X 2 + (b/a)X + (c/a). In this case, J[α] = (a, b, c)R and J[α] ⊆ p for p ∈ Dp1 (R). So α is super-primitive over R. By Theorem 7.1.15, α is super-primitive over R[1/a] and R[1/b]. Thus R[α] ∩ K ⊆ R[1/a][α] ∩ K = R[1/a]. On the other hand, we can apply Proposition 7.1.21, we have R[α] ∩ K ⊆ R[1/b] and hence R[α] ∩ K ⊆ R[1/a] ∩ R[1/b] = R because a, b is a regular sequence. Therefore R[α] ∩ K = R. (2) This follows from η2 Iη1 ⊆ R[α] ∩ K = R. Example 7.1.23 Let R be a polynomial ring k[s, t] over a field k. Let ϕα (X ) = X 2 + (t/s)X + (1/s). Then by Proposition 7.1.22(1) R[α] ∩ K = R. Example 7.1.24 Let R be the same as in Example 7.1.23. Let ϕα (X ) = X 2 + (t/s)X + (1/s 2 ). It is clear that 1/s ∈ R[α] ∩ K . Put B = R[1/s]. Then B is flat over R and α is super-primitive over R. Moreover ηd = 1/s 2 ∈ B. So R[α]∩K ⊆ B = R[1/s] by Corollary 7.1.26. The inclusion R[1/s] ⊆ R[α]∩K is obvious. Hence R[α] ∩ K = R[1/s]. Proposition 7.1.25 Let ϕα (X ) = X 2 + (1/a)X + (c/b) with a, b, c ∈ R. Then R[α] ∩ K = R[ac/b]. Proof The inclusion R[ac/b] ⊆ R[α] ∩ K is obvious. Put u = ac/b. Then R[u] ϕα (X ) = X 2 + (1/a)X + (u/a). Hence J[α] = a(1, 1/a, u/a)R[u] = R[u], which means that α is super-primitive over R[u]. So we have R[u][α] ∩ K = R[u] by Proposition 7.1.21. Thus R[u] ⊆ R[α] ∩ K ⊆ R[u][α] ∩ K = R[u], which yields R[α] ∩ K = R[u]. Example 7.1.26

We use the same notations as in Example 7.1.23.

(i) Let ϕα (X ) = X 2 + (1/s)X + (1/t). Then R[α] ∩ K = R[s/t]. (ii) Let ϕα (X ) = X 2 + (1/s)X + (1/t 2 ). Then R[α] ∩ K = R[s/t 2 ].

7.2 Exclusive Extensions and Integral Extensions

7.2

143

Exclusive Extensions and Integral Extensions

Let R be a Noetherian integral domain and R[X ] a polynomial ring. Let γ be an element of a field extension L of the quotient field K of R. If γ ∈ L is transcendental over R, then R[γ ]∩K = R, and if γ ∈ K then R[γ ]∩K = R[γ ]. So we have only to consider an element algebraic over R. We study the following problem: Problem. Let R be a Noetherian integral domain, K denote the quotient field of R, and α be an algebraic element over R of degree d. When does R[α] ∩ K = R? Now we start with the following definition. Definition 7.2.1 When R[α] ∩ K = R, we say that α is an exclusive element over R and that R[α] is an exclusive extension of R. Proposition 7.2.2

If R[α] ∩ K = R, then I˜[α] ⊆ Iηd .

Proof Note that α d + η1 α d−1 + · · · + ηd = 0. Take a ∈ I˜[α] . Then aηd = − (aα d + aη1 α d−1 + · · · + aηd−1 α) ∈ R[α] ∩ K = R. Thus a ∈ Iηd , which means that I˜[α] ⊆ Iηd . Corollary 7.2.3 Assume that α is super-primitive over R. When I˜[α] + Iηd = R, the following statements are equivalent: (i) α is exclusive over R, i.e., R[α] ∩ K = R, (ii) I˜[α] ⊆ Iηd . Proof (ii) ⇒ (i): By Proposition 7.1.12, we have R = R[ I˜[α] ηd ] = R[α] ∩ K because I˜[α] ηd ⊆ R. (i) ⇒ (ii) follows from Proposition 7.1.2. From Proposition 7.1.12, we have the following result: Corollary 7.2.4 If I[α] = R, then R[α] ∩ K = R. Moreover if α is superprimitive over R, then Iηd = R, i.e., ηd ∈ R, implies R[α] ∩ K = R. The following lemma is very elementary, but we give a proof for convenience. Lemma 7.2.5 Let f (X ) ∈ R[X ] be a monic polynomial and let g(X ) ∈ K [X ]. If f (X )g(X ) is a monic polynomial in R[X ], then g(X ) ∈ R[X ].

144

Exclusive Extensions of Noetherian Domains

Proof Put f (X ) := X n + a1 X n−1 + · · · + an with ai ∈ R and g(X ) = λ0 X m + · · · + λm with λi ∈ K . Since f (X )g(X ) is monic, we have λ0 = 1. Assume that λ0 = 1, λ2 , . . ., λ−1 ∈ R and λ ∈ R. Then the coefficient of the degree ( + n)-term is λ + a1 λ−1 + · · · + a λ0 ∈ R. Hence λ ∈ R, a contradiction. Thus λi ∈ R for all i. So we have g(X ) ∈ R. Recall the following, which has been proved in Section 2.3. Proposition 7.2.6 (cf. Theorem 2.3.2 ) Assume that R is normal. Then α is integral over R if and only if I[α] = R. Proof (⇒): Since R is normal, we have ϕα (X ) ∈ R[X ] by Theorem 2.3.2. By Theorem 2.2.9, α is super-primitive and hence anti-integral because R is a Krull domain. Hence I[α] = R follows from Theorem 2.3.2. (⇐): Since I[α] = R, we have Ker(π ) = I[α] ϕα (X )R[X ] = ϕα (X )R[X ]. Thus ϕα (α) = 0 gives rise to an integral dependence of α. So α is integral over R.

Corollary 7.2.7 Assume that R is normal. If α is integral over R, then α is an exclusive element over R. Proof We have I[α] = R by Proposition 7.2.6 and it follows that R[α]∩K = R by Corollary 7.2.6. Proposition 7.2.8 If I˜[α] = R and if α is exclusive over R, i.e., R[α]∩K = R, then R[α] is integral over R.
d Proof By Proposition 7.2.6, I˜[α] ⊆ Iηd and hence I[α] = i=1 Iηi = R. Thus η1 , . . ., ηd ∈ R, which means that ϕα (X ) ∈ R[X ] and ϕα (α) = 0. This yields that α is integral over R. In Lemma 1.1.3, we have seen that if α ∈ K is both integral and anti-integral over R then α ∈ R. For the case d ≥ 0, we have a similar result as follows. Theorem 7.2.9 If α is both anti-integral and integral over R, then α is exclusive over R, i.e., R[α] ∩ K = R. Proof Since α is integral over R, there exists a monic polynomial f (X ) ∈ R[X ] such that f (α) = 0. Since α is anti-integral over R, we have Ker(π) = I[α] ϕα (X )R[X ]. Hence f (X ) ∈ Ker(π ) = I[α] ϕα (X )R[X ]. So there exists a

7.3 An Exclusive Extension Generated by a Super-Primitive Element

145

polynomial g(X ) ∈ I[α] R[X ] such that f (X ) = ϕα (X )g(X ). Since both f (X ) and ϕα (X ) are monic, g(X ) is monic. Thus ϕα (X ) ∈ R[X ] by Lemma 7.2.5, which shows that I[α] = R. Therefore R[α] ∩ K = R by Corollary 7.2.4.

7.3

An Exclusive Extension Generated by a Super-Primitive Element

Let R be a Noetherian domain and K its quotient field. Let α be an algebraic element over K with the minimal polynomial ϕα (X ) = X d +η1 X d−1 +· · ·+ηd . In Section 7.1, we have shown that if R[α] ∩ K = R then I˜[α] ⊆ Iηd . Our objective of this section is to show the converse of this result under certain assumptions, which will be established in Theorem 7.3.7. A nonzero element α in L is called co-monic if α −1 is integral over R and a polynomial f (X ) ∈ R[X ] is co-monic if the constant term of f (X ) is 1. Proposition 7.3.1 Assume that α is anti-integral over R of degree d. Then the following conditions are equivalent: (i) α is co-monic; (ii) ηd I[α] = R; (iii) there exists an element a ∈ R such that ηd = 1/a and I[α] = a R; (iv) there exists a co-monic polynomial g(X ) ∈ R[X ] of degree d with g(α) = 0; (v) R[α −1 ] is a free R-module of rank d. Proof (i) ⇒ (ii): Take a co-monic polynomial f (X ) in R[X ] satisfying f (α) = 0. Then f (X ) ∈ I[α] ϕα (X )R[X ] and hence f (0) = 1 ∈ ηd I[α] ⊆ R. So we have ηd I[α] = R. (ii) ⇒ (iii): There exists a ∈ I[α] such that ηd a = 1. Hence ηd = 1/a and I[α] = a R. (iii) ⇒ (iv): Since I[α] = a R, g(X ) := aϕα (X ) ∈ R[X ] is a required polynomial. (iv) ⇒ (v): The polynomial f (Y ) := X −d g(X ) in R[Y ] with Y = 1/ X is monic in Y of degree d and satisfies f (α −1 ) = 0. Hence R[α −1 ] is a free R-module of rank d because α −1 is of degree d. (v) ⇒ (i) is obvious. Recall the following result shown in Theorem 4.3.2:

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Lemma 7.3.2 Assume that α is an anti-integral element over R of degree d. Put R[α]/R := { p ∈ Spec(R) | p R[α] = R[α]} and J[α] := { p ∈ Spec(R) | p + J[α] } = R. Then R[α]/R = V ( J˜[α] ) ∩ J[α] , where V ( J˜[α] ) denotes { p ∈ Spec(R) | J˜[α] ⊆ p}. Remark 7.3.3 Under the assumptions in Lemma 7.3.2, if either J˜[α] = R or grade( J˜[α] ) > 1 then p R[α] = R[α] for all p ∈ Dp1 (R). Lemma 7.3.4 Assume that α is a super-primitive element over R. The following statements are equivalent: (i) J˜[α] = R or grade( J˜[α] ) > 1 (i.e., J˜[α] contains an R-regular sequence of length > 1; see [M2] for the definition); (ii) I˜[α] ⊆ Iηd . Proof (i) ⇒ (ii): First consider the case J˜[α] = R. Suppose that there exists a ∈ I˜[α] but a ∈ Iηd . Then β := aηd ∈ R. Since β I[α] = a(ηd I[α] ) ⊆ R and βηi I[α] = (aηi )ηd I[α] ⊆ R (i = 1, . . ., d −1), we have β I[α] (1, η1 , . . ., ηd−1 ) ⊆ R. Since grade( J˜[α] ) = grade(I[α] (1, η1 , . . ., ηd−1 )) >
1, β R p = β(I[α] (1, η1 , . . ., ηd−1 )) p ⊆ R p for all p ∈ Dp1 (R). Hence β ∈ Dp1 (R) R p = R, which is a contradiction. Thus I˜[α] ⊆ Iηd . Second, suppose that J˜[α] = R. By the assumption, there is an equation 1 = a0 + a1 η1 + · · · + ad−1 ηd−1 with ai ∈ I[α] . So ηd = a0 ηd + (a1 ηd )η1 + · · · + (ad−1 ηd )ηd−1 . Take x ∈ I˜[α] . Then xηd = xa0 ηd + (a1 ηd )xη1 + · · · + (ad−1 ηd )xηd−1 ∈ R, and hence x ∈ Iηd . Hence I˜[α] ⊆ Iηd . (ii) ⇒ (i): We may assume that J˜[α] = I[α] (1, η1 , . . ., ηd−1 ) = R. Suppose that grade( J˜[α] ) = grade(I[α] (1, η1 , . . ., ηd−1 )) = 1. Then there exists p ∈ Dp1 (R) such that I[α] (1, η1 , . . ., ηd−1 ) ⊆ p. Since p ∈ Dp1 (R), there exists β ∈ K such that β ∈ R and Iβ = p. Then β I[α] (1, η1 , . . ., ηd−1 ) ⊆ R. Thus (I[α] β)ηi ⊆ R for i = 1, . . ., d − 1 and I[α] β ⊆ R, which yield I[α] β ⊆ I˜[α] ⊆ Iηd . Hence I[α] βηd ⊆ R, which shows that J[α] β ⊆ R. But since α is super-primitive over R, J[α] ⊆ p for all p ∈ Dp1 (R). Thus β ∈ R, which is a contradiction. So we conclude that grade( J˜[α] ) > 1. Proposition 7.3.5 Assume that α is super-primitive and co-monic over R. If I˜[α] ⊆ Iηd , then R[α] ∩ K = R, i.e., α is exclusive over R. Proof Suppose that R[α] ∩ K = R. Then there exists β ∈ R[α] ∩ (K \ R). Write β = c0 + c1 α + · · · + cn α n with ci ∈ R. We shall show that α ∈ Iβ R[α]. Since β, c0 , . . ., cn ∈ K and α is an algebraic element of degree d over K , we have n ≥ d. From this we have:

7.3 An Exclusive Extension Generated by a Super-Primitive Element

147

(1) (α −1 )n β = c0 (α −1 )n + c1 (α −1 )n−1 + · · · + cn ∈ R[α −1 ]. Since α is co-monic, R[α −1 ] is a free R-module of rank d by Proposition 7.3.1. So we rewrite (1) and have: (α −1 )n β = b0 (α −1 )d−1 + b1 (α −1 )d−2 + · · · + bd−1 with bi ∈ R. Hence we obtain: (2) (α −1 )n = β −1 b0 (α −1 )d−1 + β −1 b1 (α −1 )d−2 + · · · + β −1 bd−1 . On the other hand, since (α −1 )n ∈ R[α −1 ], we have: (3) (α −1 )n = a0 (α −1 )d−1 + a1 (α −1 )d−2 + · · · + ad−1 with ai ∈ R. Comparing the coefficients of (2) with those of (3), we have ai = β −1 bi (0 ≤ i ≤ d − 1), and hence βai = bi ∈ R. Thus (a0 , a1 , . . ., ad−1 )R ⊆ Iβ . It is obvious that α ∈ (a0 , . . ., ad−1 )R[α] ⊆ Iβ R[α] by (3). Take p ∈ Dp1 (R) such that Iβ ⊆ p. By (3), α −1 = a0 α n−d + a1 α n−d+1 + · · · + ad−1 α n−1 . So we get α −1 ∈ p R[α]. Thus p R[α] contains α, α −1 . So we have p R[α] = R[α], which is a contradiction because of Remark 7.3.3 and Lemma 7.3.4. Therefore R[α] ∩ K = R. Remark 7.3.6 In Section 7.2, we showed that if α is super-primitive over R with ηd ∈ R then α is exclusive. In the next theorem, the implication (i) ⇒ (ii) has been shown in Proposition 7.2.2 without any assumptions. Theorem 7.3.7 Assume that R contains an infinite field k and that α is super-primitive over R. Then the following statements are equivalent: (i) α is exclusive over R; (ii) I˜[α] ⊆ Iηd ; (iii) grade( J˜[α] ) > 1 or J˜[α] = R. Proof The implication (i) ⇒ (ii) was shown in Proposition 7.2.2. (ii) ⇔ (iii) follows from Lemma 7.3.4. So we have only to prove the implication (ii) ⇒ (i). Take p ∈ Dp1 (R). If I[α] ⊆ p, then α is integral over R p and hence R[α] ∩ K ⊆ R p [α] ∩ K = R p . Consider the case I[α] ⊆ p. Since α is super-primitive, J[α] ⊆ p. Since k is infinite, there exists λ ∈ k such that α − λ is co-monic over R p . Moreover α − λ is super-primitive by Proposition 2.2.10, Hence by Proposition 7.3.5, we have R[α] ∩ K
= R[α − λ] ∩ K ⊆ R p [α − λ] ∩ K = R p [α] ∩ K = R p . Thus R[α] ∩ K ⊆ p∈Dp1 (R) R p = R, and hence R = R[α] ∩ K , that is, R[α] is exclusive over R.

Remark 7.3.8 Assume that R contains a field k. Let x be an indeterminate S and put S := R ⊗k k(x). Then S contains the infinite field k(x). Put I[α] :=

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d

S S S S : S ηi ), J[α] := I[α] (1, η1 , . . ., ηd )S and J˜[α] := I[α] (1, η1 , . . ., ηd−1 )S. S S S Since S is faithfully flat over R, I[α] = I[α] S, J[α] = J[α] S, and J˜[α] = J˜[α] S. So if α is super-primitive (respectively anti-integral) over R, then so is α over S. Moreover α is exclusive over S if and only if so is α over R; I˜[α] ⊆ Iηd if S S and only if I˜[α] ⊆ IηSd ; grade( J˜[α] ) > 1 if and only if grade( J˜[α] ) > 1; and α is exclusive over S, i.e., S[α] ∩ K (x) = S[α] if and only if α is exclusive over R, i.e., R[α] ∩ K = R. i=1 (S

Lemma 7.3.9 (cf. Theorem 7.3.7) Assume that α is super-primitive over R. Then the following statements are equivalent: (i) I˜[α] ⊆ Iηd ; (ii) grade( J˜[α] ) > 1 or J˜[α] = R. Furthermore if R contains a field, then the following (iii) is equivalent to (i): (iii) α is exclusive over R. Proof (i) ⇔ (ii) follows from Lemma 7.3.4. Next assume that R contains a field. We may assume that R contains an infinite field. Hence our conclusion (ii) ⇔ (iii) follows from Theorem 7.3.7. Remark 7.3.10

Assume that R contains a field.

(i) When α is a super-primitive element over R, α is exclusive over R if and only if grade( J˜[α] ) > 1 by Lemma 7.3.9. (ii) The equality J˜[α−1 ] = I[α] (η1 , . . ., ηd ) holds. This follows from the similar argument of Remark 2.5.14. (iii) By Theorem 2.4.8, α is super-primitive over R if and only if so is α −1 . So by (i) and (ii) above, we have α −1 is exclusive over R if and only if grade(I[α] (η1 , . . ., ηd )) > 1. (iv) If grade(I[α] (η1 , . . ., ηd−1 )) > 1, then both α and α −1 are exclusive over R by (i), (ii), and (iii). Proposition 7.3.11 Assume that α is an anti-integral element of degree d over R. If I[α] R[α] P = R[α] P for every P ∈ Dp1 (R[α]), then R[α] ∩ K = R[η1 , . . ., ηd ]. −1 −1 −1 Proof It follows that R[α] P ⊇ I[α] I[α] R[α] P = I[α] R[α] P ⊇ I[α]

η1 , . . .,
ηd because P∈Dp1 (R) R[α] P = R[α]. Thus R[α] ∩ K ⊇ R[η1 , . . ., ηd ] =: C,

7.3 An Exclusive Extension Generated by a Super-Primitive Element

149

α is integral and anti-integral over C. Since K (C) = K and [K (α) : K ] = d, R[α] is a free C-module C + Cα + · · · + Cα d−1 . Hence R[α] ∩ K = C, as was to be shown. Theorem 7.3.12 Assume that α is a super-primitive element of degree d over R and that R contains a field. Let φ : Spec(R[α]) → Spec(R) be the canonical map induced from the inclusion R ⊆ R[α]. Then the following statements are equivalent: (i) Dp1 (R) is contained in the image of φ, i.e., Im(φ) ⊇ Dp1 (R); (ii) α is exclusive over R. Proof Im(φ) ⊇ Dp1 (R) if and only if J˜[α] p = J[α] p = R p for every p ∈ Dp1 (R) by Theorem 4.3.2. Since α is super-primitive over R, we see that grade(J[α] ) > 1. Hence we have grade( J˜[α] ) > 1, that is, J˜[α] p = R p for every p ∈ Dp1 (R). So we conclude that α is exclusive over R by Lemma 7.3.9. Conversely, the set of diffusing points (i.e., p ∈ Spec(R) such that p R[α] p =
d−1 R[α] p ) is given by i=1 V (I[α] ηi ) \ V (I[α] ηd ) = V ( J˜[α] ) \ V (I[α] ηd ) by Theorem 4.3.2. So take p ∈ Dp1 (R) such that p R[α] p = R[α] p . Then p ⊇ J˜[α] and p ⊇ I[α] ηd . Thus we have grade( J˜[α] ) = 1, which yields that α is not exclusive over R by Lemma 7.3.9. Hence q R[α]q = R[α]q for all q ∈ Im(φ). Proposition 7.3.13 Assume that α1 , . . ., αn are super-primitive elements over R and that R contains a field. Put A := R[α1 , . . ., αn ] and let φ : Spec(A) → Spec(R) be the canonical map induced from the inclusion R ⊆ A. If Dp1 (R) is contained in the image of φ, then each αi (1 ≤ i ≤ n) is exclusive over R. Proof Let ψi : Spec(A) → Spec(R[αi ]) and φi : Spec(R[αi ]) → Spec(R) be the canonical maps induced from the inclusion R[αi ] ⊆ A and R ⊆ R[αi ], respectively. Then φ = φi · ψi induces the inclusions Im(φi ) ⊇ Im(φ) ⊇ Dp1 (R). Thus our conclusion follows Theorem 7.3.12. Problem.

Is the converse statement of Proposition 7.3.13 valid?

Theorem 7.3.14 Assume that α is a super-primitive element of degree d over R. If ηd ∈ R, then the canonical map φ : Spec(R[α]) → Spec(R) is surjective. Proof Note that φ : Spec(R[α]) → Spec(R) is surjective if and only if V ( J˜[α] ) = V (J[α] ) ⊆ V (I[α] ηd ) (cf. Theorem 4.3.2). Since ηd ∈ R, we have ηd I[α] ⊆ I[α] , which implies that V ( J˜[α] ) = V (J[α] ) by the construction of J[α] . So we have our conclusion.

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Note that φ : Spec(R[α]) → Spec(R) is surjective and φ is flat if and only if φ is faithfully flat. So we obtain the following corollary. Corollary 7.3.15 Assume that ηd ∈ R. If R[α] is flat over R, then R[α] is faithfully flat over R. The following proposition gives rise to a condition for ηd ∈ R. Proposition 7.3.16 Assume that α is a super-primitive element of degree d over R and that R contains a field. Then the following statements are equivalent: (i) ηd ∈ R; (ii) α is exclusive over R and Iηd R[α] = R[α]. Proof (i) ⇒ (ii) follows from Lemma 7.3.9. (ii) ⇒ (i): Since ηd ∈ Iη−1 R[α] = Iη−1 Iηd R[α] ⊆ R[α], we conclude that d d ηd ∈ R[α] ∩ K = R. Remark 7.3.17

Let I denote an ideal of R. Then grade(I ) > 1 ⇔ I −1 = R

where I −1 := R : K I . Theorem 7.3.18 Assume that α is an anti-integral element of degree d over R. Let f : Spec(R[α]) → Spec(R) and g : Spec(R[ηd ]) → Spec(R) be the canonical maps obtained from the inclusions R ⊆ R[α] and R ⊆ R[ηd ], respectively. If grade R[α] (J[α] R[α]) > 1, grade R[α] (J[ηd ] R[α]) > 1, and grade R[ηd ] (J[α] R[ηd ]) > 1, then R[α] ∩ K = R[ηd ] and Im( f ) = Im(g). R[η ]

Proof First note that J[α] R[ηd ] ⊆ J[α] d (here we use the notation as in Remark 7.3.8; put S = R[ηd ]). So grade R[ηd ] (J[α] R[ηd ]) > 1 yields that R[η ] grade R[ηd ] (J[α] d ) > 1, so that α is super-primitive over R[ηd ]. Take P ∈ Dp1 (R[α]) and put p := P ∩ R. Since J[α] R[α] ⊆ P, we have either J[α] ⊆ p or Iηd ⊆ p. Thus R p [α] is flat over R p . So we have Iηd R p [α] = (R : R ηd )R p [α] = R p [α]
: R p [α] ηd = R p [α]. Thus ηd ∈ R p [α] by Proposition 2.5.11. Therefore ηd ∈ P∈Dp1 (R[α]) R[α] P = R[α]. As mentioned above, α is super-primitive over R[ηd ]. Since ηd ∈ R[ηd ], α is exclusive over R[ηd ]. Hence the canonical map ψ : Spec(R[α]) → Spec(R[ηd ]) is surjective by Proposition 7.3.14.

7.3 An Exclusive Extension Generated by a Super-Primitive Element

151

Consider the following commutative diagram: Spec(R[η  d ][α])  ψ Spec(R[ηd ])

===== f

−−−−→

Spec(R[α])   g Spec(R)

Here we use that ηd ∈ R[α]. Since ψ is surjective and f · ψ = g, we conclude that Im( f ) = Im(g). We say that α is an ultra-primitive element of degree d over R if grade(I[α] + C(R/R)) > 1, where R denotes the integral closure of R in K and C(R/R) denotes the conductor between R and R. Proposition 7.3.19 Assume that α is an ultra-primitive element of degree d. If grade(I[α] : R I[α−1 ] ) > 1, then ηd ∈ R. Proof Take p ∈ Dp1 (R). Then either I[α] ⊆ p or C(R/R) ⊆ p. If I[α] ⊆ p, then I[α] ⊆ Iηd ⊆ p, that is, ηd ∈ R p . If I[α] ⊆ p, then C(R/R) ⊆ p and hence R p is a normal domain. Note that I[α−1 ] p = ηd I[α] p ⊆ I[α] p (cf. Lemma 2.4.6). The latter inclusion ηd I[α] p ⊆ I[α] p , that is, ηd ∈ (I[α] : K I[α] ) p implies that ηd is integral over R p , noting that I[α] p is finitely generated over R p . So we have
ηd ∈ R p . Therefore ηd ∈ p∈Dp1 (R) R p = R. Note that ϕα (X ) = X d + η1 X d−1 + · · · + ηd with η j ∈ K , where η0 := 1, and put ϕα−1 (X ) = X d + η1 X d−1 + · · · + ηd with ηj ∈ K , where η0 := 1. Then ηj = ηd− j /ηd for every j (0 ≤ j ≤ d). (i) Put J˜[α] := I[α] (η0 , η1 , . . ., ηi−1 , ηi+1 , . . ., ηd ) for (0 ≤ i ≤ d), and put (i)   ˜J −1 := I[α−1 ] (η0 , η1 , . . ., ηi−1 , ηi+1 , . . ., ηd ) for (0 ≤ i ≤ d). It is clear that [α ] (d) (d) J˜[α] = J˜[α] and J˜[α−1 ] = J˜[α−1 ] by definition. We start with the following lemma. Lemma 7.3.20

The following equalities hold:

(i) Iηi = ηd Iηd−i for all i (0 ≤ i ≤ d); (ii) I[α−1 ] = ηd I[α] ; (iii) J[α] = J[α−1 ] ; (i) (iv) J˜[α] = J˜[α(d−i) −1 ] for each i (0 ≤ i ≤ d). Proof (i) Take x ∈ ηd Iηd−i . Then x = ηd y with y ∈ Iηd−i and xηd−i = ηd−i yηd ∈ ηd R. Hence x(ηd−i /ηd ) ∈ R, which implies that x ∈ Iηd−i /ηd = Iηi .

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  Conversely, take x ∈ Iηd−i . Hence xηd−i = x(ηi /ηd ) = (x/ηd )ηi ∈ R and so x/ηd ∈ Iηi . Thus x ∈ ηd Iηi .

(ii) I[α−1 ] = dj=0 Iηj = dj=0 ηd Iη j = ηd I[α] . (iii) J[α] = I[α] (η0 , . . ., ηd ) = ηd I[α] (η0 /ηd , . . ., ηd−1 /ηd , ηd /ηd ) = I[α−1 ] (η0 , . . ., ηd ) = J[α−1 ] . (iv) is obtained by the same way as the proof of (ii).

Lemma 7.3.21 (cf. Theorem 7.3.7 and its proof) If α is exclusive over R, i.e., R[α] ∩ K = R, then either grade( J˜[α] ) > 1 or J˜[α] = R. Recall that if α is a super-primitive element (respectively an anti-integral element) over R then so is α −1 over R (cf. Theorems 2.4.5 and 2.4.8). Proposition 7.3.22 Assume that α is exclusive over R, i.e., R[α] ∩ K = R and that R contains an infinite field. Then α (respectively α −1 ) is exclusive over R, i.e., R[α] ∩ K = R (respectively R[α −1 ] ∩ K = R) if and only if either grade( J˜[α] ) > 1 (respectively grade( J˜[α−1 ] ) > 1) or J˜[α] = R (respectively J˜[α−1 ] = R). Proof The implication (⇒) follows from Lemma 7.3.21. The reverse implication (⇐) follows from (Theorem 7.3.7). Proposition 7.3.23 Assume that α is a super-primitive element over R. Then the following statements are equivalent for each i (0 ≤ i ≤ d):
(i) j=i Iη j ⊆ Iηi ;
 (i’) j=d−i Iηj ⊆ Iηd−i ; (i) (i) (ii) grade( J˜[α] ) > 1 or J˜[α] = R;

˜ (d−i) (ii’) grade( J˜[α(d−i) −1 ] ) > 1 or J[α −1 ] = R. Proof The equivalences: (i) ⇔ (i ) and (ii) ⇔ (ii ) are shown in Lemma 7.3.20. (i) (i) (i) ⇒ (ii): We assume that J˜[α] = R. Suppose that grade( J˜[α] ) = 1. Then (i) there exists p ∈ Dp1 (R) such that J˜[α] = I[α] (η1 , . . ., ηi−1 , ηi+1 , . . ., ηd ) ⊆ p. Since p ∈ Dp1 (R), there exists an element β ∈ K such that Iβ = p. Thus (i) β J[α] = β I[α] (η1 , . . ., ηi−1 , ηi+1 ,
. . ., ηd ) ⊆ R. Hence (I[α] β)η j ⊆ R for all
j = i, which yields that I[α] β ⊆ j=i Iη j . Since j=i Iη j ⊆ Iηi , we have that I[α] βηi ⊆ R. Thus J[α] β ⊆ R, that is, J[α] ⊆ Iβ = P. But since α is superprimitive over R, we have J[α] ⊆ p, which is a contradiction. So we conclude (i) > 1. that grade( J˜[α]

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153

˜ (i) (ii) ⇒
(i): First, consider the case J[α] = R. Suppose that there exists an element a ∈ j=i Iη j but a ∈ Iηi . Then β := aηi ∈ R. So β I[α] (η0 , η1 , . . ., ηi−1 , ηi+1 , (i) . . ., ηd ) ⊆ R. Since grade( J˜[α] ) = grade(I[α] (η0 , η1 , . . ., ηi−1 , ηi+1 , . . ., ηd )) > (η0 , η1 , . . ., ηi−1 , ηi+1 , . . ., ηd )R p ⊆ R p for every 1, we have β R p = β(I[α]
p ∈ Dp1 (R). Hence β ∈ p∈Dp1 (R) R p = R, which is a contradiction. Thus
˜ (i) j=i Iη j ⊆ Iηi . Second, suppose that J[α] = R. Then there exists an equality: 1 = a0 η0 + a1 η1 + · · · + ai−1 + ai+1 ηi+1 + · · · + ad ηd with ai ∈ I[α] . So ηi = (a0 ηi
)η0 + (a1 ηi )η1 + · · · + (ai−1 ηi )ηi−1 + (ai+1 ηi )ηi+1 + · · · + (ad ηi )ηd . Take X ∈ j=i Iη j . Then xηi = x(a0 ηi )η0 + x(a1 ηi )η1 + · · · + x(ai−1 ηi )ηi−1 + x(ai+1
ηi )ηi+1 +· · ·+x(ad ηi )ηd ∈ R, and hence x ∈ Iηi . Therefore we conclude that j=i Iη j ⊆ Iηi . The equivalence (i ) ⇔ (ii ) can be shown by the same way as the proof of (i) ⇔ (ii). Remark 7.3.24 Let A be an integral domain containing R. Then the following statements are equivalent: (1) A is exclusive over R (i.e., A ∩ K = R); (2) A p is exclusive over R p (i.e.,
A p ∩ K = R p ) for all p ∈ Dp1 (R). [Indeed, we have only to recall that p∈Dp1 (R) R p = R.] Theorem 7.3.25 Assume that α is a super-primitive element of degree d over R, that R contains an infinite field, and that grade(I[α] + I[α−1 ] ) > 1. Then the following statements are equivalent: (i) R[α, α −1 ] is exclusive over R, i.e., R[α, α −1 ] ∩ K = R; (i) (i) (ii) grade( J˜[α] ) > 1 or J˜[α] = R for every i (0 ≤ i ≤ d);
(iii) j=i Iη j ⊆ Iηi for every i (0 ≤ i ≤ d); (ii’) grade( J˜[α(i)−1 ] ) > 1 or J˜[α(i)−1 ] = R for every i (0 ≤ i ≤ d);
(iii’) j=i Iηj ⊆ Iηi for every i (0 ≤ i ≤ d). Proof The equivalences (ii) ⇔ (iii), (ii ) ⇔ (iii ), and (ii) ⇔ (ii ) follow from Proposition 7.3.23.
(i) ⇒ (iii): Suppose that
j=i Iη j ⊆ Iηi for some i (0 ≤ i ≤ d). Then there exists an element a ∈ j=i Iη j with λ := aηi ∈ R. Since ϕα (α) = 0, we have aϕα (α) = aα d + aη1 α d−1 + · · · + aηi α d−i + · · · + aηd = 0, which yields −λα d−i = aη0 α d + · · · + aηi−1 α d−i+1 + aηi+1 α d−i−1 + · · · + aηd . So we have −λ = aα i + · · · + (aηi−1 )α + (aηi + 1)α −1 + · · · + (aηd )α i−d ∈ R[α, α −1 ] ∩ K = R, which is a contradiction. Therefore we conclude that
j=i Iη j ⊆ Iηi for every i (0 ≤ i ≤ d).

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(ii’) ⇒ (i): Since R[α, α −1 ] ∩ K = R ⇔ R p [α, α −1 ] ∩ K = R p for all p ∈ Dp1 (R) by Remark 7.3.24, we may assume that R is a local domain with maximal ideal p with depth(R) = 1. In this case, I[α] + I[α−1 ] = R because grade(I[α] + I[α−1 ] ) > 1, and hence I[α] = R or I[α−1 ] = R. Take λ ∈ R[α, α −1 ]∩ K . Then (∗) λ = a0 α n + · · · + an−1 α + an + b1 α −1 + · · · + bm α −m where ai , b j ∈ R. Since λ ∈ R ⇔ λ − an ∈ R, we may assume that an = 0. Put ψ(X ) := a0 X n+m + · · · + an−1 X m+1 − λX m + b1 X m−1 + · · · + bm . Then ψ(α) = 0. Since α −1 is a super-primitive element by Lemma 7.3.20, we may assume that I[α] = R by symmetry. Thus ϕα (X ) ∈ R[X ]. (a) Assume that n = 0. Then λ ∈ R[α −1 ]. Since J˜[α−1 ] = J˜[α(d)−1 ] = R, it follows that α −1 is exclusive over R by Proposition 7.3.22. Hence λ ∈ R[α −1 ]∩ K = R. (b) Assume that m ≥ d. Then deg ψ(X ) = n + m ≥ d. Since ϕα (X ) ∈ R[X ], there exists a polynomial h(X ) ∈ R[X ] such that ψ ∗ (X ) := ψ(X )−ϕα (X )h(X ) is of degree m. Note that ψ(X ) ∈ R[X ] if and only if ψ ∗ (X ) ∈ R[X ]. Hence using ψ ∗ (X ) instead of ψ(X ), we can assume that n = 0, which is done in (a). (c) Assume that m < d. Then since ϕα (X ) ∈ R[X ], there exists a polynomial h(X ) ∈ R[X ] such that ψ ∗ (X ) := ψ(X ) − ϕα (X )h(X ) is of degree less than d. Note that ψ ∗ (α) = ψ(α) − ϕα (α)h(α) = 0. It follows that ψ ∗ (X ) = 0 in R[X ] because α is an element of degree d over K . Hence ψ(X ) = ϕα (X )h(X ) ∈ R[X ], which means that λ ∈ R. The implications: (i) ⇒ (iii ) and (ii) ⇒ (i) can be proved by the similar way to the above argument. Theorem 7.3.26 Assume that grade(I[α] + I[α−1 ] ) > 1. Then R[α, α −1 ]∩ K = R if and only if R[α] ∩ K = R and R[α −1 ] ∩ K = R (i.e., both α and α −1 are exclusive over R). Proof (⇐): Take p ∈ Dp1 (R). Then I[α] ⊆ p or I[α−1 ] ⊆ p. by the assumption grade(I[α] + I[α−1 ] ) > 1. Assume that I[α] ⊆ p. Then (I[α] ) p = R p , so that ϕα (X ) ∈ R p [X ]. Since ϕα (α) = α d + η1 α d−1 + · · · + ηd = 0, we have α = −(η1 + η2 α −1 + · · · + ηd α 1−d ) ∈ R p [α −1 ]. Thus R p [α] ⊆ R p [α −1 ] and hence R p [α, α −1 ] = R p [α −1 ]. So we have R p [α, α −1 ] ∩ K = R p by the assumption α −1 is exclusive over R p (cf. Remark 7.3.24). Next assume that I[α−1 ] ⊆ p. Then in the similar way to the above case, we have R p [α, α −1 ]∩ K = is exclusive over R p (cf. Remark R p by the assumption α

7.3.24). Thus we have R ⊆ R[α, α −1 ] ∩ K ⊆ p∈Dp1 (R) R p [α, α −1 ] ∩ K = p∈Dp1 (R) R p = R. The implication (⇒) follows from the inclusions R ⊆ R[α ±1 ] ∩ K ⊆ R[α, α −1 ] ∩ K = R.

7.4 Finite Generation of an Intersection R[α] ∩ K over R

7.4

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Finite Generation of an Intersection R[α] ∩ K over R

Our objective is to investigate when the R-subalgebra R[α] ∩ K is finitely generated over R. In this section, we fix the following notation in addition to the definitions mentioned above unless otherwise specified: Let R be a Noetherian domain with quotient field K . Let L be an algebraic field extension of K and α be an element in L which is of degree d over K . Let ϕα (X ) := X d + η1 X d−1 + · · · + ηd denote the minimal polynomial of α over K (that is, ηi ∈ K ). Let
B be an intermediate ring between R d B B B and K . Put IηBi := B : B ηi , I[α] := i=1 IηBi , J[α] := I[α] (1, η1 , . . ., ηd ) and
d−1 B R R R I˜[α] := i=1 IηiB . When B = R, I[α] = I[α] , J[α] = J[α] , and I˜[α] = I˜[α] . Note B B B ˜ ˜ that if B is flat over R, then I[α] = I[α] B, I[α] = I[α] B, and J[α] = J[α] B. We start with the following lemma: Lemma 7.4.1

Let t be an element in R.

(1) I[α−t] = I[α] , J[α−t] = J[α] , and I˜[α−t] = I˜[α] . (2) Let ηd(t) denote the constant term of the minimal (monic) polynomial ϕα−t (X ) ∈ K [X ] of α − t. Then R[ I˜[α−t] ηd(t) ] = R[ I˜[α] ηd ]. Proof (1) By definition, we see that I[α] ⊆ I[α−t] (respectively J[α] ⊆ J[α−t] , I˜[α] ⊆ I˜[α−t] ). By symmetry, we also have I[α] ⊆ I[α−t] (respectively J[α−t] ⊆ J[α] , I˜[α−t] ⊆ I˜[α] ). (2) It is easy to see that ηd(t) = ϕα (t). We have R[ I˜[α] ηd ] ⊇ R[ I˜[α] ϕα (t)] = R[ I˜[α] ηd(t) ] by definition. By symmetry, we also have R[ I˜[α] ηd(t) ] ⊇ R[ I˜[α] ηd ]. An element α is called α co-monic over R if α −1 is integral over R. For a polynomial f (X ) ∈ R[X ], we call f (X ) a co-monic polynomial over R if the constant term of f (X ) is 1. The following theorem is the first main result of this section: Theorem 7.4.2 Assume that R contains an infinite field k, that J[α] = R, and that I˜[α] is an invertible ideal. Then R[α] ∩ K = R[ I˜[α] ηd ] and hence the R-algebra R[α] ∩ K is finitely generated and flat over R. Proof Take p ∈ Spec(R). Since J[α] = R, I[α] is invertible. Note that I˜[α] is invertible by assumption. So it follows that I[α] R p = a R p and I˜[α] = b R p for some a, b ∈ R. Note that R p = J[α] R p = I[α] (1, η1 , . . ., ηd )R p =

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a(1, η1 , . . ., ηd )R p . Since k is an infinite field and since J[α] = R ⊆ p, there exists an element t ∈ k such that ϕα (t) = 0 and aϕα (t) ∈ R \ p. Let ϕα−t (X ) := X d + η1(t) + · · · + ηd(t) . Since the constant term ηd(t) of the minimal polynomial ϕα−t (X ) of α − t is given by ϕα (t), α − t is co-monic over R p and in this case, ηd(t) = ϕα (t) and aϕα (t) ∈ R p \ p R p . Since I˜[α−t] = I˜[α] ⊇ I[α] = I[α−t] by Lemma 7.4.1, we have b R p ⊇ a R p and hence a = bc for some ( p) ( p) c ∈ R p . So I˜[α−t] ηd = b R p ϕα (t) = (a/c)ϕα (t)R p . Hence R p [ I˜[α−t] ηd ] = R[(a/c)ϕα (t)] = R p [1/c] ⊆ R[α] p ∩ K because aϕα (t) is a unit in R p . R p [1/c] R p [1/c] Since R p [1/c] is flat over R, we have I[α−t] = I[α−t] R p [1/c], I˜[α−t] = R p [1/c] ˜I[α−t] R p [1/c], and J[α] = J[α] R p [1/c] = R p [1/c]. Hence α is superprimitive over R p [1/c]. Since α −t is co-monic over R p [1/c], α −t is exclusive over R p [1/c], i.e., R[α] p [1/c]∩ K = R p [α −t][1/c]∩ K = R p [1/c] by Proposition 7.3.5. So we have R[α] p [1/c] ∩ K = R p [1/c] ⊇ R[α] p ∩ K ⊇ R p [1/c]. ( p) Therefore we obtain R[α] p ∩ K = R p [ I˜[α−t] ηd ] = R p [ I˜[α] ηd ] by Lemma 7.4.1. Since p ∈ Spec(R) is arbitrary, we conclude that R[α] ∩ K = R[ I˜[α] ηd ]. Moreover as was seen above, R[α] p ∩ K = R p [ I˜[α] ηd ] = R p [1/c] for each p ∈ Spec(R), which implies that R[α] ∩ K is flat over R. Remark 7.4.3 In Corollary 7.4.13, we have shown that: Assume that α is a super-primitive element over R. If P∩R ⊇ Iηd + I˜[α] for any P ∈ Dp1 (R[ I˜[α] ηd ], then R[α] ∩ K = R[ I˜[α] ηd ]. In particular, if Iηd + I˜[α] = R, then R[α] ∩ K = R[ I˜[α] ηd ]. A related result is the following theorem, which is the second main result. Theorem 7.4.4 Assume that R contains an infinite field. Assume that α is a super-primitive element over R and that R[ I˜[α] ηd ] is flat over R. Then R[α] ∩ K = R[ I˜[α] ηd ] and hence R[α] ∩ K is finitely generated over R. Proof Put B := R[ I˜[α] ηd ]. Since B is flat over R, we have IηBi = B : B B B B B ηi = Iηi B. So I[α] = I[α] B, I˜[α] = I˜[α] B, and J[α] = I[α] (1, η1 , . . ., ηd ) = I[α] (1, η1 , . . ., ηd )B = J[α] B. Since α is super-primitive over R, grade(J[α] ) > B 1. Since B is flat over R, grade(J[α] ) > 1, which means that α is super-primitive B B B B over B. Note that I˜[α] ηd ⊆ B = R[ I˜[α] ηd ]. So I˜[α] ⊆ IηBd and hence I˜[α] ⊆ I[α] . Thus α is exclusive over B by Theorem 7.4.7. Thus B[α]∩ K = B ⊇ R[α]∩ K . Take b ∈ I˜[α] . Then bηi ∈ R for all 1 ≤ i ≤ d − 1. So we have b(α d + η1 α d−1 + · · ·+ηd−1 α) = −bηd ∈ R[α]∩ K . Hence B ⊆ R[α]∩ K . The reverse inclusion is obvious. Proposition 7.4.5 Assume that R contains an infinite field. Assume that α is a super-primitive element over R and that there exists a Noetherian domain

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157

B such that R[ I˜[α] ηd ] ⊆ B ⊆ R[α] ∩ K , and that B is flat over R. Then R[α] ∩ K = B. B Proof Since B is flat over R, we have IηBi = B : B ηi = Iηi B. So I[α] = I[α] B, B B B I˜[α] = I˜[α] B, and J[α] = I[α] (1, η1 , . . ., ηd ) = I[α] (1, η1 , . . ., ηd )B. Note that B B B B I˜[α] ηd ⊆ B = R[ I˜[α] ]. So I˜[α] ⊆ IηBd and hence I˜[α] ⊆ I[α] . Thus α is exclusive over B by Theorem 7.3.7. Since α is a super-primitive element over R, we B have grade(J[α] ) > 1. The flatness of B over R implies that grade(J[α] ) = grade(J[α] B). So α is super-primitive over B because so is R[α] over R. Hence by Theorem 7.3.7, we have B[α] ∩ K = B ⊇ R[α] ∩ K ⊇ B. Therefore R[α] ∩ K = B.

The following proposition means that the associated prime divisors of D/R, where D is an integral domain containing R is determined by the birational extension D ∩ K of R. Proposition 7.4.6 Let D be a Noetherian integral domain containing R. Then Ass R (D ∩ K /R) = Ass R (D/R). Proof Put B := D ∩ K . Since R ⊆ B ⊆ D, we have Ass R (B/R) ⊆ Ass R (D/R). Conversely, take p ∈ Ass R (D/R). Then there exists β ∈ R[α] such that p = R : R β. Take a nonzero element a ∈ p. Then aβ ∈ R and hence β ∈ K , so that β ∈ D ∩ K = B. Thus p ∈ Ass R (B/R). Now recall the following lemma: Lemma 7.4.7 (cf. Lemma 1.2.11). Let C be an integral domain and D a subalgebra of the quotient field of C. Then D ⊆ C is flat if and only if for every prime ideal p of D, either pC = C or D p = C p . Proposition 7.4.8 Let D be an integral domain containing R. Then D ∩ K is flat over R if and only if p(D ∩ K ) = D ∩ K for each p ∈ Ass R (D ∩ K /R). Proof Put B := D ∩ K and note that B is a birational extension of R. (⇐) Take P ∈ Spec(R). Assume that R P = B P . Then R P ⊇ B. Choose an element b ∈ B with b ∈ R P . Then Ib := R : R b ⊆ P. Take a prime divisor p of Ib such that p ⊆ P. Then p ∈ Ass R (B/R). By assumption, p B = B and hence P B = B. By Lemma 7.4.7, B is flat over R. (⇒) Assume that B is flat over R. Then for p ∈ Ass R (B/R), there exists b ∈ B such that p = Ib . So we have b ∈ R p , which means that B p = R p . By Lemma 7.4.7, we conclude that p B = B.

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Remark 7.4.9 If there exists p ∈ Ass R (R[α]/R) such that p R[α] = R[α], then R[α] ∩ K is not flat over R. The next theorem shows that flatness implies finiteness in the anti-integral extension, which is the third main result. Theorem 7.4.10 Assume that α is an anti-integral element over R. If R[α] ∩ K is flat over R, then R[α] ∩ K is finitely generated over R. Proof Take a nonzero element a ∈ I[α] . Then α has a monic relation of degree d over R[1/a]. So R[1/a][α] is a free R[1/a]-module. Thus B ⊆ R[1/a][α] ∩ K = R[1/a], which yields that B/R ⊆ R[1/a]/R and that Ass R (B/R) ⊆ Ass(R[1/a]/R). So Ass R (B/R) is a finite set because Ass R (R[1/a]/R) ⊆ { p ∈ Spec(R)| p is a prime divisor of a R}. Put Ass R (B/R) = { p1 , . . ., p } and put I = p1 ∩ · · · ∩ p . Since B is flat over R, we have I B = B by Lemma 7.4.7 because R pi = B pi for 1 ≤ i ≤ . So we can take elements bi ∈ I  bi βi . Put C = R[β1 , . . ., β ]. The inclusion and βi ∈ B such that 1 = i=1 C ⊆ B is obvious. Take P ∈ Spec(C) and put P ∩ R = p. Then I ⊆ p and p B = B. Since B is flat over R, we have B
p = R p by Lemma 7.4.7. Thus B p = C p . Since B ⊆ C P , we have B ⊆ C P = C. Therefore B = C.

Proposition 7.4.11

Take β ∈ K . The following statements are equivalent:

(i) β is integral over R[α]; (ii) β is almost integral over R[α] ∩ K . Proof (i)⇒(ii): Since β is integral over R[α], β is almost integral over R[α]. Put I := {x ∈ R[α]|xβ  ∈ R[α] for all  > 0} ⊆ R[α], which is an ideal of R[α]. Then I = (0). Take x ∈ I \ (0). Then there exists an algebraic relation: a0 x n + · · · + an−1 x + an = 0 with an = 0, ai ∈ R[α]. In this case, an ∈ I ∩ R, which means that I ∩ R = (0). Take a ∈ I ∩ R with an = 0. Then aβ  ∈ R[α] ∩ K . So β is almost integral over R[α] ∩ K . (ii) ⇒ (i): Since β is almost integral over R[α] ∩ K , β is almost integral over R[α]. Since R[α] is Noetherian, β is integral over R[α]. An element β ∈ K is called a flat element over R if R[β] is flat over R. Remark 7.4.12 In Corollary 7.4.17, we see: Assume that α is super-primitive over R and that R[ηd ] is flat over R. Then R[α] ∩ K ⊆ R[ηd ].

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Proposition 7.4.13 Assume that R contains an infinite field and that α is a super-primitive element over R. If ηd is a flat element over R and ηd is integral over R[α], then R[α] ∩ K = R[ηd ]. Proof The inclusion R[ηd ] ⊆ R[α] ∩ K follows from the above remark. Since R[ηd ] is flat over R, α is super-primitive over R[ηd ] by Proposition 7.1.15. By Theorem 7.3.7, we have R[ηd ][α] ∩ K = R[ηd ]. Thus R[α] ∩ K ⊆ R[α][ηd ] ∩ K = R[ηd ][α] ∩ K = R[ηd ]. Therefore R[α] ∩ K = R[ηd ]. For an element β ∈ K , let Jβ denote the ideal J[β] , that is, Iβ (1, β)R. Proposition 7.4.14 Assume that α is a super-primitive element over R. Assume that ηd is integral over R[α] and is super-primitive over R and that J[α] + Jηd = R. Then ηd ∈ R[α] ∩ K . Proof Take P ∈ Spec(R[α]) and put p = P ∩ R. Then by assumption, either J[α] ⊆ p or Jηd ⊆ p. Assume that J[α] ⊆ p. R[α] Then R[α] p is flat over R p . So Jηd R[α] p ⊆ Jηd p and hence ηd is superprimitive over R[α] p . Since ηd is integral over R[α] p , we have ηd ∈ R[α] p ⊆ R[α] P . Assume now that Jηd ⊆ p. Since Jηd = Iηd + ηd Iηd , we have Iηd ⊆ p or ηd Iηd ⊆ p. If Iηd ⊆ p, then ηd ∈ R p ⊆ R[α] p ⊆ R[α] P . If ηd Iηd ⊆ p, then ηd = 1/a for some a ∈ p R p . Since 1/a is integral over R[α] p , by the same argument in the proof of Proposition 7.4.13, we have ηd = 1/a ∈ R[α] p ⊆ R[α] P . Hence ηd ∈ R[α]. Remark 7.4.15 If ηd ηi ∈ R[α] for all 1 ≤ i ≤ d − 1, then ηd is integral over R[α]. Indeed, since α d + η1 α d−1 + · · · + ηd = 0, we have ηd2 + (α d )ηd + (η1 ηd α d−1 + · · · ηd−1 ηd α) = 0. So ηd satisfies the monic relation of degree 2 over R[α]. Proposition 7.4.16 Assume that R contains an infinite field, that α is an antiintegral element over R, and that ηd is integral over R[α] and anti-integral over R. If Jηd R[α] = R[α], then ηd ∈ R[α] ∩ K . Proof Take P ∈ Spec(R[α]) and put p = P ∩ R. Then Jηd R[α] = R[α]. So Jηd ⊆ p. Since Jηd = Iηd + ηd Iηd , we have either Jηd ⊆ p or ηd Iηd ⊆ p. If Jηd ⊆ p, then ηd ∈ R p ⊆ R[α] P . If ηd Iηd ⊆ p, then ηd = 1/a for some a ∈ p R p . Since ηd is integral over R[α] p , we have ηd = 1/a ∈ R[α] p by the proof of Proposition 7.4.14. Hence ηd ∈ R[α].

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Pure Extensions

Our main objective of this section is to show the following results: (1) A pure extension of an integral domain is an exclusive extension. (2) Let R be a ring and A be a ringextension of R. Then A is flat over R and A is a pure extension of R if and only if A/R is flat over R. (3) If a simple extension R[α] of a Noetherian domain R is anti-integral and integral over R, then R[α] is a pure extension of R. (4) Finally we give some examples. We start with the following definition. Definition 7.5.1 Let R be a ring and let φ : N → M be an R-homomorphism of R-modules. We say that φ is pure if φ induces the injection φ ⊗ R 1 E : N ⊗ R E → M ⊗ R E for every R-module E. When N is an R-submodule of M, we say that N is a pure R-submodule of M if the inclusion i : N → M induces the injection i ⊗ R 1 E : N ⊗ R E → M ⊗ R E for every R-module E. Proposition 7.5.2 Let R be a ring and let φ : N → M be an R-homomorphism of R-modules. Then φ is pure over R if and only if φ ⊗ R 1 L : N ⊗ R L → M ⊗ R L is injective for each finitely generated R-module L. Proof Any R-module E is obtained as an inductive limit of an inductive system: E 1 ⊆ E 2 ⊆ · · ·, where each E i is a finitely generated R-module. Then we have N ⊗ R E = N ⊗ R lim E i = lim(N ⊗ R E i ) and M ⊗ R E = M ⊗ R lim E i = → → → lim(M ⊗ R E i ), since φ ⊗ R 1 Ei : N ⊗ R E i → M ⊗ R E i is injective for each i. → Since the functor lim is exact, it follows that lim(N ⊗ R E i ) → lim(M ⊗ R E i ) → → → is injective. Hence φ ⊗ R 1 E : N ⊗ R E → M ⊗ R E is injective, which means that φ is pure over R. Proposition 7.5.3 Let R be a ring and let φ : N → M and ψ : M → L be R-homomorphisms of R-modules N , M, and L. (i) If both φ and ψ are pure, then ψ · φ is pure. (ii) If ψ · φ is pure, then φ is pure. (iii) Let A be R-algebra. If φ is pure, then φ ⊗ R 1 A : N ⊗ R A → M ⊗ R A is pure.

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161

Proof Let E be an R-module. (i) Since φ ⊗ R 1 E and ψ ⊗ R 1 E are injective, so is (ψ ⊗ R 1 E )(φ ⊗ R 1 E ) = (ψ · φ) ⊗ R 1 E . Hence ψ · φ is pure. (ii) Since (ψ ·φ)⊗ R 1 E = (ψ ⊗ R 1 E )(φ⊗ R 1 E ) is injective, φ⊗ R 1 E : N ⊗ R E → M ⊗ R E is injective. Hence φ is pure. (iii) Let W be an A-module. Then W is regarded as an R-module naturally. Consider (φ ⊗ R 1 A ) ⊗ A 1W : (N ⊗ R A) ⊗ A W → (M ⊗ R A) ⊗ A W . Note that (φ ⊗ R 1 A ) ⊗ A 1W = φ ⊗ R 1W : N ⊗ R W → M ⊗ R W . Since φ is pure, φ ⊗ R 1W is injective. Thus φ ⊗ R 1 A is pure. Corollary 7.5.4 Let R be a ring and let φ : N → M be a pure Rhomomorphism of R-modules N into M. Let I be an ideal of R. Then φ −1 (I M) = I N. Proof Since φ is pure, φ ⊗ R 1 R/I : N /I N = N ⊗ R R/I → M ⊗ R/I = M/I M is injective. Hence φ −1 (I M) = I N . Remark 7.5.5

Let R be a ring and N ⊆ M be R-modules.

(1) If M = N ⊕ L for some R-submodule L of M. Then N is a pure Rsubmodule of M. (2) If M/N is flat over R, then N is a pure R-submodule of M. Proposition 7.5.6 Let R be a ring and let φ : N → M. Then the following statements are equivalent: (i) φ is pure; (ii) φ p : N p → M p is pure for every prime ideal p of R; (iii) φm : Nm → Mm is pure for every maximal ideal m of R. Proof (i) ⇒ (ii): In Proposition 7.5.3, we have only to take A = R p . (ii) ⇒ (iii) is obvious. (iii) ⇒ (i): Let E be an R-module. Then for every maximal ideal m of R, Ker(φ ⊗ R 1 E )m = Ker(φm ⊗ Rm 1Wm ) = 0 because φm is pure. Thus Ker(φ ⊗ R 1 E ) = 0, which implies that φ is pure. Definition 7.5.7 Let R be a ring and let A be an R-algebra. We say that A is a pure extension of R if R is a pure R-submodule of A.

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The next proposition shows that a pure extension satisfies the Lying-Over Theorem. Theorem 7.5.8 Let R be a ring and let A be an R-algebra. If A is a pure extension of R, the canonical map Spec(A) → Spec(R) is surjective. Proof Suppose that Spec(A) → Spec(R) is not surjective. Then there exists a prime ideal p of R such that p A ∩ R ⊃ p. Since the canonical homomorphism − R p → A p is pure by Proposition 7.5.6, we may replace R p (resp. A p ) by R (resp. A). In this case, p A = A. Since A is a pure extension of R, we have the injection R/ p = R ⊗ R R/ p → A ⊗ R R/ p = A/ p A = 0. Thus R/ p = 0, a contradiction. Hence Spec(A) → Spec(R) is surjective. Proposition 7.5.9 Let R be a ring and let A be a pure extension of R. If A is a field, then R is a field. Proof Suppose that there exists a nonzero prime ideal p of R. Then the canonical homomorphism R/ p = R ⊗ R R/ p → A ⊗ R R/ p = A/ p A is injective because the inclusion R → A is pure. Thus A/ p A = 0. Since A is a field and p = (0), we have p A = A, a contradiction. Hence R is a field. Let A be a ring extension of an integral domain R and K denote the quotient field of R. We say that A is an exclusive extension of R if A ∩ K = R. Let B be an intermediate ring between A and R. If A is an exclusive extension of R, then B is also an exclusive extension of R. The following proposition shows that a pure extension is an exclusive extension. Theorem 7.5.10 Let R be an integral domain with quotient field K and let A be a pure extension of R. Then A ∩ K = R. In particular, if R ⊆ A ⊆ K , then A = R. Proof Take γ ∈ A ∩ K . We shall show that γ ∈ R. Take a ∈ (R : R γ ) \ (0). Applying Corollary 7.5.4 to the case I = a R, N = R and M = A, we have a A ∩ R = a R. Since aγ ∈ a A ∩ R = a R, it follows that γ ∈ R. The converse inclusion is obvious. Corollary 7.5.11 Let R be an integral domain and let A be an exclusive extension of R. If an element a ∈ R is a unit in A, then a is a unit in R.

7.5 Pure Extensions

163

Proof If a ∈ R is a unit in A, then there exists ζ ∈ A such that aζ = 1. Since ζ = 1/a ∈ K , ζ ∈ A ∩ K = R. Hence a is a unit in R. Corollary 7.5.12 Let R be an integral domain and let A be a pure extension of R. If an element a ∈ R is a unit in A, then a is a unit in R. Theorem 7.5.13 Let R be a ring and let A be a ring extension of R. Then A is flat over R and A is a pure extension of R if and only if A/R is flat over R. Proof

Let M be an R-module. We have the exact sequence: (∗)

0 −→ R −→ A −→ A/R −→ 0

Tensoring − ⊗ R M, we have the exact sequence: ψ

(∗∗) Tor1R (M, A) → Tor1R (M, A/R) → R ⊗ R M → A ⊗ R M → · · · (⇒): Since A is flat over R, we have Tor1R (M, A) = 0. Since A is a pure extension of R, ψ is injective. Hence Tor1R (M, A/R) = 0 by (∗∗), which means that A/R is flat over R (cf. [M2]). (⇐): Since R and A/R are flat over R, the exact sequence (∗) yields that A is flat over R (cf. [M2]). Next the exact sequence (∗∗) yields that the canonical homomorphism R ⊗ R M → A ⊗ R M is injective because Tor1R (M, A/R) = 0. Thus A is a pure extension of R. In the rest of this section, we consider the following situation. Let R be a Noetherian domain and R[X ] a polynomial ring. Let α be an element of an algebraic field extension L of the quotient field K of R and π : R[X ] → R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with deg ϕα (X ) = d and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd Then ηi (1 ≤ i ≤ d) are uniquely determined by α. Let Iηi := R : R ηi
d and I[α] := i=1 Iηi , the latter of which is called a denominator ideal of α. We say that α is an anti-integral element if Ker(π ) = I[α] ϕα (X )R[X ]. For f (X ) ∈ R[X ], let c( f (X )) denote the ideal of R generated by the coefficients of f (X ). For an ideal J of R[X ], let c(J ) denote the ideal generated by the coefficients of the elements in J . If α is an anti-integral element, then c(Ker(π )) = c(I[α] ϕα (X )R[X ]) = I[α] (1, η1 , . . ., ηd ). Put J[α] = I[α] (1, η1 , . . ., ηd ). If J[α] ⊆ p for all p ∈ Dp1 (R) := { p ∈ Spec(R)|depth(R p ) = 1}, then α is called a superprimitive element. It is known that a super-primitive element is an anti-integral

164

Exclusive Extensions of Noetherian Domains

element (cf. Chapter 2). By definition, the super-primitive is characterized by the set of Dp1 (R). Proposition 7.5.14 pure extension of R.

If α is anti-integral and integral over R, then R[α] is a

Proof The extension R[α] is a flat R-module by Theorem 2.3.2, and hence R[α] = R⊕α R⊕· · ·⊕α d−1 R by Theorem 2.3.8. Thus the inclusion R → R[α] is pure by Remark 7.5.5. Theorem 7.5.15 Assume that R is a Noetherian normal domain and that α is integral over R. Then R[α] is a flat, pure extension of R. Proof Since R is a Krull domain, α is a super-primitive element over R by Theorem 2.2.9. So our conclusion follows from Proposition 7.5.14. Example 7.5.16 Let k be an infinite field and t be an indeterminate. Put R := k[t], a polynomial ring. (i) Let α be an element satisfying the following algebraic relation: 1 1 α2 + α + 2 = 0 t t Since R is a Noetherian normal domain, α is a super-primitive element by Theorem 2.2.9. It is easy to see that I[α] = t 2 R and J[α] = I[α] (1, 1/t, 1/t 2 )R = R. So R[α] is a flat extension of R by Theorem 2.3.6 but that α is not integral over R by Theorem 2.3.2. By Theorem 7.3.7, R[α] is not an exclusive extension of R, and hence R[α] is not a pure extension of R by Theorem 7.5.10. (ii) Let α be an element satisfying the following algebraic relation: 1 1 α 2 + α + = 0 · · · (∗) t t Since R is a Noetherian normal domain, α is a super-primitive element. It is easy to see that I[α] = t R and J[α] = I[α] (1, 1/t, 1/t)R = R. So R[α] is a flat extension of R but that α is not integral over R. By Lemma 7.3.7, R[α] is an exclusive extension of R. Consider the R-module R[α]/R. Since tα 2 + α + 1 = 0 in R[α], we have α = −tα 2 in R[α]/R. Thus α i R ⊆ α i+1 R for all i ≥ 1. Let Hn := α R + · · · + α n−1 R ⊆ R[α]/R.

7.5 Pure Extensions

165

Then Hn ⊆ α n R. Suppose that there exist a, b ∈ R (a = 0, b = 0) such that a(bα n ) = 0. Then abα n ∈ R and hence α n ∈ K . So by (∗), we have α n−1 ∈ K . Using (∗) successively, we have α ∈ K , which contradicts d = 2. So Hn is a torsion-free R-submodule of R[α]/R. Since R is a PID k[t], Hn is a free R-submodule of R[α]/R. Hence R[α]/R = lim Hn is a → flat R-module. Hence R[α] is a pure extension of R by Theorem 7.5.13.

Chapter 8 Ultra-Primitive Extensions and Their Generators

Let R be a Noetherian domain and R[X ] a polynomial ring. Let α be an element of an algebraic extension field L of the quotient field K of R and π : R[X ] → R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with deg ϕα (X ) = d and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd Then ηi (1 ≤ i ≤ d) are uniquely determined by α. Let Iηi := R : R ηi and
d I[α] := i=1 Iηi . If Ker(π) = I[α] ϕα (X ), we say that α is anti-integral over R. Put J[α] = I[α] (1, η1 , . . ., ηd ). Then J[α] = c(I[α] ϕα (X )), where c( ) denotes the ideal generated by the coefficients of the polynomials in ( ), that is, the content ideal of ( ). If J[α] ⊆ p for all p ∈ Dp1 (R) := { p ∈ Spec(R) | depth(R p ) = 1}, then α is called a super-primitive element over R. It is known that a superprimitive element over R is anti-integral over R (cf. Theorem 2.2.8). It is also known that any algebraic element over a Krull domain R is anti-integral over R (cf. Theorem 2.2.9). When α is an element in K , ϕα (X ) = X − α. So we have J[α] = I[α] (1, α) = Iα (1, α) = Iα + α Iα = Iα + Iα−1 , where Iα := R : R α. In this case, α ∈ K is anti-integral over R if and only if R = Rα =: Rα. Throughout this chapter, we use the following notation unless otherwise specified: Let R be a Noetherian domain with quotient field K and R denote the integral closure of R in K . Then R is a Krull domain (cf. [M2). Let C(R/R) denote the conductor ideal, i.e., {a ∈ R|a R ⊆ R}. Let L be an algebraic extension field of K and α be an element in L which is of degree d over K . Let ϕα (X ) := X d + η1 X d−1 + · · · + ηd denote the minimal polynomial of α over K (that is, ηi ∈ K ). Let I be an ideal of R. When I = R, then we define grade(I ) := the length of a maximal R-sequence in I . When I = R, let grade(I ) := ∞ by convention (cf. [M2, p.106]). Note in advance that K [α] is an algebraic extension field and K [α] = K (α).

167

168

8.1

Ultra-Primitive Extensions and Their Generators

Super-Primitive Elements and Ultra-Primitive Elements

We say that R is “N-1” if R is a finite R-module (cf. [M1, (32.A)]). Let C(R/R) denote the conductor ideal, i.e., {a ∈ R|a R ⊆ R}. If R is N-1, then C(R/R) = 0. We say that R is “N-2” if, for any finite extension L of K , the integral closure R L of R in L is a finite module. Let B be an intermediate ring between R and L. Assume that B is integral over R. If R is N-2, then B is N-1. If R is N-1 (resp. N-2), so is any localization of R (cf. [M1, (32.A)]). It is known that when R is of characteristic zero, R is N-1 if and only if R is N-2 (cf. [M1, (32.B)]). As shown in the examples below, a sum (a product) of anti-integral elements is not necessarily anti-integral. So we introduce some stronger property and we want to find a set of elements with such a property that is closed in addition and multiplication. This is our objective of this section. An element α ∈ L is called to be an ultra-primitive element over R if α satisfies grade(I[α] + C(R/R)) > 1 If R is not N-1, then C(R/R) may be (0). In this case, if α ∈ K is ultraprimitive over R, then the condition grade(I[α] + C(R/R)) > 1 implies that I[α] = R because grade(I[α] ) = 1 or I[α] = R, and hence that α ∈ R. Theorem 8.1.1 implies that an ultra-primitive element is a super-primitive element and hence an anti-integral element. Theorem 8.1.1 over R.

If α ∈ L is ultra-primitive over R, then it is super-primitive

Proof In order to prove that α is super-primitive over R, it suffices that α is super-primitive over R p for every p ∈ Dp1 (R) by Theorem 2.2.8. Take p ∈ Dp1 (R). Then I[α] ⊆ p or C(R/R) ⊆ p by assumption. When I[α] ⊆ p, α has a monic algebraic relation of degree d over R p . So α is anti-integral over R p by Propositions 2.1.2 and 2.1.3. Since I[α] ⊆ J[α] , we have J[α] ⊆ p. Thus α is super-primitive over R. When C(R/R) ⊆ p, R p = R p holds. Hence α is super-primitive over R p by Theorem 2.2.9. Remark 8.1.2 Assume that R is normal (i.e., integrally closed in K ). Then any algebraic element over R is ultra-primitive over R by definition. Example 8.1.3 Assume that R is N-1, that is, R is a finitely generated R-module. So C(R/R) = (0).

8.1 Super-Primitive Elements and Ultra-Primitive Elements

169

(1) Take a ∈ C(R/R) \ (0). Then α := 1/a is anti-integral but not ultraprimitive over R. In fact R[α] ∩ R[α −1 ] = R ∩ R[α −1 ] = R. Since I[α] + C(R/R) = Iα + C(R/R) = C(R/R), grade(I[α] + C(R/R)) = 1. (2) Take γ ∈ R \ R. Then γ is not anti-integral over R. Indeed, suppose that γ is anti-integral over R. Then R[γ ] is a free R-module of rank 1 since γ is integral over R. Thus R[γ ] = R, a contradiction. Put γ = a/b with a, b ∈ R. Then α := 1/b, and β := a. Then α and β are anti-integral over R. In fact, R[α] ∩ R[α −1 ] = R[1/a] ∩ R[a] = R[1/a] ∩ R = R and a ∈ R. But αβ = γ is not anti-integral over R. Example 8.1.4 Let k be a field and k[t] be a polynomial ring. Let R := k[t 2 , t 3 ](t 2 ,t 3 ) . Then α := (t 3 + 1)/t 2 and β := −1/t 2 are anti-integral over R. In fact, R[α] ∩ R[α −1 ] = R[1/t 2 ] ∩ R[t 2 ] = R and R[β] ∩ R[β −1 ] = R[1/t 2 ] ∩ R[t 2 ] = R[1/t 2 ] ∩ R = R. But α + β = t is not anti-integral over R. Remark 8.1.5 If α is anti-integral and integral over R, then R[α] is a free R-module R ⊕ Rα ⊕ · · · ⊕ Rα d−1 (cf. Theorem 2.3.8). Let R ⊆ S be Noetherian domains, let S be the integral closure of S in its quotient field K (S) and R[X ] and S[X ] be polynomial rings. Let f (X ), g(X ) ∈ (S) (S) K (S)[X ] and put I (S) f := S[X ] : S f (X ), I g := S[X ] : S g(X ), I f g := S[X ] : S f (X )g(X ), which are ideals of S, and I (R) f g := R[X ] : R f (X )g(X ), which is an ideal of R. Lemma 8.1.6 We employ the above notation. Assume that S is N-1 and that f (X ), g(X ) ∈ K [X ] (K = K (R)). Then the following statements hold: (S) (1) I (R) f g ⊆ I f g ∩ R;     (S) (S) (S) (2) I (S) I = I = I ∩ Ig(S) ; g f fg f   (S) (3) c(I (S) I f (X )g(X )) ⊇ c(I (R) g f f g f (X )g(X )).

Proof (1) I (R) f g = R[X ] : R f (X )g(X ) ⊆ S[X ] : R f (X )g(X ) = (S[X ] : S f (X )g(X )) ∩ R = I (S) f g ∩ R. (2) Let S denote the integral closure of S in K (S). Then S is a Noetherian normal domain and C(S/S) = (0) since S is N-1.  Step 1:

We first show:

(S) I (S) f Ig =



I (S) fg .

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Ultra-Primitive Extensions and Their Generators

(S) (S) Proof. The inclusion I (S) f I g ⊆ I f g is obvious. Take P ∈ Ht1 (S). Then S P is a (S) DVR and hence a unique factorization domain (UFD). So (I (S) ) P = (I (S) f Ig  f g )P  (S) (S) (S) (S) by Gauss’ Lemma. Since I f g is divisorial, we have I f Ig ⊇ I f g . Thus      (S) (S) (S) (S) (S) I f Ig = I f g . It is obvious that I f ∩ Ig(S) = I (S) f Ig .

 Step 2: We shall show that   I (S) I (S) fg ∩ S = fg .

I (S) f ∩ S =



 I (S) f ,

Ig(S) ∩ S =



Ig(S) , and

(S) Proof. First note that I (S) f = S[X ] : S f (X ) and I f = S[X ] : S f (X ). Hence (S) I (S) f ∩ S = S[X ] : S f (X ) ⊇ S[X ] : S f (X ) = I f . Conversely, take P ∈ (S P ) Dp1 (S). Assume that C(S/S) ⊆  P. Then S P = S P . Since (I (S) f )P = I f   (S P ) and (I (S) , we have (I (S) (I (S) f )P = I f f ) P ∩ SP = f ) P . Next assume that  the choice of P. C(S/S) ⊆ P. Note  that C(S/S)S P = P S P by  (S) Take c ∈ I f ∩ S. Then c ∈ P S P = C(S/S)S P and hence c ∈

C(S/S) P ∩ (I (S) ) . Thus c f (X ) ∈ S P [X ]. So we have c ∈ (I (S) f ) P , which f P    (S) (S) shows that ( I (S) ∩ S) ⊆ (I ) . Therefore we have I ∩ S ⊆ I (S) P P f f f f  (S) (S) (S) because I f is divisorial. So I f ∩ S = I f . By the same argument, we     have Ig(S) ∩ S = Ig(S) and I (S) ∩ S = I (S) fg fg . Step 3: 

We have

(S) I (S) f Ig

= =

   

  I (S) f ∩S  I (S) f

∩S

 Ig(S)

 ∩

 Ig(S)



∩S

=



(S) ∩S= I (S) f Ig ∩ S    (S) I ∩ S = I (S) I (S) = I (S) g f fg ∩ S = fg

=

I (S) f

 Ig(S) ∩ S

by Step 1 and Step 2. (3) follows from (1) and (2).

 I (S) f ∩

Ig(S) ∩ S

8.1 Super-Primitive Elements and Ultra-Primitive Elements

171

Lemma 8.1.7 Let S be a Noetherian N-1 domain with R ⊆ S ⊆ L and let S[X ] be a polynomial ring. Let f (X ), g(X ) ∈ K (S)[X ] be nonzero polynomials (S) such that f (X )g(X ) ∈ K [X ]. If c(I (R) f g f (X )g(X )) = R, then c(I f f (X )) = (S) c(Ig g(X )) = S. (R) Proof Since c(I (S) f g f (X )g(X )) ∩ R ⊇ c(I f g f (X )g(X )) = R by Lemma (S) (S) 8.1.6(1), we have c(I (S) f g f (X )g(X )) = S and hence c(I f I g f (X )g(X )) = S  by Lemma 8.1.6(3). Note that c( f (X )) c( f (X )g(X )) = c( f (X ))+1 c(g(X )) for a large  (cf. [G, (28.1)]). Thus (S) (S)  c(I (S) f f (X )) c(I f I g f (X )g(X )) +1 (S) = (I (S) (Ig )c( f (X )) c( f (X )g(X )) f ) +1 (S) = (I (S) (Ig )c( f (X ))+1 c(g(X )) f ) +1 = c(I (S) c(Ig(S) g(X )) f f (X ))  +1 = c(I (S) c(Ig(S) g(X )). Suppose that So we have c(I (S) f f (X )) f f (X )) (S) (S) c(I f f (X ))c(Ig g(X )) = S. Then (S) (S)  (S)  c(I (S) f f (X )) = (c(I f f (X ))c(I g g(X )))c(I f f (X )) (S) (S) yields that there exists c ∈ c(I (S) f f (X ))c(I g g(X )) such that (1 + c)c(I f  f (X )) = (0). Since S is an integral domain, we have c(I (S) f f (X )) = (0), a (S) contradiction. Hence c(I (S) f f (X )) = c(I g g(X )) = S.

Let B be a Noetherian domain such that R ⊆ B ⊆ L and K (B) denote the quotient field of B. Let ϕαK (B) (X ) := X  + η1(B) X −1 + · · · + η(B) be
 B := i=1 IηB(B) the monic minimal polynomial of α ∈ L over K (B). Put I[α] i

B B (= B[X ] : B ϕαK (B) (X )), where IηB(B) := B : B ηi(B) . Put J[α] := I[α] (1, η1(B) , . . ., i

η(B) )(= c(IαB ϕαK (B) (X ))). Let B be an integral domain containing R. Define Dp1 (B) ∩ R := {P ∩ R|P ∈ Dp1 (B)} Proposition 8.1.8 Let B be a Noetherian domain such that R ⊆ B ⊆ L and that Dp1 (B) ∩ R ⊆ Dp1 (R). Assume that R is N-2. If α is super-primitive over R, then α is super-primitive over B. Proof Take P ∈ Dp1 (B). Then p := P ∩ R ∈ Dp1 (R) by the assumption. Localizing at p, we may assume that J[α] = R because α is super-primitive

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B over R. So we have only to prove that J[α] = B. Since ϕαK (B) (X )|ϕα (X ) in K (B) K (B)[X ]. Put ϕα (X ) = ϕα (X )φ(X ) in K (B)[X ]. In this case, c(I fBg f (X ) g(X )) = c(B[X ] : B ϕα (X )) ⊇ c(R[X ] : R ϕα (X ))B = J[α] B = B and φ0 (X ) ∈ K (B)[X ] is monic. Then we can apply Lemma 8.1.7 for S := B, f (X ) := B φ(X ) ∈ K (B)[X ], and g(X ) := ϕαK (B) (X ) ∈ K (B)[X ]. Thus we have J[α] = B K (B) B K (B) ϕα (X )) and c((I[α] ϕα (X ))) = c(IgB g(X )) = B by Lemma 8.1.7 c(I[α] (R) B because c(I (R) f g f (X )g(X )) = c(Iϕα ϕα (X )) = J[α] = R, and so J[α] B K (B) = c(I[α] ϕα (X )) = B. Therefore we conclude that α is super-primitive over B.

Lemma 8.1.9 Let I , J , and T be ideals of R such that grade(I + T ) > 1 and grade(J + T ) > 1. Then grade(I J + T ) > 1. √ Proof Let a√1 , a2 (respectively b√1 , b2 ) be √ a regular sequence in I + T √ (respectively J + T ). Note that ( I + T )( J + T ) ⊆ I J + T . We have only to show√ that at least two elements in {a1 b1 , a1 b2 , a2 b1 , a2 b2 } form a regular sequence in I J + T , but this is straightforward. Let α, β be elements in L. Assume that R is N-2.   (1) Assume that Dp1 (R[α + β]) ∩ R ⊆ Dp1 (R). Then I[α] I[β] ⊆ I[α+β] .   (2) Assume that Dp1 (R[αβ]) ∩ R ⊆ Dp1 (R). Then I[α] I[β] ⊆ I[αβ] .

Lemma 8.1.10

Proof Take p ∈ Dp1 (R). Suppose that I[α] I[β] ⊆ p. Then I[α] ⊆ p and I[β] ⊆ p. Since I[α] ⊆ J[α] (respectively I[β] ⊆ J[β] ), α (respectively β) is super-primitive over R p . So α and β are both integral over R p by Theorem 2.3.2, and hence α + β and αβ are integral over R p . Since R is N-2, both R p [α + β] and R p [αβ] are N-1. Thus α is super-primitive and integral over R p [α + β] and R [γ ] R p [αβ] by Proposition 8.1.8. Put γ := α + β. Hence I[α]p = R p [γ ] and so R [γ ]

I[α]p ∩ R p = R p [γ ] ∩ R p = R p . Let ϕγK [α] (X ) ∈ K [α][X ] denote the minimal monic polynomial of γ over K [α]. We have that R p [α][β] = R p [γ ][α] is a free R p [γ ]-module R p [γ ] ⊕ R p [γ ]α⊕ · · · ⊕ R p [γ ]α −1 , where  = deg ϕαK [γ ] (X ). In particular, (∗)

K [γ ][α] = K [γ ] ⊕ K [γ ]α ⊕ · · · ⊕ K [γ ]α −1

Put ϕγK [α] (X ) = φ0 (X ) + φ1 (X )α + · · · + φd−1 (X )α −1 with φi (X ) ∈ K [X ], where φ0 (X ) is monic in K [X ] with deg φ0 (X ) ≤ deg ϕγK [α] (X ) ≤ deg ϕγ (X ). Note that φ0 (γ ) + φ1 (γ )α + · · · + φd−1 (γ )α d−1 = ϕγK [α] (γ ) = 0, which is a direct sum in (∗). Hence φ0 (γ ) = 0. So we have ϕγ (X )|φ0 (X ). Hence ϕγ (X ) =

8.1 Super-Primitive Elements and Ultra-Primitive Elements R [α]

φ0 (X ). Thus R p = I[β]p

R p : R p φ0 (X ) = R p : R p ϕγ (X ) = R [α]

R p = I[β]p R

R [α]

∩ R p = I[γ p] R I[γ p] ,

173

∩ R p = R p [α] : R p ϕγK [α] (X ) ⊆ R

R

p that is, I[α+β] = I[γ p] = R p . Similarly

R [α]

K [α] p ∩ R p ⊆ I[βα] ∩ R p = R p [α] : R p ϕβα (X ) ⊆ R p : R p ϕβα (X ) R

p p = I[βα] , that is, I[βα] = R p . So we have I[α+β] ⊆ p and I[αβ] ⊆ p, which means     that I[α] I[β] ⊆ I[α+β] and that I[α] I[β] ⊆ I[αβ] because I[α] , I[β] , I[α+β] , and I[αβ] are divisorial ideals in R by definition.

Corollary 8.1.11 Let α, β be elements in L and let a ∈ R \ (0). Assume that R is N-2. Assume that both α and β are ultra-primitive over R. (1) Assume that Dp1 (R[α + β]) ∩ R ⊆ Dp1 (R). Then grade(I[α+β] + C(R/R)) > 1, i.e., α + β is ultra-primitive over R. (2) Assume that Dp1 (R[αβ]) ∩ R ⊆ Dp1 (R). Then grade(I[αβ] + C(R/R)) > 1, i.e., αβ is ultra-primitive over R. (3) grade(I[aα] + C(R/R)) > 1, i.e., aα is ultra-primitive over R.   I[α] I[β] ⊆ I[α+β] (resp. Proof By Lemma 8.1.10, we have    I[α] I[β] ⊆ I[αβ] ). Hence grade(I[α+β] + C(R/R)) = grade( I[α+β] +  C(R/R)) ≥ grade( I[α] I[β] + C(R/R)) > 1 (resp. grade(I[αβ] + C(R/R))   = grade( I[αβ] + C(R/R)) ≥ grade( I[α] I[β] + C(R/R)) > 1) by Lemma 8.1.9. (3) follows from that I[a] I[α] = I[α] because a ∈ R (and hence I[α] = Ia = R). Lemma 8.1.12 Let R → B → A be Noetherian domains. Assume that A is integral and flat over R. Then B is flat over R. Proof Let 0 → N → M be an exact sequence of finitely generated R-modules, Let φ : N ⊗ R B → M ⊗ R B be the canonical B-homomorphism induced from N → M. Note that (N ⊗ R B) ⊗ B A = N ⊗ R A and (M ⊗ R B) ⊗ B A = M ⊗ R A. Since A is flat over R, the canonical homomorphism N ⊗ R A → M ⊗ R A is injective. Hence φ ⊗ B A is injective. Put T := Ker(φ) and let ψ : T → N ⊗ R B. Suppose that T = 0. Since φ ⊗ B A is injective, we have ψ ⊗ B A = 0 and Ann B (ψ) = B. So it follows that Ann B (ψ)A = A. But since A is integral over B, we have Ann B (ψ)A = A, a contradiction. Hence T = 0 and hence φ : N ⊗ R B → M ⊗ R B is injective, which means that B is flat over R. Theorem 8.1.13 Let B be a Noetherian domain such that R ⊆ B ⊆ L. Assume that R is N-2 and that B is flat and integral over R. Let a be an element

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in R and let α and β be elements in B, which are ultra-primitive over R. Then so are aα, α + β, and αβ. Proof We have only to show that Dp1 (R[α +β])∩ R ⊆ Dp1 (R). Take a prime ideal p of R with grade( p) ≥ 2. Then p contains a regular sequence a1 , a2 of R. Since R[α +β] is flat over R by Lemma 8.1.12, a sequence a1 , a2 is a regular sequence of R[α + β]. Hence every prime ideal P of R[α + β] lying over p is of grade ≥ 2. Hence we conclude that Dp1 (R[α +β])∩ R ⊆ Dp1 (R). Similarly, we can show that Dp1 (R[αβ]) ∩ R ⊆ Dp1 (R). Hence grade(I[α] + C(R/R)) > 1 and grade(I[β] + C(R/R)) > 1 imply that grade(I[α+β] + C(R/R)) > 1 and grade(I[αβ] + C(R/R)) > 1 by Corollary 8.1.11. Since I[a] = Ia = R for any a ∈ R, every element in R is ultra-primitive over R by the definition of grade. Note here that grade(R) = ∞ by definition. Hence we have the following results from Theorem 8.1.13. Corollary 8.1.14 Let B be a Noetherian domain such that R ⊆ B ⊆ L. Assume that R is N-2 and that B is flat and integral over R. The set D of all elements in B which are ultra-primitive over R forms an intermediate ring between R and B. Corollary 8.1.15 Assume that R is N-2 and that α is an ultra-primitive element and integral over R. Then every element in R[α] is ultra-primitive over R. Proof The ring R[α] is a free R-module by Remark 8.1.5 because an ultraprimitive element is also an anti-integral element. Our conclusion follows Theorem 8.1.13. Remark 8.1.16 If C(R/R) = (0), then an element in L which is ultraprimitive over R is in R. In fact, if α ∈ L is ultra-primitive over R, then the condition grade(I[α] +C(R/R)) > 1 implies that grade(I[α] ) > 1. But this yields I[α] = R because I[α] is divisorial. Hence α ∈ R. Proposition 8.1.17 Let B be a Noetherian domain such that R ⊆ B ⊆ L. Assume that R is N-2 and that B is flat and integral over R. Let α1 , . . ., αn ∈ B be ultra-primitive over R. Then every element in R[α1 , . . ., αn ] is ultra-primitive over R. Proof

This follows Theorem 8.1.13.

8.2 Comparisons of Subrings of Type R[aα] ∩ R[(aα)−1 ]

8.2

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Comparisons of Subrings of Type R[aα] ∩ R[(aα)−1 ]

In this section, we use the following notations unless otherwise specified: Let R be a Noetherian domain with quotient field K , R denote the integral closure of R in K , and C(R/R) denote the conductor, i.e., {a ∈ R|a R ⊆ R}. Let L be an algebraic field extension of K and α be an element in L which is of degree d over K . Let ϕα (X ) := X d + η1 X d−1 + · · · + ηd denote the minimal polynomial of α over K . Put Rα := R[α] ∩ R[α −1 ] . Let a ∈ R be a nonzero element. Let ϕaα (X ) := X d + η1(a) X d−1 + · · · + ηd(a) , the minimal monic polynomial of aα over K . Then ϕaα (X ) = X d + aη1 X d−1 + · · · + a d ηd . Hence ηi(a) = a i ηi (1 ≤ i ≤ d). So we have Iηi ⊆ Ia i ηi = Iη(a) and i hence I[α] ⊆ I[aα] . Lemma 8.2.1 Let a be a nonzero element of R such that a is not a zerodivisor on R/I[α] . Then I[aα] = I[α] . Proof I[aα] =

By definition, we have d  i=1

Ia i ηi =

d  i=1

(Iηi : R a ) ⊆ i

d  i=1

(Iηi : R a ) = d

d 

Iηi

: R a d = I[α] : a d

i=1

Thus I[α] : R a d = I[α] and hence I[aα] ⊆ I[α] because a is not a zero-divisor in R/I[α] (and hence so is a d ). Since I[aα] ⊇ I[α] is shown above, we have I[aα] = I[α] . Proposition 8.2.2 Let a be a nonzero element of R which is not a zerodivisor on R/I[α] . If α is super-primitive over R, then aα is also super-primitive over R. Proof Take p ∈ Dp1 (R). Assume that p ∈ Ass R (R/I[α] ). Since I[α] is a divisorial ideal of R, we have depth(R p ) = 1. Since α is a super-primitive element over R, J[α] ⊆ p. Put ϕaα (X ) = X d + η1(a) X d−1 + · · · + ηd(a) , which is the minimal polynomial of aα. Then ηi(a) = a i ηi (1 ≤ i ≤ d). Suppose that a i (I[α] ηi ) ⊆ p for all i. Note here that I[α] ηi ⊆ R. Then we have (I[α] )ηi ⊆ p for all i since a ∈ p, which means that J[α] ⊆ p, a contradiction. So there exists i such that a i (I[α] ηi ) ⊆ p. Hence by using Lemma 8.2.1,   J[aα] = I[aα] 1, η1(a) , . . ., ηd(a) = I[α] 1, aη1 , . . ., a d ηd ⊆ p

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Next assume that p ∈ Ass R (R/I[α] ). Then R p = (I[α] ) p ⊆ (I[aα] ) p ⊆  p. Therefore we conclude that aα is super(J[aα] ) p , which means that J[aα] ⊆ primitive over R. Recall that an element α ∈ L is called ultra-primitive over R if grade(I[α] + C(R/R)) > 1. Proposition 8.2.3 integral over R.

Assume that α is ultra-primitive over R. Then α is anti-

Proof In order to prove that α is anti-integral over R, it suffices that α is antiintegral over R p for every p ∈ Dp1 (R) by Theorem 2.4.3. Take p ∈ Dp1 (R). Then I[α] ⊆ p or C(R/R) ⊆ p by assumption. When I[α] ⊆ p, α has a monic algebraic relation of degree d over R p . So α is anti-integral over R p by Theorem 2.3.8. When C(R/R) ⊆ p, We have R p = R p . Hence α is anti-integral over R p by Theorem 2.2.9. Corollary 8.2.4 Assume that α is ultra-primitive over R. Then for any nonzero element a ∈ R, aα is anti-integral over R. Proof It is clear from the paragraph preceding Lemma 8.2.1 that I[α] ⊆ I[aα] . Then we see that α implies that grade(I[aα] + C(R/R)) > 1. Thus aα is antiintegral over R by Proposition 8.2.3. Remark 8.2.5 Assume that d = 1. If α and aα are both anti-integral over R, we have Rα = R[aα] ∩ R[(aα)−1 ] = R. Proposition 8.2.6 Let a be a nonzero element in R. Assume that d ≥ 2 and that α and aα are anti-integral over R. Then R[aα] ∩ R[(aα)−1 ] ⊆ Rα if and only if a I[aα] ⊆ I[α] . Proof Put ζi = α i + η1 α i−1 + · · · + ηi for i (1 ≤ i ≤ d − 1). Then it follows that Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 and that R[aα] ∩ R[(aα)−1 ] = R ⊕ I[aα] aζ1 ⊕ · · · ⊕ I[aα] a d−1 ζd−1 by Theorem 3.1.19 because both α and aα are anti-integral over R. Thus R[aα] ∩ R[(aα)−1 ] ⊆ Rα if and only if I[aα] a i ζi ⊆ I[α] ζi for all i. This means that it is equivalent to I[aα] a i ⊆ I[α] for all i (1 ≤ i ≤ d − 1). Hence a I[aα] ⊆ I[α] .

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177

Corollary 8.2.7 Let a be a nonzero element in R. Assume that d ≥ 2 and that α is ultra-primitive over R. Then R[aα] ∩ R[(aα)−1 ] ⊆ Rα if and only if a I[aα] ⊆ I[α] . Proof

This follows Corollary 8.2.4 and Propositions 8.2.3 and 8.2.6.

Proposition 8.2.8 Let a be a nonzero element in R. Assume that both α and aα are anti-integral over R. Assume moreover that a is not a zero-divisor on R/Iηi for any i (1 ≤ i ≤ d). Then R[aα] ∩ R[(aα)−1 ] ⊆ Rα. Proof Since a is not a zero-divisor on R/Iηi by the assumption, we have Ia i ηi = Iηi : R a i = Iηi , and hence I[aα] ⊆ I[α] . The reverse inclusion is obvious. Hence I[aα] = I[α] . In particular, a I[aα] = a I[α] ⊆ I[α] . Corollary 8.2.9 Let a be a nonzero element in R. Assume that α is ultraprimitive over R. Further assume that a is not a zero-divisor on R/Iηi for any i (1 ≤ i ≤ d). Then R[aα] ∩ R[(aα)−1 ] ⊆ Rα. Proof

This follows from Corollary 8.2.4 and Proposition 8.2.8.

Remark 8.2.10 Let a be a nonzero element
d in R. Assume
d that both αi and aα are anti-integral over R. Then I[aα] = i=1 Ia i ηi = i=1 (Iηi : R a ) ⊆
d
d d d d d i=1 (Iηi : R a ) = ( i=1 Iηi ) : R a = I[α] : R a . Hence a I[aα] ⊆ I[α] . In particular, a d−1 I[aα] ⊆ I[α] : R a. If a is not a zero-divisor on R/(I[α] : R a), then I[aα] ⊆ I[α] : R a. By the argument in the above remark, we extend Proposition 8.2.8 to the following theorem. Proposition 8.2.11 Let a be a nonzero element in R. Assume that both α and aα are anti-integral over R. Further assume that a is not a zero-divisor on R/(I[α] : Ra). Then R[aα] ∩ R[(aα)−1 ] ⊆ Rα. Proof Since a is not a zero-divisor on R/(I[α] : R a), we have a I[aα] ⊆ I[α] by the remark above. So R[aα] ∩ R[(aα)−1 ] ⊆ Rα by Proposition 8.2.8.

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Corollary 8.2.12 Let a be a nonzero element in R. Assume that α is ultraprimitive over R. Assume moreover that a is not a zero-divisor on R/(I[α] : R a). Then R[aα] ∩ R[(aα)−1 ] ⊆ Rα. Proof

This follows from Corollary 8.2.4 and Proposition 8.2.11.

Lemma 8.2.13 Let I be a nonzero ideal of R and let a be an element of R. Then a I = I if and only if a is a unit in R. Proof Suppose that a is a nonunit of R, and satisfies a I = I . Let p ∈ Spec(R) such that a ∈ p. Then a I p = I p , and so I p = (0) by Nakayama’s (NAK) Lemma (cf. [M1]). Since R is an integral domain, we have I = (0), a contradiction. Theorem 8.2.14 Let a be a nonzero element in R. Assume that d ≥ 2 and that both α and aα are anti-integral over R. Assume moreover that α is integral over R. Then the following statements are equivalent: (1) R[aα] ∩ R[(aα)−1 ] = Rα; (2) R[aα] = R[α]; (3) a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1); (4) a is a unit in R. Proof Since α is integral over R, we have I[α] = R and I[aα] = R. By Theorem 2.3.8, R[α] and R[aα] are free R-modules, that is, R[α] = R ⊕ Rα ⊕ · · · ⊕ Rα d−1 and

R[aα] = R ⊕ Raα ⊕ · · · ⊕ Ra d−1 α d−1

(2) ⇒ (4): By the above direct sum decompositions, we have Ra i α i = Rα i for all i (1 ≤ i ≤ d − 1). Hence aR = R. (4) ⇒ (2) is obvious. Since α and aα are anti-integral over R, we have: R[aα] ∩ R[(aα)−1 ] = R ⊕ a I[aα] ζ1 ⊕ · · · ⊕ a d−1 I[aα] ζd−1 and

Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1

where ζi ’s are the same as in the proof of Proposition 8.2.6. (3)⇔ (1): a i I[aα] ζi = I[α] ζi for all i (1 ≤ i ≤ d − 1) if and only if R[aα] ∩ R[(aα)−1 ] = Rα. (4) ⇒ (1) is trivial.

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179

(1) ⇒ (4): Note that a i R ⊇ a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1). Since α is integral over R, we have I[α] = R. Hence a R = R, which yields that a is a unit in R. Corollary 8.2.15 Let a be a nonzero element in R. Assume that d ≥ 2 and that α is ultra-primitive over R. Assume further that α is integral over R. Then the following statements are equivalent: (1) R[aα] ∩ R[(aα)−1 ] = Rα; (2) R[aα] = R[α]; (3) a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1); (4) a is a unit in R. Proof

This follows from Corollary 8.2.4 and Theorem 8.2.14.

Theorem 8.2.16 Let a be a nonzero element in R. Assume that both α and aα are anti-integral over R. If d ≥ 3, then the following statements are equivalent: (1) R[aα] ∩ R[(aα)−1 ] = Rα; (2) R[aα] = R[α] and R[(aα)−1 ] = R[α −1 ]; (3) a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1); (4) a is a unit in R. Proof

(1) ⇔ (3): Since α and aα are anti-integral over R, we have: R[aα] ∩ R[(aα)−1 ] = R ⊕ a I[aα] ζ1 ⊕ · · · ⊕ a d−1 I[aα] ζd−1

and Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 where ζi ’s are the same as defined in the proof of Proposition 8.2.6. Hence a i I[aα] ζi = I[α] ζi for all i (1 ≤ i ≤ d − 1). Thus a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1). The reverse argument yields (3) ⇒ (1). (4) ⇒ (2) is obvious. (4) ⇒ (3) is trivial. (3) ⇒ (4): Since d ≥ 3, I[α] = a I[aα] = a 2 I[aα] by assumption, we have I[aα] = a I[aα] . Hence a is a unit in R by Lemma 8.2.13. (2) ⇒ (1) is trivial. Theorem 8.2.17 Let a be a nonzero element in R. Assume that both α and aα are ultra-primitive over R and α is also ultra-primitive over Rα. If d ≥ 3,

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then the following statements are equivalent: (i) Raα = Rα i.e., R[aα] ∩ R[(aα)−1 ] = R[α] ∩ R[α −1 ]; (ii) R[α] = R[aα]; (iii) a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1); (iv) a is a unit in R. Proof Since α and aα are ultra-primitive over R, they are super-primitive over R by Theorem 8.1.1 and hence they are also anti-integral over R. (i) ⇔ (iii): R[aα] ∩ R[(aα)−1 ] = R[α] ∩ R[α −1 ] if and only if a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1). (iv) ⇒ (ii) is obvious. (iv) ⇒ (iii) is trivial. (iii) ⇒ (iv): Since d ≥ 3 and I[α] = a I[aα] = a 2 I[aα] , we have I[aα] = a I[aα] . Hence a is a unit in R. (ii) ⇒ (iv): Note that aα is ultra-primitive over R by Proposition 8.1.3. So α and aα are anti-integral over Rα. Since Rα and R[α] are birational, we have Rα[aα] ∩ Rα[(aα)−1 ] = Rα. It is clear that the equality Rα[α] ∩ Rα[α −1 ] = Rα always holds without the assumption that α is anti-integral over R. Since a ∈ R, we have I[α] ⊆ I[aα] . Thus aα is ultra-primitive over Rα because α is ultra-primitive over Rα. So R[(aα)−1 ] ⊆ Rα[(aα)−1 ] ⊆ R[α −1 ][(aα)−1 ] = R[(aα)−1 ], and hence Rα[(aα)−1 ] = R[(aα)−1 ]. Since R[α] ⊆ Rα[α] ⊆ R[α], we have Rα[α] = R[α]. Note that Rα[α] = Rα[aα] because R[α] = R[aα] by the assumption. Thus we have: R[aα] ∩ R[(aα)−1 ] = R[α] ∩ R[(aα)−1 ] = Rα[α] ∩ Rα[(aα)−1 ] = Rα[aα] ∩ Rα[(aα)−1 ] = Rα. Note that the minimal monic polynomial of aα is ϕaα (X ) = X d + aη1 X d−1 + · · · + a d ηd . So we obtain from Theorem 3.1.18: R ⊕ a I[aα] ζ1 ⊕ a 2 I[aα] ζ2 ⊕ · · · ⊕ a d−1 I[aα] ζd−1 =

R[aα] ∩ R[(aα)−1 ]

=

Rα

=

R ⊕ I[α] ζ1 ⊕ I[α] ζ2 ⊕ · · · ⊕ I[α] ζd−1

Hence a i I[aα] = I[α] for each i (1 ≤ i ≤ d − 1). Noting that d ≥ 3, we have a I[aα] = I[α] = a 2 I[aα] , and hence I[aα] = a I[aα] . Since I[aα] is a nonzero, finitely generated ideal of R, we conclude that a must be a unit in R by Lemma 8.2.13. Proposition 8.2.18 Let a be a nonzero element in R. Assume that d ≥ 2 and that both α and aα are anti-integral over R. Assume moreover that there exists p ∈ Spec(R) such that a ∈ p and I[α] ⊆ p. Then R[aα]⊂ R[α]. −

8.2 Comparisons of Subrings of Type R[aα] ∩ R[(aα)−1 ]

181

Proof Consider R p . Then α is anti-integral and integral over R p by Theorem 2.3.2 because I[α] ⊆ p. Hence if R[aα] = R[α], then a R p = R p by Theorem 8.2.14, which is a contradiction because a ∈ p. Corollary 8.2.19 Let a be a nonzero element in R. Assume that d ≥ 2 and that α is ultra-primitive over R. Assume moreover that there exists p ∈ Spec(R) such that a ∈ p and I[α] ⊆ p. Then R[aα]⊂ R[α]. −

Proof

This follows from Corollary 8.2.4 and Proposition 8.2.18.

We consider the behaviour of the ring Raα for a ∈ R \ (0). Lemma 8.2.20 Let a, b be elements of R. Assume that a and b are nonzero-divisors on R/I[α] . Then I[α] = I[aα] = I[bα] hold. If, in addition, α is a super-primitive element over R, then a I[α] = bI[α] implies that a R = b R. Proof First we shall show that I[α] = I[aα] = I[bα] . Since ϕaα (X ) = X d + + a d ηd is the minimal polynomial aη1 X d−1 + a 2 η2 X d−2 + · · ·

d of aα of degree
d d d over K , we have I[α] ⊆ i=1 Ia i ηi = I[aα] and I[aα] = i=1 Ia i ηi = i=1
d
d (Iηi : R a i ) ⊆ i=1 (Iηi : R a d ) = ( i=1 Iηi ) : R a d = I[α] : R a d . Since a is a non-zero-divisor on R/I[α] , it follows that I[α] = I[aα] . Next we assume that α is a super-primitive element over R. We have only to show that b/a ∈ R. In order to show that α is super-primitive over R, it suffices that b/a ∈ R p for every p ∈ Dp1 (R). If p ⊇ I[α] , then (I[α] ) p = R p , so that a R p = a(I[α] ) p = b(I[α] ) p = b R p . Hence b/a ∈ R p . If p ⊇ I[α] , then I[α] R p = x R p for some x ∈ I[α] by Theorem 2.3.10 because α is super-primitive. Thus a I[α] R p = ax R p = bx R p = bI[α] R p , and hence a R p = b R p . Therefore b/a ∈ R p for all
p ∈ Dp1 (R), which means that b/a ∈ p∈Dp1 (R) R p = R by [Y]. Lemma 8.2.21 Let a be an element of R. Assume that a is a non-zero-divisor on R/I[α] . Assume that α is a super-primitive element of degree d over R. Then aα is a super-primitive element over R. Proof

By Lemma 8.2.20, we have J[aα] = I[aα] (1, aη1 , . . ., a d ηd ) = I[α] (1, aη1 , . . ., a d ηd )

Suppose that p ⊇ J[aα] for some p ∈ Dp1 (R). Then I[α] ⊆ I[α] (1, aη1 , . . ., a d ηd ). Since α is super-primitive over R, p is a prime divisor of I[α] . Hence a ∈ p by assumption. Since a i ηi I[α] ⊆ p (1 ≤ i ≤ d), it holds ηi I[α] ⊆ p.

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Therefore we obtain J[α] = I[α] (1, η1 , . . ., ηd ) ⊆ p. This contradicts that α is super-primitive over R. Hence we see that p ⊇ J[aα] for any p ∈ Dp1 (R). This yields that aα is super-primitive over R. Assume that α is an anti-integral element of degree d over R. Then it is known that Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 where ζi ; = α i + η1 α i−1 + · · · + ηi (1 ≤ i ≤ d − 1), by Theorem 3.1.18. Let Raα := R[aα] ∩ R[(aα)−1 ] for a ∈ R \ (0), where R \ (0) denotes R \ {0}. Theorem 8.2.22 Assume that α is a super-primitive element of degree d(≥ 2) over R and let a, b ∈ R. Assume that a and b are non-zero-divisors on R/I[α] . Then the following statements are equivalent: (1) Raα = Rbα; (2) a = ub for some u ∈ U (R). Proof (1)⇒(2): I[α] = I[aα] = I[bα] by Lemma 8.2.20. Since α is superprimitive over R and a, b are non-zero-divisors on R/I[α] , it follows that aα, bα are super-primitive over R by Lemma 8.2.21, and hence anti-integral over R. Therefore we have the direct sum decompositions of Raα, Rbα as follows: Raα

Rbα

=

R ⊕ I[aα] aζ1 ⊕ · · · ⊕ I[aα] a d−1 ζd−1

=

R ⊕ I[α] aζ1 ⊕ · · · ⊕ I[α] a d−1 ζd−1

=

R ⊕ I[bα] aζ1 ⊕ · · · ⊕ I[bα] a d−1 ζd−1

=

R ⊕ I[α] bζ1 ⊕ · · · ⊕ I[α] bd−1 ζd−1

Note that ζi = α i + η1 α i−1 + · · · + ηi (1 ≤ i ≤ d − 1). Since Raα = Rbα and d ≥ 2, we conclude I[α] aζ1 = I[α] bζ1 , and hence a I[α] = bI[α] . Therefore a/b ∈ R by Lemma 8.2.20, and thus a = ub for some u ∈ U (R). (2) ⇒ (1): Since a = ub for some u ∈ U (R), we conclude that I[α] a i ζi = I[α] bi ζi for all i (1 ≤ i ≤ d − 1). Hence Raα = Rbα by the above equalities.

Corollary 8.2.23 Let S := R \ pi ( pi ∈ AssR(R/I[α] ). Consider a mapping : S → {Raα | a ∈ S} sending a ∈ S to Raα. Then Ker( ) = U (R) and it induces the bijection : S/U (R) ∼ = {Raα | a ∈ S}. Lemma 8.2.24 Assume that α is an ultra-primitive element over R. Then, for any a ∈ R, aα is ultra-primitive over R, and therefore aα is an anti-integral element over R.

8.3 Subrings of Type R[H α] ∩ R[(H α)−1 ]

183

Proof Since α is ultra-primitive over R, then α is anti-integral over R by Theorem 8.1.1. Hence α is anti-integral over R p for any p ∈ Dp1 (R) by Theorem 2.4.3. If p ⊇ C(R/R), then R p is normal, and hence aα is antiintegral by Theorem 2.2.9. If p ⊇ I[α] , then α is integral over R p by Theorem 2.3.2, and thus aα is integral over R p . Note that aα has the monic relation of degree d. Therefore aα is anti-integral over R p . Recall that R denotes the integral closure of R in K . Theorem 8.2.25 Assume that α is an ultra-primitive element of degree d (≥ 3) over R and that R is a finite R-module. Then for elements a, b ∈ R \ (0), the following statements are equivalent: (1) Raα = Rbα; (2) a = ub for some u ∈ U (R). Proof (2) ⇒ (1): Since α is ultra-primitive over R, aα and bα are anti-integral over R by Lemma 8.2.24. The rest of this implication can be seen in the same way as in the proof of Theorem 8.2.22. (1) ⇒ (2): Since aα and bα are anti-integral over R by Lemma 8.2.24, we have the decompositions: Raα

=

R ⊕ I[aα] aζ1 ⊕ · · · ⊕ I[aα] a d−1 ζd−1

Rbα

=

R ⊕ I[bα] bζ1 ⊕ · · · ⊕ I[bα] bd−1 ζd−1

Since d ≥ 3, we have a I[aα] = bI[bα] . It follows that b2 I[bα] = a 2 I[aα] = abI[bα] , and hence bI[bα] = a I[bα] . Therefore a = ub for some u ∈ U (R) by Lemma 8.2.20. Corollary 8.2.26 If α is an ultra-primitive element over R, then the mapping : R\(0)/U (R) → {Raα | a ∈ R\(0)} defined by (a mod U (R)) = Raα for each a ∈ R \ (0), is bijective.

8.3

Subrings of Type R[H α] ∩ R[(H α)−1 ]

In this section, we treat a subring R[H α] ∩ R[(H α)−1 ] with an invertible ideal H . Let B be an intermediate ring between R and K . Put IηBi := B : B ηi and
d B B B := i=1 IηBi , J[α] := I[α] (1, η1 , . . ., ηd ). Note that if B is flat over R, then I[α] B B I[α] = I[α] B and J[α] = J[α] B. In particular, take a nonzero element t ∈ R.

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If α is anti-integral over R, then α is also anti-integral over R[1/t]. Moreover if grade(I[α] + C(R/R)) > 1, then grade(I[α] [1/t] + C(R/R)[1/t]) > 1, which will be used in this section. Let H denote an invertible ideal in R. Then for each prime ideal p of R, H p = a( p)R p for some a( p) ∈ H . Let the minimal (monic) polynomial of (a( p)) d−1 (a( p)) X + · · · + ηd . Then a( p)α be denoted by ϕa( p)α (X ) := X d + η1
d ϕa( p)α (X ) = X d + a( p)η1 X d−1 + · · · + a( p)d ηd . So I[a( p)α] = i=1 Iη(a( p)) = i
d i=1 Ia( p)i ηi . Define I[H α] :=



I[a( p)α]

p∈Spec(R)

Since H is an invertible ideal of R, H is finitely generated. So there exist ak ∈ H (1 ≤ k ≤ ) and tk ∈ R (1 ≤ k ≤ ) such that H [1/tk ]
= ak R[1/tk ] (1 ≤ k ≤ ) and that (t1 , . . ., t )R = R. So we have I[H α] := k=1 I[ak α] , and we R[1/t ] have I[H α] k = I[H α] [1/tk ] = I[ak α] [1/tk ] because R[1/tk ] is flat over R. In this case, H = (a1 , . . ., a )R. Let b1 , . . ., bm be a set of generators of H . Then si for each j, there exist i j such b j ∈ ai j R[1/ti j ]. In this case, b j = ai j ci j /ti j j R[1/ti ]

R[1/ti ]

R[1/ti ]

j for some ci j ∈ R and si j ∈ N. Since I[ai α] j ⊆ I[ai ci α] ⊆ I[b j α] j , we j j j



R[1/ti ] have k=1 I[ak α] ⊆ mj=1 I[b j α] j ⊆ mj=1 I[b j α] (R[1/ti j ]) ⊆ mj=1 I[b j α] . By

symmetry, we have k=1 I[ak α] ⊇ mj=1 I[b j α] . Hence I[H α] does not depend on a choice of generators of H . In this section, we fix the generators a1 , . . ., a of H and the elements t1 , . . ., t ∈ R with the above properties if necessary. Put ϕak α (X ) := X d + η1(ak ) X d−1 + · · · + ηd(ak ) . Then ϕak α (X ) = X d + ak η1 X d−1 + · · · + akd ηd . So we have ηi(ak ) = aki ηi (1 ≤ i ≤ d) and (1 ≤ k ≤ ). Hence I[α] ⊆ I[H α] by definition.

Lemma 8.3.1 If H ⊆ P for every P ∈ Ass R (R/I[α] ), then ak ∈ P for every P ∈ Ass R[1/tk ] (R[1/tk ]/I[α] [1/tk ]) for every k (1 ≤ k ≤ ). Proof Take P ∈ Ass R[1/tk ] (R[1/tk ]/I[α] [1/tk ]) such that H [1/tk ] ⊆ P . Note that P ∩ R ∈ Ass R (R/I[α] ). Thus H [1/tk ] ⊆ P yields H ⊆ P ∩ R. Since H [1/tk ] = ak R[1/tk ], we have ak ∈ P ∩ R. Hence ak ∈ P . Remark 8.3.2 Assume that d = 1. If α and ak α (1 ≤ k ≤ ) are anti-integral over R, then R[H α] ∩ R[(H α)−1 ] = Rα = R. Theorem 8.3.3 Assume that d ≥ 2 and that α is ultra-primitive over R. Then R[H α] ∩ R[(H α)−1 ] ⊆ Rα if and only if H I[H α] ⊆ I[α] .

8.3 Subrings of Type R[H α] ∩ R[(H α)−1 ]

185

Proof By Corollary 8.2.4, α and ak α (1 ≤ k ≤ ) are anti-integral over R. We have only to show R[ak α][1/tk ] ∩ R[(ak α)−1 ][1/tk ] ⊆ R[α][1/tk ] ∩ R[α −1 ][1/tk ] if and only if ak I[H α] [1/tk ] ⊆ I[α] [1/tk ] for all k (1 ≤ k ≤ ). R[1/t ] Since R[1/tk ] is flat over R, we have ak I[ak α] [1/tk ] = H [1/tk ]I[H α] k and R[1/t ] I[α] [1/tk ] = I[α] k . Hence we need only to prove that R[ak α][1/tk ] R[1/t ] ∩R[(ak α)−1 ][1/tk ] ⊆ R[α][1/tk ] ∩ R[α −1 ][1/tk ] if and only if ak I[H α] k ⊆ R[1/t ] I[α] k for all k (1 ≤ k ≤ ). Since α is anti-integral over R, α is anti-integral over R[1/tk ]. Hence replacing R[1/tk ] (respectively ak ) by R (respectively a), we have only to show that R[aα]∩R[(aα)−1 ] ⊆ Rα if and only if a I[aα] ⊆ I[α] . This follows Proposition 8.2.6. Corollary 8.3.4 Assume that d ≥ 2 and that R is normal i.e., integrally closed in K . Then R[H α] ∩ R[(H α)−1 ] ⊆ Rα if and only if H I[H α] ⊆ I[α] . Proof Since R is a Noetherian normal domain, α is anti-integral over R and ak α is anti-integral over R[1/tk ] for each k by Theorem 2.2.9. So the conclusion follows from Proposition 8.2.6. Corollary 8.3.5 Assume that α is super-primitive over R and that H ⊆ P for every P ∈ Ass R (R/I[α] ). Then R[H α] ∩ R[(H α)−1 ] ⊆ Rα if and only if H I[H α] ⊆ I[α] . Proof Note first that ak α is super-primitive over R by Lemma 8.3.1 and Proposition 8.2.2. Since a super-primitive element is an anti-integral element by Theorem 2.2.8, the conclusion follows from Proposition 8.2.2. Theorem 8.3.6 Assume that H ⊆ P

d Assume that α is ultra-primitive over R.−1 for every P ∈ i=1 Ass R (R/Iηi ). Then R[H α] ∩ R[(H α) ] ⊆ Rα. Proof By Corollary 8.2.4, α and ak α (1 ≤ k ≤ ) are anti-integral over R. We have only to show R[ak α][1/tk ] ∩ R[(ak α)−1 ][1/tk ] ⊆ R[α][1/tk ] ∩ R[α −1 ][1/tk ] for each k (1 ≤ k ≤ ). Note that ak is not a zero-divisor on R/I ηi for all k and i. We have only to show that R[aα] ∩ R[(aα)−1 ] ⊆ Rα by replacing R[1/tk ] (resp. ak ) by R (resp. a). This follows from Proposition 8.2.8. Corollary that H ⊆ P for each

d 8.3.7 Assume that R is normal. Assume P ∈ i=1 Ass R (R/Iηi ). Then R[H α] ∩ R[(H α)−1 ] ⊆ Rα. Proof Both α and ak α are anti-integral over R by Theorem 2.2.9. Our conclusion follows from Proposition 8.2.8.

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Corollary 8.3.8 Assume that α is super-primitive over R and that H ⊆ P for every P ∈ Ass R (R/I[α] ). Then R[H α] ∩ R[(H α)−1 ] ⊆ Rα. Proof Note that ak α is super-primitive over R by Lemma 8.3.1 and Proposition 8.2.2. Since a super-primitive element is an anti-integral element by Theorem 2.2.8, the conclusion follows from Proposition 8.2.8. In the rest of this section, we examine some conditions under which the equality R[H α] ∩ R[(H α)−1 ] = Rα holds. Theorem 8.3.9 Assume that d ≥ 2 and that α is ultra-primitive over R. Assume moreover that α is integral over R. Then the following statements are equivalent: (1) R[H α] ∩ R[(H α)−1 ] = Rα; (2) R[H α] = R[α]; (3) H i I[H α] = I[α] for all i (1 ≤ i ≤ d − 1); (4) H = R. Proof By Corollary 8.2.4, α and ak α (1 ≤ k ≤ ) are anti-integral over R. By the choice of ak and tk (1 ≤ k ≤ ), we have the following equivalences: (1) ⇔ R[ak α][1/tk ] ∩ R[(ak α)−1 ][1/tk ] = R[α][1/tk ] ∩ R[α −1 ][1/tk ] for all k; (2) ⇔ R[ak α][1/tk ] = R[α][1/tk ] for all k; (3) ⇔ aki I[H α] [1/tk ] = I[α] [1/tk ] for all k and for all i (1 ≤ k ≤ , 1 ≤ i ≤ d−1) R[1/t ] R[1/t ] ⇔ aki I[ak α] k = I[α] k for all k and for all i (1 ≤ k ≤ , 1 ≤ i ≤ d − 1); (4) ⇔ ak R[1/tk ] = R[1/tk ] for all k. By the same argument as in the proof of Proposition 8.2.6, replacing R[1/tk ] (resp. ak ) by R (resp. a ) if necessary, we need only to show the following equivalences among (1 ), (2 ), (3 ), and (4 ): (1 ) R[aα] ∩ R[(aα)−1 ] = Rα; (2 ) R[aα] = R[α] and R[(aα)−1 ] = R[α −1 ]; (3 ) a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1); (4 ) a R = R, that is, a is a unit in R. These equivalences follow from Theorem 8.2.14. Corollary 8.3.10 Assume that R is normal. Assume further that d ≥ 2 and that α is integral over R. Then the following statements are equivalent: (1) R[H α] ∩ R[(H α)−1 ] = Rα; (2) R[H α] = R[α];

8.3 Subrings of Type R[H α] ∩ R[(H α)−1 ]

187

(3) H i I[H α] = I[α] for all i (1 ≤ i ≤ d − 1); (4) H = R. Proof Since R is a Noetherian normal domain, α is anti-integral over R and ak α is anti-integral over R[1/tk ] for each k by Theorem 2.2.9. Our conclusion follows from Theorem 8.2.14. Corollary 8.3.11 Assume that α is super-primitive over R and that H ⊆ P for every P ∈ Ass R (R/I[α] ). Assume further that d ≥ 2 and that α is integral over R. Then the following statements are equivalent: (1) R[H α] ∩ R[(H α)−1 ] = Rα; (2) R[H α] = R[α]; (3) H i I[H α] = I[α] for all i (1 ≤ i ≤ d − 1); (4) H = R. Proof Note that ak α is super-primitive over R by Lemma 8.3.1 and Proposition 8.2.2. Since a super-primitive element is an anti-integral element by Theorem 2.2.8, the conclusion follows from Theorem 8.2.14. Theorem 8.3.12 Assume that α is ultra-primitive over R. Assume moreover that d ≥ 3. Then the following statements are equivalent: (1) R[H α] ∩ R[(H α)−1 ] = Rα; (2) R[H α] = R[α] and R[(H α)−1 ] = R[α −1 ]; (3) H i I[H α] = I[α] for all i (1 ≤ i ≤ d − 1); (4) H = R. Proof By Corollary 8.2.4, α and ak α (1 ≤ k ≤ ) are anti-integral over R. By the choice of ak and tk (1 ≤ k ≤ ), we have the following equivalences: (1) ⇔ R[ak α][1/tk ] ∩ R[(ak α)−1 ][1/tk ] = R[α][1/tk ] ∩ R[α −1 ][1/tk ] for all k; (2) ⇔ R[ak α][1/tk ] = R[α][1/tk ] and R[(ak α)−1 ][1/tk ] = R[α −1 ][1/tk ] for all k; (3) ⇔ aki I[H α] [1/tk ] = I[α] [1/tk ] for all k and for all i (1 ≤ k ≤ , 1 ≤ i ≤ d−1) R[1/t ] R[1/t ] ⇔ aki I[ak α] k = I[α] k for all k and for all i (1 ≤ k ≤ , 1 ≤ i ≤ d − 1); (4) ⇔ ak R[1/tk ] = R[1/tk ] for all k. By the same argument as in the proof of Theorem 8.3.9, replacing R[1/tk ] (resp. ak ) by R (resp. a ) if necessary, we need only to show the following equivalences among (1’), (2’), (3’), and (4’):

188

Ultra-Primitive Extensions and Their Generators

(1’) R[aα] ∩ R[(aα)−1 ] = Rα; (2’) R[aα] = R[α] and R[(aα)−1 ] = R[α −1 ]; (3’) a i I[aα] = I[α] for all i (1 ≤ i ≤ d − 1); (4’) a R = R, that is, a is a unit in R. These equivalences follow from Theorem 8.2.16. Corollary 8.3.13 Assume that R is normal. Assume moreover that d ≥ 3. Then the following statements are equivalent: (1) R[H α] ∩ R[(H α)−1 ] = Rα; (2) R[H α] = R[α] and R[(H α)−1 ] = R[α −1 ]; (3) H i I[H α] = I[α] for all i (1 ≤ i ≤ d − 1); (4) H = R. Proof Both α and ak α are anti-integral over R by Theorem 2.2.9. Our conclusion follows from Theorem 8.2.16. Corollary 8.3.14 Assume that α is super-primitive over R and that H ⊆ P for every P ∈ Ass R (R/I[α] ). Assume further that d ≥ 3. Then the following statements are equivalent: (1) R[H α] ∩ R[(H α)−1 ] = Rα; (2) R[H α] = R[α] and R[(H α)−1 ] = R[α −1 ]; (3) H i I[H α] = I[α] for all i (1 ≤ i ≤ d − 1); (4) H = R. Proof Note that ak α is super-primitive over R by Lemma 8.3.1 and Proposition 8.2.2. Since a super-primitive element is an anti-integral element by Theorem 8.2.8, the conclusion follows from Theorem 8.2.16. Theorem 8.3.15 Assume that d ≥ 2 and that α is ultra-primitive over R. Assume moreover that there exists p ∈ Spec(R) such that H ⊆ p and I[α] ⊆ p. Then R[H α]⊂ R[α]. −

Proof By Corollary 8.2.4, α and ak α (1 ≤ k ≤ ) are anti-integral over R. Consider R p . Then α is anti-integral and integral over R p because I[α] ⊆ p. Hence if R[H α] = R[α], then H R p = R p by Proposition 8.2.14, which is a contradiction because H ⊆ p. Corollary 8.3.16 Assume that R is normal. Assume moreover that there exists p ∈ Spec(R) such that H ⊆ p and I[α] ⊆ p. Then R[H α]⊂ R[α]. −

8.4 A Linear Generator of an Ultra-Primitive Extension R[α]

189

Proof Since R is a Noetherian normal domain, α is anti-integral over R and ak α is anti-integral over R[1/tk ] for each k by Theorem 2.2.9. Our conclusion follows from Proposition 8.2.18. Corollary 8.3.17 Assume that α is super-primitive over R and that H ⊆ P for every P ∈ Ass R (R/I[α] ). Assume further that there exists p ∈ Spec(R) such that H ⊆ p and I[α] ⊆ p. Then R[H α]⊂ R[α]. −

Proof Note that ak α is super-primitive over R by Lemma 8.3.1 and Proposition 8.2.2. Since a super-primitive element is an anti-integral element by Theorem 2.2.8, the conclusion follows from Proposition 8.2.18.

8.4

A Linear Generator of an Ultra-Primitive Extension R[α]

In this section, we investigate some conditions for an element λα + µ to be a generator of a simple ring extension R[α], where λ, µ ∈ K and α is an ultra-primitive element over R. Let Rα := R[α] ∩ R[α −1 ]. More definitely, our objective is to show the following result: Assume that α is ultra-primitive of degree d ≥ 3 over R, that α is also ultra-primitive over Rα, and that α is exclusive over R, i.e., R[α] ∩ K = R. Then for λ, µ ∈ K , R[α] = R[λα + µ] if and only if λ ∈ U (R) and µ ∈ R, where U (R) denotes the set of the units in R. An element α ∈ L is called to be an ultra-primitive element over R if α satisfies grade(I[α] + C(R/R)) > 1 We say that an extension R[α] is an ultra-primitive extension of R if α is ultra-primitive over R. Note that C(R/R) may be (0). In this case, if α ∈ K is ultra-primitive over R, then the condition grade(I[α] + C(R/R)) > 1 implies that I[α] = R because grade(I[α] ) = 1 or I[α] = R, and hence that α ∈ R. Namely, if C(R/R) = (0), an ultra-primitive extension of R is R itself. Remark 8.4.1 Assume that R is normal (i.e., integrally closed in K ). Then any algebraic element over R is ultra-primitive over R by definition. Remark 8.4.2 If α is anti-integral and integral over R, then R[α] is a free R-module R ⊕ Rα ⊕ · · · ⊕ Rα d−1 (cf. Theorem 2.3.8 and Corollary 2.3.5).

190

Ultra-Primitive Extensions and Their Generators

Proposition 8.4.3 Let a be an element in R. Assume that α is an ultraprimitive element of degree d ≥ 3 over R and that α is also ultra-primitive over Rα. Then R[α] = R[aα] if and only if a ∈ U (R). Proof

This is seen in Theorem 8.2.17.

Example 8.4.4 sarily valid.

For d ≤ 2, the conclusion of Proposition 8.4.3 is not neces-

(1) We construct an example in the case d = 1. Let k be the field of the rational numbers Q and R be a polynomial ring k[t]. Let α := 1/t 2 and a := t. Then aα = 1/t, ϕα (X ) = X − 1/t 2 , and ϕaα (X ) = X − 1/t. Hence I[α] = t 2 R and I[aα] = t R. We have R[α] = k[t, 1/t 2 ] = k[t, 1/t], and R[aα] = k[t, 1/t], which means that R[α] = R[aα] but a ∈ U (R). (2) We construct an example in the case d = 2. Let k be the field of the rational numbers Q and R be a ring k[s 2 , t], which is a subring of a polynomial ring k[s, t]. Let α := 1/(st 2 ) and a := t. Then aα = 1/(st). We have ϕα (X ) = X 2 − (1/(st 2 ))2 and ϕaα (X ) = X 2 − 1/(st)2 . In this case, I[α] = s 2 t 4 R and I[aα] = s 2 t 2 R, and hence I[α] = t 2 I[aα] . Note that R[α] = k[s 2 , t, 1/(st 2 )] = k[s 2 , t, 1/s, 1/(st 2 )] = k[s, t, 1/s, 1/(st 2 )] = k[s, t, 1/s, 1/t] and that R[aα] = k[s 2 , t, 1/(st)] = k[s, t, 1/s, 1/t], which shows that R[α] = R[aα]. But I[α] = s 2 t 4 R = s 2 t 2 R = I[aα] and hence I[α] = t 2 I[aα] . It is clear that a = t ∈ U (R). Recall that α ∈ L is said to be exclusive over R if R[α] ∩ K = R. Proposition 8.4.5 Assume that α is an ultra-primitive element of degree d ≥ 3 over R, that α is also ultra-primitive over Rα, and that α is exclusive over R. Let λ ∈ R and let µ ∈ K . Then R[α] = R[λα + µ] if and only if λ ∈ U (R) and µ ∈ R. Proof The implication (⇐) is obvious. Since λα + µ ∈ R[α], we have µ ∈ R[α] ∩ K = R. Thus R[α] = R[λα] and hence λ ∈ U (R) by Proposition 8.4.3. R Let IλR denote the ideal R : R λ for λ ∈ K and I[α] denote




d

R i=1 Iηi .

 Lemma 8.4.6 (1) Let λ ∈ K \ (0). If IλR = IλR−1 , then λ ∈ U (R). (2) Assume that α is an ultra-primitive element of degree d ≥ 3 over R, that α is

8.4 A Linear Generator of an Ultra-Primitive Extension R[α]

191

also ultra-primitive over Rα, and that α is exclusive over R. If R[α] = R[λα] for λ ∈ K \ (0), then λ ∈ R. Proof (1) Since R is Noetherian, R is a Krull domain (cf. [M1]). Take P ∈ Ht1 (R). Then R P is a DVR. Assume that λ ∈ R P . Then 1/λ ∈ R P and hence (IλR ) P = (IλR−1 ) P = R P . Next assume that λ ∈ R P . Then (IλR ) P = R P . In either   case, since ( IλR ) P = ( IλR−1 ) P = R P , it follows that (IλR ) P = (IλR−1 ) P = R P . Since IλR and IλR−1 are divisorial ideals of R, IλR = IλR−1 = R, and hence λ, λ−1 ∈ R. Thus λ ∈ U (R). (2) We assume that IλR = R. Note that α is ultra-primitive   over R by definition because C(R/R) = R. We shall show that

IλR =

IλR−1 . Take P ∈ Dp1 (R)

such that IλR ⊆ P. Then λ ∈ R P and R P [α] = R P [λα], which is induced from R[α] = R[λα]. Since α is ultra-primitive over R, we have λ ∈ U (R P ) by Proposition 8.4.5. Thus λ−1 ∈ R P . Hence IλR−1 ⊆ P. Therefore noting that   IλR−1 and IλR are divisorial ideals, we have IλR ⊆ IλR−1 . Conversely, take P ∈ Dp1 (R) such that IλR−1 ⊆ P. Then λ−1 ∈ R P . So we have R P [α] = R P [λ−1 α]. P. Thus λ−1 ∈ U (R P ) by Proposition 8.4.5. Thus λ ∈ R P , that is, IλR ⊆ 

Therefore noting that IλR−1 and IλR are divisorial ideals, we have   IλR . Therefore IλR = IλR−1 . The conclusion follows from (1).

IλR−1 ⊆

Proposition 8.4.7 Assume that α is an ultra-primitive element of degree d ≥ 3 over R, that α is also ultra-primitive over Rα, and that α is exclusive over R. Then for λ ∈ K and µ ∈ R, R[α] = R[λα +µ] if and only if λ ∈ U (R). Proof Since µ ∈ R, we may assume that µ = 0, that is, R[α] = R[λα]. The implication (⇐) is obvious. We shall show the converse implication. Case I: When C(R/R) = (0), I[α] = R by definition of grade and the assumption. Thus R[α] = R[λα] is a free R-module R ⊕ Rα ⊕ · · · ⊕ Rα d−1 . Since 1, α, . . ., α d−1 are linearly independent over K , we have λα = bα for some b ∈ R. Thus λ = b ∈ R and hence λ ∈ U (R) by Proposition 8.4.5. We are done. Case II: Now consider the case that C(R/R) = (0). Step 1: 8.4.5.

When λ ∈ R, i.e., Iλ = R, λ ∈ R and hence λ ∈ U (R) by Proposition

Step 2:

Suppose that λ ∈ R, i.e., Iλ = R. By Lemma 8.4.6(2), we have λ ∈ R.

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Let p be a prime divisor of Iλ (= (0)). Then p ∈ Dp1 (R) and 0 = C(R/R) ⊆ p because λ ∈ R. So I[α] ⊆ p since α is ultra-primitive over R. Thus R p [α] is a free R p -module R p ⊕ R p α ⊕ · · · ⊕ R p α d−1 . Note that R p [λα] is a finitely generated R p -module and is generated by 1, λα, . . ., (λα)n−1 for some n ∈ N. Thus R p [λα] = R p ⊕ R p (λα) ⊕ · · · ⊕ R p (λα)n−1 . Since R p [α] = R p [λα] and they are free R p -modules of rank d and 1, λα, . . ., (λα)d−1 are linearly independent over K . So we have d = n. Hence R p α = R p (λα); R p = R p λ, so that λ ∈ R p , that is, Iλ R p = R p , which contradicts the assumption Iλ ⊆ p. Therefore λ ∈ R. By Proposition 8.4.5, we conclude that λ ∈ U (R). This completes the proof. Proposition 8.4.8 Let β be an element in R[α] = R + Rα + · · · + Rα d−1 . Then β has a monic algebraic relation of degree d over R.  Proof Let βα i = λi+1, j+1 α j (0 ≤ i ≤ d − 1, 0 ≤ j ≤ d − 1) with λi, j ∈ R. In this case, det(βδi j − λi, j ) = 0, which is a monic algebraic relation in β. Hence β has a monic algebraic relation of degree d over R. Lemma 8.4.9 Let λ ∈ K \ (0). Assume that grade(Iλ + Iλ−1 ) = 1. Then √ C(R/R) ⊆ Iλ . Proof If C(R/R) = (0), then C(R/R) ⊆ C(R/R) = (0) in the sequel.

√

Iλ . So we assume that

Step 1: Take p ∈ Dp1 (R). If p ⊇ Iλ , then p ⊇ Iλ−1 . Indeed, If p ⊇ Iλ−1 , then grade(I grade(Iλ ) √ = 1, we√have √ λ + Iλ−1 ) ≥ 2, a contradiction. √ Since √ √ Iλ−1 ⊆ Iλ . By symmetry, we also have Iλ−1 ⊇ Iλ . Thus Iλ−1 = Iλ . √ Step 2: We shall show that if C(R/R) ⊆ Iλ , then grade(IλR + IλR−1 ) = 1. √ Suppose that grade(IλR + IλR−1 ) ≥ 2 and C(R/R) ⊆ Iλ . Let q be a prime √ √ divisor of √ Iλ−1 = Iλ such that C(R/R) ⊆ q. Then grade(q) = 1. Note that √ ( Iλ )R = ( Iλ−1 )R. By Lying-Over Theorem, q R = R. Moreover C(R/R) = C(R/R)R ⊆ q R. So there exists a prime ideal P of R lying over q such that C(R/R) ⊆ P . Take a nonzero element a ∈ (IλR ∩ R) ∩ q. Then there exists a prime divisor P of a R such that P ⊆ P . In this case, P ∩ R ⊆ P ∩ R = q. Let {P1 , . . ., P } be the set of all prime divisors of a R. Let P := P1 . Note that Pi ∈ Ht1 (R) for all i (1 ≤ i ≤ ). Since grade(IλR + IλR−1 ) ≥ 2 and Iλ ⊆ IλR , there exists β ∈ IλR−1 such that a, β form a regular sequence in R. Note here that β ∈ Pi for all i. Since β is integral over R, there is an integral relation:

 β s + a1 β s−1 + · · · + as = 0 with ai ∈ R (as = 0). Since IλR−1 ⊆ i=1 Pi , there

 R exists e ∈ Iλ−1 ∩ R such that e ∈ i=1 Pi . Consider eβ instead of β, we can

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193

assume that ai ∈ IλR−1 ∩ R for all (1 ≤ i ≤ s). Let {at1 , at2 , . . ., atk } (t1 < t2 < · · · < tk = s) be the set of the nonzero coefficients of β s +a1 β s−1 +· · ·+as = 0.

 If as ∈ i=1 Pi , then a, as form a regular sequence in R with a ∈ IλR , as ∈ IλR−1 . If as ∈ Pi , then β tk−1 + a1 β tk−1 −1 + · · · + atk−1 ∈ Pi . If atk−1 ∈ Pi , then

a, atk−1 form a regular sequence in R with a ∈ IλR , atk−1 ∈ IλR−1 . If atk−1 ∈ Pi ,

tk−2 −1 then β tk−2 a +· · ·+atk−2 ∈ Pi . Continuing this process, we find at j such 1β that at j ∈ Pi (and hence at j ∈ q because of grade(q) = 1) because β ∈ Pi . In this case, a, at j form a regular sequence in R with a ∈ IλR ∩ R, at j ∈ IλR−1 ∩ R. Since C(R/R) ⊆ q, we can choose an element c ∈ C(R/R)\q. By construction, at j ∈ IλR−1 ∩ R, that is, at j λ−1 ∈ R. Put b := at j c. Then bλ−1 = c(at j λ−1 ) ∈ R. Hence b ∈ R : R λ−1 = Iλ−1 ⊆ q. But since c, at j ∈ q, we have b = cat j ∈ q, a √ contradiction. Therefore if C(R/R) ⊆ Iλ , then grade(IλR + IλR−1 ) = 1. Step 3: We conclude that if grade(Iλ + Iλ−1 ) = 1, then C(R/R) ⊆ grade(IλR + IλR−1 ) = 1.

√

Iλ or

  Step 4: Assume that grade(IλR +IλR−1 ) = 1. In this case, we have IλR−1 = IλR by the same argument as in (Step 1). Hence λ ∈ U (R) by Lemma 8.4.6 (1). This means that λC(R/R) ⊆ R and hence that C(R/R) ⊆ Iλ . These steps complete the proof. Theorem 8.4.10 Assume that α is an ultra-primitive element of degree d ≥ 3 over R, that α is also ultra-primitive over Rα, and that α is exclusive over R, i.e., R[α] ∩ K = R. Then for λ, µ ∈ K , R[α] = R[λα + µ] if and only if λ ∈ U (R) and µ ∈ R. Proof (⇐) is obvious. (⇒): When C(R/R) = (0), I[α] = R by definition of grade and the assumption. So R[α] is a free R-module R ⊕ Rα ⊕ · · · ⊕ Rα d−1 by Remark 8.4.2. Since R[α] = R[λα + µ], we have λα + µ = b0 + b1 α + · · · + bd−1 α d−1 for some bi ∈ R. Since 1, α, . . ., α d−1 are linearly independent over K . Hence λ = b1 ∈ R and µ = b0 ∈ R. Thus λ ∈ U (R) by Proposition 8.4.7 and µ ∈ R. We are done. Now we consider the case C(R/R) = (0). We divide a proof into three steps.  √ Step 1: First, we show that Iλ = Iµ . Take a prime ideal p of R. Assume that Iλ ⊆ p. Then λ ∈ R p . So we have µ ∈ R p [α] ∩ K . Since α is exclusive, R p [α] ∩ K = R p . Hence µ ∈ R p , which means that Iµ ⊆ p. Conversely, assume that Iµ ⊆ p. Then µ ∈ R p . So  λ ∈ R p by Proposition 8.4.7, √ we have which yields that Iλ ⊆ p. Therefore Iλ = Iµ .

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 √ Step 2: Next, take a prime ideal p of R. Assume that Iλ = Iµ ⊆  p. −1 ⊆ hence λ ∈ U (R ) by Proposition 8.4.5. Thus I  p. Then λ, µ √ ∈ R p , and p λ √ Therefore Iλ−1 ⊇ Iλ . Step 3: Now we shall prove Iλ−1 = R. Suppose that Iλ−1 = R. Let q be a prime divisor of Iλ−1 . Then grade(q) = 1 because Iλ−1 is a divisorial ideal of R. Since √ Iλ ⊆ q by (Step 2), we have Jλ = Iλ + Iλ−1 ⊆ q. Hence grade(Jλ ) = 1. Thus we have (0) = C(R/R) ⊆ q by Lemma 8.4.9. Since depth(Rq ) = 1 and α is ultra-primitive over Rq , it follows that I[α] ⊆ q. Thus Rq [α] is a free Rq -module Rq ⊕ Rq α ⊕· · ·⊕ Rq α d−1 by Remark 8.4.2. Since Rq [α] = Rq [λα +µ], we can choose b0 , b1 , . . ., bd−1 ∈ Rq such that λα + µ = b0 + b1 α + · · · + bd−1 α d−1 . Since 1, α, . . ., α d−1 are linearly independent over K in K [α], we have λ = b1 ∈ Rq and µ = b0 ∈ Rq . Hence Iλ R√ q = Ib1 Rq = Rq , which means that Iλ ⊆ q, a contradiction, because Iλ ⊆ Iλ ⊆ q as was seen before. Hence Iλ−1 = R, that is, λ−1 ∈ R. Next we shall prove λ ∈ R. Put a := λ−1 ∈ R. Then λ = 1/a. Note that R[α + µ/λ] = R[α + aµ] ⊆ R[λα + µ] = R[α], where the last equality follows from the assumption. Thus α + aµ ∈ R[α] yields that aµ ∈ R[α]∩ K = R. So we have µ = b/a for some b ∈ R. In this case, we have λ = 1/a and µ = b/a. Hence R[λα + µ] = R[α/a + b/a] = R[(α + b)/a]. Put β := α + b. Then R[β] = R[(1/a)β]. Noting that b ∈ R, it follows that R[α] = R[β], that β is exclusive over R because α is exclusive over R, and that β is ultra-primitive over R because I[α] = I[β] . Considering β instead of α, we conclude that λ = 1/a ∈ R by Proposition  8.4.7, which means that λ ∈ U (R). √ In addition, µ ∈ R because R = Iλ = Iµ .

8.5

Two Generators of Simple Extensions

Let R be a Noetherian domain, R[X ] be a polynomial ring, and α be an element of an algebraic extension field of the quotient field K of R. We investigate in this section the following equality: R[ f (α), g(α)] = R[α], where f (X ), g(X ) ∈ R[X ]. More precisely we consider the following equalities: (1) R[α 2 , α 2+1 ] = R[α] for some/all  > 0; (2) R[α 2 ] = R[α]; (3) R[ f (α), g(α)] = R[α], where f (X ) (respectively g(X )) ∈ R[X ] is a monic polynomial of degree 2 (respectively of odd degree  ≥ 3). We start with the following lemma.

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Lemma 8.5.1 Take β, γ ∈ K and assume that the fractional ideal (β, γ )R contains R. Then (i) ((β, γ )R)−1 = Iβ ∩ Iγ , where ((β, γ )R)−1 := R : K (β, γ )R; (ii) if (β, γ )R is invertible, then ((β, γ )R)−1 = (a, b)R and aβ + bγ = 1 for some a, b ∈ R. Proof (i) Since I := ((β, γ )R)−1 ⊆ R, we have I = ((β, γ )R)−1 ∩ R = (R : K (β, γ )R)∩ R = (R : K β)∩(R : K γ )∩ R = (R : R β)∩(R : R γ ) = Iβ ∩Iγ . (ii) Since I (β, γ )R = R, there exist a, b ∈ I such that aβ + bγ = 1. By the same reason as above, β I ⊆ R and γ I ⊆ R. Take any x ∈ I . Then x = axβ + bxγ ∈ (a, b)R, whence I ⊆ (a, b)R. The reverse inclusion is trivial. Proposition 8.5.2 equivalent:

Assume that d ≥ 2. Then the following statements are

(a) I[α] = Iηd−1 ∩ Iηd and (ηd−1 , ηd )R is an invertible ideal such that R ⊆ (ηd−1 , ηd )R; (b) there exist a, b ∈ I[α] such that aηd−1 + bηd = 1; (c) I[α] = (a, b)R and aηd−1 + bηd = 1 for some a, b ∈ R; (d) I[α] (ηd−1 , ηd )R = R; (e) J[α] = R and 1, η1 , · · ·, ηd−2 ∈ (ηd−1 , ηd )R. Proof The implications: (a) ⇒ (d) ⇒ (c) ⇒ (b) are obvious. (b) ⇒ (a): Since I[α] (ηd−1 , ηd )R ⊆ R, it follows that 1 = aηd−1 + bηd ∈ −1 ⊇ R. I[α] (ηd−1 , ηd )R, and R = I[α] (ηd−1 , ηd )R. Thus (ηd−1 , ηd )R = I[α] −1 Therefore I[α] = ((ηd−1 , ηd )R) = Iηd−1 ∩ Iηd by Lemma 8.5.1. (a) ⇒ (e): Note first that I[α] = Iηd−1 ∩ Iηd = ((ηd−1 , ηd )R)−1 by Lemma 8.5.1. Since (ηd−1 , ηd )R is invertible, we have ((ηd−1 , ηd )R)−1 (ηd−1 , ηd )R = R, and hence ((ηd−1 = (a, b)R with aηd−1 + bηd−1 = 1 for some a, b ∈ R by Lemma 8.5.1. In this case, we have a, b ∈ ((ηd−1 , ηd )R)−1 = I[α] . So I[α] = (a, b)R. Therefore we conclude that J[α] = R and 1, . . ., ηd−2 ∈ (ηd−1 , ηd )R. (e) ⇒ (a): Note that R = J[α] = I[α] (1, η1 , . . ., ηd )R = I[α] (ηd−1 , ηd )R. −1 Hence (ηd−1 , ηd )R is invertible and (ηd−1 , ηd )R = I[α] ⊇ R. We have I[α] = −1 ((ηd−1 , ηd )R) = Iηd−1 ∩ Iηd by Lemma 8.5.1. Theorem 8.5.3 Assume that α is an anti-integral element of degree d ≥ 2 over R. Then the following statements are equivalent: (1) R[α 2 , α 2+1 ] = R[α] for any positive integer ; (2) R[α 2 , α 2+1 ] = R[α] for some positive integer ;

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(3) I[α] = Iηd−1 ∩ Iηd and (ηd−1 , ηd )R is an invertible ideal such that R ⊆ (ηd−1 , ηd )R; (4) there exist a, b ∈ I[α] such that aηd−1 + bηd = 1; (5) I[α] = (a, b)R and aηd−1 + bηd = 1 for some a, b ∈ R; (6) I[α] (ηd−1 , ηd )R = R; (7) R[α] is flat over R and 1, η1 , . . ., ηd−2 ∈ (ηd−1 , ηd )R; (8) J[α] = R and 1, η1 , . . ., ηd−2 ∈ (ηd−1 , ηd )R. Proof The equivalences: (3) ⇔ (4) ⇔ (5) ⇔ (6) ⇔ (8) follows from Proposition 8.5.2. Since α is anti-integral over R, J[α] = R if and only if R[α] is flat over R by Theorem 2.3.6. So (7) ⇔ (8) is valid. (2) ⇒ (4): Since R[α] = R[α 2 , α 2+1 ], we can write:  (#) α = a0 + a10 (α 2 ) + a01 (α 2+1 ) + ai j (α 2 )i (α 2+1 ) j i+ j≥2

for some ai j ∈ R. Let (∗) f (X ) = a0 − X + a10 (X 2 ) + a01 (X 2+1 ) +



ai j (X 2 )i (X 2+1 ) j

i+ j≥2

Then f (α) = 0 by (#). Since α is anti-integral over R, it follows that f (X ) ∈ Ker(π ) = I[α] ϕα (X )R[X ]. So we can write: (∗∗) f (X ) = ϕα (X )(b0 + b1 X + · · · + bm X m ) for some bi ∈ I[α] . Thus comparing (∗) and (∗∗), we have an equality: b1 ηd−1 + b0 ηd = −1, and hence a := −b1 , b := −b0 satisfy (4). (5) ⇒ (2): If n ≥ 2 + 1 and n is odd, then n = 2k + (2 + 1) for some nonnegative integer k, so that α n ∈ R[α 2 , α 2+1 ]. If n ≥ 2 + 1 and n is even, then obviously α n ∈ R[α 2 , α 2+1 ]. Thus α n ∈ R[α 2 , α 2+1 ] for all n ≥ 2 + 1. By the assumption, there exist b0 , b1 ∈ I[α] such that b1 ηd−1 + b0 ηd = −1. Fix n ≥ 2 + 1 for a while. Now construct b0 , b1 , b2 , . . ., bn ∈ I[α] as follows: First let b0 , b1 be the same as above. Let i ≥ 3 be odd. Suppose that α i ∈ R[α 2 , α 2+1 ] and there exist b2 , b3 , . . ., bi ∈ I[α] such that b0 ηd−i + b1 ηd−i+1 + · · · + bi ηd = 0. The case of α i+2 ∈ R[α 2 , α 2+1 ]: Put −c = b0 ηd−i + b1 ηd−i−1 + · · · + bi ηd−2 . Then −c = (b0 ηd−1 + b1 ηd )c, and hence b0 ηd−i−2 + b1 ηd−i−1 + · · · + bi ηd−2 = −(cb0 )ηd−1 − (cb1 )ηd

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Thus putting bi+1 := cb0 and bi+2 := cb1 , we have b0 ηd−i−2 + b1 ηd−i−1 + · · · + bi ηd−2 + bi+1 ηd−1 + bi+2 ηd = 0 The case of α i+2 ∈ R[α 2 , α 2+1 ]: Put bi+1 := 0 and bi+2 := 0. Let f (X ) := ϕα (X )(b0 + b1 X + · · · + bn X n ). Then by construction of bi ’s, we have 0 = f (α) = b0 ηd + (b0 ηd−1 + b1 ηd )α + γ = b0 ηd − α + γ with some γ ∈ R[α 2 , α 2+1 ]. Hence we conclude that α ∈ R[α 2 , α 2+1 ]. Corollary 8.5.4 Assume that α is an anti-integral element of degree d = 2 over R. Then R[α 2 , α 2+1 ] = R[α] for some/all  ≥ 1 if and only if R[α] is flat over R. Corollary 8.5.5 Assume that R is normal and that d = 2. Then R[α 2 , α 2+1 ] = R[α] for some/all  ≥ 1 if and only if R[α] is flat over R. Corollary 8.5.6 Assume that α is an anti-integral element of degree d ≥ 2 over R. If R[α 2 ] = R[α], then I[α] (ηd−1 , ηd )R = R and R[α] is flat over R. Proof Since R[α 2 ] ⊆ R[α 2 , α 3 ] ⊆ R[α] = R[α 2 ], we have R[α 2 , α 3 ] = R[α]. Hence I[α] (ηd−1 , ηd )R = R and R[α] is flat over R by Theorem 8.5.3.

Proposition 8.5.7 R[[α]]. Proof

Assume that d ≥ 2. If I[α] (ηd−1 , ηd )R = R, then R[[α 2 ]] =

By the assumption, there exist b0 , b1 ∈ I[α] satisfying: (∗) b0 ηd−1 + b1 ηd = −1

Put c1 := b0 ηd−3 + b1 ηd−2 ∈ R. Considering (∗) × c1 , we have (c1 b0 )ηd−1 + (c1 b1 )ηd = −c1 = −(b0 ηd−3 + b1 ηd−2 ). Hence putting b2 := c1 b0 , b3 := c1 b1 ∈ I[α] , we obtain b0 ηd−3 + b1 ηd−2 + b2 ηd−1 + b3 ηd = 0 In the sequel, consider ηi for i ≤ d and put η0 = 1, ηi = 0 for i < 0. Next put b0 ηd−5 + b1 ηd−4 + b2 ηd−3 + b3 ηd−2 =: −c2 ∈ R. Considering (∗) × c2 , we are in the similar situation. Put b4 := c2 b0 , b5 := c2 b1 ∈ I[α] . Then we have the similar equality: b0 ηd−5 + b1 ηd−4 + b2 ηd−3 + b3 ηd−2 + b4 ηd−1 + b5 ηd = 0

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Repeating this process, we obtain the sequence b0 , b1 , b2 , b3 , . . . in I[α] . Let b(X ) := b0 + b1 X + b2 X 2 + b3 X 3 + · · · ∈ R[[X ]]. Then ϕα (X )b(X ) = a0 − X + a2 X 2 + a4 X 4 + · · · + a2n X 2n + · · · ∈ R[[X ]]. So α = a0 + a2 (α 2 ) + a4 (α)2 + · · · + a2n (α 2 )n + · · · ∈ R[[α]]. Proposition 8.5.8 Assume that α is an anti-integral element of degree d ≥ 2 over R and let f (X ) denote a monic polynomial in X 2 R[X ] of odd degree ≥ 3. If R[α 2 , f (α)] = R[α], then I[α] (ηd−1 , ηd ) = R. 2 Proof It is obvious , α 2+1 ] ⊆ R[α 2 , α 3 ] for all  ≥ 1. So we have  that R[α 2 2 2+1 ] = R[α 2 , α 3 ]. Hence R[α 2 , α 3 ] = R[α] R[α , f (α)] ⊆ ≥1 R[α , α since R[α 2 , f (α)] = R[α]. Thus we obtain I[α] (ηd−1 , ηd )R = R by Theorem 8.5.3.

Proposition 8.5.9 R[α] and R[α 2 , f (α)] are birational to each other for any monic polynomial f (X ) of odd degree  ≥ 1. Proof If  = 1, then trivially K (α) = K (α 2 , f (α)). Next suppose that  ≥ 3. We may assume that f (α) = u 1 α + u 3 α 3 + · · · + u  α  in R[α 2 , f (α)]. Hence we have f (α) = α(u 1 + u 3 α 2 + · · · + u  α −1 ) ∈ K (α 2 , f (α)). Since u 1 + u 3 α 2 + · · · + u  α −1 ∈ R[α 2 ] \ (0), it follows that α ∈ K (α 2 , f (α)), and hence K (α) = K (α 2 , f (α)). Remark 8.5.10 Let f (X ) ∈ R[X ] be a monic polynomial of odd degree  ≥ 3 and g(X ) = X 2 +t X +s ∈ R[X ]. Let B denote the subring R[g(α), f (α)]. Take a polynomial h(X ) of degree n. If n is even, then n = 2k for some nonnegative integer k. Thus the polynomial h(X ) − ag(X )k is of degree ≤ n − 1 for some a ∈ R. If n ≥  and n is odd, then n =  + 2k for some nonnegative integer k. So h(X ) − a f (X )g(X )k is of degree ≤ n − 1 for some a ∈ R. Hence any element β ∈ R[α] is expressed as u 1 α + u 3 α 3 + · · · + u −2 α −2 + γ , where u i ∈ R and γ ∈ B. Thus the R-module R[α]/R[g(α), f (α)] is finitely generated. Remark 8.5.11 Assume that (R, m) is a local domain and that d ≥ 2, and let f (X ) denote a monic polynomial X  + u 1 X −1 + · · · + u −1 X + u  in R[X ] of odd degree  ≥ 3 such that u 2i ∈ m (1 ≤ i ≤ ( − 1)/2). Then for any integer n ≥ , α n ∈ R[α 2 , α  ] + m R[α] = R[α 2 , f (α)] + m R[α] and R[α 2 , α  ] + m R[α] = R[g(α), f (α)] + m R[α]. Theorem 8.5.12 Assume that (R, m) is a local domain and that α is an anti-integral element of degree d ≥ 2, and let f (X ) denote a monic polynomial

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X  + u 1 X −1 + · · · + u −1 X + u  in R[X ] of odd degree  ≥ 3 such that u 2i ∈ m (1 ≤ i ≤ ( − 1)/2). Then R[α 2 , f (α)] = R[α] if and only if there exist b0 , b1 ∈ I[α] such that b0 ηd + b1 ηd−1 = 1. Proof Put  = 2k + 1, (k ≥ 1). Then we have that α  = f (α) + c, where c ∈ R[α 2 , f (α)] + m R[α] because α 2i ∈ R[α 2 , f (α)] and u 2i ∈ m (1 ≤ i ≤ ( − 1)/2) (cf. Remark 8.5.1). Hence R[α 2 , f (α)] + m R[α] = R[α 2 , α  ] + m R[α]. (⇒): Assume that R[α] = R[α 2 , f (α)]. Then R[α] = R[α 2 , f (α)]+m R[α] = R[α 2 , α  ] + m R[α]. Since the R-module R[α]/R[α 2 , α  ] is finitely generated by Remark 8.5.11, we have R[α] = R[α 2 , α  ] by NAK (cf. [M1]). So we conclude that b1 ηd−1 + b0 ηd = 1 for some b0 , b1 ∈ I[α] by Theorem 8.5.2. (⇐): Assume that b1 ηd−1 + b0 ηd = 1 for some b0 , b1 ∈ I[α] . Then we have R[α] = R[α 2 , α  ] by Theorem 8.5.2. So it follows that R[α] = R[α 2 , α  ] + m R[α] = R[α 2 , f (α)] + m R[α]. Since the R-module R[α]/R[α 2 , f (α)] is finitely generated over R by Remark 8.5.11, we have R[α] = R[α 2 , f (α)] by NAK Lemma (cf. [M1]). Corollary 8.5.13 Assume that (R, m) is a local domain and that α is an anti-integral element of degree d ≥ 2. Let g(X ) = X 2 + t X + s ∈ R[X ] with t ∈ m and let f (X ) be a polynomial X  + u 1 X −1 + · · · + u −1 X + u  ∈ R[X ] of odd degree  ≥ 3 with u 2 , u 4 , . . ., u −1 ∈ m. Then R[g(α), f (α)] = R[α] if and only if there exist b0 , b1 ∈ I[α] such that b1 ηd−1 + b0 d = 1. Proof From Theorem 8.5.12, we have only to note that R[g(α), f (α)] + m R[α] = R[α 2 , α  ] + m R[α] and that the R-module R[α]/R[g(α), f (α)] is finitely generated, which follow from Remarks 8.5.10 and 8.5.11. Remark 8.5.14 Assume that R contains a field k of char k = 2. Let g(X ) = X 2 + t X + s with s, t ∈ R and f (X ) be a monic polynomial in R[X ] of degree  > 2, where  is odd. Then g(X ) = (X + 12 t)2 + s − ( 12 t)2 . Put α ∗ := α + 12 t ∈ R[α]. Then R[α ∗ ] = R[α] and R[g(α), f (α)] = R[α ∗2 , α ∗ + u −2 α ∗−2 + · · · + u 1 α ∗ ] for some u j ∈ R. Note here that if α is anti-integral over R then so is α ∗ by Proposition 3.1.13.

Chapter 9 Flatness and Contractions of Ideals

Let R be a Noetherian domain with quotient field K . Let L be an algebraic field extension of K and α ∈ L. Let A be an integral domain containing R. Assume that A and L are in some large fixed field. Let K (A) denote the quotient field of A. Let π (A) : A[X ] −→ A[α] be the canonical A-algebra homomorphism sending X to α. Let ϕα(A) (X ) denote the minimal polynomial of α over K (A) and deg ϕα(A) (X ) = d A . Write ϕα(A) (X ) = X d A + η1 X d A −1 + · · · + ηd A d A A A where ηi ∈ K (A). Let IηAi := A : A ηi and I[α] := i=1 Iηi . It is easy to see that A (A) A I[α] = A[X ] : A ϕα (X ). If A = R, we put d := d A , Iηi := IηAi , and I[α] := I[α] . A It is easy to see that I[α] (respectively I[α] ) is an ideal of R (respectively A). An element α ∈ L is called an anti-integral element over R of degree d if the equality Ker(π) = I[α] ϕα (X )R[X ] holds. For a polynomial f (X ) ∈ R[X ], c( f (X )) denotes the ideal of R generated by the coefficients of f (X ). Let J be an ideal of R[X ]; we denote by c(J ) the ideal of R generated by the coefficients of the polynomials in J . If α ∈ L is anti-integral, it follows that c(Ker(π )) = c(I[α] ϕα (X )R[X ]) = I[α] (1, η1 , . . ., ηd ). We put J[α] := I[α] (1, η1 , . . ., ηd ), which is an ideal of R. It is not hard to see that J[α] = c(I[α] ϕα (X )). Similarly A A let J[α] := I[α] (1, η1 , . . ., ηd A ). When α is an element in K , ϕα (X ) = X − α. So we have J[α] := I[α] (1, η1 , . . ., ηd ) = I[α] + α I[α] . Throughout this chapter, we use the above notations unless otherwise specified.

9.1

Flatness of a Birational Extension

In Theorem 9.1.4, it will be shown that if β1 , . . ., βn ∈ K are all strongly flat elements, then R[β1 , . . ., βn ] is flat over R. Example 9.1.5 shows that this assertion does not always hold if βi ∈ K for some i, that is, the assertion above is not necessarily valid in the non-birational case. In this section, we treat this problem in the birational case.

201

202

Flatness and Contractions of Ideals

Theorem 9.1.1 Let {αi |1 ≤ i ≤ n} be a finite subset of K . Then A := R[α1 , . . ., αn ] is flat over R if and only if Iα1 · · · Iαn A = A, or equivalently Iαi A = A for all i. Proof (⇐): Lemma 1.2.11 shows that A is flat over R if and only if Iα A = A for every α ∈ A. This is applied to α = αi for each i. (⇒): Consider each p ∈ Spec(R). If Iα1 · · · Iαn ⊆ p, then Iαi is not contained in p for all i, and hence αi ∈ R p for all i. Therefore it follows that A p = R p in this case. Thus by Lemma 1.2.11, we see that A is flat over R. The following is an immediate consequence of Theorem 9.1.1. Corollary 9.1.2 Let {αi |1 ≤ i ≤ n} be a finite subset of K , and assume that A := R[α1 , . . ., αn ] is flat over R. Then we have Spec(A) ∼ = Spec(R) \

n !

V (Iαi )

i=1

Corollary 9.1.3 Iα B = B.

For α ∈ K , B := R[α, α −1 ] is flat over R if and only if

Proof If B is flat over R, then Iα Iα−1 B = B by Theorem 9.1.1. Note that Iα−1 = α Iα . Hence we have α(Iα )2 B = B. Since α is a unit in B, we have Iα−1 B = α Iα B = B. Thus B is flat over R by Theorem 9.1.1. Theorem 9.1.4 Let {αi |1 ≤ i ≤ n} be a finite subset of K . If R[αi ] (1 ≤ i ≤ n) are flat over R, then R[α1 , . . ., αn ] is flat over R. Proof Since αi (1 ≤ i ≤ n) are strongly flat elements over R, by definition, we have R[αi ] are flat over R, and hence Iαi R[αi ] = R[αi ] by Theorem 9.1.1. Thus Iαi R[α1 , . . ., αn ] = R[α1 , . . ., αn ] for all i (1 ≤ i ≤ n). Therefore Iα1 · · · Iαn R[α1 , . . ., αn ] = R[α1 , . . ., αn ] This shows that R[α1 , . . ., αn ] is flat over R by Theorem 9.1.1. The following example shows that Theorem 9.1.4 does not hold in general for non-birational extensions. Example 9.1.5 Let R ⊆ A be a non-birational extension of Noetherian domains, and K and L be the quotient fields of R and A, respectively. Assume

9.2 Flatness of a Non-Birational Extension

203

that there are elements α ∈ L \ K and η ∈ K such that L = K [α], α 2 = a ∈ R, both η and η−1 ∈ R p for some p ∈ Spec(R), and that aη2 ∈ R. Hence we see that R[α] and R[β] are free R-modules. Furthermore we can show that A = R[α, β] is not a flat R-module. Indeed, let us assume that R[α, β] is a flat R-module. By localizing at p, we may assume that R is a local ring with maximal ideal m. Note that A = R + Rα + Rβ + Rαβ. Since A is a free R-module of rank 2, A/m A = R/m + (R/m)α + (R/m)β + (R/m)αβ, α, β ∈ A/m A is a two-dimensional vector space over R/m. Since 1 ∈ A/m A is a linearly independent over R/m, we have the following possibilities: (1) A/m A = R/m + (R/m)α, (2) A/m A = R/m + (R/m)β, (3) A/m A = R/m + (R/m)αβ. Therefore, by Nakayama’s lemma, we have A = R + Rα, A = R + Rβ, or A = R + Rαβ. If α ∈ R + Rαβ, then we have α = x + yαβ for some elements x, y ∈ R. Since η and η−1 ∈ R by assumption, we notice that x, y are nonzero elements of R. Now, it holds that R # y 2 α 2 β 2 = (α − x)2 = α 2 − 2xα + x 2 , and α = (a + x 2 − y 2 α 2 β 2 )/2x ∈ K . This is a contradiction. Therefore α ∈ R + Rαβ. In the same way, we see that β ∈ R + Rα. Thus it follows that α ∈ R + Rαβ or β ∈ R + Rα. Suppose that β ∈ R + Rα. Then β = x + yα for some elements x, y ∈ R. Note that x, y are nonzero elements of R. Since β 2 = x 2 +2x yα+α 2 +a, we have 2x yα = β 2 −ax 2 ∈ R. Since x y = 0, it holds that α ∈ K , a contradiction. Therefore A is not a flat R-module. Finally, we construct the example satisfying these conditions. Let k be a field and X, Y, Z be the indeterminates over k. Let R = k[X, Y ]/(Y 2 − X 2 Y + X 2 ) = k[x, y], where x, y denote the residue classes of X, Y in R. Let A = R[Z ]/(Z 2 − (y − 1)) = R[z], where z denotes the residue class of Z in A. Now, put α = z, η = y/x, and p = (x, y) ∈ Spec(R). Then we see that α ∈ K . Now assume that η ∈ R p . Then we have y ∈ R p , and hence p R p = x R p . Thus R p is a discrete valuation ring (DVR). This is a contradiction. Therefore we have η ∈ R p . Similarly, we see that η−1 ∈ R p . Further, we have η2 = y 2 /x 2 = y − 1 = a ∈ R and aη2 = (y − 1)2 ∈ R. Therefore it follows that R[z, yz/x] is not a flat R-module but R[z] and R[yz/x] are flat R-modules by the preceding argument.

9.2

Flatness of a Non-Birational Extension

In this section, we treat the case that some of β1 , . . ., βn are not contained in K and show that the similar assertion to Theorem 9.1.4 holds even in the non-birational case if β1 , . . ., βn satisfy certain conditions.

204

Flatness and Contractions of Ideals

It is known that if α is anti-integral and integral over R of degree d, then R[α] is a free R-module of rank d (cf. Corollary 2.3.5, Theorem 2.3.8). We start with the following definition. Definition 9.2.1 Assume that α ∈ L is anti-integral and integral over R of degree d. For an element β ∈ K (α) = K [α], putting: β = λ0 + λ1 α + · · · + λd−1 α d−1 with λi ∈ K , let Tβ :=

d−1

Iλi = R[α] : R β

i=0

Lemma 9.2.2 Let A be a Noetherian domain containing R. Assume that α ∈ L is anti-integral over A of degree d A . Put F(A) := { f (X ) ∈ Ker π (A) | deg f (X ) = } for a nonnegative integer , where π (A) : A[X ] −→ A[α] is the canonical A-algebra homomorphism sending X to α. If  ≥ d A := [K (A)(α) : K (A)], (A) then c(F(A) ) = J[α] . Proof

Since α is an anti-integral element of degree d A over A, we have A (A) Ker(π (A) ) = I[α] ϕα (X )A[X ]

and hence A A (A) J[α] = c(Ker π (A) ) = c(I[α] ϕα (X )A[X ]) A (A) Take f (X ) ∈ I[α] ϕα (X )A[X ] with deg f (X ) = d A . For an integer  ≥ d A , −d A we see that X f (X ) ∈ F(A) . Thus c( f (X )) = c(X −d A f (X )) ⊆ c(F(A) ). So A we obtain that J[α] ⊆ c(F(A) ). Conversely, take f (X ) ∈ F(A) . Then f (X ) ∈ A (A) Ker π (A) = I[α] ϕα (X )A[X ]. Hence we can write:

f (X ) =



f i (X )gi (X )

with deg f i (X ) = d A , f i (α) = 0. Hence, we have c( f (X )) ⊆ A A A J[α] . Thus c(F(A) ) ⊆ J[α] . So we conclude that J[α] = c(F(A) ).



c( f i (X ))A ⊆

Lemma 9.2.3 Let β ∈ L be an anti-integral element over R and let A be an Noetherian domain containing R. If β is an anti-integral element over A of A . degree e (≤ d A ), then J[β] A ⊆ J[β]

9.2 Flatness of a Non-Birational Extension

205

Proof Take f (X ) ∈ I[β] ϕβ (X )R[X ] with deg f (X ) = d A . Consider the natural exact sequence: π (A)

0 −→ Ker(π (A) ) −→ A[X ] −→ A[β] −→ 0 A Then f (X ) ∈ Ker π (A) . So by Lemma 9.2.2, we have c( f (X )) ⊆ c(Fd(A) ) = J[β] , A A which yields that J[β] A ⊆ J[β] .

Making these preparations, we have the following proposition. Proposition 9.2.4 Assume that α ∈ L is an anti-integral element over R of degree d. Let β ∈ K [α]. Assume that β is anti-integral over both R and R[α]. If R[β] is flat over R, then R[α, β] is flat over R[α]. A Proof Put A = R[α]. By Theorem 2.3.6, we have J[β] = R. Hence J[β] =A by Lemma 9.2.3. Using Theorem 2.3.6 again, we have that A[β] is flat over A. Hence R[α, β] is flat over R[α].

Remark 9.2.5 In Theorem 2.4.5, it is shown that α ∈ L is anti-integral over R if and only if so is α −1 . Proposition 9.2.6 Let L be an algebraic extension of a finite degree over K . Assume that α, β1 , . . ., βn ∈ L, [K (α) : K ] = d, L = K [α], and that R[α] is flat over R. Consider A = R[α, β1 , . . ., βn ]. Then we can write βi = ηi,0 + ηi,1 α + · · · + ηi,d−1 α d−1 with ηi, j ∈ K (1 ≤ i ≤ n, 1 ≤ j ≤ d − 1). If Tβ1 · · · Tβn A = A, then A is flat over R. Proof Since R[α] is flat over R, we have only to show that A is flat over R[α]. Let P ∈ Spec(R[α]) and P ∩ R = p. If p ⊇ Tβ1 · · · Tβn , then p A = A. This contradicts the fact that P ⊇ p A. Hence p does not contain Tβ1 · · · Tβn , so that p does not contain Tβi for all i. Thus by the definition of Tβi , it holds that p does not contain Iηi, j for all i, j. Hence we see that βi ∈ R p [α]. Therefore it follows that A p = R p [α]. This implies that A p is flat over R p , and hence A is flat over R. Corollary 9.2.7 Under the circumstances, if Tβi R[α, βi ] = R[α, βi ] for all i, we have R[α, β1 , . . ., βn ] is flat over R. Proof Since Tβ1 · · · Tβn R[α, β1 , . . ., βn ] = R[α, β1 , . . ., βn ], the assertion is easily seen by Proposition 9.2.6. Now we show when β ∈ L is flat over R[α] in terms of the ideal Tβ . For this purpose, we study some properties of the ideal Tβ .

206

Flatness and Contractions of Ideals

Proposition 9.2.8 Assume that α ∈ L is anti-integral and integral over R of degree d. Then the following statements hold: (i) For β1 , . . ., βn ∈ K [α], put A := R[α, β1 , . . ., βn ]. If (Tβ1 ∩. . .∩Tβn )A = A, then A is flat over R. R[α] (ii) For β ∈ K [α], the equality I[β] ∩ R = Tβ holds.

R[α] R[α] Proof (i) By the definitions of Tβ and I[β] , we have Tβ ⊆ I[β] . So the R[α] assumption yields Tβi A = A (1 ≤ i ≤ n). Thus we have I[βi ] A = A (1 ≤ i ≤ n). From this, it follows that (I[βR[α] ∩ · · · ∩ I[βR[α] )A = A, and hence A is flat 1] n] over R[α]. Since R[α] is flat over R, A is flat over R. R[α] (ii) Note that β ∈ K (α). Since I[β] = R[α] : R[α] β and Tβ = R[α] : R β, we R[α] have I[β] ∩ R = Tβ .

Remark 9.2.9 (i) For β ∈ K [α], the rings R[α] and R[α][β] are birational. So it holds that R[α] I[βR[α] −1 ] = β I[β] . (ii) By use of (i) and Proposition 9.2.8 (ii), we have the following: R[α] R[α] R[α] R[α] J[β] = I[β] + β I[β] = I[β] + I[βR[α] −1 ] ⊇ (Tβ + Tβ −1 )R[α]

and R[α] J[β] ∩ R ⊇ Tβ + Tβ −1

The next theorem is our main result in this section. Theorem 9.2.10 Assume that α is anti-integral and integral over R of degree d. Let β1 , . . ., βn be elements in K [α]. If J[βi ] = R and each βi is anti-integral over R[α] (1 ≤ i ≤ n), then R[α, β1 , . . ., βn ] is flat over R. Proof Since β1 , . . ., βn ∈ K [α], A := R[α, β1 , . . ., βn ] is birational over R[α]. By Theorem 2.3.6, each R[βi ] (1 ≤ i ≤ n) is flat over R. Thus A is flat over R[α]. Since R[α] is flat over R by the assumption, A is flat over R by Theorem 9.1.4. Proposition 9.2.11 Assume that α is anti-integral and integral over R. Let β1 , . . ., βn ∈ K [α]. If (Tβi +Tβi−1 )R[α] = R[α] for each i, then R[α, β1 , . . ., βn ] is flat over R.

9.2 Flatness of a Non-Birational Extension Proof

207

By using Remark 9.2.9, we have J[βR[α] = (Tβi + Tβi−1 )R[α] = R[α] i]

for each i because R[α] is faithfully flat over R. So J[βR[α] = R[α]. Thus each i] βi is a super flat (s-flat) element over R[α]. Since βi ∈ K (α), R[α, β1 , . . ., βn ] is flat over R[α] by Theorem 9.1.4. Hence R[α, β1 , . . ., βn ] is flat over R. In the preceding results, the condition that β is anti-integral over R[α] works effectively. We attempt to characterize this condition in terms of the ideal Tβ . Let Dp1 (R) := { p ∈ Spec(R)|depth(R p ) = 1}. Let α be an element in an algebraic field extension L of K . We say that α is a super-primitive element of degree d if J[α] ⊆ p for all ∈ Dp1 (R). Let A be a Noetherian domain containing R. An element β ∈ K [α] is a super-primitive element over A if and only if A grade(J[β] ) > 1. It is known in Theorem 2.2.8 that the super-primitive elements are anti-integral. In the next proposition, we give a condition for α to be a super-primitive element over R[α]. Proposition 9.2.12 Assume that α is anti-integral and integral over R of degree d. Let β be an element in K [α]. (i) β is super-primitive over R[α] if grade(Tβ + Tβ −1 ) > 1. (In particular, if grade(Tβ + Tβ −1 ) > 1, then β is anti-integral over R[α].) (ii) β is a flat element over R[α] if Tβ + Tβ −1 = R. Proof (i) Since R[α] is faithfully flat and integral over R, grade(Tβ + Tβ −1 ) > R[α] R[α] 1 ⇒ grade(J[β] ∩ R) > 1 ⇒ grade(J[β] ) > 1. Our conclusion follows. R[α] (ii) Since β is an s-flat element over R[α], we have J[β] = R[α] and hence R[α] J[β] ∩ R = R. So we have Tβ + Tβ −1 = R by the same way as in (i), and hence R[α] J[β] = R[α] by Remark 9.2.9. So we conclude that β is a flat element over R[α] by Theorem 2.3.6. We close this section by investigating some relationships among the ideals R[α] , and Tβ . I[β] , I[β] Proposition 9.2.13 Assume that α is anti-integral and integral over R of degree d. Let β ∈K [α] which is anti-integral over both R and R[α]. Then   R[α] I[β] = Tβ = I[β] ∩ R. R[α] Proof First we shall show the inclusion I[β] ⊆ I[β] ∩ R. Let ϕβ (X ) = d d−1 + · · · + ζd with ζi ∈ K . Then for a nonzero element a ∈ I[β] , there X + ζ1 X

208

Flatness and Contractions of Ideals

exists an algebraic dependence: aβ d + a1 β d−1 + · · · + ad = 0, ai = aζi ∈ R. From this, we have ad (β −1 )d + · · · + a1 β −1 + a = 0. Put f (X ) := ad X d + · · · + a1 X + a. Then f (β −1 ) = 0. By the assumption, β is anti-integral over both R and R[α]. Hence β −1 is also  anti-integral over both R and R[α] by Remark 9.2.5. So we have f (X ) = f i (X )gi (X ), where deg f i (X ) = 1, f i (β −1 ) = 0 with f i (X ), gi (X ) ∈ R[α][X ]. Consider the constant term, we see R[α] R[α] R[α] −1 that a ∈ β −1 I[βR[α] × β I[β] = I[β] . Thus I[β] ⊆ I[β] ∩ R. But since −1 ] = β   R[α] I[β] ∩ R = Tβ by Proposition 9.2.8, we have I[β] ⊆ Tβ . Thus I[β] ⊆ Tβ . Conversely, take a ∈ Tβ . Then aβ ∈ R[α]. Since R[α] is a free R-module of rank d, aβ has a monic relation of degree d: (aβ)d +b1 (aβ)d−1 +· · ·+bd = 0, bi ∈ R. d From this, we have: β d + (b1 /a)β d−1 + · · · + bd /a d = 0. Thus  a ∈  I[β] . N Therefore for a sufficiently large integer N , Tβ ⊆ I[β] and hence Tβ ⊆ I[β] . The second equality follows from Proposition 9.2.8(ii). Next we discuss certain relationships between flat extensions and unramified extensions. Let β be an anti-integral element in K . Then the universal module of differentials  R (R[β]) of R[β] over R is given by  R (R[β]) = R[β]/I[β] ϕβ (β)R[β] Since ϕβ (X ) = X − β (β ∈ K ), we have I[β] ϕβ (β) = I[β] . Thus R[β] is unramified over R

⇐⇒

 R (R[β]) = (0)

⇐⇒

I[β] R[β] = R[β]

⇐⇒

R[β] is flat over R

(see Theorems 5.1.1, 5.1.4 for details). This means that, in the birational case, the flatness is equivalent to the unramifiedness. But in the non-birational case, there exists a counterexample to this assertion as follows: Example 9.2.14 Let β ∈ L satisfy β d = a ∈ R and a ∈ U (R) with [K (β) : K ] = d for d > 1. Then ϕβ (X ) = X d −a and ϕβ (β) = dβ d−1 ∈ U (R[β]). Thus R[β] is integral and hence flat over R by Corollary 2.3.5, but not unramified over R. Theorem 9.2.15 Assume that α is anti-integral and integral over R of degree d. Let β ∈ K [α] and assume that β is anti-integral over both R and R[α]. (i) R[β] is flat over R if and only if R[α, β] is unramified over R[α]. (ii) If R[β] is unramified over R, then R[β] is flat over R.

9.3 Anti-Integral Elements and Coefficients of its Minimal Polynomial 209 Proof (i) By Theorem 9.2.10, R[β] is flat over R if and only if R[α, β] is flat over R[α]. Since β ∈ K [α], that is, R[α, β] and R[α] are birational, we have the equivalence: R[α, β] is flat over R[α] if and only if R[α, β] is unramified over R[α]. (ii) Assume that R[β] is unramified over R. Then { f  (X )| f (X ) ∈ I[β] ϕβ (X ) R[X ]} generates the unit ideal in R[β]. Since f (β) = 0 and β is anti-integral R[α] (R[α]) over R[α], we have f (X ) ∈ I[β] ϕβ (X )R[α][X ], where ϕβ(R[α]) (X ) denotes the monic minimal polynomial of β over K (α). Hence f (X ) =



R[α] (R[α]) Fi (X )G i (X ), Fi (X ) ∈ I[β] ϕβ (X )R[α][X ]

R[α] where G i (X ) ∈ R[α][X ], deg Fi (X ) = 1. Noting that Fi (β) ∈ I[β] , we have  R[α] R[α]   f (β) = Fi (β)G i (β) ∈ I[β] R[α, β]. Thus since I[β] R[α, β] = R[α, β], R[α, β] is flat over R[α]. So by Proposition 9.2.4, we conclude that R[β] is flat over R.

Theorem 9.2.15 yields the following corollary. Corollary 9.2.16 Under the same assumptions as in Theorem 9.2.15, if R[β] is unramified over R then R[α, β] is unramified over R[α].

9.3

Anti-Integral Elements and Coefficients of its Minimal Polynomial

Let R be a Noetherian domain with quotient field K and L be an algebraic field extension of K . Let α be an element in L which is of degree d over K and ϕα (X ) := X d + η1 X d−1 + · · · + ηd denote the minimalpolynomial of α d over K (that is, ηi ∈ K ). Put Iηi := R : R ηi and I[α] := i=1 Iηi , and put J[α] = I[α] (1, η1 , . . ., ηd ) and J˜[α] := I[α] (1, η1 , . . ., ηd−1 ) as before. The objective of this section is to investigate some relations between the extensions, R[α] over R and R[η1 , . . ., ηd ]/R. Lemma 9.3.1 (cf. Theorem 2.3.6 and Proposition 4.4.1) Assume that α is anti-integral over R. Then (1) R[α] is flat over R if and only if J[α] = R; (2) R[α] is faithfully flat over R if and only if J˜[α] = R. Lemma 9.3.2 Assume that α is anti-integral over R. If I[α] R[α] = R[α], then R[α] is flat over R.

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Proof Since J[α] ⊇ I[α] , the equality I[α] R[α] = R[α] induces J˜[α] R[α] = R[α]. Hence R[α] is flat over R by Lemma 9.3.1. Proposition 9.3.3 Assume that α is an anti-integral element over R. If I[α] R[α] = R[α], then R → R[η1 , . . ., ηd ] is an open immersion. Proof Since I[α] R[α] = R[α], R[α] is flat over R by Lemma 9.3.3. So we have J[α] = I[α] (1, η1 , . . ., ηd )R = R by Lemma 9.3.1. Take p ∈ Spec(R). If I[α] ⊆ p, then ηi ∈ R p for all i. Thus R[η1 , . . ., ηd ] p = R p . Assume that I[α] ⊆ p. Then I[α] (1, η1 , . . ., ηd )R p = R p and hence I[α] ηi R p = R p for some i. Thus we have I[α] R p = Iηi R p . It follows that Iηi R[η1 , . . ., ηd ] p = R[η1 , . . ., ηd ] p and p R[η1 , . . ., ηd ] p = R[η1 , . . ., ηd ] p . Hence R[η1 , . . ., ηd ] p is flat over R p by [Y, (1.1)]. Since R[η1 , . . ., ηd ] and R are birational, R → R[η1 , . . ., ηd ] is an open immersion. Theorem 9.3.4 Assume that α is anti-integral over R. If R[α] is flat over R, then R[η1 , . . ., ηd ] and R[η1 , . . ., ηd ][α] are flat over R, and R[η1 , . . ., ηd ][α] is flat over R[η1 , . . ., ηd ]. Proof Since R[α] is flat over R, we have J[α] = I[α] (1, η1 , . . ., ηd ) = R. Since η1 , . . ., ηd ∈ R[η1 , . . ., ηd ], we have I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]. Take p ∈ Spec(R). If p ⊇ I[α] , we have p R[η1 , . . ., ηd ] = R p because η1 , . . ., ηd ∈ R p . If p ⊇ I[α] , we have p R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]. Hence R[η1 , . . ., ηd ] is flat over R by [Y, (1.1)]. [Since R[η1 , . . ., ηd ][α] is a free R[η1 , . . ., ηd ]module of rank d.] Thus R[η1 , . . ., ηd ] ⊆ R[η1 , . . ., ηd ][α] is a flat extension. Therefore R ⊆ R[η1 , . . ., ηd ][α] is also a flat extension. Example 9.3.5 Let R be a polynomial ring k[a, b] over a field k. An element α is a root of an irreducible polynomial ϕα (X ) = X 2 + (1/a)X + 1/b. Then ϕα (X ) is the minimal polynomial of α and α is an anti-integral element over R because R is a Noetherian normal domain. We have J[α] = (a, b)R = R and R[η1 , . . ., ηd ] = R[1/a, 1/b]. Since R[η1 , . . ., ηd ] is obtained by localizations, R[η1 , . . ., ηd ] is flat over R. But since J[α] = R, R[α] is not flat over R. Thus the converse statement of Theorem 9.3.4 is not always valid. Theorem 9.3.6

The following statements are equivalent:

(1) R[α] is integral over R; (2) R[η1 , . . ., ηd ] is integral over R.

9.3 Anti-Integral Elements and Coefficients of its Minimal Polynomial 211 Proof Let R denote the integral closure of R in K . (2) ⇒ (1): Since R[η1 , . . ., ηd ] is integral over R, we have R[η1 , . . ., ηd ] ⊆ R. Since α is integral over R[η1 , . . ., ηd ], α is integral over R. So α is integral over R. (1)  ⇒ (2): Since R is a Noetherian domain, R is a Krull domain. So R = R P , P ∈ Ht1 (R), where R P is a DVR. Since α is anti-integral and integral over a DVR R P , we have ϕα (X ) ∈ R P [X ]. Hence ηi ∈ R P for all i. So ηi ∈ R, which implies that R[η1 , . . ., ηd ] is integral over R. Lemma 9.3.7 (cf. The proof of Theorem 5.1.1). Assume that α is an antiintegral element over R. Then  R (R[α]) ∼ = R[α]/I[α] ϕα (α)R[α]. Theorem 9.3.8 Assume that α is an anti-integral element over R. If R[α] is unramified over R, then R[η1 , . . ., ηd ][α] is unramified over R[η1 , . . ., ηd ] and R[η1 , . . ., ηd ] is unramified over R. ∼ R[α]/I[α] ϕ  (α)R[α] by Lemma 9.3.7. Since Proof Note that  R (R[α]) = α R[α] is unramified over R, we have I[α] ϕα (α)R[α] = R[α]. Thus I[α] ϕα (α)R[α] [η1 , . . ., ηd ] = R[α][η1 , . . ., ηd ] = R[η1 , . . ., ηd ][α]. Since ϕα (α) ∈ R[α] [η1 , . . ., ηd ], ϕα (α) is an invertible element in R[η1 , . . ., ηd ][α]. Hence R[η1 , . . ., ηd ][α] is unramified over R[η1 , . . ., ηd ]. Note R[η1 , . . ., ηd ][α] = R[η1 , . . ., ηd ] [X ]/ϕα (X )R[η1 , . . ., ηd ][X ]. So R[η1 , . . ., ηd ][α] is flat over R[η1 , . . ., ηd ]. Moreover R[η1 , . . ., ηd ][α] = I[α] ϕα (α)R[η1 , . . ., ηd ][α] ⊆ I[α] R[η1 , . . ., ηd ] [α] ⊆ R[η1 , . . ., ηd ][α]. Hence I[α] R[η1 , . . ., ηd ][α] = R[η1 , . . ., ηd ][α]. Since R[η1 , . . ., ηd ][α] is flat over R[η1 , . . ., ηd ] and R[η1 , . . ., ηd ][α] is integral over R[η1 , . . ., ηd ], R[η1 , . . ., ηd ][α] is faithfully flat over R[η1 , . . ., ηd ]. So I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]. Thus R → R[η1 , . . ., ηd ] is an open immersion by Proposition 9.3.3 and hence unramified. Now we characterize the ring R[α]∩ K under the condition I[α] R[α] = R[α]. Lemma 9.3.9 Let p be a prime ideal of R. If p R[α] = R[α], then α −1 is integral over R p . Proof Since p R[α] = R[α], we have a0 + a1 α + · · · + a α  = 1 for some ai ∈ p. Thus α satisfies the equation: a α  + · · · + a1 α + (a0 − 1) = 0. Hence α −1 satisfies the equation: a + a1 α −1 + · · · + a1 α −1 + (a0 − 1)(α −1 ) = 0. Since a0 − 1 is a unit in R p . Thus we can conclude that α −1 is integral over R p .

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Proposition 9.3.10 Assume that α is an anti-integral element over R. Consider the following statements: (1) I[α] R[α] = R[α]; (2) I[α] + I[α−1 ] = R; (3) I[α] = Iηd and Iηd (1, ηd )R = R. Then the following implications hold: (1) =⇒ (2) ⇐⇒ (3). Proof (1) ⇒ (2): There exists p ∈ Spec(R) such that I[α] + I[α−1 ] ⊆ p. Since I[α] R[α] = R[α], α −1 is integral over R p by Lemma 9.3.9. Since α is anti-integral over R, so is α −1 by Theorem 2.4.5. Hence α −1 is anti-integral and integral over R p . Thus ϕα−1 (X ) ∈ R p [X ] and hence I[α−1 ] R p = R p , which contradicts the assumption I[α−1 ] ⊆ p. (2) ⇔ (3): Since I[α−1 ] = ηd I[α] by Lemma 2.4.6. we have I[α] + I[α−1 ] = I[α] (1, ηd )R = R. So we have I[α] = Iηd and Jηd = Iηd (1, ηd )R = R. The converse implication can be seen by tracing the above argument backward.

Example 9.3.11 The following example shows that the implication (2) ⇒ (1) is not valid in general. Let R be a polynomial ring k[a, b] over a field k. Let α be a solution of the equation: ϕα (X ) := X 2 + (b/a 2 )X + ((a − 1)/a)2 = 0. Then α is anti-integral over R because R is a Noetherian normal domain. We have I[α] = a 2 R, ϕα−1 (X ) = X 2 + (b/(a − 1)2 )X + (a/(a − 1))2 , and I[α−1 ] = = R and J˜[α] = (a − 1)2 R. Thus I[α] + I[α−1 ] = R. Moreover we have J[α]   2 2 2 a (1, b/a )R = (a , b)R. Since grade( J˜[α] ) > 1, we have J˜[α] = I[α] . Hence I[α] R[α] = R[α], which implies that the implication (2) ⇒ (1) does not always hold. Recall that an element α ∈ L is called exclusive over R if R[α]∩ K = R. Now we study the exclusiveness for a while. We start with recalling the following lemma. Lemma 9.3.12 (Theorem 7.2.7) Assume that R contains an infinite field k and that α is a super-primitive element over R. Then the following statements are equivalent: (1) α is exclusive over R; (2) I˜[α] ⊆ Iηd ; (3) grade( J˜[α] ) > 1 or J˜[α] = R.

9.3 Anti-Integral Elements and Coefficients of its Minimal Polynomial 213 Proposition 9.3.13 Assume that α is super-primitive over R and that R contains an infinite field. If either grade( J˜[α] ) > 1 or J˜[α] = R, then both α and α −1 are exclusive, i.e., R[α] ∩ K = R[α −1 ] ∩ K = R. Proof By Lemma 9.3.12, we have the following equivalences: (a) α is exclusive over R ⇔ grade( J˜[α] ) > 1; (b) α −1 is exclusive over R ⇔ grade(I[α−1 ] (η1 /ηd , . . ., ηd−1 /ηd , 1)) > 1 ⇔ grade(I[α] (η1 , . . ., ηd )) > 1; where the last equivalence follows from Lemma 2.4.6. These equivalence induce our conclusion. Proposition 9.3.14 Assume that α is a super-primitive element over R. If R[α] is faithfully flat over R, then α is exclusive. Proof From Lemma 9.3.1, it follows the equivalence: R[α] is faithfully flat over R ⇔ J˜[α] = R. So we have our conclusion by Lemma 9.3.12. Proposition 9.3.15 Assume that α is super-primitive over R and that R contains an infinite field. If R[α] p is faithfully flat over R p for each p ∈ Dp1 (R), then α is exclusive, i.e., R[α] ∩ K = R. Proof By Lemma 9.3.12, note that R[α] p is faithfully flat over R p for each p ∈ Dp1 (R) ⇔ grade( J˜[α] ) > 1 or J˜[α] = R by Lemma 9.3.1. The latter condition gives rise to the statement that α is exclusive over R by Lemma 9.3.12. Lemma 9.3.16 Assume that α is super-primitive over R. If I[α] R[α] = R[α], then R[η1 , . . ., ηd ] ⊆ R[α]. Proof Since I[α] R[α] = R[α], R[α] is flat over R by Lemma 9.3.2. Take P ∈ Dp1 (R[α]) and put p := P ∩ R. Then p ∈ Dp1 (R). Since α is superprimitive over R, the ideal I[α] R p is a principal ideal. So there exists a ∈ I[α] such that I[α] R p = a R p . Hence a R[α] p = R[α] p by the assumption I[α] R[α] = R[α]. [Since I[α] ⊆ Iηi by definition, putting ηi = bi /a with bi ∈ R.] Since a is an  invertible element in R[α] p , we have ηi ∈ R[α] p ⊆ R[α] P . Thus ηi ∈ P∈Dp1 (R[α]) R[α] P = R[α]. Therefore R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ] ⊆ R[α]. Now we give an equivalent condition to I[α] R[α] = R[α].

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Theorem 9.3.17 Assume that α is a super-primitive element over R. The following statements are equivalent: (1) I[α] R[α] = R[α]; (2) R[η1 , . . ., ηd ] ⊆ R[α] and I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]. If the condition (2) holds, R[η1 , . . ., ηd ] is flat over R. Proof (1) ⇒ (2): The first statement is shown in Lemma 9.3.16. The assumption I[α] R[α] = R[α] implies that R[α] is flat over R by Theorem 9.3.4 and R[η1 , . . ., ηd ] is flat over R by Lemma 9.3.2. Hence J[α] = R by Proposition 2.3.6. Since α is anti-integral over R[η1 , . . ., ηd ] (cf. Theorem 7.1.15 and since R[η ,...,η ] α is integral over R[η1 , . . ., ηd ], it follows that I[α] 1 d = R[η1 , . . ., ηd ], R[η1 ,...,ηd ] where I[α] = R[η1 , . . ., ηd ] : R[η1 ,...,ηd ] ϕα (X ). Thus I[α] R[η1 , . . ., ηd ] = R[η ,...,η ] I[α] 1 d because R[η1 , . . ., ηd ] is flat over R. Moreover we have R[η1 , . . ., ηd ] ⊆ R[α] by Lemma 9.3.16. (2) ⇒ (1): Since R[η1 , . . ., ηd ] ⊆ R[α] and I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ], we have I[α] R[α] = R[α]. Proposition 9.3.18 Assume that α is a super-primitive element over R and that R contains an infinite field. If R[ηd ] is flat over R, then R[α] ∩ K ⊆ R[ηd ]. Proof Since R and R[ηd ] have the same quotient α is d fieldR[ηKd ,] the element R[η ] d] Iηi , where IηR[η := of degree d over both R and R[ηd ]. Put I[α] d := i=1 i R[ηd ] R[ηd ] R[ηd ] : R[ηd ] ηi . Then I[α] ⊆ I[α] , so that J[α] = I[α] (1, η1 , . . ., ηd ) ⊆ J[α] = R[η ] R[η ] I[α] d (1, η1 , . . ., ηd ), where J[α] d := [α] R[ηd ] (1, η1 , . . ., ηd ). Since α is superprimitive over R, we have grade(J[α] ) > 1. Since R[ηd ] is flat over R, we have R[η ] grade(J[α] R[ηd ]) > 1 and hence grade(J[α] d ) > 1. So α is super-primitive over R[η ] d] R[ηd ]. Since ηd ∈ R[ηd ], we have I˜[α] d ⊆ IηR[η . So applying Lemma 9.3.12 d to the extension R[α] of R[ηd ], we obtain R[α] ∩ K ⊆ R[ηd ][α] ∩ K = R[ηd ].

Theorem 9.3.19 Assume that α is super-primitive over both R and R[ηd ] and that R contains an infinite field. Consider the following statements: (1) I[α] R[α] = R[α]; (2) R[ηd ] ⊆ R[α], I[α] = Iηd and R[ηd ] is flat over R; (3) R[α] ∩ K = R[ηd ] = R[η1 , . . ., ηd ]. Then the implications (1) ⇔ (2) ⇒ (3) hold.

9.3 Anti-Integral Elements and Coefficients of its Minimal Polynomial 215 Proof (1) + (2) ⇒ (3): (1) implies that R[η1 , . . ., ηd ] ⊆ R[α] by Lemma 9.3.16. Since R[ηd ] is flat over R, R[ηd ] ⊇ R[α] ∩ K by Proposition 9.3.18. Hence we have R[α] ∩ K = R[ηd ] ⊇ R[η1 , . . ., ηd ] = R[η1 , . . ., ηd−1 ]. (1) ⇒ (2): We have R[ηd ] ⊆ R[η1 , . . ., ηd ] ⊆ R[α] by Lemma 9.3.16, and I[α] = Iηd by Proposition 9.3.10. Since R → R[ηd ] → R[η1 , . . ., ηd ] is an open immersion by Proposition 9.3.3, R → R[ηd ] is flat. d] (2) ⇒ (1): Since R[ηd ] is flat over R, we have IηR[η = Iηd R[ηd ]. Thus the fact d R[ηd ] ηd ∈ R[ηd ] implies that Iηd = R[ηd ]. So it follows that Iηd R[α] = R[α] because R[ηd ] ⊆ R[α]. Since I[α] = Iηd , we conclude I[α] R[α] = R[α].

Remark 9.3.20 Assume that α is anti-integral over R and that ηd ∈ R. Then R[α] is faithfully flat over d R if and only if R[α] is flat over R. Indeed, since ηd ∈ R, we have I[α] = i=1 Iηi = I˜[α] and hence J[α] = I[α] (1, η1 , . . ., ηd ) = ˜ I[α] (1, η1 , . . ., ηd−1 ) = J[α] . Hence J˜[α] = R if and only if J[α] = R. Thus our conclusion follows from Lemma 9.3.1. Proposition 9.3.21 Assume that α is a super-primitive element of degree d over R. Assume that the polynomial ψ(X ) := X d−1 + η1 X d−2 + · · · + ηd−1 is irreducible in K [X ] and let β be a solution of ψ(X ) = 0. Further assume that ηd ∈ R. Then β is anti-integral over R, and R[α] is flat over R if and only if R[β] is flat over R. Proof Since ηd ∈ R, noting that J[α] = I[α] (1, η1 , . . ., ηd ) by definition, we d d−1 conclude that I[α] = i=1 Iηi = i=1 = I[β] and hence J[β] = I[β] (1, η1 , . . ., ηd−1 ) = I[α] (1, η1 , . . ., ηd−1 ) = J[α] . Theorem 9.3.22 Assume that K contains a field of characteristic zero and that ηd ∈ R. Let β be a solution of ϕα (X ) = 0, where ϕα (X ) = d X d−1 + (d − 1)η1 X d−2 + · · · + ηd−1 . Then (1) if α is super-primitive over R, so is β; (2) R[α] is flat over R if and only if R[β] is flat over R. Proof By the similar argument in the proof of Proposition 9.3.21, we have J[α] = J[β] . Example 9.3.23 Consider the case d = 2. Put ϕα (X ) := X 2 + ηX + a with a ∈ R. Let α be a solution of an equation ϕα (X ) = 0. Then R[α] is flat over

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R if and only if R[η] is flat over R because α is super-primitive over R. In this case, α is characterized by η. Lemma 9.3.24 element over R.

If I[α] is an invertible ideal of R, then α is a super-primitive

Proof For each p ∈ Spec(R), (I[α] ) p is a principal ideal of R p . So the conclusion follows Theorem 2.3.10. We say that an element α ∈ L is an s-flat element over R if J[α] = R. Proposition 9.3.25 (1) An element α ∈ L is an s-flat element over R if and only if so is α −1 . (2) α ∈ L is s-flat over R if and only if α is anti-integral/super-primitive over R and R[α] is flat over R. Proof (1) Since J[α] = J[α−1 ] by Proposition 2.4.7, we conclude that J[α] = R if and only if J[α−1 ] = R. (2) (⇒): J[α] = R by definition, and hence J[α] ⊆ p for every p ∈ Dp1 (R), which means that α is super-primitive over R. A super-primitive element is an anti-integral element by Theorem 2.2.8. In this case, J[α] = R implies that R[α] is flat over R by Theorem 2.2.6. (⇐): Assume that α is anti-integral over R and R[α] is flat over R. Then we have J[α] = R by Theorem 2.3.6. Thus α is an s-flat element over R. Proposition 9.3.26 When α ∈ K is s-flat over R if and only if both R[α] and R[α −1 ] are flat over R. Proof

This follows from Proposition 1.2.12 and Theorem 2.3.6.

Proposition 9.3.27 For some i (1 ≤ i ≤ d), assume that I[α] = Iηi and that ηi is an s-flat element over R. Then α is an s-flat element over R. Moreover if i = d, then R[α] is faithfully flat over R. Proof Since ηi is an s-flat element over R, we have Jηi = Iηi (1, ηi ) = R by definition. Since I[α] = Iηi and J[α] = I[α] (1, η1 , . . ., ηd ) ⊇ Jηi = R, we have J[α] = R. So α is an s-flat element over R. Assume that i = d.

9.3 Anti-Integral Elements and Coefficients of its Minimal Polynomial 217 Then J˜[α] ⊇ Iηi (1, ηi ) = R and hence J˜[α] = R. Thus our conclusion follows from Lemma 9.3.1. Theorem 9.3.28 Assume that (R, m) is a local ring. Then the following statements are equivalent: (i) α is an s-flat element over R; (ii) for some i (1 ≤ i ≤ d), the following conditions hold: ηi is an s-flat element over R and I[α] = Iηi . Proof (ii) ⇒ (i) is shown in Proposition 9.3.27. (i) ⇒ (ii): We have only to show this in the case I[α] ⊆ m. We have J[α] = I[α] (1, η1 , . . ., ηd ) = R because α is an s-flat element over R. So there exists i such that ηi I[α] = R because R is local. Thus I[α] = Iηi and Jηi = Iηi (1, ηi ) ⊇ Iηi ηi = I[α] ηi = R, which yields that ηi is an s-flat element over R. Remark 9.3.29 Let (R, m) be a local ring. If there exists a prime ideal p of R such that none of η1 , . . ., ηd is a strongly flat element over R p , then R[α] is not flat over R. Such p is the one not containing J[α] . Example 9.3.30 Let R be a local ring k[a, b](a,b) , where k[a, b] is a polynomial ring over a field k. (1) Let α be a solution of the equation: ϕα (X ) := X 2 + (b/a)X + a/b = 0. Then ϕα (X ) is a minimal polynomial of α and α is super-primitive over R because R is a Noetherian normal domain. We have I[α] = ab R and J[α] = I[α] (1, b/a, a/b)R = ab(b/a, a/b)R = (b2 , a 2 )R = R. So R[α] is not flat over R. We see that η1 := b/a and η2 := a/b and that neither Iη1 nor Iη2 is equal to R. Note here that I[α] = Iη1 and I[α] = Iη2 . (2) Let α be a solution of the equation: ϕα (X ) := X 3 + (b/a)X 2 + (a/b)X + 1/a = 0. Then α is super-primitive over R as in (1). It follows that 1/a is an s-flat element. But I[α] = ab R is equal to none of Ib/a , Ia/b , and I1/a . Since J[α] = R, R[α] is not flat over R. Theorem 9.3.31 Assume that I[α] is an invertible ideal of R. If R[α] is flat over R, then for each p ∈ Spec(R), there exists i such that each ηi is an s-flat element over R p and that I[α] R p = Iηi R p . Proof Since I[α] is an invertible ideal, α is super-primitive over R by Lemma 9.3.24. So R[α] is flat over R if and only if J[α] = R. Take p ∈ Spec(R). Localizing at p, we may assume that R is a local ring with maximal ideal m.

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Since I[α] is invertible, we have I[α] = a R and ηi = bi /a for some a, bi ∈ R. Assume first that a ∈ m. Then ηi ∈ R and hence ηi is an s-flat element over R for all i (1 ≤ i ≤ d). Assume next that a ∈ m. Then J[α] = I[α] (1, η1 , . . ., ηd ) = (a, b1 , . . ., bd ) = R and hence there exists i such that bi ∈ m. So ηi = bi /a is an s-flat element and I[α] = a R = Iηi .

9.4

Denominator Ideals and Flatness (II)

Let α be an anti-integral element of degree d over R. Consider the following statements: (i) I[α] R[α] = R[α]; (ii) R[α] is flat over R; (iii) R[α] is unramified over R. We have shown that the statements above are all equivalent if d = 1, i.e., α ∈ K (cf. Theorem 5.2.8 It seems natural to ask whether the above equivalences are valid in the case d > 1. Our objective is to investigate the implications among (i), (ii), and (iii). We begin with the following known result. Proposition 9.4.1 Let α be an anti-integral element of degree d over R. Then the implications : (i) ⇒ (ii), (iii) ⇒ (ii), and (iii) ⇒ (i) hold. Proof The implication (i) ⇒ (ii) is seen in Theorem 3.4.1. The implications (iii) ⇒ (ii) and (iii) ⇒ (i) are seen in Theorem 5.1.4 and Remark 5.1.3, respectively. The following examples and the remark show that the reverse implications in Proposition 9.4.1 are not always valid. Example 9.4.2 We construct an example which shows that the implication (ii) ⇒ (i) is not always valid. Let k be a field and let R := k[x, y] be a polynomial ring. Let α be an algebraic element whose minimal polynomial is ϕα (X ) = X 3 +(1/x)X 2 +(1/y)X +(1/x y) ∈ k(x, y)[X ]. Then α is anti-integral over R of degree 3 (cf. Theorems 2.2.9 and 2.2.8. We have I[α] = x y R and hence I[α] R[α] = x y R[α] = R[α]. It is easy to see that J[α] = R. Hence R[α] is flat over R by Theorem 2.3.6.

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Example 9.4.3 In view of Example 9.4.14, we know that the implication (ii) ⇒ (iii) is not always valid as follows. Let β ∈ L satisfy β d = a ∈ R and a ∈ U (R) with [K (β) : K ] = d for d > 1. Then ϕβ (X ) = X d − a and ϕβ (β) = dβ d−1 ∈ U (R[β]). Thus R[β] is integral and hence flat over R by Corollary 2.3.5 but not unramified over R. Remark 9.4.4 If I[α] = R, then R[α] is integral over R by Theorem 2.3.2. In this case, I[α] R[α] = R[α]. But in general, we cannot say that R[α] is unramified over R. In fact, when I[α] R[α] = R[α], R[α] is unramified over R if and only if ϕα (α)R[α] = R[α] (cf. Theorem 5.1.1). But ϕα (α) is not always a unit in R[α] even if R[α] is integral over R. So the implication (i) ⇒ (iii) does not always hold. Let α be an element algebraic over R. For an ideal N of R, put V (N ) := { p ∈ Spec(R)|N ⊆ p}. Let  R[α]/R := { p ∈ Spec(R)| p R[α] = R[α]} and  J[α] := { p ∈ Spec(R)| p + J[α] = R}. Proposition 9.4.5 Assume that α is an anti-integral element over R of degree d. Then the following statements are equivalent: (1) I[α] R[α] = R[α];   (2) J˜[α] = I[α] and J[α] = R. If (2) holds, then Im[Spec(R[α]) → Spec(R)] = Spec(R) \ V (I[α] ), an open subset in Spec(R). Proof (1) ⇒ (2): Since R[α] is flat over R by Proposition 9.4.1, we have J[α] = R by Theorem 2.3.6. Thus  R[α]/R = V ( J˜[α] ) ∩  J[α] = V ( J˜[α] ) (cf. Theorem 4.3.2) and hence I[α] ⊆ J˜[α] . So we have  R[α]/R ⊆ V (I[α] ). Since I[α] R[α] = R[α] by the assumption, we have ) ⊆  R[α]/R . Hence  V (I[α] ˜ ˜ V (I[α] ) =  R[α]/R = V ( J[α] ), which means that J[α] = I[α] . (2) ⇒ (1): From the assumption, it follows that  R[α]/R = V ( J˜[α] ) ∩  J[α] = V ( J˜[α] ) = V (I[α] ). Hence I[α] R[α] = R[α]. Next assume (2) holds. We know Im[Spec(R[α]) → Spec(R)] = (Spec(R) \ V ( J˜[α] )) ∪ V (J[α] ) by Theorem 4.3.2. So (Spec(R) \ V ( J˜[α] )) ∪ V (J[α] ) = Spec(R) \ V (I[α] ) by (2). Remark 9.4.6 Assume that α is anti-integral over R of degree d. If I[α] R[α] = R[α], then I[α] is an invertible ideal of R because R = J[α] = I[α] (1, η1 , . . ., ηd ) by Proposition 9.4.5.

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Proposition 9.4.7 Assume that α is an anti-integral element over R of degree d. Consider the following statements: (1) I[α] R[α] = R[α]; (2) I[α] + I[α−1 ] = R; (3) I[α] = Iηd and Jηd = R. Then the implications (1) ⇒ (2) ⇔ (3) hold. Proof (1) ⇒ (2): Suppose that I[α] + I[α−1 ] = R. Take p ∈ Spec(R) such that I[α] + I[α−1 ] ⊆ p. Then I[α] R p = a R p for some a ∈ I[α] by Remark 9.4.6. Put d d−1 + · · · + ad = 0. a0 = a, aηi = ai (1 ≤ i ≤ d). Then we havea0 α + a1 α ˜ Since I[α] (1, η1 , . . ., ηd ) = R and J[α] = I[α] by Proposition 9.4.5, ad is   an invertible element in R p . Moreover we have a0 , · · ·, ad−1 ∈ J˜[α] = I[α] . So we can take a large integer n such that (ad )n = (−(a0 α d + · · · + ad−1 α))n = a(b N α N + · · · + bn α n ) for some bi ∈ R p . Thus we have an algebraic relation: c N α N + · · · + c1 α +

1 =0 a

for some ci ∈ R p . This means that α is comonic over R p , i.e., α −1 is integral over R p . Since α is anti-integral over R, α −1 is also anti-integral over R. Hence α −1 is integral and anti-integral over R p . So we have I[α−1 ] R p = R p , which contradicts the assumption I[α−1 ] ⊆ p. (2) ⇔ (3): Since I[α−1 ] = ηd I[α] , (2) implies that R = I[α] + I[α−1 ] = I[α] (1, ηd ). Hence I[α] = Iηd and Jηd = Iηd (1, ηd ) = R. The converse implication is easy to see. Example 9.4.8 Let R be a polynomial ring k[a, b] over a field k and α be a root of the equation: a a−1 b X2 + X+ =0 a−1 a(a − 1) a Then the minimal polynomial of α is given by b ϕα (X ) = X + 2 X + a 2

a−1 a

2

In this case, I[α] = a 2 R; ϕα−1 (X ) = X 2 + (b/(a − 1)2 )X + (a/(a − 1))2 , and I[α−1 ] = (a − 1)2 R. Thus I[α] + I[α−1 ] = R, J[α] J˜ = a 2 (1, b/a 2 )R = = R and  [α] 2 (a , b)R. Hence grade( J˜[α] ) > 1. Therefore J˜[α] = I[α] , which means that

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I[α] R[α] = R[α]. The implication (2) ⇒ (1) in Proposition 9.4.7 is not always valid. Next, under the condition I[α] R[α] = R[α], we get a condition which is equivalent to the statement that R[α] is unramified over R. Theorem 9.4.9 Assume that α is an anti-integral element over R of degree d and that I[α] R[α] = R[α]. Then R[α][η1 , . . ., ηd ] is unramified over R[η1 , . . ., ηd ] if and only if R[α] is unramified over R. Proof (⇒): Take P ∈ Spec(R[α]) and put p = P∩R. Since I[α] R[α] = R[α], we have I[α] ⊆ p. Thus η1 , . . ., ηd ∈ R p . Since R[α][η1 , . . ., ηd ] is unramified over R[η1 , . . ., ηd ], it follows that R p [α] = R p [α][η1 , . . ., ηd ] is unramified over R[η1 , . . ., ηd ] = R p , So we see that I[α] ϕα (α)R p [α][η1 , . . ., ηd ] = I[α] ϕα (α)R p [α] = R p [α] = R p [α][η1 , . . ., ηd ] by Lemma 9.3.7. Hence ϕα (α) is a unit in R p [α][η1 , . . ., ηd ] = R p [α] because I[α] ⊆ p. Therefore R p [α] is unramified over R p . So we conclude that R[α] is unramified over R. (⇐): Since I[α] R[α] = R[α], we have only to show that ϕα (α) is a unit in R[α][η1 , . . ., ηd ] by Lemma 9.3.7. Take P  ∈ Spec(R[α][η1 , . . ., ηd ]), and put P := P  ∩ R[α] and p := P ∩ R. Since I[α] R[α] = R[α], we have p ⊇ I[α] . Hence η1 , . . ., ηd ∈ R p and R p [α][η1 , . . ., ηd ] = R p [α]. Since R[α] is unramified over R by the assumption, it follows that ϕα (α)−1 ∈ R p [α], which means that ϕα (α) is a unit in R p [α]. Therefore we conclude that ϕα (α) is a unit in R[α][η1 , . . ., ηd ]. Remark 9.4.10 Under the same assumptions as in Theorem 9.4.9, R[α] [η1 , . . ., ηd ] is unramified over R[η1 , . . ., ηd ] if and only if ϕα (α) is a unit in R[α][η1 , . . ., ηd ]. (See Remark 9.4.4.) Theorem 9.4.11 Assume that α is an anti-integral element over R of degree d. Consider the following statements: (1) I[α] R[α] = R[α]; (2) I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]; (3) R[η1 , . . ., ηd ] is flat over R. Then the implications (1) ⇒ (2) ⇒ (3) hold.

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Proof (1) ⇒ (2): The assumption I[α] R[α] = R[α] yields I[α] R[α][η1 , . . ., ηd ] = R[α][η1 , . . ., ηd ]. Since α is integral over R[η1 , . . ., ηd ] with a monic relation ϕα (X ) ∈ R[η1 , . . ., ηd ][X ]. So we have I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]. (2) ⇒ (3): Since I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ], we have Iηi R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ] for each i. Thus Iη1 · · · Iηd R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ]. Hence R[η1 , . . ., ηd ] is flat over R by Theorem 9.1.1. Theorem 9.4.12 Assume that α is an anti-integral element over R. If R[α] is flat over R, then R[η1 , . . ., ηd ] is flat over R. Proof Since R[α] is flat over R, we have J[α] = I[α] (1, η1 , . . ., ηd )R = R by Theorem 2.3.6. Hence (1, η1 , . . ., ηd )R is an invertible ideal of R. Localizing at p ∈ Spec(R), we can assume that (1, η1 , . . ., ηd )R p = (1/a)R p for some a ∈ R. Hence R[η1 , . . ., ηd ] p = R p [1/a]. Thus R[η1 , . . ., ηd ] is flat over R.

Example 9.4.13 In Theorem 9.4.11, the implication (2) ⇒ (1) does not always hold. Let R := k[x, y] be a polynomial ring over a field k. Let α be an element satisfying the minimal polynomial ϕα (X ) := X 2 + (1/x)X + 1/y ∈ k(x, y)[X ]. Then α is anti-integral over R. It is easy to see that I[α] = x y R and J[α] = (x, y)R. Hence R[α] is not flat over R (Theorem 2.3.6). But both 1/x and 1/y are strongly flat elements over R. So R[1/x, 1/y] is flat over R. In this case, I[α] R[1/x, 1/y] = R[1/x, 1/y] but I[α] R[α] = R[α]. Indeed, J[α] = R and I[α] R[α] ⊆ J[α] R[α] = R[α]. Next we investigate the relationship between R[α] and R[η1 , . . ., ηd ]. Let R denote the integral closure of R in K . Then R is a Krull domain (cf. [M1]). Theorem 9.4.14 gral over R.

R[α] is integral over R if and only if R[η1 , . . ., ηd ] is inte-

Proof (⇒): Since R is a Krull domain, α is anti-integral over R (cf. Theorem 2.2.9). By the assumption, we have α is integral over R, and hence α is integral over R. Thus ϕα (X ) ∈ R[X ] (cf. [M2]). So we have η1 , . . ., ηd are integral over R. Therefore R[η1 , . . ., ηd ] is integral over R. (⇐): Note that α satisfies the monic relation ϕα (X ) = 0 over R[η1 , . . ., ηd ]. Since R[η1 , . . ., ηd ] is integral over R, α is integral over R. Lemma 9.4.15 Assume that α is a super-primitive element over R. If I[α] R[α] = R[α], then R[η1 , . . ., ηd ] ⊆ R[α].

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Proof Since I[α] R[α] = R[α], R[α] is flat over R by Proposition 9.4.1. Take P ∈ Dp1 (R[α])(:= {P ∈ Spec(R)|depth(R P ) = 1}) and put p := P ∩ R. Then p ∈ Dp1 (R). Since α is super-primitive over R, we have J[α] R p = I[α] (1, η1 , . . ., ηd )R p = R p . So I[α] R p is an invertible ideal, and hence I[α] R p is a principal ideal of R p . Thus there exists an element a ∈ I[α] such that I[α] R p = a R p . Since I[α] R[α] = R[α], it follows that a R p [α] = R p [α]. So since I[α] ⊆ Iηi , we have aηi = bi for some bi ∈R p . Hence ηi ∈ R p [α] ⊆ R[α] P because a is a unit in R p [α]. Therefore ηi ∈ P∈Dp1 (R[α] R[α] P = R[α], which means that R[η1 , . . ., ηd ] ⊆ R[α].

Theorem 9.4.16 Assume that α is a super-primitive element over R. Then I[α] R[α] = R[α] if and only if R[η1 , . . ., ηd ] is flat over R and R[η1 , . . ., ηd ] ⊆ R[α]. Proof (⇒): From I[α] R[α] = R[α], it follows that I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ] and R[η1 , . . ., ηd ] is flat over R by Theorem 9.4.11. So we have 9.4.15. R[η1 , . . ., ηd ] ⊆ R[α] by d Theorem R[η ,...,η ] 1 ,...,ηd ] 1 ,...,ηd ] (⇐): Let I[α] 1 d := i=1 IηR[η , where IηR[η := {b ∈ R[η1 , . . ., ηd ] | i i bηi ∈ R[η1 , . . ., ηd ]}. Since α is an anti-integral element over R, α is also anti1 ,...,ηd ] integral over R[η1 , . . ., ηd ] and IηR[η = I[α] R[η1 , . . ., ηd ]. Since R[α] is i integral over R[η1 , . . ., ηd ] with the integral dependence ϕα (α) = 0, we have 1 ,...,ηd ] IηR[η = R[η1 , . . ., ηd ]. Thus I[α] R[η1 , . . ., ηd ] = R[η1 , . . ., ηd ], and hence i I[α] R[α] = R[α] because R[η1 , . . ., ηd ] ⊆ R[α]. Corollary 9.4.17 Assume that α is a super-primitive element over R. If I[α] R[α] = R[α], then R[α] ∩ K = R[η1 , . . ., ηd ]. Proof Note that K (η1 , . . ., ηd ) = K . By Theorem 9.4.15, R[η1 , . . ., ηd ] ⊆ R[α]. Since R[α] is integral over R[η1 , . . ., ηd ] and α is anti-integral over R[η1 , . . ., ηd ], R[α] is free over R[η1 , . . ., ηd ] of rank d. So R[α] = R[η1 , . . ., ηd ] + R[η1 , . . ., ηd ]α + · · · + R[η1 , . . ., ηd ]α d−1 ⊆ K + K α + · · · + K α d−1 , which means that R[α] ∩ K = R[η1 , . . ., ηd ]. Theorem 9.4.18 Assume that α is a super-primitive element over R of degree d. Then R[α] ∩ K = R if and only if R[α] ∩ R[η1 , . . ., ηd ] = R. Proof The implication (⇒) is obvious because R[η1 , . . ., ηd ] ⊆ K . We must prove the converse implication. Take a ∈ I˜[α] . Then aηd = −(aα d +aη1 α d−1 + · · · + aηd−1 α) ∈ R[α]. Thus I˜[α] ηd ⊆ R[α] ∩ R[η1 , . . ., ηd ] = R, which implies that I˜[α] ⊆ Iηd . So we conclude that α is exclusive by Corollary 7.1.3.

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Proposition 9.4.19 If there exists p ∈ Spec(R) such that R[η1 , . . ., ηd ] p is integral over R p and R[η1 , . . ., ηd ] p = R p , then α is not anti-integral over R. Proof Suppose that α is anti-integral over R. Then α is anti-integral over R p . Since R[η1 , . . ., ηd ] p is integral over R p , it follows that α is integral over R p by Theorem 9.4.14. Thus ϕα (X ) ∈ R p [X ] (cf. Theorem 9.3.2), which contradicts the assumption R[η1 , . . ., ηd ] p = R p . Recall that C(R/R) denotes the conductor between R and R. Proposition 9.4.20 Assume that R is a finite R-module. If there exists a prime divisor p of I[α] such that p ⊇ I[α] : R C(R/R), then α is not anti-integral over R. Proof Localizing at p, we may assume that (R, p) is a local ring with p ∈ Dp1 (R). We have C(R/R) ⊆ I[α] ⊆ Iηi . Thus ηi ∈ C(R/R)−1 = R for 1 ≤ i ≤ d. But since p ⊇ I[α] , we have R[η1 , . . ., ηd ] = R. Our conclusion follows from Proposition 9.4.19. Remark 9.4.21 If α is ultra-primitive over R, i.e., grade(I[α] + C(R/R)) > 1, then α is an anti-integral element over R. Thus “ultra-primitive ⇒ superprimitive ⇒ anti-integral” (cf. Proposition 7.5.1 and Theorem 2.2.8). We close this section by giving the following result. Proposition 9.4.22 Assume that α is an anti-integral element over R of degree d. Let β1 , · · ·, βn be elements in K , each of which is a super-primitive element over R. If R[α] is flat over R, then β1 , · · ·, βn are super-primitive elements over R[α]. Proof Note that I[βi ] = Iβi (:= R : R βi ) and J[βi ] = Iβi + βi Iβi = Iβi + Iβi−1 because βi ∈ K . Since βi is super-primitive over R, we have grade(Iβi + Iβi−1 ) = grade(J[βi ] ) > 1 by definition. Let Tβi := R[α] : R βi . Since R[α] is flat over R, we have Iβi + Iβi−1 ⊆ Tβi + Tβi−1 , and hence we have grade(Tβi + Tβi−1 ) > 1. Thus βi is super-primitive over R[α] by Proposition 9.4.12.

9.5

Contractions of Principal Ideals and Denominator Ideals

Let R be a Noetherian domain, R[X ] be a polynomial ring, α be an element of an algebraic extension field of the quotient field K of R, and let A := R[α].

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225

Our main aims of this section are to investigate the contraction β A ∩ R of a principal fractional ideal β A with an element β ∈ K (α) = K [α], the contraction (A : A β) ∩ R of a denominator ideal A : A β, and to show what conditions are required for β A = A or β A ∩ R = R. In what follows, we use the following notations unless otherwise specified: Let R be a Noetherian domain (which is commutative and has a unit), R[X ] be a polynomial ring, α be an element of an algebraic extension field of the quotient field K of R, and π : R[X ] → R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with deg ϕα (X ) = d and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Then ηi ∈ K (1 ≤ i ≤ d) are uniquely determined by α. Put Iηi := R : R ηi , d I[α] := i=1 Iηi , E (α) := K er (π ) , and E K(α) := E (α) ⊗ R K = ϕα (X )K [X ]. For an element γ in K [α], we can write γ = τ0 +τ1 α +· · ·+τd−1 α d−1 uniquely with τi ∈ K . Put γ∗ (X ) := τ0 + τ1 X + · · · + τd−1 X d−1 ∈ K [X ] and α (γ ) := E K(α) + γ∗ (X ). The ideal I[α] is also expressed as R[X ] : R ϕα (X ) := {a ∈ R|aϕα (X ) ∈ R[X ]} and I[α] ϕα (X )R[X ] ⊆ E (α) . Since π (respectively π ⊗ R K ) is surjective, every element in A (respectively K [α]) is expressed as f (α) for some f (X ) in R[X ] (respectively K [X ]). We have γ∗ (α) = γ ∈ L and α (γ ) = {ψ(X ) ∈ K [X ]|ψ(α) = γ }. For any φ(X ) ∈ α (γ ), α (γ ) = E K(α) + φ(X ). For γ ∈ L \ {0}, we have that γ ∈ A ⇔ α (γ ) ∩ R[X ] = ∅ and γ ∈ L \ A ⇔ α (γ ) ∩ R[X ] = ∅. Lemma 9.5.1 Let β ∈ L = K [α]. Then β A ∩ R = (E K(α) + ψ(X )R[X ]) ∩ R for every ψ(X ) ∈ α (β). Proof Take a ∈ β A ∩ R. Then a = βg(α) for some g(α) ∈ A. Since a −βg(α) = a −ψ(α)g(α) = 0, we have a −ψ(X )g(X ) ∈ E K(α) = ϕα (X )K [X ]. Thus a − ψ(X )g(X ) = ϕα (X )τ (X ) for some τ (X ) ∈ K [X ], which means that a = ϕα (X )τ (X ) + ψ(X )g(X ) ∈ (E K(α) + ψ(X )R[X ]) ∩ R. Thus β A ∩ R ⊆ (E K(α) + ψ(X )R[X ]) ∩ R. Conversely, take a ∈ (E K(α) + ψ(X )R[X ]) ∩ R with

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ψ(X ) ∈ α (β). Then a = ϕα (X )τ (X ) + ψ(X )g(X ) for some τ (X ) ∈ K [X ] and g(X ) ∈ R[X ]. Hence a = ψ(α)g(α) = βg(α) ∈ β A ∩ R. So we have (E K(α) + ψ(X )R[X ]) ∩ R ⊆ β A ∩ R. Therefore we have β A ∩ R = (E K(α) + ψ(X )R[X ]) ∩ R. Corollary 9.5.2 Assume that β ∈ A. Then α (β)∩ R[X ] = ∅ and β A∩ R = (E (α) , f (X ))R[X ] ∩ R for every f (X ) ∈ α (β) ∩ R[X ]. Proof We have seen that α (β) ∩ R[X ] = ∅ because β ∈ A. We have that (E K(α) + f (X )R[X ]) ∩ R = β A ∩ R for every f (X ) ∈ α (β) by Lemma 9.5.1. Take f (X ) ∈ α (β) ∩ R[X ] = ∅. Then β A ∩ R = (E K(α) + f (X )R[X ]) ∩ R = (E K(α) + f (X )R[X ]) ∩ R[X ] ∩ R = ((E K(α) ∩ R[X ]) + f (X )R[X ]) ∩ R = (E (α) , f (X ))R[X ] ∩ R. Lemma 9.5.3 Assume that β = 0 in L. Then there exist ψ1 (X ) and ψ2 (X ) in K [X ] such that ϕα (X )ψ1 (X ) + β∗ (X )ψ2 (X ) = 1 in K [X ] and that deg ψ2 < d. Furthermore, such polynomials ψ1 (X ) and ψ2 (X ) are uniquely determined. Proof Since ϕα (X ) is irreducible polynomial in K [X ] and since β = 0, we have (ϕα (X ), β∗ (X ))K [X ] = K [X ]. So there exist ψ1 (X ), ψ2 (X ) ∈ K [X ] such that ϕα (X )ψ1 (X ) + β∗ (X )ψ2 (X ) = 1. Since ϕα (X ) is a monic polynomial in K [X ], we can take ψ2 (X ) such as deg ψ2 (X ) < d. Next we show that ψ1 (X ) and ψ2 (X ) are uniquely determined. Suppose that there exist λ1 (X ), λ2 (X ) ∈ K [X ] which satisfy the same conditions as for ψ1 (X ), ψ2 (X ). Then ϕα (X )ψ1 (X ) + β∗ (X )ψ2 (X ) = 1 and ϕα (X )λ1 (X ) + β∗ (X )λ2 (X ) = 1. Hence ϕα (X )(ψ1 (X ) − λ1 (X )) + β∗ (X )(ψ2 (X ) − λ2 (X )) = 0. Since (ϕα (X ), β∗ (X ))K [X ] = K [X ], we have ψ2 (X ) − λ2 (X ) = ϕα (X )µ(X ) for some µ(X ) ∈ K [X ]. We see that deg ψ2 (X ) < d and deg λ2 (X ) < d imply ψ2 (X ) − λ2 (X ) = 0, that is, ψ2 (X ) = λ2 (X ). Thus we have also that ψ1 (X ) = λ1 (X ). We have [K (α) : K ] = d and L = K ⊕ K α ⊕· · ·⊕ K α d−1 . Since ϕα (α) = 0, we have ψ2 (α) = β −1 ∈ L. So we write: β −1 = β∗ (α)−1 = λ0 + λ1 α + · · · + λd−1 α d−1 . Noting that deg ψ2 (X ) < d, we have ψ2 (X ) = λ0 + λ1 X + · · · + λd−1 X d−1 ∈ K [X ]. R For ψ(X ) ∈ K [X ], let Iψ(X ) := {a ∈ R|aψ(X ) ∈ R[X ]}. Proposition 9.5.4 Under the same notation as in Lemma 9.5.3. If I[α] = R, i.e., ϕα (X ) ∈ R[X ], then β A ∩ R = IψR2 (X ) . Proof Take a ∈ IψR2 (X ) . Then a = ϕα (X )(aψ1 (X ))+β∗ (X )(aψ2 (X )) ∈ (E K(α) + β∗ (X )R[X ]) ∩ R = β A ∩ R by Lemma 9.5.1. Conversely, take a ∈ β A ∩ R.

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Then a = ϕα (X )ψ(X ) + β∗ (X )h(X ) for some ψ(X ) ∈ K [X ] and some h(X ) ∈ R[X ] by Lemma 9.5.1. Since ϕα (X ) is a monic polynomial in R[X ], we can assume that deg h(X ) < d. So we have 1 = ϕα (X )(a −1 g(X )) + β∗ (X )(a −1 h(X )). Thus a −1 h(X ) = ψ2 (X ) by Lemma 9.5.3, which means that a ∈ IψR2 (X ) .

Remark 9.5.5 [OY15] For β ∈ K (α), β R[α] ⊇ R[α] ⇔ β −1 ∈ R[α] ⇔ β R[α] ∩ R = R. When β ∈ R[α], β R[α] = R[α] ⇔ β R[α] ∩ R = R. Corollary 9.5.6 Assume that β = 0 in L and that ϕα (X ) ∈ R[X ]. Then β R[α] ∩ R = R if and only if ψ2 (X ) ∈ R[X ]. Assume furthermore that β ∈ R[α]. Then β R[α] = R[α] if and only if ψ2 (X ) ∈ R[X ]. Proof By Proposition 9.5.4, β R[α]∩R = R ⇔ IψR2 (X ) = R ⇔ ψ2 (X ) ∈ R[X ]. Next, when β ∈ R[α], β R[α] ∩ R = R ⇔ β R[α] = R[α by Remark 9.5.5.

Definition 9.5.7 Let β ∈ K (α)\{0} and β −1 = λ0 +λ1 α +· · ·+λd−1 α d−1 ∈ K (α) = K ⊕ K α ⊕ · · · ⊕ K α d−1 . We define: H[β] :=

d−1

Iλi

i=0

Note that deg ψ2 (X ) < d, β∗−1 (X ) = ψ2 (X ) ∈ K [X ], and ψ2 (X ) = λ0 + λ1 X + · · · + λd−1 X d−1 as the same notation as before. Remark 9.5.8 It is easy to see that H[β] = IψR2 (X ) for β ∈ K (α) \ {0} by definition. Hence for β ∈ K (α) \ {0}, β R[α] ∩ R = H[β] by Proposition 9.5.4 if I[α] = R. Remark 9.5.9 (1) For λ ∈ K , Iλ := R : R λ is divisorial. In fact, it is easy to see that Iλ = λ−1 R ∩ R, which is the intersection of fractional principal ideals. So Iλ is divisorial by [F, p.12]. Thus the ideal H[β] is a divisorial ideal of R. (2) If I be a nontrivial divisorial ideal of R, then grade(I ) = 1. Indeed, suppose that grade (I ) > 1. Then I contains a regular sequence a, b. Consider I −1 := R : K I . Take γ ∈ I −1 . Then γ a, γ b ∈ R and hence a, b ∈ γ −1 R ∩ R. Put γ := c/d with c, d ∈ R. Then ac, bc ∈ d R.

228

Flatness and Contractions of Ideals So ac = de, bc = d f for some e, f ∈ R. Thus ac f = de f = bce and hence a f = be. Since a, b is a regular sequence, it follows that e ∈ a R. Let e = ag with sg ∈ R. Then a f = bag and so f = bg and bc = d f = dbg, which implies that c = dg. Thus γ = c/d = dg/d = g ∈ R. So we have I −1 ⊆ R, which induces (I −1 )−1 ⊇ R ⊃ I . This contradicts the assumption that I is divisorial, i.e., (I −1 )−1 = I .

The following lemma follows from the definition. Lemma 9.5.10

Under the above notation, we have

β R[α] ∩ R = {a ∈ R|aψ2 (α) ∈ R[α]} = R[α] : R β −1 The following result is a strong version of Proposition 9.5.4 (Remark 9.5.8). Theorem 9.5.11 Let β be a nonzero element in L. If grade(H[β] + I[α] ) > 1, then β R[α] ∩ R = H[β] . Proof Let p be a prime divisor of the ideal H[β] . Since H[β] is a divisorial ideal by Remark 9.5.9(1), we have depth(R p ) = 1 by Remark 9.5.9(2). So by the assumption, we see that I[α] ⊆ p. Localizing at p, we have ϕα (X ) ∈ R p [X ]. Thus (β R[α] ∩ R) p ⊆ H[β] R p by Proposition 9.5.4 and Remark 9.5.8, and hence β R[α] ∩ R ⊆ H[β] . Conversely, take a ∈ H[β] . Then aλi ∈ R for all i, and hence aψ2 (α) ∈ R[α]. Thus a ∈ β R[α] ∩ R by Lemma 9.5.10. Remark 9.5.12 Let β ∈ K (α) \ {0}. Then β R[α] ∩ R = (R[α] : R[α] β −1 ) ∩ R = R[α] : R β −1 . Corollary 9.5.13 Let β be an element of K (α) \ {0}. If grade(H[β −1 ] + I[α] ) > 1, then (R[α] : R[α] β) ∩ R = R[α] : R β = H[β −1 ] . Proof Replacing β by β −1 in Theorem 9.5.11, we have β −1 R[α]∩ R = H[β −1 ] . Since β −1 R[α] ∩ R = (R[α] : R[α] β) ∩ R = R[α] : R β by Remark 9.5.12, we come to the conclusion. Corollary 9.5.14

Let β be a nonzero element in K (α). Assume that grade(H[β] + I[α] ) > 1

Then β R[α] ∩ R = R if and only if ψ2 (X ) ∈ R[X ]. Assume furthermore that β ∈ R[α]. Then β R[α] = R[α] if and only if ψ2 (X ) ∈ R[X ].

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We now give an example which shows that the condition on “grade” in Theorem 9.5.11 is effective. Example 9.5.15 Let k be a field and R := k[t] be a polynomial ring whose quotient field is denoted by K . Let α be an element 1/t. Then ϕα (X ) = X −1/t. It is easy to see that I[α] = t R. Put β = α −1 = t. Then β −1 = 1/t. We have that H[β] = I[α] = t R = R and grade(H[β] + I[α] ) = grade(I[α] ) = 1. In this case, we have β R[α] ∩ R = t R[1/t] ∩ R ⊇ R. Hence β R[α] ∩ R = R = H[β] . Proposition 9.5.16

If grade(I[α] + I[α−1 ] ) > 1, then α R[α] ∩ R = I[α−1 ] .

Proof Since ϕα (α) = α d +ηαd−1 +· · ·+ηd = 0, we have α −1 = −(ηd−1 α d−1 + ηd−1 η1 α d−2 + · · · + ηd−1 ηd−1 ). So I[α−1 ] = Iηd−1 ∩ Iηd−1 η1 ∩ · · · ∩ Iηd−1 ηd−1 . On the other hand, we have (α −1 )d + ηd−1 ηd−1 (α −1 )d−1 + · · · + ηd−1 η1 (α −1 ) + ηd−1 = 0. Hence H[α] = Iηd−1 ∩ Iηd−1 η1 ∩· · ·∩ Iηd−1 ηd−1 = I[α−1 ] = ηd I[α] . Thus our conclusion follows from Theorem 9.5.11. Corollary 9.5.17 a R[α] ∩ R = a R.

Let a ∈ R and assume that grade(a R + I[α] ) > 1. Then

n Let { p1 , . . ., pn } be the set of prime divisors of I[α] and T := R \ i=1 pi . Then T is a multiplicatively closed subset of R. Let RT denote the localization of R with respect to T . Proposition 9.5.18 Assume that α is a super-primitive element of degree d over R and take λ ∈ K . Then grade(Iλ + I[α] ) > 1 if and only if λ ∈ RT . Proof (⇐): Since λ ∈ RT , there exists a ∈ T such that aλ ∈ R. Hence a ∈ Iλ . By the definition of T , we have a ∈ p1 ∪ · · · ∪ pn . Let p be a depth 1 prime ideal containing I[α] . Since α is super-primitive over R, (I[α] ) p is a principal ideal in R p . Hence p is a prime divisor of I[α] . Thus p = pi for some i. We conclude that grade(Iλ + I[α] ) > 1. (⇒): Let pi (1 ≤ i ≤ n) be a prime divisor of I[α] . Then pi is a depth 1 prime ideal of R because I[α] is a divisorial ideal. Hence Iλ ⊆ pi for all i by the assumption. Thus there exists an element a ∈ I[α] ∩ T , which means that λ ∈ RT . Remark 9.5.19 Put U := RT + RT α + · · · + RT α d−1 . Note that U is not always a ring. Let β be an element in L.

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(1) grade(H[β] + I[α] ) > 1 if and only if β −1 ∈ U . In fact, putting β −1 := d−1 λ0 + λ1 α + · · · + λd−1 α d−1 with λi ∈ K . Then H[β] = i=0 Iλi . So if grade(H[β] + I[α] ) > 1, then H[β] ∩ T = ∅ and hence λi ∈ RT for all i. Thus β −1 ∈ U . Conversely, if β −1 ∈ U , then λi ∈ RT for all i. Hence H[β] RT = RT . Since I[α] is a divisorial ideal, we conclude that grade(H[β] + I[α] ) > 1. (2) R[α] ⊆ U if and only if I[α] = R and, in this case, U is a ring. Since “if” part is obvious, we have only to show “only if” part. Since α d + η1 α d−1 + · · · + ηd = 0, we have −α d = η1 α d−1 + · · · + ηd ∈ A ⊆ R[α] ⊆ U , which yields that −α d = λ1 α d−1 + · · · + λd with λi ∈ RT . Hence η1 α d−1 + · · · + ηd = λ1 α d−1 + · · · + λd . Since [K (α) : K ] = d, we have λi = ηi for all i and hence ηi ∈ RT . But we know that Iηi ⊇ I[α] and either grade(Iηi ) = 1 or Iηi = R. So Iηi = R (Proposition 9.5.18). Thus I[α] = R. It follows that α d + η1 α d−1 + · · · + ηd = 0 with ηi ∈ RT , which means that U is a ring. Proposition 9.5.20 Let α be an algebraic element of degree d over R and let β be a nonzero element in Rα. Then β Rα ∩ R = Rα : R β −1 Proof Take a ∈ Rα ∩ R. Then a = βy with some y ∈ Rα. Hence y ∈ aβ −1 ∈ Rα. Thus a ∈ Rα : Rβ −1 . Conversely, take x ∈ R such that xβ −1 ∈ Rα. Then x ∈ β Rα ∩ R. Remark 9.5.21 Under the same notation as in Proposition 9.5.20, it is easy to see that γ R[α] ∩ R = R[α] : R γ −1 for a nonzero element γ ∈ R[α]. Now recall the following structure theorem of the subring Rα := R[α] ∩ R[α −1 ]. Lemma 9.5.22 (cf. Theorem 3.1.18) Put ζi := α i + η1 α i−1 + · · · + ηi for 1 ≤ i ≤ d − 1 and ηi ∈ K . Assume that α is an anti-integral element of degree d over R. Then Rα = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 . Corollary 9.5.23 Assume that α is an anti-integral element of degree d over R. Then Rα ∩ K = R (i.e., Rα is exclusive over R). Remark 9.5.24 Take β ∈ R[α]. Note that K (α) = K [α] = K + K α + · · · + K α d−1 , and put β −1 := λ0 + λ1 α + · · · + λd−1 α d−1 with λi ∈ K . We showed

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231

d−1 in Theorem 9.5.11, that if grade( i=0 Iλi + I[α] ) > 1 ( of course, I[α] = R is d−1 sufficient), then β R[α] ∩ R = i=0 Iλi . The following theorem is similar to the known result (cf. Remark 9.5.24). But as forthe extension Rα of R, we need not require the assumption such d−1 as grade( i=0 Iλi + I[α] ) > 1. Let ζi be the same as in Lemma 9.5.22. Then K [α] = K (α) = K (ζ1 , . . ., ζd−1 ) = K [ζ1 , . . ., ζd−1 ]. Theorem 9.5.25 Assume that α is an anti-integral element of degree d over R. Let β be a nonzero element in Rα and let β −1 := λ0 +λ1 ζ1 +· · ·+λd−1 ζd−1 with λi ∈ K . Then β Rα ∩ R = Iλ0 ∩

d−1

(I[α] : R λi )

i=1

Proof Since α is super-primitive over R, Rα = R ⊕ I[α] ζ1 +⊕ · · ·⊕ I[α] ζd−1 by Lemma 9.5.22. Hence using Proposition 9.5.20, we have β Rα ∩ R = Rα : R β −1 = (R⊕ I[α] ζ1 +⊕ · · ·⊕ I[α] ζd−1 ) : R (λ0 +λ1 ζ1 +· · ·+λd−1 ζd−1 ) = d−1 Iλ0 ∩ i=1 (I[α] : R λi ). Lemma 9.5.26 Assume that α is an anti-integral element of degree d over R. If a, b ∈ R is a regular sequence on R, then it also is Rα. Proof Take P ∈ Dp1 (Rα) such that a, b ∈ P. Then a, b ∈ p := P ∩ R. But it is known that p ∈ Dp1 (R) (cf. Lemma 9.5.22), which is a contradiction.

Proposition 9.5.27 Assume that α is an anti-integral element of degree d over R. Let β1 , β2 ∈ Rα. Assume also that grade(I[α] + (βi Rα ∩ R)) > 1 for i = 1, 2. If grade((β1 Rα ∩ R) + (β2 Rα ∩ R)) > 1, then β1 , β2 form a regular sequence on Rα. Proof Take P ∈ Dp1 (Rα) such that β1 , β2 ∈ P. Put p := P ∩ R. Then p ∈ Dp1 (R) by Lemma 9.5.26. By construction, we have p ⊇ (β1 Rα ∩ R) + (β2 Rα ∩ R), which shows that grade( p) ≥ 2, a contradiction. Theorem 9.5.28 Assume that α is an anti-integral  element of degree d over k (βi Rα ∩ R) = R. R. Let β1 , . . ., βk ∈ Rα. Assume further that J[α] + i=1 k If grade( i=1 (βi Rα ∩ R)) ≥ n, then grade((β1 , . . ., βk )Rα) ≥ n.

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Proof Take P ∈ Spec(Rα) such that (β1 , . . ., betak ) ⊆ P. Put p := P ∩ R. Then J[α] ⊆ P by the assumption. So R p [α] is flat over R p and hence (Rα) p is flat over R p , for (J[α] ) p = R p implies that (I[α] ) p is an invertible ideal of R p and (Rα) p = R p ⊕ (I[α] ) p ζ1 ⊕ · · · ⊕ (I[α] ) p ζd−1 by Lemma 9.5.22. Thus the flatness guarantees that the grade does not decrease.

Chapter 10 Anti-Integral Ideals and Super-Primitive Polynomials

Let R be a Noetherian integral domain, R[X ] be a polynomial ring, and K be the quotient field of R. Let α be an element of an algebraic field extension L of K and π : R[X ] −→ R[α] be the R-algebra homomorphism sending X to α. Let ϕα (X ) be the monic minimal polynomial of α over K with  deg ϕα (X ) = d d (R : R ηi ) and write ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Let I[α] := i=1 (= R[X ] : R ϕ(X )). For f (X ) ∈ R[X ], let c( f (X )) denote the ideal generated by the coefficients of f (X ). Let J[α] := I[α] c(ϕα (X )), which is an ideal of R and contains I[α] . The element α is called an anti-integral element of degree d over R if Ker(π ) = I[α] ϕα (X )R[X ]. When α is an anti-integral element over R, R[α] is called an anti-integral extension of R. In the case K (α) = K , an anti-integral element α is the same as an anti-integral element (i.e., R = Rα) defined in Chapter 1. The element α is called a super-primitive element of degree d over R if J[α] ⊂ p for all primes p of depth 1. As was seen in the above, Chapter 3 concerned a simple extension with certain properties by use of the ideal Ker(π ). This chapter deals with an ideal H of R[X ] such that P ∩ R = (0) for all P ∈ Ass R[X ] (R[X ]/H ) (i.e., exclusive) or a monic polynomial ϕ(X ) in K [X ]. We define the super-primitiveness, anti-integrality and flatness of the ideal H or a polynomial ϕ(X ). If necessary, we can consider a simple ring extension R[X ]/H or R[X ]/(ϕ(X )K [X ] ∩ R[X ]). When H is a prime ideal or ϕ(X ) ∈ K [X ] is an irreducible polynomial, we come back to the case in Chapter 2. When H is not a prime ideal or ϕ(X ) ∈ K [X ] is not an irreducible polynomial, we can extend the super-primitiveness, anti-integrality and flatness to a simple extension which is not necessarily an integral domain. We use the following notations throughout this section unless otherwise specified: Let R be a Noetherian integral domain, R[X ] be a polynomial ring, and K be the quotient field of R. Let H be an ideal of R[X ] and ϕ(X ) be a monic polynomial in K [X ].

233

234

10.1

Anti-Integral Ideals and Super-Primitive Polynomials

Anti-Integral Ideals and Super-Primitive Ideals

Let R[X ] be a polynomial ring over R and H denote an ideal of R[X ]. We say that H is an exclusive ideal if P ∩ R = (0) for every prime ideal P ∈ Ass R[X ] (R[X ]/H ). When H is exclusive, HK := H ⊗ R K ⊂ K [X ]. Since HK =

is a principal ideal of K [X ], we can write HK = ϕ H (X )K [X ] for some monic polynomial ϕ H (X ) ∈ K [X ]. Note that the monic polynomial ϕ H (X ) is uniquely determined by H . Let d =deg ϕ H (X ) and put ϕ H (X ) = X d +η1 X d−1 +· · ·+ηd d with ηi ∈ K . Let I H := i=1 Iηi (= R[X ] : R ϕ H (X )) and J H := c(I H ϕ H (X )). Then I H ϕ H (X )R[X ] ⊆ R[X ]. For f (X ) ∈ K [X ], c( f (X )) denotes the content ideal of f (X ), the fractional ideal of R generated by the coefficients of f (X ). Definition 10.1.1

Let H be an exclusive ideal of R[X ].

(1) The ideal H is called an anti-integral ideal of R[X ] or is of anti-integral type if H = I H ϕ H (X )R[X ]. (2) The ideal H is called a super-primitive ideal of R[X ] or is of superprimitive type if grade(I H c(ϕ H (X ))) > 1. Remark 10.1.2 Assume that the ideal H of R[X ] is exclusive. Then HK ∩ R[X ] = H . In particular, the inclusion, I H ϕ H (X )R[X ] ⊆ H holds. Indeed since HK ⊇ H asserts the inclusion, H ⊆ HK ∩ R[X ]. Conversely, take any f (X ) ∈ ϕ H K [X ] ∩ R[X ]. Then f (X ) = ϕ H (X )ζ (X ) for some ζ (X ) ∈ K [X ]. Since f (X ) ∈ HK , there exists a nonzero a ∈ R such that a f (X ) ∈ H . Since H is exclusive over R, we have that a ∈ P for each P ∈ Ass R[X ] (R[X ]/H ). Thus f (X ) ∈ H , which implies that H ⊇ HK ∩ R[X ]. Proposition 10.1.3 Assume that H is an exclusive ideal of R[X ]. If H is generated by some polynomials of the least degree, then H is of anti-integral type. Proof Let d denote the least degree of a polynomial in H . Then ϕ H (X ) = X d + η1 X d−1 + · · · + ηd (ηi ∈ K ). Since aϕ H (X ) ∈ R[X ] for some a ∈ R \ {0}. Take f (X ) = a0 X d + a1 X d−1 + · · · + ad ∈ H with ai ∈ R. Since f (X ) ∈ H , we have ai /a0 = ηi (1 ≤ i ≤ d). So a0 ∈ I H and hence f (X ) = a0 ϕ H (X ). Thus H ⊆ I H ϕ H (X )R[X ]. Since I H ϕ H (X )R[X ] ⊆ H by Remark 10.1.2, we have H = I H ϕ H (X )R[X ], which means that H is an anti-integral ideal. Let f (X ) = a0 X n + a1 X n−1 + · · · + an be a polynomial in R[X ]. We say that f (X ) is a Sharma polynomial in R[X ] if there does not exist t ∈ R with t ∈ a0 R such that tai ∈ a0 R for 1 ≤ i ≤ n.

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235

We require the following two lemmas seen in Chapter 3 (Theorem 2.1.2 and Theorem 2.1.8): Lemma 10.1.4 Let f (X ) ∈ R[X ]. Then f (X ) is a Sharma polynomial if and only if c( f (X )) ⊆ P for any P ∈ Dp1 (R). Lemma 10.1.5 Assume that an ideal H is exclusive and that deg ϕ H (X ) = d. The following statements are equivalent: (i) H is a principle ideal of R[X ]; (ii) I H is a principal ideal of R; (iii) there exists a Sharma polynomial in H of degree d. If one of the above conditions holds, then H is generated by a Sharma polynomial. Proof (iii) ⇒ (i): Let f (X ) be a Sharma polynomial in H of degree d. Since deg ϕ H (X ) = d, this Sharma polynomial has the least degree. So by [Sh], H is principal. (i) ⇒ (ii): Let H = f (X )R[X ]. Then f (X )R[X ] ⊇ I H ϕ H (X )R[X ]. Note that H ⊗ R K = f (X )K [X ] = ϕ H (X )K [X ] and hence deg f (X ) = deg ϕ H (X ) = d. Take a ∈ I H . Then aϕ H (X ) = b f (X ) for some b ∈ R. Let f (X ) = a0 X d + · · · + ad with ai ∈ R. Then a = ba0 , so that I H ⊇ a0 R for some b ∈ R. Since ba0 ηi = aηi = bai (1 ≤ i ≤ d), we have a0 ηi = ai ∈ R. Hence a0 ∈ I H , which implies that I H = a0 R. (ii) ⇒ (iii): Let I H = b R. Then I H ϕ H (X )R[X ] = bϕ H (X )R[X ] ⊆ H and bηi ∈ R (1 ≤ i ≤ d). Suppose that there exists t ∈ b R with tbηi ∈ b R (1 ≤ i ≤ d). Then tηi ∈ R and hence t ∈ I H = b R, a contradiction. Thus bϕ H (X ) ∈ R[X ] is a Sharma polynomial of degree d. Proposition 10.1.6 Assume that H is an exclusive ideal. If H is a superprimitive ideal of R[X ], then H is an anti-integral ideal of R[X ]. Proof Since H is exclusive, H ⊇ I H ϕ H (X )R[X ] by Remark 10.1.2. Take f (X ) ∈ H . Since f (X ) ∈ HK , there exists a nonzero element a ∈ R such that a f (X ) ∈ I H ϕ H (X )R[X ]. Take an arbitrary prime ideal p ∈ Dp1 (R). Then I H c(ϕ H (X )) ⊆ p because grade(I H ϕ H (X )R[X ]) > 1. So there exists a Sharma polynomial in H p of the least degree. By Lemma 10.1.4, we have H p = g(X )R p [X ] for some g(X ) = bϕ H (X ) and some b ∈ R. Since f (X ) ∈ H p , there exists c ∈ R \ p such that c f (X ) ∈ I H ϕ H (X )R[X ]. Thus f (X ) ∈ I H ϕ H (X )R[X ].

236

Anti-Integral Ideals and Super-Primitive Polynomials

Remark 10.1.7 If H is a super-primitive ideal of R[X ], then grade(c(H )) > 1. But the converse statement is not necessarily valid as can be seen in the example below. Example 10.1.8 Consider an integral domain R satisfying the condition: R⊂ R, where R denotes the integral closure of R in K . Take α ∈ R but α ∈ R. =

Then we have an exact sequence: 0 → H → R[X ] → R[α] → 0 Note that H is a prime ideal of R[X ] because R[α] ∼ = R[X ]/H is an integral domain. In this case, H contains a monic polynomial (i.e., a polynomial giving an integral dependence of α). Hence c(H ) = R. Suppose that H is a superprimitive ideal, then H is an anti-integral ideal by Proposition 10.1.6. Since H = I[α] ϕα (X )R[X ], we have J[α] = I[α] c(ϕα (X )) = c(H ) = R. Thus R[α] is flat over R (cf. Theorem 2.3.6). Since α is integral and flat in K , R[α] = R, that is, α ∈ R, which is a contradiction. Therefore H is not a super-primitive ideal. Theorem 10.1.9

Assume that H is an exclusive ideal of R[X ].

(1) If H is of anti-integral type and grade(c(H )) > 1, then H is of superprimitive type. (2) If H is of anti-integral type and H contains a Sharma polynomial, then H is of super-primitive type. Proof Note that H = I H ϕ H (X )R[X ] by the assumption. Since c(H ) = I H c(ϕ H (X )), (1) is valid. The statement (2) follows from Lemma 10.1.4.

Proposition 10.1.10 Let f (X ) be a Sharma polynomial. Then f (X )R[X ] does not have any embedded prime divisor. Proof Let P be a prime divisor of f (X )R[X ]. Then depth(R[X ] P ) = 1. Suppose that p := P ∩ R = (0). Then depth(R p ) = 1 and P = p R[X ]. Thus f (X ) ∈ P = p R[X ], which implies that c( f (X )) ⊆ p. Hence grade(c( f (X ))) = 1. This contradicts the assumption that f (X ) is a Sharma polynomial. So we have P ∩ R = (0). Since P = PK ∩ R[X ] and ht(P) = 1, f (X )R[X ] has no embedded prime divisor. Remark 10.1.11 Let f (X ) be a Sharma polynomial in R[X ] and P be a prime divisor of f (X )R[X ]. It does not necessarily follow that P is of

10.1 Anti-Integral Ideals and Super-Primitive Ideals

237

super-primitive type. But if P is of anti-integral type, then P is of super-primitive type. Indeed, consider the exact sequence: 0 → P → R[X ] → R[α] → 0 where α denotes X mod.P. In this case, since P contains a Sharma polynomial f (X ), α is a super-primitive element over R by Theorem 2.2.8. Next consider the case R = R. If P ∩ R = (0), then P is of super-primitive type. Let H be an exclusive ideal of R[X ]. Define J H := I H c(ϕ H (X )), an ideal of R. Theorem 10.1.12

Let H be an exclusive ideal of R[X ].

(1) If J H = R, then R[X ]/H is flat over R. (2) If J H = R, then H is of super-primitive type and I H is an invertible ideal of R. Proof (1) Take p ∈ Spec(R). Since R = J H = I H c(ϕ H (X )), there exists b ∈ I H such that (I H ) p is a principal ideal b R p . So (I H ) p = bϕ H (X )R p [X ]. It follows that R p [X ]/(I H ) p = R p [X ]/bϕ H (X )R p [X ]. Thus R p [X ]/bϕ H (X ) R p [X ] is flat over R p by [M1, (20.F)] because R p = (J H ) p = c(bϕ H (X )) p . Hence R[X ]/H is flat over R. (2) Since R = J H = I H c(ϕ H (X )), I H is an invertible ideal of R and H is a super-primitive ideal by definition. Corollary 10.1.13 Let H be an exclusive ideal of R[X ]. If J H = R, then H = I H ϕ H (X )R[X ] and H is an invertible ideal of R[X ]. Furthermore H  ( > 0) is also an invertible ideal of R[X ]. Proof By Theorem 10.1.12(2), H is of super-primitive type and hence H is of anti-integral type by Proposition 10.1.6. So H = I H ϕ H (X )R[X ], which is an invertible ideal of R[X ]. Remark 10.1.14 Assume that H satisfies the condition in Corollary 10.1.13. Let J := {a ∈ R | aϕ H (X ) ∈ R[X ]}, an ideal of R. Then H  = J ϕ H (X ) R[X ] and J = (I H ) . Proposition 10.1.15 Let f (X ) ∈ R[X ] be a Sharma polynomial. Assume that f (X ) is irreducible in K [X ]. Then f (X )R[X ] is a prime ideal in R[X ]. Proof Let P be a prime divisor of f (X )R[X ]. Then f (X ) ∈ ϕ P (X )K [X ]. Since f (X ) is irreducible, there exists a ∈ R such that f (X ) = aϕ P (X ) (a ∈ P).

238

Anti-Integral Ideals and Super-Primitive Polynomials

Since grade(c( f (X ))) > 1, P is of super-primitive type. So P is of anti-integral type by Proposition 10.1.6 and hence P = I P ϕ P (X )R[X ]. Thus f (X )R[X ] = P. Proposition 10.1.16 Let f (X ) be a Sharma polynomial in R[X ]. Let P be a prime ideal in R[X ] such that P is a unique prime divisor of the ideal f (X )R[X ]. Then P is of super-primitive type. Proof By the assumption, f (X )K [X ] = ϕ P (X ) K [X ]. Since f (X ) is a Sharma polynomial, we have f (X )K [X ] ∩ R[X ] = f (X )R[X ] (cf. [Sh]). Hence f (X )R[X ] ⊆ (I P ϕ P (X )) R[X ]. So, since grade(c( f (X ))) > 1, we have grade(I P ϕ P (X )) > 1. Hence P is of super-primitive type. d−1 Lemma 10.1.17 Let ϕ(X ) = X d +η1 X +· · · ηd (ηi ∈ K ) be a monic irred R ducible polynomial in K [X ]. Put Iϕ(X ) := i=1 Iηi (= R[X ] : R ϕ(X )). If it holds R R that grade(Iϕ(X c(ϕ(X ))) > 1, then ϕ(X )K [X ] ∩ R[X ] = Iϕ(X ) ) ϕ(X )R[X ] ∈ Spec(R[X ]).

Proof Since ϕ(X )K [X ] ∈ Spec(K [X ]), we have P := ϕ(X )K [X ] ∩ R[X ] ∈ R Spec(R[X ]). By construction, P ⊇ Iϕ(X ) R[X ]. By the assumption, R grade(Iϕ(X ) c(ϕ(X ))) > 1 R Thus P is of super-primitive type. So we obtain P = Iϕ(X ) ϕ(X )R[X ] ∈ Spec(R[X ]).

Proposition 10.1.18 Let f (X ) be a Sharma polynomial in R[X ] and let f (X ) = aϕ1 (X )e1 · · · ϕt (X )et be a product of irreducible polynomials ϕi (X ) ∈ K [X ] and a ∈ R. Then f (X )R[X ] = (ϕ1 (X )e1 K [X ] ∩ R[X ]) ∩ · · · ∩ (ϕt (X )et K [X ] ∩ R[X ]) is a primary decomposition. Proof

Since f (X ) is a Sharma polynomial, we have

f (X )R[X ]

= =

f (X )K [X ] ∩ R[X ] ϕ1 (X )e1 · · · ϕt (X )et K [X ] ∩ R[X ]

= =

(ϕ1 (X )e1 K [X ] ∩ · · · ∩ (ϕt (X )et K [X ]) ∩ R[X ] (ϕ1 (X )e1 K [X ] ∩ R[X ]) ∩ · · · ∩ (ϕt (X )et K [X ] ∩ R[X ])

10.2 Super-Primitive Polynomials and Sharma Polynomials

239

Since f (X ) ∈ ϕi (X )K [X ] ∩ R[X ], the prime ideal ϕi (X )K [X ] ∩ R[X ] is of super-primitive type. So ϕi (X )K [X ] ∩ R[X ] = IϕRi (X ) R[X ] by Lemma 10.1.17. ϕi (X )ei K [X ] ∩ R[X ] is an IϕRi (X ) R[X ]-primary ideal. Proposition 10.1.19 Let I ⊆ J be ideals of R[X ]. Assume that I ⊗ R K = J ⊗ R K and that, for each p ∈ Dp1 (R), J R p contains a Sharma polynomial over R p of the least degree. Then I = J . Proof Take f (X ) ∈ I . Then f (X ) ∈ I ⊆ I ⊗ R K = J ⊗ R K . Hence there exists a ∈ R such that a f (X ) ∈ J . Let p be a prime divisor of a R. Then p ∈ Dp1 (R). By the assumption, J p = g(X )R p [X ], c(g(X )) = R p for some g(X ) ∈ J . Put f (X ) = g(X )ζ (X ) with ζ (X ) ∈ R p [X ]. Then there exists b ∈ R \ p such that b f (X ) ∈ J . Since a, b is a regular sequence, we have f (X ) ∈ J .

10.2

Super-Primitive Polynomials and Sharma Polynomials

R Definition 10.2.1 Let ϕ(X ) be a monic polynomial in K [X ] and Iϕ(X ) := R[X ] : R ϕ(X ). The polynomial ϕ(X ) is called a super-primitive polynomial if grade(IϕR c(ϕ(X ))) > 1.

Let R denote the integral closure of R in K and C(R/R) denote the conductor ideal between R and R. For an element η ∈ K , we put Iη := {a ∈ R | aη ∈ R}. Remark 10.2.2

Assume that R⊂ R. −

(1) There exists a polynomial in K [X ], which is not super-primitive. In fact, take η ∈ R \ R and ϕ(X ) := X − η. Then ϕ(X ) is not super-primitive (cf. Example 10.2.9). (2) When η ∈ K satisfies grade(Iη + C(R/R)) > 1, the polynomial X − η is super-primitive. More generally, let ϕ(X ) = X d + η1 X d−1 + ·d · · + ηd (ηi ∈ K ) satisfy grade(Iηi +C(R/R)) > 1 for all i. Then grade( i=1 Iηi + R C(R/R)) = grade(Iϕ(X + C(R/R)) > 1. Hence ϕ(X ) is super-primitive ) (cf. Proposition 10.2.10).

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Proposition 10.2.3 Let B be an ideal of R. Assume that every prime divisor of B is contained in Dp1 (R). Let a, b be a regular sequence and f ∈ R. If a f, b f ∈ B, then f ∈ B. Proof

Consider a primary decomposition of B: B = q1 ∩ · · · ∩ qn

√ where qi = pi . Then pi ∈ Dp1 (R) for all 1 ≤ i ≤ n. Take pi . Then a, b is an R-regular sequence. So a ∈ pi or b ∈ pi . If a ∈ pi , then f ∈ qi R pi ∩ R = qi . A similar argument is applicable to the case b ∈ pi . Hence we conclude that f ∈ B. Proposition 10.2.4 Let ϕ(X ) ∈ K [x] be an irreducible super-primitive polyR R nomial. Let I := Iϕ(X ) and J := Iϕ(X ) ( > 0). Then I ϕ(X )R[X ] is a prime  ideal of R[X ] and J ϕ(X ) R[X ] is I ϕ(X )R[X ]-primary ideal. Proof Since ϕ(X ) K [X ] is ϕ(X )K [X ]-primary, the ideal ϕ(X ) K [X ]∩ R[X ] is ϕ(X )K [X ] ∩ R[X ]-primary. ϕ(X ) is super-primitive, and hence anti-integral. So ϕ(X )K [X ] ∩ R[X ] = I ϕ(X )R[X ]. Thus we have only to prove that ϕ(X ) K [X ] ∩ R[X ] = J ϕ(X ) R[X ]. The implication (⊇) is obvious. We shall show that the implication (⊆). Since J is a denominator ideal in R, J is a divisorial ideal. So J R[X ](⊆ R[X ]) is also a divisorial ideal. Hence J ϕ(X ) R[X ](⊆ R[X ] is a divisorial ideal. Thus any prime divisor of J ϕ(X ) R[X ] is of depth 1. Take f (X ) ∈ ϕ(X ) K [X ] ∩ R[X ]. Then f (X )/ϕ(X ) ∈ K [X ]. We must show that f (X )/ϕ(X ) ∈ J R[X ]. Let p be any prime divisor of J . Then p ∈ Dp1 (R). In this case, we have to show f (X )/ϕ(X ) ∈ (J R[X ]) p . Since (I ϕ(X )) R[X ] ⊆ J ϕ(X ) R[X ], we have grade(J c(ϕ(X ) )) > 1. Hence J p = a R p for some a ∈ J . Put g(X ) = aϕ(X ) . Then g(X ) ∈ J and c(g(X ))R p = R p , which shows f (X ) ∈ g(X )R p [X ]. Thus f (X )/ϕ(X ) ∈ (g(X/ϕ(X ) )R p [X ] = a R p [X ] = (J R[X ]) p . From this, we get ϕ(X ) K [X ] ∩ R[X ] = J ϕ  R[X ]. By Lemma 10.2.17, I ϕ(X )R[X ] ∈ Spec(R[X ]). Therefore we have J ϕ(X ) R[X ] is an I ϕ(X )R[X ]-primary ideal. Theorem 10.2.5 Let ϕ1 (X ), . . ., ϕt (X ) ∈ K [X ] be irreducible superprimitive polynomials. Let Ii := IϕRi (X ) (1 ≤ i ≤ t) and let Ji := IϕRi (X )ei (1 ≤ i ≤ t, ei > 0). Then ϕ1 (X )e1 · · · ϕt (X )et K [X ] ∩ R[X ] =

t

(Ji ϕi (X )ei R[X ])

i=1

and this is a primary decomposition of the left side ideal, where Ji ϕi (X )ei R[X ] is an Ii ϕi (X )R[X ]-primary ideal for each 1 ≤ i ≤ t.

10.2 Super-Primitive Polynomials and Sharma Polynomials Proof

241

It follows that ϕ1 (X )e1 · · · ϕt (X )et K [X ] ∩ R[X ] =

(ϕ1 (X )e1 K [X ] ∩ · · · ∩ ϕt (X )et K [X ]) ∩ R[X ]

=

(ϕ1 (X )e1 K [X ] ∩ R[X ]) ∩ · · · ∩ (ϕt (X )et K [X ] ∩ R[X ] ∩ R[X ])

=

J1 ϕ1 (X )e1 R[X ] ∩ · · · ∩ Jt ϕ(X )et R[X ]

by Lemma 10.2.17. By Proposition 10.2.4, Ji ϕi (X )ei R[X ] is an Ii ϕi (X )R[X ]primary ideal for each 1 ≤ i ≤ t. Proposition 10.2.6 Let ϕ1 (X ), . . ., ϕt (X ) ∈ K [X ] be irreducible superprimitive polynomials. Then ϕ1 (X )e1 · · · ϕt (X )et K [X ] ∩ R[X ] p is a principal ideal at each p ∈ Dp1 (R). Proof Let J := R[X ] : R ϕ1 (X )e1 · · · ϕt (X )et and Ii := R[X ] : R ϕi (X ). Since ϕi (X ) is a super-primitive polynomial, we have grade(Ii c(ϕi (X ))) > 1 for each 1 ≤ i ≤ t. Since (I1 )e1 · · · (It )et ⊆ J , we have grade(J c(ϕ1 (X )e1 · · · ϕt (X )et ) > 1. Take p ∈ Dp1 (R). There exists f i (X ) ∈ Ii ϕi (X ) such that c( f i (X )) ⊆ p for 1 ≤ i ≤ t. Put f (X ) := f 1 (X )e1 · · · f t (X )et . Then c( f i (X )) ⊆ p. Since f (X ) ∈ ϕ1 (X )e1 · · · ϕt (X )et K [X ] ∩ R[X ] and f (X ) has the least degree, we have (ϕ1 (X )e1 · · · ϕt (X )et K [X ] ∩ R[X ]) p = f (X )R[X ] p by Corollary 2.1.6. R Remark 10.2.7 Let ϕ(X ) ∈ K [X ] be a monic polynomial and I := Iϕ(X ) . If I ϕ(X )R[X ] contains a Sharma polynomial, then ϕ(X ) is a super-primitive polynomial. Indeed, take a Sharma polynomial f (X ) ∈ I ϕ(X )R[X ]. Suppose there exists p ∈ Dp1 (R) such that I c(ϕ(X )) ⊆ p. Then I ϕ(X )R[X ] ⊆ p R[X ] and hence f (X ) ∈ p R[X ]. Thus c( f (X )) ⊆ p, which contradicts the assumption that f (X ) is a Sharma polynomial.

Theorem 10.2.8 Let ϕ(X ) ∈ K [X ] be a monic polynomial. Then ϕ(X ) is R a super-primitive polynomial if and only if Iϕ(X ) ϕ(X )R[X ] contains a Sharma polynomial. Proof The implication (⇐) follows from Remark 10.2.7. R (⇒) Take a nonzero element a ∈ Iϕ(X ) and put f 1 (X ) = aϕ(X ). Let p1 , . . ., pn ∈ Dp1 (R) such that pi ⊇ c( f 1 (X )). Since ϕ(X ) is a super-primitive polynomial, R R we have Iϕ(X ) c(ϕ(X )) ⊆ p1 , . . ., pn . There exists gi (X ) ∈ Iϕ(X ) ϕ(X )R[X ]

242

Anti-Integral Ideals and Super-Primitive Polynomials

such that c(gi (X )) ⊆ pi (1 ≤ i ≤ n). Put f (X ) = f 1 (X )gi (X ) · · · gn (X ). In this case, c( f (X )) ⊆ c( f 1 (X )), c(gi (X )) (1 ≤ i ≤ n). Thus for each p ∈ Dp1 (R), c( f (X )) ⊆ p. Hence f (X ) is a Sharma polynomial and f (X ) ∈ R Iϕ(X ) ϕ(X )R[X ]. The following example shows that an irreducible factor ϕ(X ) in K [X ] of a Sharma polynomial f (X ) is not always a super-primitive polynomial. Example 10.2.9

Assume that R⊂ R. Take α ∈ R \ R. Since α is integral =

over R, there exists a monic polynomial in R[X ] such that f (α) = 0. Note that c( f (X )) = R because f (X ) is a monic. Thus f (X ) is a Sharma polynomial. It is obvious that X − α is a factor of f (X ) in K [X ]. Suppose X − α is anti-integral. Then f (X ) ∈ (X − α)K [X ] ∩ R[X ] = Iα (X − α)R[X ]. So R = c( f (X )) ⊆ Iα c(X − α), which means that R[α] ∼ = R[X ]/(X − α) is flat over R (cf. Theorem 2.3.6). Since α is integral over R, we have R[α] = R, that is, α ∈ R, which is a contradiction. Therefore the polynomial X − α is not an anti-integral polynomial, and hence X − α is not a super-primitive polynomial. Proposition 10.2.10 (1) (2)

Let ϕ(X ) be a monic polynomial in K [X ].

R R If grade(Iϕ(X ) + C(R/R)) > 1, then grade(Iϕ(X ) c(ϕ(X ))) > 1. R R grade(Iϕ(X ) c(ϕ(X ))) > 1 if and only if Iϕ(X ) ϕ(X )R[X ] contains a Sharma

polynomial. R (3) If R = R, then grade(Iϕ(X ) c(ϕ(X ))) > 1.

R R R Proof (1) Take p ∈ Dp1 (R). If Iϕ(X ) ⊆ p, then Iϕ(X ) ⊆ Iϕ(X ) c(ϕ(X )). So R R Iϕ(X ) ⊆ p. On the other hand, if Iϕ(X ) ⊆ p, then C(R/R) ⊆ p by the assumption R and hence R p is a DVR. Let v p ( ) denote the valuation of R p . If v p (Iϕ(X ) ) = e, R e then (Iϕ(X ) ) p = p R p and there exists i such that V p (ηi ) = −e > 0, where R ϕ(X ) = X d + η1 X d−1 + · · · + ηd with η j ∈ K . Thus v p (Iϕ(X ) ηi ) = 0. R R Hence (Iϕ(X ) ηi ) p ⊆ p R p . Thus (Iϕ(X ) c(ϕ(X ))) p ⊆ p R p . This shows that R R Iϕ(X ) c(ϕ(X )) ⊆ p Therefore we obtain that grade(Iϕ(X ) c(ϕ(X ))) > 1. R (2)(⇒): Take a ∈ Iϕ(X ) and put f (X ) = aϕ(X ). If f (X ) is a Sharma polynomial, then there is nothing to prove. Suppose that f (X ) is not a Sharma polynomial. Then there exists a prime ideal p in Dp1 (R) such that c( f (X )) ⊆ p. Take a nonzero element b ∈ c( f (X )). Then p is a prime divisor of the ideal b R because p ∈ Dp1 (R). Since the number of prime divisors of b R is finite, the number of prime ideals in Dp1 (R) containing c( f (X )) is also finite. Let p1 , . . ., pn be the R prime ideals containing c( f (X )). Since Iϕ(X ) c(ϕ(X )) ⊆ pi for all 1 ≤ i ≤ n

10.3 Anti-Integral, Super-Primitive, or Flat Polynomials

243

R by the assumption, there exists gi (X ) ∈ Iϕ(X ) ϕ(X ) such that gi (X ) ∈ pi R[X ] d for each 1 ≤ i ≤ n. Put g(X ) := f (X ) + X g1 (X ) + · · · + X nd gn (X ). Then n c(g(X )) = i=1 (c(gi (x)) + c( f (X )) ⊆ p for all p ∈ Dp1 (R). Thus g(X ) is a Sharma polynomial. R (2)(⇐): Let f (X ) be a Sharma polynomial in Iϕ(X ) ϕ(X )R[X ]. Then we can ex R press f (X ) = f i (X )gi (X ) with f (X ) ∈ Iϕ(X ) ϕ(X ) and gi (X ) ∈ R[X ]. R Suppose that there exists p ∈ Dp1 (R) such that Iϕ(X ) c(ϕ(X )) ⊆ p. Then f i (X ) ∈ p R[X ] and hence f (X ) ∈ p R[X ], which contradicts the assumption that f (X ) is a Sharma polynomial. (3) follows from C(R/R) = R and (1).

In Proposition 10.2.10, the converse implication of (1) is not always valid as can be seen in the following example. Example 10.2.11 Let k be a field and t be an indeterminate. Let R := k[t 2 , t 3 ]. R Then R = k[t]. Let ϕ(X ) := X 2 + X + 1/t 2 ∈ K [X ] := k(t)[X ]. Then Iϕ(X ) = 2 R 2 2 R t R and Iϕ(X ) c(ϕ(X )) = t (1, 1/t )R = R. So we have grade(Iϕ(X ) c(ϕ(X ))) = R 2 grade(R) > 1, but grade(Iϕ(X ) +C(R/R)) = grade(t R+t R) = grade(t R) = 1. Remark 10.2.12

Consider a Noetherian domain R satisfying R⊂ R and let −

ϕ(X ) = X + η1 X + · · · + ηd ∈ K [X ]. Assume that every ηi ∈ R and R that η j ∈ R for some j. Then Iϕ(X ) ϕ(X )R[X ] does not contain any Sharma polynomial. d

10.3

d−1

Anti-Integral, Super-Primitive, or Flat Polynomials

Definition 10.3.1 Let ϕ(X ) ∈ K [X ] be a monic polynomial. We say that R ϕ(X ) is an anti-integral polynomial if ϕ(X )K [X ] ∩ R[X ] = Iϕ(X ) ϕ(X )R[X ]. We say that ϕα (X ) is a super-primitive polynomial if ϕα (X ) is anti-integral and R grade(Iϕ(X ) c(ϕ(X ))) > 1

Proposition 10.3.2 Let ϕ(X ) ∈ K [X ] be a monic polynomial. If the ideal R R Iϕ(X ) ϕ(X )R[X ] contains a monic polynomial in R[X ], then Iϕ(X ) = R, i.e., ϕ(X ) ∈ R[X ].

244

Anti-Integral Ideals and Super-Primitive Polynomials

R Proof Let f (X ) ∈ Iϕ(X )R[X ] be a monic polynomial in R[X ]. Then ) ϕ(X   R f (X ) = ai ϕ(X )gi (X ) = ( ai gi (X ))ϕ(X ) with some ai ∈ Iϕ(X ) and some gi (X ) ∈ R[X ]. Comparing the coefficients of the leading terms, we conclude R that Iϕ(X ) = R.

Corollary 10.3.3 Let ϕ(X ) ∈ K [X ] be a monic polynomial. Assume that ϕ(X ) is a factor of a monic polynomial in R[X ] and that ϕ(X ) is an antiintegral polynomial. Then ϕ(X ) ∈ R[X ]. Proof By the assumption, there exists a monic polynomial f (X ) ∈ ϕ(X )K [X ] ∩R[X ]. So we have f (X ) ∈ ϕ(X )K [X ] ∩ R[X ] = Iϕ(X ) ϕ(X )R[X ]. Hence ϕ(X ) ∈ R[X ] by Proposition 10.3.2. The following example shows that every polynomial ϕ(X ) ∈ K [X ] is not always an anti-integral polynomial. Assume that R⊂ R and take η ∈ R \ R. Then ϕ(X ) := − X − η is not an anti-integral polynomial. Indeed, Proposition 10.3.2 shows that ϕ(X ) = X − η is an anti-integral polynomial if and only if η ∈ R. So if ϕ(X ) is an anti-integral polynomial, then η ∈ R, which is a contradiction.

Example 10.3.4

It is easy to see the following remarks by definition. Remark 10.3.5 Let ϕ(X ) ∈ K [X ] be a monic polynomial and H := ϕ(X )K [X ] ∩ R[X ], an ideal of R[X ]. (1) H is an exclusive ideal in R[X ]. (2) ϕ(X ) ∈ K [X ] is an anti-integral polynomial if and only if H is an antiintegral ideal of R[X ]. (3) ϕ(X ) ∈ K [X ] is a super-primitive polynomial if and only if H is a super-primitive ideal of R[X ]. Remark 10.3.6 When H is a prime ideal, H is an anti-integral (resp. a super-primitive) ideal if and only if the extension R[α] of R is an anti-integral (resp. a super-primitive) extension of R as in Chapter 2, where α denotes the residue class of X in R[X ]/H . Proposition 10.3.7 Let ϕ(X ) ∈ K [X ] be a monic polynomial. If ϕ(X ) is a super-primitive polynomial, then ϕ(X ) is an anti-integral polynomial.

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Proof Since ϕ(X ) is super-primitive, H := ϕ(X )K [X ] ∩ R[X ] is a superprimitive ideal of R[X ]. Note that the ideal H is exclusive. Hence the conclusion follows from Proposition 10.1.6. Lemma 10.3.8 If both f (X ) and g(X ) ∈ R[X ] are Sharma polynomials, then f (X )g(X ) is a Sharma polynomial. Proof Suppose that there exists p ∈ Dp1 (R) such that f (X )g(X ) ∈ p R[X ]. Then f (X ) ∈ p R[X ] or g(x) ∈ p R[X ] and hence c( f (X )) ⊆ p or c(g(X )) ⊆ p, which is a contradiction by Lemma 10.1.6. Theorem 10.3.9 mials. Then

Let ϕ1 (X ), . . ., ϕn (X ) ∈ K [X ] be super-primitive polyno-

(1) ϕ(X ) := ϕ1 (X ) · · · ϕn (X ) ∈ K [X ] is a super-primitive polynomial; (2) if ϕ(X ) ∈ R[X ], ϕi (X ) ∈ R[X ] for each 1 ≤ i ≤ n. Proof (1) Since ϕi (X ) is super-primitive, we have IϕRi (X ) c(ϕi (X )) ⊆ p for every p ∈ Dp1 (R). Take p ∈ Dp1 (R). Then there exists ai ∈ IϕRi (X ) such that c(ai ϕi (X )) ⊆ p. Thus ai ϕi (X ) is a Sharma polynomial in R p [X ]. Put a := R a1 · · · an . Then a ∈ Iϕ(X ) and aϕ(X ) = (a1 ϕ1 (X )) · · · (an ϕn (X )). So by Lemma 10.3.8, aϕ(X ) is a Sharma polynomial in R p [X ] and hence c(aϕ(X )) ⊆ p. R Therefore we have grade(Iϕ(X ) c(ϕ(X ))) > 1. So aϕ(X ) is a super-primitive polynomial. (2) follows from Proposition 10.3.7 and Corollary 10.3.3. Question. Is the similar statement to Proposition 10.3.9 valid for anti-integral polynomials? The following proposition asserts that a partial answer to the above question is valid. Proposition 10.3.10 Let ϕ1 (X ), . . ., ϕn (X ) ∈ K [X ] be anti-integral polynomials. If ϕ(X ) := ϕ1 (X ) · · · ϕn (X ) ∈ R[X ] is a monic polynomial, then ϕi (X ) ∈ R[X ] for all 1 ≤ i ≤ n. Proof Since ϕi (X ) is an anti-integral polynomial, ϕ(X ) ∈ ϕi (X )K [X ] ∩ R[X ] = IϕRi (X ) ϕi (X )R[X ]. So IϕRi (X ) = R by Proposition 10.3.2. Thus we conclude that ϕi (X ) ∈ R[X ].

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Example 10.3.11 Let R = k[t 2 , t 3 ], where k is a field and t is an indeterminate. Let ϕ1 (X ) := X − t, ϕ2 (X ) := X + t ∈ K [X ]. Then neither ϕ1 (X ) nor ϕ2 (X ) is an anti-integral polynomial. But ϕ(X ) := ϕ1 (X )ϕ2 (X ) = (X − t)(X + t) = X 2 − t 2 is an anti-integral polynomial. Proposition 10.3.12 Let ϕ(X ) ∈ K [X ] and let ϕ(X ) = ϕ1 (X )e1 · · · ϕn (X )en be the irreducible decomposition in K [X ]. Assume that ϕi (X ) is a superR R e1 primitive polynomial for each 1 ≤ i ≤ n. Then Iϕ(X ) = (((Iϕ1 (X ) ) · · · (IϕRn (X ) )en )−1 )−1 . R R Proof (IϕR1 (X ) )e1 · · · (IϕRn (X ) )en ⊆ Iϕ(X ) is obvious. Since Iϕ(X ) is a divisorial R e1 R en −1 −1 R ideal, we have (((Iϕ1 (X ) ) · · · (Iϕn (X ) ) ) ) ⊆ Iϕ(X ) . Note that both sides of this implication are divisorial. Consider the localization at a prime ideal of depth 1. We may assume that (R, m) is a local domain with the maximal ideal m ∈ Dp1 (R). Since ϕi (X ) is a super-primitive polynomial, there exists ai ∈ IϕRi (X ) such that ai ϕi (X ) is a Sharma polynomial by Lemma 10.1.4. Put a := a1 · · · an . Then (a1 ϕ1 (X ))e1 · · · (an ϕn (X ))en = aϕ1 (X )e1 · · · ϕn (X )en is a Sharma R polynomial by Lemma 10.3.8. Thus Iϕ(X ) ϕ(X )R[X ] = aϕ(X )R[X ] and hence R R R Iϕ(X ) = a R. Therefore Iϕ(X ) = (((Iϕ1 (X ) )e1 · · · (IϕRn (X ) )en )−1 )−1 .

Definition 10.3.13 Let ϕ(X ) ∈ K [X ] be a monic polynomial. We say that R ϕ(X ) is a flat polynomial if Iϕ(X ) c(ϕ(X )) = R. Theorem 10.3.14

Let ϕ(X ), ψ(X ) be flat polynomials. Then

(1) ϕ(X )ψ(X ) is also a flat polynomial; R R R (2) Iϕ(X )ψ(X ) = Iϕ(X ) Iψ(X ) ; (3) ϕ(X ) is a super-primitive polynomial. Proof

Note that R R R R R Iϕ(X ) · Iψ(X ) ⊆ Iϕ(X ) ∩ Iψ(X ) ⊆ Iϕ(X )ψ(X )

R R R Hence Iϕ(X ) Iψ(X ) (ϕ(X )ψ(X ))R[X ] ⊆ Iϕ(X )ψ(X ) ϕ(X )ψ(X )R[X ]. R R R (1) Hence R = Iϕ(X ) Iψ(X ) c(ϕ(X )ψ(X )) ⊆ Iϕ(X )ψ(X ) c(ϕ(X )ψ(X )) ⊆ R. So ϕ(X )ψ(X ) is a flat polynomial. (2) By the argument in the proof of (1), c(ϕ(X )ψ(X )) is an invertible ideal of R R R R. Thus Iϕ(X )ψ(X ) = Iϕ(X ) Iψ(X ) . (3) follows from the definition.

10.3 Anti-Integral, Super-Primitive, or Flat Polynomials

247

Proposition 10.3.15 Let ϕ(X ) ∈ K [X ] be a monic polynomial and let ϕ(X ) = ϕ1 (X )e1 · · · ϕn (X )en be an irreducible decomposition in K [X ]. If ϕi (X ) R R e1 R en is a flat polynomial for each 1 ≤ i ≤ n, then Iϕ(X ) = (Iϕ1 (X ) ) · · · (Iϕn (X ) ) and n R R ei ei Iϕ(X ) ϕ(X )R[X ] = ( i=1 (Iϕi (X ) ) ϕi (X ) )R[X ] is a primary decomposition. Proof

This follows form Proposition 10.3.12 and Proposition 10.3.14 (1)(2).

Definition 10.3.16 Let ϕ(X ) ∈ K [X ] be a monic polynomial. We say that R ϕ(X ) is an ultra-primitive polynomial if grade(Iϕ(X ) + C(R/R)) > 1. Theorem 10.3.17 Let ϕ(X ), ψ(X ) ∈ K [X ] be monic polynomials. If both ϕ(X ) and ψ(X ) are ultra-primitive polynomials, then so is ϕ(X )ψ(X ). R R R R R R Proof Note that Iϕ(X ) · Iψ(X ) ⊆ Iϕ(X ) ∩ Iψ(X ) ⊆ Iϕ(X )ψ(X ) . Since grade(Iϕ(X ) + R R R C(R/R)) > 1 and grade(Iψ(X ) + C(R/R)) > 1, we have 1 < grade(Iϕ(X ) Iψ(X ) + R C(R/R)) ≤ grade(Iϕ(X )ψ(X ) + C(R/R)).

Remark 10.3.18 (1) We have the following implications: an ultra-primitive polynomial ⇒ a super-primitive polynomial ⇒ an anti-integral polynomial. The reverse implications are not valid in general. (2) Take η ∈ K . (i) X −η is a super-primitive polynomial if and only if grade(Iη (1, η)) > 1. (ii) X − η is an ultra-primitive polynomial if and only if grade(Iη + C(R/R)) > 1. (iii) X − η is a flat polynomial if and only if Iη (1, η) = R.

Definition 10.3.19

Let η ∈ K .

1) η is called a super-primitive element if grade(Iη (1, η)) > 1. 2) η is called an ultra-primitive element if grade(Iη + C(R/R)) > 1. 3) η is called an s-flat element if and only if Iη (1, η) = R. Remark 10.3.20 The above definition is the same as in Chapter 2, that is, η ∈ K is super-primitive over R if the extension R[η] is a super-primitive

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extension of R as in Chapter 2. An element η ∈ K is an s-flat element over R if and only if η is anti-integral over R and R[η] is a flat extension of R. Proposition 10.3.21 Let ϕ(X ) = X d + η1 X d−1 + · · · + ηd ∈ K [X ]. If ηi is an ultra-primitive element for each 1 ≤ i ≤ d, then ϕ(X ) is an ultra-primitive polynomial. d Proof Note that Iη1 · · · Iηd ⊆ i=1 Iηi . Since grade(Iηi +C(R/R)) > 1 by defd inition, grade(Iη1 · · · Iηd +C(R/R)) > 1 and hence grade( i=1 I +C(R/R)) > d ηi R R 1. Therefore grade(Iϕ(X ) + C(R/R)) > 1 because Iϕ(X ) = i=1 Iηi , which implies that ϕ(X ) is an ultra-primitive polynomial. In the notations as in Proposition 10.3.21, even if ηi is a super-primitive (respectively s-flat) element for each 1 ≤ i ≤ d, ϕ(X ) is not necessarily a super-primitive (respectively s-flat) polynomial. We see this in the following example. Example 10.3.22 Note first that an s-flat element (respectively an s-flat polynomial) is a super-primitive element (respectively a super-primitive polynomial). Let R := k[t 2 , t 3 ], where k is a field, and t be an indeterminate. Put η1 := 1/t 3 , η2 := 1/t 4 , and ϕ(X ) := X 2 +η1 X +η2 = X 2 +(1/t 3 )X +(1/t 4 ). Then Iη1 = t 3 R, Iη2 = t 4 R. Hence Iη1 (1, η1 ) = Iη2 (1, η2 ) = R. Thus η1 and η2 R are s-flat elements and hence they are super-primitive elements. We have Iϕ(X ) = 3 4 6 7 R R Iη1 ∩ Iη2 = t R ∩t R = (t , t )R. So we have Iϕ(X ) c(ϕ(X ) = Iϕ(X ) (1, η1 , η2 ) = R (t 6 , t 7 )(1, 1/t 3 , 1/t 4 )R = (t 2 , t 3 )R⊂ R. Thus grade(Iϕ(X ) c(ϕ(X ))) = 1, which − means that ϕ(X ) is not a super-primitive polynomial.

Chapter 11 Semi Anti-Integral and Pseudo-Simple Extensions

This chapter contains some supplementary results.

11.1

Anti-Integral Extensions of Polynomial Rings

Let R be a Noetherian domain. In Chapters 1 and 2, we investigated some properties of a simple algebraic extension R[α] in terms of ideals of R. Our objective of this section is to study about a simple birational extension R[X ][α] of a polynomial ring R[X ] by means of ideals of R, and hence α is an element in K (X ). We observe mainly what conditions are to be required for α being super-primitive over R[X ] or R[X ][α] being flat over R[X ]. In what follows, we fix the following notations unless otherwise specified: Let R denote a Noetherian domain, K denote its quotient field, and R[X ] denote a polynomial ring over R. Let η(X ) be an element in the quotient field K (X ) of R[X ] and put η(X ) = ψ(X )/φ(X ), where ψ(X ), φ(X ) ∈ K [X ] with (φ(X ), ψ(X ))K [X ] = K [X ] and φ(X ) monic. Note that expression η(X )/φ(X ) is unique. Let deg φ(X ) =: . We start with the following definition. Definition 11.1.1 We say that η(X ) satisfies the condition (F) over R if the ideal R[X ] : R[X ] η(X ) := { f (X ) ∈ R[X ] | f (X )η(X ) ∈ R[X ]} of R[X ] is generated by some polynomials of degree . For ζ (X ) ∈ K [X ], put IζR(X ) := {a ∈ R | aζ (X ) ∈ R[X ]}, which is an ideal of R. Proposition 11.1.2 The element η(X ) satisfies the condition (F) if and only R R if R[X ] : R[X ] η(X ) = (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ].

249

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Semi Anti-Integral and Pseudo-Simple Extensions

Proof ( ⇐ ) is trivial. ( ⇒ ) Since η(X ) satisfies the condition (F), R[X ] : R[X ] η(X ) is generated by some polynomials of degree . Take h(X ) ∈ R[X ] : R[X ] η(X ) with deg h(X ) = . Put g(X ) = h(X )η(X ) ∈ R[X ] ⊆ K [X ]. Then we have h(X )ψ(X ) ∈ φ(X )K [X ]. Since (φ(X ), ψ(X ))K [X ] = K [X ], it follows that h(X ) ∈ φ(X ) K [X ]. So deg h(X ) = deg φ(X ) =  implies that h(X ) = ζ φ(X ) with ζ ∈ K . Since φ(X ) is monic, the equality ζ φ(X ) = h(X ) ∈ R[X ] yields ζ ∈ R R Iφ(X ) ⊆ R. Note that g(X ) = h(X )η(X ) = ζ ψ(X ) ∈ R[X ]. Hence ζ ∈ Iψ(X ) . R R R R Thus ζ ∈ Iφ(X ) ∩ Iψ(X ) . Therefore we have h(X ) ∈ (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ]. R R Conversely the implication (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ] ⊆ R[X ] : R[X ] η(X ) is obvious. For f (X ) = a0 X n + a1 X n−1 + · · · + an ∈ K [X ], let c( f (X )) denote the fractional ideal (a0 , a1 , . . . , an )R of f (X ) in K . Lemma 11.1.3 Let f (X ), g(X ) ∈ R[X ] which are co-prime in K [X ], i.e., ( f (X ), g(X ))K [X ] = K [X ]. Then grade(( f (X ), g(X ))R[X ]) > 1 ⇔ grade((c( f (X )), c(g(X )))R) > 1 Proof ( ⇒ ) Suppose that (c( f (X )), c(g(X )))R ⊆ p for some p ∈ Dp1 (R). Then ( f (X ), g(X ))R[X ] ⊆ p R[X ] = P ∈ Dp1 (R[X ]), a contradiction. ( ⇐ ) Suppose that ( f (X ), g(X ))R[X ] ⊆ P for some P ∈ Dp1 (R[X ]). When P ∩ R = (0), there exists an irreducible polynomial ζ (X ) ∈ K [X ] such that P = ζ (X )K [X ] ∩ R[X ], which contradicts the assumption that f (X ), g(X ) are co-prime in K [X ]. Next consider the case P ∩ R = (0). Then P ∈ Dp1 (R[X ]) implies that p := P ∩ R ∈ Dp1 (R) and P = p R[X ]. In this case, we have (c( f (X )), c(g(X )))R ⊆ p, a contradiction. The following proposition shows that the condition (F) is determined by Dp1 (R). Proposition 11.1.4 η(X ) satisfies the condition (F) over R if and only if η(X ) satisfies the condition (F) over R p for each p ∈ Dp1 (R). Proof By Proposition 11.1.2, η(X ) satisfies the condition (F) over R if and R R only if R[X ] : R[X ] η(X ) = (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ]. Since R[X ] : R[X ] η(X ) R R and (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ] are divisorial ideals of R[X ] (and then contractions in R are of depth ≤ 1), this equality holds if and only if it holds by localizing at each p ∈ Dp1 (R).

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251

Lemma 11.1.5 Let φ(X ) ∈ K [X ] and let H be a divisorial ideal of R R satisfying H ⊆ Iφ(X ) . If grade(H c(φ(X ))) > 1, then φ(X )K [X ] ∩ R[X ] = H φ(X )R[X ]. R Proof Since H ⊆ Iφ(X ) , we have φ(X )K [X ]∩ R[X ] ⊇ H φ(X )R[X ]. We shall show the converse inclusion. For this, we have only to prove (1/φ(X ))(φ(X ) K [X ] ∩ R[X ]) ⊆ H R[X ]. Since H is a divisorial ideal of R, H R[X ] is also a divisorial ideal of R[X ]. So localizing at each p ∈ Dp1 (R), we need to prove the desired inclusion. Hence we may assume that (R, m) is a local ring of depth(R) = 1. By the assumption, H c(φ(X )) = R. Thus H is an invertible ideal and hence a principal ideal H = a R. Put f (X ) = aφ(X ). Then c( f (X )) = R. By [KY7], we have φ(X )K [X ] ∩ R[X ] = f (X )R[X ] = aφ(X )R[X ] = H φ(X )R[X ]. R Remark 11.1.6 It is shown that if grade(Iφ(X ) c(φ(X ))) > 1 then φ(X )K [X ]∩ R R[X ] = Iφ(X ) φ(X )R[X ] in [KY7]. In the above proof of Lemma 11.1.5, we used a little more generalized form of this result.

In the next proposition, we shall give a sufficient condition for η(X ) to satisfy the condition (F). R R Proposition 11.1.7 If grade((Iφ(X ) ∩ Iψ(X ) )c(φ(X ))) > 1, then η(X ) satisR fies the condition (F) over R. Consequently, R[X ] : R[X ] η(X ) = (Iφ(X ) ∩ R Iψ(X ) )φ(X )R[X ].

Proof By Proposition 11.1.2, we have only to show that R[X ] : R[X ] η(X ) = R R (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ]. The inclusion ⊇ is obvious. We shall show the converse inclusion. Take f (X ) ∈ R[X ] : R[X ] η(X ). Then g(X ) := f (X )η(X ) ∈ R[X ]. Hence f (X )ψ(X ) = g(X )φ(X ). Since φ(X ) and ψ(X ) are co-prime in R R K [X ], we have f (X ) ∈ φ(X )K [X ]∩ R[X ]. Since grade((Iφ(X ) ∩ Iψ(X ) )c(φ(X ))) R R > 1, we have φ(X )K [X ]∩ R[X ] = (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ] by Lemma 11.1.5.

R R Proposition 11.1.8 If grade((Iφ(X ) ∩ Iψ(X ) )c(ψ(X ))) > 1, then η(X ) satisfies the condition (F) over R.

Proof By Proposition 11.1.2, we need to show that R[X ] : R[X ] η(X ) = R R (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ]. Take f (X ) ∈ R[X ] : R[X ] η(X ). Then f (X )η(X ) =: g(X ) ∈ R[X ]. Since f (X )ψ(X ) = g(X )φ(X ) and (φ(X ), ψ(X ))K [X ] = K [X ], we have g(X ) ∈ ψ(X )K [X ] ∩ R[X ]. It follows that ψ(X )K [X ] ∩

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R R R R[X ] = (Iφ(X ) ∩ Iψ(X ) )ψ(X )R[X ] by Lemma 11.1.5 because grade((Iφ(X ) ∩ R R R Iψ(X ) )c(ψ(X ))) > 1. Thus g(X ) = bψ(X ) j(X ) for some b ∈ Iφ(X ) ∩ Iψ(X ), j(X ) ∈ R[X ]. So substituting this in f (X )ψ(X ) = g(X )φ(X ), we obtain that R R f (X ) = bφ(X ) j(X ) ∈ (Iφ(X ) ∩ Iψ(X ) )φ(X )R[X ].

Let A be a Noetherian domain and A[Y ] a polynomial ring. Let α be a nonzero element of an algebraic field extension L of the quotient field K (A) of A and π (A) : A[Y ] → A[α] be the A-algebra homomorphism sending Y to α. Let ϕα (Y )(A) be the monic minimal polynomial of α over K (A) with deg ϕα (Y )(A) = d and write ϕα (Y )(A) = Y d + η1 Y d−1 + · · · + ηd Then ηi (1 ≤ i ≤ d) are uniquely determined by α. Let IηAi := A : A ηi and
d A I[α] := i=1 IηAi , the latter of which is called a generalized denominator ideal of α. We say that α is an anti-integral element if and only if Ker(π (A) ) = A (A) I[α] ϕα (Y )A[Y ]. The concept of anti-integrality is given in Chapter 1 in the birational case, and the higher degree case appears in Chapter 2. For f (Y ) ∈ A[Y ], let c( f (Y )) denote the ideal of A generated by the coefficients of f (Y ). For an ideal J of A[Y ], let c(J ) denote the ideal generated by the coefficients of the elements in J . If α is an anti-integral element over A, then c(Ker(π (A) )) = A (A) A A A ϕα (Y )A[Y ]) = I[α] (1, η1 , . . . , ηd ). Put J[α] = I[α] (1, η1 , . . . , ηd ). If c(I[α] A J[α] ⊆ p for all p ∈ Dp1 (A) := { p ∈ Spec(A) | depth(R p ) = 1}, then α is called a super-primitive element over A. It is known that a super-primitive element is an anti-integral element (cf. Chapter 2). Remark 11.1.9 In the above notation, consider the case: A = R[X ] and A A A A α = η(X ). Then J[η(X )] = Iη(X ) + η(X )Iη(X ) = Iη(X ) (1, η(X ))R[X ], where A A I[η(X )] = Iη(X ) := R[X ] : R[X ] η(X ). If η(X ) satisfies the condition (F), we A R R A R have J[η(X )] = (Iφ(X ) ∩ Iψ(X ) )(φ(X ), ψ(X ))R[X ] because Iη(X ) = (Iφ(X ) ∩ R Iψ(X ) )φ(X )R[X ] by Proposition 11.1.2. Theorem 11.1.10

Assume that

R R R R grade((Iφ(X ) ∩ Iψ(X ) )c(φ(X )) + (Iφ(X ) ∩ Iψ(X ) )c(ψ(X ))) > 1

Then (1) η(X ) satisfies the condition (F) over R; (2) η(X ) is a super-primitive element over R[X ]; R R (3) ((Iφ(X ) ∩ Iψ(X ) )(φ(X ), ψ(X )))R[X ] = R[X ] if and only if R[X, η(X )] is flat over R[X ].

11.2 Subrings of R[α] Associated with Ideals of R

253

R R Proof (1) For each p ∈ Dp1 (R), either (Iφ(X ) ∩ Iψ(X ) )c(φ(X )) ⊆ p or R R (Iφ(X ) ∩ Iψ(X ) )c(ψ(X )) ⊆ p. In the former case (respectively the latter case), η(X ) satisfies the condition (F) over R p by proposition 11.1.7 (respectively Proposition 11.1.8). Thus we have our conclusion by Proposition 11.1.4. R[X ] R (2) By (1), η(X ) satisfies the condition (F) over R. So we have J[η(X )] = (Iφ(X ) ∩ R[X ] R Iψ(X ) )(φ(X ), ψ(X ))R[X ] by the preceding remark. Hence J[η(X )] ⊆ p R[X ] for each p ∈ Dp1 (R). Considering (φ(X ), ψ(X ))K [X ] = K [X ], we have R[X ] J[η(X )] ⊆ P for each P ∈ Dp1 (R[X ]) with P ∩ R = (0). Therefore we conclude that grade(J[η(X )] ) > 1. (3) By Theorem 2.3.6, we see that R[X, η(X )] is flat over R[X ] if and only R[X ] R R if J[η(X )] = R[X ]. Since J[η(X )] = (Iφ(X ) ∩ Iψ(X ) )(φ(X ), ψ(X ))R[X ] by the preceding remark, we have our conclusion.

For the next theorem, we introduce the notation: Let α be an algebraic element over K (X ) and the minimal (monic) polynomial of α be ϕα (Y ) = Y d +η1 (X )Y d−1 +· · ·+ηd (X ) with ηi (X ) ∈ K (X ) and η0 (X ) = 1. Put ηi (X ) = ψi (X )/φi (X ), where φi (X ), ψi (X ) ∈ K [X ] with (φi (X ), ψi (X ))K [X ] = K [X ] and φi (X ) is monic. Theorem 11.1.11 Under the above circumstances, assume that each ηi (X ) (0 ≤ i ≤ d) satisfies the condition (F). 
d (1) If grade( dj=0 ( i=1 (IφRi (X ) ∩ IψRi (X ) )φi (X ))η j (X )R[X ]) > 1, then α is a super-primitive element over R[X ]. 
d (2) If dj=0 ( i=1 (IφRi (X ) ∩ IψRi (X ) )φi (X ))η j (X )R[X ] = R[X ], then R[X, α] is flat over R[X ]. Proof Since each ηi (X )(0 ≤ i ≤ d) satisfies the condition (F), we have R[X ] I R[X ] = (I R ∩ I R )φi (X )R[X ] by Proposition 11.1.2. Thus I[α] =
ηid(X ) R[X ] φi (X )
d ψi (XR) R[X ] R i=1 Iηi (X ) = i=1 (Iφi (X ) ∩ Iψi (X ) )φi (X )R[X ]. Since I[α] (1, η1 (X ), . . . , R[X ] ) > 1, which means that α is super-primitive ηd (X )) = J[α] , we have grade(J[α] over R[X ]. Since the flat obstruction ideal of an anti-integral extension R[X ][α] R[X ] of R[X ] is J[α] , the last part follows. (See the proof of Theorem 11.1.10(3)).

11.2

Subrings of R[α] Associated with Ideals of R

Assume now that α is an anti-integral element of degree d ≥ 2 over R. We have seen in Theorem 3.1.18 that R α = R ⊕ I[α] ζ1 ⊕ · · · ⊕ I[α] ζd−1 as an R-module, where ζi := α i + η1 α i−1 + · · · + ηi (1 ≤ i ≤ d). Note that

254

Semi Anti-Integral and Pseudo-Simple Extensions

ζd = ϕα (α) = 0. Put ζ0 := 1 and η0 := 1 for convenience. It is easy to see that αζi = ζi+1 − ηi+1 (0 ≤ i ≤ d − 1) by definition. Note that R α and R[α] are birational and their quotient fields are equal to K (α). Let H be an ideal of R and put C H := R + H ζ1 + · · · + H ζd−1 , which is an R-submodule in K (α) = K ⊕ K ζ1 ⊕ · · · ⊕ K ζd−1 . The following lemma is seen immediately. Lemma 11.2.1 Assume that α is an anti-integral element of degree d over R. Let H denote an ideal of R. Then C H forms a subring of K (α) if and only if H 2 ζi ζ j ⊆ C H for all i, j ∈ {1, 2, . . . , d − 1} with j ≤ i. Proposition 11.2.2 Let H denote an ideal of R. Assume that α is an antiintegral element over R of degree d = 2. Then C H forms a subring of K (α) if and only if H 2 η1 ⊆ H and H 2 η2 ⊆ R. Proof Note ζ1 = α + η1 and hence ζ12 = ζ1 (α + η1 ) = αζ1 + η1 ζ1 = ζ2 − η2 + η1 ζ1 = η1 ζ1 − η2 (here ζ2 = 0). Assume that C H is a subring of K (α). Then H 2 ζ12 ∈ C H yields that for any h ∈ H , h 2 η1 ζ1 − h 2 η2 ∈ R + H ζ1 ⊆ K ⊕ K ζ1 . Consequently, H 2 η1 ⊆ H and H 2 η2 ⊆ R. Conversely, take a + h 1 ζ1 , b + h 2 ζ1 ∈ C H with h 1 , h 2 ∈ H and a, b ∈ R. Then (a + h 1 ζ1 )(b + h 2 ζ1 ) = ab + (ah 2 + bh 1 )ζ1 + h 1 h 2 ζ12 ∈ R + H ζ1 because h 1 h 2 ζ12 ∈ H 2 ζ12 = H 2 (η1 ζ1 − η2 ) ⊆ R + H ζ1 = C H . We consider the case d = 3, and compute ζi ζ j by a definite calculation. Example 11.2.3 Assume that α is an anti-integral element of degree d = 3 over R. Note ζ1 = α + η1 , ζ2 = αζ1 + η2 and ζ3 = αζ2 + η3 . Thus ζ12 = ζ1 (α + η1 ) = αζ1 + η1 ζ1 = ζ2 − η2 + η1 ζ1 , ζ2 ζ1 = ζ2 (α + η1 ) = αζ2 + ζ2 η1 = ζ3 − η3 + η1 ζ2 = −η3 + η1 ζ2 , and ζ22 = ζ2 (αζ1 + η2 ) = (αζ2 )ζ1 + ζ2 η2 = (ζ3 − η3 )ζ1 + ζ2 η2 = −η3 ζ1 + η2 ζ2 because ζ3 = 0. Assume that C H is a subring of K (α). Then we have H 2 η1 ⊆ H , H 2 η2 ⊆ H , and H 2 η3 ⊆ H since ζ12 , ζ2 ζ1 , and ζ22 are in C H . Conversely, if H 2 η1 ⊆ H , H 2 η2 ⊆ H and H 2 η3 ⊆ H , then ζ12 , ζ2 ζ1 and ζ22 are in C H . So we conclude that C H is a subring of K (α) by Lemma 11.2.1.

11.2 Subrings of R[α] Associated with Ideals of R

255

Lemma 11.2.4 Assume that α is an anti-integral element of degree d(≥ 3) over R and that 1 ≤ j ≤ i < d. (i) If i + j < d, then ζi ζ j = ζi+ j −

j 

ηi+t ζ j−t +

t=1

j−1 

η j−s ζi+s

s=0

(ii) If i + j ≥ d, then ζi ζ j = −

d−i 

ηi+t ζ j−t +

t=1

d−i−1 

η j−s ζi+s

s=0

(Note that d − i > 0 and d − i − 1 ≥ 0 because 1 ≤ j ≤ i < d.) Proof Note first that η0 := 1, ζ0 := 1, ζd = 0, and ζi+1 = αζi + ηi+1 . Now we compute ζi ζ j as follows: ζi ζ j = ζi (αζ j−1 + η j ) = αζi ζ j−1 + η j ζi = (ζi+1 − ηi+1 )ζ j−1 + η j ζi = ζi+1 ζ j−1 − ηi+1 ζ j−1 + η j ζi = ζi+2 ζi−2 − ηi+2 ζ j−2 + η j−1 ζi+1 − ηi+1 ζ j−1 + η j ζi = ζi+2 ζi−2 − (ηi+2 ζ j−2 + ηi+1 ζ j−1 ) + (η j ζi + η j−1 ζi+1 ) ········· ········· (i) Repeating the above process, we have ζi ζ j = ζi+ j ζ0 −

j 

ηi+t ζ j−t +

t=1

that is, ζi ζ j = ζi+ j −

j 

η j−s ζi+s

s=0

ηi+t ζ j−t +

t=1

j−1 

j−1 

η j−s ζi+s

s=0

(ii) Put  := i + j − d. Then j ≤ i < d yields  < d − 1 and j −  ≥ 1. Thus continuing the above process yields ζi ζ j = ζd ζ −

j−  t=1



j−−1

ηi+t ζ j−t +

s=0

η j−s ζi+s

256

Semi Anti-Integral and Pseudo-Simple Extensions that is, ζi ζ j = −

d−i 

ηi+t ζ j−t +

t=1

d−i−1 

η j−s ζi+s

s=0

Remark 11.2.5 Assume that α is an anti-integral element of degree d(≥ 3) ( j) over R and that 1 ≤ j ≤ i < d. For (i, j) with i + j < d, let 1 := { j −t | t = (i, j) 1, . . . , j} and 1 := {i + s | s = 0, . . . , j − 1}. For (i, j) with i + j ≥ d, let ( j,i) 2 := { j − t | t = 1, . . . , d − i} and (i) 2 := {i + s | s = 0, . . . , d − i − 1}. (i, j)

( j)

(i) If i + j < d, then 1 ∩ 1 = ∅ because j − t < i + s, and  (i, j) ( j) ∪ 1 ) = {0, 1, 2, . . . , d − 1}. i+ j
(ii) Assume i + j ≥ d. Then (i) = ∅ because j − t < i + s. Put 2 ∩ 2 ( j,i) ek := −1 if k ∈ 2 and ek := 1 if k ∈ (i) 2 . Then  ζi ζ j = ek ηi+ j−k ζk ( j,i)

k∈ 2

∪ (i) 2

2 Consider ζd−1 . Then we have 2 ζd−1 = ζd−1 (αζd−2 + ηd−1 )

= αζd−1 ζd−2 + ηd−1 ζd−1 = (ζd − ηd )ζd−2 + ηd−1 ζd−1 = −ηd ζd−2 + ηd−1 ζd−1 2 Hence if ζd−1 ∈ C H , then H 2 ηd ⊆ H .

Theorem 11.2.6 Assume that α is an anti-integral element of degree d(≥ 3) over R. Let H be an ideal of R. Then C H := R + H ζ1 + · · · + H ζd−1 is a subring of K (α) if and only if H 2 ηi ⊆ H for all i (1 ≤ i ≤ d). Proof

Our conclusion follows Lemmas 11.2.1 and 11.2.4 and Remark 11.2.5.

Corollary 11.2.7 Assume that α is an anti-integral element of degree d(≥ 3) over R. Let H be an ideal of R. If C H is a subring of K (α), then every element in H ηi is integral over R for all i (1 ≤ i ≤ d). Proof We have H 2 ηi ⊆ H for all i (1 ≤ i ≤ d) by Theorem 11.2.6. So H ηi ⊆ H : K H . Since H is a finitely generated ideal, any element in H : K

11.2 Subrings of R[α] Associated with Ideals of R

257

H is integral over R. Thus every element of H ηi is integral over R for all i (1 ≤ i ≤ d). Corollary 11.2.8 Assume that R is normal and that α is of degree d (≥ 3) over R. Let H be an ideal of R. If C H is a subring of K (α), then H ⊆ I[α] and hence C H ⊆ R α . Proof Since R is normal, the element α is anti-integral over R by Theorem 2.2.9. We have H ηi ⊆ H : K H = R for all i (1 ≤ i ≤ d), that is, H ⊆ I[α] .

Theorem 11.2.9 Let H be an ideal of R. Assume that α is an anti-integral element of degree d (≥ 2) over R. (1) If C H is a subring of K (α), then H 2 ⊆ I[α] . (2) Assume that grade(H ) > 1. Then C H is a subring of K (α) if and only if ηi ∈ R for all i, if and only if α is integral over R. (3) If I[α] = R and if C H is a subring of K (α), then grade(H ) ≤ 1. (4) When H is an invertible ideal and d ≥ 3, H ⊆ I[α] if and only if C H is a subring of K (α). Proof (1) Assume that d ≥ 3. Since C H is a ring, we have H 2 ηi ⊆ H for all (1 ≤ i ≤ d) by Theorem 11.2.6. Hence H 2 ηi ⊆ H ⊆ R, which means that H 2 ⊆ Iηi for all (1 ≤ i ≤ d). Thus H 2 ⊆ I[α] . Next assume that d = 2. Then Proposition 11.2.2 yields our conclusion. (2) Assume that C H is a subring of K (α). Then H 2 ⊆ I[α] by (1). Hence 1 ≤ grade(H ) ≤ grade(I[α] ), which means that I[α] = R because I[α] is a divisorial ideal of R. So we have Iηi = R for all i (1 ≤ i ≤ d), that is, ηi ∈ R for all i (1 ≤ i ≤ d). Furthermore α is integral over R by Theorem 2.3.2. Conversely, assume that α is integral over R. Then I[α] = R, and hence ηi ∈ R for all (1 ≤ i ≤ d). So, if d ≥ 3, H 2 ηi ⊆ H yields that C H is a ring by Theorem 11.2.6. If d = 2, C H is a ring by Proposition 11.2.2. (3) Our conclusion follows from the result (2). (4) Since d ≥ 3, C H is a subring if and only if H 2 ηi ⊆ H for all i (1 ≤ i ≤ d). Since H is invertible, H 2 ηi ⊆ H for all i (1 ≤ i ≤ d) if and only if H ηi ⊆ R for all i (1 ≤ i ≤ d) if and only if H ⊆ I[α] . Theorem 11.2.10 Assume that α is an anti-integral element of degree d(≥ 2) over R. Let H be an ideal of R. If H ⊆ I[α] , then C H is a subring of K (α).

258

Semi Anti-Integral and Pseudo-Simple Extensions

Proof The inclusion H ⊆ I[α] yields that H 2 ηi ⊆ H (⊆ R) for all i (1 ≤ i ≤ d). Hence C H is a ring by Proposition 11.2.2 and Theorem 11.2.6.

Corollary 11.2.11 Assume that R is a locally factorial domain and that α is an element of degree d(≥ 2). The collection := {C H | H is an ideal of R and C H is a subring of K (α)} has the maximum member C I[α] (= R α ). Proof Note first that α is anti-integral over R because R is a Krull domain (cf. Theorem 2.2.9). Take C H ∈ . Then H 2 ⊆ I[α] by Theorem 11.2.9(1). Since I[α] is a divisorial ideal, we have grade(H ) = grade(H 2 ) = 1. Since R is locally factorial, every nonzero ideal of grade 1 is invertible. So H is invertible. Hence H ⊆ I[α] by Theorem 11.2.9(4). Finally we close this section by giving the following results concerned with the ring R α . Proposition 11.2.12 Assume that α is an anti-integral element of degree d. If an element a in R is a non-zero-divisor on R/I[α] , then R[aα] ∩ R[α −1 ] = R + I[α] (aζ1 ) + · · · + I[α] (a d−1 ζd−1 ). Proof We have only to show the inclusion (⊆) because a i I[α] ζi ⊆ R[aα] ∩ R[α −1 ]. Take an element β ∈ R[aα] ∩ R[α −1 ] and write β = xn (aα)n + · · · + x1 (aα) + x0 = y0 + y1 (α −1 ) + · · · + ym (α −1 )m with xi , y j ∈ R. Then we have α m β = xn a n α n+m + · · · + x0 α m = y0 α m + y1 α m−1 + · · · + ym . Put f (X ) := xn a n X n+m + · · · + x0 X m − (y0 X m + y1 X m−1 + · · · + ym ) ∈ Ker(π), where π : R[X ] → R[α] denotes the canonical R-homomorphism. Since α is an anti-integral element over R, Ker(π ) = I[α] ϕα (X )R[X ]. Hence xn a n ∈ I[α] . Since a is a nonzero-divisor of R/I[α] , we have xn ∈ I[α] . Put β∗ (X ) := xn a n X n + · · · + x1 a X + x0 . Since xn ϕα (X ) ∈ R[X ], considering deg(β∗ (X ) − xn ϕα (X )X n−d ) < n if n ≥ d, we can use the induction on n and assume that n < d. So, considering β − xn a n ζn instead of β, we can conclude that β ∈ R + I[α] (aζ1 ) + · · · + I[α] (a d−1 ζd−1 ) by induction on n. Proposition 11.2.13 Assume that α is an anti-integral element of degree d over R. If an element a in R is a non-zero-divisor on R/I[α−1 ] , then R[a −1 α] ∩ R[α −1 ] = R α . Proof We have only to show the inclusion (⊆). Take an element β ∈ R[a −1 α]∩ R[α −1 ] and write β = xn (a −1 α)n + · · · + x1 (a −1 α) + x0 = y0 + y1 (α −1 ) + · · ·

11.3 Semi Anti-Integral Elements

259

+ ym (α −1 )m with xi , y j ∈ R. Then a n β = xn α n + · · · + a n x0 = a n y0 + a n y1 (α −1 ) + · · · + a n ym (α −1 )m . Since α is anti-integral over R, so is α −1 (cf. Theorem 2.4.5). Putting f (X ) := a n ym X m+n +· · ·+a n X n −(a n x0 X n +· · ·+xn ), we have f (α −1 ) = 0 and hence f (X ) ∈ I[α−1 ] ϕα−1 (X ). Thus we have a n ym ∈ I[α−1 ] . Since a is a nonzero-divisor on R/I[α−1 ] , we have ym ∈ I[α−1 ] . Hence ym ϕα−1 (X ) ∈ R[X ]. Put β∗ (X ) := y0 + y1 X + · · · + ym X m . Considering deg(β∗ (X ) − ym ϕα−1 (X )) < m, we can assume that m < d by induction. Let ϕα−1 (X ) := X d + η1 X d−1 + · · · + ηd be the monic minimal polynomial of α −1 over K and ζi := (α −1 )i + η1 (α −1 )i−1 + · · · + ηi (1 ≤ i ≤ d). Considering β − ym ζm instead of β, we obtain that β − ym ζm ∈ R + I[α−1 ] ζ1 + · · · +  I[α−1 ] ζd−1 = R[α −1 ] ∩ R[α] (cf. Theorem 3.1.18) by induction on m. Thus −1 β ∈ R[α ] ∩ R[α]. Therefore we have R[a −1 α] ∩ R[α −1 ] ⊆ R α .

11.3

Semi Anti-Integral Elements

Let R be a Noetherian domain with quotient field k. Let L be an extension of a field K and α be an element of L which is algebraic over K . Put d = [K (α) : K ]. We consider the canonical exact sequence 0 −→ I −→ R[X ] −→ R[α] −→ 0. If I is generated by polynomials of degree d, then α is said to be an anti-integral element over R (cf. Proposition 2.2.3). Let A = R[α]. It is well known that if α is integral and anti-integral over R, then A is a free R-module of rank d. Note that α has a unique monic relation of degree d over K . Let ϕα (X ) = X d + η1 X d−1 + · · · + ηd (ηi ∈ K ) be the minimal
d polynomial of α over K and Iηi = R : R ηi = {a ∈ R | aηi ∈ R} and I[α] = i=1 Iηi . Then we have: Proposition 11.3.1

The following statements are equivalent:

(1) α is an anti-integral element over R; (2) R[X ] : K [X ] ϕα (X ) ⊆ I[α] R[X ]. Proof (1) ⇒ (2): For any element g(X ) of R[X ] : K [X ] ϕα (X , we have g(X )ϕα (X ) = f (X ) for some f (X ) ∈ R[X ]. Let π : R[X ] −→ R[α] be the natural ring homomorphism. Then f (X ) ∈ Ker(π ). We consider the natural exact sequence 0 −→ Ker(π ) −→ R[X ] −→ R[α] −→ 0.

260

Semi Anti-Integral and Pseudo-Simple Extensions

 Since α is anti-integral over R, Ker(π) = I[α] , and hence f (X ) = i (h i ϕα (X )) ti (X ) forsome ti (X ) ∈ I[α] , ti (X ) ∈ R[X ]. Canceling ϕα (X ), we see that g(X ) = i h i ti (X ) ∈ I[α] R[X ]. (2) ⇒ (1): For any element f (X ) ∈ Ker(π ), we have f (X ) = g(X )ϕα (X ) for some g(X ) ∈ K [X ]. Thus g(X ) ∈  R[X ] : K [X ] ϕα (X ) ⊆ I[α] R[X ] by assumption. Hence we can write g(X ) = i h i ti (X ), h i ∈ I[α] , ti (X ) ∈ R[X ].  Therefore we obtain f (X ) = i (h i ϕα (X ))ti (X ) ∈ I[α] ϕα (X )R[X ]. Thus α is anti-integral over R. This observation leads to the following definition. Definition 11.3.2 If R[X ] : K [X ] ϕα (X ) ⊆ R[X ], α is said to be a semi anti-integral element over R. If α is an anti-integral element over R, it is a semi anti-integral element over R by Proposition 11.3.1. Remark 11.3.3 Let R be the integral closure of R in K . By Gauss’ Lemma, we have easily seen ηi ∈ R for all i, if α is integral over R. Next we shall show that if α is an integral and a semi anti-integral element over R, then α is an anti-integral element over R. For the purpose of this proof, we need the following proposition. If α is an anti-integral element over R, we put Hα := R[X ] : K [X ] ϕα (X ) Proposition 11.3.4

The following statements are equivalent:

(1) α is an anti-integral element over R; (2) Hα = I[α] R[X ]. Proof (1) ⇒ (2): Let g(X ) be an arbitrary element of Hα . Then, from the proof (1) ⇒ (2) of Proposition 11.3.1, we can write g(X ) = i Hi ti (X ), h i ∈ I[α] , ti (X ) ∈ R[X ]. Hence we see that g(X ) ∈ I[α] R[X ]. Since one inclusion is obvious, we have the desired result. (2) ⇒ (1): Assume that f (α) = 0 for some f (X ) ∈ R[X ]. Then we have f (X ) = g(X )ϕα (X ) for someg(X ) ∈ K [X ]. Thus g(X ) ∈ R[X ] : K [X ] ϕα (X ) = Hα . Hence we have g(X ) = i gi X i , with gi ∈ I[α] . Note that gi ϕα (X ) ∈ R[X ] and deg(gi ϕα (X )) = d for all i. Therefore f (X ) is generated by I[α] ϕα (X ). Thus α is anti-integral over R. As a consequence we have the following result.

11.3 Semi Anti-Integral Elements

261

Theorem 11.3.5 If α is an integral and a semi anti-integral element over R, then α is an anti-integral element over R, and hence A = R[α] is a free R-module of rank d. Proof Since α is integral over R, there exists a monic polynomial f (X ) ∈ R[X ] such that f (α) = 0. First, we consider the case of deg f (X ) = d. In this case, α is anti-integral over R, and hence there is nothing to prove. Next we assume that n = deg f (X ) > d. Let f (X ) = (X n−d + λ1 X n−d−1 + · · · + λn−d )ϕα (X ) = (X n−d + λ1 X n−d−1 + · · · + λn−d )(X d + η1 X d−1 + · · · + ηd ) λi , η j ∈ K . Since α is semi anti-integral over R, we see λ1 , λ2 , . . . , λn−d ∈ R by the definition. If we compare the coefficients of X n−1 in both sides, we see that ηi + λ j ∈ R, and hence ηi ∈ R. Proceeding in this way, we get ηi ∈ R for all i. Thus ϕα (X ) ∈ R[X ], so that α is anti-integral over R. Remark 11.3.6 Let R be a Noetherian domain with quotient field K . Let L be a field-extension of K and α be an element of L which is algebraic over K . Put d = [K (α) : K ]. Assume that α is integral over R. Then there exists a monic polynomial f (X ) ∈ R[X ] such that f (α) = 0. Then we have: deg f (X ) = d ⇔ R[α] is a free R-module of rank d ⇔ ϕα (X ) ∈ R[X ] ⇔ f (X ) = ϕα (X ). Thus we obtain the following proposition. Proposition 11.3.7 Spec(R) | p ⊇ I[α] }.

{ p ∈ Spec(R) | R p [X ] is not flat over R p } = { p ∈

We assume that there exists a monic polynomial f (X ) ∈ R[X ] of degree d + 1 such that f (α) = 0. Then we have f (X ) = (X + λ)ϕα (X ), λ ∈ K . It follows that λ ∈ R. Thus we have the following proposition. Proposition 11.3.8

The following statements are equivalent:

(1) λ ∈ R; (2) ϕα (X ) ∈ R[X ]. Therefore { p ∈ Spec(R) | R p [α] is not flat over R p } = { p ∈ Spec(R) | p ⊇ Iλ }. Proof Let f (X ) = X d+1 + a1 X d + · · · + ad and ϕα (X ) = X d + η1 X d−1 + · · · + ηd . Since f (X ) = (X + λ)ϕα (X ), we deduce that a1 = λ + η1 , ai =

262

Semi Anti-Integral and Pseudo-Simple Extensions

ληi−1 + ηi (2 ≤ i ≤ d), and ad+1 = ληd . Note that ai ∈ R (1 ≤ i ≤ d). Thus we see that λ ∈ R is equivalent to ηi ∈ R. The second half is easily seen from the fact that R p [α] is flat over R p if and only if ϕα (X ) ∈ R[X ]. This completes the proof. Remark 11.3.9

Consider the canonical exact sequence: 0 −→ P −→ R[X ] −→ R[α] −→ 0

Assume that there exists a monic polynomial f (X ) ∈ R[X ] of degree d + 1 with f (α) = 0. Then we have P = I[α] ϕα (X )R[X ] + f (X )R[X ]. Finally, we have the following theorem. Theorem 11.3.10 Let R be a Noetherian domain with quotient field K . Let L be an extension of a field K and let α be an element of L which is algebraic over K . Let d = [K (α) : K ] and ϕα (X ) = X d + η1 X d−1 + · · · + ηd , ηi ∈ K be a monic relation of α over K . Assume that α is integral over R. Let B be an intermediate ring between R and R. Then the following statements are equivalent: (1) R[X ] : K [X ] ϕα (X ) ⊆ B[X ]; (2) B[α] is flat over B and α is anti-integral over B.

11.4

Pseudo-Simple Extensions

Let R be a Noetherian integral domain, K denote the quotient field of R, and A be an R-algebra of finite type contained in K . Then A = R[F] for some fractional ideal F of R. We can write F = I α for some ideal I of R and some element α in K . So A is of the form R[I α]. In general, it seems natural to consider an extension like R[I α], where I is an ideal of R and α is an element in some extension of R, where α does not necessarily belong to K . We say that R[I α] is a pseudo-simple extension of R defined by I and α if depth(R P ) > 1 for any prime ideal P containing I . Our objective of this section is to study the ring like R[I α]. In particular, we are interested in the case α is anti-integral over R. The anti-integrality is studied in Chapters 1, 2, 3, and 4. Throughout this section, we use the following notations unless otherwise specified: let R be a Noetherian domain, K the quotient field of R, and R the

11.4 Pseudo-Simple Extensions

263

integral closure of R in K . For a nonzero element α of K , put R : R α := {a ∈ R | aα ∈ R}(= α −1 R ∩ R) We start with recalling the following definition which is seen in Chapter 1: Let R α := R[α] ∩ R[α −1 ] in K . We say that α is anti-integral over R if R α = R. Theorem 11.4.1 Assume that R is a local domain with the maximal ideal m and let α be a nonzero element in K . Assume that α is anti-integral over R and that R[α] = R[mα]. Then α ∈ R or α −1 ∈ R. Proof

Since α ∈ R[mα], we can write α = b 0 + b1 α + · · · + b n α n

where bi ∈ m i for all i > 0 and b0 ∈ R.

Case I: Assume that b0 ∈ m. Then 1 − b0 (α −1 ) = b1 + b2 α + · · · + bn α n−1 ∈ R α = R α = R. Hence 1 − b0 (α −1 ) ∈ R. Since b0 is a unit in R, we have α −1 ∈ R. Case II: Assume that b0 ∈ m. Then we have b0 + (b1 − 1)α + · · · + bn α n = 0. Put f (X ) = bn X n + · · · + b2 X 2 + (b1 − 1)X + b0 . Then b1 − 1 ∈ m. Thus c( f (X )) = R, where c( ) denotes the content ideal. Take h(X ) ∈ R[X ] which satisfies h(α) = 0, c(h(X )) = R, and deg h(X ) is minimal among such ones. Write h(X ) = cd X d + · · · + c0 with ci ∈ R and

c d α d + · · · + c0 = 0

(∗)

If c0 is a unit in R, then by the same argument as in Case I, we have that α −1 ∈ R. Assume that c0 ∈ m. Then there exists s (0 < s ≤ d) such that cs is a unit in R. From (∗), we have c0 α d−s+1 +· · ·+cs α +cs−1 = −(cs−2 α −1 +· · ·+c0 α −s+1 ) ∈ R α = R. Thus there exists v ∈ R such that cd α d−s+1 + · · · + cs α + v = 0. By the minimality of deg h(X ), we get d − s + 1 = d, that is, s = 1. Hence c1 is a unit in R. Consider the equality cd α d−1 + · · · + c1 = −(α −1 )c0 ∈ R α = R. Put t = cd α d−1 + · · · + c1 = −(α −1 )c0 . If c1 − t ∈ m, then the polynomial cd X d−1 +· · ·+c1 −t ∈ R[X ] gives a contradiction to the minimality of deg h(X ).

264

Semi Anti-Integral and Pseudo-Simple Extensions

So c1 − t ∈ m. Hence t ∈ m because c1 is a unit in R. Thus t = −(α −1 )c0 implies αt + c0 = 0 and hence α ∈ R. Corollary 11.4.2 Assume that (R, m) is a normal local domain and let α be a nonzero element in K such that neither α nor α −1 belongs to R. Then R[α] = R[mα]. Proof Since R is normal and R α is integral over R, R α = R, that is, any α ∈ K is anti-integral over R. So our conclusion follows from Theorem 11.4.1.

Corollary 11.4.3 Let R be a Noetherian domain, let I be an ideal of R, and let α be a nonzero element in K which is anti-integral over R. If R[α] = R[I α], then R p [α] is flat over R p for any p ∈ Spec(R) with p ⊇ I . Proof We may assume that R is a local domain with the maximal ideal m. Since R[α] = R[I α] implies R[α] = R[mα], R[α] is flat over R by Theorem 11.4.1. Corollary 11.4.4 Assume that (R, m) is a local domain. Then the following statements are equivalent: (a) R is a DVR; (b) every α in K \ {0} is anti-integral over R and R[α] = R[mα]. Proof (a) ⇒ (b) follows from Corollary 11.4.2. (b) ⇒ (a): For any nonzero element α in K , α ∈ R or α −1 ∈ R by Theorem 11.4.1, which asserts that R is a valuation domain. Since R is Noetherian, R is a DVR. Lemma 11.4.5 Let I be an ideal of R and P a prime ideal containing I . Then if depth(R P ) > 1, then grade(I R P ) > 1. Proof We may assume that (R, m) is a local ring. Take a ∈ I√ , a = 0 and let a R = Q 1 ∩ · · · ∩ Q n be a primary decomposition with Pi = Q i . In this case depth(R P ) = 1 for all P = Pi . Since Pi  ⊇I by the assumption, we have I ⊆ P1 ∪ · · · ∪ Pn . Hence there exists b ∈ I \ Pi . It is easy to see that {a, b} is a regular sequence. Thus grade(I ) > 1.

11.4 Pseudo-Simple Extensions

265

Definition 11.4.6 Let I be an ideal of R and α be an element which belongs to some extension of R which is not necessarily equal to K . Consider the extension A = R[I α] of R. We say that A is a pseudo-simple extension of R defined by I and α if I = R or grade(I ) > 1. Any simple extension R[α] (where α belongs to some extension of R) is a pseudo-simple extension of R. In the next theorem, we use the following fact: for α ∈ K , R : R α is a divisorial ideal of R, that is, R : K (R : K (R : R α)) = R : R α. This follows from the facts α(R : R α) = α R ∩ R and R : K (R : R α) contains α. Theorem 11.4.7 Let α be an element in some extension of R. Assume that A = R[I α] is a pseudo-simple extension of R defined by I and α. If A is flat over R, then A = R[α], that is, A is a simple extension generated by α. Proof We may assume that (R, m) is a local ring. By Lemma 11.4.5, there exist a, b ∈ I which form a regular sequence. It is obvious that aα, bα ∈ A. Since A is flat over R, either a, b form an A-regular sequence or (a, b)A = A. In the former case, A : A α contains a, b and hence A : A α = A because A : A α is a divisorial ideal of A. Thus α ∈ A. In the latter case, (a, b)A = A implies that 1 = aβ + bγ for some β, γ ∈ A. Thus α = (aα)β + (bα)γ ∈ A. Therefore in any case, we conclude α ∈ A. In the following examples, let k denote a field and k[x, y] denote a polynomial ring. Example 11.4.8 Let R = k[x, y], I = (x, y) and α = 1/x y. Since α −1 ∈ R, R = R α and hence α is anti-integral over R. By definition, R[I α] is a pseudo-simple extension of R and R[I α] = R[(x, y)(1/x y)] = R[1/x, 1/y] = R[1/x y] = R[α], a simple extension. Moreover since α −1 ∈ R, R = R[α −1 ] → R[α −1 , α] = R[α] is obtained by localization and hence is flat. This shows that R[I α] = R[α] does not always imply I = R. Example 11.4.9 Let R = k[x, y] and A = R[x/y, y/x] = R[(x 2 , y 2 )(1/x y)], which is a pseudo-simple extension of R. It is not hard to see that A is not a simple extension of R. Since α −1 = x y ∈ R, α is anti-integral over R. By Theorem 11.4.7, A is not flat over R. Example 11.4.10 Let I be an ideal (x, y) of R = k[x, y] and α is an element in some extension of R which satisfies y 2 α 2 + (x − 1)α + 1 = 0. Then [K (α) : K ] = 2, that is, α ∈ K . R[I α] is a pseudo-simple extension of R and R[I α] =

266

Semi Anti-Integral and Pseudo-Simple Extensions

R[xα, yα]. Since α ∈ R[I α], we have R[I α] = R[α]. Since α −1 is a zero of the single monic polynomial and α −1 ∈ R[α], R → R[α −1 ] is flat and R[α −1 ] → R[α −1 , α] = R[α] is a localization. Hence R → R[α] is flat. Lemma 11.4.11 Let a be an element of R and let α be a nonzero element which is algebraic over R and let ϕα (X ) = X d + µ1 X d−1 + · · · + µd be the minimal polynomial of α in K [X ]. Assume that the kernel of the canonical homomorphism
R[X ] → R[α] is generated by some polynomials of degree d. Put I[α] = 1≤i≤d (R : R µi ). Then (i) if a ∈ I[α] , aα is integral over R;  (ii) if aα is integral over R, a ∈ I[α] . Proof (i) Put bi = aµi ∈ R. Then aα d + bi α d−1 + · · · + bd = 0 is a relation of α over R. From this, we get (aα)d + abi (aα)d−1 + · · · + a d bd = 0. So aα is integral over R. (ii) Let (aα)n + c1 (aα)n−1 + · · · + cn = 0 be a relation of aα over R and put h(X ) =  a n X n + c1 a n−1 X n−1 + · · · + cn . Then h(α) = 0 and we can write h(X ) = i f i (X )gi (X ) with f i (α) = 0, deg f i (X ) = d for all i. Since every  coefficient of f i (X ) belongs to I[α] , a n ∈ I[α] . Thus a ∈ I[α] . Proposition 11.4.12 Let α be a nonzero element which is algebraic over R and let A = R[I α] be a pseudo-simple extension of R defined by I and α. Assume that the kernel of the canonical homomorphism R[X ] → R[α] is generated by some polynomials of degree d. If A is integral over R, then α is integral over R. Proof We have only to show that α is integral over R P for all P ∈ Spec(R). If P does not contain I , A P = R P [α] implies that α is integral over R P . So we may assume that R is a local domain and I = R. Since A is a pseudo-simple extension, we have gradeI > 1, and hence there exists a regularsequence a, b in I . By assumption aα and bα are integral over R. Thus a, b ∈ I[α] by Lemma   11.4.11. But I[α] is a divisorial ideal, which implies that I[α] = R and hence I[α] = R. Thus ϕα (X ) ∈ R[X ]. Therefore α is integral over R. Proposition 11.4.13 Let A = R[I α] be a pseudo-simple extension of R defined by I and α ∈ K . Assume that α is anti-integral over R. Then the following statements are equivalent: (a) A is integral over R; (b) α is integral over R;

11.4 Pseudo-Simple Extensions

267

(c) A = R; (d) R = R[α]; (e) I ⊆ R : R α. Proof (a) ⇒ (b) follows from Proposition 11.4.12. (b) ⇒ (d) follows from Lemma 11.4.11. (d) ⇒ (c), (d) ⇒ (b), (e) ⇒ (c), and (b) ⇒ (a) are trivial.

Corollary 11.4.14 Any proper pseudo-simple extension R[I α] of R defined by an ideal I and an anti-integral element α is not integral over R.

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Index

( )0 , 125

A α, 23, 122, 135 altitude formula, 10, 127 Ann R ( ), 113 Ant(α), 62 anti-integral, 1 anti-integral element, 2, 8, 20, 27, 30, 112, 126, 135, 201, 252 anti-integral extension, 1, 5–12, 27, 233 different modules, 114–120 elements, 99–103, 112 flatness, 33–37, 97 simple birational extension and, 15, 116 subrings, 49–95 unit groups, 126–129 anti-integral ideal, 234 anti-integral polynomial, 239 anti-integral type, 230 * A R, 14

d A , 197 D(α) , 60 (α), 50 d A/R , 117  A/R , 20 denominator ideal, 97, 163 dimension formula, 10, 68, 128 Dp1 (B) ∩ R, 171 Dp1 (R), 31, 149

E E (α) , 24, 49, 120, 123 E˜ (α) , 120 (α) E B , 140 (α) E K , 225 excellent element, 99 exclusive element, 143, 144 exclusive extension, 143, 162 exclusive ideal, 234, 237

B

F

birational-integral extension, 15 blowing-up, 5, 29, 43 blowing-up point, 29, 43, 100

flat element, 158, 159. See also s-flat element flat polynomial, 246, 247, 248

C

G

C(R/R), 73, 152 c( ), 23, 49, 97, 120, 125, 135, 167, 201, 233, 250 co-monic, 145 condition (F), 249 content ideal, 24 contracted, 6, 7

 J[α] , 108, 147, 219 generalized denominator ideal, 49, 97, 252

D d, 24, 135 D( ), 20, 109, 117

H H, 233, 235 η, 49, 175 Hα , 183, 260 H[β] , 227 HK , 234 Ht1 (R), 24

275

276

Index

I

N

I∞ , 64 Itx , 150, 228 Iα , 135 I[α] , 24, 49, 64, 131, 135, 195 I H , 234 I[H α] , 186 In , 64 Ind , 49 A , 201 I[α]

N, 114 N-1, 168 N-2, 168 N[ α] , 66

(S) I f , 169 B , 139 I[α] Rα Iα , 57, 103 S Iβ , 125 IηBi , 139 R , 239 Iϕ(X )

integral, 2, 46 integral closure, 1, 14, 33, 54, 65–80, 127, 167, 183, 211, 222, 236, 239, 260, 263 I˜[α] , 150 invertible ideal, 237

O obstruction of anti-integrality, 62, 253 obstruction of ideal integrality, 84 obstruction of integrality, 34 obstruction of super-primitiveness, 62 R, 152 J , 71 R, 1

P π (A) , 197 pseudo-simple extensions, 262–267 pure extensions, 160–165

Q J

Q, 9, 25, 112, 117, 190

J −1 , 13 Jα , 17 J[α] , 136, 195 A , 201 J[α] B J[α] , 139 Rα

Jα , 103 J˜ [α] , 24, 108, 135 (i) J˜[α] , 152 J H , 234

K K , vi, 1, 23, 132, 135 k( p), 24

L L , 24, 111, 132, 135 α(γ ), 225 LCM-stability, 103, 125

M Mα, 138 MαB , 139

R R, vi, 1, 22, 135 R(α), 2, 36, 101, 116, 164, 171 R[α] ∩ K , 136

S semi anti-integral element, 259–262 s-flat, 212 s-flat element, 207, 216, 217, 218, 247, 248 Sharma polynomial, 24–27, 30, 31, 36, 37, 234, 235, 236, 237, 239–243 strict closure, 14–15 Sup(α), 62, 63, 64 super-flat. See s-flat element super-primitive element, 31, 34, 36, 49, 55, 56, 60, 63, 69, 73, 78, 97, 105, 112, 131, 136, 145–154, 167, 207, 233, 252 super-primitive extension, 27, 130, 244 super-primitive ideal, 234 super-primitive polynomial, 239, 240, 243 super-primitive type, 236, 237

Index

277

T

V

Tβ , 200 The Ring, 2–5. See also R terms integral closedness, 65–80

V ( ), 20, 21, 30, 34, 43, 98, 108, 109, 219 ϕα (X ), 24, 49 ϕaα (X ), 176 K (B) ϕα (X ), 171 (A) ϕα (X ), 201 ϕ H (X ), 234 vanishing point, 42–47

U U ( ), vi, 125, 126, 128, 129, 130, 131, 132, 182, 183, 189, 190, 191, 192, 193, 194, 208, 219 ultra-primitive element, 151, 168, 174, 182, 183, 189, 190, 191, 193, 247 ultra-primitive polynomial, 247 unramified, 108, 111, 112, 113, 114, 115, 118, 119, 120, 208, 218, 221 unramified element, 108, 118 unramified extension, 108, 115, 116, 117, 118 upper-primary, 19, 20 upper-prime, 15, 16, 17, 18 upper-quasi-primary, 19, 20 Ur( ), 113, 114

W

B (A), 108

R (A), 111, 120

R (R[α]), 114

R (S), 108, 112

Z Z, 19, 87, 90, 120, 125, 128, 129 Zariski Main Theorem, 1, 10 ζi , 57, 254