wydawnictwo uniwersytetu Łódzkiego...poem: shreeram s. abhyankar , wielomiany i szeregi potęgowe...

Tadeusz Krasiński – University of Łódź Department of Algebraic Geometry and Theoretical Computer Science

Faculty of Mathematics and Computer Science 90-238 Łódź, 22 Banacha St.

[email protected]

Stanisław Spodzieja – University of Łódź Department of Analytic Functions and Differential Equations

Faculty of Mathematics and Computer Science 90-238 Łódź, 22 Banacha St. [email protected]

INITIATING EDITOR

Beata Koźniewska

REVIEWERS Piotr Jędrzejewicz, Tadeusz Krasiński, Arkadiusz Płoski, Kamil Rusek

Stanisław Spodzieja, Michał Szurek, Piotr Tworzewski

TYPESETTING Stanisław Spodzieja

TECHNICAL EDITOR Leonora Wojciechowska

COVER DESIGN Michał Jankowski

Printed directly from camera-ready materials provided to the Łódź University Press by Department of Analytic Functions and Differential Equations

© Copyright by Authors, Łódź 2017

© Copyright for this edition by Uniwersytet Łódzki, Łódź 2017

Published by Łódź University Press First edition. W.08359.17.0.K

ISBN 978-83-8088-922-4

e-ISBN 978-83-8088-923-1

Printing sheets 11.75

Łódź University Press 90-131 Łódź, 8 Lindleya St.

www.wydawnictwo.uni.lodz.pl e-mail: [email protected]

tel. (42) 665 58 63

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Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

DEDICATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1. Arkadiusz Płoski

Photo of Arkadiusz Płoski . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Arkadiusz Płoski – Scientific biography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2. Kamil Rusek

Photo of Kamil Rusek . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Kamil Rusek – Scientific biography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3. Krzysztof Kurdyka

Photo of Krzysztof Kurdyka . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Krzysztof Kurdyka – Scientific biography . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4. Shreeram Shankar Abhyankar

Photo of Shreeram Shankar Abhyankar . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Note about Shreeram Shankar Abhyankar (1930–2012) . . . . . . . . . . . . . . 25

Poem: Shreeram S. Abhyankar , Polynomials and power series . . . . . 27

Poem: Shreeram S. Abhyankar , Wielomiany i szeregi potęgowe

(translation into Polish by Michał Szurek) . . . . . . . . . . . . . . . . . . . 29

Poem: Shreeram S. Abhyankar , Wielomiany i szeregi potęgowe

(translation into Polish by Tadeusz Krasiński) . . . . . . . . . . . . . . . 33

SCIENTIFIC ARTICLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35

5. Szymon Brzostowski, Tadeusz Krasiński, and Justyna Walewska

A short proof that equisingular plane curve singularities

are topologically equivalent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

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6

6. Szymon Brzostowski and Tomasz Rodak

The Łojasiewicz exponent via the valuative Hamburger-Noether

process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7. Stanisław Janeczko, Zbigniew Jelonek, and Maria Aparecida

Soares Ruas

On a generic symmetry defect hypersurface . . . . . . . . . . . . . . . . . . . . . . . . 67

8. Zbigniew Jelonek, Wojciech Kucharz, and Krzysztof Kurdyka

Vector bundles and blowups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9. Piotr Jędrzejewicz, Łukasz Matysiak, and Janusz Zieliński

A note on square-free factorizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

10.Wojciech Kucharz and Krzysztof Kurdyka

Rationality of semialgebraic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

11. Jean Moulin Ollagnier and Andrzej Nowicki

Rational constants of cyclotomic derivations . . . . . . . . . . . . . . . . . . . . . . . 97

12. Andrzej Nowicki

Divergence-free polynomial derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

13. Beata Osińska-Ulrych and Grzegorz Skalski

The Cauchy-Kowalevski Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

14. Arkadiusz Płoski

Formal and convergent solutions of analytic equations . . . . . . . . . . . . . 161

15. Arkadiusz Płoski and Maciej Sękalski

Bezout’s inequality for real polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

16. Stanisław Spodzieja and Anna Szlachcińska

Łojasiewicz exponent of overdetermined semialgebraic mappings . . 179

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DOI: http://dx.doi.org/10.18778/8088-922-4.01

Preface

Annual Conferences in Analytic and Algebraic Geometry have been organizedby Faculty of Mathematics and Computer Science of the University of Łódź since1980. Until now, proceedings of these conferences (mainly in Polish) have com-prised educational materials describing current state of branches of mathematicsmentioned in the conference title, new approaches to known topics, and new proofsof known results (see the Internet page: http://konfrogi.math.uni.lodz.pl/).

The content of this volume consists of new results and survey articles concerningreal and complex algebraic geometry, singularities of curves and hypersurfaces,invariants of singularities, algebraic theory of derivations and other topics.

This volume is dedicated to three mathematicians celebrating in 2017 the jubileesof 70th and 60th birth anniversaries: Arkadiusz Płoski (born in 1947), Kamil Rusek(born in 1947) and Krzysztof Kurdyka (born in 1957). All of them are closelyconnected with our Conferences in Analytic and Algebraic Geometry. The first onehas been participating in the conferences since 1982, the second one since 1983and third one since 2008. Thanks to their mathematical vigor and stimulation theconferences become more interesting and fruitful. On next pages we provide shortscientific biographies of each of them.

Additionally, we would like to commemorate the fifth anniversary of ShreeramShankar Abhyankar death whose views on mathematics we share. The artistic ma-nifest of his approach to mathematics is his poem written in 1970. In this volumewe present this poem along with two translations into Polish.

We would like to thank many people for the help in preparing the volume. Inparticular, Michał Jankowski for designing the cover, referees for preparing reportsof all the articles and all participants of the Conferences for their good humor andenthusiasm during the conferences.

Tadeusz KrasińskiStanisław Spodzieja

November 2017, Łódź

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DEDICATIONS

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Professor Arkadiusz Płoski

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Analytic and Algebraic Geometry 2Łódź University Press 2017, 13–13


ARKADIUSZ PŁOSKISCIENTIFIC BIOGRAPHY

Professor Arkadiusz Płoski was born in Kielce on February 5, 1947. He graduatedfrom the Maria Curie-Skłodowska University in Lublin in 1970 and continued hismathematical studies as postgraduate student at the Institute of Mathematics ofthe Polish Academy of Sciences. In 1974 he received PhD on the basis of thethesis „On formal and convergent solutions of analytic equations” written under thesupervision of Professor Stanisław Łojasiewicz. Then he began working at the MariaCurie-Skłodowska University in Lublin, including one year break in the academicyear 1976/1977 when he worked at Paris-Sud University in France. He obtainedhabilitation degree from the Faculty of Mathematics and Physics of the JagiellonianUniversity in Kraków in 1987. His habilitation thesis was entitled „Multiplicity andthe Łojasiewicz exponent”. He became a professor of mathematics ten years later.

He specializes in complex analytic geometry and signularity theory and he isthe author of about seventy papers published in international journals and dozensof surveys and popular science papers. Since 1980 he has been working at KielceUniversity of Technology keeping close scientific relations with Polish and foreigncenters. In Poland he cooperates mostly with centers in Łódź and Kraków. Hewas often invited abroad as a visiting professor to France (Universities in Borde-aux, Lille and Grenoble), to Germany (Max Planck Institute in Bonn, Universityin Keiserslautern), to Italy (University of Calabria) and to Spain (Universidad dela Laguna). He was awarded the Polish Mathematical Society Award for YoungMathematicians (1974) for his doctoral thesis, the III grade Minister Award (1974,1988) and the Award of the Faculty of Mathematical, Physical and Chemical Scien-ces of Polish Academy of Sciences (1987) for his habilitation thesis.

He has been an active participant in the Conferences Analytic and AlgebraicGeometry organized by the Faculty of Mathematics and Computer Science of theUniversity of Łódź since 1982. In 2013 the received a Universitatis Lodziensis AmicoMedal.

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Professor Kamil Rusek

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KAMIL RUSEKSCIENTIFIC BIOGRAPHY

Kamil Rusek was born on October 1, 1947, at Aleksandrówka (near Częstocho-wa). He studied mathematics at the Jagiellonian University in Kraków and in 1970after obtaining his Masters degree started there post-graduated studies for thenext 3 years. Rusek was awarded PhD in 1973 for the thesis „Criteria of analytici-ty of mappings with values in some function spaces” written under supervision ofProfessor Józef Siciak. He joined the Institute of Mathematics of the JagiellonianUniversity and worked there until his retirement in 2012.

In 1989 he earned post-doctoral degree (the habilitation) on the basis of thedissertation „Polynomial automorphisms” and in 1999 he obtained the rank ofprofessor.

Between 1978 and 1996 he lectured as a visiting professor at the Universities ofLille, Rennes, Chambery and Albuquerqee.

In 1998–2005 he combined his obligations at the Jagiellonian University withthe job at the State Higher Vocational School at Nowy Sącz.

In 2012 – the year of his retirement – he decided to accept the position of full-time professor at the Institute of Mathematics of the Pedagogical University inKraków for 5 years.

Rusek’s field of scientific interests fluctuated from the theory of analytic func-tions in topological vector spaces through analytic, algebraic and semialgebraicgeometry to commutative algebra and matrix theory. For almost 30 years the cen-tral theme of his research have been polynomial mappings, especially questionsconcerning the Jacobian Conjecture and polynomial automorphisms, the problemsby which he became captivated. From this scope he published, as the author orthe co-author, 10 research papers and 2 survey articles. Among his results in thisrange there are an exact estimate of the degree of the inverse of a polynomial au-tomorphism (with T. Winiarski), the invertibility of injective endomorphisms ofquasi-projective varieties (with S. Cynk), criteria for surjectivity of injective real

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18 KAMIL RUSEK SCIENTIFIC BIOGRAPHY

algebraic mappings (with K. Kurdyka), inveritibility of cubic linear mappings oflow index of nilpotency (with L. M. Drużkowski), an identification of some essentialobstacles in a geometric approach to the Jacobian Conjecture and a proof of theMeisters-Olech Conjecture.

Teaching has always been a very important factor in Rusek’s carrier. For thelast 30 years he devoted a lot of his effort to teaching algebra at various levels.Thanks to his dedication, as well as that of other people working with him in thisarea (mainly K. Nowak) the interest in algebra, especially commutative one, hasincreased significantly amongst faculty and students communities in Kraków.

Kamil Rusek was also deeply involved in education of young scholars. Duringthe years 1990–2008 he was a supervisor of four PhD thesis and some of his formerPhD students have now significant achievements in mathematics.

He has been an active participant in the Conferences Analytic and AlgebraicGeometry organized by the Faculty of Mathematics and Computer Science of theUniversity of Łódź since 1983.

It should be also mentioned that Rusek’s interests reach far beyond mathematics.He is a fan of literature and classical music, and is very keen on hiking and downhillskiing.

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Professor Krzysztof Kurdyka

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KRZYSZTOF KURDYKASCIENTIFIC BIOGRAPHY

Professor Krzysztof Kurdyka was born in Kraków on February 28, 1957. He rece-ived his MSc in 1981 and PhD degree in mathematics at the Jagiellonian Universityin Kraków in 1984. His PhD thesis written under the supervision of Professor Stani-sław Łojasiewicz was entitled „The regular points of the subanalytic sets”. Between1984 and 1986 he worked as an assistant in the Department of Real Functions atthe Institute of Mathematics of the Jagiellonian University. In 1986/87 he wasa postdoc at the Fourier Institute in Grenoble. Then he worked at the UniversityRennes I, the Jagiellonian University and the University of Savoie. In 1994 he wasappointed as a Professor at the University of Savoie and in 2007 he received theProfessor degree of the highest rank in France (classe exceptionelle). Between 2001and 2005 he was a Director of the Institute (Laboratory) of Mathematics of theUniversity of Savoie, which obtained at that time the status of CNRS unit (theNational Centre for Scientific Research).

Professor Krzysztof Kurdyka is a valued and appreciated specialist in analy-tic and algebraic geometry, singularity theory and theory of o-minimal structures.He cooperates with many leading scientific centers worldwide and is an author ofmore than sixty scientific papers, considerably contributing to the developmentof real and complex analytic and algebraic geometry. Among his many importantcontributions to these branches of mathematics are a proof of the Rene Thom’s gra-dient conjecture, research on subanalytic stratifications and on the concept of arcanalytic functions arc-symmetic sets which he introduced to the real geometry. Aninequality discovered by Professor Kurdyka, thence called the Kurdyka-Łojasiewiczinequality, found a great resonance in numerous scientific studies, namely in opti-mization problems.

Since 2008 Professor Krzysztof Kurdyka has been an active participant in Analy-tic and Algebraic Geometry Conferences organized by the Faculty of Mathematicsand Computer Science of the University of Łódź and in 2015 he was awarded

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22 KRZYSZTOF KURDYKA SCIENTIFIC BIOGRAPHY

a Universitatis Lodziensis Amico Medal. During the International Congress of Ma-thematicians 2018 in Rio de Janeiro he will give an invited talk, jointly with Pro-fessor Wojciech Kucharz from the Jagiellonian University. Actually, some resultson continuous rational functions, which they will present were obtained during theconferences in Łódź.

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Professor Shreeram Shankar Abhyankar (1930–2012)in the Conference in Hanoi (2006)

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NOTE ABOUTSHREERAM SHANKAR ABHYANKAR (1930–2012)

Shreeram Shankar Abhyankar, famous American mathematician, was born inIndia in 1930, died in 2012. He, as a pupil of Oscar Zariski, specialized in algebraicgeometry. He proved the desingularization theorem for 3-dimensional varieties inpositive characteristic. He was also excellent expert in two-dimensional jacobianconjecture (which is still open). In proofs of his results he used classical methodslike Tchirnhausen transformations, elimination theory, Newton diagrams etc. Sincewe like and still use such methods in our researches (PhD thesis by Brzostowskion approximation roots of non-characteristic degrees, Newton diagrams in manypapers by Płoski, Lenarcik, Oleksik, Brzostowski, Walewska, elimination theoryused by Chądzyński, Płoski, Krasiński) we owe a lot to him. He expressed his pas-sion for the classic approach to mathematics and his opposition against formalismin the Bourbaki style in a poem written in 1970 at the International Conferencein Nice, France. The inspiration was van der Waerdens historical lecture on thedevelopment of algebraic geometry. We have obtained the permission of his wife,Yvonne Abhyankar, to publish this poem. Since we agree with his views we presenthis poem and two translations into Polish. One can find more remembrances onShreeram Shankar Abhyankar in Notices of AMS in November 2014.

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Analytic and Algebraic Geometry 2

Lodz University Press 2017, 27–28


POLYNOMIALS AND POWER SERIES

SHREERAM SHANKAR ABHYANKAR

Polynomials and power seriesMay they forever rule the world.

Eliminate, eliminate, eliminateEliminate the eliminators of elimination theory.

As you must resist the superbourbaki coupSo must you fight the little bourbakis too.

Kronecker, Kronecker, Kronecker above allKronecker, Mertens, Macauley, and Sylvester.

Not the theology of HilbertBut the constructions of Gordan.

Not the surface of RiemannBut the algorithm of Jacobi.

Ah! the beauty of the identity of Rogers and RamanujanCan it be surpassed by Dirichlet and his principle?

Germs, viruses, fungii, and functorsStacks and sheaves of the lotFear them notWe shall be the victors.

Come ye forward who dare represent a functorWe shall eliminate youBy resultants, discriminants, circulants, and alternantsGiven to us by Kronecker, Mertens, Macaulay, and Sylvester.

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28 SHREERAM SHANKAR ABHYANKAR

Let not here enter the omologists, homologistsAnd their cohorts the cohomologists crystalline

For this ground is sacred.

Onward soldiers! defend your fortressFight the Tor with a determinant long and tallBut shun the Ext above all.

Morphic injectives, toxic projectivesEtal, eclat, devious devisage

Arrows poisonous large and smallMay the armour of Tschirnhausen

Protect us from the scourge of them all.

You cannot conquer us with rings of ChowAnd shrieks of Chern

For we too are armed with polygons of NewtonAnd algorithms of Perron.

To arms, to arms, fractions continued or notFear not the scheming ghost of GrothendieckFor the power of power series is with you

May they converge or notMay they be polynomials or notMay they terminate or not.

Can the followers of G by mere smooth talkEver make the tiniest singularity simpleLong live Oscar Zariski and Karl Weierstrass.

What need have we for rings japanese, excellent or badWhen, in person, Nagata himself is on our side.

What need to tensorizeWhen you can uniformize

What need to homologizeWhen you can desingularize

(Is Hironaka on our side?).

Alas! Princeton and fair Harvard you tooReduced to satellites in the Bur-Paris Zoo.

Approval for publication by members of the family of Professor Abhyankar

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WIELOMIANY I SZEREGI POTĘGOWE


(TRANSLATION BY MICHAŁ SZUREK)

Wielomiany!Tak! Oraz

Potęgowe szeregiŚwiatem niech rządzą,

Aż po czasu brzegi!

Eliminujmy,Eliminujmy,

Bez cienia krępacji,Eliminujących teorię

eliminacji!

SuperbourbaczyNiech sczeźnie puczysta!

I bourbaczątkaWyplenim do czysta!

Kronecker, Kronecker,a potem ta kolej:

Kronecker i Mertens,Sylvester, Macalauy.

Powierzchnie Riemanna a co nam do tego?Algorytm liniowy ty weź Jacobiego.

Teologia Hilberta i po diabła ona?Wolimy grupowe konstrukcje Gordona.

Tożsamość Rogersa i Ramanujana! Ją ceni esteta.Niech się przy niej zasada schowa Dirichleta!

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30 SHREERAM SHANKAR ABHYANKAR (TRANSLATION BY MICHAŁ SZUREK)

Tam kiełki i sterty, funktory i źdźbła.I stosy i snopy na polach.

To nic, bo wiktoria ku nam będzie szła!Wygramy, bo silna w nas wola.

Fora ze dwora reprezentanty funktoraMy mamy coś na was, coś anty:

To rezultanty i alterantyKroneckera, Mertensa, Macaulaya i Sylwestra

Dyskryminanty!

My prastare plemię. I homologiczneNie wejdą metody, choćby krystaliczneW święte ziemie.

Naprzód, żołnierze! Dla was honory!Wyznacznikiem rozkręcim dziś Tory.Twierdza nie padnie, na ustach tekst„Do diabła z Ext!

Morficzne injekcje, toksyczne projekcje,I strzałki zatrute w kojądra.Etale, globale, nie dają żyć wcale,Gdzież rada jest mądra? Jej cena?Odpowiem: weź Tschirnhausena!

Władcom pierścieni ChowMy odpór damy: „ Łoł !Przed Chernem lęk nam nieznany!My algorytmy mamy Perrona!Newtona wielomiany!

Do broni, ułamki! Gdzie wasze łańcuchyZabrzęczą Grothendieck nie sięga!Nie straszne schematy i duch jego kruchyBo z wami szeregów potęga!

Zbieżne czy nie,Wszak bez nich jest źle.Niech wiją się, wiją, jak wstęga.

Gadka-szmatka, choćby gładkanie uprości

Żadnej,Kolego,

Osobliwości!


WIELOMIANY I SZEREGI POTĘGOWE 31

I na tym się zasadza krasaPodejścia Weierstrassa.

Pierścienie japońskie to czasu strata.Wszak z nami jest nawet Nagata!

Czemu mamy tensorować?Można uniformizować!Czemu homologizować?Ha! Desingularizować!

Czy Hironaka się zbrata?

I tak to i Princeton i Harvard klasycznyPowoli się zmienią w park zoologiczny!

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WIELOMIANY I SZEREGI POTĘGOWE


(TRANSLATION BY TADEUSZ KRASIŃSKI)

Potęgowe szeregi i wielomianyNiech zawsze rządzą światem.

Eliminujmy, eliminujmy, eliminujmy,Eliminujmy eliminatorów teorii eliminacji.

Tak jak superbourbakistom musisz stawić opórTak małych bourbakistów powstrzymuj napór.

Bo,Ponad wszystkich Kronecker, Kronecker, Kronecker,A z nim Mertens, Macaulay i Sylwester.

Nie dla formalnej Hilberta teologiiLecz dla konstrukcyj Gordana victorii.Nie dla powierzchni RiemannaLecz dla twórcy „jakobiana”.

Ach! Ileż piękna w tożsamości Rogersa i RamanujanaCzy może je przewyższyć Dirichleta zasada.

Kiełków, stogów i funktorów,Snopów, grzybów i wirusówNie obawiaj się ni trochęBo zwycięstwo jest za progiem.

Wystąp ty, co funktory reprezentujeszKażdy z nas cię wyeliminujeRezultantami, wyróżnikami, cyrkulantami i alternantamiKroneckerami, Mertensami, Macaulayami i Sylwestrami.

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34 SHREERAM SHANKAR ABHYANKAR (TRANSLATION BY TADEUSZ KRASIŃSKI)

Niech nie wchodzą tutaj omologiści i homologiściA nawet kryształowi kohomologiści,Bo ta ziemia jest święta, matematyczni ateiści.

Naprzód rycerze! Obrońcie swą twierdzęPokonajcie Tor wszelkimi wyznacznikamiI Ext każdymi sposobami.Morficzne injekcje, toksyczne projekcjeEtalne, eklatne pokrętne aberracjeTrujące strzałki długie i krótkieUkryte w sile diagramów magicznej.Tylko Tchirnhausena zbroja złotaOchroni nas przed tą górą błota.

Nie pokonacie nas Chowa pierścieniamiAni Cherna krzykamiBo bronią naszą wielokąty NewtonaI algorytmy Perrona.

Do broni, do broni, ułamki łańcuchaNie bójcie się schematów Grothendiecka duchaBo potęga szeregów potęgowych jest z nami

Nawet jeśli są zbieżne lub nie,Nawet jeśli są wielomianowe lub nieNawet jeśli się kończą lub nie.

Czy naśladowcy Grothendicka samą „gładką gadką”Uczynią kiedykolwiek najmniejszą osobliwość gładką.Niech długo i wiecznie żyją Oscar ZariskiI Nullstellensatz Weierstrassa nam bliski.

Po co nam pierścienie japońskie, wspaniałe lub nieGdy sam Nagata jest po naszej stronie.

Po co tensorowaćGdy można uniformizować

Po co homologizowaćGdy można desyngularyzować

(czy Hironaka na pomoc przybywa?).

Niestety! Princeton i sławny Harvard ty teżDo ZOO paryskiego się zredukujesz.

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SCIENTIFIC ARTICLES




A SHORT PROOF THAT EQUISINGULAR PLANE CURVESINGULARITIES ARE TOPOLOGICALLY EQUIVALENT

SZYMON BRZOSTOWSKI, TADEUSZ KRASIŃSKI, AND JUSTYNA WALEWSKA

Abstract. We prove that if two plane curve singularities are equisingular,then they are topologically equivalent. The method we will use is P. For-tuny Ayuso’s who proved this result for irreducible plane curve singularities.

1. Introduction

Let Γ, Γ be two plane curve singularities (shortly singularities) at 0 ∈ C2. Wetreat a singularity as the germ of an 1-dimensional analytic set passing through 0or as a representative of such a germ. Among many possible equivalences betweenΓ and Γ: topological, analytic, bilipschitz, etc. the most natural is, in retrospect,the topological one. Γ and Γ are topologically equivalent if and only if there existneighbourhoods U1 and U2 of 0 ∈ C2 and a homeomorphism Φ : U2 → U1 suchthat Γ ∩ U1 = Φ(Γ ∩ U2). It is known that the equivalence classes of this relationposses complete, discrete sets of invariants. Such are, for instance:

1. the Puiseux characteristic sequences of branches of Γ together with intersectionmultiplicities between them,

2. the sequences of multiplicities of branches occurring during the desingulariza-tion process of Γ together with an appropriate relation,

3. the dual weighted graph encoding the desingularization process,4. the Enriques diagrams,5. the semigroups of branches and intersection multiplicities between them,

and many others (see [2], [6], [15]). To establish that each of these data sets isa complete set of topological invariants of Γ is a difficult task. The usual references

2010 Mathematics Subject Classification. Primary 32S05; Secondary 14H20.Key words and phrases. Plane curve singularity, equisingular singularities, topologically equiv-

alent singularities, blow up.37


38 S. BRZOSTOWSKI, T. KRASIŃSKI, AND J. WALEWSKA

here are the classical papers of Brauner [1], Kähler [9], Burau [3], [4], but theseare difficult to follow and full of the theory of knots (a new approach, thoughin the same spirit, can be found in Wall [15], see also [10], [11], [12]). In turn,to establish that any two of these data sets are “equivalent”, i.e. determine oneanother, is relatively easier. Thus, it is natural and widely accepted to define: twosingularities Γ, Γ are equisingular if and only if they have the same data sets oftype 1 or 2 or 3 or 4 or 5.

Recently, in the case of branches (=irreducible singularities), P. Fortuny Ayuso[8] gave a new and simple proof of the implication that equisingularity impliestopological equivalence, in which he completely eliminated knot theory. He usedonly desingularization process.

In the article we extend this result to arbitrary singularities (with many branches)using the same idea as P. Fortuny Ayuso. For a proof of the inverse implication(also without knot theory) for bilipschitz equivalency see the recent preprint byA. Fernandes, J. E. Sampaio and J. P. Silva [7]. We also recommend the paperby W. D. Neumann and A. Pichon [14]. Since Ayuso’s method involves desingu-larization process which itself may be described in many ways (see the ways 1, 2,3, 4, 5), we opt for one of these – the Enriques diagrams language – as the mostconvenient for our purposes.

In Section 2 we recall briefly the desingularization process, the Enriques dia-grams and their properties. Section 3 is devoted to the main result.

2. Desingularization and the Enriques diagrams

The basic construction in desingularization process is the blowing-up. Sincedesingularization of plane curve singularities leads naturally to blowing-ups of com-plex manifolds, we recall this notion right away for manifolds. One can find thedetails in many sources [2], [5], [6], [13].

Let M be a 2-dimensional complex manifold and P ∈M. The blowing-up of Mat P is a 2-dimensional manifold M and a holomorphic mapping π : M →M withproperties:

1. E := π−1(P ) is biholomorphic to the 1-dimensional projective space P of linesin C2 passing through 0 (E is called the exceptional divisor of the blowing-up π),

2. π|M\E : M \ E →M \ P is a biholomorphism,

3. for a neighbourhood U of P the mapping π|π−1(U) is biholomorphic to a localstandard blowing-up of a neighbourhood of 0 ∈ C2, where by the standardblowing-up we mean πst : B → C2, where B = (z, l) ∈ C2 × P : z ∈ l andπst(z, l) := z.

The blowing-up M at P always exists and is uniquely defined up to a biholo-morphism. The points in E are called infinitely near to P. Since the blowing-up ofM at P leads to a manifold M , we may repeat the process, this time blowing-up


EQUISINGULAR PLANE CURVE SINGULARITIES 39

M at points of M, in particular at points in E. In this case the points in consec-utive exceptional divisors are also called infinitely near to P. Since the blowing-upπ : M → M is a proper mapping, the image π(Γ) of an arbitrary analytic subsetΓ ⊂ M is an analytic subset of M.

Let Γ be a plane curve singularity at P ∈M. We define the proper preimage ofΓ as the closure π−1(Γ \ P) and denote it by Γ. The analytic set Γ is obtainedby adding to the set π−1(Γ)\E its accumulation points on E. By an analysis of anequation of Γ in local coordinates in M it follows that the number of such points isequal to the number of tangent lines to Γ at P ; in particular, there are only finitelymany of them. All these points are said to lie on or belong to Γ. If Q is sucha point, then the germ of the proper preimage of Γ to which the point Q belongsis denoted by ΓQ. In particular, if Γ is an irreducible singularity, then this is justone point (as an irreducible singularity has only one tangent line). We continuethe process of blowing-ups during desingularization of Γ through blowing-ups atconsecutive points which are infinitely near to P and belong to Γ, until we get

a manifold M and π :M →M such that:

1. the proper preimage Γ of Γ by π is non-singular,2. in each point of Γ∩E the germ

Γ transversally intersects the exceptional divisorE (it means Γ and E at such a point are nonsingular and their tangent linesare different).

Two singularities Γ and Γ at 0 ∈ C2 are equisingular if they “have the same desingu-larization process”. To describe accurately what it means we have to pay attentionto mutual positions of consecutive proper preimages of the singularity with respectto “newly pasted” projective spaces. One of such descriptions is the Enriques’ dia-gram E(Γ) of the resolution of a singularity Γ. It is a graph (precisely a tree) witha distinguished root and two kinds of edges: straight and curved. We outline itsconstruction for an irreducible singularity; for a reducible singularity with manybranches we construct the Enriques diagrams for each branch separately and nextwe “glue” these diagrams by identifying vertices representing the same infinitelynear points and, if neccessary, prolonging blowing-ups to separate branches.

Let Γ be an irreducible singularity at 0 ∈ C2 and π : M → (C2, 0) its resolution.The vertices of E(Γ) are all the points belonging to Γ in this process of desingu-larization (including the point 0 ∈ C2 and all points infinitely near to 0 that lieon Γ). These are centers of consecutive blowing-ups including also the last one inwhich the process is finished. This last point is a maximal point with respect tothe partial ordering in the set of points lying on Γ, induced by successive blowing-ups. The point 0 ∈ C2 is a root of E(Γ). The edges of E(Γ) connect successivecenters. So, for an irreducible singularity E(Γ) is a bamboo (in the language ofgraph theory). However, there are two kinds of edges: straight and curved. Theyare drawn according to the following rules.



Let P and Q be two successive points that belong to Γ, and Q is infinitely nearto P :

1. if ΓP is not tangent to the exceptional divisor at P , then edge PQ is curvedand moreover it has at P the same tangent as the edge ending at P (Figure 1).

2. if ΓP is tangent to the exceptional divisor at P , then edge PQ is straight but:

a) if ΓP is tangent to the “last-pasted” projective space, then this straight edgeis perpendicular to the edge ending at P (Figure 2).

b) if ΓP is tangent to the “earlier-pasted” projective space, then this straightedge is an extension of the previous one, which is also necessarily straight(Figure 3).

The above discussion describes all possible cases that can occur, and thus yieldsthe construction of E(Γ).

Figure 1. ΓP is not tangent to the exceptional divisor. Possi-ble cases: (a) P = 0 is the root of E(Γ), (b) P belongs to onlyone component of E, (c) P belongs to two components of E, (d)corresponding edge in E(Γ).

Remark 1. Notice that, by the very construction of E(Γ), both its first and its lastedge are always curved.



Figure 2. ΓP is tangent to the last-pasted component of Γ. Pos-sible cases: (a) P belongs to only one component of E, (b) Pbelongs to two components of E.

Figure 3. ΓP is tangent to the “earlier-pasted” component of E.

Examples 1. Let Γ = (x, y) : x2 − y3 = 0. Then E(Γ) is as in Figure 4(a).

2. Let Γ = (x, y) : x2 − y5 = 0. Then E(Γ) is as in Figure 4(b).

3. Let Γ = (x, y) : x3 − y5 = 0. Then E(Γ) is as in Figure 4(c).



Figure 4. The Enriques diagrams of singularities in Example 1.

As we stated above, if Γ is a reducible singularity with k branches Γ1, . . . ,Γk,then E(Γ) is formed in the following way: first we construct E(Γ1), . . . , E(Γk) andthen we identify all their vertices representing one and the same infinitely nearpoint (in particular, the tree root 0 is a common point of all E(Γi), i = 1, . . . , k).If the Enriques diagrams of two (or more) branches end at the same infinitely nearpoint, we prolong the process of blowing-ups to separate them. These branchesare already non-singular and transversal to the exceptional divisor; so we add onlycurved edges. It is ilustrated by the following example.Example 4. Let Γ1 = (x, y) : x2 − y3 = 0, Γ2 = (x, y) : x2 − y3 − y4 = 0,Γ = Γ1 ∪ Γ2 = (x, y) : (x2 − y3)(x2 − y3 − y4) = 0. Their Enriques diagrams aredrawn in Figure 5.

Figure 5. (a) The Enriques diagram of both Γ1 and Γ2, (b) theEnriques diagram of Γ.

It is interesting to observe that E(Γ) does not have to be weighted. All neces-sary data needed to recognize the equisingularity class of Γ can be read off from



E(Γ). In particular, there is a formula for the multiplicity µP (Γ) of successiveproper preimages of Γ (see [5], Theorem 3.5.3) at vertices of E(Γ). Moreover, theintersection multiplicity i(Γ, Γ) of two singularities Γ and Γ can also be read offfrom E(Γ) and E(Γ). This is the famous Noether formula (see [5], Theorem 3.3.1).

Theorem 1 (Noether’s formula). If Γ, Γ are two singularities at 0 ∈ C2, then

i(Γ, Γ) =∑

P∈V (E(Γ∪Γ))

µP (ΓP ) · µP (ΓP ),

where by V (E(Γ)) we denote the set of vertices of E(Γ).

After these preparations, we can finally give a precise definition of equisingular-ity.

Definition 1. Two plane curve singularities Γ, Γ are equisingular if their Enriquesdiagrams E(Γ) and E(Γ) are isomorphic (it means there exists a graph isomorphismE(Γ) with E(Γ) which preserves the shapes and angles between edges).

If Γ and Γ are reducible, then the equisingularity of Γ to Γ can be equivalentlyexpressed in the terms of their branches and intersection multiplicities (see [5],Theorem 3.8.6).

Theorem 2. If Γ has k branches γ1, . . . , γk and Γ has k branches γ1, . . . , γk thenΓ and Γ are equisingular if and only if k = k and, after renumbering branches,

1. E(γi) ∼= E(γi), i = 1, . . . , k,2. i(γi, γj) = i(γi, γj), i, j = 1, . . . , k, i 6= j.

3. The main result

We prove the following theorem

Theorem 3. If Γ, Γ are two equisingular plane curve singularities, then Γ and Γare topologically equivalent.

First we need several lemmas.

Lemma 1. If π : B → C2 is the standard blowing-up of C2 at 0 and Φ : B → B isa homeomorphism which keeps the exceptional divisor E = π−1(0) invariant (i.e.Φ(E) = E), then the mapping Φ : C2 → C2 defined by

Φ(x, y) :=

π Φ π−1(x, y) if (x, y) 6= (0, 0)(0, 0) if (x, y) = (0, 0)

is a homeomorphism of C2. We will call Φ the projection of Φ.

Proof. The Lemma is obvious as Φ is a bijection and π is a closed mapping (π iseven a proper mapping).



Of course Lemma 1 can be extended to any sequence of blowing-ups.

Lemma 2. If π : B → B, B, B – complex 2-manifolds, is a composition of blowing-ups and Φ : B → B is a homeomorphism which keeps the exceptional divisor Einvariant, then the projection Φ of Φ is a homeomorphism of B.

Lemma 3 (Ayuso [8]). Let Γ, Γ be two nonsingular branches transverse to bothaxes Ox and Oy. Then there exists a homeomorphism Φ : C2 → C2 of C2 which isthe identity outside any given ball with center at 0, keeps axes Ox and Oy invariant,is biholomorphic in a neighbourhood of 0 ∈ C2, and Φ(Γ) = Γ.

Proof. (Ayuso) We may assume that Γ is the germ of the line L : y = x and Γ is thegerm of the parametric curve y = s(x) = ax+ h.o.t. with a 6= 0. Take the verticalsmooth vector field X = (0, log s(x)

x · y) in a sufficiently small neighbourhood of0 ∈ C2, so that a branch of log s(x)

x exists, and extend it to a smooth verticalvector field on the whole of C2 by gluing it with the zero vector field outside anygiven ball with center at 0. The flow (φt)t∈R for X, consisting of diffeomorphisms,is defined for all t ∈ R (because the support of X is compact). The diffeomorphismΦ := φ1 satisfies all required conditions. In fact, since X is vertical and X = (0, 0)on Ox, Φ keeps axes Ox and Oy invariant. Moreover, for small x ∈ C

Φ(x, x) = φ1(x, x) = c(x,x)(1) = (x, s(x))

where c(x,y)(t), t ∈ R, is the unique integral curve for X satisfying c(x,y)(0) = (x, y);precisely: c(x,y)(t) = (x, yet log s(x)/x) for sufficiently small (x, y) and t. SinceX is holomorphic in a neighbourhood of 0, Φ = φ1 also is biholomorphic ina neighbourhood of 0.

Before we state the next lemma, we introduce a new notion. Let L : y = ax,a ∈ C, be a line in C2 and r > 0. By a cone surrounding L with radius r wemean the set Cr(L) consisting of all lines y = (a+ z)x, |z| < r without the origin.Clearly, Cr(L) is an open set in C2.

Lemma 4. Let L1, . . . , Lm and L1, . . . , Lm, m ≥ 1, be two systems of different linesin C2 passing through 0 ∈ C2. Then there exists a homeomorphism Φ : C2 → C2 ofC2, such that:

1. Φ is the identity outside arbitrary small ball with center at 0,

2. Φ transforms the germs of Li at 0 onto the germs of Li at 0, for i = 1, . . . ,m,

3. Φ transforms biholomorphically some disjoint cones Ci surrounding Li ontocones surrounding Li in a small neighbourhood of 0, and each of these biholo-morphisms Φ|Ci is the restriction of a biholomorphism of a neighbourhood of 0.

In the case Li = Li we may choose this biholomorphic restriction to be identity.

Proof. For simplicity we first assume m = 1. We may arrange things so thatL : y = ax, L : y = bx, a, b ∈ C, a · b 6= 0. The linear mapping Ψ : C2 → C2,



Ψ(x, y) := (x, ab y) is a biholomorphism of C2 which transforms L onto L and more-over maps any cone Cr(L) onto the cone Cr|a|/|b|(L). We will define Φ on eachcomplex plane Cx := x×C ⊂ C2 separately. Note that the trace of Cr(L) on Cxis the disk D(bx, r |x|) with center at bx and radius r |x|, and similarly the traceof Cr|a|/|b|(L) on Cx is the disk D(ax, r |a||b| |x|). The restriction Ψ|Cx maps the

disk D(bx, r |x|) onto the disk D(ax, r |a||b| |x|). Obviously, there exists an extension

Ψx of Ψ|D(bx, r |x|) to a homeomorphism of the whole Cx which is identity out-side an open ball Dx properly containing both D(bx, r |x|) and D(ax, r |a||b| |x|) (i.e.

D(bx, r |x|), D(ax, r |a||b| |x|) ⊂ D); see Figure 6.

Figure 6. A schematic representation of the homeomorphism Ψx

on Cx for small |x|.

Of course, we may choose Dx and Ψx so that they also depend continuously onx. Then we define Φ : C2 → C2 as follows:

1. for small |x| we put Φ|Cx := Ψx,2. for big |x| we put Φ|Cx := Id |Cx,3. for intermediate |x| we continuously join Ψx to Id .

The mapping Φ satisfies all conditions in the assertion of the lemma.The case m ≥ 2 is similar. We should only choose radii r so that the cones

surrounding Li and their images under (x, y) 7→ (x, aibi y) be disjoint.

Now we may pass to the proof of the main theorem.



Proof of the main theorem. Let Γ, Γ be two equisingular plane curve singularities.Hence their Enriques diagrams E(Γ) and E(Γ) are isomorphic. In particular, Γ

and Γ have the same number of branches, say k. After renumbering them we mayassume that γ1, . . . , γk and γ1, . . . , γk are branches of Γ and Γ, respectively, andE(γi) ∼= E(γi), i(γi, γj) = i(γi, γj), i, j = 1, . . . , k, i 6= j.

The vertices of E(Γ) represent points infinitely near to 0 ∈ C2 in the process ofdesingularization π : B → (C2, 0) of Γ. They are centres of consecutive blowing-ups.If we apply this process of desingularization π to Γ, then some of these points willoccur also in E(Γ). For instance, 0 ∈ C2 is a common point of E(Γ) and E(Γ) – itrepresents the root of E(Γ) and E(Γ).We will prove that Γ and Γ are topologicallyequivalent by induction on the sum n of numbers of non-common points in E(γi)and E(γi) for i = 1, . . . , k, where E(γi) (respectively E(γi)) means the subdiagramin E(Γ) (resp. E(Γ)) representing points belonging to γi (resp. γi). Notice E(γi)and E(γi) may differ (see Example 4), but only in points of multiplicity one.

1. n = 0. This means that the process of desingularization of Γ is exactly thesame as of Γ. The centres of consecutive blowing-ups are exactly the same.

Consider one of the maximal points P of desingularization – it represents a leafin E(Γ) and simultaneously in E(Γ). It belongs to only one of branches γ1, . . . , γkand only one of γ1, . . . , γk. Since these branches are equisingular, we may assumeP ∈ γ1 and P ∈ γ1 (see Figure 7).

Figure 7. The case of nonsingular branches transversal to E.

If we denote the proper preimages of γ1 and γ1 passing through P by γP1 andγP1 , then they are non-singular and transversal to E. By Lemma 3, there existsa homeomorphism of B which transforms γP1 onto γP1 in arbitrarily small neigh-bourhood of P , keeps the exceptional divisor invariant, and is the identity outside



another neighbourhood of P. Doing the same for all maximal points of desingu-larization we see that all these homeomorphisms glue to a homeomorphism of Bwhich transforms the proper preimage of Γ by π onto the proper preimage of Γby π while keeping the exceptional divisor invariant. By Lemma 2, its projectiongives a homeomorphism of (C2, 0) which transforms Γ on Γ.

2. Assume the theorem holds for any pair of equisingular singularities for whichthe number of non-common points in all branches in the desingularization processof Γ is equal to (n− 1), n ≥ 1.

Take now singularities Γ, Γ for which this sum is equal to n. Since n ≥ 1, thereexist equisingular branches, say γ1 and γ1, of Γ and Γ such that in E(γ1) thereexist points which do not belong to E(γ1). Take the last common point P in E(γ1)and E(γ1). In this point the proper preimages γP1 and γP1 of both branches γ1 andγ1 have different tangent lines L1 and L1. Moreover, γP1 and γP1 are not tangentto any component of the exceptional divisor (as γ1 and γ1 are equisingular i.e.E(γ1) ∼= E(γ1) and in consequence E(γ1) ∼= E(γ1)); see Figure 8.

Figure 8. The general case of branches with different tangentlines and not tangent to components of the exceptional divisor.

It may happen there exist branches of Γ whose proper preimages at P havethe same tangent line L1 as γP1 . Assume these are γP2 , . . . , γPr , r ≤ k. Then, ofcourse, γP2 , . . . , γPr share the same tangent line L1 as γP1 . Moreover, there mayexist other branches of Γ whose proper preimages also pass through P. Assumethese are γPr+1, . . . , γ

Ps , s ≤ k. Their tangent lines are different from L1. Denote

all their different tangent lines by L2, . . . , Lm. Then, by equisingularity of γi toγi, the branches γPr+1, . . . , γ

Ps also pass through P and have also (m− 1) different

tangent lines, say L2, . . . , Lm.



Now we apply Lemma 4 to the manifold B containing P to get a new singularityΓ′ which will be equisingular and topologically equivalent to Γ and which will haveless non-common points with Γ in desingularization process of Γ than Γ does. Weconsider two cases:

(a) among L2, . . . , Lm there is no L1. Then in Lemma 4 we take the systems oflines L1, . . . , Lm and L1, L2, . . . , Lm. We obtain a homeomorphism Φ of B whichmaps L1 together with branches γP2 , . . . , γPr tangent to it, respectively, onto L1

and some new branches Φ(γP1 ), . . . ,Φ(γPr ) tangent to L1, and which leaves the re-maining branches γPr+1, . . . , γ

Ps passing through P unchanged. Moreover, we may

assume that Φ leaves the exceptional divisor unchanged (in appropriate local co-ordinates at P the exceptional divisor may be represented as additional lines inthe above systems of lines). The projections of Φ(γP1 ), . . . ,Φ(γPr ) to (C2, 0) arenew branches at 0 ∈ C2. Denote them by γ′1, . . . , γ′r. These branches together withγ′r+1 := γr+1, . . . , γ

′k := γk define a new plane curve singularity Γ′. We claim Γ′ is

equisingular and topologically equivalent to Γ. In fact, regarding equisingularity,we notice E(γ′1), . . . , E(γ′r) are isomorphic to E(γ1), . . . , E(γr) because desingu-larization process up to P is the same for γ′i and γi, i = 1, . . . , r, and at P thebranches (γ′1)

P= Φ(γP1 ), . . . , (γ′r)

P= Φ(γPr ) and γP1 , . . . , γ

Pr are biholomorphic

and not tangent to any components of the exceptional divisior passing throughP . Hence obviously E(γ′1), . . . , E(γ′r) are isomorphic to E(γ1), . . . , E(γr). Sinceγ′r+1 = γr+1, . . . , γ

′k = γk, obviously E(γ′r+1) = E(γr+1), . . . , E(γ′k) = E(γk).

Moreover, the equalities i(γ′i, γ′j) = i(γi, γj) i, j = 1, . . . , k, i 6= j, hold for the same

reasons and because of the Noether’s formula. Topological equivalence of Γ′ andΓ is obvious as Φ is a homeomorphism of B which leaves the exceptional divisorunchanged.

(b) among L2, . . . , Lm there is L1, say L2 = L1. Then in Lemma 4 we take thesystems of lines L1, L2, L3 . . . , Lm and L1, L

′, L3, . . . , Lm, where L′ is a new linedifferent from L1, L3, . . . , Lm. The same reasoning as in item (a) also gives a newsingularity Γ′ which is equisingular and topologically equivalent to Γ and whichhas less non-common points with Γ in desingularization process of Γ than Γ does.

In each case we get Γ′ which is equisingular to Γ, and hence to Γ, and which hasless non-common points with Γ in desingularization process of Γ than Γ does. Byinduction hypothesis, Γ is topologically equivalent to Γ′ and hence to Γ. This endsthe proof.

Problem 1. As we know the topological equivalence of plane curve singularities isthe same as their bilipschitz equivalence [14], we pose the problem to find, using theAyuso’s method, a bilipschitz homeomorphism.

References

[1] K. Brauner, Das Verhalten der Funktionen in der Umgebung ihrer Verzweigungsstellen, Abh.Math. Sem. Univ. Hamburg 6(1) (1928), 1–55.



[2] E. Brieskorn and H. Knörrer, Plane algebraic curves, Birkhäuser Verlag, Basel, 1986. Trans-lated from the German by John Stillwell.

[3] W. Burau, Kennzeichnung der Schlauchknoten, Abh. Math. Sem. Univ. Hamburg 9(1)(1933), 125–133.

[4] W. Burau, Kennzeichnung der Schlauchverkettungen, Abh. Math. Sem. Univ. Hamburg,10(1) (1934), 285–297.

[5] E. Casas-Alvero, Singularities of plane curves, volume 276 of London Mathematical SocietyLecture Note Series. Cambridge University Press, Cambridge, 2000.

[6] T. de Jong and G. Pfister, Local analytic geometry, Advanced Lectures in Mathematics.Friedr. Vieweg & Sohn, Braunschweig, 2000, Basic theory and applications.

[7] A. Fernandes, J. E. Sampaio, and J. P. Silva, Hölder equivalence of complex analytic curvesingularities, ArXiv e-prints, April 2017.

[8] P. Fortuny Ayuso, A short proof that equisingular branches are isotopic, ArXiv e-prints,March 2017.

[9] E. Kähler, Über die Verzweigung einer algebraischen Funktion zweier Veränderlichen in derUmgebung einer singulären Stelle, Math. Z. 30(1) (1929), 188–204.

[10] T. Krasiński, Curves and knots I. Torus knots of first order, In XXXII Conference andWorkshop “Analytic and Algebraic Geometry”, 23–45. University of Łódź Press, 2011, (inPolish).

[11] T. Krasiński, Curves and knots II. Torus knots of higher order, In XXXIII Conference andWorkshop “Analytic and Algebraic Geometry”, 33–49. University of Łódź Press, 2012, (inPolish).

[12] T. Krasiński, Curves and knots III. Knots of analytic irreducible curves, In XXXIV Con-ference and Workshop “Analytic and Algebraic Geometry”, 15–25. University of Łódź Press,2013, (in Polish).

[13] S. Łojasiewicz, Geometric desingularization of curves in manifolds, In Analytic and Alge-braic Geometry, 11–32. Faculty of Mathematics and Computer Science. University of Łódź,2013. Translated from the Polish by T. Krasiński.

[14] W.D. Neumann and A. Pichon, Lipschitz geometry of complex curves, J. Singul., 10 (2014),225–234.

[15] C.T.C. Wall, Singular points of plane curves, volume 63 of London Mathematical SocietyStudent Texts. Cambridge University Press, Cambridge, 2004.

Faculty of Mathematics and Computer Science, University of Łódź,ul. Banacha 22, 90-238 Łódź, Poland

E-mail address, Szymon Brzostowski: [email protected]

E-mail address, Tadeusz Krasiński: [email protected]

E-mail address, Justyna Walewska: [email protected]

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THE ŁOJASIEWICZ EXPONENT VIA THE VALUATIVEHAMBURGER-NOETHER PROCESS

SZYMON BRZOSTOWSKI AND TOMASZ RODAK

Abstract. Let k be an algebraically closed field of any characteristic. Weapply the Hamburger-Noether process of successive quadratic transformationsto show the equivalence of two definitions of the Łojasiewicz exponent L(a) ofan ideal a ⊂ k[[x, y]].

1. Introduction

Let k be an algebraically closed field of arbitrary characteristic. Let Ξ denotethe set of pairs of formal power series ϕ ∈ k[[t]]2 such that ϕ 6= 0 and ϕ(0) = 0.We call the elements of Ξ parametrizations. We say that a parametrization ϕis a parametrization of a formal power series f ∈ k[[x, y]] if f ϕ = 0. Forϕ = (ϕ1, . . . , ϕn) ∈ k[[t]]n we put ordϕ := minj ordϕj , where ordϕj stands forthe order of the power series ϕj . Let a ⊂ k[[x, y]] be an ideal. We consider theŁojasiewicz exponent of a defined by the formula

(1.1) L(a) := supϕ∈Ξ

(inff∈a

ord f ϕordϕ

).

Such concept was introduced and studied by many authors in different contexts.Lejeune-Jalabert and Teissier [10] observed that, in the case of several complexvariables, L(a) is the optimal exponent r > 0 in the Łojasiewicz inequality

∃C,ε>0∀||x||<ε maxj|fj(x)| > C||x||r,

where (f1, . . . , fk) is an arbitrary set of generators of a. Moreover, they provedthat, with the help of the notion of integral closure of an ideal, the number L(a)may be seen algebraically. This is what we generalize below (see Theorem 1) partly

2010 Mathematics Subject Classification. Primary 14B05, 13B22; Secondary 13H05, 13F30.Key words and phrases. Łojasiewicz exponent, quadratic transformation, valuation, integral

closure.51


52 S. BRZOSTOWSKI AND T. RODAK

answering [3, Question 2]. D’Angelo [6] introduced L(a) independently, as an orderof contact of a. He showed that this invariant plays an important role in complexfunction theory in domains in Cn.

There has been some interest in understanding the nature of the curves that‘compute’ L(a). In fact, the supremum in (1.1) may be replaced by maximum.A more exact result in this direction says that if a = (f1, . . . , fm)k[[x, y]] is an(x, y)-primary ideal, then there exists a parametrization ϕ of f1 × · · · × fm suchthat

L(a) = inff∈a

ord f ϕordϕ

.

For holomorphic ideals, this was proved by Chądzyński and Krasiński [5], andindependently by McNeal and Némethi [12]. The case of ideals in k[[x, y]], where kis as above, is due to the authors [3]. De Felipe, García Barroso, Gwoździewicz andPłoski [7] gave a shorter proof of this result; moreover, they answered [3, Question1], by showing that L(a) is always a Farey number, i. e. a rational number of theform N + b/a, where N , a, b are integers such that 0 < b < a < N .

2. Methods and results

Once and for all we agree that all the rings considered in the paper are commu-tative with unity. Let a denote the integral closure of an ideal a (see Section 4).Our main result is

Theorem 1. Let a ⊂ k[[x, y]] be an ideal. Then

(2.1) L(a) = inf

p

q: (x, y)pk [[x, y]] ⊂ aq

.

The general idea of the proof is the following. It is easy to see, that the righthand side of (2.1) is equal to

supν

ν (a)

ν ((x, y) k [[x, y]]),

where ν runs through the set of all rank one discrete valuations with center(x, y) k [[x, y]]. This is a consequence of the well-known valuative criterion of inte-gral dependence (see Theorem 5). On the other hand, there is a correspondancebetween valuations of the field k(C) and parametrizations centered at points ofa given irreducible curve C (see [14, Chapter V §10]). A mathematician’s ba-sic instinct, then, lead us to believe that the same reasoning could be repeatedfor parametrizations in place of valuations. For this we need a version of crite-rion of integral dependence which is based on parametrizations (well-known in thecomplex analytic setting). This is where the Hamburger-Noether process comesin. Namely, if (R,m) is a local regular two-dimensional domain, then using Ab-hyankar theorem (Theorem 15) we may find for any given valuation ν with center ma sequence of quadratic transformations of R producing rings and their associatedvaluations which, respectively, approximate the valuation ring of ν and ν itself.


THE ŁOJASIEWICZ EXPONENT 53

The aforementioned valuations, given by the process, are in fact expressible ina quite explicit form even in the case R = k [[x1, . . . , xn]] (see Lemmas 19 and 20);however, the unique feature of Abhyankar theorem is the ‘approximation phenome-non’, which for non-divisorial valuations only holds in the two-dimensional case (cf.Example 18). Altogether, the above observations plus the usual valuative criterionof integral dependence allows us to prove a parametric version of the criterion overk [[x, y]].

The structure of the paper is as follows. Sections 3 and 4 are of introductorynature. In Section 5 we give detailed description of the concept of the quadratictransformation of a local regular domain. This notion was developed and used byZariski and Abhyankar in the 50’s in the framework of valuation theory and theresolution of singularities problem. A sequence of successive quadratic transforma-tions starting from a local regular domain containing an algebraically closed fieldleads to an inductive construction called the Hamburger-Noether process. This isdescribed in Section 6. In this setting Hamburger-Noether process may be consid-ered as a generalization of a classical construction of the normalization of a planealgebroid curve (see [4, 13]) to the case of valuations [8]. Finally, in Sections 7 and8 we prove the aforementioned parametric criterion of integral dependence and asa result obtain Theorem 1.

3. Valuations

An integral domain V is called a valuation ring if every element x of its field offractions K satisfies

x /∈ V =⇒ 1/x ∈ V.We say that V is a valuation ring of K. The set of ideals of a valuation ring V istotally ordered by inclusion. In particular, V is a local ring. In general, this ringneed not be Noetherian, nevertheless its finitely generated ideals are necessarilyprincipal.

A valuation of a field K is a group homomorphism ν : K∗ → Γ, where Γ isa totally ordered abelian group (written additively), such that for all x, y ∈ K∗, ifx+ y 6= 0 then

ν (x+ y) > min ν (x) , ν (y) .Occasionally, when convenient, we will extend ν to K setting ν (0) := +∞. Theimage of ν is called the value group of ν and is denoted Γν . Set

Rν := x ∈ K : x = 0 or ν (x) > 0 ,mν := x ∈ K : x = 0 or ν (x) > 0 .

Then Rν is a valuation ring of K and mν is its maximal ideal.Let Γ be an ordered abelian group. A subgroup Γ′ ⊂ Γ is called isolated if the

relations 0 6 α 6 β, α ∈ Γ, β ∈ Γ′ imply α ∈ Γ′. The set of isolated subgroups ofΓ is totally ordered by inclusion. The number of proper isolated subgroups of Γ iscalled the rank of Γ, and written rk Γ. If ν is a valuation of a field K, then we say



that ν is of rank rk ν := rk Γν . It is well known that the rank of ν is equal to theKrull dimension of Rν [2, VI.4.5 Proposition 5].

If V is a valuation ring of K, then there exists a valuation ν of K such thatV = Rν . If ν1, ν2 are valuations of K then Rν1 = Rν2 if and only if there exists anorder-preserving group isomorphism ϕ : Γν1 → Γν2 satisfying ν2 = ϕ ν1. In sucha case we say that valuations ν1 and ν2 are equivalent.

Let R be an integral domain with field of fractions K. The valuation ν of Kis said to be centered on R if R ⊂ Rν . In this case the prime ideal p = mν ∩ Ris called the center of ν on R. Quite generally, if A ⊂ B is a ring extension, q isa prime ideal of B and p = q ∩ A then we have a natural monomorphism A/p →B/q. Consequently, the residue field of p, that is the field of fractions of A/p,may be considered as a subfield of the residue field of q. In this setting we havethe following important dimension inequality due to I. S. Cohen. We write belowtr.degAB for the transcendence degree of the field of fractions of B over that ofA, where A ⊂ B is an extension of integral domains.

Theorem 2 ([11, Theorem 15.5]). Let A be a Noetherian integral domain, and Ban extension ring of A which is an integral domain. Let q be a prime ideal of Band p = q ∩A; then we have

ht q + tr. degA/pB/q 6 ht p + tr. degAB.

In what follows we will be interested in the case where (R,m, k) is a local Noe-therian domain with residue field k and ν is a valuation with center m on R. Weset tr.degk ν := tr.degk Rν/mν . Directly from the above theorem we get:

Proposition 3. Let (R,m, k) be a local Noetherian domain and let ν be a valuationwith center m on R. Then

rk ν + tr. degk ν 6 dimR.

In particular, tr.degk ν 6 dimR− 1.

Definition 4. Let (R,m, k) be a local Noetherian domain and let ν be a valuationwith center m on R. If tr.degk ν = dimR− 1 then we say that ν is divisorial withrespect to R (or is a prime divisor for R).

4. Integral closure of ideals

Let a be an ideal in a ring R. We say that an element x ∈ R is integral over aif there exist N > 1 and a1 ∈ a, a2 ∈ a2, . . . , aN ∈ aN such that

xN + a1xN−1 + · · ·+ aN = 0.

The set of elements of R that are integral over a is called the integral closure ofa and is denoted a. It turns out that the integral closure of an ideal is always anideal.

Next theorem is the celebrated valuative criterion of integral dependence.



Theorem 5 ([9, Proposition 6.8.4]). Let a be an ideal in an integral Noetheriandomain R. Let V be the set of all discrete valuation rings V of rank one between Rand its field of fractions for which the maximal ideal of V contracts to a maximalideal of R. Then

a =⋂V ∈V

aV ∩R.

5. Quadratic transformation of a ring

Definition 6. Let (R,m) be a local regular domain and let x ∈ m \ m2. SetS = R

[mx

]and let p be a prime ideal in S containing x. Then the ring Sp is called

a (first) quadratic transform of R. If ν is a valuation with center m on R andxRν = mRν then Sp, where p := R ∩mν , is called a (first) quadratic transform ofR along ν.

Remark 7. Keep the notations from the above definition. Then xS = mS and forany k ∈ N, xkS ∩R = mkS ∩R = mk. Indeed, the equalities xS = mS, xkS = mkSand the inclusion mk ⊂ mkS ∩ R are clear. Take r ∈ mkS ∩ R. Then there existl > 0 and aj ∈ mk+j , j = 0, . . . , l, such that

a0 +a1

x+ · · ·+ al

xl= r.

Thus xlr ∈ mk+l. On the other hand, (R,m) is a local regular domain, hence theassociated graded ring grmR is an integral domain (as isomorphic to the ring ofpolynomials R

m [Y1, . . . , Yn]). We have(xl + ml+1

)·(r + mk

)= xlr + mk+l, which

is zero in grmR. Consequently, since xl /∈ ml+1 we must have r ∈ mk.

Remark 8. It is clear from the definition, that if (T, n) is a quadratic transformationof (R,m) along ν then ν has center n on T .

Proposition 9. Let (R,m) be a local regular domain of dimension n > 1. Setx1, . . . , xn as the generators of m. Let R [Y ], where Y = (Y2, . . . , Yn), be apolynomial ring in n − 1 variables over R. If ϕ : R [Y ] → S := R

[x2

x1, . . . , xn

x1

]is an R-homomorphism given by ϕ (Yj) := xj/x1, j = 2, . . . , n, then kerϕ =(x1Y2 − x2, . . . , x1Yn − xn)R [Y ].

Proof. Take f ∈ kerϕ. Using successive divisions with remainder we may write fin the form

f (Y ) = A2 ·(Y2 −

x2

x1

)+ · · ·+An ·

(Yn −

xnx1

)+B,

where A2, . . . , An ∈ S [Y ], B ∈ R[x2

x1, . . . , xn

x1

]. We must have B = 0, since



f ∈ kerϕ. There exists N such that

(5.1) xN1 f (Y ) = A′

2 · (x1Y2 − x2) + · · ·+A′

n · (x1Yn − xn) ,

where A′

2, . . . , A′

n ∈ R [Y ].Now, observe that R [[Y ]] is a regular local ring of dimension 2n − 1 and

x1Y2 − x2, . . . , x1Yn − xn, x1, Y2, . . . , Yn is its regular system of parameters [11,Theorems 15.4, 19.5]. Thus R [[Y ]] / (x1Y2 − x2, . . . , x1Yn − xn) is a regular lo-cal domain and, consequently (x1Y2 − x2, . . . , x1Yn − xn)R [[Y ]] is a prime ideal.Thus (x1Y2 − x2, . . . , x1Yn − xn)R [Y ] is also prime. Moreover, this ideal doesnot contain x1 since x1, . . . , xn minimally generates m. This and (5.1) givesf ∈ (x1Y2 − x2, . . . , x1Yn − xn)R [Y ].

Proposition 10. Under the notations from Proposition 9 we have:

1) S is regular,2) if p ⊂ S is a prime ideal containing x1 then Sp is a regular local ring and

tr.degR/m Sp/pSp = dimR− dimSp,

3) if p = mν ∩S, where ν is a valuation with center m on R such that ν (x1) 6ν (xj), j = 2, . . . , n, then

tr.degR/m ν − tr.degSp/pSpν = dimR− dimSp.

Proof. Let p ⊂ S be a prime ideal. We have R ⊂ S ⊂ Rx1 , so Rx1 = Sx1 . Thus, ifx1 /∈ p then

Sp = (Sx1)pSx1

= (Rx1)pRx1

= Rp,

hence Sp is a regular local ring.Now, assume that x1 ∈ p. Let R [Y ], Y = (Y2, . . . , Yn), be a polynomial ring.

Put b := (x1Y2 − x2, . . . , x1Yn − xn)R [Y ]. We have S ' R [Y ] /b by Proposition9. Let p? := p/x1S, S? := S/x1S. Since b ⊂ mR [Y ] and x1S = mS,

(5.2) S? =S

mS' R [Y ]

mR [Y ]' R

m[Y ] .

The ring S? is regular, as a ring of polynomials over a field, thus there existy2, . . . , yk+1 ∈ S, such that p?S?p? = (y2, . . . , yk+1)S?p? and ht p? = k. Moreover

dimSp = ht pSp = ht pS = ht p? + 1 = k + 1

and pSp = (x1, y2, . . . , yk+1)Sp. Consequently, Sp is a regular local ring. Thisproves 1).

Using the identifications (5.2), we have

tr.degR/mSp

pSp= tr.degR/m

(Rm [Y ]

p? Rm [Y ]

)0

= dimRm [Y ]

p? Rm [Y ]

= dimR

m[Y ]− ht p? = n− 1− k = dimR− dimSp.

This gives 2).



Since R/m ⊂ Sp/pSp ⊂ Rν/mν , the proof of 3) follows from 2) and from theequality

tr.degR/mRν/mν = tr.degSp/pSpRν/mν + tr. degR/m Sp/pSp.

Lemma 11. Let (T, n) be a quadratic transformation of R. Then

1) nk ∩R = mk for any k ∈ N,2) if xT = mT for some x ∈ R, then x ∈ m\m2 and T = Sp, where S := R

[mx

]and p := S ∩ n.

Proof. By the definition of the quadratic transformation there exist x′ ∈ m \ m2

and a prime ideal p′ in S′ := R[mx′

]such that x′ ∈ p′, T = S′p′ , n = p′T .

We have

mk ⊃ nk ∩R =(nk ∩ S′

)∩R ⊃ p′k ∩R ⊃ x′kS′ ∩R = mkS′ ∩R = mk.

This gives the first assertion.For the proof of the second one, observe that x ∈ mT∩R ⊂ n∩R = m. Moreover,

if x ∈ m2, then m = xT ∩ R ⊂ m2T ∩ R ⊂ n2 ∩ R = m2, which is a contradiction.Thus x ∈ m \m2.

Set S := R[mx

]. Since xT = mT = x′T , the element x/x′ is invertible in T .

Hence S ⊂ T . Let p := n∩S. Clearly Sp ⊂ T . On the other hand, the localizationsS x′

xand S′x

x′are equal; denote them by Q. Since p′Q = n ∩Q and pQ ⊂ n ∩Q,

T = S′p′ = Qp′Q = Qn∩Q ⊂ QpQ = Sp.

Definition 12. Let (R,m) be a local regular domain and let f ∈ R, f 6= 0. Thenwe write ordR f for the greatest l > 0 such that f ∈ ml. As usually, we also putordR 0 := +∞. We will call ordR the order function on R. Moreover, for an ideala ⊂ R we put ordR a := minf∈a ordR f .

Corollary 13. Let (R,m) be a local regular domain. Then the order function ordRis a valuation of the field of fractions of R. Moreover, if x ∈ m \m2, S := R

[mx

]and p := xS, then T := Sp is a valuation ring of the order function on R.

Proof. Since as in the proof of Proposition 10, S/xS is isomorphic with the ring ofpolynomials with coefficients in R/m, the ideal xS is prime and htxS = 1. Thus,again by Proposition 10, T is a local regular one-dimensional domain. Hence it isa discrete valuation ring of rank one with valuation given by ordT . By Lemma 11,nr ∩R = mr, so

(nr \ nr+1

)∩R = mr \mr+1 and we get that ordT restricted to R

is equal to ordR. Consequently, ordR extends to a valuation of the field of fractionsof R with valuation ring equal to T .



From Proposition 10 we infer that the quadratic transformation Sp of R is againa regular local domain. If ht p > 1 then dimSp > 1, thus we may set R′ = Sp andconsider a quadratic transformation of R′. This leads to an inductive process, whereat each step we must choose the ‘center’ of the next quadratic transformation. Thisprocess is finite exactly when at some point as the ‘center’ we take a height oneprime ideal. In this case we end up with a discrete valuation ring of rank one.

In what follows we will be interested in the situation in which the above processis driven by a certain valuation ν with center m on R. Here, at each step as thenext ‘center’ we take the ideal Ri ∩ mν . As a result we get a sequence (finite ornot) of quadratic transformations along ν:

(5.3) R = R0 ⊂ R1 ⊂ · · · ⊂ Rν .

Remark 14. Actually, the sequence 5.3 is uniquely determined by the valuation ν.To see this it is enough to check that a local quadratic transformation (T, n) of(R,m) along ν is unique. Let x, x′ ∈ m \ m2 be such that xRν = mRν = x′Rν .Set S := R

[mx

], p := mν ∩ S, T := Sp and similarly S′ := R

[mx′

], p′ := mν ∩ S′,

T ′ := S′p′ . Since x′/x ∈ S \ p, x′/x is invertible in T . Hence x′T = xT = n andS′ ⊂ T , where we set n := pT . Moreover, n ∩ S′ = (mν ∩ T ) ∩ S′ = mν ∩ S′ = p′.Thus T = T ′ by Lemma 11.

Theorem 15 ([1, Proposition 3, Lemma 12]). The sequence (5.3) is finite if andonly if ν is a divisorial valuation with respect to R. In this case there exists m > 1such that

R = R0 ⊂ R1 ⊂ · · · ⊂ Rm−1 ⊂ Rm = Rν .

Moreover, if dimR = 2 and the sequence (5.3) is infinite, then

Rν =⋃i

Ri and mν =⋃i

mi,

where mi stands for the maximal ideal of Ri.

Lemma 16. Let (R,m) be a two-dimensional local regular domain and let ν bea valuation with center m on R. Assume that (5.3) is a sequence of quadratictransformations along ν. Let F ⊂ Rν \ 0 be a finite set and let h ∈ Rν \ 0 besuch that for every f ∈ F we have f/h ∈ mν . Then there exists i > 0 such thatdimRi = 2 and minf∈F ordRi f > ordRi h.

Proof. By Theorem 15 there exists i such that f/h ∈ mi for any f ∈ F . Henceminf∈F ordRi

f > ordRih. Thus, we get the assertion if dimRi = 2. So, assume

that dimRi = 1. This means that the sequence (5.3) is necessarily finite andRi = Rν is a valuation ring of ordRi−1

. It follows that ordRi−1= ordRi

. SincedimRi−1 = 2, we get the assertion.



6. Hamburger-Noether expansion

Let (R,m) be an n-dimensional local regular domain, n > 1. We will assume inthis section that there exists an algebraically closed field k ⊂ R such that k → R/mis an isomorphism.

Lemma 17. Let (T, n) be a quadratic transformation of R. Then the followingconditions are equivalent:

1. dimT = n,2. tr.degR/m T/n = 0,3. the natural homomorphism k → T/n is an isomorphism,4. for every regular system of parameters x1, . . . , xn of R there exist j ∈1, . . . , n and a1, . . . aj−1, aj+1, . . . , an ∈ k such that

x1

xj− a1, . . . ,

xj−1

xj− aj−1, xj ,

xj+1

xj− aj+1, . . . ,

xnxj− an

is a regular system of parameters of T .

Proof. 1. =⇒ 2. Follows from Proposition 10.2. =⇒ 3. By the assumptions the field R/m is algebraically closed and the field

extension k = R/m ⊂ T/n is algebraic. Hence, the last inclusion is in fact equality.Consequently, the field k ⊂ T is isomorphic with the residue field of T .

3. =⇒ 4. The ideal mT is principal, hence without loss of generality we mayassume that mT = x1T . Choose ai ∈ k as the image of xi/x1 in T/n. PutS := R

[mx1

], p := n ∩ S. Then by Lemma 11 we have T = Sp, n = pT . Every

f ∈ S may be written in the form

f = f0 +A

(x2

x1− a2, . . . ,

xnx1− an

),

where f0 ∈ R and A ∈ R [Y2, . . . , Yn] is a polynomial without constant term. Wehave f ∈ n if and only if f0 ∈ m, hence

p =

(x1,

x2

x1− a2, . . . ,

xnx1− an

)R

[m

x1

].

Thusp

x1S' (Y2 − a2, . . . , Yn − an)

R

m[Y ] ,

by Proposition 9. Consequently dimT = dimSp = ht p = n.4. =⇒ 1. Obvious.

Example 18. Set

ν (x) := (0, 0, 1) ,

ν (y) := (0, 1, 0) ,

ν (z) := (1, 0, 0)



and for any f ∈ k [[x, y, z]] \ 0 put as ν (f) the lexicographic minimum of

aν (x) + bν (y) + cν (z) : (a, b, c) ∈ supp f ,

where supp f denotes the set of (a, b, c) ∈ Z3 such that the monomial xaybzc appearsin the expansion of f with non-zero coefficient. It is easy to see that ν extends toa valuation with center (x, y, z) k [[x, y, z]]. The value group Γν is equal to Z3 withlexicographical ordering. Let

k [[x, y, z]] =: R0 ⊂ R1 ⊂ · · · ⊂ Rν

be the sequence of successive quadratic transformations of k [[x, y, z]] along ν. Ob-serve that ν (z/y) = (1,−1, 0) > 0, hence z/y ∈ Rν . Nevertheless, we claimthat z/y /∈

⋃∞i=0Ri. Indeed, set S := R0 [y/x, z/x] and notice that, since

ν (x) < ν (y) < ν (z), we have

p := mν ∩ S =(x,y

x,z

x

)S

is a maximal ideal in S. Thus R1 = (R0)p and x1 := x, y1 := y/x, z1 := z/x

is the regular system of parameters in R1, where again ν (x1) < ν (y1) < ν (z1).Obviously z/y = z1/y1 /∈ R1 and in the same way z/y /∈ R2 and so on. This provesthat the second statement in the Theorem 15 does not hold in the multidimensionalcase.

Lemma 19. Let (T, n) be an n-dimensional local regular domain such that thereexists a sequence

(6.1) R = R0 ⊂ R1 ⊂ · · · ⊂ Rm = T,

where for each i = 1, . . . ,m, Ri is a quadratic transformation of Ri−1. Setx1, . . . , xn as the generators of m. Then there exists a regular system of pa-rameters y1, . . . , yn of T and polynomials A1, . . . , An ∈ k [Y1, . . . , Yn] such thatxj = Aj (y1, . . . , yn), j = 1, . . . , n.

Proof. Induction with respect to m. The case m = 0 is trivial. Assume that theassertion is true for some m− 1 > 0. Consider the sequence (6.1). By Proposition10 we have dimR0 > dimR1 > · · · > dimRm. Thus, for each i = 0, . . . ,m,dimRi = n. By the induction hypothesis there exist a regular system of parametersy′1, . . . , y

′n of Rm−1 and polynomials A′1, . . . , A′n ∈ k [Y1, . . . , Yn] such that xj =

A′j (y′1, . . . , y′n), j = 1, . . . , n. On the other hand, by Lemma 17, there exist j0,

a regular system of parameters y1, . . . , yn of Rm and a1, . . . aj0−1, aj0+1, . . . , an ∈ k



such that

y′1 = yj0 (y1 + a1) ,

...

y′j0−1 = yj0 (yj0−1 + aj0−1) ,

y′j0 = yj0 ,

y′j0+1 = yj0 (yj0+1 + aj0+1) ,

...

y′n = yj0 (yn + an) .

Now, according to the above equalities we may easily define polynomialsA1, . . . , An.

Let R := k [[x1, . . . , xn]] be the ring of formal power series and let f ∈ R \ 0.We will write in f for the initial form of f , which is the lowest degree non-zerohomogeneous form in the expansion of f . Clearly, ordR f is equal to the degree ofthe initial form of f . For the ring of formal power series R as above we will oftenwrite ord(x1,...,xn) instead of ordR.

Lemma 20. Let R := k [[x1, . . . , xn]] be a ring of formal power series. Let (T, n)be an n-dimensional local regular domain between R and field of fractions of R.Assume that there exists a regular system of parameters y1, . . . , yn of T and poly-nomials A1, . . . , An ∈ k [Y1, . . . , Yn] such that xj = Aj (y1, . . . , yn), j = 1, . . . , n.Then for every non-zero f ∈ R we have

ordT f = ord(Y1,...,Yn) f (A1 (Y1, . . . , Yn) , . . . , An (Y1, . . . , Yn)) .

Proof. Set Φ := (A1, . . . , An). Take f ∈ R, f 6= 0.First, assume that f is a polynomial. We have

f (x1, . . . , xn) = f (Φ (Y1, . . . , Yn))|Y1=y1,...,Yn=yn.

Thus f (Φ (Y1, . . . , Yn)) is a non-zero polynomial. Let P := in f (Φ (Y1, . . . , Yn)).Since y1, . . . , yn is a regular system of parameters of T ,

ordT f = ordT P (y1, . . . , yn) = degP = ord(Y1,...,Yn) f (Φ (Y1, . . . , Yn)) ,

which gives the assertion in this case.If f is an arbitrary non-zero power series then, cutting the tail in the power

series expansion of f , we find a polynomial f ∈ R such that ordT f = ordT f andord(Y1,...,Yn) f = ord(Y1,...,Yn) f . By the case considered above we have ordT f =

ord(Y1,...,Yn) f .



7. Parametric criterion of integral dependence

Let R = k [[x, y]], ∆ = k [[t]] be the rings of formal power series over an alge-braically closed field k. Let m and d be the maximal ideals of R and ∆ respectively.For any ϕ ∈ d× d we have a natural local k-homomorphism ϕ∗ : R → ∆ given bythe substitution.

Theorem 21. Let a be an ideal in R and let h ∈ R. Then h is integral over a ifand only if ϕ∗h ∈ ϕ∗a for any ϕ ∈ d× d.

Proof. Assume that h is integral over a. There exist an integer N and the elementsaj ∈ aj , j = 1, . . . , N , such that

hN + a1hN−1 + · · ·+ aN = 0.

Take parametrization ϕ ∈ d2. Let r := ordR a. Then

Nord∆ϕ∗h > min

j(rj + (N − j) ord∆ϕ

∗h) .

This gives ord∆ϕ∗h > r, hence ϕ∗h ∈ ϕ∗a.

Assume now, that h is not integral over a. Since the case a = 0 is clear, inwhat follows we will assume that a 6= 0. By the valuative criterion of integraldependence (Theorem 5) there exists a valuation ν with center m on R such thath /∈ aRν . Consider the sequence of successive quadratic transformations of R alongν:

R = R0 ⊂ R1 ⊂ · · · ⊂ Rν .

Denote by mi the only maximal ideal of Ri, i > 0. Let F ⊂ R\0 be any finite setof generators of a. Then f/h ∈ mν for any f ∈ F . Hence, by Lemma 16 there existsi > 0 such that dimRi = 2 and minf∈F ordRi

f > ordRih. By Lemmas 19 and

20, there exist polynomials A,B ∈ k [X,Y ] such that for any g ∈ R, ordRig =

ord(X,Y ) g (A (X,Y ) , B (X,Y )). Set Pg (X,Y ) := in g (A (X,Y ) , B (X,Y )) forg ∈ R. Then degPg = ordRi g. Let (a, b) ∈ k2 be such that Ph (a, b) 6= 0 andPf (a, b) 6= 0 for f ∈ F . Put ϕ := (A (at, bt) , B (at, bt)). Clearly ord∆ ϕ∗h = degPhand ord∆ ϕ∗f = degPf for f ∈ F . Hence ord∆ ϕ∗h < minf∈F ord∆ ϕ∗f =minf∈a ord∆ ϕ∗f , so ϕ∗h /∈ ϕ∗a.

Example 22. Let R = k[[x, y]], where k is an algebraically closed field. Considera := (x2 + y3, x3), h := y4, f := x2 + y3. Let ϕ := (t3,−t2) ∈ d × d. Notice thatϕ∗f = 0. Now, for any g ∈ R\0 we define ν(g) := (k, ord∆ ϕ∗g′), where g = fkg′

and gcd(f, g′) = 1. It is easy to check that ν extends to a valuation with center(x, y)R on R. We will find the Hamburger-Noether expansion along ν. Using thiswe will show that h is not integral over a.

First step.: We have ν(x) = (0, 3), ν(y) = (0, 2), so we put x1 := xy , y1 := y.

Second step.: Now ν(x1) = (0, 1), ν(y1) = (0, 2), so let x2 := x1, y2 := y1x1.



Continuing in the above manner we get

Recursiveformula forxi, yi

Valuation xi, yi in terms of x, y x, y in terms of xi, yi

x1 := xy,

y1 := y

ν(x1) = (0, 1),ν(y1) = (0, 2)

x1 = xy,

y1 = y

x = x1y1,y = y1

x2 := x1,y2 := y1

x1

ν(x2) = (0, 1),ν(y2) = (0, 1)

x2 = xy,

y2 = y2

x

x = x22y2,y = x2y2

x3 := x2,y3 := y2

x2+ 1

ν(x3) = (0, 1),ν(y3) = (1,−6)

x3 = xy,

y3 = y3+x2

x2

x = x33(y3 − 1),y = x23(y3 − 1)

x4 := x3,y4 := y3

x3

ν(x4) = (0, 1),ν(y4) = (1,−7)

x4 = xy,

y4 = (y3+x2)y

x3

x = x34(x4y4 − 1),y = x24(x4y4 − 1)

x5 := x4,y5 := y4

x4

ν(x5) = (0, 1),ν(y5) = (1,−8)

x5 = xy,

y5 = (y3+x2)y2

x4

x = x35(x25y5 − 1),

y = x25(x25y5 − 1)

x6 := x5,y6 := y5

x5

ν(x6) = (0, 1),ν(y6) = (1,−9)

x6 = xy,

y6 = (y3+x2)y3

x5

x = x36(x36y6 − 1),

y = x26(x36y6 − 1)

......

......

xi := xi−1,yi :=

yi−1

xi−1

ν(xi) = (0, 1),ν(yi) = (1,−i−3)

xi =xy,

yi =(y3+x2)yi−3

xi−1

x = x3i (xi−3i yi − 1),

y = x2i (xi−3i yi − 1)

Successive steps of the Hamburger-Noether algorithm.

Hence, aRi = (x6i (x

i−3i yi− 1)2 +x6

i (xi−3i yi− 1)3, x9

i (xi−3i yi− 1)3)Ri = x9

iRi andhRi = x8

iRi for i > 6. Thus h /∈⋃i>6 aRi = aRν . Observe also that y5 ∈ a \ a.

8. The main result

We keep the notations from the previous section. In particular R = k [[x, y]], kis algebraically closed and for an ideal a ⊂ R we have

L (a) = sup06=ϕ∈d×d

(inff∈a

ord∆ ϕ∗f

ord∆ ϕ∗ (x, y)R

)= sup

06=ϕ∈d×d

ord∆ ϕ∗a

ord∆ ϕ∗ (x, y)R.

Recall that we want to prove the following

Theorem 1. Let a ⊂ R be an ideal. Then

(8.1) L(a) = inf

p

q: (x, y)

pR ⊂ aq

.

Proof. The cases a = R or a = 0 are trivial. Assume that a is a proper ideal andht a = 1. Then, clearly, the right hand side of (8.1) is equal to ∞. Let p ⊂ R be



a height one prime ideal such that a ⊂ p. By [13, Appendix C] there exists f ∈ Rsuch that p = fR. Hence, one can find ϕ ∈ d× d such that ϕ∗f = 0 [13, Theorem2.1]. Consequently L (a) =∞.

Now, assume that ht a = 2, so that a is (x, y)R-primary.‘6’ Fix any p > 0, q > 0 such that (x, y)

pR ⊂ aq. Take ϕ ∈ d× d. Without loss

of generality we may assume that ord∆ ϕ∗x 6 ord∆ ϕ∗y. Since xp ∈ aq, Theorem21 asserts that ord∆ ϕ∗xp > ord∆ ϕ∗aq. This easily gives

p

q>

ord∆ ϕ∗a

ord∆ ϕ∗x=

ord∆ ϕ∗a

ord∆ ϕ∗ (x, y)R.

Hence p/q > L (a) and consequently we get the desired inequality.‘>’ Take any p > 0, q > 0 such that p/q > L (a). Then, for every ϕ ∈ d × d,

ϕ 6= 0, we havep

q>

ord∆ ϕ∗a

ord∆ ϕ∗ (x, y)R

or, what amounts to the same thing, ord∆ ϕ∗ (x, y)pR > ord∆ ϕ∗aq. Hence, for any

h ∈ (x, y)pR we have ord∆ ϕ∗h > ord∆ ϕ∗aq. Thus, (x, y)

pR ⊂ aq, by Theorem

21. As a result, we get the inequality ‘>’ in (8.1).

References

[1] S. Abhyankar, On the valuations centered in a local domain, Amer. J. Math., 78 (1956),321–348.

[2] N. Bourbaki, Elements of mathematics. Commutative algebra, Hermann, Paris; Addison-Wesley Publishing Co., Reading, Mass., 1972. Translated from the French.

[3] S. Brzostowski and T. Rodak, The Łojasiewicz exponent over a field of arbitrary character-istic, Rev. Mat. Complut., 28 (2) (2015), 487–504.

[4] A. Campillo, Algebroid curves in positive characteristic, Lecture Notes in Mathematics 813,Springer, Berlin, 1980.

[5] J. Chądzyński and T. Krasiński, The Łojasiewicz exponent of an analytic mapping of twocomplex variables at an isolated zero, in: Singularities (Warsaw, 1985), Banach Center Publ.20, PWN, Warsaw, 1988, 139–146.

[6] J.O. D’Angelo, Real hypersurfaces, orders of contact, and applications, Annals of Mathe-matics, 115 (3) (1982), 615–637.

[7] A.B. de Felipe, E.R. García Barroso, J. Gwoździewicz and A. Płoski, Łojasiewicz exponentsand Farey sequences, Rev. Mat. Complut., 29 (3) (2016), 719–724.

[8] C. Galindo, Intersections of 1-forms and valuations in a local regular surface, J. Pure Appl.Algebra, 94 (3) (1994), 307–325.

[9] C. Huneke and I. Swanson, Integral closure of ideals, rings, and modules, London Mathe-matical Society Lecture Note Series 336, Cambridge University Press, Cambridge, 2006.

[10] M. Lejeune-Jalabert and B. Teissier, Clôture intégrale des idéaux et équisingularité, Ann.Fac. Sci. Toulouse Math. (6) 17 (2008), no. 4, 781–859.

[11] H. Matsumura, Commutative ring theory, Cambridge Studies in Advanced Mathematics, 8,Cambridge University Press, Cambridge, second edition, 1989. Translated from the Japaneseby M. Reid.

[12] J.D. McNeal and A. Némethi, The order of contact of a holomorphic ideal in C2, Math. Z.250 (4) (2005), 873–883.



[13] A. Płoski, Introduction to the local theory of plane algebraic curves, in: Analytic and al-gebraic geometry (Łódź 2013, Faculty of Mathematics and Computer Science. University ofŁódź, Łódź, 2013, 115–134.

[14] R.J. Walker. Algebraic curves. Dover Publications, Inc., New York, 1962.

Faculty of Mathematics and Computer Science, University of Łódź,ul. S. Banacha 22, 90-238 Łódź, Poland

E-mail address, Szymon Brzostowski: [email protected]

E-mail address, Tomasz Rodak: [email protected]

“17˙Janeczko˙Jelonek˙Ruas” — 2017/12/1 — 20:44 — page 66 — #1





ON A GENERIC SYMMETRY DEFECT HYPERSURFACE

STANIS LAW JANECZKO, ZBIGNIEW JELONEK,AND MARIA APARECIDA SOARES RUAS

Abstract. We show that symmetry defect hypersurfaces for two generic

members of the irreducible algebraic family of n−dimensional smooth irre-ducible subvarieties in general position in C2n are homeomorphic and they

have homeomorphic sets of singular points. In particular symmetry defect

curves for two generic curves in C2 of the same degree have the same numberof singular points.

1. Introduction

Let Xn ⊂ C2n be a smooth algebraic variety. In [7] we have investigated thecentral symmetry of X (see also [1], [2], [3]). For p ∈ C2n we have introduceda number µ(p) of pairs of points x, y ∈ X, such that p is the center of the intervalxy. Recall that the subvariety Xn ⊂ C2n is in a general position if there existpoints x, y ∈ Xn such that TxX ⊕ TyX = C2n.

We have showed in [7] that if X is in general position, then there is a closedalgebraic hypersurface B ⊂ C2n, called symmetry defect hypersurface of X, suchthat the function µ is constant (non-zero) exactly outside B. Here we prove that thesymmetry defect hypersurfaces for two generic members of an irreducible algebraicfamily of n−dimensional smooth irreducible subvarieties in general position in C2n

are homeomorphic.

Moreover, by a version of Sard theorem for singular varieties (see [4]), we havethat the symmetry defect hypersurfaces for two generic members of an irreducible

2010 Mathematics Subject Classification. 14D06, 14Q20.Key words and phrases. Center symmetry set, affine algebraic variety, family of algebraic sets,

bifurcation set of a polynomial mapping.Z. Jelonek and M. A. S. Ruas are partially supported by the grant of Narodowe Centrum

Nauki, grant number 2015/17/B/ST1/02637.

67


68 S. JANECZKO, Z. JELONEK, AND M.A.S. RUAS

algebraic family of n−dimensional smooth irreducible subvarieties in general po-sition in C2n have homeomorphic sets of singular points. In particular symmetrydefect curves for two generic curves in C2 of the same degree have the same numberof singular points.

2. Bifurcation set

Let X be an irreducible affine variety. Let Sing(X) denote the set of singularpoints of X. Let Y be another affine variety and consider a dominant morphismf : X → Y. If X is smooth then by Sard’s Theorem a generic fiber of f is smooth.In a general case the following theorem holds (see [4]):

Theorem 2.1. Let f : Xk → Y l be a dominant polynomial mapping of affinevarieties. For generic y ∈ Y we have Sing(f−1(y)) = f−1(y) ∩ Sing(X).

Recall the following (see [5], [6]):

Definition 2.2. Let f : X → Y be a generically-finite (i.e. a generic fiber is finite)

and dominant (i.e. f(X) = Y ) polynomial mapping of affine varieties. We say thatf is finite at a point y ∈ Y, if there exists an open neighborhood U of y such thatthe mapping f |f−1(U): f

−1(U)→ U is proper.

It is well-known that the set Sf of points at which the mapping f is not finite,is either empty or it is a hypersurface (see [5], [6]). We say that the set Sf is theset of non-properness of the mapping f.

Definition 2.3. Let X,Y be smooth affine n−dimensional varieties and let f :X → Y be a generically finite dominant mapping of geometric degree µ(f). Thebifurcation set of the mapping f is the set

B(f) = y ∈ Y : #f−1(y) 6= µ(f).

We have the following theorem (see [7]):

Theorem 2.4. Let X,Y be smooth affine complex varieties of dimension n. Letf : X → Y be a polynomial dominant mapping. Then the set B(f) is either empty(so f is an unramified topological covering) or it is a closed hypersurface.

3. A super general position

In this section we describe some properties of a variety Xn ⊂ C2n which impliesthat X is in a general position. Recall that the subvariety Xn ⊂ C2n is in a generalposition if there exist points x, y ∈ Xn such that TxX ⊕ TyX = C2n.

Definition 3.1. Let Xn ⊂ C2n be a smooth algebraic variety. We say that X isin very general position if there exists a point x ∈ X such that the set TxX ∩ Xhas an isolated point (here we consider TxX as a linear subspace of C2n).


ON A GENERIC SYMMETRY DEFECT HYPERSURFACE 69

We consider also a slightly stronger property:

Definition 3.2. Let Xn ⊂ C2n be a smooth algebraic variety and let S = X \X ⊂π∞ be the set of points at infinity of Xn. We say that X is in super general positionif there exists a point x ∈ X such that TxX ∩ S = ∅ (here we consider TxX asa linear subspace of P2n = C2n ∪ π∞).

We have the following:

Proposition 3.3. If X is in a super general position, then it is in a very generalposition.

Proof. Let x ∈ X be a point such that TxX ∩S = ∅. Take R = TxX ∩X. Then theset R is finite, since otherwise the point at infinity of R belongs to TxX∩S = ∅.

We have also:

Proposition 3.4. Let X ⊂ C2n be in a super general position. Then for a genericpoint x ∈ X we have TxX ∩ S = ∅.

Proof. It is easy to see that the set Γ = (s, x) ∈ S ×X : s ∈ TxX is an algebraicsubset of S×X. Let π : Γ 3 (s, x)→ x ∈ X be a projection. It is a proper mapping.Since the variety X is in a very general position, we see that at least one pointx0 ∈ X is not in the image of π. Thus almost every point of X is not in the imageof π, because the image of π is a closed subset of X.

Finally we have:

Theorem 3.5. If X ⊂ C2n is in a very general position, then it is in a generalposition, i. e., there exist points x, y ∈ X such that TxX ⊕ TyX = C2n. In fact forevery generic pair (x, y) ∈ X ×X we have TxX ⊕ TyX = C2n.

Proof. Let x0 ∈ X be the point such that the set Tx0X ∩ X has an isolated

point. The space Tx0X is given by n linear equations li = 0. Let F : X 3

x → (l1(x), ..., ln(x)) ∈ Cn. By the assumption the fiber over 0 of F has an iso-lated point, in particular the mapping F is dominant. Now by the Sard Theo-rem almost every point x ∈ X is a regular point of F. This means that TxX iscomplementary to Tx0

X, i.e., Tx0X ⊕ TxX = C2n. If we consider the mapping

Φ : X ×X 3 (x, y)→ x+ y ∈ C2n, we see that it has the smooth point (x0, x). Inparticular almost every pair (x, y) is a smooth point of F, which implies that forevery generic pair (x, y) ∈ X ×X we have TxX ⊕ TyX = C2n.

We shall use in the sequel the following:

Proposition 3.6. Let Xn ⊂ C2n be a generic smooth complete intersection ofmulti-degree d1, ..., dn. If every di > 1, then X is in a super general position.



Proof. We can assume that X is given by n smooth hypersurfaces fi = ai+fi1+...+fidi (where fik is a homogenous polynomial of degree k), which have independentall coefficients (see section below). The tangent space is described by polynomialsfi1, i = 1, ..., n and the set S of points at infinity of X is described by polynomialsfidi , i = 1, ..., n. Since these two families of polynomials have independent coeffi-cients, we see that generically the zero sets at infinity of these two families aredisjoint. In particular such a generic X is in a super general position.

4. Algebraic families

Now we introduce the notion of an algebraic family.

Definition 4.1. Let M be a smooth affine algebraic variety and let Z be a smoothirreducible subvariety of M × Cn . If the restriction to Z of the projection π :M × Cn → M is a dominant map with generically irreducible fibers of the samedimension, then we call the collection Σ = Zm = π−1(m)m∈M an algebraic familyof subvarieties in Cn. We say that this family is in a general position if a genericmember of Σ is in a general position in Cn.

We show that the ideals I(Zm) ⊂ C[x1, ..., xn] of a generic member of Σ dependin a parametric way on m ∈M.

Lemma 4.2. Let Σ be an algebraic family given by a smooth variety Z ⊂M × Cn. The ideal I(Z) ⊂ C[M ][x1, ...., xn] is finitely generated, let the poly-nomials f1(m,x), ..., fs(m,x) form its set of generators. The ideal I(Zm) ⊂C[x1, ..., xn] of a generic member Zm := π−1(m) ⊂ Cn of Σ is equal to I(Zm) =(f1(m,x), ..., fs(m,x)).

Proof. Let dim Z = p and dim M = q. Thus the variety M × Cn has dimensionn+ q. Choose local holomorphic coordinates on M. Since the variety Z is smoothwe have

rank

∂f1∂m1

(m,x) . . . ∂f1∂mq

(m,x) ∂f1∂x1

(m,x) . . . ∂f1∂xn

(m,x)...

......

...∂fs∂m1

(m,x) . . . ∂fs∂mq

(m,x) ∂fs∂x1

(m,x) . . . ∂fs∂xn

(m,x)

= n+ q − p

on Z. Let us consider the projection π : Z 3 (m,x) 7→ m ∈ M. By Sard’s theorema generic m ∈ M is a regular value of the mapping π. For such a regular value mwe have that dim ker d(m,x)π ∩ T(m,x)Z = p− q for every x such that (m,x) ∈ Z.In local coordinates on M this is equivalent to



rank

1 . . . 0 0 . . . 0...

......

...0 . . . 1 0 . . . 0

∗ . . . ∗ ∂f1∂x1

(m,x) . . . ∂f1∂xn

(m,x)...

......

...

∗ . . . ∗ ∂fs∂x1

(m,x) . . . ∂fs∂xn

(m,x)

= n+ 2q − p.

Consequently for (m,x) ∈ Z and m a regular value of π we have

rank

∂f1∂x1

(m,x) . . . ∂f1∂xn

(m,x)...

...∂fs∂x1

(m,x) . . . ∂fs∂xn

(m,x)

= n+ q − p.

Note that n + q − p = codim Zm (in Cn). This means that the ideal(f1(m,x), ..., fs(m,x)) locally coincide with I(Zm), because it contains local equa-tions of Zm. Hence it also coincides globally, i.e., (f1(m,x), ..., fs(m,x)) =I(Zm).

Remark 4.3. This can be also obtained by a computation of a scheme theoreticfibers of π and using the fact that such generic fibers are reduced.

Example 4.4. a) Let N :=(n+dd

)and let Z ⊂ CN × Cn be given by equations

Z = (a, x) ∈ CN × Cn :∑|α|≤d aαx

α = 0. The projection π : Z 3 (a, x) → a ∈CN determines an algebraic family of hypersurfaces of degree d in Cn. If n = 2 andd > 1 this family is in general position in C2.

b) More generally let N1 :=(n+d1d1

), N2 :=

(n+d2d2

), Nn :=

(n+dndn

)and let Z ⊂

CN1 × CN2 ... × CNn × C2n be given by equations Z = (a1, a2, ..., an, x) ∈ CN1 ×CN2 ... × CNn × Cn :

∑|α|≤d1 a1αx

α = 0,∑|α|≤d2 a2αx

α = 0, ...,∑|α|≤dn anαx

α =

0. The projection π : Z 3 (a1, a2, ..., an, x) → (a1, a2, ..., an) ∈ CN1 × CN2 ... ×CNn determines an algebraic family Σ(d1, d2, ..., dn, 2n) of complete intersectionsof multi-degree d1, d2, ...., dn in C2n. If d1, d2, ..., dn > 1, then this family is ingeneral position in C2n. This follows from Proposition 3.6.

5. Defect of symmetry

Let us recall that a following result is true (see e.g. [7]):

Lemma 5.1. Let X,Y be complex algebraic varieties and f : X → Y a polynomialdominant mapping. Then two generic fibers of f are homeomorphic.



Proof. Let X1 be an algebraic completion of X. Take X2 = graph(f) ⊂ X1 × Y ,where Y is a smooth algebraic completion of Y. We can assume that X ⊂ X2. LetZ = X2 \X. We have an induced mapping f : X2 → Y , such that fX = f.

There is a Whitney stratification S of the pair (X2, Z). For every smooth strata

Si ∈ S let Bi be the set of critical values of the mapping f |Si.Take B =

⋃Bi.

Take X3 = X2 \ f−1(B) and Z1 = Z \ f−1(B). The restriction of the stratificationS to X3 gives a Whitney stratification of the pair (X3, Z1). We have a propermapping f1 : X3 → Y \ B which is submersion on each strata. By the Thomfirst isotopy theorem there is a trivialization of f1, which preserves the strata. Itis an easy observation that this trivialization gives a trivialization of the mappingf : X \ f−1(B)→ Y \B.

Definition 5.2. Let X be an affine variety. Let us define Singk(X) := Sing(X)for k := 1 and inductively Singk+1(X) := Sing(Singk(X)).

As a direct application of the Lemma 5.1 and Theorem 2.1 we have:

Theorem 5.3. Let f : Xn → Y l be a dominant polynomial mapping of affinevarieties. If y1, y2 are sufficiently general then f−1(y1) is homeomorphic to f−1(y2)and Sing(f−1(y1)) is homeomorphic to Sing(f−1(y2)). More generally, for everyk we have Singk(f−1(y1)) is homeomorphic to Singk(f−1(y2)).

Now we are ready to prove:

Theorem 5.4. Let Σ be an algebraic family of n−dimensional algebraic subvari-eties in C2n in general position. Symmetry defect hypersurfaces B1, B2 for genericmembers C1, C2 ∈ Σ are homeomorphic and they have homeomorphic singularparts i.e., Sing(B1) ∼= Sing(B2). More generally, for every k we have Singk(B1)is homeomorphic to Singk(B2).

Proof. Let Σ be given by a variety Z ⊂ M × C2n. The ideal I(Z) ⊂C[M ][x1, ...., x2n] is finitely generated. Choose a finite set of generatorsf1(m,x), ..., fs(m,x).

By Sard Theorem we can assume that all fibers of π : Z → M are smooth andfor every m ∈ M we have I(Zm) = f1(m,x), ..., fs(m,x) (see Lemma 4.2). Letus define

R = (m,x, y) ∈M × C2n × C2n : fi(m)(x) = 0, i = 1, ..., s & fi(m)(y) = 0,

i = 1, ..., s.

The variety R is a smooth irreducible subvariety of M × C2n × C2n of codi-mension 2n. Indeed, for given (m,x, y) ∈ M × C2n × C2n choose polynomials

fi1 , ..., fin and fj1 , ..., fjn such that rank [∂fil∂xs

(m,x)]l=1,...,n;s=1,...,n = n and rank

[∂fjl∂xs

(m,x)]l=1,...,n;s=1,...,n = n. Since Z is a smooth variety of dimension dimM+n,

we have that Z locally near (m,x) is given by equations fi1 , ..., fin and near (m, y)



is given by equations fj1 , ..., fjn . Hence the variety R near the point (m,x, y) isgiven as

(m,x, y) ∈M × C2n × C2n : fil(m)(x) = 0, l = 1, ..., n & fjl(m)(y) = 0,

l = 1, ..., s.

In particular R is locally a smooth complete intersection, i.e., R is smooth.

Moreover we have a projection R→M with irreducible fibers which are productsZm × Zm, m ∈ M . This means that R is irreducible. Note that R is an affinevariety. Consider the following morphism

Ψ : R 3 (m,x, y) 7→ (m,x+ y

2) ∈M × C2n.

By the assumptions the mapping Ψ is dominant. Indeed for every m ∈M the fiberZm is in a general position in C2n and consequently the set Ψ(R) ∩ m × C2n isdense in m× C2n.

We know by Theorem 2.4 that the mapping Ψ has constant number of points inthe fiber outside the bifurcation set B(Ψ) ⊂M ×C2n. This implies that B(Zm) =m×C2n ∩B(Ψ). In particular the symmetry defect hypersurface of the variety Zmcoincide with the fiber over m of the projection π : B(Ψ) 3 (m,x) 7→ m ∈M. Nowwe conclude the proof by Theorem 5.3.

Corollary 5.5. Symmetry defect sets B1, B2 for generic curves C1, C2 ⊂ C2 of thesame degree d> 1 are homeomorphic and they have the same number of singularpoints.

Corollary 5.6. Let C1, C2 be two smooth varieties, which are generic complete in-tersection of multi-degree d1, d2, ..., dn in C2n (where all di > 1). Then symmetrydefect hypersurfaces B1, B2 of C1, C2, are homeomorphic and they have homeomor-phic singular parts (i.e., Sing(B1) ∼= Sing(B2)). More generally, for every k wehave Singk(B1) is homeomorphic to Singk(B2).

References

[1] P.J. Giblin and P.A. Holtom, The centre symmetry set, in: Geometry and Topology of

Caustics (Warsaw 1998), Banach Center Publ. Vol. 50, ed. S. Janeczko and V.M. Zakalyukin,

Warsaw, 1999, 91–105.[2] P.J. Giblin and S. Janeczko, Geometry of curves and surfaces through the contact map,

Topology Appl. 159 (2012), 466–475.

[3] S. Janeczko, Bifurcations of the Center of Symmetry, Geom. Dedicata 60, (1996), 9–16.[4] Z. Jelonek, On semi-equivalence of generically finite polynomial mappings, to appear.

[5] Z. Jelonek, The set of points at which a polynomial map is not proper, Ann. Polon. Math.58 (1993), 259–266.

[6] Z. Jelonek, Testing sets for properness of polynomial mappings, Math. Ann. 315 (1999),1–35.

[7] S. Janeczko, Z. Jelonek and M.A.S. Ruas, Symmetry defect of algebraic varieties, AsianJ. Math. 18 (2014), 525–544.



(Stanis law Janeczko) Instytut Matematyczny, Polska Akademia Nauk, Sniadeckich 8,

00-956 Warszawa, Poland, Wydzia l Matematki i Nauk Informacyjnych, Politechnika

Warszawska, Pl. Politechniki 1, 00-661 Warszawa, Poland

E-mail address: [email protected]

(Zbigniew Jelonek) Instytut Matematyczny, Polska Akademia Nauk, Sniadeckich 8,

00-956 Warszawa, Poland


(Maria Aparecida Soares Ruas) Departamento de Matematica, ICMC-USP, Caixa Postal

668, 13560-970 Sao Carlos, S.P., Brasil


“18˙Jelonek˙Kucharz˙Kurdyka” — 2017/12/1 — 20:45 — page 75 — #1




VECTOR BUNDLES AND BLOWUPS

ZBIGNIEW JELONEK, WOJCIECH KUCHARZ, AND KRZYSZTOF KURDYKA

Abstract. Let X be a nonsingular quasi-projective complex algebraic variety

and let E be an algebraic vector bundle on X of rank r ≥ 2. The pullback

of E by the blowup of X at a suitably chosen nonsingular subvariety of X of

codimension r contains a line subbundle that can be explicitly described.

1. Introduction

Kleiman [2, Problem 1] considers the problem of splitting vector bundles on

a nonsingular quasi-projective variety V over an infinite field k: For any vector

bundle G on V of rank at least 2, Kleiman [2, Theorem 4.7] proves that the pullback

of G by the blowup of a suitably chosen nonsingular subvariety contains a line

bundle. Henceforth we assume that k = C and obtain Kleiman’s theorem as

Corollary 1.3, which is a special case of Corollary 1.2 derived from Theorem 1.1.

It does not seem possible to deduce Theorem 1.1 and Corollary 1.2 directly from

[2]. Furthermore, the proof of Theorem 1.1 is short and very simple. In fact the

main virtues of our note are its simplicity and brevity.

Let X be a nonsingular quasi-projective complex algebraic variety. For any

closed nonsingular (not necessarily irreducible ) subvariety Z of X, let

π(X,Z) : B(X,Z)→ X

denote the blowup of X at Z. As usual, the line bundle determined by the excep-

tional divisor D := π(X,Z)−1(Z) will be denoted by O(D). If Z is empty, then

B(X,Z) = X and π(X,Z) is the identity map, D = 0 and O(D) is the standard

2010 Mathematics Subject Classification. 14J60, 14E15 .

Key words and phrases. Complex algebraic variety, vector bundle, blowup.

75


76 Z. JELONEK, W. KUCHARZ, AND K. KURDYKA

trivial line bundle on X. For an algebraic vector bundle E on X and a section u

of E, the zero locus of u will be denoted by Z(u),

Z(u) := x ∈ X : u(x) = 0.

If u is transverse to the zero section, then Z(u) is a closed nonsingular subvariety

of X which is either empty or of codimension equal to the rank of E.

The main result, whose proof is postponed until Section 2, is the following:

Theorem 1.1. Let E be an algebraic vector bundle on X of rank r ≥ 2. If s

is a section of E which is transverse to the zero section and Z := Z(s), then the

pullback vector bundle π(X,Z)∗E on B(X,Z) contains an algebraic line subbundle

isomorphic to O(D), where D is the exceptional divisor of the blowup π(X,Z) :

B(X,Z)→ X.

Of course, E may not have a section that is transverse to the zero section.

However, if E is generated by global sections s1, . . . , sk, then for a general point

(t1, . . . , tk) ∈ Ck, the section

s = t1s1 + · · ·+ tksk

is transverse to the zero section. There is always an algebraic line bundle L on X

such that the vector bundle E ⊗ L is generated by global sections. It suffices to

take as L a high tensor power of an ample line bundle on X, cf. [1].

Corollary 1.2. Let E be an algebraic vector bundle on X of rank r ≥ 2. Let

L be an algebraic line bundle on X such that the vector bundle E ⊗ L admits

a section v transverse to the zero section, and let Z := Z(v). Then the pull-

back vector bundle π(X,Z)∗E on B(X,Z) contains an algebraic line subbundle

isomorphic to O(D)⊗π(X,Z)∗L∨, where D is the exceptional divisor of the blowup

π(X,Z) : B(X,Z)→ X and L∨ stands for the dual line bundle to L.

Proof. According to Theorem 1.1, the pullback vector bundle π(X,Z)∗(E ⊗L) on

B(X,Z) contains an algebraic subbundle isomorphic to O(D). The vector bundle

π(X,Z)∗E is isomorphic to

π(X,Z)∗(E ⊗ L)⊗ π(X,Z)∗L∨,

and hence it contains a line subbundle isomorphic to O(D)⊗ π(X,Z)∗L∨.

Since for a suitably chosen line bundle L, the vector bundle E ⊗ L admits

a section transverse to the zero section, the next result follows immediatly.

Corollary 1.3. Let E be an algebraic vector bundle on X of rank r ≥ 2. Then

there exists a closed nonsingular subvariety Z of X, either empty or of codimesion

r, such that the pullback vector bundle π(X,Z)∗E on B(X,Z) contains an algebraic

line subbundle.


VECTOR BUNDLES AND BLOWUPS 77

Corollary 1.3 is not a new result. It is proved (for varieties over an arbitrary

infinite field) in Kleiman’s paper [2].

2. Proof of Theorem 1.1

For any nonsingular complex algebraic variety Y , denote by TY its tangent

bundle. Let X be a nonsingular quasi-projective complex algebraic variety and let

Z be a closed nonsingular subvariety of X with dimZ < dimX − 1. Consider the

blowup

π(X,Z) : B(X,Z)→ X

of X at Z. As a point set B(X,Z) is the union of X \Z and the projective bundle

P(NZX) on Z associated with the normal bundle

NZX := (TX |Z)/TZ

to Z in X. The map π(X,Z) is the identity on X \ Z and the bundle projection

P(NZX)→ Z on P(NZX).

Proof of Theorem 1.1. By abuse of notation, the total space of the vector bundle E

will also be denoted by E. RegardX as subvariety of E, identifying it with its image

by the zero section. Furthermore, identify the normal bundle to X in E with the

vector bundle E. Thus as a point set the space B(E,X) is the union of E \X and

the projective bundle P(E) associated with E, while π(E,X) : B(E,X)→ E is the

identity on E\X and the bundle projection P(E)→ X on P(E). If p : E → X is the

bundle projection, then the pullback vector bundle (p π(E,X))∗E on B(E,X)

contains an algebraic line subbundle L defined as follows. The fiber of L over

a point e ∈ (E\X) is the line e×Ce, and the restriction L|P(E) is the tautological

line bundle on P(E). Note that u : B(E,X) → L, defined by u(e) = (e, e) for

e ∈ (E \X) and u|P(E) = 0, is a section of L, transverse to the zero section and

satisfying Z(u) = P(E).

Since the section s is transverse to X in E, for each point z in Z, the differential

dsz : TX,z → TE,z induces a linear isomorphism

dsz : (NZX)z → (NXE)z = Ez

between the fibers over z of the normal bundle to Z in X and the normal bundle

to X in E. Define s : B(X,Z) → B(E,X) by s(x) = s(x) for x ∈ X \ Z and

s(l) = dsz(l) for l ∈ P(NZX)z with z ∈ Z. Thus s(l) is in P(Ez). By construction,

s is an algebraic morphism satisfying

p π(E,X) s = π(X,Z).


78 Z. JELONEK, W. KUCHARZ, AND K. KURDYKA

Hence the pullback s∗L is an algebraic line subbundle of

s∗((p π(E,X))∗E) = (p π(E,X) s)∗E = π(X,Z)∗E.

It remains to prove that the line bundles s∗L and O(D) are isomorphic. By

construction, s is transverse to P(E) in B(E,X) and s−1(P(E)) = π(X,Z)−1(Z).

Since the section u : B(E,X) → L is transverse to the zero section and Z(u) =

P(E), the pullback section s∗u : B, (X,Z) → s∗L is also transverse to the zero

section and Z(s∗u) = π(X,Z)−1(Z) = D. Consequently, the vector bundle s∗L is

isomorphic to O(D), as required.

References

[1] R. Hartshorne, Algebraic Geometry, Springer 1977.

[2] S. Kleiman, Geometry on Grassmanians and applications to splitting bundles and smoothing

cycles, Inst. Hautes Etudes Sci. Publ. Math. 36 (1969), 281–297.

(Zbigniew Jelonek) Instytut Matematyczny, Polska Akademia Nauk, Sniadeckich 8,

00-956 Warszawa,Poland


(Wojciech Kucharz) Institute of Mathematics, Faculty of Mathematics and Computer

Science, Jagiellonian University, ul. Lojasiawicza 6, 30-348 Krakow, Poland


(Krzysztof Kurdyka) Laboratoire de Mathematiques, UMR 5127 du CNRS, Universite

Savoie Mont Blanc, Campus Scientifique, 73 376 Le Bourget–du–Lac Cedex, France


“19˙Jedrzejewicz˙Matysiak˙Zielinski” — 2017/12/1 — 20:43 — page 79 — #1



A NOTE ON SQUARE-FREE FACTORIZATIONS

PIOTR JĘDRZEJEWICZ, ŁUKASZ MATYSIAK, AND JANUSZ ZIELIŃSKI

Abstract. We analyze properties of various square-free factorizations in gre-atest common divisor domains (GCD-domains) and domains satisfying theascending chain condition for principal ideals (ACCP-domains).

1. Introduction

Throughout this article by a ring we mean a commutative ring with unity. Bya domain we mean a ring without zero divisors. By R∗ we denote the set of allinvertible elements of a ring R. Given elements a, b ∈ R, we write a ∼ b if a andb are associated, and a | b if b is divisible by a. Furthermore, we write a rpr b if aand b are relatively prime, that is, have no common non-invertible divisors. If R isa ring, then by Sqf R we denote the set of all square-free elements of R, where anelement a ∈ R is called square-free if it can not be presented in the form a = b2cwith b ∈ R \R∗, c ∈ R.

In [4] we discuss many factorial properties of subrings, in particular involvingsquare-free elements. The aim of this paper is to collect various ways to presentan element as a product of square-free elements and to study the existence anduniqueness questions in larger classes than the class of unique factorization do-mains. In Proposition 1 we obtain the equivalence of factorizations (ii) – (vii) forGCD-domains. We also prove the existence of factorizations (i) – (iii) in Propo-sition 1 for ACCP-domains, but their uniqueness we obtain in Proposition 2 forGCD-domains. Recall that a domain R is called a GCD-domain if the intersectionof any two principal ideals is a principal ideal. Recall also that a domain R is calledan ACCP-domain if it satisfies the ascending chain condition for principal ideals.

2010 Mathematics Subject Classification. Primary 13F15, Secondary 13F20.Key words and phrases. Square-free element, factorization, pre-Schreier domain, GCD-

domain, ACCP-domain.

79


80 P. JĘDRZEJEWICZ, Ł. MATYSIAK, AND J. ZIELIŃSKI

We refer to Clark’s survey article [1] for more information about GCD-domainsand ACCP-domains.

It turns out that some preparatory properties (Lemma 2) hold in a largerclass than GCD-domains, namely pre-Schreier domains. A domain R is called pre-Schreier if every non-zero element a ∈ R is primal, that is, for every b, c ∈ R suchthat a | bc there exist a1, a2 ∈ R such that a = a1a2, a1 | b and a2 | c. Integrallyclosed pre-Schreier domains are called Schreier domains. The notion of Schreierdomain was introduced by Cohn in [2]. The notion of pre-Schreier domain was in-troduced by Zafrullah in [6], but this property had featured already in [2], as wellas in [3] and [5]. The reason why we consider pre-Schreier domains in Lemma 2is that we were looking for a minimal condition under which a product of pairwi-se relatively prime square-free elements is square-free. For further information onpre-Schreier domains we refer the reader to [6].

2. Preliminary lemmas

Note the following easy lemma.

Lemma 1. Let R be a ring. If a ∈ Sqf R and a = b1b2 . . . bn, then b1, b2, . . . ,bn ∈ Sqf R and bi rpr bj for i 6= j.

In the next lemma we obtain the properties we will use in the proofs of Pro-positions 1 b) and 2 (i). Recall that every GCD-domain is pre-Schreier ([2], The-orem 2.4).

Lemma 2. Let R be a pre-Schreier domain.

a) Let a, b, c ∈ R, a 6= 0. If a | bc and a rpr b, then a | c.b) Let a, b, c, d ∈ R. If ab = cd, a rpr c and b rpr d, then a ∼ d and b ∼ c.c) Let a, b, c ∈ R. If ab = c2 and arpr b, then there exist c1, c2 ∈ R such that a ∼ c21,b ∼ c22 and c = c1c2.

d) Let a1, . . . , an, b ∈ R. If ai rpr b for i = 1, . . . , n, then a1 . . . an rpr b.

e) Let a1, . . . , an ∈ R. If a1, . . . , an ∈ Sqf R and ai rpr aj for all i 6= j, thena1 . . . an ∈ Sqf R.

Proof. a) If a | bc, then a = a1a2 for some a1, a2 ∈ R \ 0 such that a1 | b anda2 | c. If, moreover, a rpr b, then a1 ∈ R∗. Hence, a ∼ a2, so a | c.b) Assume that ab = cd, a rpr c and b rpr d. If a = 0 and R is not a field, thenc ∈ R∗, so d = 0 and then b ∈ R∗. Now, let a, d 6= 0.

Since a | cd and a rpr c, we have a | d by a). Similarly, since d | ab and d rpr b, weobtain d | a. Hence, a ∼ d, and then b ∼ c.c) Let ab = c2 and a rpr b. Since c | ab, there exist c1, c2 ∈ R \ 0 such that c1 | a,c2 | b and c = c1c2. Hence, a = c1d and b = c2e for some d, e ∈ R, and we obtainde = c1c2. We have d rpr c2, because d | a and c2 | b, analogously e rpr c1, so d ∼ c1and e ∼ c2, by b). Finally, a ∼ c21, b ∼ c22.


A NOTE ON SQUARE-FREE FACTORIZATIONS 81

d) Induction. Let ai rpr b for i = 1, . . . , n+1. Put a = a1 . . . an. Assume that a rpr b.Let c ∈ R \ 0 be a common divisor of aan+1 and b. Since c | aan+1, there existc1, c2 ∈ R \ 0 such that c1 | a, c2 | an+1 and c = c1c2. We see that c1, c2 | b, soc1, c2 ∈ R∗, and then c ∈ R∗.e) Induction. Take a1, . . . , an+1 ∈ Sqf R such that ai rpr aj for i 6= j. Put a =a1 . . . an. Assume that a ∈ Sqf R. Let aan+1 = b2c for some b, c ∈ R \ 0.

Since c | aan+1, there exist c1, c2 ∈ R \ 0 such that c = c1c2, c1 | a andc2 | an+1, so a = c1d and an+1 = c2e, where d, e ∈ R. We obtain de = b2. By d) wehave a rpr an+1, so d rpr e. And then by c), there exist b1, b2 ∈ R such that d ∼ b21,e ∼ b22 and b = b1b2. Since a, an+1 ∈ Sqf R, we infer b1, b2 ∈ R∗, so b ∈ R∗.

3. Square-free factorizations

In Proposition 1 below we collect possible presentations of an element asa product of square-free elements or their powers. We distinct presentations (ii)and (iii), presentations (iv) and (v), and presentations (vi) and (vii), because (ii),(iv) and (vi) are of a simpler form, but in (iii), (v) and (vii) the uniqueness will bemore natural (in Proposition 2).

Proposition 1. Let R be a ring. Given a non-zero element a ∈ R \ R∗, considerthe following conditions:

(i) there exist b ∈ R and c ∈ Sqf R such that a = b2c,

(ii) there exist n > 0 and s0, s1, . . . , sn ∈ Sqf R such that a = s2n

n s2n−1n−1 . . . s

21s0,

(iii) there exist n > 1, s1, s2, . . . , sn ∈ (Sqf R) \ R∗, k1 < k2 < . . . < kn, k1 > 0,and c ∈ R∗ such that a = cs2

kn

n s2kn−1n−1 . . . s

2k22 s

2k11 ,

(iv) there exist n > 1 and s1, s2, . . . , sn ∈ Sqf R such that si | si+1 for i =1, . . . , n− 1, and a = s1s2 . . . sn,

(v) there exist n > 1, s1, s2, . . . , sn ∈ (Sqf R) \R∗, k1, k2, . . . , kn > 1, and c ∈ R∗such that si | si+1 and si 6∼ si+1 for i = 1, . . . , n− 1, and a = csk11 s

k22 . . . s

knn ,

(vi) there exist n > 1 and s1, s2, . . . , sn ∈ Sqf R such that si rpr sj for i 6= j, anda = s1s22s

33 . . . s

nn,

(vii) there exist n > 1, s1, s2, . . . , sn ∈ (Sqf R) \ R∗, k1 < k2 < . . . < kn, k1 > 1,and c ∈ R∗ such that si rpr sj for i 6= j, and a = csk11 s

k22 . . . s

knn .

a) In every ring R the following holds:

(i)⇐ (ii)⇔ (iii), (iv)⇔ (v)⇒ (vi)⇔ (vii).

b) If R is a GCD-domain, then all conditions (ii) – (vii) are equivalent.

c) If R is an ACCP-domain, then conditions (i) – (iii) hold.

d) If R is a UFD, then all conditions (i) – (vii) hold.



Proof. a) Implication (i) ⇐ (ii) and equivalencies (ii) ⇔ (iii), (iv) ⇔ (v), (vi) ⇔(vii) are obvious, so it is enough to prove implication (iv)⇒ (vi).

Assume that a = s1s2 . . . sn, where s1, s2, . . . , sn ∈ Sqf R and si | si+1 fori = 1, . . . , n − 1. Let si+1 = siti+1, where ti+1 ∈ R, for i = 1, . . . , n − 1. Putalso t1 = s1. Then si = t1t2 . . . ti for each i. Since sn ∈ Sqf R, by Lemma 1we obtain that t1, t2, . . . , tn ∈ Sqf R and ti rpr tj for i 6= j. Moreover, we haves1s2 . . . sn = tn1 t

n−12 . . . tn.

b) Let R be a GCD-domain.

(vi)⇒ (iv) Assume that a = s1s22s33 . . . s

nn, where s1, s2, . . . , sn ∈ Sqf R and si rpr sj

for i 6= j. We see that

s1s22s33 . . . s

nn = sn(snsn−1)(snsn−1sn−2) . . . (snsn−1 . . . s2)(snsn−1 . . . s2s1).

Since R is a GCD-domain, snsn−1 . . . si ∈ Sqf R for each i by Lemma 2 e).

(vi) ⇒ (ii) Let a = s1s22s33 . . . s

nn, where s1, s2, . . . , sn ∈ Sqf R, and si rpr sj for

i 6= j. For every k ∈ 1, 2, . . . , n put k =∑ri=0 c

(k)i 2i, where c(k)i ∈ 0, 1. Then

a =n∏k=1

skk =n∏k=1

s

∑r

i=0c(k)i2i

k =n∏k=1

r∏i=0

sc(k)i2i

k =r∏i=0

( n∏k=1

sc(k)i

k

)2i,

where∏nk=1 s

c(k)i

k ∈ Sqf R for each i by Lemma 2 e).

(ii) ⇒ (vi) Let a = s2n

n s2n−1n−1 . . . s

21s0, where s0, s1, . . . , sn ∈ Sqf R. For every

k ∈ 1, 2, . . . , 2n+1 − 1 put k =∑ni=0 c

(k)i 2i, where c(k)i ∈ 0, 1. Let t′k = gcd(si :

c(k)i = 1), t′′k = lcm(si : c

(k)i = 0) and t′k = gcd(t′k, t

′′k) · tk, where tk ∈ R (by

[2], Theorem 2.1, in a GCD-domain least common multiples exist). Then tk isthe greatest among these common divisors of all si such that c(k)i = 1, which arerelatively prime to all si such that c(k)i = 0. In particular, tk | si for every k, isuch that c(k)i = 1, and tk rpr si for every k, i such that c(k)i = 0. In each case,

gcd(si, tk) = tc(k)i

k . Moreover, tk rpr tl for every k 6= l.Since si | t1t2 . . . t2n+1−1, we obtain

si = gcd(si,2n+1−1∏k=1

tk) =2n+1−1∏k=1

gcd(si, tk) =2n+1−1∏k=1

tc(k)i

k ,

son∏i=0

(si)2i

=n∏i=0

2n+1−1∏k=1

(tc(k)i

k

)2i=2n+1−1∏k=1

n∏i=0

tc(k)i2i

k =2n+1−1∏k=1

t

∑n

i=0c(k)i2i

k =2n+1−1∏k=1

tkk.

Moreover, tk ∈ Sqf R, because for k ∈ 1, 2, . . . , 2n+1 − 1 there exists i suchthat c(k)i = 1, and then tk | si.c) Let R be an ACCP-domain. In this proof we follow the idea of the second proofof Proposition 9 from [1], p. 7, 8.


A NOTE ON SQUARE-FREE FACTORIZATIONS 83

(i) If a 6∈ Sqf R, then a = b21c1, where b1 ∈ R \ R∗, c1 ∈ R. If c1 6∈ Sqf R, thenc1 = b22c2, where b2 ∈ R \R∗, c2 ∈ R. Repeating this process, we obtain a stronglyascending chain of principal ideals Ra $ Rc1 $ Rc2 $ . . ., so for some k we willhave ck−1 = b2kck, bk ∈ R \R∗, and ck ∈ Sqf R. Then a = (b1 . . . bk)2ck.

(iii) If a 6∈ Sqf R, then by (i) there exist a1 ∈ R \ R∗ and s0 ∈ Sqf R such thata = a21s0. If a1 6∈ Sqf R, then again, by (i) there exist a2 ∈ R \ R∗ and s1 ∈ Sqf Rsuch that a1 = a22s1. Repeating this process, we obtain a strongly ascending chain ofprincipal ideals Ra $ Ra1 $ Ra2 $ . . ., so for some k we will have ak−1 = a2ksk−1,ak ∈ (Sqf R) \R∗, sk−1 ∈ Sqf R. Putting sk = ak we obtain:

a = a21s0 = a22

2 s21s0 = . . . = s2

n

n . . . s222 s21s0.

d) This is a standard fact following from the irreducible decomposition.

4. The uniqueness of factorizations

The following proposition concerns the uniqueness of square-free decompositionsfrom Proposition 1. In (i) – (iii) we assume that R is a GCD-domain, in (iv) – (vii)R is a UFD.

Proposition 2. (i) Let b, d ∈ R and c, e ∈ Sqf R. If

b2c = d2e,

then b ∼ d and c ∼ e.(ii) Let s0, s1, . . . , sn ∈ Sqf R and t0, t1, . . . , tm ∈ Sqf R, n 6 m. If

s2n

n s2n−1n−1 . . . s

21s0 = t2

m

m t2m−1m−1 . . . t

21t0,

then si ∼ ti for i = 0, . . . , n and, if m > n, then ti ∈ R∗ for i = n+ 1, . . . ,m.

(iii) Let s1, s2, . . . , sn ∈ (Sqf R) \R∗, t1, t2, . . . , tm ∈ (Sqf R) \R∗, k1 < k2 < . . . <kn, l1 < l2 < . . . < lm and c, d ∈ R∗. If

cs2kn

n s2kn−1n−1 . . . s

2k22 s

2k11 = dt2

lm

m t2lm−1m−1 . . . t

2l22 t2l11 ,

then n = m, si ∼ ti and ki = li for i = 1, . . . , n.

(iv) Let s1, s2, . . . , sn ∈ Sqf R, t1, t2, . . . , tm ∈ Sqf R, n 6 m, si | si+1 for i =1, . . . , n− 1, and ti | ti+1 for i = 1, . . . ,m− 1. If

s1s2 . . . sn = t1t2 . . . tm,

then si ∼ ti+m−n for i = 1, . . . , n and, if m > n, then ti ∈ R∗ for i = 1, . . . ,m−n.(v) Let s1, s2, . . . , sn ∈ (Sqf R) \ R∗, t1, t2, . . . , tm ∈ (Sqf R) \ R∗, k1, k2, . . . , kn> 1, l1, l2, . . . , lm > 1, c, d ∈ R∗, si | si+1 and si 6∼ si+1 for i = 1, . . . , n − 1,ti | ti+1 and ti 6∼ ti+1 for i = 1, . . . ,m− 1. If

csk11 sk22 . . . s

knn = dtl11 t

l22 . . . t

lmm ,




(vi) Let s1, s2, . . . , sn ∈ Sqf R, t1, t2, . . . , tm ∈ Sqf R, n 6 m, si rpr sj for i 6= jand ti rpr tj for i 6= j. If

s1s22s33 . . . s

nn = t1t22t

33 . . . t

mm,

then si ∼ ti for i = 1, . . . , n and, if m > n, then ti ∈ R∗ for i = n+ 1, . . . ,m.

(vii) Let s1, s2, . . . , sn ∈ (Sqf R) \R∗, t1, t2, . . . , tm ∈ (Sqf R) \R∗, 1 6 k1 < k2 <. . . < kn, 1 6 l1 < l2 < . . . < lm, c, d ∈ R∗, si rpr sj for i 6= j, and ti rpr tj fori 6= j. If

csk11 sk22 . . . s

knn = dtl11 t

l22 . . . t

lmm ,


Proof. (i) Assume that b2c = d2e. Put f = gcd(b, d), g = gcd(c, e), b = fb0,d = fd0, c = gc0, and e = ge0, where b0, c0, d0, e0 ∈ R. We obtain b20c0 = d20e0,gcd(c0, e0) = 1 and gcd(b0, d0) = 1, so also gcd(b20, d

20) = 1. By Lemma 2 b), we

infer b20 ∼ e0 and c0 ∼ d20, but c0, e0 ∈ Sqf R by Lemma 1, so b0, d0 ∈ R∗, and thenc0, e0 ∈ R∗.Statements (ii), (iii) follow from (i).

Statements (iv) – (vii) are straightforward using the irreducible decomposition.

Acknowledgements

The authors would like to thank the referee for helpful remarks.

References

[1] P.L. Clark, Factorizations in integral domains, math.uga.edu/~pete/factorization2010.pdf.[2] P.M. Cohn, Bezout rings and their subrings, Proc. Camb. Phil. Soc. 64 (1968), 251–264.[3] W. Dicks, E.D. Sontag, Sylvester domains, J. Pure Appl. Algebra 13 (1978), 243–275.[4] P. Jędrzejewicz, Ł. Matysiak, J. Zieliński, On some factorial properties of subrings, to appearin Univ. Iagel. Acta Math., arXiv:1606.06592.

[5] S. McAdam, D.E. Rush, Schreier rings, Bull. London Math. Soc. 10 (1978), 77–80.[6] M. Zafrullah, On a property of pre-Schreier domains, Comm. Algebra 15 (1987), 1895–1920.

Nicolaus Copernicus University, Faculty of Mathematics and Computer Science,87-100 Toruń, Poland

E-mail address, Piotr Jędrzejewicz: [email protected]

E-mail address, Łukasz Matysiak: [email protected]

E-mail address, Janusz Zieliński: [email protected]

“20˙Kucharz˙Kurdyka” — 2017/12/1 — 20:45 — page 85 — #1




RATIONALITY OF SEMIALGEBRAIC FUNCTIONS

WOJCIECH KUCHARZ AND KRZYSZTOF KURDYKA

Abstract. Let X be an algebraic subset of Rn, and f : X → R a semialgebraic

function. We prove that if f is continuous rational on each curve C ⊂ X then:1) f is arc-analytic, 2) f is continuous rational on X. As a consequence we

obtain a characterization of hereditarily rational functions recently studied by

J. Kollar and K. Nowak.

1. Introduction

Our goal is to give a short introduction to some results on real rational functions.The interested reader may consult [4] and [5] for a more comprehensive treatment.We strive for simplicity of our presentation. Keeping this in mind, we deal onlywith semialgebraic functions. Furthermore, we explain in detail the role of Bertini’stheorem in establishing a criterion for rationality of such functions.

Throughout this section, X denotes an algebraic subset of Rn. Recall thata function f : X → R is said to be regular at x ∈ X if there exist two polynomialsp, q ∈ R[x1, . . . , xn] such that q(x) 6= 0 and f = p/q in a Zariski open neighborhoodof x in X. Furthermore, f is regular on a subset of X if it is regular at each pointof this subset. We say that the function f : X → R is rational if it is regular ona Zariski open dense subset of X (this minor deviation from the standard definitionis justified, f being defined everywhere on X). Obviously, f is rational if and onlyif its restriction to each irreducible component of X is rational. We also say thatthe function f is continuous rational if it is continuous (in the strong topology) onX and rational in the sense just defined.

2010 Mathematics Subject Classification. 14P05, 14P10, 26C15.Key words and phrases. Semialgebraic functions, continuous rational functions, Bertini

theorem.The first named author was partially supported by the National Science Centre (Poland),

under grant number 2014/15/B/ST1/00046.The work was partially supported by ANR project STAAVF (France).

85


86 W. KUCHARZ AND K. KURDYKA

Example 1.1. Let C = x3 = y2 ⊂ R2, and let f : C → R be the function definedby f = y/x for (x, y) 6= (0, 0) and f(0, 0) = 0. Then f is continuous rational on Cbut is not regular at (0, 0).

Note that Example 1.1 makes sense also in the complex setting.

Example 1.2. The function f : R2 → R, defined by f(x, y) = x3

x2+y2 for (x, y) 6=(0, 0) and f(0, 0) = 0, is continuous rational on R2 but is not regular at (0, 0).

Example 1.2 is specific to real algebraic geometry. Indeed, in the complex setting,a continuous rational function on a nonsingular algebraic set is actually regular.

Continuous rational functions on nonsingular real algebraic sets are studied bythe first named author [6, 7, 8] in the context of approximation of continuous mapsinto spheres. Both authors initiated a theory of vector bundles [9] on real algebraicvarieties, in which continuous rational functions are used to define morphisms.Continuous rational functions, under the name fonctions regulues, are the objectof investigation in [3]. An interesting phenomenon discovered by J. Kollar is recalledbelow.

Example 1.3. The algebraic surface

S := x3 − (1 + z2)y3 = 0 ⊂ R3

is an analytic submanifold, and the function f : S → R defined by f(x, y, z) =(1 + z2)1/3 is analytic and semialgebraic. Furthermore, f is continuous rational onS since f(x, y, z) = x/y on S without the z-axis. On the other hand, f restrictedto the z-axis is not a rational function, and f cannot be extended to a continuousrational function on R3, cf. [5].

In order to avoid such a pathology Kollar and Nowak [5] proposed the followingnotion. A function f : X → R is said to be hereditarily rational if for every algebraicset Z ⊂ X the restriction f |Z is a rational function on Z. According to the mainresult of [5], any continuous and hereditarily rational function on X ⊂ Rn can beextended to a continuous and hereditarily rational function on Rn. Moreover, ifthe algebraic set X is nonsingular, then any continuous rational function on X ishereditarily rational [5].

Now we introduce a crucial notion for this paper. We say that a function f : X →R is curve-rational if for every irreducible algebraic curve C ⊂ X the restrictionf |C is rational on C.

Our main result, whose proof is given in Section 3, is the following.

Theorem A. For a function f : X → R on an algebraic subset X of Rn, thefollowing conditions are equivalent:

(a) f is hereditarily rational.(b) f can be extended to a hereditarily rational function on Rn.(c) f is semialgebraic and curve-rational.


RATIONALITY OF SEMIALGEBRAIC FUNCTIONS 87

As demonstrated by Example 1.2, a semialgebraic continuous rational functionneed not be curve-rational.

We are now heading towards our second result, which is proved in Section 4. Wesay that a function f : X → R is continuously curve-rational if for every irreduciblealgebraic curve C ⊂ X the restriction f |C is continuous rational on C.

In the 1980’s the notion of arc-analytic function was introduced by the secondnamed author [10]. A function f : V → R, defined on a real analytic set V , is saidto be arc-analytic if f γ is analytic for every analytic arc γ : (−1, 1) → V . Anarc-analytic function on Rn need not be continuous [1] and may have a nondiscretesingular set [11].

Theorem B. Any semialgebraic, continuously curve-rational function on an alge-braic subset of Rn is continuous and arc-analytic.

As an immediate consequence of Theorems A and B we obtain the followingcharacterization of continuous hereditarily rational functions.

Corollary 1.4. A function on an algebraic subset of Rn is continuous and hered-itarily rational if and only if it is semialgebraic and continuously curve-rational.

In Section 2, which is crucial for the proof of Theorem A, we recall some resultson semialgebraic sets, prove a suitable version of Bertini’s theorem, and analyzethe Zariski closure of a Nash submanifold of Rn.

2. Preliminaries

We will use the notion of dimension in various situations. If Y is an algebraicsubset of Rn or Cn, then by dimY we mean the Krull dimension of the ring ofpolynomial R-valued or C-valued, respectively, functions on Y . Recall that a Nashsubmanifold of Rn is an analytic submanifold that is also a semialgebraic set. Fora semialgebraic subset A of Rn,

dimA := max dimN,

where N runs through the collection of Nash submanifolds of Rn contained in A.

The Zariski closure of an arbitrary subset S of Rn will be denoted by SZ

. Hence

SZ

is the smallest algebraic subset of Rn containing S. If X is an algebraic subsetof Rn, we denote by XC its complexification, that is, the smallest algebraic subsetof Cn containing X. For A as above,

dimA = dimAZ

= dim(AZ

)C,

cf. [2, Section 2.8]. We will frequently use these equalities without explicitlyreferring to them.

In the sequel we will also use the following standard facts on rational functions.Let X be an irreducible algebraic subset of Rn. Recall that each nonempty Zariskiopen subset of X is Zariski dense. Let f : X → R be a rational function that is



regular in a nonempty Zariski open set X0 ⊂ X. We denote by G(f) the Zariskiclosure in Rn × R of the graph of f |X0

, that is,

G(f) := graph(f |X0)Z.

SinceX is irreducible, so areX0 and graph(f |X0). Consequently, G(f) is irreducible

as well. We have

dimG(f) = dim(graph(f |X0)) = dimX0 = dimX.

It readily follows that G(f) does not depend on the choice of X0. By complexifyingX ⊂ Rn and G(f) ⊂ Rn×R, we obtain irreducible complex algebraic sets XC ⊂ Cn

and G(f)C ⊂ Cn×C. For our purposes, the key property of G(f)C is the following.

Lemma 2.1. Let π : Cn × C → Cn be the canonical projection. With notation asabove,

G(f)C ∩ π−1(x) = (x, f(x))for all x ∈ X0.

Proof. We can choose polynomials p, q ∈ R[x1, . . . , xn] with q(x) 6= 0 and f(x) =p(x)/q(x) for all x ∈ X0, cf. [2, p. 62]. Set

U := z ∈ XC | q(z) 6= 0

and define g : U → C by g(z) = p(z)/q(z) for z ∈ U . It suffices to prove that

(i) G(f)C ∩ (U × C) = graph(g).

To this end denote by G the Zariski closure of graph(g) in Cn × C. Then G isirreducible with

dimG = dimXC = dimX = dimG(f)C.

Since graph(g) is Zariski closed in U × C, it follows that

(ii) G ∩ (U × C) = graph(g).

Clearly, graph(f |X0) ⊂ graph(g), which implies that G(f)C ⊂ G. Consequently,

(iii) G(f)C = G,

both G(f)C and G being complex algebraic sets of the same dimension. Hence (i)follows by combining (ii) and (iii).

Next we study the Zariski closure of Nash manifolds. One readily checks that theZariski closure of a connected Nash submanifold of Rn is an irreducible algebraicset. It is convenient to introduce the following notion.

We say that two Nash submanifolds A and B of Rn are compatible if for anynonempty open subsets (in the relative strong topology) A′ ⊂ A, B′ ⊂ B there existpoints a ∈ A′, b ∈ B′ and an irreducible algebraic curve C ⊂ Rn with the followingproperties: a is an accumulation point of A ∩ (C \ a) and b is an accumulationpoint of B ∩ (C \ b). In that case, both a and b belong to C.



Proposition 2.2. If two connected Nash submanifolds A and B of Rn, with

dimA = dimB, are compatible, then AZ

= BZ

.

Proof. Note that AZ

and BZ

are irreducible algebraic subsets of the same di-

mension. Suppose that A and B are compatible, but AZ 6= B

Z. Then the sets

A′ := A \BZ and B′ := B \AZ are nonempty and open in A and B, respectively.Let a ∈ A′, b ∈ B′, and C be as in the definition of compatible Nash submanifoldsgiven above. Since a is an accumulation point of A ∩ (C \ a), it follows that theintersection A ∩ C is an infinite set and hence the irreducibility of C implies the

inclusion C ⊂ AZ . Thus we get a contradiction since a, b ∈ C.

We will need an affine version of the classical theorem of Bertini. For the sakeof completeness we include a full proof of it.

Theorem 2.3 (Bertini). Let π : Cn × Ck → Cn be the canonical projection, Y ⊂Cn×Ck an irreducible algebraic set, and X the Zariski closure of π(Y ). If dimY =dimX = d ≥ 2, then the set of affine (n − d + 1)-planes in Cn contains a Zariskiopen dense subset B such that for every L ∈ B the intersection Y ∩ π−1(L) is anirreducible curve.

Proof. We will repeat almost word by word the proof of Theorem 3.3.1 (a projectiveversion of Bertini’s theorem) from the excellent survey of R. Lazarsfeld [12]. Firstwe fix a general affine (n − d)-plane Λ such that π−1(Λ) cuts Y transversally infinitely many smooth points. By a translation we may assume that Λ is actuallya linear subspace of Cn. The space of linear (n− d+ 1)-planes which contain Λ isparametrized by a projective space T = Pd−1. Given t ∈ T we denote by Lt thecorresponding linear (n− d+ 1)-plane. Consider the set

V := (y, t) ∈ Y × T | π(y) ∈ Lt.

The issue is to establish the irreducibility of a general fiber of the second projection

p : V → T.

To this end note that the first projection V → Y is actually the blowing up ofY along the finite set Y ∩ π−1(Λ). Hence V is irreducible. Furthermore, if we fixa point y0 ∈ Y ∩ π−1(Λ), then the mapping s : T → V , defined by s(t) = (y0, t),defines a global section of p whose image is away from the singular locus of V .Moreover, p is a submersion at each point s(t) = (y0, t), t ∈ T .

Let Z ⊂ V be the union of singular points of V and critical points of p. ClearlyZ is nowhere dense in V , hence V \Z is connected and p is a submersion on V \Z.By a result of Verdier [14, Corollary 5.1], there exists a Zariski closed nowheredense set R ⊂ T such that

p′ : V ′ → T ′



is a locally trivial fibration (for the strong topology), where T ′ = T \ R, V ′ =p−1(T ′), and p′ stands for the restriction of p to V ′. Clearly s restricted to T ′ isa section of p′.

We claim that for each t ∈ T ′ the fiber p′−1(t) is connected, which impliesthat p−1(t) is irreducible. Indeed, let Wt be the connected component of p′−1(t)containing s(t) and set

W :=⋃t∈T ′

Wt.

The fiber p′−1(t) has finitely many connected components, and hence Wt is openand closed in it. Moreover, since p′ is a locally trivial fibration, it follows thatW is open and closed in V ′. Hence W = V ′ and Wt = p′−1(t). Thus p′−1(t) isconnected.

We will make use of Theorem 2.3 to study real algebraic sets. First we recallthat an algebraic subset V of Cn is said to be defined over R if it is the set ofcommon zeros (in Cn) of a collection of polynomials with real coefficients. In otherwords, V is required to be stable under complex conjugation. In that case,

V (R) := V ∩ Rn

is called the set of real points of V . If V (R) is Zariski dense in V , then

V (R)C = V.

Proposition 2.4. Let X be a d-dimensional irreducible algebraic subset of Rn. LetA and B be Nash submanifolds of Rn, both of dimension d and contained in X.Let A′ ⊂ A and B′ ⊂ B be nonempty open subsets (in the relative strong topology).Then there exist points a ∈ A′, b ∈ B′ and an affine (n − d + 1)-plane M in Rn

such that C := X ∩M is an irreducible curve for which a is an accumulation pointof A ∩ (C \ a) and b is an accumulation point of B ∩ (C \ b). In particular, Aand B are compatible.

Proof. It suffices to consider the case d ≥ 2. Set r := n− d+ 1. If X0 is the set ofregular points of X, then the subsets A′ ∩X0 ⊂ A and B′ ∩X0 ⊂ B are nonemptyand open. By Theorem 2.3 (with k = 0), for a general affine r-plane L in Cn theintersection XC ∩ L is an irreducible complex curve. We can choose such an L sothat it is defined over R and the affine r-plane M := L(R) in Rn cuts A and Btransversally at some points a ∈ A′∩X0 and b ∈ B′∩X0. Then the complex curveXC ∩ L is defined over R, and hence

C := (XC ∩ L)(R) = X ∩M

is an irreducible real curve satisfying the required conditions.



3. Proof of Theorem A and related results

The following result will play the key role.

Theorem 3.1. Let X be an irreducible algebraic subset of Rn of dimension d ≥2, and f : X → R a semialgebraic function. Let D be a Nash submanifold ofRn, contained in X and of dimension d. Let C be the collection of all irreduciblealgebraic curves in X of the form X ∩M for some affine (n − d + 1)-plane M inRn with D ∩M 6= ∅. Assume that for every curve C ∈ C the restriction f |C isa rational function. Then the function f is rational.

Proof. Let A and B be connected Nash submanifolds of Rn, both of dimensiond with A ⊂ D and B ⊂ X, for which the restrictions f |A and f |B are analyticfunctions. Such A and B exist, the function f being semialgebraic, cf. [2].

Claim 1. The graphs F := graph(f |A) and G := graph(f |B) have the sameZariski closure in Rn × R.

Note that F and G are connected Nash submanifolds of Rn × R of dimensiond. By Proposition 2.2, it suffices to show that F and G are compatible. To thisend let A′ and B′ be nonempty open subsets of A and B, respectively. Accordingto Proposition 2.4, there exist points a ∈ A′, b ∈ B′ and a curve C ∈ C such thata is an accumulation point of A ∩ (C \ a) and b is an accumulation point ofB ∩ (C \ b). Since f |C is a rational function, there exists a set C0 ⊂ C such thatits complement C \ C0 is finite, f |C is regular on C0, and

G(f |C) = graph(f |C0)Z

is an irreducible algebraic curve in Rn × R. The points a and b may not be in C0,but it does not matter. By construction, (a, f(a)) and (b, f(b)) are accumulationpoints of

F ∩ (graph(f |C0) \ (a, f(a))) and G ∩ (graph(f |C0

) \ (b, f(b))),respectively. This argument shows that F and G are compatible, which completesthe proof of Claim 1.

Note that the algebraic subset

Y := FZ

of X × R is of dimension d and irreducible.

Claim 2. There exists a nonempty Zariski open set X0 ⊂ X such thatgraph(f |X0) ⊂ Y .

Since the function f is semialgebraic, there is a decomposition

X = E ∪A1 ∪ . . . ∪Ak

into disjoint semialgebraic sets such that dimE < d, and for i = 1, . . . , k the Ai isa d-dimensional connected Nash submanifold of Rn for which the restriction f |Ai



is an analytic function, cf. [2]. By Claim 1, the Zariski closure of graph(f |Ai) is

equal to Y . Hence Claim 2 holds for X0 := X \ EZ .

Let π : Cn × C → Cn be the canonical projection. Obviously, the complexi-fications YC ⊂ Cn × C and XC ⊂ Cn are irreducible complex algebraic sets ofdimension d. We have π(YC) ⊂ XC since π(Y ) ⊂ X and π is continuous in theZariski topology. It follows that the restriction π0 : YC → XC of π is genericallyfinite-to-one. Hence there exist a positive integer l and a nonempty Zariski openset U ⊂ XC such that for every point x ∈ U the set π−1

0 (x) = YC ∩ π−1(x) consistsof l distinct points.

Claim 3. The map π0 : YC → XC is generically one-to-one, that is, l = 1.

Set r := n − d + 1. By Theorem 2.3 (with k = 0), the set of affine r-planesin Cn contains a Zariski open and dense subset B such that for every L ∈ B theintersection XC ∩ L is an irreducible complex curve. Shrinking B if necessary andmaking use of Theorem 2.3 (with k = 1), we may assume that YC ∩ π−1(L) is alsoan irreducible complex curve. Now we choose an affine r-plane M in Rn such thatMC ∈ B and M cuts X transversally at some regular point contained in A∩X0∩U .Then XC ∩MC is an irreducible complex curve defined over R, and

Γ := X ∩Mis its set of real points. By construction, Γ is an irreducible real curve with

ΓC = XC ∩MC.

Since the restriction f |Γ is a rational function, there exists a set Γ0 ⊂ Γ such thatits complement Γ \ Γ0 is finite, f |Γ is regular on Γ0, and

G(f |Γ) = graph(f |Γ0)Z

is an irreducible algebraic curve in Rn × R. By Claim 2,

graph(f |Γ0∩X0) ⊂ Y ∩ π−1(M).

Since the intersection Γ0 ∩ X0 is nonempty, we obtain G(f |Γ) ⊂ Y ∩ π−1(M).Taking the complexifications we get G(f |Γ)C ⊂ YC ∩ π−1(MC). The last inclusionimplies that

G(f |Γ)C = YC ∩ π−1(MC)

since both G(f |Γ)C and YC ∩ π−1(MC) are irreducible complex curves. Accordingto Lemma 2.1, the equality

G(f |Γ)C ∩ π−1(x) = (x, f(x))holds for all x ∈ Γ0. It follows that

π−10 (x) = YC ∩ π−1(x) = (x, f(x))

for all x ∈ Γ0. Since the intersection Γ0 ∩ U is nonempty, we conclude that l = 1,which proves Claim 3.

We are ready to complete the proof of the theorem. Obviously, YC ∩ π−1(U) isa constructible set, which according to Claim 3 is the graph of some function



g : U → C. By Zariski’s theorem on constructible graph (see for example Lojasiewicz [13, p. 449]), there exist a nonempty Zariski open set U ′ ⊂ U andcomplex polynomial functions p, q on Cn such that

q(z) 6= 0 and g(z) =p(z)

q(z)for all z ∈ U ′.

It follows from Claim 2 that g(x) = f(x) ∈ R for all x ∈ X0 ∩ U ′. In particular,using the standard notation for complex conjugation, we get

(∗) f(x) =p(x)

q(x)=p(x)

q(x)for all x ∈ X0 ∩ U ′.

The polynomials

p1(z) := p(z)q(z) + p(z)q(z) and q1(z) := 2q(z)q(z)

satisfy p1(z) = p1(z) and q1(z) = q1(z), which implies that they have real coeffi-cients. In view of (∗) we get

f(x) =p1(x)

q1(x)for all x ∈ X0 ∩ U ′.

Hence f is a rational function on X, as required.

As an immediate consequence of Theorem 3.1 we obtain the following criterionfor rationality of semialgebraic functions on Rn.

Corollary 3.2. Let U be a nonempty open subset (in the strong topology) of Rn.A semialgebraic function on Rn is rational, provided that its restriction to everyaffine line passing through a point in U is a rational function.

Let us note that the hypothesis in Corollary 3.2 cannot be relaxed too much.

Example 3.3. The function f : R2 → R, defined by f(x, y) = (x4 + y4)12 , is

semialgebraic and arc-analytic. The restriction of f to an affine line L ⊂ R2 isa rational function if and only if L passes through the origin. Obviously, f is nota rational function.

It is convenient to record the following observation (cf. [5, p. 91]).

Remark 3.4. Let X be an algebraic subset of Rn, and f : X → R a function.Then f is hereditarily rational if and only if there exists a sequence of algebraicsets

X = X0 ⊃ X1 ⊃ · · · ⊃ Xm = ∅such that for i = 0, . . . ,m − 1 the restriction f |Xi

is regular on Xi \ Xi+1. Inparticular, every hereditarily rational function on X is semialgebraic. Indeed, setX0 := X. If f is rational, then there exists an algebraic subset X1 ⊂ X0 such thatdimX1 < dimX0 and f is regular on X0 ⊂ X1. If f is hereditarily rational, wecan repeat this process with f |X1 , and so on, which yields a sequence of algebraicsets with the required properties and shows that f is semialgebraic. The conversereadily follows.



Next we deal with the extension problem for hereditarily rational functions.

Proposition 3.5. Let W ⊂ X be algebraic subsets of Rn, and f : X → Ra hereditarily rational function that is regular on X \W . Then f can be extendedto a hereditarily rational function on Rn that is regular on Rn \W .

Proof. We use induction on dimX. The case dimX is obvious.

Since f is a rational function, it is regular on a Zariski open dense set X0 ⊂ X.We may assume that W ⊂ X \ X0 so, in particular, dimW < dimX. Hence, byinduction, f |W can be extended to a hereditarily rational function ϕ : Rn → R thatis regular on Rn \W . The function g := f − ϕ|X on X is hereditarily rational,regular on X \ W and vanishes on W . It suffices to extend g to a hereditarilyrational function G : Rn → R that is regular on Rn \ W . This can be done asfollows. Since g is regular on X \W , we can find polynomial functions p, q on Rn

satisfying

q(x) 6= 0 and g(x) =p(x)

q(x)for all x ∈ X \W,

cf. [2, p. 62]. Let r be a polynomial function on Rn vanishing precisely in X. SetP := pq, Q := q2 + r2, and define G : Rn → R by

G(x) =P (x)

Q(x)for x ∈ Rn \W and G(x) = 0 for x ∈W.

By construction, G is regular on Rn \W and G|X = g. Furthermore, G is heredi-tarily rational in view of Remark 3.4. The proof is complete.

Now we can prove the main result of this paper.

Proof of Theorem A. It is clear that (b) ⇒ (a), whereas (a) ⇒ (b) follows fromProposition 3.5. By Remark 3.4, (a)⇒ (c). Thus it remains to show that (c)⇒ (a).

Suppose that condition (c) holds. We want to prove that for every algebraicset Z ⊂ X the restriction f |Z is a rational function. It suffices to do it for Zirreducible of dimension at least 2. In that case, however, the assertion followsfrom Theorem 3.1.

4. Nash arcs and meromorphic functions

Throughout this section, X denotes an algebraic subset of Rn. A mapγ : (−1, 1)→ X that is analytic and semialgebraic is called a Nash arc in X.

Lemma 4.1. Let f : X → R be a continuously curve-rational function, andγ : (−1, 1)→ X a Nash arc. Then the function f γ is analytic.

Proof. We may assume that γ is a nonconstant map. Then γ((−1, 1)) is a semi-algebraic curve, and hence its Zariski closure C is an irreducible algebraic curvein X. By assumption, f |C is a continuous rational function. In particular, there



exist two polynomials p, q ∈ R[x1, . . . , xn] with q(x) 6= 0 and f(x) = p(x)/q(x) forall x ∈ C \ C1, where C1 is a finite set. The function

f(γ(t)) = (f |C)(γ(t)) =p(γ(t))

q(γ(t))

is continuous and meromorphic on (−1, 1), and hence it is analytic.

Lemma 4.2. Let f : X → R be a semialgebraic function. Assume that for everyNash arc γ : (−1, 1)→ X the function f γ is analytic. Then f is continuous.

Proof. Let S1 ⊂ R2 be the unit circle. We compactify R via the embedding R→ S1,which is the inverse of the stereographic projection from the north pole, and regardF := graph(f) as a subset of Rn × R2. Fix a point x0 ∈ X and let l ∈ S1 be anypoint such that (x0, l) belongs to the closure of F in Rn × R2. It suffices to provethat f(x0) = l. By the Nash curve selection lemma [2, Proposition 8.1.13], thereexists a Nash arc ϕ = (ψ, g) : (−1, 1)→ Rn × R2 with ϕ(0) = (ψ(0), g(0)) = (x0, l)and ϕ((0, 1)) ⊂ F . In particular, g(t) = f(ψ(t)) for t ∈ (0, 1). Consequently,g(t) = f(ψ(t)) for t ∈ (−1, 1) since both g and f ψ are analytic functions. Hence

l = g(0) = f(x0) = limt→0

f(ψ(t)),

which proves the continuity of f at x0.

Let M be a connected real analytic manifold. We say that a function λ : M → Ris meromorphic if there exist two analytic functions α : M → R and β : M → Rsuch that the set M0 := y ∈M | β(y) 6= 0 is nonempty and λ(y) = α(y)/β(y) forall y ∈M0.

Proposition 4.3. Let f : X → R be a hereditarily rational function. Then for anyconnected real analytic manifold M and any analytic map ϕ : M → X the functionf ϕ is meromorphic.

Proof. We first note that the Zariski closure Z of ϕ(M) is an irreducible algebraicsubset of X (possibly with dimZ > dimM). By assumption, the restriction f |Zis a rational function, and hence there exist two polynomials p, q ∈ R[x1, . . . , xn]with q(x) 6= 0 and f(x) = p(x)/q(x) for all x ∈ Z \ Z1, where Z1 ⊂ Z is analgebraic set, Z1 6= Z. Consequently, f ϕ is a meromorphic function since the setM0 := ϕ−1(Z \Z1) is nonempty and f(ϕ(y)) = p(ϕ(y))/q(ϕ(y)) for all y ∈M0.

Proof of Theorem B. Let X be an algebraic subset of Rn, and f : X → R a semial-gebraic function that is continuously curve-rational. According to Theorem A, f ishereditarily rational. Furthermore, by Lemmas 4.1 and 4.2, f is continuous. Now,let η : (−1, 1)→ X be an analytic arc. In view of Proposition 4.3, f η is a mero-morphic function. Thus, f η is analytic since it is continuous and meromorphicon (−1, 1). Consequently, f is arc-analytic.



References

[1] E. Bierstone, P. Milman and A. Parusinski, A function which is arc-analytic but not contin-uous, Proc. Amer. Math. Soc. 113, 2 (1991), 419–424.

[2] J. Bochnak, M. Coste and M.-F. Roy, Real algebraic geometry, Springer-Verlag, Berlin, 1998.[3] G. Fichou, J. Huisman, F. Mangolte and J.-Ph. Monnier, Fonctions regulues, J. Reine Angew.

Math. 718 (2016), 103–151.

[4] J. Kollar, W. Kucharz and K. Kurdyka, Curve-rational functions, Math. Ann. (2017), DOI10.1007/s00208-016-1513-z.

[5] J. Kollar and K. Nowak, Continuous rational functions on real and p-adic varieties, Math.

Z. 279 (2015), 85–97.[6] W. Kucharz, Rational maps in real algebraic geometry, Adv. Geom. 9 (2009), 517–539.

[7] W. Kucharz, Continuous rational maps into the unit 2-sphere, Arch. Math. (Basel) 102

(2014), no. 3, 257–261.[8] W. Kucharz, Approximation by continuous rational maps into spheres, J. Eur. Math. Soc.

(JEMS) 16 (2014), no. 8, 1555–1569.

[9] W. Kucharz and K. Kurdyka, Stratified-algebraic vector bundles, J. Reine Angew. Math.(2016), DOI 10.1515/crelle-2015-0105.

[10] K. Kurdyka, Ensembles semi-algebriques symetriques par arcs, Math. Ann. 281 no. 3 (1988),445–462.

[11] K. Kurdyka, An arc-analytic function with nondiscrete singular set, Ann. Polon. Math. 59,

3 (1994), 251–254.[12] R. Lazarsfeld, Positivity in algebraic geometry. I. Classical setting: line bundles and linear

series. Ergebnisse der Mathematik un ihrer Grenzgebiete 3. Folge. A series of Modern Surveys

in Mathematics 48. Springer-Verlag, Berlin, 2004.[13] S. Lojasiewicz, Introduction to complex analytic geometry, Birkhuser Verlag, Basel, 1991.

[14] J.-L. Verdier, Stratifications de Whitney et theoreme de Bertini-Sard, Invent. Math. 36

(1976), 295–312.

(Wojciech Kucharz) Institute of Mathematics, Faculty of Mathematics and ComputerScience, Jagiellonian University, Lojasiewicza 6, 30-348 Krakow, Poland


(Krzysztof Kurdyka) Laboratoire de Mathematiques (LAMA), UMR 5127 CNRS

Universite Savoie Mont Blanc, Campus Scientifique, 73 376 Le Bourget-du-Lac Cedex,France


“21˙Ollagnier˙Nowicki” — 2017/12/1 — 20:48 — page 97 — #1




RATIONAL CONSTANTS OF CYCLOTOMIC DERIVATIONS

JEAN MOULIN OLLAGNIER AND ANDRZEJ NOWICKI

1. Introduction

Let K(X) = K(x0, . . . , xn−1) be the field of rational functions in n > 3variables over a field K of characteristic zero. Let d be the cyclotomic derivationof K(X), that is, d is the K-derivation of K(X) defined by

d(xj) = xj+1, for j ∈ Zn.We denote by K(X)d the field of constants of d, that is, K(X)d = f ∈K(X); d(f) = 0.

We are interested in algebraic descriptions of the field K(X)d. However, weknow that such descriptions are usually difficult to obtain. Fields of constantsappear in various classical problems; for details we refer to [2], [3], [12], [9] and[11].

We already know (see [10]) that if K contains the n-th roots of unity, thenK(X)d is a field of rational functions over K and its transcendence degree over Kis equal to m = n − ϕ(n), where ϕ is the Euler totient function. In our proof ofthis fact the assumption concerning n-th roots plays an important role. We do notknow if the same is true without this assumption. What happens, for example,when K = Q ?

In this article we give a partial answer to this question, for arbitrary field K ofcharacteristic zero.

We introduce a class of special positive integers, and we prove (see Theorem 9.1)that if n belongs to this class, then the mentioned result is also true for arbitraryfield K of characteristic zero, without the assumption concerning roots of unity.

2010 Mathematics Subject Classification. Primary 12H05; Secondary 13N15.Key words and phrases. Derivation, cyclotomic polynomial, Darboux polynomial, Euler to-

tient function, Euler derivation, factorisable derivation, Jouanolou derivation, Lotka-Volterra

derivation.

97


98 J. MOULIN OLLAGNIER AND A. NOWICKI

Moreover, we construct a set of free generators of K(X)d, which are polynomialswith integer coefficients. Thus, if the number n is special, then

K(X)d = K (F0, . . . , Fm−1) ,

for some, algebraically independent, polynomials F0, . . . , Fm−1 belonging to thepolynomial ring Z[X] = Z [x0, . . . , xn−1], and where m = n − ϕ(n). Note that inthe segment [3, 100] there are only 3 non-special numbers: 36, 72 and 100. We donot know if the same is true for non-special numbers, for example when n = 36.

In our proofs we use classical properties of cyclotomic polynomials, and an im-portant role play some results ([4], [5], [16], [17] and others) on vanishing sums ofroots of unity.

2. Notations and preparatory facts

Throughout this paper n > 3 is an integer, ε is a primitive n-th root ofunity, and Zn is the ring Z/nZ. Moreover, K is a field of characteristic zero,K[X] = K[x0, . . . , xn−1] is the polynomial ring over K in variables x0, . . . , xn−1,and K(X) = K(x0, . . . , xn−1) is the field of quotients of K[X]. The indexes of thevariables x0, . . . , xn−1 are elements of the ring Zn. The cyclotomic derivation d isthe K-derivation of K(X) defined by d(xj) = xj+1 for j ∈ Zn.

For every sequence α = (α0, α1, . . . , αn−1), of integers, we denote by Hα(t) thepolynomial from Z[t] defined by

Hα(t) = α0 + α1t1 + α2t

2 + · · ·+ αn−1tn−1.

An important role in our paper will play two subsets of Zn denoted by Gn andMn. The first subset is the set of all sequences α = (α0, . . . , αn−1) such thatα0, . . . , αn−1 are integers and

α0 + α1ε1 + α2ε

2 + · · ·+ αn−1εn−1 = 0.

The second subset Mn is the set of all such sequences α = (α0, . . . , αn−1) whichbelong to Gn and the integers α0, . . . , αn−1 are nonnegative, that is, they belongto the set of natural numbers N = 0, 1, 2, . . . . To be precise,

Gn = α ∈ Zn; Hα(ε) = 0 , Mn = α ∈ Nn; Hα(ε) = 0 = Gn ∩ Nn.

If α, β ∈ Gn, then of course α±β ∈ Gn, and if α, β ∈Mn, then α+β ∈Mn. ThusGn is an abelian group, and Mn is an abelian monoid with zero 0 = (0, . . . , 0).

Let us recall that ε is an algebraic element over Q, and its monic minimalpolynomial is equal to the n-th cyclotomic polynomial Φn(t). Recall also (seefor example [6] or [7]) that Φn(t) is a monic irreducible polynomial with integercoefficients of degree ϕ(n), where ϕ is the Euler totient function. This implies thefollowing proposition.

Proposition 2.1. Let α ∈ Zn. Then α ∈ Gn if and only if there exists a polynomialF (t) ∈ Z[t] such that Hα(t) = F (t)Φn(t).


RATIONAL CONSTANTS OF CYCLOTOMIC DERIVATIONS 99

Put e0 = (1, 0, 0, . . . , 0), e1 = (0, 1, 0, . . . , 0), . . . , en−1 = (0, 0, . . . , 0, 1), and let

e =∑n−1i=0 ei = (1, 1, . . . , 1). Since

∑n−1i=0 ε

i = 0, the element e belongs to Mn.

The monoidMn has an order >. If α, β ∈ Gn, the we write α > β, if α−β ∈ Nn,that is, α > β ⇐⇒ there exists γ ∈ Mn such that α = β + γ. In particular,α > 0 for any α ∈ Mn. It is clear that the relation > is reflexive, transitive andantisymmetric. Thus Mn is a poset with respect to >.

Let α ∈ Mn. We say that α is a minimal element of Mn, if α 6= 0 and there isno β ∈ Mn such that β 6= 0 and β < α. Equivalently, α is a minimal element ofMn, if α 6= 0 and α is not a sum of two nonzero elements of Mn.

We denote by ζ, the rotation of Zn given by ζ(α) = (αn−1, α0, α1, . . . , αn−2) ,for α = (α0, α1, . . . , αn−1) ∈ Zn. The mapping ζ is a Z-module automorphism ofZn. Note that ζ−1(α) = (α1, . . . , αn−1, α0), for all α = (α0, α1, . . . , αn−1) ∈ Zn. Ifa, b ∈ Z and a ≡ b (mod n), then ζa = ζb. Moreover, ζ(ej) = ej+1 for all j ∈ Zn,and ζ(e) = e.

Let us recall from [10] some basic properties of Mn and Gn.

Proposition 2.2 ([10]).

(1) If α ∈ Gn, then there exist β, γ ∈Mn such that α = β − γ.

(2) The poset Mn is artinian, that is, if α(1) > α(2) > α(3) > . . . is a sequenceof elements from Mn, then there exists an integer s such that α(j) = α(j+1) for allj > s.

(3) The set of all minimal elements of Mn is finite.

(4) For any 0 6= α ∈ Mn there exists a minimal element β such that β 6 α.Moreover, every nonzero element of Mn is a finite sum of minimal elements.

(5) Let α ∈ Zn. If α ∈ Gn, then ζ(α) ∈ Gn. If α ∈ Mn, then ζ(α) ∈ Mn.Moreover, α is a minimal element of Mn if and only if ζ(α) is a minimal elementof Mn.

Look at the cyclotomic polynomial Φn(t). Assume that Φn(t) = c0 + c1t+ · · ·+cϕ(n)t

ϕ(n). All the coefficients c0, . . . , cϕ(n) are integers, and c0 = cϕ(n) = 1. Putm = n− ϕ(n) and

γ0 =(c0, c1, . . . , cϕ(n), 0, . . . , 0︸︷︷︸

m−1

).

Note that γ0 ∈ Zn, and Hγ0(t) = Φn(t). Consider the elements γ0, γ1, . . . , γm−1defined by γj = ζj(γ0), for j = 0, 1, . . . ,m− 1. Observe that Hγj (t) = Φn(t) · tj forall j ∈ 0, . . . ,m− 1. Since Φn(ε) = 0, we have Hγj (ε) = 0, and so, the elementsγ0, . . . , γm−1 belong to Gn. Moreover, we proved in [10], that they form a basisover Z, which is the following theorem.

Theorem 2.3 ([10]). Gn is a free Z-module, and the elements γ0, . . . , γm−1, wherem = n− ϕ(n), form its basis over Z.



3. Standard minimal elements

Assume that p is a prime divisor of n, and consider the sequences

m(p, r) =

p−1∑i=0

er+inp ,

for r = 0, 1, . . . , np − 1. Observe that each m(p, r) is equal to ζr (m(p, 0)). Each

m(p, r) is a minimal element of Mn (see [10] for details). We say that m(p, r) is

a standard minimal element of Mn. In [10] we used the notation E(p)r instead of

m(p, r). It is clear that if r1, r2 ∈ 0, 1, . . . , np − 1 and r1 6= r2, then m(p, r1) 6=m(p, r2).

If α = (α0, . . . , αn−1) ∈ Zn, then we denote by |α| the sum α0 + · · · + αn−1.Observe that, for every r, we have |m(p, r)| = p. This implies, that if p 6= q areprime divisors of n, then m(p, r1) 6= m(q, r2) for all r1 ∈ 0, . . . , np − 1, r2 ∈0, 1, . . . , nq − 1. Note the following two obvious propositions.

Proposition 3.1.

np−1∑r=0

m(p, r) = (1, 1, . . . , 1) = e.

Proposition 3.2. If p is a prime divisor of n, then the standard elements m(p, 0),m(p, 1), . . . , m(p, np − 1) are linearly independent over Z.

The following two propositions are less obvious and deserve a proof.

Proposition 3.3. Let n = pqN , where p 6= q are primes and N is a positiveinteger. Then

p−1∑k=0

m(q, kN) =

q−1∑k=0

m(p, kN).

which, for any shift r, is easily extended to

p−1∑k=0

m(q, kN + r) =

q−1∑k=0

m(p, kN + r).

Proof. If m is a positive integer, then we denote by [m] the set 0, 1, . . . ,m − 1.First observe that

k + ip; k ∈ [p], i ∈ [q]

=k + iq; k ∈ [q], i ∈ [p]

= [pq].

Hence,

p−1∑k=0

m(q, kN) =

p−1∑k=0

q−1∑i=0

ekN+inq=

p−1∑k=0

q−1∑i=0

eN(k+ip) =

pq−1∑k=0

eNk;

q−1∑k=0

m(p, kN) =

q−1∑k=0

p−1∑i=0

ekN+inp=

q−1∑k=0

p−1∑i=0

eN(k+iq) =

pq−1∑k=0

eNk.

Thus,p−1∑k=0

m(q, kN) =pq−1∑k=0

ekN =q−1∑k=0

m(p, kN).



Proposition 3.4. Let p be a prime divisor of n. Let 0 6 r < np , and a ∈ Z. Then

ζa(m(p, r)

)= m(p, b), where b = (a+ r)

(mod

n

p

)Proof. Put w = n

p , and [p] = 0, 1, . . . , p − 1. Let a + r = cw + b, where c, b ∈ Z

with 0 6 b < w. Observe thatb+ (c+ i)w (mod n); i ∈ [p]

=b+ iw; i ∈ [p]

.

Hence,

ζa(m(p, r)

)= ζa

(p−1∑i=0

er+iw

)=

p−1∑i=0

ζa (er+iw) =

p−1∑i=0

ea+r+iw

=

p−1∑i=0

eb+cw+iw =

p−1∑i=0

eb+(c+i)w =

p−1∑i=0

eb+iw = m(p, b),

and b = (a+ r) (mod w).

We will apply the following theorem of Redei, de Bruijn and Schoenberg.

Theorem 3.5 ([13], [1], [15]). The standard minimal elements of Mn generatethe group Gn.

Known proofs of the above theorem used usually techniques of group rings. Lamand Leung [5] gave a new proof using induction and group-theoretic techniques.

We know (see for example [10]) that if n is divisible by at most two distinctprimes, then every minimal element ofMn is standard. It is known (see for example[5], [17], [14]) that in all other cases always exist nonstandard minimal elements.

4. The sets Ij

Let n > 3 be an integer, and let n = pα11 · · · pαs

s , where p1, . . . , ps are distinctprimes and α1, . . . , αs are positive integers. Put nj = n

pjfor j = 1, . . . , s. Let

I1, . . . , Is be sets of integers defined as follows:

I1 =r ∈ Z; 0 6 r < n1

,

I2 =r ∈ Z; 0 6 r < n2, gcd(r, p1) = 1

,

I3 =r ∈ Z; 0 6 r < n3, gcd(r, p1p2) = 1

,

...

Is =r ∈ Z; 0 6 r < ns, gcd(r, p1p2 · · · ps−1) = 1

.

That is, I1 = r ∈ Z; 0 6 r < n1 and Ij = r ∈ Z; 0 6 r <nj , gcd(r, p1 · · · pj−1) = 1 for j = 2, . . . , s. This definition depends of the fixedsuccession of primes. We will say that the above I1, . . . , Is are the n-sets of type[p1, . . . , ps].



Let for example n = 12 = 223. Then I1 = 0, 1, 2, 3, 4, 5, I2 = 1, 3 are the12-sets of type [2, 3], and I1 = 0, 1, 2, 3, I2 = 1, 2, 4, 5 are the 12-sets of type[3, 2].

Example 4.1. The 30-sets of a a given type:

type I1 I2 I3[2, 3, 5] 0, 1, 2, . . . , 14 1, 3, 5, 7, 9 1, 5[2, 5, 3] 0, 1, 2, . . . , 14 1, 3, 5 1, 3, 7, 9[3, 2, 5] 0, 1, 2, . . . , 9 1, 2, 4, 5, 7, 8, 10, 11, 13, 14 1, 5[3, 5, 2] 0, 1, 2, . . . , 9 1, 2, 4, 5 1, 2, 4, 7, 8, 11, 13, 14[5, 2, 3] 0, 1, 2, 3, 4, 5 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14 1, 3, 7, 9[5, 3, 2] 0, 1, 2, 3, 4, 5 1, 2, 3, 4, 6, 7, 8, 9 1, 2, 4, 7, 8, 11, 13, 14

Now we calculate the cardinality of the sets I1, . . . , Is. We denote by |X| thenumber of all elements of a finite set X. First observe that if a, b are relatively primepositive integers, then in the set 1, 2, . . . , ab there are exactly ϕ(a)b numbersrelatively prime to a. In fact, let u ∈ 1, 2, . . . , ab. Then u = ka + r, where0 6 k 6 b and 0 6 r < a, and gcd(u, a) = 1 ⇐⇒ gcd(r, a) = 1. Thus, every suchu, which is relatively prime to a, is of the form ka+ r with 1 6 r < a, gcd(r, a) = 1and where k is an arbitrary number belonging to 0, 1, . . . , b− 1. Hence, we haveexactly b such numbers k, and so, the number of integers in 1, . . . , ab, relativelyprime to a, is equal to ϕ(a)b. As a consequence of this fact we obtain

Lemma 4.2. Let a > 2, b > 2 be relatively prime integers. Then there are exactlyϕ(a)b such integers belonging to 0, 1, . . . , ab− 1 which are relatively prime to a.

Let us recall that ϕ(n) = n(

1− 1p1

)· · ·(

1− 1ps

). Now we are ready to prove

the following proposition.

Proposition 4.3. |I1| = n1, and |Ij | = nj

(1− 1

p1

)(1− 1

p2

)· · ·(

1− 1pj−1

), for

all j = 2, 3, . . . , s.

Proof. The case |I1| = n1 is obvious. Let j > 2, and put a = pα11 · · · p

αj−1

j−1 , b =

pαj−1j p

αj+1

j+1 · · · pαss . Then gcd(a, b) = 1, nj−1 = ab−1, and if r ∈ 0, 1, . . . , nj−1,

then r ∈ Ij ⇐⇒ gcd(r, a) = 1. Hence, by Lemma 4.2, we have

|Ij | = ϕ(a)b = pα11 · · · p

αj−1

j−1

(1− 1

p1

)· · ·(

1− 1pj−1

)b

= pα11 · · · p

αj−1

j−1

(1− 1

p1

)· · ·(

1− 1pj−1

)pαj−1j p

αj+1

j+1 · · · pαss

= npj

(1− 1

p1

)· · ·(

1− 1pj−1

)= nj

(1− 1

p1

)· · ·(

1− 1pj−1

).

This completes the proof.

Lemma 4.4. Consider some nonzero numbers z1, . . . , zs. Define w1 by w1 = 1z1

and wj by wj = 1zj

(1− 1

z1

)(1− 1

z2

)· · ·(

1− 1zj−1

)for j = 2, . . . , s. Then

w1 + w2 + · · ·+ ws = 1−(

1− 1z1

)(1− 1

z2

)· · ·(

1− 1zs

).



Proof. The case s = 1 is obvious. Assume now that it is true for an integer s > 1,and consider nonzero numbers z1, . . . , zs+1. Then we have

1−(

1− 1z1

)· · ·(

1− 1zs+1

)=(

1−(

1− 1z1

)· · ·(

1− 1zs

))+ 1

zs+1

(1− 1

z1

)· · ·(

1− 1zs

)= w1 + · · ·+ ws + ws+1.

Proposition 4.5. |I1|+ |I2|+ · · ·+ |Is| = n− ϕ(n).

Proof. We know, by Proposition 4.3, that |Ij | = nwj , for j = 1, . . . , s, where

w1 = 1p1

and wj = 1pj

(1− 1

p1

)(1− 1

p2

)· · ·(

1− 1pj−1

)for j = 2, . . . , s. Thus, by

Lemma 4.4,

|I1|+ |I2|+ · · ·+ |Is| = n (w1 + · · ·+ ws)

= n(

1−(

1− 1p1

)(1− 1

p2

)· · ·(

1− 1ps

))= n− n

(1− 1

p1

)(1− 1

p2

)· · ·(

1− 1ps

)= n− ϕ(n).


Let us recall the following well-known lemma where ε is a primitive n-th root ofunity.

Lemma 4.6. Let c be an integer and let U =n−1∑r=0

(εc)r. If n - c then U is equal to

0, and in the other case, when n | c, this sum is equal to n.

Using this lemma we may prove the following proposition.

Proposition 4.7. If c ∈ Z then, for any j ∈ 1, . . . , s, the sum Wj =∑r∈Ij

(εpjc)r

is an integer.

Proof. First consider the case j = 1. Let η = εp1 . Then η is a primitive n1-th root

of unity, and W1 =n1−1∑r=0

(ηc)r. It follows from Lemma 4.6 that W1 is an integer.

Now assume that j > 2. Put X = 0, 1, . . . , nj − 1, and Di = r ∈ X; pi | rfor i = 1, . . . , j − 1. Then Ij = X r (D1 ∪ · · · ∪Dj−1), and then Wj = U − V ,where

U =∑r∈X

(εpjc)r, V =

∑r∈D1∪···∪Dj−1

(εpjc)r.

Observe that U =nj−1∑r=0

(ηc)r, where η = εpj is a primitive nj-root of unity. Thus,

by Lemma 4.6, U is an integer. Now we will show that V is also an integer. For



this aim first observe that

V =

j−1∑k=1

(−1)k+1∑

i1<···<ik

∑r∈Di1...ik

(εpjc)r,

where the sum∑

i1<···<ikruns through all integer sequences (i1, . . . , ik) such that

1 6 i1 < · · · < ik 6 j − 1, and where Di1...ik = Di1 ∩ · · · ∩Dik .

Let 1 6 i1 < · · · < ik 6 j − 1 be a fixed integer sequence. Then we have∑r∈Di1...ik

(εpjc)r

=u−1∑r=0

(ηc)r,

where η = εpj ·pi1 ···pik , and u =nj

pi1 ···pik= n

pj ·pi1 ···pik. Since η is a primitive u-th

root of unity, it follows from Lemma 4.6 that the last sum is an integer. Hence,every sum of the form

∑r∈Di1...ik

(εpjc)r

is an integer, and consequently, V is an

integer. We already know that U is an integer. Therefore, Wj = U − V is aninteger.

5. Special numbers

As in the previous section, let n = pα11 · · · pαs

s , where p1, . . . , ps are distinctprimes and α1, . . . , αs are positive integers. Put nj = n

pjfor j = 1, . . . , s. Assume

that [p1, . . . , pn] is a fixed type, and I1, . . . , Is are the n-sets of type [p1, . . . , ps].If j ∈ 1, . . . , s and 0 6 r < nj , then we have the standard minimal element

m(pj , r) =∑pj−1i=0 er+inj

. Let us recall that each m(pj , r) belongs to the monoidMn, and it is a minimal element of Mn. Moreover, nj = n

pjfor j = 1, . . . , s.

The main role in this section will play the sets A1, . . . ,As, which are subsets ofthe monoid Mn. We define these subsets as follows

Aj =m(pj , r); r ∈ Ij

,

for all j = 1, . . . , s. We denote by A the union A = A1 ∪ · · · ∪ As. Note that theabove setsA andA1, . . . ,As are determined by the fixed succession P = [p1, . . . , qn]of the primes p1, . . . , ps. In our case we will say that A is the n-standard set oftype P .

Observe that the sets A1, . . . ,As are pairwise disjoint, and as a consequence ofProposition 4.5 we have the equality |A| = n− ϕ(n).

Let us recall (see Theorem 2.3) that the group Gn is a free Z-module, and itsrank is equal to n−ϕ(n), so this rank is equal to |A|. We are interested in findingconditions for A to be a basis of Gn. First we need A to be linearly independentover Z.



Special numbers will then be convenient to prove Theorem 9.1. We will say thatthe number n is special of type P if the n-standard set A of type P is linearlyindependent over Z. Moreover, we will say that the number n is special if thereexists a type P for which n is special of type P . We will say that the number n isabsolutely special if it is special with respect to any type P .

Example 5.1. Let n = 12 = 223 and consider the type [2, 3]. In this case we have:s = 2, p1 = 2, p2 = 3, n1 = 6, n2 = 4, I1 = 0, 1, 2, 3, 4, 5 and I2 = 1, 3. The12-standard set A of type [2, 3] is the set of the following 8 sequences:

m(2, 0) = (1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0),m(2, 1) = (0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0),m(2, 2) = (0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0),m(2, 3) = (0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0),m(2, 4) = (0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0),m(2, 5) = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1),m(3, 1) = (0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0),m(3, 3) = (0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1).

Observe that m(2, 1) +m(2, 3) +m(2, 5) = m(3, 1) +m(3, 3). Hence, the set Ais not linearly independent over Z. This means, that 12 is not a special number oftype [2, 3].

Now consider n = 12 and the type [3, 2]. In this case p1 = 3, p2 = 2, n1 = 4,n2 = 6, I1 = 0, 1, 2, 3 and I2 = 1, 2, 2, 5. The 12-standard set A of type [3, 2]is in this case the set of the following 8 sequences:

m(3, 0) = (1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0),m(3, 1) = (0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0),m(3, 2) = (0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0),m(3, 3) = (0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1),m(2, 1) = (0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0),m(2, 2) = (0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0),m(2, 4) = (0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0),m(2, 5) = (0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1).

It is easy to check that in this case the set A is linearly independent over Z. Thus,12 is a special number of type [3, 2], and 12 is not a special number of type [2, 3].

We will prove that the number n is absolutely special if and only if either n issquare-free or n is a power of a prime number. Moreover, we will prove that thenumber n is special if and only if n = p1p2 · · · ps−1pαs

s , where p1, . . . , ps are distinctprimes and αs > 1.

Proposition 5.2. Every power of a prime is an absolutely special number.



Proof. Let n = pm, where p is a prime and m > 1. Then s = 1, n1 = pm−1,I1 = 0, 1, . . . , pm−1 − 1 and there is only one type P = [p]. Thus, A = A1 and,by Proposition 3.2, the set A is linearly independent over Z.

Lemma 5.3. Let p be a prime number, and let N > 2 be an integer such that p - N .Then, for every integer r, there exists a unique cr ∈ 0, 1, . . . , p− 1 such that thenumber r + crN is divisible by p. Moreover, all numbers of the form r + crN with0 6 r < N are pairwise different.

Proof. Let r ∈ Z. Consider the integers r, r + N, r + 2N, . . . , r + (p − 1)N , andobserve that these numbers are pairwise noncongruent modulo p. Thus, thereexists a unique cr ∈ 0, 1, . . . , p− 1 such that r+ crN = 0 (mod p). Assume thatr1 + cr1N = r2 + cr2N for some r1, r2 ∈ 0, 1, . . . , N − 1. Then N | r1 − r2 andso, r1 = r2.

Despite the fact that we need the full Theorem 5.10 (A generates Gn), we firststate and prove the following Proposition (A is linearly independent over Z) fora better understanding. This Proposition is not equivalent, as A could generatea subgroup of Gn of finite index.

Proposition 5.4. Let n = p1 · · · ps−1 · pαs , where s > 2, α > 1, and p1, . . . , psare distinct primes. Then n is a special number of every type of the form[pσ(1), . . . , pσ(s−1), ps

],where σ is a permutation of 1, . . . , s− 1.

Proof. Let P be a fixed type with ps at the end. Without loss of generality, wemay assume that P = [p1, . . . , ps−1, ps]. Let I1, . . . , Is be n-sets of type P , andassume that

(a)s∑j=1

∑r∈Ij

γ(j)r m(pj , r)

= (0, 0, . . . , 0),

where each γ(j)r is an integer. We will show that γ

(j)r = 0 for all j, r.

Note, that every standard element u = m(pj , r) is a sequence (u0, u1, . . . , un−1),where all u0, . . . , un−1 are integers belonging to 0, 1. We will denote by S(u) the

support of u, that is, S(u) =k ∈ 0, 1, . . . , n− 1; uk = 1

.

Consider the case j = 1. Put p = p1 and N = n1 = np = p2p3 . . . ps−1 · pαs .

Observe that p - N , and all the numbers n2, . . . , ns are divisible by p. Let u =m(pj , r) with r ∈ Ij , where j > 2. Then p - r, and

S(u) = r, r + nj , r + 2nj , . . . , r + (pj − 1)nj,and hence, all the elements of S(u) are not divisible by p.

Look at the support of m(p1, r) with r ∈ I1. We have S(m(p1, r)

)=r, r +

N, r+2N, . . . , r+(p−1)N. It follows from Lemma 5.3 that in this support there

exists exactly one element divisible by p. Let us denote this element by r + crN .



We know also from the same lemma, that all the elements r+ crN with r ∈ I1 arepairwise different. These arguments imply, that in the equality (a) all the integers

γ(1)r , with r ∈ I1, are equal to zero.

Now let 2 6 j0 < s, and assume that we already proved the equalities γ(j)r = 0

for all j < j0 and r ∈ Ij . Then the equality (a) is of the form

(b)s∑

j=j0

∑r∈Ij

γ(j)r m(pj , r)

= (0, 0, . . . , 0),

We will show that γ(j0)r = 0 for all r ∈ Ij0 .

Put p = pj0 and N = nj0 = np . Observe that p - N , and all the numbers nj with

j > j0 are divisible by p. Let u = m(pj , r) with r ∈ Ij , where j > j0. Then p - r,and

S(u) = r, r + nj , r + 2nj , . . . , r + (pj − 1)nj,and hence, all the elements of S(u) are not divisible by p.

Look at the support of m(pj0 , r) with r ∈ Ij0 . We have S(m(pj0 , r)

)=r, r+

N, r+2N, . . . , r+(p−1)N. It follows from Lemma 5.3 that in this support there

exists exactly one element divisible by p. Let us denote this element by r + crN .We know also from the same lemma, that all the elements r+ crN with r ∈ Ij0 arepairwise different. These arguments imply, that in the equality (b) all the integers

γ(j0)r , with r ∈ Ij0 , are equal to zero.

Hence, by the induction hypothesis, the equality (b) reduces to the equality∑r∈Is

γ(s)r m(ps, r) = (0, 0, . . . , 0),

where each γr(s) is an integer. Now we use Proposition 3.2 and we have γr(s) = 0for all r ∈ Is. Thus, we proved that in the equality (a) all the integers of theform γjr , where j ∈ 1, . . . , s and r ∈ Ij , are equal to zero. This means that then-standard set A of type P is linearly independent over Z. Therefore, n is a specialnumber of type P .

Using the above proposition for α = 1 we obtain

Proposition 5.5. Every square-free integer n > 2 is absolutely special.

Lemma 5.6. Let n = pα11 · · · pαs

s , where s > 2, p1, . . . , ps are distinct prime num-bers and α1, . . . , αs are positive integers. Let P = [p1, . . . , ps]. If α1 > 2, then n isnot a special number of type P .

Proof. Put p = p1, q = p2, u = np2 , v = n

pq , a =u−1∑k=0

m(p, pk+ 1), b =v−1∑k=0

m(q, pk+

1). Observe that a is a sum of elements from A1, and b is a sum of elements from



A2. Moreover, n1 = np = pu, n2 = n

q = pv,

a =u−1∑k=0

p−1∑i=0

epk+1+in1 =u−1∑k=0

p−1∑i=0

epk+1+ipu =u−1∑k=0

p−1∑i=0

ep(k+iu)+1 =

n1−1∑j=0

epj+1,

b =v−1∑k=0

q−1∑i=0

epk+1+in2=v−1∑k=0

q−1∑i=0

epk+1+ipv =v−1∑k=0

q−1∑i=0

ep(k+iv)+1 =

n1−1∑j=0

epj+1.

Hence, a =

n1−1∑j=0

epj+1 = b. This implies that the n-standard set A of type P is not

linearly independent over Z. Thus, n is not a special number of type P .

Lemma 5.7. Let n = pα11 · · · pαs

s , where s > 2, p1, . . . , ps are distinct prime num-bers and α1, . . . , αs are positive integers. Let P = [p1, . . . , ps]. If there existsj0 ∈ 1, 2, . . . , s− 1 such that αj0 > 2, then n is not a special number of type P .

Proof. If j0 = 1 then the assertion follows from Lemma 5.6. Assume that j0 > 2,and let A1, . . . ,As be the n-standard sets of type P . Put N = pα1

1 · · · pαj0−1

j0−1 ,

p = pj0 , q = pj0+1, u = nNp2 , v = n

Npq , w = npN , a =

u−1∑k=0

m(p, pNk + 1), and

b =v−1∑k=0

m(q, pNk + 1). Observe that a is a sum of elements from Aj0 , and b is

a sum of elements from Aj0+1. Moreover, nj0 = np = pNu, nj0+1 = n

q = pNv,

a =

u−1∑k=0

p−1∑i=0

epNk+1+inj0=

u−1∑k=0

p−1∑i=0

epNk+1+ipNu

=u−1∑k=0

p−1∑i=0

epN(k+iu)+1 =w−1∑j=0

epNj+1,

b =

v−1∑k=0

q−1∑i=0

epNk+1+inj0+1 =

v−1∑k=0

q−1∑i=0

epNk+1+ipNv

=v−1∑k=0

q−1∑i=0

epN(k+iv)+1 =w−1∑j=0

epNj+1.

Hence, a =w−1∑j=0

epNj+1 = b, where w = npN . This implies that the n-standard set

A of type P is not linearly independent over Z. Thus, n is not a special number oftype P .

As a consequence of the above facts we obtain the following theorems.

Theorem 5.8. An integer n > 2 is special if and only if n = p1p2 · · · ps−1pαss ,

where p1, . . . , ps are distinct primes and αs > 1.



Theorem 5.9. An integer n > 2 is absolutely special if and only if either n issquare-free or n is a power of a prime number.

The smallest non-special positive integer n > 2 is n = 36. In the segment [2, 100]there are 3 non-special numbers: 36, 72 and 100.

Let us recall that if n is a special number, then its n-standard set A is linearlyindependent over Z. Now we will show that, in this case, the set A is a basis ofGn. Let us denote by A the subgroup of Gn generated by A. Every element of Ais a finite combination over Z of some elements of A.

We already know (see Theorem 3.5) that the group Gn is generated by all thestandard minimal elements of Mn. Thus, for a proof that A is a basis of Gn, itsuffices to prove that every standard minimal element of Mn belongs to A.

Theorem 5.10. Let n = p1 · · · ps−1pαs , where s > 1, α > 1, and p1, . . . , ps arepairwise different primes. Let P = [p1, . . . , ps], and let A be the n-standard set oftype P . Then every standard minimal element of Mn belongs to A.

Proof. First, all p1-standard elements m(p1, r) with 0 6 r < np1

belong to A1 and

thus to A.

To go further, for j > 1, we will use the relations given in Proposition 3.3 andwe define therefore the height of a pj-standard element (that may not belong toAj) as the number of primes among p1, · · · , pj−1 that divide r and denote it byh(m(pj , r)). Elements of Aj have height 0. A pj-standard element has an heightat most j − 1.

By definition all standard elements of height 0 belong to A and thus to A.

To achieve the proof by induction, we use the following fact.

Key fact. For j > 1, let m(pj , r) be a pj-standard element with a non-zero height.Then some of the pi, 1 ≤ i < j divide r. Let then denote by p one of them and pjby q.

As all prime factors but the last have exponent 1 in the decomposition of n, whenwe apply Proposition 3.3, N = n/pq is coprime with p and a multiple of all pl, 1 ≤l < j, l 6= i.

For any k, 1 ≤ k ≤ p − 1, r + kN is coprime with p and keeps the same otherdivisors among the other pl, 1 ≤ l < j, l 6= i : the height h(m(pj , r + lN)) is thenh(m(pj , r))− 1.

Whence the following relation we get from Proposition 3.3

m(q, r) =

q−1∑k=0

m(p, kN + r)−p−1∑k=1

m(q, kN + r).



which means

m(pj , r) =

q−1∑k=0

m(pi, kN + r)−p−1∑k=1

m(pj , kN + r).

and m(pj , r) is a Z-linear combination of some m(pj , r′) with a strictly smaller

height and of some m(pi, r′′) for an index i < j.

The proof is now a double induction with the following steps.

Let j > 1 and suppose that all m(pi, r) have been proven to belong to A for alli < j.

All m(pj , r) with a 0 height belong to Aj and then to A.

For any h′, 1 ≤ h′ < j, if we know that all m(pj , r) with h(m(pj , r)) < h′ belong

to A, then the same is true for all m(pj , r) with h(m(pj , r)) = h′ according to theprevious key fact.

6. The cyclotomic derivation d

Throughout this section n > 3 is an integer, K is a field of characteristic zero,K[X] = K[x0, . . . , xn−1] is the polynomial ring over K in variables x0, . . . , xn−1,and K(X) = K(x0, . . . , xn−1) is the field of quotients of K[X]. We denote by Znthe ring Z/nZ. The indexes of the variables x0, . . . , xn−1 are elements of Zn. Wedenote by d the cyclotomic derivation of K[X], that is, d is the K-derivation ofK[X] defined by

d(xj) = xj+1, for j ∈ Zn.

We denote also by d the unique extension of d to K(X). We denote by K[X]d andK(X)d the K-algebra of constants of d and the field of constants of d, respectively.Thus,

K[X]d = F ∈ K[X]; d(F ) = 0, K(X)d = f ∈ K(X); d(f) = 0.

Now we recall from [10] some basic notions and facts concerning the derivationd. As in the previous sections, we denote by ε a primitive n-th root of unity, andfirst we assume that ε ∈ K.

The letters % and τ we book for two K-automorphisms of the field K(X), definedby

%(xj) = xj+1, τ(xj) = εjxj for all j ∈ Zn.Observe that %d%−1 = d. We denote by u0, u1, . . . , un−1 the linear forms, belongingto K[X], defined by

uj =n−1∑i=0

(εj)ixi, for j ∈ Zn.



Then we have the equalities

xi =1

n

n−1∑j=0

(ε−i)juj ,

for all i ∈ Zn. Thus, K[X] = K[u0, . . . , un−1], K(X) = K(u0, . . . , un−1), and theforms u0, . . . , un−1 are algebraically independent over K. Moreover,

τ(uj) = uj+1, %(uj) = ε−juj , d(uj) = ε−juj ,

for all j ∈ Zn.

It follows from the last equality that d is a diagonal derivation of the polynomialring K[U ] = K[u0, . . . , un−1] which is equal to the ring K[X].

If α = (α0, . . . , αn−1) ∈ Zn, then we denote by uα the rational monomialuα00 · · ·u

αn−1

n−1 . Recall (see Section 2) that Hα(t) is the polynomial α0 + α1t1 +

· · ·+ αn−1tn−1 belonging to Z[t]. Since d(uj) = ε−juj for all j ∈ Zn, we have

d(uα) = Hα(ε−1)uα, for all α ∈ Zn.

Note that ε−1 is also a primitive n-th root of unity. Hence, by Proposition 2.1, wehave the equivalence Hα(ε−1) = 0 ⇐⇒ Hα(ε) = 0, and so, we see that if α ∈ Zn,then d(uα) = 0 ⇐⇒ α ∈ Gn, and if α ∈ Nn, then d(uα) = 0 ⇐⇒ α ∈ Mn.

Moreover, if F = b1uα(1)

+ · · ·+ bruα(r)

, where b1, . . . , br ∈ K and α(1), . . . , α(r) are

pairwise different elements of Nn, then d(F ) = 0 if and only if d(biu

α(i))

= 0 for

every i = 1, . . . , r. In [10] we proved the following proposition.

Proposition 6.1 ([10]). If the primitive n-th root ε belongs to K, then:

(1) the ring K[X]d is generated over K by all elements of the form uα withα ∈Mn;

(2) the ring K[X]d is generated over K by all elements of the form uβ, where βis a minimal element of the monoid Mn;

(3) the field K(X)d is generated over K by all elements of the form uγ withγ ∈ Gn;

(4) the field K(X)d is the field of quotients of the ring K[X]d.

Let m = n − ϕ(n), and let γ0, . . . , γm−1 be the elements of Gn introduced inSection 2. We know (see Theorem 2.3) that these elements form a basis of thegroup Gn. Consider now the rational monomials w0, . . . , wm−1 defined by

wj = uγj for j = 0, 1, . . . ,m− 1.

It follows from Proposition 6.1, that these monomials belong to K(X)d and theygenerate the field K(X)d. We proved in [10] that they are algebraically independentover K. Moreover, in [10] proved the following theorem.



Theorem 6.2. If the primitive n-th root ε belongs to K, then the field of constantsK(X)d is a field of rational functions over K and its transcendental degree over Kis equal to m = n− ϕ(n), where ϕ is the Euler totient function. More precisely,

K(X)d = K(w0, . . . , wm−1

),

where the elements w0, . . . , wm−1 are as above.

7. The polynomials Sp,m

In this section we use the notations from the previous section, and we againassume that K is a field of characteristic zero containing ε. Let us recall that if pis a prime divisor of n and 0 6 r 6 n

p − 1, then m(p, r), is the standard minimal

element of the monoidMn defined by m(p, r) =p−1∑i=0

er+inp . Observe that if a, b are

integers such that a ≡ b (mod np ), then

p−1∑i=0

ea+inp =p−1∑i=0

eb+inp . Thus, we may define

m(p, a) :=

p−1∑i=0

ea+inp , for a ∈ Z.

Note, that if a ∈ Z, then m(p, a) = m(p, r), where r is the remainder of division of a

by np . Moreover, ζ

np

(m(p, b)

)= m(p, b) for b ∈ Z, and more general, ζa

(m(p, b)

)=

m(p, a+ b) for all a, b ∈ Z (see Proposition 3.4).

For every integer a, we define

Sp,a := um(p,a) =

p−1∏i=0

ua+inp .

Observe that Sp,a = Sp,r, where r is the remainder of division of a by np . Each Sp,a

is a monomial belonging to K[U ] = K[u0, . . . , un−1]. Since m(p, a) ∈ Mn ⊂ Gn,each Sp,a belongs to the constant field K(X)d.

Recall (see Section 6) that % is the K-automorphism of the field K(X), definedby

%(xj) = xj+1, for j ∈ Zn.We have %(uj) = ε−juj for j ∈ Zn. In particular, %(u0) = u0. The proof of thefollowing proposition is an easy exercise.

Proposition 7.1. If a ∈ Z, then % (Sp,a) = ε−bSp,a, where b = pa + (p−1)n2 . In

particular, if p is odd then % (Sp,a) = ε−apSp,a. If p = 2, then n is even and

% (S2,a) = ε−(2a+n2 )S2,a.

Recall the following well known lemma, which appears in many books of linearalgebra.



Lemma 7.2. For any integer n > 2,

u0u1 . . . un−1 =

∣∣∣∣∣∣∣∣∣x0 x1 · · · xn−1xn−1 x0 · · · xn−2

......

...x1 x2 · · · x0

∣∣∣∣∣∣∣∣∣ .In particular, the product u0u1 . . . un−1 is a polynomial belonging to Z[X].

Using this lemma we obtain the following proposition.

Proposition 7.3. The polynomial Sp,0 belongs to Z[X].

Proof. Put b = np , η = εb, and vi = uib, yi =

b−1∑j=0

xi+jp for all i = 0, 1, . . . , p−1, Then

η is a primitive p-th root of unity, and vi =p−1∑k=0

(ηi)kyk, for all i = 0, 1, . . . , p− 1.

Now we use Lemma 7.2, and we have

Spj ,0 = v0v1 · · · vp−1 =

∣∣∣∣∣∣∣∣∣y0 y1 · · · yp−1yp−1 y0 · · · yp−2

......

...y1 y2 · · · y0

∣∣∣∣∣∣∣∣∣ .Thus, Spj ,0 ∈ Z[X].

Let n = pα11 · · · pαs

s , where p1, . . . , ps are distinct primes and α1, . . . , αs arepositive integers. Let nj = n

pjfor j = 1, . . . , s. Assume that P = [p1, . . . , pn] is

a fixed type, and I1, . . . , Is are the n-sets of type P .

For every j ∈ 1, . . . , s we denote by Vj the K-subspace of K[U ] generated byall the monomials Spj ,r with r ∈ Ij . Let us remember

Vj =⟨Spj ,r; r ∈ Ij

⟩, for j = 1, . . . , s.

We will say that V1, . . . ,Vs are n-spaces of type P . As a consequence of Propositions4.3 and 4.5 we obtain the following proposition.

Proposition 7.4. If V1, . . . ,Vs are n-spaces of type P = [p1, . . . , ps], then

dimK V1 = n1, and dimK Vj = nj

(1− 1

p1

)(1− 1

p2

)· · ·(

1− 1pj−1

), for all j =

2, 3, . . . , s. Moreover,

dimK (V1 ⊕ · · · ⊕ Vs) = n− ϕ(n).

Let A be the n-standard set of type P . Let us recall (see Section 5) thatA = A1 ∪ · · · ∪ As, where Aj = p(pj , r); r ∈ Ij for j = 1, . . . , s. Hence, for eachj we have the equality Vj = 〈ua; a ∈ Aj〉. Let S the set of all the monomials ua

with a ∈ A, that is,

S =Spj ,r; j ∈ 1, . . . , s, r ∈ Ij

.



Proposition 7.5. If the number n is special of type P , then the above set S isalgebraically independent over K, and K(X)d = K (S).

Proof. Assume that n is special of type P . Let γ0, . . . , γm−1 be the elementsof Gn defined in Section 2, and let wi = uγi for i = 0, . . . ,m − 1. Recall thatm = n − ϕ(n). Put Γ = γ0, . . . , γm−1, and W = w0, . . . , wm−1. We know(see Theorem 2.3) that Γ is a basis of Gn. Since n is special, the set A is alsoa basis of Gn. This implies that K (S) = K(W ). But, by Theorem 6.2, the set W isalgebraically independent over K and K(W ) = K(X)d. Moreover, |S| = |W | = mHence, the set S is also algebraically independent over K, and we have the equalityK(X)d = K (S).

In the above proposition we assumed that n is special of type P . This as-sumption is very important. Consider for example n = 12 and P = [2, 3]. Weknow (see Example 5.1) that 12 is not special of type P . In this case the set Sis not algebraically independent over K. In fact, we have the polynomial equalityS2,1S2,3S2,5 = S3,1S3,3.

8. The polynomials Tp,m

Let n = pα11 · · · pαs

s , where p1, . . . , ps are distinct prime numbers andα1, . . . , αs are positive integers. Let nj = n

pjfor j = 1, . . . , s. Assume that

P = [p1, . . . , pn] is a fixed type, and I1, . . . , Is are the n-sets of type P .

Now assume that j is a fixed element from the set 1, . . . , s, and a is an integer.Put

Tpj ,a =∑r∈Ij

(ε−apj

)rSpj ,r.

Observe that Tpj ,a = Tpj ,m, where m is the remainder of division of a by nj . Letus recall that ε ∈ K. Thus, every Tpj ,a is a polynomial from K[U ] belonging tothe subspace Vj .

Proposition 8.1. For every j = 1, . . . , s, all the polynomials Tpj ,m with 0 6 m <nj, generate the K-space Vj.

Proof. Let q ∈ Ij and consider the sum H =nj−1∑m=0

(εqpj )mTpj ,m. Put η = εpj . Then

η is a primitive nj-th root of unity, and we have

H =nj−1∑m=0

(εqpj )m

( ∑r∈Ij

εrpjmSpj ,r

)=∑r∈Ij

(nj−1∑m=0

ε(q−r)pjm

)Spj ,r

=∑r∈Ij

(nj−1∑m=0

η(q−r)m

)Spj ,r = njSpj ,q.



In the last equality we used Lemma 4.6. Thus, if q ∈ Ij , then Spj ,q =

1nj

nj−1∑m=0

(εqpj )mTpj ,m. But ε ∈ K, so now it is clear that all Tpj ,m with 0 6 m < nj ,

generate the K-space Vj .

Now we will prove that every polynomial Tpj ,a belongs to the ring Z[X]. Forthis aim first recall (see Section 6) that τ is a K-automorphism of K(X) definedby

τ(xj) = εjxj for all j ∈ Zn.

Since τ(ui) = ui+1 for all i ∈ Zn, we have

Spj ,r = τ r(Spj ,0

)for j ∈ 1, . . . , s and r ∈ Z (in particular, for r ∈ Ij). We say (us in [10]) thata rational function f ∈ K(X) is τ -homogeneous, if f is homogeneous in the ordinarysense and τ(f) = εcf for some c ∈ Zn. In this case we say that c is the τ -degree off and we write degτ (f) = c. Note that degτ (f) is an element of Zn. Every rationalmonomial xα = xα0

0 · · ·xαn−1

n−1 , where α = (α0, . . . , αn−1) ∈ Zn, is τ -homogeneous

and its τ -degree is equal ton−1∑i=0

iαi (mod n).

Let j be a fixed number from 1, . . . , s and consider the polynomial Spj ,0.We know by Proposition 7.3 that this polynomial belongs to Z[X]. Hence, wehave the unique determined polynomials B0, . . . , Bn−1 ∈ Z[X] such that Spj ,0 =B0 + · · ·+Bn−1, and each Bi is τ -homogeneous of τ -degree i.

Put Ci = τnj (Bi), for all i = 0, . . . , n − 1. Since τ(Bi) = εiBi, we have Ci =εinjBi, and this implies that τ(Ci) = εiCi. In fact,

τ(Ci) = τ (τnj (Bi)) = τ(εinjBi

)= εinjτ(Bi) = εinj · εiBi = εi · εinjBi = εiCi.

Thus, every polynomial Ci is τ -homogeneous of τ -degree i. Observe that

τnj(Spj ,0

)= Spj ,0.

But τnj(Spj ,0

)=

n−1∑i=0

Ci, so Ci = τnj (Bi) = Bi and so, εinjBi = Bi, for all

i = 0, . . . , n− 1. Thus, if Bi 6= 0, then n | inj . But n = pjnj so, if Bi 6= 0, then iis divisible by pj . Therefore,

Spj ,0 =

nj−1∑k=0

Bkpj ,



where each Bkpj is τ -homogeneous polynomial from Z[X] of τ -degree kpj . Hence,for every m ∈ 0, . . . , n− 1, we have

Tpj ,m =∑r∈Ij

ε−rpjmSpj ,r =∑r∈Ij

ε−rpjmτ r(Spj ,0

)=

∑r∈Ij

ε−rpjmτ r

(nj−1∑k=0

Bkpj

)=∑r∈Ij

ε−rpjm

(nj−1∑k=0

τ r(Bkpj )

)

=∑r∈Ij

ε−rpjm

(nj−1∑k=0

εkpjrBkpj

)=nj−1∑k=0

Bkpj

( ∑r∈Ij

εrpj(k−m)

).

Observe that, by Proposition 4.7, every sum∑r∈Ij

εrpj(k−m) is an integer. Moreover,

every polynomial Bkpj belongs to Z[X]. Hence, Tpj ,m ∈ Z[X].

Recall that Tpj ,a = Tpj ,m, where m is the remainder of division of a by nj . Thus,we proved the following proposition.

Proposition 8.2. For any j ∈ 1, . . . , s and a ∈ Z, the polynomial Tpj ,m belongsto the polynomial ring Z[X].

Now we will prove some additional properties of the polynomials Tpj ,a.

Proposition 8.3. Assume that s > 2, and let i, j ∈ 1, . . . , s, i < j. Then

pi−1∑k=0

Tpj , k npipj

= 0.

Proof. Put p = pi, q = pj , and N = npq . Then we have

pi−1∑k=0

Tpj , k npipj

=p−1∑k=0

Tq, kN =p−1∑k=0

∑r∈Ij

(ε−kNq

)rSq,r =

∑r∈Ij

(p−1∑k=0

(ε−

np r)k)

Sq,r.

Let η = ε−np . Then η is a primitive p-th root of unity. If r ∈ Ij , then p - r and, by

Lemma 4.6, we havep−1∑k=0

(ε−

np r)k

=

p−1∑k=0

ηrk = 0.

Thus,pi−1∑k=0

Tpj , k npipj

=∑r∈Ij

(p−1∑k=0

(ε−

np r)k)

Sq,r. =∑r∈Ij

0 · Sq,r = 0.

Proposition 8.4. For any integer a, we have

%(Tpj , a

)=

Tpj , a+1, when pj 6= 2,

−Tpj , a+1, when pj = 2.



Proof. First assume that pj is odd. In this case (see Proposition 7.1), %(Spj r

)=

ε−pjrSpj r for any r ∈ Z. Hence,

%(Tpj , a

)=

∑r∈Ij

(ε−apj )r%(Spj r

)=∑r∈Ij

(ε−apj )rε−pjrSpj r

=∑r∈Ij

(ε−(a+1)pj

)rSpj r = Tpj , a+1.

Now let pj = 2. Then, by Proposition 7.1, %(Spj r

)= ε−(pjr+n

2 )Spj , r for any

r ∈ Z. Moreover, ε−n2 = −1. Thus, we have

%(Tpj , a

)=

∑r∈Ij

(ε−apj )r%(Spj r

)=∑r∈Ij

(ε−apj )rε−(pjr+n

2 )Spj , r

=∑r∈Ij

ε−n2

(ε−(a+1)pj

)rSpj r = −

∑r∈Ij

(ε−(a+1)pj

)rSpj r = −Tpj , a+1.


Proposition 8.5. Assume that s > 2. Let i, j ∈ 1, . . . , s, i < j, and let a ∈ Z.Then

Tpj , a = −pi−1∑k=1

Tpj , a+k npipj

.

Proof. It follows from Proposition 8.4 that Tpj , a = (−1)pj−1%a(Tpj , 0

). Hence,

using Proposition 8.3, we obtain

Tpj , a = (−1)pj−1%a(Tpj , 0

)= (−1)pj−1%a

(−pi−1∑k=1

Tpj , k npipj

)= (−1)pj

pi−1∑k=1

%a(Tpj , k n

pipj

)= (−1)pj

pi−1∑k=1

(−1)pj−1Tpj , a+k npipj

= −pi−1∑k=1

Tpj , a+k npipj

.


For any j ∈ 1, . . . , s, let us denote by Wj the Z-module generated by all thepolynomials Tpj , r with r ∈ Ij . It is clear that every polynomial Tp1, a, for arbitraryinteger a, belongs to W1.

Theorem 8.6. If the number n is special, then for all j ∈ 1, . . . , s and a ∈ Z,the polynomial Tpj , a belongs to Wj.

Proof. Let n = p1 · · · ps−1 · pαs , where s > 1, α > 1, and p1, . . . , ps are distinctprimes. Let nj = n

pjfor j = 1, . . . , s. Assume that P = [p1, . . . , pn] is a fixed type,

and I1, . . . , Is are the n-sets of type P .

Let j be a fixed element from 1, . . . , s. If s = 1 or j = 1, then we are done.Assume that s > 2, j > 2, and a is an integer. Since Tpj , a = Tpj ,m, where m is



the remainder of division of a by nj , we may assume that 0 6 a < nj . We use thefollowing notations:

M := p1, p2, . . . , pj−1 , q := pj , Bc := Tpj , c for c ∈ Z.We will show that Ba ∈ Wj . If gcd(a, p1 · · · pj−1) = 1, then a ∈ Ij and so,Ba ∈ Wj . Now let gcd(a, p1 · · · pj−1) > 2. In this case, a is divisible by someprimes belonging to M .

Step 1. Assume that a is divisible by exactly one prime number pi belongingto M . Then i < j and, by Proposition 8.5, we have the equality

Ba = −pi−1∑k=1

Ba+k npiq.

Let k ∈ 1, . . . , pi − 1, and consider c := a+ k npiq

. Since n is special, the number

k npiq

is not divisible by pi. But pi | a, so pi - c. If p ∈M and p 6= pi, then p - a and

p | k npiq

, so p - c. Hence, the numbers c and p1 · · · pj−1 are relatively prime. This

implies that the element c (mod nj) belongs to Ij , and so, Bc ∈ Wj . Therefore, bythe above equality, Ba ∈ Wj .

We see that if s = 2 or j = 2, then we are done. Now suppose that s > 3 andj > 3.

Step 2. Let 1 6 t 6 j − 2, and assume that we already proved that Bc ∈ Wj

for every integer c which is divisible by exactly t primes belonging to M . Assumethat a is divisible by exactly t+ 1 distinct primes m1, . . . ,mt+1 from M . We have:mi | a for i = 1, . . . , t+ 1, and m - a for m ∈M r m1, . . . ,mt+1. Put p = mt+1.It follows from Proposition 8.5, that have the following equality:

Ba = −p−1∑k=1

Ba+k npq.

Let k ∈ 1, . . . , p − 1, and consider c := a + k npq . Since n is special, the number

k npq is not divisible by p. But p | a, so p - c, and consequently, mt+1 - c. It is clear

that mi | c for all i = 1, . . . , t, and m - c for all m ∈M rm1, . . . ,mt. This meansthat c is divisible by exactly t primes from M . Thus, by our assumption, Bc ∈ Wj .Therefore, by the above equality, Ba ∈ Wj .

Now we use a simple induction and, by Steps 1 and 2, we obtain the proof ofour theorem.

9. The main theorem

Assume that n > 3 is a special number of a type P . Let I1, . . . , Is be then-sets of type P , let A be the n-standard set of type P , and let

S =Spj ,r; j ∈ 1, . . . , s, r ∈ Ij

, T =

Tpj ,r; j ∈ 1, . . . , s, r ∈ Ij

.



Since n is special, we have the following sequence of important properties.

(1) A is a basis of the group Gn (Theorems 5.8, 3.5 and 5.10).

(2) S is algebraically independent over K, and K(X)d = K (S) (Proposition7.5).

(3) K (S) = K (T ) (Proposition 8.1 and Theorem 8.6).

We know also (see Proposition 8.2) that each element of T is a polynomialbelonging to Z[X]. Moreover, |T | = |S| = |A| = n − ϕ(n). In particular, theset T is algebraically independent over K. Put an order on the set T . LetT = F0, F1, . . . , Fm−1 where m = n − ϕ(n). Thus, if the number n is special,then K(X)d = K(F0, . . . , Fm−1), where F0, . . . , Fm−1 are polynomials belongingto Z[X], and these polynomials are algebraically independent over Q.

Let us recall, that K is a field of characteristic zero containing ε (where ε isa primitive n-th root of unity). But the polynomials F0, . . . , Fm−1 have integercoefficients, and they are constants of d. They are not dependent from the fieldK. Since the polynomials d(x0), . . . , d(xn−1) belong to Z[X], we see that we mayassume that K is a field of characteristic zero, without the assumption concerningε. Thus, we proved the following theorem.

Theorem 9.1. Let K be an arbitrary field of characteristic zero, n > 3 an integer,and K[X] = K [x0, . . . , xn−1] the polynomial ring in n variables over K. Letd : K[X] → K[X] be the cyclotomic derivation, that is, d is a K-derivation ofK[X] such that

d(xi) = xi+1 for i ∈ Zn.

Assume that n = p1p2 · · · ps−1pαs , where s > 1, α > 1 and p1, . . . , ps are distinctprimes. Let m = n− ϕ(n), where ϕ is the Euler totient function. Then

K(X)d = K (F0, . . . , Fm−1) ,

where F0, . . . , Fm−1 are algebraically independent over Q polynomials belonging toZ[X].

More exactly, F0, F1, . . . , Fm−1 =Tpj , r; j ∈ 1, . . . , s, r ∈ Ij

, where

I1, . . . , Is are the n-sets of type [p1, . . . , ps].

We end this article with several examples illustrating the above theorem.

Example 9.2. If n = 4, then K(X)d = K (F0, F1), where F0 = x20 − 2x1x3 + x22,and F1 = %(F0).

Example 9.3. If n = 8, then K(X)d = K (F0, F1, F2, F3), where F1 = %(F0),F2 = %2(F0), F3 = %3(F0) and F0 = x20 + x24 − 2x3x5 − 2x7x1 + 2x2x6.



Example 9.4. If n = 9, then K(X)d = K (F0, F1, F2), where F1 = %(F0),F2 = %2(F0),

F0 = 3x1x24 + 3x28x2 + 3x8x

25 − 3x0x4x5 − 3x1x0x8 − 3x2x4x3 − 3x2x7x0

−3x8x6x4 + 3x22x5 + 3x27x4 + 3x21x7 + x36 + x30 − 3x1x3x5 + 6x0x6x3−3x8x7x3 − 3x2x1x6 − 3x5x7x6 + x33.

Example 9.5. If n = 6 and P = [2, 3], then K(X)d = K (F0, F1, F2, F3), where

F0 = x20 − 2x1x5 + 2x2x4 − x23,F3 = (x21 + x4x3 − 2x1x4 + x0x1 + x25 − x5x3 + x2x3 − 2x2x5 + x0x5

−2x0x3 − x0x2 − x4x0 + x24 − x1x3 + x22 + x4x5 + x1x2 + x20−x1x5 − x4x2 + x23)(x0 − x1 + x2 − x3 + x4 − x5),

and F1 = %(F0), F2 = %2(F0).

Example 9.6. If n = 6 and P = [3, 2], then K(X)d = K (F0, F1, F2, F3), where

F0 = x30 + x32 + x34 + 3x0x23 + 3x2x

25 + 3x4x

21 − 3x0x2x4 − 3x5x0x1

−3x1x2x3 − 3x3x4x5,

F2 = 2x21 + x22 − x23 − 2x24 − x25 + x20−2x1x3 + 2x2x4 + 4x3x5 + 2x4x0,−2x5x1 − 4x2x0.

and F1 = %(F0), F3 = %(F2).

Example 9.7. If n = 12, then K(X)d = K (F0, . . . , F7), where

F0 = −3x6x2x4 − 3x6x8x10 − 3x4x0x8 + x30 + 3x26x0 − 3x1x8x3 + 3x23x6+3x29x6 + x38 − 3x1x11x0 + 6x5x11x8 − 3x1x5x6 + 3x27x10 + 3x210x4+3x211x2 + 3x21x10 + 3x25x2 + 3x22x8 + 6x3x0x9 + 6x1x7x4 − 3x7x11x6−3x7x5x0 − 3x10x11x3 − 3x10x5x9 − 3x4x11x9 − 3x4x5x3 − 3x1x2x9−3x7x2x3 − 3x7x8x9 + x34 − 3x10x2x0,

F4 = 4x6x8 + x23 − 2x10x8 + 2x7x3 + 2x7x11 − 2x10x0 − 2x4x2 − 2x4x6+2x1x9 + 2x1x5 + 4x0x2 − 2x0x6 − 4x3x11 − 2x21 + x211 + x25 + 4x4x10−2x2x8 − 2x27 + x29 − 4x9x5,

and F1 = %(F0), F2 = %2(F0), F3 = %3(F0), F5 = %(F4), F6 = %3(F4), F7 = %4(F4).

References

[1] N.G. de Bruijn, On the factorization of cyclic groups, Indag. Math. 15 (1953), 370–377.

[2] A. van den Essen, Polynomial Automorphisms and the Jacobian Conjecture, Progress inMathematics vol. 190, 2000.

[3] G. Freudenburg, Algebraic Theory of Locally Nilpotent Derivations, Encyclopedia of Math-

ematical Sciences 136, Springer, 2006.[4] T.Y. Lam and K.H. Leung, On the cyclotomic polynomial Φpq(x), American Mathematical

Monthly, 103(7) (1996), 562–564.

[5] T.Y. Lam and K.H. Leung, On vanishing sums of roots of unity, J. Algebra, 224 (2000),91–109,

[6] S. Lang, Algebra, Second Edition, Addison-Wesley Publishing Company, 1984.[7] S. Lang, Undergraduate Algebra, Second Edition, Springer, 1990.



[8] J. Moulin Ollagnier and A. Nowicki, Derivations of polynomial algebras without Darboux

polynomials, J. Pure Appl. Algebra, 212 (2008), 1626–1631.

[9] J. Moulin Ollagnier and A. Nowicki, Monomial derivations, Communications in Algebra, 39(2011), 3138–3150.

[10] J. Moulin Ollagnier and A. Nowicki, Constants of cyclotomic derivations, J. Algebra 394

(2013), 92–119.[11] A. Nowicki, Polynomial derivations and their rings of constants, N. Copernicus University

Press, Torun, 1994.[12] A. Nowicki and M. Nagata, Rings of constants for k–derivations in k[x1, . . . , xn], J. Math.

Kyoto Univ., 28 (1988), 111–118.

[13] L. Redei, Ein Beitrag zum Problem der Faktorisation von endlichen Abelschen Gruppen,Acta Math. Hungar, 1 (1950), 197–207.

[14] A. Satyanarayan Reddy, The lowest 0, 1-polynomial divisible by cyclotomic polynomial,

arXiv: 1106.127v2 [math.NT] 15Nov 2011.[15] I.J. Schoenberg, A note on the cyclotomic polynomial, Mathematika, 11 (1964), 131–136.

[16] J.P. Steinberger, The lowest-degree polynomial with nonnegative coefficients divisible by the

n-th cyclotomic polynomial, The electronic journal of combinatorics 19(4) (2012), #P1[17] J.P. Steinberger, Minimal vanishing sums of roots of unity with large coefficients, Proc.

Lond. Math. Soc. (3) 97 (2008), 689–717.

(Jean Moulin Ollagnier) Laboratoire LIX, Ecole Polytechnique, F 91128 PalaiseauCedex, France


(Andrzej Nowicki) Nicolaus Copernicus University, Faculty of Mathematics andComputer Science, ul. Chopina 12/18, 87-100 Torun, Poland


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“22˙Nowicki2” — 2017/12/1 — 20:47 — page 123 — #2




DIVERGENCE-FREE POLYNOMIAL DERIVATIONS

ANDRZEJ NOWICKI

Abstract. In this paper we present some new and old properties of diver-

gences and divergence-free derivations.

Throughout the paper all rings are commutative with unity. Let k be a ring andlet d be a k-derivation of the polynomial ring k[X] = k[x1, . . . , xn]. We denote byd? the divergence of d, that is,

d? =∂d(x1)

∂x1+ · · ·+ ∂d(xn)

∂xn.

The derivation d is said to be divergence-free if d? = 0.

1. Preliminaries

Let k be a ring, and let R be a k-algebra. A k-linear mapping d : R→ R is saidto be a k-derivation of R if

d(ab) = ad(b) + d(a)b,

for all a, b ∈ R. We denote by Derk(R) the set of all k-derivations of R. If d, d1, d2 ∈Derk(R) and x ∈ R, then the mappings xd, d1 + d2 and [d1, d2] = d1d2 − d2d1 arealso k-derivations of R. Thus, the set Derk(R) is an R-module which is also a Liealgebra.

We denote by Rd the kernel of d, that is,

Rd =a ∈ R; d(a) = 0

This set is a subring of R, called the ring of constants of R (with respect to d). IfR is a field, then Rd is a subfield of R.

2010 Mathematics Subject Classification. Primary 12H05; Secondary 13N15.Key words and phrases. Derivation, divergence, Darboux polynomial, jacobian derivation,

Jacobian Conjecture, homogeneous derivation.

123

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124 A. NOWICKI

Now let k[X] = k[x1, . . . , xn] be a polynomial ring in n variables over a ring k.For each i ∈ 1, . . . , n the partial derivative ∂

∂xiis a k-derivation of k[X]. It is

a unique k-derivation d of k[X] such that d(xi) = 1 and d(xj) = 0 for all j 6= i. Iff1, . . . , fn are polynomials belonging to k[X], then the mapping

f1∂

∂x1+ · · ·+ fn

∂

∂xn

is a k-derivation of k[X]. It is a k-derivation d of k[X] such that d(xj) = fj forall j = 1, . . . , n. It is not difficult to show that every k-derivation of k[X] is ofthe above form. As a consequence of this fact we know that Derk(k[X]) is a freek[X]-module on the basis ∂

∂x1, . . . , ∂

∂xn. If d ∈ Derk(k[X]) and f ∈ k[X], then

d(f) =∂f

∂x1d(x1) + · · ·+ ∂f

∂xnd(xn).

Now assume that k is a domain containing Q and d is a k-derivation of k[X].We say that F ∈ k[X] is a Darboux polynomial of d if F 6= 0 and d(F ) = ΛF,for some Λ ∈ k[X]. In this case such Λ is unique and it is said to be the cofactorof F . Every nonzero element belonging to the ring of constants k[X]d is of coursea Darboux polynomial. If F1, F2 ∈ k[X] r 0 are Darboux polynomials of d thenthe product F1F2 is also a Darboux polynomial of d. The cofactor of F1F2 is inthis case the sum of the cofactors of F1 and F2. If F ∈ k[X] r k is a Darbouxpolynomial of d, then all factors of F are also Darboux polynomials of d. Thus,looking for Darboux polynomials of d reduces to looking for irreducible ones.

For a discussion of Darboux polynomial in a more general setting, the reader isreferred to [15], [19], [13], [14].

A k-derivation d of k[X] is called homogeneous of degree s if all the polynomialsd(x1), . . . , d(xn) are homogeneous of degree s. In particular, each partial derivative∂∂xi

is homogeneous of degree 0. The zero derivation is homogeneous of everydegree. The sum of homogeneous derivations of the same degree s is homogeneousof degree s. Note some basic properties of homogeneous derivations (see [19] forproofs and details).

Proposition 1.1. Let d be a homogeneous k-derivation of k[X] and let F ∈ k[X].If F ∈ k[X]d, then each homogeneous component of F belongs also to k[X]d. Inparticular, the ring k[X]d, is generated over k by homogeneous polynomials.

Proposition 1.2. Let d be a homogeneous k-derivation of k[X], where k isa domain containing Q, and let 0 6= F ∈ k[X] be a Darboux polynomial of d withthe cofactor Λ ∈ k[X]. Then Λ is homogeneous, and all homogeneous componentsof F are also Darboux polynomials with the common cofactor equal to Λ.

Note that Darboux polynomials of a homogeneous derivation are not necessarilyhomogeneous. Indeed, let n = 2, d(x1) = x1, d(x2) = 2x2, and let F = x21 + x2.Then d is homogeneous, F is a Darboux polynomial of d (because d(F ) = 2F ), andF is not homogeneous.

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DIVERGENCE-FREE POLYNOMIAL DERIVATIONS 125

2. Basic properties of divergences

Let k be a ring and let d be a k-derivation of the polynomial ring k[X] =k[x1, . . . , xn]. Let us recall that we denote by d? the divergence of d, that is,

d? =∂d(x1)

∂x1+ · · ·+ ∂d(xn)

∂xn.

We say that the derivation d is divergence-free if d? = 0. For example, everypartial derivative ∂

∂xiis a divergence-free k-derivation of k[X]. It is clear that

(d + δ)? = d? + δ? for all d, δ ∈ Derk(k[X]). Thus, the sum of divergence-freederivations is also a divergence-free derivation.

Proposition 2.1. If d ∈ Derk(k[X]) and r ∈ k[X], then:

(rd)? = rd? + d(r).

Proof. (rd)? =n∑p=1

∂rd(xp)∂xp

=n∑p=1

(r∂d(xp)∂xp

+ ∂r∂xp

d(xp))

= rn∑p=1

∂d(xp)∂xp

+n∑p=1

∂r∂xp

d(xp) = rd? + d(r).

Thus, if d is a divergence-free k-derivation of k[X] and r ∈ k[X]d, then thederivation rd is divergence-free.

Proposition 2.2. Let d, δ ∈ Derk(k[X]) and let [d, δ] = dδ − δd. Then

[d, δ]? = d(δ?)− δ(d?).

Proof. Put fi = d(xi), gi = δ(xi) for i = 1, . . . , n, and observe thatn∑p=1

n∑i=1

∂gp∂xi

∂fi∂xp

=n∑p=1

n∑i=1

∂fp∂xi

∂gi∂xp

.

Thus, we have

[d, δ]∗ =n∑p=1

∂∂xp

((dδ − δd)(xp)

)=

n∑p=1

∂∂xp

(d(gp)− δ(fp)

)=

n∑p=1

∂∂xp

(n∑i=1

∂gp∂xi

fi −n∑i=1

∂fp∂xi

gi

)=

n∑p=1

n∑i=1

(∂∂xp

∂gp∂xi· fi +

∂gp∂xi

∂fi∂xp− ∂

∂xp

∂fp∂xi· gi − ∂fp

∂xi

∂gi∂xp

)=

n∑p=1

n∑i=1

(∂∂xp

∂gp∂xi· fi − ∂

∂xp

∂fp∂xi· gi)

=n∑p=1

n∑i=1

(∂∂xi

∂gp∂xp· fi − ∂

∂xi

∂fp∂xp· gi)

=n∑p=1

(d(∂gp∂xp

)− δ

(∂fp∂xp

))= d

(n∑p=1

∂gp∂xp

)− δ

(n∑p=1

∂fp∂xp

)= d (δ?)− δ (d?) .

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126 A. NOWICKI


The above propositions imply that the set of all divergence-free derivations ofk[X] is closed under the sum and the Lie product.

Let d be a k-derivation of k[X]. Given a polynomial f ∈ k[X], we denoteby Vf , the k-submodule of k[X] generated by the set

f, d(f), d2(f), d3(f), . . .

.

The derivation d is called locally finite, if every module Vf , for all f ∈ k[X], isa finitely generated over k. The derivation d is called locally nilpotent, if for everyf ∈ k[X] there exists a positive integer m such that dm(f) = 0. Every locallynilpotent derivation is locally finite. There exist, of course, locally finite derivationswhich are not locally nilpotent. Locally finite and locally nilpotent derivations wasintensively studied from a long time; see for example [7], [6], [12], [19], where manyreferences on this subject can be found.

The following result is due to H. Bass, G. Meisters [2] and B. Coomes,V. Zurkowski [4]. Another its proof is given in [19] (Theorem 9.7.3).

Theorem 2.3. Let k be a reduced ring containing Q. If d is a locally finite k-derivation of k[X] = k[x1, . . . , xn], then d?, the divergence of d, is an elementof k.

Recall that a ring k is called reduced if k has no nonzero nilpotent elements. Ifk is non-reduced then the above property does not hold, in general.

Example 2.4. Let k = Q[y]/(y2) and let d be the k derivation of k[x] (a polynomialring in a one variable) defined by d(x) = ax2, where a = y + (y2). Since d2(x) =2a2x3 = 0, d is locally finite. But d? = 2ax 6∈ k.

Note the following important property of locally nilpotent derivations.

Theorem 2.5. ([19], [6]). If k is a reduced ring containing Q, then every locallynilpotent k-derivation of k[X] is divergence-free.

The derivation d from Example 2.4 is locally nilpotent. This means that if k isnon-reduced then there exist locally nilpotent k-derivations of k[X] with a nonzerodivergence.

In the paper of Berson, van den Essen, and Maubach [3] is quoted the followingresult, which is related to their investigation of the Jacobian Conjecture.

Theorem 2.6. ([3]). Let k be any commutative Q-algebra, and let d be a k-derivation of k[x, y]. If d is surjective and divergence-free, then d is locally nilpotent.

This result was shown earlier by Stein [21] in the case k is a field.

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3. Divergences and jacobians

If h1, . . . , hn are polynomials belonging to k[X] = k[x1, . . . , xn], then wedenote by [h1, . . . , hn] the jacobian of h1, . . . , hn, that is,

[h1, . . . , hn] =

∣∣∣∣∣∣∣∣∣∣∣∣

∂h1

∂x1

∂h2

∂x1· · · ∂hn

∂x1

∂h1

∂x2

∂h2

∂x2· · · ∂hn

∂x2

......

...

∂h1

∂xn

∂h2

∂xn· · · ∂hn

∂xn

∣∣∣∣∣∣∣∣∣∣∣∣.

Proposition 3.1. Let d be a k-derivation of k[X] and let h1, . . . , hn ∈ k[X]. Then

d(

[h1, . . . , hn])

= − [h1, . . . , hn] d? +n∑p=1

[h1, . . . , d(hp), . . . , hn] .

Proof. Put fi = d(xi), fij = ∂fi∂xj

, hij = ∂hi

∂xj, for all i, j ∈ 1, . . . , n, and let Sn

denote the group of all permutations of 1, . . . , n. Observe that

(a) d(hσ(p)p) =∂

∂xpd(hσ(p))−

n∑q=1

hσ(p)qfqp,

for all σ ∈ Sn and p ∈ 1, . . . , n, and

(b)

∑σ∈Sn

(−1)|σ|hσ(1)1 · · ·hσ(p−1)(p−1)hσ(p)qhσ(p+1)(p+1) · · ·hσ(n)n

= [h1, . . . , hn]δpq,

for all p, q ∈ 1, . . . , n, where |σ| is the sign of σ, and δpq is the Kronecker delta.The above determines that

d([h1, . . . , hn]) =n∑p=1

∑σ∈Sn

(−1)|σ|hσ(1)1 · · · d(hσ(p)p) · · ·hσ(n)n

(a)=

n∑p=1

∑σ∈Sn

(−1)|σ|hσ(1)1 · · · (∂

∂xpd(hσ(p))−

n∑q=1

hσ(p)qfpq) · · ·hσ(n)n

(b)=

n∑p=1

[h1, . . . , d(hp), . . . , hn]−n∑p=1

n∑q=1

fpq[h1, . . . , hn]δpq

=n∑p=1

[h1, . . . , d(hp), . . . , hn]−n∑p=1

fpp[h1, . . . , hn]

=n∑p=1

[h1, . . . , d(hp), . . . , hn]− [h1, . . . , hn]d?.


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128 A. NOWICKI

As a consequence of the above proposition we obtain the following propositionfor divergence-free derivations.

Proposition 3.2. If d is a divergence-free k-derivation of k[X] and h1, . . . , hn arepolynomials belonging to k[X], then

d(

[h1, . . . , hn])

=n∑p=1

[h1, . . . , d(hp), . . . , hn] .

Consider the case n = 2. Put x = x1 and y = x2. If f ∈ k[x, y], then we denote:

fx = ∂f∂x , fy = ∂f

∂y . Observe that for every f ∈ k[x, y] we have the equality

[fx, x] + [fy, y] = 0.

In fact, [fx, x] + [fy, y] =

∣∣∣∣ fxx 1fxy 0

∣∣∣∣+

∣∣∣∣ fyx 0fyy 1

∣∣∣∣ = −fxy + fyx = 0.

In the case n = 3 we have a similar equality. If f, g ∈ k[x, y, z], then

[fx, g, x] + [fy, g, y] + [fz, g, z] = 0.

Let us check: [fx, g, x] + [fy, g, y] + [fz, g, z]

=

∣∣∣∣∣∣fxx gx 1fxy gy 0fxz gz 0

∣∣∣∣∣∣+

∣∣∣∣∣∣fyx gx 0fyy gy 1fyz gz 0

∣∣∣∣∣∣+

∣∣∣∣∣∣fzx gx 0fzy gy 0fzz gz 1

∣∣∣∣∣∣=

∣∣∣∣ fxy gyfxz gz

∣∣∣∣− ∣∣∣∣ fyx gxfyz gz

∣∣∣∣+

∣∣∣∣ fzx gxfzy gy

∣∣∣∣= (fxygz − fxzgy)− (fyxgz − fyzgx) + (fzxgy − fzygx)

= fxy(gz − gz) + fxz(gy − gy) + fyz(gx − gx) = 0.

The same we have for every n > 2.

Proposition 3.3. If f, g1, g2, . . . , gn−2 are polynomials belonging to k[x1, . . . , xn],then

n∑p=1

[∂f

∂xp, g1, g2, . . . , gn−2, xp

]= 0.

Proof. Put fp = ∂f∂xp

, fp,j =∂fp∂xj

= ∂2f∂xpxj

, and

Ap = [fp, g1, g2, . . . , gn−2, xp] , Gj =

(∂g1∂xj

,∂g2∂xj

, . . . ,∂gn−2∂xj

),

for all p, j ∈ 1, . . . , n. Note, that Ap is the jacobian of fp, g1, . . . , gn−2, xp, and Gjis a sequence of n− 2 polynomials from k[X]. Observe that, for every p = 1, . . . , n,

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we have

Ap =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

fp,1 G1 0...

......

fp,p−1 Gp−1 0fp,p Gp 1fp,p+1 Gp+1 0...

......

fp,n Gn 0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= (−1)n+pDp, where Dp =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

fp,1 G1

......

fp,p−1 Gp−1fp,p+1 Gp+1

......

fp,n Gn

∣∣∣∣∣∣∣∣∣∣∣∣∣∣.

Consider the n× (n− 2) matrix

M =

G1

G2

...Gn

.

If p, q are different elements of 1, . . . , n, then denote by Bp,q the determinant ofthe (n − 2) × (n − 2) matrix that results from deleting the p-th row and the q-throw of the matrix M . It is clear that Bp,q = Bq,p for all p 6= q.

Now consider the Laplace expansions with respect to the first column for all thedeterminants D1, . . . , Dn. Let p, q ∈ 1, . . . , n, p < q. We have

Dp =p−1∑i=1

(−1)i+1fp,iBp,i +n∑

j=p+1

(−1)jfp,jBp,j ,

Dq =q−1∑i=1

(−1)i+1fq,iBq,i +n∑

j=q+1

(−1)jfq,jBq,j .

In the first equality appears the component (−1)qfp,qBp,q, and in the second equal-ity appears the component (−1)p+1fq,pBq,p. But fp,q = fq,p, Bp,q = Bq,p, andmoreover

n∑r=1

Ar =n∑r=1

(−1)n+rDr.

Hence, in the sum∑nr=1Ar the polynomial fp,q appears exactly two times, and we

have

(−1)p+n(−1)qfp,qBp,q + (−1)q+n(−1)p+1fp,qBp,q

=(

(−1)n+p+q + (−1)n+p+q+1)fp,qBp,q

= 0 · fp,qBp,q = 0.

Therefore,n∑p=1

[∂f∂xp

, g1, g2, . . . , gn−2, xp

]=

n∑p=1

Ap = 0.

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130 A. NOWICKI

4. Jacobian derivations in two variables

Now assume that n = 2. If f ∈ k[x, y], then we denote by ∆f the k-derivationof k[x, y] defined by

∆f (g) = [f, g],

for all g ∈ k[x, y]. We say that a k-derivation d of k[x, y] is jacobian, if there existsa polynomial f ∈ k[x, y] such that d = ∆f . Note, that

∆f (x) = −fy, ∆f (y) = fx.

If f ∈ k[x, y] is a homogeneous polynomial of degree m, then ∆f is a homogeneousk-derivation of degree m− 1.

Proposition 4.1. Let f, g ∈ k[x, y], and a ∈ k. Then:

(1) ∆f+g = ∆f + ∆g;

(2) ∆af = a∆f ;

(3) ∆fg = f∆g + g∆f ;

(4) [∆f ,∆g] = ∆[f,g].

Proof. The conditions (1) and (2) are obvious. Let h ∈ k[x, y]. Then we have

∆fg(h) = [fg, h] = −[h, fg] = −∆h(fg) = − (f∆h(g) + g∆h(f))

= −f [h, g]− g[h, f ] = f [g, h] + g[f, h] = f∆g(h) + g∆f (h)

= (f∆g + g∆f ) (h).

Thus, we proved (3). We now check (4):

[∆f ,∆g] (x) = (∆f∆g −∆g∆f ) (x) = ∆f (−gy)−∆g (−fy)

= −gyx (−fy)− gyyfx + fyx (−gy) + fyygx

= (gyxfy + gxfyy)− (gyyfx + gyfyx)

= (gxfy)y − (fxgy)y = (gxfy − fxgy)y = −[f, g]y = ∆[f,g](x);

[∆f ,∆g] (y) = (∆f∆g −∆g∆f ) (y) = ∆f (gx)−∆g (fx)

= −gxxfy + gxyfx + fxxgy − fxygx= (gxyfx + gyfxx)− (gxxfy + gxfxy)

= (gyfx)x − (fygx)x = (fxgy − fygx)x = [f, g]x = ∆[f,g](y).

Thus, we proved that [∆f ,∆g] and ∆[f,g] are k-derivations of k[x, y] such that

[∆f ,∆g] (x) = ∆[f,g](x), [∆f ,∆g] (y) = ∆[f,g](y).

This implies that [∆f ,∆g] = ∆[f,g].

Let us recall the following result of the author [18].

Theorem 4.2. Let k be a field of characteristic zero, and let f, g ∈ k[x, y] r k. If[f, g] = 0, then there exist a polynomial h ∈ k[x, y] and polynomials u(t), v(t) ∈ k[t]such that f = u(h) and g = v(h).

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If d and δ are k-derivations of k[x, y], then we write d ∼ δ in the case when ad =bδ, for some nonzero a, b ∈ k[x, y]. It is clear that if d ∼ δ, then k[x, y]d = k[x, y]δ

and k(x, y)d = k(x, y)δ. As a consequence of Theorem 4.2 we get

Proposition 4.3. Let k be a field of characteristic zero, and let f, g ∈ k[x, y] r k.Then [f, g] = 0 if and only if ∆f ∼ ∆g.

Proof. Let us observe that if u(t) ∈ k[t] r k, then ∂u∂t (f) 6= 0 and ∆f ∼ ∆u(f),

because

∆u(f) =∂u

∂t(f) ·∆f .

Assume that [f, g] = 0. It follows from Theorem 4.2 that f = u(h) and g = v(h),for some u, v ∈ k[t] and some h ∈ k[x, y]. Since f 6∈ k and g 6∈ k, we have u 6∈ knad h 6∈ k. Hence, ∆f = ∆u(h) ∼ ∆h ∼ ∆v(h) = ∆g, and hence ∆f ∼ ∆g.

Now suppose that ∆f ∼ ∆g. Let a∆f = b∆g, for some nonzero a, b ∈ k[x, y].Then we have afx = a∆f (y) = b∆g(y) = bgx and afy = −a∆f (x) = −b∆g(x) =bgy. Hence, fx = ugx and fy = ugy, where u = b/a. Therefore,

[f, g] = fxgy − fygx = ugxgy − ugygx = 0.


Every ∆f is a divergence-free k-derivation of k[x, y]. Indeed:

∆∗f = ∆f (x)x + ∆f (y)y = −fyx + fxy = 0.

We now show that if k contains Q, then the converse of this fact is also true. Themain role in our proof plays the following lemma.

Lemma 4.4. If Q ⊂ k and f, g ∈ k[x, y], then the following conditions are equiv-alent:

(a) there exists H ∈ k[x, y] such that Hx = f and Hy = g;

(b) fy = gx.

Proof. (a) ⇒ (b) follows from the equality ∂x∂y = ∂y∂x.

(b) ⇒ (a). Let

f =∑α,β

a(α, β)xαyβ , g =∑α,β

b(α, β)xαyβ ,

where all a(α, β), b(α, β) belong to k. If α > 1 and β > 1, then 1αa(α − 1, β) =

1β b(α, β − 1). Put

F =∑α,β

c(α, β)xαyβ ,

where c(0, 0) = 0 and, if α > 1 then c(α, β) = 1αa(α − 1, β), and if β > 1 then

c(α, β) = 1β b(α, β − 1). It is easy to check that Hx = f and Hy = g.

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132 A. NOWICKI

Proposition 4.5. If Q ⊂ k and d is a divergence-free k-derivation of k[x, y], thenthere exists a polynomial h ∈ k[x, y] such that d = ∆h.

Proof. Let d(x) = P , d(y) = Q and suppose that Px + Qy = 0. Put f = Qand g = −P . Then fy = gx and hence, by Lemma 4.4, there exists a polynomialh ∈ k[x, y] such that hx = f and hy = g, that is, d = ∆h.

Thus, we have

Proposition 4.6. Let Q ⊂ k, and let d be a k-derivation of k[x, y]. Then d isjacobian if and only if d is divergence-free .

Theorem 4.7. If Q ⊂ k and d is a nonzero k-derivation of k[x, y] then the fol-lowing two conditions are equivalent:

(1) k[x, y]d 6= k;

(2) d ∼ δ, where δ is a divergence-free k-derivation of k[x, y].

Proof. Since k[x, y]d = k[x, y]hd for every nonzero polynomial h in k[x, y], we mayassume that the polynomials d(x) and d(y) are relatively prime.

(1)⇒ (2). Suppose k[x, y]d 6= k and let F ∈ k[x, y]drk. Put d(x) = P , d(y) = Qand h = gcd(Fx, Fy). Then PFx +QFy = 0, h 6= 0 and there exist relatively primepolynomials A,B ∈ k[x, y] such that Fx = Ah and Fy = Bh. Hence AP = −BQand hence, A | Q, Q | A, B | P and P | B. This implies that there exists anelement a ∈ k r 0 such that aA = Q and aB = −P . Let δ = hd. Then d ∼ δand δ is divergence-free . Indeed,

δ? = (hP )x + (hQ)y = −(ahB)x + (ahA)y = −aFyx + aFxy = 0.

The implication (2) ⇒ (1) is obvious.

Now it is easy to prove the following theorem (see [19] Theorem 7.2.13).

Theorem 4.8. Let Q ⊂ k, and let d and δ be k-derivations of k[x, y] such thatk[x, y]d 6= k and k[x, y]δ 6= k. Then k[x, y]d = k[x, y]δ if and only if d ∼ δ.

5. Jacobian derivations in n variables

Assume that n > 2. Let F = (f1, . . . , fn−1), where f1, . . . , fn−1 are polyno-mials belonging to k[X] = k[x1, . . . , xn]. We denote by ∆F the mapping from k[X]to k[X] defined by

∆F (h) = [f1, . . . , fn−1, h] ,

for all h ∈ k[X]. This mapping is a k-derivation of k[X]. We say that it isa jacobian derivation of k[X]. If n = 2, then ∆F = ∆f1 is the jacobian k-derivationfrom the previous section. If the polynomials f1, . . . , fn−1 are homogeneous of de-grees m1, . . . ,mn−1, respectively, then the derivation ∆F is homogeneous of degree

(m1 + · · ·+mn−1)− (n− 1), provided rank[∂fi∂xj

]= n− 1.

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Now assume that n = 3. In this case F = (f, g) is a sequence of two polynomialsf, g from k[X] = k[x, y, z], and ∆(f,g) is a k-derivation of k[x, y, z] such that

∆(f,g)(x) = fygz − fzgy, ∆(f,g)(x) = fzgx − fxgz, ∆(f,g)(x) = fxgy − fygx.It is easy to check that ∆(f,g) is a divergence-free k-derivation of k[x, y, z]. Ingeneral, for any n > 2, we have the following theorem.

Theorem 5.1. Every jacobian k-derivation of k[x1, . . . , xn] is divergence-free .

Proof. Consider a jacobian k-derivation ∆F with F = (f1, . . . , fn−1), where f1, . . . ,fn−1 are polynomials belonging to k[X] = k[x1, . . . , xn]. Since every partial deriv-ative of k[X] is a divergence-free k-derivation, we have (see Proposition 3.2) theequalities of the form

∂

∂xp[f1, . . . , fn−1, xp] = [f1, . . . , fn−1, 1] +

n−1∑i=1

[f1, . . . ,

∂fi∂xp

, . . . , fn−1, xp

],

for all p = 1, . . . , n. Note that [f1, . . . , fn−1, 1] = 0. Using Proposition 3.3 weobtain also the equalities of the form

n∑p=1

[f1, . . . ,

∂fi∂xp

, . . . , fn−1, xp

]= 0,

for all i = 1, . . . , n− 1. We now have:

(∆F )?

=n∑p=1

∂∂xp

∆F (xp) =n∑p=1

∂∂xp

[f1, . . . , fn−1, xp]

=n∑p=1

([f1, . . . , fn−1, 1] +

n−1∑i=1

[f1, . . . ,

∂fi∂xp

, . . . , fn−1, xp

])=

n∑p=1

n−1∑i=1

[f1, . . . ,

∂fi∂xp

, . . . , fn−1, xp

]=

n−1∑i=1

(n∑p=1

[f1, . . . ,

∂fi∂xp

, . . . , fn−1, xp

])=n−1∑i=1

0 = 0.

Therefore, the derivation ∆F is divergence-free .

Other proofs of the above theorem appear in Connell and Drost [5], Theorem2.3; in Makar-Limanow [12]; and in Freudenburg’s book [7], Lemma 3.8.

Let k be a field of characteristic zero and let f1, . . . , fn be polynomials in k[X] =k[x1, . . . , xn]. Denote by w the jacobian of (f1, . . . , fn), that is, w = [f1, . . . , fn].It is well known and easy to be proved that if k[f1, . . . , fn] = k[X], then w isa nonzero element of k. The famous Jacobian Conjecture states that the converseof this fact is also true: if w ∈ k r 0 then k[f1, . . . , fn] = k[X]. The problem isstill open even for n = 2. There exists a long list of equivalent formulations of thisconjecture (see for example [22], [1], [6]). One of the equivalent formulations of theJacobian Conjecture is as follows.

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134 A. NOWICKI

Conjecture 5.2. Let k be a field of characteristic zero, and let F = (f1, . . . , fn−1),where f1, . . . , fn−1 are polynomials belonging to k[X] = k[x1, . . . , xn]. If there existsg ∈ k[X] such that ∆F (g) = 1, then the jacobian derivation ∆F is locally nilpotent.

It is difficult to prove that the above ∆F is locally nilpotent. Let us recall (seeTheorem 2.5) that every locally nilpotent derivation is divergence-free. Thus, bytheorem 5.1 we already know that ∆F is divergence-free.

We know that Derk(k[X]) is a free k[X]-module on the basis ∂∂x1

, . . . , ∂∂xn

. This

basis is commutative. We say that a basis d1, . . . , dn is commutative, if di dj =dj di for all i, j ∈ 1, . . . , n. A basis d1, . . . , dn is called locally finite (resp.locally nilpotent) if each di is locally finite (resp. locally nilpotent). Note thefollowing results of the author.

Theorem 5.3. ([17]). If k is a field of characteristic zero, then the followingconditions are equivalent.

(1) The Jacobian Conjecture is true in the n-variable case.

(2) Every commutative basis of the k[X]-module Derk(k[X]) is locally finite.

(3) Every commutative basis of the k[X]-module Derk(k[X]) is locally nilpotent.

Theorem 5.4. ([19] Theorem 2.5.5). Let k be a reduced ring containing Q. Ifd1, . . . , dn is commutative basis of the k[X]-module Derk(k[X]), then each deriva-tion di is divergence-free.

Note also some results of E. Connell, J. Drost [5] and L. Makar-Limanow [12].

Theorem 5.5. ([5]). Let D be a k-derivation of k[X] = k[x1, . . . , xn], where k isa field of characteristic zero. If tr.degkk[X]D = n − 1, then there exists g ∈ k[X]such that the derivation gD is divergence-free.

A k-derivation D of k[X] is called irreducible, if gcd (D(x1), . . . , D(xn)) = 1.

Theorem 5.6. ([12]). Let D be an irreducible locally nilpotent k-derivation ofk[X] = k[x1, . . . , xn], where k is a field of characteristic zero. Let f1, . . . , fn−1 ben − 1 algebraically independent elements of k[X]D, and set F = (f1, . . . , fn−1).Then there exists g ∈ k[X]D such that ∆F = gD. In particular, the derivation ∆F

is locally nilpotent.

6. The ideal I(d) for homogeneous derivations

In this section k is a field of characteristic zero, k[X] = k[x1, . . . , xn] isa polynomial ring over k, and d : k[X] → k[X] is a homogeneous k-derivationof degree s > 0. Put

gij = xid(xj)− xjd(xi),

for all i, j ∈ 1, . . . , n. Each gij is a homogeneous polynomial of degree s+ 1. Inparticular, gii = 0 and gji = −gij for all i, j. Moreover, for all i, j, p ∈ 1, . . . , n,

xigjp + xjgpi + xpgij = 0.

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We denote by I(d) the ideal in k[X] generated by all the polynomials gij withi, j ∈ 1, . . . , n.

Proposition 6.1. The ideal I(d) is differential, that is, d(I(d)) ⊂ I(d).

Proof. Put f1 = d(x1), . . . , fn = d(xn). Since f1, . . . , fn are homogeneous polyno-mials of degree s, we have the Euler formulas:

x1∂fi∂x1

+ · · ·+ xn∂fi∂xn

= sfi

for all i = 1, . . . , n. Thus, we have

d (gij) = d (xifj − xjfi)= fifj + xid(fj)− fjfi − xjd(fi) = xid(fj)− xjd(fi)

= xi

(∂fj∂x1

f1 + · · ·+ ∂fj∂xn

fn

)− xj

(∂fi∂x1

f1 + · · ·+ ∂fi∂xn

fn

)=

(x1

∂fj∂x1

+ · · ·+ xn∂fj∂xn

)fi −

(x1

∂fi∂x1

+ · · ·+ xn∂fi∂xn

)fj + a

= (sfj) fi − (sfi) fj + a = a,

where a is a polynomial belonging to I(d). Thus, d (gij) ∈ I(d) for all i, j, and thisimplies that d(I(d)) ⊂ I(d).

We denote by E the Euler derivation of k[X], that is,

E = x1∂

∂x1+ x2

∂

∂x2+ · · ·+ xn

∂

∂xn.

This derivation is homogeneous of degree 1. If 0 6= F ∈ k[X] is a homogeneouspolynomial of degree s, then E(F ) = sF . Thus, every nonzero homogeneouspolynomial of degree s is a Darboux polynomial of E with cofactor s.

Proposition 6.2. The ideal I(d) is equal to 0 if and only if d = h · E for someh ∈ k[X].

Proof. Suppose that d = hE with h ∈ k[X], Then d(xi) = xih for i = 1, . . . , n.Thus, gij = xi(xjh) = xj(xih) = 0 and so, I(d) = 0.

Now let I(d) = 0. Put fi = d(xi) for all i. Then, for all i, j ∈ 1, . . . , n, we havethe equality xifj = xjfi so, each xi divides fi. Thus, fi = uixi for i = 1, . . . , n,where ui ∈ k[X]. Put h = u1. Observe that ui = h for all i = 1, . . . , n. Therefore,d = hE.

Proposition 6.3. Let d : k[X] → k[X] be a homogeneous k-derivation of degrees > 1 and let h ∈ k[X] be a homogeneous polynomial of degree s − 1. ThenI(d) = I(d− hE).

Proof. Put δ = d− hE. Then, for all i, j ∈ 1, . . . , n, we have

xiδ(xj)− xjδ(xi) = xi (d(xj)− xjh)− xj (d(xi)− xih) = xid(xj)− xjd(xi).

Thus, the ideals I(d) and I(δ) are generated by the same elements.

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136 A. NOWICKI

Proposition 6.4. Let d : k[X] → k[X] be a homogeneous derivation of degree s.Then there exists a homogeneous k-derivation δ : k[X] → k[X], of degree s, suchthat I(d) = I(δ) and δ(xn) ∈ k[x1, . . . , xn−1].

Proof. Let d(xn) = Axn + B, where A ∈ k[X] and B ∈ k[x1, . . . , xn−1]. Putδ = d − AE. Then I(d) = I(δ) (by Proposition 6.3) and δ(xn) = d(xn) − Axn =B ∈ k[x1, . . . , xn−1].

Let us recall that all the polynomials gij are homogeneous of degree s+1, gii = 0and xigjp + xjgpi + xpgij = 0, for all i, j, p ∈ 1, . . . , n.

Proposition 6.5. Let wij ; i, j = 1, . . . , n be a family of polynomials in k[X].Suppose that:

(1) all the polynomials wij are homogeneous of degree s+ 1;

(2) wii = 0 for i = 1, . . . , n;

(3) xiwjp + xjwpi + xpwij = 0, for all i, j, p ∈ 1, . . . , n.Then there exist homogeneous of degree s polynomials f1, . . . , fn ∈ k[X] such that

wij = xifj − xjfi,for all i, j ∈ 1, . . . , n.

Proof. Let Yi =n∑j=1

∂wij∂xj

, for i = 1, . . . , n. Then, for i, j,∈ 1, . . . , n, we have:

xiYj − xjYi = xin∑p=1

∂wjp

∂xp− xj

n∑p=1

∂wip

∂xp

= xi∂wji

∂xi− xj ∂wij

∂xj+ xi

∑p6=i

∂wjp

∂xp− xj

∑p6=j

∂wip

∂xp

= xi∂wji

∂xi− xj ∂wij

∂xj+ xi

∑p6=i, p6=j

∂wjp

∂xp− xj

∑p6=j, p6=i

∂wip

∂xp

= xi∂wji

∂xi− xj ∂wij

∂xj+

∑p6=i, p6=j

∂∂xp

(xiwjp − xjwip)

= xi∂wji

∂xi− xj ∂wij

∂xj+

∑p6=i, p6=j

∂∂xp

(−xpwij)

= xi∂wji

∂xi+ xj

∂wji

∂xj−

∑p6=i, p6=j

xp∂wij

∂xp−

∑p6=i, p6=j

wij

= −∑np=1 xp

∂wij

∂xp− (n− 2)wij = −(s+ 1)wij − (n− 2)wij

= −(s+ n− 1)wij .

Thus, xiYj − xjYi = −(s + n − 1)wij . Let fi = − 1s+n−1Yi, for i = 1, . . . , n. Then

we havewij = xifj − xjfi,

for all i, j ∈ 1, . . . , . It is clear that the polynomials f1, . . . , fn are homogeneousof degree s.

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Proposition 6.6. Let wij ; i, j = 1, . . . , n be a family of polynomials in k[X] such

as in Proposition 6.5, and let Yi =n∑j=1

∂wij

∂xj, for i = 1, . . . , n. Then

n∑i=1

∂Yi

∂xi= 0.

Proof. Put A =n∑i=1

∂Yi

∂xi. Then we have:

A =n∑i=1

∂∂xi

(n∑j=1

∂wij

∂xj

)=

n∑i=1

n∑j=1

∂2wjp

∂xi∂xj= −

n∑i=1

n∑j=1

∂2wji

∂xj∂xi= −A.

Thus, A = 0.

Theorem 6.7. Let k be a field of characteristic zero, and let d : k[X] → k[X]be a homogeneous k-derivation of degree s. Then there exists a divergence-free k-derivation δ : k[X]→ k[X] such that δ is homogeneous of degree s and I(d) = I(δ).

Proof. Let wij = xid(xj) − xjd(xi) for i, j ∈ 1, . . . , n. The polynomials wijsatisfy the properties (1)− (3) of Proposition 6.5. Put

Yi =n∑j=1

∂wij∂xj

, fi = − 1

s+ n− 1Yi,

for i = 1, . . . , n. Then wij = xifj − xjfi (see the proof of Proposition 6.5). Letδ : k[X]→ k[X] be the k-derivation defined by δ(xi) = fi, for i = 1, . . . , n. Then δis homogeneous of degree s and I(d) = I(δ). Moreover, it follows from Proposition6.6 that the divergence δ∗ is equal to zero.

7. Polynomials Md in two variables

In this section we assume that n = 2 and k is a field of characteristic zero.Given a homogeneous k-derivation d of k[X] we studied in the previous section thedifferential ideal generated by all polynomials of the form xid(xj)−xjd(xi). In thecase n = 2 this ideal is generated only by one polynomial

Md = xd(y)− yd(x).

If d is homogeneous derivation of degree s, then Md is a homogeneous polynomialand degMd = s + 1. If d is the Euler derivation E, then Md = 0. It is easy tocheck that Md = 0 if and only if d = h · E for some h ∈ k[x, y].

Proposition 7.1. If d is a homogeneous k-derivation of k[x, y] and Md 6= 0, thenMd is a Darboux polynomial of d and its cofactor is equal to the divergence d∗, thatis,

d(Md) = d∗Md.

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138 A. NOWICKI

Proof. Put f = d(x), g = d(y). Since d is homogeneous, we have xfx + yfy = sfand xgx + ygy = sg, where s is the degree of d. So, we have,

d(Md)− d∗Md = d(xg − yf)− (fx + gy)(xg − yf)

= fg + x(gxf + gyg)− gf − y(fxf + fyg)− (fx + gy)(xg − yf)

= xgxf + xgyg − yfxf − yfyg − xfxg + yfxf − xgyg + ygyf

= (xgx + ygy)f − (xfx + yfy)g

= sgf − sfg = 0,

and hence, Md is a Darboux polynomial with cofactor d∗

The above property does not hold when d(x), d(y) are homogeneous of differentdegrees. Let for example, d(x) = 1, d(y) = x. Then Md = x2 − y, d∗ = 0 andd(Md) = d(x2−y) = 2x−x = x 6= 0·(x2−y). The above property also does not holdwhen deg d(x) = deg d(y) and the polynomials d(x), d(y) are not homogeneous. Letd(x) = x+ 1, d(y) = y. Then Md = −y, d∗ = 2, d(Md) = −y 6= −2y.

We say that a Darboux polynomial f is said to be essential if f 6∈ k.

Proposition 7.2. Every homogeneous k-derivation of k[x, y] has an essential Dar-boux polynomial f ∈ k[x, y] r k.

Proof. If Md 6= 0 then, by the previous proposition, Md is a Darboux polynomial.If Md = 0, then x− y is a Darboux polynomial.

The following examples show that the above property does not hold when d isnot homogeneous, and when d is a homogeneous derivations in three variables. Letus recall that k is a field of characteristic zero.

Example 7.3. ([10], [19], [20]). The derivation ∂x + (xy + 1)∂y has no essentialDarboux polynomial.

Example 7.4. ([8]). The derivation (1 − xy)∂x + x3∂y has no essential Darbouxpolynomial.

Example 7.5. ([9]). Let d be the k-derivation of k[x, y, z] defined by:

d(x) = y2, d(y) = z2, d(z) = x2.

Then d is homogeneous, divergence-free , and d has no essential Darboux polyno-mial.

Proposition 7.6. Let d : k[x, y] → k[x, y] be a homogeneous k-derivation, andlet f = d(x), g = d(y). If h, λ ∈ k[x, y] are homogeneous polynomials such thatd(h) = λh, then

Mdhx = (yλ−mg)h, Mdhy = (mf = xλ)h,

where m = deg h.

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Proof. We have the following sequences of equalities:

fhx + ghy = λh,

yfhx + yghy = yλh,

yfhx + g(mh− xhx) = yλh,

(xg − yf)hx = (yλ−mg)h,

Mdhx = (yλ−mg)h.

fhx + ghy = λh,

xfhx + xghy = xλh,

f(mh− yhy) + xghy = xλh,

(xg − yf)hy = (mf − xλ)h,Mdhy = (mf = xλ)h.

We used the Euler formula.

Proposition 7.7. If d : k[x, y] → k[x, y] is a nonzero homogeneous k-derivation,then every irreducible Darboux polynomial of d is a divisor of the polynomial Md.

Proof. Let h ∈ k[x, y] r k be an irreducible Darboux polynomial of d, and letλ be its cofactor. Thus, d(h) = λh. We know, by Proposition 1.2, that λ ishomogeneous. Since h 6∈ k, we have either hx 6= 0 or hy 6= 0. Let us suppose thathx 6= 0. Then the polynomials hx and h are relatively prime and (by Proposition7.6) Mdhx = (yλ−mg)h. Thus, h divides Md. In the case hy 6= 0 we do the sameprocedure,

The Euler derivation E : k[x, y]→ k[x, y] is a nonzero homogeneous derivation,and every nonzero homogeneous polynomial from k[x, y] is a Darboux polynomialof E. Thus, E has infinitely many homogeneous irreducible Darboux polynomials,The same property has every derivation hE with a nonzero homogeneous h ∈k[x, y]. Let us recall that in this case the polynomial Md is equal to zero. Thefollowing proposition states that other homogeneous derivations have only finitelymany homogeneous irreducible Darboux polynomials.

Theorem 7.8. Let k be a field of characteristic zero, and let d : k[x, y]→ k[x, y] bea nonzero homogeneous k-derivation of degree s such that Md 6= 0. Then d has atmost s+ 1 pairwise nonassociated irreducible homogeneous Darboux polynomials.

Proof. It follows from Proposition 7.7, because Md is a nonzero homogeneouspolynomial of degree s+ 1.

In the above theorem we were interested in irreducible homogeneous Darbouxpolynomials. Without the word ”homogeneous” such property does not hold, ingeneral. Let for example, d = x∂x + 2y∂y. Then d(x2 + ay) = 2(x2 + ay) for everya ∈ k and hence, d is a nonzero homogeneous k-derivation and d has infinitelymany, pairwise nonassociated, irreducible Darboux polynomials,

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140 A. NOWICKI

8. Sums of jacobian derivations

In this section k is always a commutative ring containing Q.

We know (see Proposition 4.6) that every divergence-free k-derivation of k[x, y]is a jacobian derivation. A similar property for n > 3 variables does not hold ingeneral. Let, for example, d be the k-derivation of k[x, y, z], defined by: d(x) =y2, d(y) = z2, d(z) = x2 (as in Example 7.5). Then d is divergence-free . It isknown that k[x, y, z]d = k (see [9] or [15], [19]) so, d is not jacobian. There existmany similar examples for arbitrary n > 3 (see [11], [23], [19]). In this section wewill show that every divergence-free k-derivation of k[X] = k[x1, . . . , xn] is a finitesum of some jacobian derivation.

Let f be a polynomial from k[X], and let i, j ∈ 1, . . . , n. We denote by Ωfi,jthe k-derivation of k[X] defined by

Ωfi,j(g) =

∣∣∣∣∣∂f∂xi

∂g∂xi

∂f∂xj

∂g∂xj

∣∣∣∣∣ = fxigxj− fxj

gxi

for all g ∈ k[X]. It is clear that Ωfi,i = 0 and Ωfj,i = −Ωfi,j for all i, j ∈ 1, . . . , n.If i 6= j, then we have

Ωfi,j(xp) =

0, if p 6= i, p 6= j,

− ∂f∂xj

, if p = i,

∂f∂xi

, if p = j,

for all p = 1, . . . , n. Note the following obvious proposition.

Proposition 8.1. Every derivation of the form Ωfi,j is divergence-free .

Another common notation for Ωfi,j , is Ωfxi,xj. If n = 2 and f ∈ k[x, y], then

Ωfx,y = ∆f , where ∆f is the jacobian derivation of k[x, y] from a previous section.If n = 3 and f ∈ k[x, y, z], then we have three k-derivations of the above forms:Ωfx,y, Ωfx,z and Ωfy,z.

Proposition 8.2. Let d be a k-derivation of k[x, y, z], where k is a commutativering containing Q. If d is divergence-free , then there exist polynomials u, v ∈k[x, y, z] such that

d = Ωux,y + Ωvy,z.

Proof. Put f = d(x), g = d(y), h = d(z) and R = k[x, y, z]. Since d is divergence-free , we have the equality fx + gy + hz = 0. Since the partial derivative ∂

∂y is

a surjective mapping from R to R, there exists a polynomial H ∈ R such thath = Hy. Let

f = f, g = g +Hz,

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and consider the k[z]-derivation d of R = k[z][x, y] defined by d(x) = f and d(y) =g. Observe that the derivation d is divergence-free . Indeed,(

d)∗

= fx + gy = fx + gy +Hzy = fx + gy +Hyz = fx + gy + hz = 0.

It follows from Proposition 4.5, that there exists a polynomial F ∈ R such thatd = ∆F . Hence, d(x) = −Fy and d(y) = Fx and hence, f = −Fy, g = Fx −Hz.Put u = F , v = H and δ = Ωux,y + Ωvy,z. Then we have:

δ(x) =

∣∣∣∣ ux 1uy 0

∣∣∣∣ = −uy = −Fy = f,

δ(y) =

∣∣∣∣ ux 0uy 1

∣∣∣∣+

∣∣∣∣ vy 1vz 1

∣∣∣∣ = ux − vz = Fx −Hz = g,

δ(z) =

∣∣∣∣ vy 0vz 1

∣∣∣∣ = vy = Hy = h.

Therefore, d = δ = Ωux,y + Ωvy,z.

Example 8.3. Let d = ys ∂∂x + zs ∂∂y + xs ∂∂z , where s > 1. Then d = Ωux,y + Ωvy,zfor u = zsx− 1

s+1ys+1 and v = xsy.

Proposition 8.4. Let d be a k-derivation of k[x, y, z], where k is a commutativering containing Q. If d is divergence-free , then there exist polynomials A,B,C ∈k[x, y, z] such that

d = ΩAx,y + ΩBy,z + ΩCz,x.

In other words, there exist polynomials A,B,C ∈ k[x, y, z] such that

d(x) = Cz −Ay, d(y) = Ax −Bz, d(z) = By − Cx.

Proof. Let u, v ∈ k[x, y, z] as in Proposition 8.2. Put A = u, B = v and C = 0.Then d = ΩAx,y + ΩBy,z + ΩCz,x.

Example 8.5. Let d = ys ∂∂x + zs ∂∂y + xs ∂∂z , where s > 1. Then d = ΩAx,y +

ΩBy,z + ΩCz,x where A = 12

(zsx− 1

s+1ys+1), B = 1

2

(xsy − 1

s+1zs+1)

and C =

12

(ysz − 1

s+1xs+1).

Example 8.6. If f, g ∈ k[x, y, z], then ∆(f,g) = ΩAx,y + ΩBy,z + ΩCz,x, where

A = fzg, B = fxg, C = fyg.

Quite recently, Piotr Jedrzejewicz generalizes Propositions 8.2 and 8.4 for arbi-trary n > 3. Such generalizations seem to be well-known, although we could notfind a reference.

Theorem 8.7 (Jedrzejewicz). Let d be a k-derivation of k[X] = k[x1, . . . , xn],

where n > 3 and k is a commutative ring containing Q. If d is divergence-free,

then there exist polynomials u1, . . . , un−1 ∈ k[X] such that

d = Ωu11,2 + Ωu2

2,3 + · · ·+ Ωun−1

n−1,n.

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142 A. NOWICKI

In particular, we have the following equalities

(∗)

d(x1) = −(u1)x2,

d(x2) = (u1)x1− (u2)x3

,d(x3) = (u2)x2 − (u3)x4 ,

...d(xn−1) = (un−2)xn−2 − (un−1)xn ,d(xn) = (un−1)xn−1 .

Proof. By induction on n. For n = 3 it follows from Proposition 8.2. Let n > 3 andsuppose that our assertion is true for this n. Let d be a divergence-free k-derivationof R = k [x1, . . . , xn+1]. Put fi = d(xi) for all i = 1, . . . , n+1. We have the equality∑n+1i=1 (fi)xi

= 0. Since the partial derivative ∂∂xn

is a surjective mapping from Rto R, there exists a polynomial P ∈ R such that fn+1 = Pxn . Let

g1 = f1, g2 = f2, . . . , gn−1 = fn−1, gn = fn + Pxn+1 ,

and consider the k [xn+1]-derivation d of R defined by d(xi) = gi for all i = 1, . . . , n.Observe that the derivation d is divergence-free . Indeed,

(d)∗

=

n∑i=1

(gi)xi=

n−1∑i=1

(fi)xi+ (fn)xn

+ Pxnxn+1 =

n+1∑i=1

(fi)xi= 0,

because Pxnxn+1 = (fn+1)xn+1. By induction there exist polynomials v1, . . . , vn−1 ∈

R satisfying the equalities (∗) for the derivation d, that is,

g1 = d(x1) = − (v1)x2, gn = d(xn) = (vn−1)xn−1

and gi = d(xi) = (vi−1)xi−1− (vi)xi+1

for i = 2, . . . , n − 1. Let us recall that

gn = fn + Pxn+1 Put ui = vi for i = 1, . . . , n− 1, and un = P . Then d(x1) = f1 =−(u1)x2

, and d(xi) = −(ui−1)xi−1for i = 2, . . . , n− 1. Moreover,

d(xn) = fn = gn − Pxn+1= (vn−1)xn−1

− Pxn+1= (un−1)xn−1

− (un)xn+1

and d (xn+1) = fn+1 = Pxn= uxn

. This means that d = Ωu11,2 +Ωu2

2,3 + · · ·+Ωunn,n+1,

and this completes the proof.

Theorem 8.8. Let d be a k-derivation of k[x1, . . . , xn], where n > 3 and k is

a commutative ring containing Q. If d is divergence-free, then there exist polyno-

mials A1, . . . , An ∈ k[x1, . . . , xn] such that

d = ΩA11,2 + ΩA2

2,3 + · · ·+ ΩAn−1

n−1,n + ΩAnn,1.

In particular, d(xi) = (Ai−1)xi−1− (Ai)xi+1

for all i ∈ Zn.

Proof. Let u1, . . . , un−1 ∈ k[x1, . . . , xn] be as in Theorem 8.7. Put Ai = ui fori = 1, . . . , n− 1 and An = 0. Then our assertion follows from Theorem 8.7.

“22˙Nowicki2” — 2017/12/1 — 20:47 — page 143 — #22


Example 8.9. Let d be the k-derivation of k[x1, . . . , xn] defined by d(xi) = xsi+1 fori = 1, . . . , n, where k is a commutative ring containing Q, s > 0, and xn+1 = x1,

x0 = xn. Then d is divergence-free, and d = ΩA11,2 + Ωu2

2,3 + · · ·+ ΩAn−1

n−1,n+ ΩAnn,1. with

Ai =1

2

(xsi+2xi −

1

s+ 1xs+1i+1

)for all i = 1, . . . , n.

Acknowledgments. The author would like to thank Jean Moulin Ollagnierand Piotr Jedrzejewicz for many valuable scientific discussions and comments.

References

[1] H. Bass, E.H. Connell and D. Wright, The jacobian conjecture: Reduction of degree and

formal expansion of the inverse Bull. Amer. Math. Soc., 7 (1982), 287–330.[2] H. Bass and G.H. Meisters, Polynomial flows in the plane, J. Algebra, 55 (1985), 173–208.

[3] J. Berson, A. van den Essen and S. Maubach, Derivations having divergence zero on R[x, y],

Israel J. Math., 124 (2001), 115–124.[4] B.A. Coomes and V. Zurkowski, Linearization of polynomial flows and spectra of derivations,

J. Dynamics and Diff. Equations, 3 (1991), 29–66.

[5] E. Connell and J. Drost, Conservative and divergence free algebraic vector fields, Proc. Amer.Math. Soc., 87 (1983), 607–612.

[6] A. van den Essen, Polynomial automorphisms and the Jacobian Conjecture, Progress inMathematics 190, 2000.

[7] G. Freudenburg, Algebraic Theory of Locally Nilpotent Derivations, Encyclopedia of Math-

ematical Sciences, Springer, 2006.[8] D.A. Jordan, Differentially simple rings with no invertible derivatives, Quart. J. Math. Ox-

ford, 32 (1981), 417–424.

[9] J.-P. Jouanolou, Equations de Pfaff algebriques, Lect. Notes in Math. 708, Springer-Verlag,Berlin, 1979.

[10] I. Kaplansky, An Introduction to Differential Algebra, Hermann, Paris, 1976.

[11] A. Maciejewski, J.M. Ollagnier, A. Nowicki and J.-M. Strelcyn, Around Jouanolou non-integrability theorem, Indag. Mathem., 11(2) (2000), 239–254.

[12] L. Makar-Limanov, Locally nilpotent derivations, a new ring invariant and applications,

preprint 1998.[13] J. Moulin Ollagnier and A. Nowicki, Derivations of polynomial algebras without Darboux

polynomials, J. Pure Appl. Algebra, 212 (2008), 1626–1631.[14] J. Moulin Ollagnier and A. Nowicki, Monomial derivations, Comm. Algebra, 39(9) (2011),

3138–3150.

[15] J. Moulin Ollagnier, A. Nowicki and J.-M. Strelcyn, On the non-existence of constants ofderivations: The proof of a theorem of Jouanolou and its development, Bull. Sci. Math., 119

(1995), 195–233.

[16] P. Nousiainen and M.E. Sweedler, Automorphisms of polynomial and power series rings,J. Pure Appl. Algebra 29 (1983), 93–97.

[17] A. Nowicki, Commutative basis of derivations in polynomial and power series rings,

J. Pure Appl. Algebra, 40 (1986), 279–283.[18] A. Nowicki, On the jacobian equation J(f, g) = 0 for polynomials in two variables, Nagoya

J. Math., 109 (1988), 151–157.

[19] A. Nowicki, Polynomial derivations and their rings of constants, Nicolaus Copernicus Uni-versity Press, Torun, 1994.

“22˙Nowicki2” — 2017/12/1 — 20:47 — page 144 — #23

144 A. NOWICKI

[20] A. Nowicki, An example of a simple derivation in two variables, Colloq. Math., 113 (2008),

25–31.

[21] Y. Stein, On the density of image of differential operators generated by polynomials,J. Analyse Math., 52 (1989), 291–300.

[22] S.S-S. Wang, A jacobian criterion for separability, J. Algebra 65 (1980), 453–494.

[23] H. Zo ladek, Multidimensional Jouanolou system, J. Reine Angew. Math 556 (2003), 47–78.

Nicolaus Copernicus University, Faculty of Mathematics and Computer Science,

ul. Chopina 12/18, 87-100 Torun, Poland


“23_Osinska_Skalski-kopia” — 2017/12/1 — 20:49 — page 145 — #1

Analytic and Algebraic Geometry 2Łódź University Press 2017, 145 – 159


THE CAUCHY-KOWALEVSKI THEOREM

BEATA OSIŃSKA-ULRYCH AND GRZEGORZ SKALSKI

Abstract. We give a recursive description of polynomials with non-negativerational coefficients, which are coefficients of expansion in a power series so-lutions of partial differential equations in Cauchy-Kowalevski theorem.

1. Introduction

In recent time we can observe the renewed interest in the algorithms associatedwith the solution of partial differential equations using power series (see: for ex-ample [8]). This study initiated by the famous theorem of Cauchy-Kowalevski1

(see original work [9], Theorem 2.1 and Proposition 2.1 in this article, compare [2],[3]) were later generalized by Riquier [11] for a wide class of orthonomic passivesystems. In both theorems, the proof of the existence and uniqueness consistedof demonstration, in a first step, the existence and uniqueness of formal solutions,and in the second step of its convergence. The work of Riquier for polynomialnonlinear differential equations was complemented by Ritt [12]. The proof usedthe method of characteristic set. Since that time many algorithms for determiningthe formal solution of partial differential equations was stated. It is well knownthat coefficients of such a formal solution are polynomials depending on coefficientsoccurring in the power series expansion of right-hand side functions in partial dif-ferential equations (see: for example [1], [4], [8], see also [13]). Moreover, thesepolynomials have non-negative rational coefficients. The aim of this paper is to

2010 Mathematics Subject Classification. 35-XX, 35A24, 35R01, 12Hxx.Key words and phrases. Partial differential equations, Cauchy problem, Cauchy-Kowalevski

Theorem.1After G. B. Folland, [3] the problem of how to spell this name is vexed not only by the usual

lack of a canonical scheme for transliterating from the Cyrillic alphabet to the Latin one but alsoby the question of whether to use the feminine ending (-skaia instead of -ski). The spelling usedhere is the one preferred by Kowalevski herself in her scientific works.

145


146 B. OSIŃSKA-ULRYCH AND G. SKALSKI

give a recursive description of these polynomials (Theorem 2.5), which is not givenexplicitly in textbooks.

Multi-indexes and partial derivatives. The n-element sequence α =(α1, . . . , αn) of non-negative integers2 will be called multi-index of dimension n.We introduce the following notations:

|α| =n∑j=1

αj , α! = α1! · · ·αn!,

and for x ∈ Rn,xα = xα1

1 · · ·xαnn .

In general, we will use the shortcut

∂j = ∂xj =∂

∂xj

for the partial derivative in Rn. For the partial derivatives of higher order it ismore convenient to use multi-index

∂α = ∂αx =

n∏j=1

(∂

∂xj

)αj=

∂|α|

∂xα11 · · · ∂x

αnn.

In particular, we note that for α = 0, ∂α it is the identity operator. Let I be anynon-empty set containing an element j. Then 1j designates a system (δi)i∈I , whereδi = 1 for i = j and δi = 0 for i 6= j. With the above descriptions it is easily tonote that the partial derivatives can be defined by induction in the following way:

(1) ∂0 = id,(2) ∂α+1j = (∂α)1j = ∂j∂

α for j ∈ 1, 2, . . . , n and all α ∈ Nn.

Let us order the set of multi-indices. We write that α 6 β, if αi 6 βi for all i.For the given complex numbers aα for |α| 6 k, by (aα)|α|6k we denote the elementof CN(k) given by ordering the α’s in this fashion, where N(k) is the number ofelements in the set α ∈ Nn : |α| 6 k. Similarly, if A ⊂ α : |α| 6 k, then wecan consider the elements of space CN of the form (aα)α∈A, where N = #A.

2. The Cauchy-Kowalevski Theorem

Let k be a positive integer and let S be an analytic hypersurface of form

S = (x, t) = (x1, . . . , xn−1, t) ∈ Rn : t = 0 .

Let F : Ω→ R be an analytic function in some neighbourhood Ω ⊂ Rn ×RN(k) ofthe origin, where

N(k) =

(n+ k

k

)= (α, j) = (α1, . . . , αn−1, j) ∈ Nn : |α|+ j 6 k.

2Here the set of non-negative integers is denoted by N.


THE CAUCHY-KOWALEVSKI THEOREM 147

If ϕ0, . . . , ϕk−1 are the real analytic functions at the origin of Rn−1, then theanalytic Cauchy problem is to look for the solution u of system (2.1) analytic atthe origin of Rn

(2.1)

F(x, t,

(∂αx ∂

jt u)|α|+j6k

)= 0,

∂jt u(x, 0) = ϕj(x), 0 6 j < k.

We assume that the equation F = 0 can be solved for ∂kt u to yield ∂kt u as ananalytic function G of the remaining variables. We do this because of the badbehaviour that can occur when this condition is not satisfied (see examples i andii, page 43 in [3]). The Cauchy problem then takes the form

(2.2)

∂kt u = G

(x, t,

(∂αx ∂

jt u)|α|+j6k, j<k

),

∂jt u(x, 0) = ϕj(x), 0 6 j < k.

This problem has at most one analytic solution (see [3, Proposition 1,21]):

Proposition 2.1. Assume that G,ϕ0, . . . , ϕk−1 are analytic functions near theorigin.Then there is at most one analytic function u satisfying (2.2).

Proof. Functions ϕ0, . . . , ϕk−1 together with (2.2) determine all the partial deriva-tives of function u of order 6 k on S. Since G is analytic, by differentiating (2.2)with respect to t we have

∂k+1t u =

∂G

∂t+

∑|α|+j6k, j<k

∂G

∂u(α,j)

(x, t,

(∂αx ∂

jt u)|α|+j6k, j<k

)∂αx ∂

j+1t u.

All the quantities on the right are known on S, so is ∂k+1t u; hence we know all

derivatives of u of order 6 k + 1 on S. Applying ∂t more times, we obtain higherderivatives. All the partial derivatives of the function u at zero are therefore knownand determine u uniquely.

In our article we focus on the following fundamental existence theorem (see [9],compare [2, Theorem 2 in paragraph 4.6.3], [3, Theorem 1.25]).

Theorem 2.2 (The Cauchy-Kowalevski Theorem). Assume that G,ϕ0, . . . , ϕk−1

are analytic functions near the origin. Then there is a neighborhood of the originon which the Cauchy problem (2.2) has a unique analytic solution.

Uniqueness of solution was proved in Proposition 2.1. It’s proof suggests theconstruction of solution: determine all the derivatives of u at the origin by differ-entiating

∂kt u = G

(x, t,

(∂αx ∂

jt u)|α|+j6k, j<k

)and plug the results into Taylor’s formula. The problem is to show that the resultingpower series converges. To this end, it is convenient to replace our k-th orderequation by a first order system of differential equations.



Theorem 2.3. The Cauchy problem (2.2) is equivalent to the Cauchy problem fora certain first order quasi-linear system of partial differential equations of the form

(2.3)

∂tY =n−1∑j=1

Aj(x, t, Y )∂xjY +B(x, t, Y ),

Y (x, 0) = Φ(x),

i.e., a solution to one problem can be read off from a solution to the other. HereY , B, and Φ are vector-valued functions, the Aj’s are matrix-valued functions, andAj, B, and Φ are explicitly determined by the functions in (2.2).

Proof. Let Y = (yαj)06|α|+j6k, where yαj will stand for ∂αx ∂jt u as an independent

variable in G. Moreover, for multi-index α 6= 0, let i = i(α) denote the smallestindex i, for which αi 6= 0 and let 1i = (δ1, . . . , δn−1), where

δj =

1 for j = i,

0 for j 6= i.

The first order system we are looking for is

(2.4)

∂tyαj = yα(j+1) for |α|+ j < k,

∂tyαj = ∂xi(α)y(α−1i(α))(j+1) for |α|+ j = k, j < k, |α| 6= 0,

∂ty0k = ∂G∂t +

∑|α|+j<k

∂G∂yαj

yα(j+1) +∑

|α|+j=kj<k

∂G∂yαj

∂xi(α)y(α−1i(α))(j+1),

and the initial conditions are

(2.5)

yαj(x, 0) = ∂αxϕj(x) for j < k,

y0k(x, 0) = G(x, 0, (∂αxϕj(x))|α|+j6k, j<k

).

Obviously, if u is a solution of (2.2), then the functions yαj = ∂αx ∂jt u satisfy (2.4)

and (2.5). Conversely, if the Y = (yαj)06|α|+j6k is a solution of (2.4) and (2.5),then u = y00 satisfies (2.2). This involves the initial conditions in an essential way.

Observe, that the equation ∂tyαj = yα(j+1) of system (2.4) implies that

(2.6) yα(j+l) = ∂ltyαj for j + l 6 k.

Then the equation ∂tyαj = ∂xi(α)y(α−1i(α))(j+1) of system (2.4) implies

∂tyαj = ∂t∂xiy(α−1i)j, for |α|+ j = k, j < k.

Thereforeyαj(x, t) = ∂xiy(α−1i)j(x, t) + cαj(x)

for some function cαj . But by the first equation of (2.5),

yαj(x, 0) = ∂αxϕj(x) = ∂xi∂α−1ix ϕj(x) = ∂xiy(α−1i)j(x, 0),



hence cαj = 0 and we have,

(2.7) yαj = ∂xiy(α−1i)j for |α|+ j = k, j < k.

Then, from the third equation of (2.4), (2.6) and (2.7), we have

∂ty0k =∂G

∂t+

∑|α|+j6kj<k

∂G

∂yαj

∂yαj∂t

=∂

∂t(G (x, t, (yαj))) ,

whencey0k(x, t) = G (x, t, (yαj(x, t))) + c0k(x)

for some function c0k. But by (2.5),

y0k(x, 0) = G (x, 0, (∂αxϕj(x))) = G (x, 0, (yαj(x, 0))) ,

hence again c0k = 0 and we have

(2.8) y0k = G(x, t, (yαj)|α|+j6k, j<k

).

Finally, by induction on p = k − j − |α|, we will prove that

(2.9) yαj = ∂xiy(α−1i)j for α 6= 0.

For p = 0, i.e. when |α| + j = k the above is true from (2.7). From the firstequation in (2.4), from (2.6) and from the inductive hypothesis we have

∂tyαj = yα(j+1) = ∂xiy(α−1i)(j+1) = ∂t∂xiy(α−1i)j ,

henceyαj(x, t) = ∂xiy(α−1i)j(x, t) + cαj(x).

But by the first equation in (2.5),

yαj(x, 0) = ∂αxϕj(x) = ∂xi∂α−1ix ϕj(x) = ∂xiy(α−1i)j(x, 0).

Therefore cαj = 0 and we get (2.9).Finally, applying (2.6) and (2.9) repeatedly we obtain that

yαj = ∂αx ∂jt y00,

and then by (2.8) and the first equation in (2.5) we find that u = y00 satisfies(2.2).

We still need a little simplification.

Theorem 2.4. The Cauchy problem (2.3) is equivalent to another problem of thesame form in which Φ = 0 i A1, . . . , An−1 and B do not depend on t.

Proof. To eliminate Φ we set U(x, t) = Y (x, t)−Φ(x). Then Y satisfies (2.3) if andonly if U satisfies:

∂tU =n−1∑i=1

Ai(x, t, U)∂xiU + B(x, t, U), U(x, 0) = 0,



where

Ai(x, t, U) =Ai(x, t, U + Φ),

B(x, t, U) =B(x, t, U + Φ) +n−1∑i=1

Ai(x, t, U + Φ)∂xiΦ.

To eliminate variable t from Ai and B we add to U an extra component u0 satisfyingthe equation ∂tu0 = 1 and the initial condition u0(x, 0) = 0. Then we replace t byu0 in Ai and B, by adding the extra equation and initial condition.

Let us assume the following designations: (x, Y ) ∈ Rn−1 × RN and (x, t) ∈Rn−1 × R, where x = (x1, . . . , xn−1), Y = (y1, . . . , yN ). Since the constructions inthese theorems preserve analyticity, we have reduced the Cauchy-Kowalevski theo-rem to the following theorem. This theorem is well known, but we add a recursivedescription coefficients of solution as polynomials of the coefficients occurring inthe series in the partial differential equation.

Theorem 2.5. Suppose that B = [bm]Nm=1 is a real analytic vector-valued function

and Ai =[aiml]Nm,l=1

, i ∈ 1, . . . , n− 1, are real analytic matrix-valued functionsdefined on a neighborhood of the origin in Rn−1×RN . Then there is a neighborhoodU of the origin in Rn, on which the Cauchy problem

(2.10)

∂tY =n−1∑i=1

Ai(x, Y )∂xiY +B(x, Y ),

Y (x, 0) = 0

has a unique analytic solution Y = (y1, . . . , yN ) : U 3 (x, t) 7→ Y (x, t) ∈ RN .Furthermore, if

aiml(x, y1, . . . , yN ) =∑σ,τ

ai;στml xσY τ , bm(x, y1, . . . , yN ) =

∑σ,τ

bστm xσY τ ,

then coefficients cαjm of ym =∑α,j c

αjm x

αtj depends polynomially on coefficients ofaiml and bm. The dependance is defined inductively in the following way:

cα0m = 0,

cαj+1m =

1

j + 1

∑i,l

∑µ+ν=αg+h=j

Paiml(µ,g) · (νi + 1)c(ν+1i)h

m + P bm(α,j)

,

where

Paiml(α,j)(c

βλk ) = P(α,j)

((ai;στml

)|σ|+|τ |6|α|+j

,(cβλk

)β6α,λ6j,k6N

),

P bm(α,j)(cβλk ) = P(α,j)

((bστm )|σ|+|τ |6|α|+j ,

(cβλk

)β6α,λ6j,k6N

)



and P(α,j) are polynomials in (X(σ,τ))|σ|+|τ |6|α|+j and (Yk(β,λ))k6N,β6α,λ6j definedinductively by the following conditions:

1. P(0,0) = X(0,0),2. P(α+1p,j) =

=1

αp + 1

[ ∑|σ|+|τ |6|α|+j

1

σ!τ !

∂P(α,j)

∂X(σ,τ)·

(n−1∑k=1

X(σ+1k,τ)Yk(1p,0) +N∑k=1

X(σ,τ+1k)Yk(1p,0)

)

+∑

k6n−1+Nβ6α,λ6j

1

β!λ!

∂P(α,j)

∂Yk(β,λ)Yk(β+1p,λ)

]

and

P(α,j+1) =

=1

j + 1

[ ∑|σ|+|τ |6|α|+j

1

σ!τ !

∂P(α,j)

∂X(σ,τ)·

(n−1∑k=1

X(σ+1k,τ)Yk(0,1) +

N∑k=1

X(σ,τ+1k)Yk(0,1)

)

+∑

k6n−1+Nβ6α,λ6j

1

β!λ!

∂P(α,j)

∂Yk(β,λ)Yk(β,λ+1)

].

The theorem will be preceded by two lemmas.

Lemma 2.6. Let f(x) =∑α∈Nn aαx

α be an analytic function in a neighbour-hood of 0 ∈ Rn and let gk(ξ) =

∑β∈Nm bk;βξ

β, k = 1, . . . , n, be an analyticfunctions in a neighbourhood of 0 ∈ Rm such that gk(0) = 0. Then the func-tion F (ξ) = f(g1(ξ), . . . , gn(ξ)) is analytic in a neighbourhood of 0 ∈ Rm, and it’sTaylor expansion takes a form

F (ξ) =∑γ∈Nm

Pγ((aα)|α|6|γ|, (bk;β)β6γ,k6n

)ξγ ,

where Pγ ∈ Z[(Xα)|α|6|γ|, (Ykβ)β6γ,16k6n], γ ∈ Nm are polynomials with non-negative integer coefficients defined by the following induction conditions:

(1) P0(X0, Y10, . . . , Yn0) = X0, where X0 = X(0,...,0) and 0 = (0, . . . , 0) ∈ Nn.(2) If the polynomial Pγ = Pγ

((Xα)|α|6|γ|, (Ykβ)β6γ,16k6n

), then the polyno-

mial Pγ+1j is of the form

Pγ+1j = Pγ+1j

((Xα)|α|6|γ|+1, (Ykβ)β6γ+1j ,16k6n

),

where

Pγ+1j =1

γj + 1

∑|α|6|γ|

(1

α!

∂Pγ∂Xα

·n∑k=1

Xα+1kYk1j

)+∑β6γ

n∑k=1

1

β!

∂Pγ∂Ykβ

· Ykβ+1j

.



Proof. Obviously,

(2.11) aα =∂αf(0)

α!, bk;β =

∂βgk(0)

β!for 1 6 k 6 n.

Let F (ξ) =∑γ cγξ

γ . Then

(2.12) cγ =∂γF (0)

γ!.

Let 1α = (δκ)κ∈Nn , where δα = 1 and δκ = 0 for κ 6= α. Clearly

∂γ =m∏j=1

(∂1j)γj

.

The above lemma arises from the fact that

∂1jF =n∑k=1

∂1kf · ∂1jgk

by induction on |γ|. Indeed, it suffices to show that, for every γ there existsa polynomial Qγ with variables Xα, |α| 6 |γ| and Ykβ , 1 6 k 6 n, β 6 γ withnon-negative integer coefficients such that

(2.13) ∂γF (ξ) = Qγ

((∂αf(g(ξ)))|α|6|γ| ,

(∂βgk(ξ)

)β6γ,16k6n

)and degQγ 6 |γ|+ 1. For |γ| = 0 that is, for γ = 0 we have

∂0F (ξ) = f(g1(ξ), . . . , gn(ξ)),

hence we set Q0(X0, Y10, . . . , Yn0) = X0, where degQ0 = 1. If (2.13) holds for|γ| = p, then for |γ| = p + 1 multi-index γ can be written as γ = γ + 1j , where|γ| = p for some j ∈ 1, . . . ,m. Therefore, induction hypothesis implies

∂γF (ξ) = ∂1j∂γF (ξ) =∑|α|6|γ|

(∂1αQγ ·

n∑k=1

∂α+1kf (g(ξ)) ∂1jgk(ξ)

)+

+∑β6γ

n∑k=1

(∂1βQγ · ∂β+1jgk(ξ)

).

The right-hand side of this equation is a polynomial with non-negative integercoefficients of variables (∂αf(g(ξ)))|α|6|γ| and

(∂βgk(ξ)

)β6γ,k6n

. It’s degree is6 |γ|+ 1. Thus, it is a searched polynomial Qγ for |γ| = p+ 1. Induction ends theabove reasoning. By (2.13),(2.11) and (2.12) we obtain

cγ =1

γ!Qγ

((∂αf(0))|α|6|γ| ,

(∂βgk(0)

)β6γ,k6n

)=

=1

γ!Qγ((α!aα)|α|6|γ|, (β!bk;β)β6γ,k6n

).



Then the right-hand side of the above formula is a searched polynomial Pγ . More-over, from the above formula, we can easily read the inductive conditions describingpolynomial Pγ of variables (Xα)|α|6|γ|, (Ykβ)β6γ,k6n:

P0 = X0,

Pγ+1j =1

γj + 1

∑|α|6|γ|

(1

α!

∂Pγ∂Xα

·n∑k=1

Xα+1kYk1j

)+∑k6nβ6γ

1

β!

∂Pγ∂Ykβ

· Ykβ+1j

because Pγ

((Xα)|α|6|γ|, (Ykβ)β6γ,k6n

)= 1

γ!Qγ((α!Xα)|α|6|γ|, (β!Yk;β)β6γ,k6n

)and

Qγ+1j

((α!Xα)|α|6|γ|+1, (β!Ykβ)β6γ+1j ,k6n

)=∑|α|6|γ|

(γ!

α!

∂Pγ∂Xα

·n∑k=1

Xα+1kYk1j

)+

+∑k6nβ6γ

γ!

β!

∂Pγ∂Ykβ

· Ykβ+1j .

We say that a power series∑aα(x−x0)α with non-negative coefficients majorize

power series∑bα(x − x0)α, if |bα| 6 aα for every multi-index α. In this case the

series∑bα(x− x0)α is absolutely convergent everywhere the series

∑aα(x− x0)α

is absolutely convergent. We say that the series a =∑aα(x− x0)α is a majorant

of series b =∑bα(x − x0)α and we write a b, after Poincaré. Similarly, for

A = [ai]i∈I and B = [bi]i∈I symbol A B means that ai bi for every i ∈ I.

Lemma 2.7. Suppose that the series∑aαx

α is convergent in

TR = x :n

maxj=1|xj | < R.

Then for every positive number r < R end every M > sup|aα|r|α| : α ∈ Nn, thegeometric series ∑

α∈Nn

M |α|!α!r|α|

xα

is convergent in Tr/n = x : max |xj | < r/n to the function

Tr/n 3 x 7→Mr

r − (x1 + . . .+ xn)∈ R

and majorize series∑aαx

α.

Proof. Let r be a positive number less than R. Then the series∑aαr|α| is conver-

gent and for every M > 0 such that |aαr|α|| 6M for all α we have

|aα| 6M

r|α|6M |α|!α!r|α|

.



On the other hand, function

f(x) =Mr

r − (x1 + . . .+ xn)

is analytic in Tr/n and for x ∈ Tr/n

f(x) = M∞∑k=0

(x1 + . . .+ xn)k

rk=∑|α|>0

M |α|!α!r|α|

xα.

This ends the proof.

Let’s move on to the proof of the Theorem 2.5.

Proof of Theorem 2.5. We are looking for the solution Y = (y1, . . . , yN ) of theCauchy problem (2.10), where

(2.14) ym =∑α,j

cαjm xαtj for 1 6 m 6 N.

Obviously,

cαjm =∂αx ∂

jt ym(0, 0)

α!j!.

The initial conditions implies that cα0m = 0 for every α andm. In order to determine

the coefficients cαjm for j > 0, we substitute (2.14) to the differential equations

(2.15) ∂tym =∑i,l

aiml(x, y1, . . . , yN )∂xiyl + bm(x, y1, . . . , yN ).

Let

(2.16)

aiml(x, y1, . . . , yN ) =∑σ,τ

ai;στml xσY τ ,

bm(x, y1, . . . , yN ) =∑σ,τ

bστm xσY τ ,

∂xiyl =∑α,j

(αi + 1)c(α+1i)jl xαtj .

Lemma 2.6 implies that aiml is a power series in x and t, whose coefficients of xαtj

are polynomials with non-negative rational coefficients in (ai;στml )|σ|+|τ |6|α|+j and(cβλk )β6α,λ6j,k6N . Moreover, the coefficients of the terms in which t occurs to thej-th power only involve the cβλk with λ 6 j. The same is true for the series obtainedfrom bm and ∂xiyl, and multiplying aiml by ∂xiyl still preserves these properties.

Roughly speaking, on the right side of (2.15) we obtain an expression of the form∑α,j

Pαjm

((ai;στml , b

στm

)|σ|+|τ |6|α|+j,i6n−1,l6N

,(cβλk

)β6α,λ6j,k6N

)xαtj ,



where Pαjm is a polynomial with non-negative coefficients. On the left side, we have

∂tym =∑α,j

(j + 1)cα(j+1)m xαtj .

Hence,

cα(j+1)m =

Pαjm

((ai;στml , b

στm

)|σ|+|τ |6|α|+j,i6n−1,l6N

,(cβλk

)β6α,λ6j,k6N

)j + 1

,

so if we know that cβλk with λ 6 j, we can determine the cβλk with λ = j + 1.Proceeding inductively, we determine all the cαjm and we find that

cαjm = Qαjm

((ai;στml , b

στm

)|σ|+|τ |6|α|+j,i6n−1,l6N

,(cβλk

)β6α,λ<j,k6N

)=

= Qαjm

(ai;στml , b

στm , cβλk

),

where Qαjm is a polynomial with nonnegative coefficients in cβλk , where λ < j. Moreprecisely the coefficients of xαtj of power series aiml i bm are polynomials withnon-negative rational coefficients of the form:

Paiml(α,j)(c

βλk ) = P(α,j)

((ai;στml

)|σ|+|τ |6|α|+j

,(cβλk

)β6α,λ6j,k6N

),

P bm(α,j)(cβλk ) = P(α,j)

((bστm )|σ|+|τ |6|α|+j ,

(cβλk

)β6α,λ6j,k6N

),

where P(α,j) is a polynomial in (X(σ,τ))|σ|+|τ |6|α|+j and (Yk(β,λ))k6N,β6α,λ6j de-fined inductively by the following conditions:

1. P(0,0) = X(0,0),2. P(α+1p,j) =

=1

αp + 1

[ ∑|σ|+|τ |6|α|+j

1

σ!τ !

∂P(α,j)

∂X(σ,τ)·

(n−1∑k=1

X(σ+1k,τ)Yk(1p,0) +N∑k=1

X(σ,τ+1k)Yk(1p,0)

)

+∑

k6n−1+Nβ6α,λ6j

1

β!λ!

∂P(α,j)

∂Yk(β,λ)Yk(β+1p,λ)

]



and

P(α,j+1) =

=1

j + 1

[ ∑|σ|+|τ |6|α|+j

1

σ!τ !

∂P(α,j)

∂X(σ,τ)·

(n−1∑k=1

X(σ+1k,τ)Yk(0,1) +N∑k=1

X(σ,τ+1k)Yk(0,1)

)

+∑

k6n−1+Nβ6α,λ6j

1

β!λ!

∂P(α,j)

∂Yk(β,λ)Yk(β,λ+1)

].

Therefore,

Pαjm

((ai;στml , b

στm

)|σ|+|τ |6|α|+j,i6n−1,l6N

,(cβλk

)β6α,λ6j,k6N

)=

=∑i,l

∑µ+ν=αg+h=j


m + P bm(α,j),

thereby cα0m = Qα0

m = 0 and

cαj+1m = Qαj+1

m

(ai;στml , b

στm , cβλk

)=

=1

j + 1

∑i,l

∑µ+ν=αg+h=j


m + P bm(α,j)

.

Now, to show convergence of Y , it suffices to find the Cauchy problem∂tY =n−1∑i=1

Ai(x, Y )∂xi Y + B(x, Y ),

Y (x, 0) = 0,

(where Ai and B are analytic equivalents of Ai and B respectively), for which:

a) there exists the analytic solution Y nearby (0, 0);

b) Ai Ai and B B.

Indeed, the solution Y = (y1, . . . , yN ) of this problem has the form ym =∑cαjm x

αtj , m = 1, . . . , N , where

cαjm = Qαjm

(ai;στml , b

στm , cβλk

),

and Qαjm are polynomials defined for the preceding Cauchy problem. Since Qαjm hasnon-negative coefficients and depends only on cβλk , where λ < j, then we can easily



show by induction that:

|cαjm | =∣∣Qαjm (ai;στml , b

στm , cβλk

) ∣∣ 6 Qαjm (|ai;στml |, |bστm |, |c

βλk |)

6 Qαjm(ai;στml , b

στm , cβ,λk

)= cαjm .

Therefore Y majorize Y which gives convergence of Y in some neighbourhood of(0, 0).

We will construct such a majorizing system. LetM > 0 be sufficiently large andr > 0 sufficiently small so that by Lemma 2.7 series for Ai and B are all majorizedby the series for

Mr

r − (x1 + . . .+ xn−1)− (y1 + . . .+ yN ).

Thus we consider the following Cauchy problem: for m = 1, . . . , N,

(2.17)

∂tym = Mrr−

∑xi−

∑yl

(∑i

∑l

∂xiyl + 1

),

ym(x, 0) = 0.

To determine the solution of this Cauchy problem it is enough to solve theCauchy problem consisting of one equation

(2.18)

∂tu = Mr

r−s−Nu (N(n− 1)∂su+ 1) ,

u(s, 0) = 0,

for if we will put

yj(x, t) = u(x1 + . . .+ xn−1, t) (j = 1, . . . , N),

we obtain that Y = (y1, . . . , yN ) satisfies (2.17). We will transform (2.18) to

(r − s−Nu)∂tu−MrN(n− 1)∂su = Mr,

and will solve this by method of characteristics:dt

dτ= r − s−Nu, ds

dτ= −MrN(n− 1),

du

dτ= Mr

with the initial conditions:

t(0) = 0, s(0) = σ, u(0) = 0.

The solution of the above is given by the formulas:

t =1

2MrN(n− 2)r2 + (r − σ)τ, s = −MrN(n− 1)τ + σ, u = Mrτ.

The elimination of σ and τ yields

u(s, t) =r − s−

√(r − s)2 − 2MrNnt

Mn.

Clearly this is analytic for s and t near 0, so the proof is complete.



There remains the question of whether the Cauchy problem (2.2) might admitnon-analytic solutions as well. In the linear case, the answer is negative: this is theHolmgren uniqueness theorem. The proof can be found in John [5], Hörmander [6],[7, vol I], or Treves [14].

A major drawback of the Cauchy-Kowalevski theorem is that it gives little con-trol over the dependence of the solution on the Cauchy data.

Example 2.8. Consider the following example in R2, due to Hadamard:∂2

1u+ ∂22u = 0

u(x1, 0) = 0, ∂2u(x1, 0) = ke−√k sin kx1,

where k > 0. One easily checks that the solution is

u(x1, x2) = e−√k(sin kx1)(sinh kx2).

As k → ∞, the Cauchy data and their derivatives of all orders tend uniformlyto zero since e−

√k decays faster than polynomially. But if x2 > 0, then

limk→∞

e−√k sinh kx2 = +∞.

The solution for the limiting case k = +∞ is of course u ≡ 0. This exampleshows that the solution of the Cauchy problem may not depend continuously onthe Cauchy data in most of the usual topologies on functions.

Acknowledgement. We would like to thank Professor Stanisław Spodzieja formany talks and valuable advice.

References

[1] R.P. Brent and H.T. Kung, Fast algorithms for manipulating formal power series, J. Assoc.Comput. Mach. 25 (1978), no. 4, 581–595

[2] L.C. Evans, Partial differential equations, AMS Press, 2010.[3] G.B. Folland, Introduction to partial differential equations, Princeton University Press, N. J.,

1975.[4] K.O. Geddes, Convergence behaviour of the Newton iteration for first-order differential equa-

tions. Symbolic and algebraic computation, pp. 189–199, Lecture Notes in Comput. Sci., 72,Springer, Berlin-New York, 1979.

[5] F. John, Parial Differential Equations (4th ed.), Springer-Verlag, New York, 1982.[6] L. Hörmander, Linear Parial Differential Operators, Springer-Verlag, Berlin, 1963.[7] L. Hörmander, The Analysis of Linear Partial Differential Operators (4 vols.), Springer-

Verlag, Berlin, 1983-85.[8] E. Hubert and N. Le Roux, Computing power series solutions of a nonlinear PDE sys-

tem, Proceedings of the International Symposium on Symbolic and Algebraic Computation,J.R. Sendra, ed., ACM Press, New York, 2003, pp. 148–155.

[9] S. von Kowalevsky, Zur Theorie der partiellen Differentialgleichung, J. Reine Angew. Math.80 (1875), 1–32, http://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002156059(accessed March 15, 2017).

[10] H. Marcinkowska, Wstęp do teorii równań różniczkowych cząstkowych, PWN, 1972.[11] C. Riquier, Les systèmes d’équations aux dérivées partielles, Gauthier-Villars, Paris, 1910.



[12] J.F. Ritt, Differential Algebra. American Mathematical Society Colloquium Publications,Vol. XXXIII, American Mathematical Society, New York, N. Y., 1950. viii+184 pp.

[13] C.J. Rust, G.J. Reid and A.D. Wittkopf, Existence and uniqueness theorems for formalpower series solutions of analytic differential systems, Proceedings of the 1999 InternationalSymposium on Symbolic and Algebraic Computation (Vancouver, BC), 105–112 (electronic),ACM, New York, 1999.

[14] F. Treves, Basic Linear Partial Differential Equations, Academic Press, New York, 1975.

Faculty of Mathematics and Computer Science, University of Łódź,ul. S. Banacha 22, 90-238 Łódź, Poland

E-mail address, Beata Osińska-Ulrych: [email protected]

E-mail address, Grzegorz Skalski: [email protected]

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FORMAL AND CONVERGENT SOLUTIONS OF ANALYTIC

EQUATIONS

ARKADIUSZ P LOSKI

Impressed by the power of the Preparation Theorem – indeed, it prepares usso well! – I considered “Weierstrass Preparation Theorem and its immediateconsequences” as a possible title for the entire book.

Sheeram S. Abhyankar, Preface to [1]

Abstract. We provide the detailed proof of a sharpened version of theM. Artin Approximation Theorem.

1. Introduction

The famous Approximation Theorem of M. Artin [2] asserts that any formalsolution of a system of analytic equations can be approximated by convergentsolutions up to a given order. In my PhD thesis [7] I was able by analysis ofthe argument used in [2] to sharpen the Approximation Theorem: any formalsolution can be obtained by specializing parameters in a convergent parametricsolution. The theorem was announced with a sketch of proof in [8]. The aimof theses notes is to present the detailed proof of this result. It is based on theWeierstrass Preparation Theorem. The other tools are: a Jacobian Lemma whichis an elementary version of the Regularity Jacobian Criterion used in [2], the trickof Kronecker (introducing and specializing variables) and a generalization of theImplicit Function Theorem due to Bourbaki [4] and Tougeron [10]. All thesesingredients are vital in the proofs of some other results of this type (see [3], [11]).

2010 Mathematics Subject Classification. 14B12.

Key words and phrases. Formal solution, parametric solution.This article is an updated version of my notes “Les solutions formelles et convergentes des

equations analytiques” prepared for the Semestre de Singularites a Lille, 1999.

161

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162 A. P LOSKI

For more information on approximation theorems in local analytic geometry werefer the reader to Teissier’s article [9] and to Chapter 8 of the book [5].

Let K be a field of characteristic zero with a non-trivial valuation. We putK[[x]] = K[[x1, . . . , xn]] the ring of formal power series in variables x = (x1, . . . , xn)with coefficients in K. If f =

∑k≥p fk is a nonzero power series represented as the

sum of homogeneous forms with fp 6= 0 then we write ord f = p. Additionally weput ord 0 = +∞ and use the usual conventions on the symbol +∞. The constantterm of any series f ∈ K[[x]] we denote by f(0). A power series u ∈ K[[x]] isa unit if uv = 1 for a power series v ∈ K[[x]]. Note that u is a unit if and only ifu(0) 6= 0. The non-units of K[[x]] form the unique maximal ideal mx of the ringK[[x]]. The ideal mx is generated by the variables x1, . . . , xn. One has f ∈ mc

x,where c > 0 is an integer, if and only if ord f ≥ c. Recall that if g1, . . . , gn ∈ K[[y]],y = (y1, . . . , yn) are without constant term then the series f(g1, . . . , gn) ∈ K[[y]]is well-defined. The mapping which associates with f ∈ K[[x]] the power seriesf(g1, . . . , gn) is the unique homomorphism sending xi for gi for i = 1, . . . , n. LetKx be the subring of K[[x]] of all convergent power series. Then Kx is a localring. If g1, . . . , gn ∈ Ky then f(g1, . . . , gn) ∈ Ky for any f ∈ Kx.

In what follows we use intensively the Weierstrass Preparation and DivisionTheorems. The reader will find the basic facts concerning the rings of formal andconvergent power series in [1], [6] and [12].

Let f(x, y) = (f1(x, y), . . . , fm(x, y)) ∈ Kx, ym be convergent power series inthe variables x = (x1, . . . , xn) and y = (y1, . . . , yN ) where m,n,N are arbitrarynon-negative integers. The theorem quoted below is the main result of [2].

The Artin Approximation Theorem. Suppose that there exists a sequence offormal power series y(x) = (y1(x), . . . , yN (x)) without constant term such that

f(x, y(x)) = 0 .

Then for any integer c > 0 there exists a sequence of convergent power seriesy(x) = (y1(x), . . . , yN (x)) such that

f(x, y(x)) = 0 and y(x) ≡ y(x) (modmcx) .

The congruence condition means that the power series yν(x)−yν(x) are of order ≥ ci.e. the coefficients of monomials of degree < c agree in yν(x) and yν(x). We willdeduce the Artin Approximation Theorem from the following result stated witha sketch of proof in [8].

Theorem. With the notation and assumptions of the Artin theorem there existsa sequence of convergent power series y(x, t) = (y1(x, t), . . . , yN (x, t)) ∈ Kx, tN ,y(0, 0) = 0, where t = (t1, . . . , tS) are new variables, S ≥ 0, and a sequence offormal power series t(x) = (t1(x), . . . , tS(x)) ∈ K[[x]]S, t(0) = 0 such that

f(x, y(x, t)) = 0 and y(x) = y(x, t(x)) .

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FORMAL AND CONVERGENT SOLUTIONS OF ANALYTIC EQUATIONS 163

The construction of the parametric solution y(x, t) depends on the given formalsolution y(x). To get the Artin Approximation Theorem from the stated aboveresult fix an integer c > 0. Let y(x, t) and t(x) be series such as in the theoremand let t(x) = (t1(x), . . . , tS(x)) ∈ KxS be convergent power series such thatt(x) ≡ t(x) modmc

x. Therefore y(x, t(x)) ≡ y(x, t(x)) modmcx and it suffices to

set y(x) = y(x, t(x)).

Before beginning the proof of the theorem let us indicate two corollaries of it.

Corollary 1. Assume that m = N , f(x, y(x)) = 0 and

detJ(f1, . . . , fN )

J(y1, . . . , yN )(x, y(x)) 6= 0 .

Then the power series y(x) are convergent.

Proof. Let y(x, t) and t(x) be power series without constant term such thatf(x, y(x, t)) = 0 and y(x) = y(x, t(x)). It is easy to check by differentiation of equal-ities f(x, y(x, t)) = 0 that (∂yν/∂tσ)(x, t) = 0 for ν = 1, . . . , N and σ = 1, . . . , S.Therefore the series y(x, t) are independent of t and the series y(x) are convergent.

Corollary 2. If f(x, y) ∈ Kx, y is a nonzero power series of n + 1 variables(x, y) = (x1, . . . , xn, y) and y(x) is a formal power series without constant termsuch that f(x, y(x)) = 0 then y(x) is a convergent power series.

Proof. By Corollary 1 it suffices to check that there exists a power series g(x, y) ∈Kx, y such that g(x, y(x)) = 0 and (∂g/∂y)(x, y(x)) 6= 0. Let I = g(x, y) ∈Kx, y : g(x, y(x)) = 0. Then I 6= Kx, y is a prime ideal of Kx, y. Assumethe contrary, that is, that for every g ∈ I: (∂g/∂y) ∈ I. Then we get by differ-entiating the equality g(x, y(x)) = 0 that (∂g/∂xi) ∈ I for i = 1, . . . , n and, byinduction, all partial derivatives of g lie in I. Consequently g = 0 for every g ∈ Ii.e. I = (0). A contradiction since 0 6= f ∈ I.

2. Reduction to the case of simple solutions

We keep the notation introduced in Introduction. We will call a sequence offormal power series y(x) ∈ K[[x]], y(0) = 0 a simple solution of the system ofanalytic equations f(x, y) = 0 if f(x, y(x)) = 0 and

rankJ(f1, . . . , fm)

J(y1, . . . , yN )(x, y(x)) = m .

Thus, in this case, m ≤ N .

In what follows we need

The Jacobian Lemma. Let I be a nonzero prime ideal of the ring Kx,x = (x1, . . . , xn). Then there exist an integer r: 1 ≤ r ≤ n and covergent powerseries h1, . . . , hr ∈ I such that

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164 A. P LOSKI

(i) rankJ(h1, . . . , hr)

J(x1, . . . , xn)(mod I) = r ,

(ii) ∀h ∈ I, ∃a /∈ I such that ah ∈ (h1, . . . , hr)Kx.

Before proving the above lemma let us note that it is invariant with respect toK-linear nonsingular transformations. If Φ is an authomorphism of Kx definedby

Φ(f(x1, . . . , xn)) = f

n∑j=1

c1jxj , . . . ,n∑j=1

cnjxj

with det(cij) 6= 0 then the Jacobian Lemma is true for I if and only if it is true forΦ(I).

Proof of the Jacobian Lemma (by induction on the number n of variables xi). Ifn = 1 then I = (x1)Kx1 and h1 = x1. Suppose that n > 1 and that the lemma istrue for prime ideals of the ring of power series in n− 1 variables. Using a K-linearnonsingular transformation we may assume that the ideal I contains a power seriesxn-regular of order k > 0 i.e. such that the term xkn appears in the power serieswith a non-zero coefficients. Therefore, by the Weierstrass Preparation Theorem Icontains a distinguished polynomial

w(x′, xn) = xkn + a1(x′)xk−1n + · · ·+ ak(x′), where x′ = (x1, . . . , xn−1) .

By the Weierstrass Division Theorem every power series h = h(x) is of the form

h(x) = q(x)w(x′, xn)+r(x′, xn) where r(x′, xn) is an xn-polynomial (of degree < k).Therefore, the ideal I is generated by the power series which are polynomials in xnand to prove the Jacobian Lemma it suffices to find power series h1, . . . , hr suchthat (i) holds and (ii) is satisfied for h ∈ I ∩Kx′[xn].

Let I ′ = I ∩Kx′ and consider the set I \ I ′[xn]. Clearly w(x′, xn) ∈ I \ I ′[xn].Let

h1(x′, xn) = c0(x′)xln + c1(x′)xl−1n + · · ·+ cl(x

′)

be a polynomial in xn of the minimal degree l, l ≥ 0, which belongs to I \ I ′[xn].Since the degree l ≥ 0 is minimal, we have

l > 0 ,

c0(x′) /∈ I ′ ,∂h1

∂xn∈ I .

Let h(x′, xn) ∈ I be a polynomial in xn. Dividing h(x′, xn) by h1(x′, xn) (Euklid’s

division) we get

(E) c0(x′)p h(x′, xn) = q(x′, xn)h1(x′, xn) + r1(x′, xn),

where xn-degree of r1(x′, xn) is less than l and p ≥ 0 is an integer. Since thexn-degree of r1(x′, xn) is < l then all coefficients of r1(x′, xn) lie in I ′. If I ′ = (0)then r1(x′, xn) = 0 and (E) proves the Jacobian Lemma.

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If I ′ 6= (0) then by the induction hypothesis there exists series h2, . . . , hr ∈ I ′such that

(i′) rankJ(h2, . . . , hr)

J(x1, . . . , xn−1)(mod I ′) = r − 1 ,

(ii′) ∀h′ ∈ I ′, ∃a′ /∈ I ′ such that a′h′ ∈ (h2, . . . , hr)Kx′.

We claim that h1, . . . , hr satisfy (i) and (ii) of the Jacobian Lemma. To check (i)observe that

detJ(h1, . . . , hr)

J(xi1 , . . . , xir−1 , xn)= det

J(h2, . . . , hr)

J(xi1 , . . . , xir−1)· ∂h1

∂xn,

where i1, . . . , ir−1 ∈ 1, . . . , n− 1 and use (i′). Applying (ii′) to the coefficients ofr1(x′, xn) we find a power series a′(x′) such that a′(x′) r1(x′, xn) ∈ (h2, . . . , hr)Kx.By (E) we get a(x′)h(x′, xn) ∈ (h1, . . . , hr)Kx where a(x′) = a′(x′) c0(x′)p /∈ Iwhich proves (ii).

Now, we can check

Proposition 2.1. Let f(x, y) = (f1(x, y), . . . , fm(x, y)) ∈ Kx, ym, f(x, y) 6= 0,y(x) = (y1(x), . . . , yN (x)) ∈ K[[x]], y(0) = 0, be formal power series such thatf(x, y(x)) = 0. Then there exist convergent power series h(x, y) = (h1(x, y), . . . ,hr(x, y)) ∈ Kx, yr such that

(i) h(x, y(x)) = 0,

(ii) rankJ(h1, . . . , hr)

J(y1, . . . , yN )(x, y(x)) = r ,

(iii) suppose that there exist formal power series y(x, t) = (y1(x, t), . . . , yN (x, t)),y(0, 0) = 0 and t(x) = (t1(x), . . . , tS(x)), t(0) = 0, such that h(x, y(x, t)) =0 and y(x) = y(x, t(x)). Then f(x, y(x, t)) = 0.

Proof. Consider the prime ideal

I = g(x, y) ∈ Kx, y : g(x, y(x)) = 0 .

Clearly f1(x, y), . . . , fm(x, y) ∈ I and I 6= (0). By the Jacobian Lemma there existformal power series h1(x, y), . . . , hr(x, y) ∈ I such that

• rankJ(h1, . . . . . . . . . , hr)

J(x1, . . . , xn, y1, . . . , yN )(x, y(x)) = r ,

• ∀g ∈ I, ∃a /∈ I such that a(x, y) g(x, y) ∈ (h1, . . . , hr)Kx, y.

We claim that h1, . . . , hr satisfy the conditions (i), (ii), (iii). Condition (i) holdssince h1, . . . , hr ∈ I. To check (ii) it suffices to observe that

(J) rankJ(h1, . . . . . . . . . , hr)

J(x1, . . . , xn, y1, . . . , yN )(x, y(x)) = rank

J(h1, . . . , hr)

J(y1, . . . , yN )(x, y(x)) .

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166 A. P LOSKI

Indeed, differentiating the equations hi(x, y(x)) = 0, i = 1, . . . , r, we get

∂hi∂xj

(x, y(x)) +N∑ν=1

∂hi∂yν

(x, y(x))∂yν∂xj

= 0 for j = 1, . . . , n

and (J) follows. To check (iii) let us write

ai(x, y)fi(x, y) =r∑

k=1

ai,k(x, y)hk(x, y) in Kx, y ,

where ai(x, y) /∈ I for i = 1, . . . ,m. Thus ai(x, y(x)) 6= 0 and ai(x, y(x, t)) 6= 0since y(x) = y(x, t(x)) and (iii) follows.

3. The Bourbaki–Tougeron implicit function theorem

Let f(x, y) = (f1(x, y), . . . , fm(x, y)) ∈ Kx, ym be convergent power series invariables x = (x1, . . . , xn) and y = (y1, . . . , yN ). Suppose that m ≤ N and put

J(x, y) =J(f1, . . . , fm)

J(yN−m+1, . . . , yN )and δ(x, y) = detJ(x, y) .

Let M(x, y) be the adjoint of the matrix J(x, y). Thus we have

M(x, y)J(x, y) = J(x, y)M(x, y) = δ(x, y)Im

where Im is the identity matrix of m rows and m columns. Let g(x, y) = (g1(x, y),. . . , gm(x, y)) ∈ Kx, ym be convergent power series defined by

g1(x, y)...

gm(x, y)

= M(x, y)

f1(x, y)

...

fm(x, y)

.

It is easy to see that

(a) gi(x, y) ∈ (f1(x, y), . . . , fm(x, y))Kx, y for i = 1, . . . ,m

and

(b) δ(x, y)fi(x, y) ∈ (g1(x, y), . . . , gm(x, y))Kx, y for i = 1, . . . ,m.

Now, we can state

The Bourbaki-Tougeron implicit function theorem. Suppose that there ex-ists a sequence of formal power series y0(x) = (y0

1(x), . . . , y0N (x)), y0(0) = 0, such

that

gi(x, y0(x)) ≡ 0 mod δ(x, y0(x))2mx for i = 1, . . . ,m .

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Then

I. Let yν(x, t) = y0ν(x) + δ(x, y0(x))2tν for ν = 1, . . . , N − m where t =

(t1, . . . , tN−m) are new variables. Then there exists a unique sequence offormal power series u(x, t) = (uN−m+1(x, t), . . . , uN (x, t)) ∈ K[[x, t]]m,u(0, 0) = 0, such that if we let yν(x, t) = y0

ν(x) + δ(x, y0(x))uν(x, t) forν = N −m+ 1, . . . , N and y(x, t) = (y1(x, t), . . . , yN (x, t)) then

f(x, y(x, t)) = 0 in K[[x, t]] .

If the series y0(x) are convergent then u(x, t) and y(x, t) are covergent aswell.

II. For every sequence of formal power series y(x) = (y1(x), . . . , yN (x)), y(0) =0, the following two conditions are equivalent

(i) there exists a sequence of formal power series t(x) = (t1(x), . . . ,tN−m(x)), t(0) = 0, such that y(x) = y(x, t(x)),

(ii) f(x, y(x)) = 0 and

yν(x)≡ y0ν(x) mod δ(x, y0(x))2mx for ν = 1, . . . , N −m

yν(x)≡ y0ν(x) mod δ(x, y0(x))mx for ν = N −m+ 1, . . . , N .

Remark. In what follows we call

y0(x) an approximate solution of the system f(x, y) = 0,

y(x, t) a parametric solution determined by the approximate solution y0(x)

y(x) satifying (i) or (ii) a subordinate solution to the approximate solution

y0(x)

Proof. Let v = (v1, . . . , vN ) and h = (h1, . . . , hn) be variables. Taylor’s formulareads

f1(x, v + h)...

fm(x, v + h)

=

f1(x, v)

...

fm(x, v)

+J(f1, . . . , fm)

J(y1, . . . , yN−m)(x, v)

h1

...

hN−m

+ J(x, v)

hN−m+1

...

hN

+

P1(u, v, h)

...

Pm(u, v, h)

(T)

where Pi(x, v, h) ∈ (h1, . . . , hN )2Kx, v, h for i = 1, . . . ,m. Let u = (uN−m+1, . . . ,uN ) be variables and put

Fi(x, t, u) = fi(x, y1(x, t), . . . , yN−m(x, t), y0N−m+1(x) + δ(x, y0(x))uN−m+1,

. . . , y0N (x) + δ(x, y0(x))uN ) .

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168 A. P LOSKI

Substituting in Taylor’s formula (T) vi = y0i (x) for i = 1, . . . , N , hi = δ(x, y0(x))2ti

for i = 1, . . . , N −m and hi = δ(x, y0(x))ui for i = N −m+ 1, . . . , N we getF1(x, t, u)

...

Fm(x, t, u)

=

f1(x, y0(x))

...

fm(x, y0(x))

+ δ(x, y0(x))2 J(f1, . . . , fm)

J(y1, . . . , yN−m)(x, y0(x))

t1...

tN−m

+ δ(x, y0(x))J(x, y0(x))

uN−m+1

...

uN

+ δ(x, y0(x))2

Q1(x, t, u)

...

Qm(x, t, u)

where Qi(x, t, u) ∈ (t, u)2Kx, t, u for i = 1, . . . ,m. Multiplying the above identityby the matrix M(x, y0(x)) and taking into account that M(x, y0(x))J(x, y0(x)) =δ(x, y0(x))Im and gi(x, y

0(x)) ≡ 0 (mod δ(x, y0(x))2mx) for i = 1, . . . ,m, we get

(*) M(x, y0(x))

F1(x, t, u)

...

Fm(x, t, u)

= δ(x, y0(x))2

G1(x, t, u)

...

Gm(x, t, u)

where Gi(0, 0, 0) = 0 for i = 1, . . . ,m. Differentiating (*) we obtain

M(x, y0(x))J(F1, . . . , Fm)

J(uN−m+1, . . . , uN )(x, t, u) = δ(x, y0(x))2 J(G1, . . . , Gm)

J(uN−m+1, . . . , uN )(x, t, u)

which implies

(**) detJ(F1, . . . , Fm)

J(uN−m+1, . . . , uN )(x, t, u)

= δ(x, y0(x))m+1detJ(G1, . . . , Gm)

J(uN−m+1, . . . , uN )(x, t, u)

since detM(x, y0(x)) = δ(x, y0(x))m−1. On the other hand

J(F1, . . . , Fm)

J(uN−m+1, . . . , uN )(x, 0, 0) = δ(x, y0(x))J(x, y0(x))

and

detJ(F1, . . . , Fm)

J(uN−m+1, . . . , uN )(x, 0, 0) = δ(x, y0(x))m+1 .

Therefore we get from (**)

detJ(G1, . . . , Gm)

J(uN−m+1, . . . , uN )(x, 0, 0) = 1 ,

in particular

detJ(G1, . . . , Gm)

J(uN−m+1, . . . , uN )(0, 0, 0) = 1 .

By the Implicit Function Theorem there exist formal power series

u(x, t) = (uN−m+1(x, t), . . . , uN (x, t))

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such that

(G1(x, t, u), . . . , Gm(x, t, u))K[[x, t, u]]

= (uN−m+1 − uN−m+1(x, t), . . . , uN − uN (x, t))K[[x, t, u]] .

If y0(x) are convergent then G(x, t, u) and u(x, t) are convergent as well. In particu-lar G(x, t, u(x, t)) = 0 and by (*) F (x, t, u(x, t)) = 0 which implies f(x, y(x, t)) = 0where yν(x, t) = y0

ν(x) + δ(x, y0(x))uν(x, t) for ν = N − m + 1, . . . , N . Letu(x, t) = (uN−m+1(x, t), . . . , uN (x, t)), u(0, 0) = 0, be power series such thatf(x, y(x, t)) = 0 where

y(x, t) = (y1(x, t), . . . , yN−m(x, t), y0N−m+1(x) + δ(x, y0(x))uN−m+1(x, t),

. . . , y0N (x) + δ(x, y0(x))uN (x, t)) .

Then F (x, t, u(x, t)) = 0 and by (*) G(x, t, u(x, t)) = 0. Thus we get u(x, t) =u(x, t). This proves the first part of the Bourbaki-Tougeron Implicit FunctionTheorem. To check the second part it suffices to observe that for any formal powerseries t(x) = (t1(x), . . . , tN−m(x)) and u(x) = (uN−m+1(x), . . . , uN (x)) withoutconstant term G(x, t(x), u(x)) = 0 if and only if u(x) = u(x, t(x)).

4. Approximate solutions

We keep the notions and assumptions of Section 3. Let x′ = (x1, . . . , xn−1).

Proposition 4.1. Let y(x) = (y1(x), . . . , yN (x)), y(0) = 0, be a formal solu-tion of the system of analytic equations f(x, y) = 0, such that the power seriesδ(x, y(x)) is xn-regular of strictly positive order p > 0. Then there exists an approx-imate solution v(x) ∈ K[[x′]][xn]N of the system f(x, y) = 0 and such that y(x) isa solution of f(x, y) = 0 subordinate to v(x).

Proof. By the Weierstrass Preparation Theorem δ(x, y(x)) = a(x) · unit where

a(x) = xpn +

p∑j=1

aj(x′)xp−jn

is a distinguished polynomial. Using the Weierstrass Division Theorem we get

yν(x) =

2p−1∑j=0

vν,j(x′)xjn + a(x)2(cν + tν(x)) for ν = 1, . . . , N −m

and

yν(x) =

p−1∑j=0

vν,j(x′)xjn + a(x)(cν + uν(x)) for ν = N −m+ 1, . . . , N

where cν ∈ K for ν = 1, . . . , N , while

t(x) = (t1(x), . . . , tN−m(x)) and u(x) = (uN−m+1(x), . . . , uN (x))

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170 A. P LOSKI

are formal power series without constant term. Let

vν(x) =

2p−1∑j=0

vν,j(x′)xjn + a(x)2cν for ν = 1, . . . , N −m

and

vν(x) =

p−1∑j=0

vν,j(x′)xjn + a(x)cν for ν = N −m+ 1, . . . , N .

Clearly v(x) = (v1(x), . . . , vN (x)) ∈ K[[x′]][xn]N .

Property 1. δ(x, v(x)) = a(x) · unit

Proof. From y(x) ≡ v(x) (mod a(x)mx) we get δ(x, y(x)) ≡ δ(x, v(x)) (mod a(x)mx)and Property 1 follows since δ(x, y(x)) = a(x) · unit.

Property 2. gi(x, v(x)) ≡ 0 (mod a(x)2mx) for i = 1, . . . ,m.

Proof. Substituting in Taylor’s formula (T) v = v(x), hν = a(x)2tν(x) for ν =1, . . . , N −m and hν = a(x)uν(x) for ν = N −m+ 1, . . . , N we get

0...

0

=

f1(x, v(x))

...

fm(x, v(x))

+ a(x)2 J(f1, . . . , fm)

J(y1, . . . , yN−m)(x, v(x))

t1(x)

...

tN−m(x)

+ a(x)J(x, v(x))

uN−m+1(x)

...

uN (x)

+ a(x)2

Q1(x)

...

Qm(x)

.

Multiplying the above identity by M(x, v(x)) and taking into account the formula

M(x, v(x))J(x, v(x)) = δ(x, v(x))Im

we get Property 2.

From Properties 1 and 2 it follows that

gi(x, v(x)) ≡ 0 (mod δ(x, v(x))2mx) for i = 1, . . . ,m

i.e. v(x) ∈ K[[x′]][xn] is an approximate solution of the system f(x, y) = 0. Since

yν(x) ≡ vν(x) mod δ(x, v(x))2mx for i = 1, . . . , N −m

and

yν(x) ≡ vν(x) mod δ(x, v(x))mx for i = N −m+ 1, . . . , N

y(x) is a subordinate solution to the approximate solution v(x).

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Proposition 4.2. Let (c0ν,j), ν = 1, . . . , N , j = 0, 1, . . . , D, be a family of constants

such that c0ν,0 = 0 for ν = 1, . . . , N . Suppose that D∑j=0

c01,jxjn, . . . ,

D∑j=0

c0N,jxjn

is an approximate solution of the system of equations f(0, xn, y) = 0 such that

ord δ

0, xn,D∑j=0

c01,jxjn, . . . ,

D∑j=0

c0N,jxjn

= p , 0 < p < +∞ .

Let V 0 = (V 0ν,j), ν = 1, . . . , N , j = 0, 1, . . . , D be variables. Then there exists

a sequence

F (x′, V 0) = (F1(x′, V 0), . . . , FM (x′, V 0)) ∈ Kx′, V 0M

such that for any family (v0ν,j(x

′)) of formal power series without constant term thefollowing two conditions are equivalent

(i)

D∑j=0

(c01,j + v01,j(x

′))xjn, . . . ,

D∑j=0

(c0N,j + v0N,j(x

′))xjn

is an approximate solution of the system f(x, y) = 0,

(ii) F (x′, (v0ν,j(x

′))) = 0 in K[[x′]].

Proof. Let

vν(xn) =

D∑j=0

(c0ν,j + V 0ν,j)x

jn, v(xn) = (v1(xn), . . . , vN (xn)) .

It is easy to check that δ(x, v(xn)) is xn-regular of order p. By the WeierstrassDivision Theorem

gi(x, v(xn)) = Qi(x, V0)δ(x, v(xn))2 +

2p−1∑j=0

Ri,j(x′, V 0)xjn

for i = 1, . . . ,m. Let

vν(x) =D∑j=0

(c0ν,j + v0ν,j)x

jn, v(x) = (v1(x), . . . , vN (x))

where (v0ν,j(x

′)) is a family of formal power series without constant term. Thus weget

gi(x, v(x)) = Qi(x, v(x))δ(x, v(x))2 +

2p−1∑j=0

Ri,j(x′, v0

ν,j(x′))xjn for i = 1, . . . ,m .

By the uniqueness of the remainder in the Weierstrass Division Theorem we havethat v(x) is an approximate solution of the system of analytic equations f(x, y) = 0

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172 A. P LOSKI

if and only if Ri,j(x′, (v0

ν,j(x′))) = 0 for i = 1, . . . ,m and j = 0, 1, . . . , 2p − 1 in

K[[x′]]. This proves the proposition.

5. Proof of the theorem (by induction on the number n ofvariables x)

The theorem is trivial for n = 0. Suppose that n > 0 and that the theorem istrue for n − 1. By Proposition 2.1 we may suppose that y(x) is a simple solutionof the system f(x, y) = 0. Let

δ(x, y) = detJ(f1, . . . , fm)

J(yN−m+1, . . . , yN ).

Without diminishing the generality we may suppose that δ(x, y(x)) 6= 0. If δ(0, 0) 6=0 then the theorem follows from the Implicit Function Theorem. Suppose thatδ(0, 0) = 0. After a linear change of the variables x1, . . . , xn we may assume thatδ(x, y(x)) is xn-regular of order p > 0. By Proposition 4.1 the system of equationsf(x, y) = 0 has an approximate solution v(x) = (v1(x), . . . , vN (x)) ∈ K[[x′]][xn]N

such that the solution y(x) is subordinate to v(x). Write

vν(x) =D∑j=0

(c0ν,j + v0ν,j(x

′))xjn, D ≥ 0 an integer

where (vν,j(x′)) is a family of formal power series without constant term. It is easy

to check that D∑j=0

c01,jxjn, . . . ,

D∑j=0

c0N,jxjn

is an approximate solution of the system f(0, xn, y) = 0 such that

ord δ

0, xn,D∑j=0

c01,jxjn, . . . ,

D∑j=0

c0N,jxjn

= p .

By Proposition 4.2 there exist convergent power series F (x′, V 0) ∈ Kx′, V 0Msuch that F (x′, (v0

ν,j(x′))) = 0. By induction hypothesis there exist convergent

power series (V 0ν,j(x

′, s)) in Kx′, s, where s = (s1, . . . , sq) are new variables andformal power series s(x′) = (s1(x′), . . . , sq(x

′)) without constant term such that

F (x′, (V 0ν,j(x

′, s))) = 0, V 0ν,j(x

′, s(x′)) = v0ν,j(x

′) .

Let

vν(x, s) =

D∑j=0

(c0ν,j + V 0ν,j(x

′, s))xjn for ν = 1, . . . , N

and v(x, s) = (v1(x, s), . . . , vN (x, s)). Thus vν(x) = vν(x, s(x′)) for ν = 1, . . . , N .Again by Proposition 4.2 v(x, s) is an approximate solution of the system

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f(x, y) = 0. By the Bourbaki-Tougeron Implicit Function Theorem the systemf(x, y) = 0 has the parametric solution determined by v(x, s):

yν(x, s, t) = vν(x, s) + δ(x, v(x, s))2tν for ν = 1, . . . , N −myν(x, s, t) = vν(x, s) + δ(x, v(x, s))uν(x, s, t) for ν = N −m+ 1, . . . , N .

On the other hand

yν(x, t) = vν(x) + δ(x, v(x))2tν for ν = 1, . . . , N −myν(x, t) = vν(x) + δ(x, v(x))uν(x, t) for ν = N −m+ 1, . . . , N

is the parametric solution determined by v(x). Since the formal solution y(x)is subordinate to the approximate solution v(x) there exist formal power seriest(x) = (t1(x), . . . , tN−m(x)), t(0) = 0, such that y(x) = y(x, t(x)). We have

yν(x, , s(x′), t) = vν(x) + δ(x, v(x))2tν for ν = 1, . . . , N −myν(x, s(x′), t) = vν(x) + δ(x, v(x))uν(x, s(x′), t) for ν = N −m+ 1, . . . , N

By the uniqueness of the parametric solution determined by the approximate so-lution v(x) we get

y(x, s(x′), t(x)) = y(x, t(x)) = y(x).

References

[1] S.S. Abhyankar, Local analytic geometry, Academic Press 1964.[2] M. Artin, On the solutions of analytic equations, Invent. math. 5 (1968), 277–291.

[3] M. Artin, Algebraic approximations of structures over complete local rings, Inst. Hautes

Etudes Sci. Publ. Math. 36 (1969), 23–58.[4] N. Bourbaki, Algebre commutative, Fasc. XXVIII, Chap. III, §4, No. 5 (1961).

[5] T. de Jong and G. Pfister, Local Analytic Geometry, vieweg 2000.[6] S. Lefschetz, Algebraic Geometry, Princeton NJ, 1953.

[7] A. P loski, Rozwi ιazania formalne i zbiezne rownan analitycznych, Ph D thesis, Institute of

Mathematics, Polish Academy of Sciences (1973) (in Polish).[8] A. P loski, Note on a theorem of M. Artin, Bull. Acad. Polonaise Sci. Ser. Math. 22 (1974),

1107–1109.

[9] B. Teissier, Resultats recents sur l’approximation des morphismes en algebre commutative(d’apres Andre, Artin, Popescu et Spivakovsky). Seminaire BOURBAKI, 16 eme annee (1993-

94), n784, 259–281.

[10] J.C. Tougeron, Ideaux de fonctions differentiables, New York: Springer 1972.[11] John J. Wavrik, A Theorem on Solutions of Analytic Equations with Applications to Defor-

mations of Complex Structures, Math. Ann. 216 (1975), 127–142.[12] O. Zariski and P. Samuel, Commutative Algebra, vol. II, Van Nostrand, Princeton, New

Jersey, 1960.

Department of Mathematics and Physics, Kielce University of Technology,AL. 1000 L PP 7, 25-314 Kielce, Poland


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BEZOUT’S INEQUALITY FOR REAL POLYNOMIALS

ARKADIUSZ PŁOSKI AND MACIEJ SĘKALSKI

Abstract. Let F (X,Y ), G(X,Y ) be polynomials of degrees m,n > 0 respec-tively. We prove, that the set (x, y) ∈ R2 : F (x, y) = G(x, y) = 0 has atmost mn connected components.

Classical Bezout’s theorem says that the number of complex solutions of a systemof n complex polynomial equations with n variables does not exceed the product ofdegrees of the polynomials, provided it is finite. An elementary proof of the theoremfor n = 2 can be found in [3], chapter X, §3.2. For real polynomials such a bounddoesn’t hold; here is the example given by Fulton [1]: the system of equations

m∏i=1

(x− i)2 +m∏j=1

(y − j)2 = 0, xz = 0, yz = 0

hasm2 solutions in R3, while the product of equations degrees is equal to 2m·2·2 =8m < m2 for m > 8.

Our aim is to show that there is no such an example in the case of two polynomialequations with two unknowns.

We will prove the following

Theorem. If polynomials F (X,Y ), G(X,Y ) ∈ R[X,Y ] have degrees m,n > 0respectively, then the set of solutions of a system of equations F (X,Y ) = G(X,Y ) =0 has at most mn connected components.

2010 Mathematics Subject Classification. Primary: 14A25. Secondary: 14P05.Key words and phrases. Polynomial, Bezout’s theorem.The article was published (in Polish) in the proceedings of XXVI Workshop on Analytic and

Algebraic Complex Geometry.

175


176 A. PŁOSKI AND M. SĘKALSKI

The proof is based on two lemmas

Lemma 1. If polynomials F,G with degrees m,n > 0 respectively are coprime thenthe system F (X,Y ) = G(X,Y ) = 0 has at most m · n real solutions.

Proof: If F,G are coprime in R[X,Y ] then they are also coprime in C[X,Y ]; hencethe system F = G = 0 has at most mn solutions in C2 and in particular in R2.

Lemma 2. If P ∈ R[X,Y ] is not constant then the set (x, y) : P (x, y) = 0 hasat most (degP )2 connected components.

Proof. It is sufficient to prove Lemma 2 for P irreducible.

In fact: suppose that P = P1 · · · · · Ps where Pi, i = 1, . . . , s are irreducible andassume that Lemma 2 is true for every Pi. Then the number of components of theset P = 0 does not exceed the sum of the numbers of component of the sets Pi = 0.Hence the number of connected components of the set P = 0 is less than or equalto∑si=1(degPi)2 ¬ (

∑si=0 degPi)

2 = (degP )2.

Suppose that P is an irreducible polynomial of positive degree and take a point(a, b) such that P (a, b) 6= 0. Put Q(X,Y ) = (X − a)2 + (Y − b)2 and consider theJacobian determinant J(P,Q) of polynomials P,Q.

If J(P,Q) = 2(Y − b)PX − 2(X − b)QY 6= 0 in R[X,Y ] then the polynomials P ,J(P,Q) are coprime otherwise we would have J(P,Q) = const P since P is prime,which is impossible because P (a, b) 6= 0 and J(P,Q)(a, b) = 0.

We will show that any connected component M of the set P = 0 intersectsthe curve J(P,Q) = 0. Let (x0, y0) be a point of M in which the polynomialQ reaches its minimum on M . If (x0, y0) ∈ M is a critical point of P then ofcourse J(P,Q)(x0, y0) = 0. If it is not a critical point then J(P,Q)(x0, y0) = 0 bythe method of Lagrange multipliers, [2], p. 152. Hence the number c of connectedcomponents of the set P = 0 is not greater than the number of solutions of thesystem P = J(P,Q) = 0. We have c ¬ (degP )2 by Lemma 1.

Let us consider the case J(P,Q) = 0 in R[X,Y ]. We have

Property. If P ∈ R[X,Y ] and J(P,Q) = 0 in R[X,Y ] then P (X,Y ) =P0(Q(X,Y )) for some P0(T ) ∈ R[T ].

Proof of the property. Put U = X − a, V = Y − b. Then the assumption of theproperty can be rewritten in the form J(P,Q) = 2(V PU − UPV ) = 0 in R[U, V ].Let us put DF = V FU − UFV for any polynomial F ∈ R[U, V ].

We have

1) if F (U, V ) is a homogeneous polynomial of degree n > 0 than DF also,2) if F = (U2 + V 2)kF then DF = (U2 + V 2)kDF ,3) if F 6= const is a homogeneous form and DF = 0 then U2 + V 2 divides F .


BEZOUT’S INEQUALITY FOR REAL POLYNOMIALS 177

To check 3) let us note that the conditions V FU −UFV = 0 and UFU +V FV =(degF )F imply the equality (U2+V 2)FU = (degF )UF . Since polynomials U2+V 2,(degF )U are coprime we have that U2 + V 2 divides F .

Now let P ∈ R[U, V ] be such that DP = 0. If P =∑Pj with Pj homogeneous

of degree j then DPj = 0. The conditions 2) and 3) give that Pj = cj(U2 + V 2)j2

for j even and Pj = 0 for j odd.

To complete the proof of Lemma 2 in the case J(P,Q) = 0 in R[X,Y ] notethat by Property we have P = P0(Q), where P0 is a polynomial of one variable.Therefore the set P = 0 consists of a finite number of circles. The number of circlesdoes not exceed degP0 < degP .

Proof of Theorem. If F,G are coprime then Theorem is true by Lemma 1. Supposethat P = GCD(F,G) is of positive degree. Then

F = G = 0 = FP

=G

P= 0 ∪ P = 0.

Put k = degP . By Lemma 1 the set FP = GP = 0 has at most (m − k)(n − k)

connected components. Hence the set under consideration has at most

(m− k)(n− k) + k2 = mn− k(m− k + n− k) ¬ mnconnected components.

References

[1] W. Fulton, Intersection Theory, Springer, Berlin 1984.[2] F. Leja, Rachunek różniczkowy i całkowy, Tenth edition, PWN, Warszawa 1969 (in Polish).[3] A. Mostowski and M. Stark, Elementy algebry wyższej, Ninth edition, PWN, Warszawa 1974

(in Polish).

Department of Mathematics and Physics, Kielce University of Technology,AL. 1000 L PP 7, 25-314 Kielce, Poland

E-mail address, Arkadiusz Płoski: [email protected]

E-mail address, Maciej Sękalski: [email protected]

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ŁOJASIEWICZ EXPONENT OF OVERDETERMINEDSEMIALGEBRAIC MAPPINGS

STANISŁAW SPODZIEJA AND ANNA SZLACHCIŃSKA

Abstract. We prove that both local and global Łojasiewicz exponent ofa continuous overdetermined semialgebraic mapping F : X → Rm on a closedsemialgebraic set X ⊂ Rn (i.e. m > dimX) are equal to the Łojasiewiczexponent of the composition L F : X → Rk for the generic linear mappingL : Rm → Rk, where k = dimX.

1. Introduction

Łojasiewicz inequalities are an important and useful tool in differential equa-tions, singularity theory and optimization (see for instance [12, 13] in the local caseand [18, 19] at infinity). In these considerations, estimations of the local and globalŁojasiewicz exponents play a central role (see for instance [11, 14, 17, 18, 25] inthe local case and [9, 16] at infinity). In the complex case, essential estimations ofthe Łojasiewicz exponent at infinity of a polynomial mapping F = (f1, . . . , fm) :CN → Cm (see Section 2.3) denoted by LC

∞(F ), was obtained by J. Chądzyński[4], J. Kollár [10], E. Cygan, T. Krasiński and P. Tworzewski [6] and E. Cygan [5].

We recall the estimation of Cygan, Krasiński and Tworzewski. Let deg fj = dj ,j = 1, . . . ,m, d1 ≥ . . . ≥ dm > 0 and let

B(d1, . . . , dm; k) =

d1 · · · dm for m 6 k,

d1 · · · dk−1dm for m > k.

2010 Mathematics Subject Classification. 14P20, 14P10, 32C07.Key words and phrases. Łojasiewicz exponent, semialgebraic set, semialgebraic mapping, poly-

nomial mapping.This research was partially supported by the Polish National Science Centre, grant

2012/07/B/ST1/03293.179


180 S. SPODZIEJA AND A. SZLACHCIŃSKA

Then for arbitrary m ≥ N , under the assumption #F−1(0) <∞, we have

(CKT) LC∞(F ) ≥ dm −B(d1, . . . , dm;N) +

∑b∈F−1(0)

µb(F ),

where #A denotes the cardinality of a set A, and µb(F ) is the intersection multiplic-ity (in general improper) in the sense of R. Achilles, P. Tworzewski and T. Winiarskiof graphF and Cn × 0 at the point (b, 0) (see [1]). A generalization of (CKT)for regular mappings was obtained by Z. Jelonek [7, 8].

In the proof of (CKT) the following theorem was used (see [20, Corollary 1], inPolish).

Theorem 1.1. Let m > N > 1, and let #F−1(0) < ∞. Then there existsa polynomial mapping G = (g1, . . . , gN ) : CN → CN of the form

gi = fi +m−1∑j=n

αj,ifj for i = 1, ..., N − 1, gN = fm,

where αj,i ∈ C, such that#G−1(0) <∞,

andLC∞(F ) ≥ LC

∞(G).

The above theorem has been generalized for complex polynomial mappings in[21, Theorem 2.1] and in the local case in [22, Theorem 2.1], and for real polynomialmappings in [24, Theorems 1–3] both at infinity and in the local case.

The purpose of the article is a generalization of the above fact to continuoussemialgebraic mappings. More precisely, we prove that both: local and global Ło-jasiewicz exponent of an overdetermined semialgebraic mapping F : X → Rm ona closed semialgebraic set X ⊂ RN (i.e. m > dimX) are equal to the Łojasiewiczexponent of the composition L F : X → Rk for the generic linear mappingL : Rm → Rk, where k = dimX (see Theorems 2.2, and 2.3). For more detailedinformations about semialgebraic sets and mappings, see for instance [2]. Moreover,we prove a version of the above fact for an analytic mapping with isolated zero (seeTheorem 2.1).

A mapping F : KN → Km, where K = R or K = C, is called overdetermined ifm > N .

2. Results

2.1. Notations. Let K = R or K = C. By the dimension dimK,0X at 0 of a setX ⊂ KN we mean the infimum of the dimensions over K at 0 of local analytic sets0 ∈ V ⊂ KN such that X ∩ U ⊂ V for some neighbourhood U ⊂ Kn of 0.

By the dimension dimRX of a setX ⊂ RN we mean the infimum of dimensions oflocal analytic sets V ⊂ RN such that X ⊂ V . In particular, if X is a semialgebraic


ŁOJASIEWICZ EXPONENT OF OVERDETERMINED SEMIALGEBRAIC MAPPINGS 181

set, dimRX is the infimum of dimensions of algebraic sets V ⊂ RN such thatX ⊂ V .

We will write "for the generic x ∈ A" instead of "there exists an algebraic set Vsuch that A \ V is a dense subset of A and for x ∈ A \ V ".

By LK(m, k) we shall denote the set of all linear mappings Km → Kk (we identifyK0 with 0). Let m ≥ k. By ∆K(m, k) we denote the set of all linear mappingsL ∈ LK(m, k) of the form L = (L1, ..., Lk),

Li(y1, ..., ym) = yi +m∑

j=k+1

αi,jyj , i = 1, ..., k,

where αi,j ∈ K.

2.2. The Łojasiewicz exponent at a point. Let X ⊂ KN be a closed suban-alytic set. If K = C we consider X as a subset of R2N . We will assume that theorigin 0 ∈ KN belongs to X and it is an accumulation point of X. We denote byF : (X, 0) → (Km, 0) a mapping of a neighbourhood U ⊂ X of the point 0 ∈ KNinto Km such that F (0) = 0, where the topology of X is induced from KN .

Let F : (X, 0)→ (Km, 0) be a continuous subanalytic mapping, i.e. the graph ofF is a closed subanalytic subset of (X ∩U)×Km for some neighbourhood U ⊂ KNof the origin. If K = C, we consider KN as R2N and Km as R2m. Then there arepositive constants C, η, ε such that the following Łojasiewicz inequality holds:

(Ł0) |F (x)| ≥ C dist(x, F−1(0) ∩X)η if x ∈ X, |x| < ε,

where | · | is the Euclidean norm in Kn, respectively in KN , and dist(x, V ) is thedistance of x ∈ KN to the set V ⊂ KN (dist(x, V ) = 1 if V = ∅). The smallestexponent η in (Ł0) is called the Łojasiewicz exponent of F on the set X at 0 andis denoted by LK

0 (F |X). If X contains a neighbourhood U ⊂ KN of 0 we willcall it the Łojasiewicz exponent of F at 0 and denote by LK

0 (F ). It is known thatLK0 (F |X) is a rational number and (Ł0) holds with any η ≥ LK

0 (F |S) and somepositive constants C, ε, provided 0 is an accumulation point of X \ F−1(0) (see[3, 23]). If 0 is not an accumulation point of X \ F−1(0), we have LK

0 (F |X) = 0.In Section 3 we will prove (cf [22, Theorem 2.1] and [24, Theorem 1])

Theorem 2.1. Let F = (f1, . . . , fm) : (X, 0) → (Rm, 0) be an analytic mappingwith isolated zero at the origin, where X ⊂ RN is a closed semialgebraic set and0 ∈ X. Let dimR,0X = n, and let n ≤ k ≤ m. Then for any L ∈ LR(m, k) suchthat the origin is an isolated zero of L F |X, we have

(2.1) LR0 (F |X) ≤ LR

0 (L F |X).

Moreover, for the generic L ∈ LR(m, k) the origin is an isolated zero of L F |Xand

(2.2) LR0 (F |X) = LR

0 (L F |X).

In particular, for the generic L ∈ ∆R(m, k) the origin is an isolated zero of LF |Xand (2.2) holds.



The above theorem gives a method for reduction of the problem of calculating theŁojasiewicz exponent of overdetermined mappings to the case where the domainand codomain are equidimmensional. It is not clear to the authors whether theabove statement is true if the origin is not isolated zero of f or the set X issubanalytic instead of semialgebraic.

If F : X → Km is a semialgebraic mapping then without any assumptions onthe set of zeroes of F we will prove in Section 4 the following

Theorem 2.2. Let F : (X, 0) → (Km, 0) be a continuous semialgebraic mapping,X ⊂ KN be a closed semialgebraic set of dimension dimR,0X = n, and let n ≤ k ≤m. Then for any L ∈ LK(m, k) such that

(2.3) F−1(0) ∩ UL = (L F )−1(0) ∩ UL for a neighbourhood UL ⊂ X of 0

we have

(2.4) LK0 (F |X) ≤ LK

0 (L F |X).

Moreover, for the generic L ∈ LK(m, k) the condition (2.3) holds and

(2.5) LK0 (F |X) = LK

0 (L F |X).

In particular, for the generic L ∈ ∆K(m, k) the conditions (2.3) and (2.5) hold.

2.3. The Łojasiewicz exponent at infinity. The second aim of this article is toobtain a similar results as in the previous section but for the Łojasiewicz exponentat infinity.

By the Łojasiewicz exponent at infinity of a mapping F : X → Km, whereX ⊂ Kn is an unbounded set, we mean the supremum of the exponents ν in thefollowing Łojasiewicz inequality :

(Ł∞) |F (x)| ≥ C|x|ν for x ∈ X, |x| ≥ Rfor some positive constants C, R; we denote it by LK

∞(F |X). If X = KN we callthe exponent LK

∞(F |X) the Łojasiewicz exponent at infinity of F and denote byLK∞(F ).In Section 5 we will prove the following version of Theorem 2.1 for the Łojasiewicz

exponent at infinity (cf [21, Theorem 2.1], [24, Theorem 3]).

Theorem 2.3. Let F = (f1, . . . , fm) : X → Rm be a continuous semialgebraicmapping having a compact set of zeros, where X ⊂ RN is a closed semialgebraicset, dimX = n, and let n ≤ k ≤ m. Then for any L ∈ LR(m, k) such that(L F )−1(0) ∩X is compact, we have

(2.6) LR∞(F |X) ≥ LR

∞(L F |X).

Moreover, for the generic L ∈ LK(m, k) the set (L F )−1(0) is compact and

(2.7) LR∞(F |X) = LR

∞(L F |X).

In particular, (2.7) holds for the generic L = (L1, ..., Lk) ∈ ∆R(m, k) and deg fj =degLj F for j = 1, . . . , k, provided deg f1 ≥ . . . ≥ deg fm > 0.



The above theorem gives a method of reduction of the problem of calculatingthe Łojasiewicz exponent at infinity of overdetermined semialgebraic mappings tothe case where the dimensions of domain and codomain are equal.


Let k ∈ Z, n ≤ k ≤ m. Take a closed semialgebraic set Z ⊂ RN of dimensiondimR Z = n, and let

π : Z 3 (x, y) 7→ y ∈ Rm.Then the set π(Z) is semialgebraic with dimR π(Z) ≤ n. Denote by Y ⊂ Cm thecomplex Zariski closure of π(Z). So, Y is an algebraic set of complex dimensiondimC Y ≤ n.

Assume that 0 ∈ Y . Let C0(Y ) ⊂ Cm be the tangent cone to Y at 0 in thesense of Whitney [26, p. 510]. It is known that C0(Y ) is an algebraic set anddim CC0(Y ) ≤ n. So, we have

Lemma 3.1. For the generic L ∈ LK(m, k),

L−1(0) ∩ C0(Y ) ⊂ 0.

In the proofs of Theorems 2.1, 2.2 and 2.3 we will need the following

Lemma 3.2. If L ∈ LK(m, k) satisfies L−1(0) ∩ C0(Y ) ⊂ 0, then there existε, C1, C2 > 0 such that for z ∈ Z, |π(z)| < ε we have

(3.1) C1|π(z)| ≤ |L(π(z))| ≤ C2|π(z)|.

Proof. It is obvious that for C2 = ||L|| we obtain |L(π(z))| ≤ C2|π(z)| for z ∈ Z.This gives the right hand side inequality in (3.1).

Now, we show the left hand side inequality in (3.1). Assume to the contrary,that for any ε, C1 > 0 there exists z ∈ Z such that

C1|π(z)| > |L(π(z))| and |π(z)| < ε.

In particular, for ν ∈ N, C1 = 1ν , ε = 1

ν there exists zν ∈ Z such that

1

ν|π(zν)| > |L(π(zν))| and |π(zν)| < 1

ν.

Thus |π(zν)| > 0 and

(3.2)1

ν>

1

|π(zν)||L(π(zν))| =

∣∣∣∣L( 1

|π(zν)|π(zν)

)∣∣∣∣ .Let λν = 1

|π(zν)| for ν ∈ N. Then |λνπ(zν)| = 1 so, by choosing subsequence, ifnecessary, we may assume that λνπ(zν)→ v when ν →∞, where v ∈ Cm, |v| = 1and π(zν)→ 0 as ν →∞, thus v ∈ C0(Y ) and v 6= 0. Moreover, by (3.2), we haveL(v) = 0. So v ∈ L−1(0)∩C0(Y ) ⊂ 0. This contradicts the assumption and endsthe proof.



We will also need the following lemma (cf. [15, 22]). Let X ⊂ RN be a closedsemialgebraic set such that 0 ∈ X.

Lemma 3.3. Let F, G : (RN , 0) → (Rm, 0) be analytic mappings, such thatord0(F − G) > LR

0 (F |X). If 0 is an isolated zero of F |X then 0 is an isolatedzero of G|X and for some positive constants ε, C1, C2,

(3.3) C1|F (x)| ≤ |G(x)| ≤ C2|F (x)| for x ∈ X, |x| < ε.

In particular, LR0 (F |X) = LR

0 (G|X).

Proof. Since F is a Lipschitz mapping in a neighbourhood of 0, then 1 ≤LR0 (F |X) <∞ and for some positive constants ε0, C,

(3.4) |F (x)| ≥ C|x|LR0(F |X) for x ∈ X, |x| < ε0.

From the assumption ord0(F − G) > LR0 (F |X) it follows that there exist η ∈ R,

η > LR0 (F |X) and ε1 > 0 such that ||F (x)| − |G(x)|| ≤ |x|η for x ∈ X, |x| < ε1.

Assume that (3.3) fails. Then for some sequence xν ∈ X such that xν → 0 asν →∞, we have

1

ν|F (xν)| > |G(xν)| or

1

ν|G(xν)| > |F (xν)| for ν ∈ N.

So, in the both above cases, by (3.4) for ν ≥ 2, we have

C

2|xν |L

R0(F |X) ≤ 1

2|F (xν)| < |F (xν)−G(xν)| ≤ |xν |η,

which is impossible. The last part of the assertion follows immediately from (3.3).

Proof of Theorem 2.1. We prove the assertion (2.1) analogously as Theorem 2.1 in[22]. We will prove the second part of the assertion.

Let G = (g1, . . . , gm) : (RN , 0) → (Rm, 0) be a polynomial mapping such thatordR

0 (F − G) > LR0 (F |X). Obviously, such a mapping G does exist. By Lemma

3.3, LR0 (F |X) = LR

0 (G|X) and 0 is an isolated zero of G|X. Taking, if necessary,intersection of X with a ball B centered at zero, we may assume that dimR,0X =dimRX. So, by Lemmas 3.1 and 3.2 for the generic L ∈ LR(m, k) we have thatL G|X has an isolated zero at 0 ∈ Rn, LR

0 (G|X) = LR0 (L G|X), and

ord0(L G− L F ) = ord0 L (G− F ) ≥ ord0(G− F )

>LR0 (F |X) = LR

0 (G|X) = LR0 (L G|X),

so, by Lemma 3.3, LR0 (L F |X) = LR

0 (L G|X) = LR0 (F |X). This gives (2.2). The

particular part of the assertion is proved analogously as in [22, Proposition 2.1].




Let X ⊂ RN be a closed semialgebraic set dimRX = n, and let 0 ∈ X. Taking,if necessary, intersection of X with a ball B centered at zero, we may assume thatdimR,0X = dimRX.

From [21, Proposition 1.1] we immediately obtain

Proposition 4.1. Let G = (g1, ..., gm) : X → Km be a semialgebraic mapping,gj 6= 0 for j = 1, ...,m, where m ≥ n ≥ 1, and let k ∈ Z, n ≤ k ≤ m.

(i) For the generic L ∈ LK(m, k),

(4.1) #[(L G)−1(0) \G−1(0)] <∞.

(ii) For the generic L ∈ ∆K(m, k),

(4.2) #[(L G)−1(0) \G−1(0)] <∞.

Proof. Let Y ⊂ CN × Cm be the Zariski closure of the graph of G, and letπ : Y 3 (x, y) 7→ y ∈ Cm. Then for (x, y) ∈ Y such that x ∈ X and y ∈ Kmwe have y = G(x). Let us consider the case n = k. Let

U = L ∈ LC(m,n) : #[(L π)−1(0) \ π−1(0)] <∞.By Proposition 1.1 in [21], U contains a non-empty Zariski open subset of LC(m,n).Then U contains a dense Zariski open subset W of LR(m,n). This gives the asser-tion (i) in the case n = k.

Let now k > n. Since for L = (L1, . . . , Lk) ∈ LK(m, k),

(L π)−1(0) ⊂ ((L1, . . . , Ln) π)−1(0),

then the assertion (i) follows from the previous case. We prove the assertion (ii)analogously as [21, Proposition 1.1].

Proof of Theorem 2.2. Without loss of generality we may assume that F 6= 0. Bythe definition, there exist C, ε > 0 such that for x ∈ X, |x| < ε we have

(4.3) |F (x)| ≥ C dist(x, F−1(0))LK0 (F |X),

and LK0 (F |X) is the smallest exponent for which the inequality holds. Let L ∈

LK(m, k) be such that F−1(0) ∩ UL = (L F )−1(0) ∩ UL for some neighbourhoodUL ⊂ KN of 0. Diminishing ε and the neighbourhood UL, if necessary, we mayassume that the equality dist(x, F−1(0)) = dist(x, F−1(0) ∩ UL) holds for x ∈ X,|x| < ε. Obviously L 6= 0, so, ||L|| > 0, and |F (x)| ≥ 1

||L|| |L(F (x))|. Then by (4.3)we obtain LK

0 (F |X) ≤ LK0 (L F |X), and (2.4) is proved.

By Proposition 4.1 and Lemmas 3.1 and 3.2, for the generic L ∈ LK(m, k) wehave that F−1(0) ∩ UL = (L F )−1(0) ∩ UL for some neighbourhood UL ⊂ KN of0 and there exist ε, C1, C2 > 0 such that for x ∈ X, |x| < ε,

(4.4) C1|F (x)| ≤ |L(F (x))| ≤ C2|F (x)|.



This and (4.3) gives (2.5) and ends the proof of Theorem 2.2.


The argument of Lemma 2.2 from [21] gives

Lemma 5.1. Let F : X → Rm with m ≥ n = dimRX be a semialgebraic mapping,where X ⊂ RN , and let n ≤ k ≤ m. Then there exists a Zariski open and densesubset U ⊂ LR(m, k) such that for any L ∈ U and any ε > 0 there exist δ > 0 andr > 0 such that for any x ∈ X,

|x| > r ∧ |L F (x)| < δ ⇒ |F (x)| < ε.

Proof. (cf. proof of Lemma 2.2 in [21]). Let us consider the case k = n. LetW ⊂ CNbe the Zariski closure of F (X). Then dimCW ≤ n. In the case dimCW < n,by Lemma 2.1 in [21] we easily obtain the assertion. Assume that dimW = n.We easily see that for an algebraic set V ⊂ W , dimC V ≤ n − 1, the mappingF |X\F−1(V ) : X \ F−1(V ) → W \ V is proper. By Lemma 2.1 in [21] there existsa Zariski open and dense subset U1 ⊂ LR(m, k) such that for any L ∈ U1 and forany ε > 0 there exists δ > 0 such that for z ∈ V ,

(5.1) |L(z)| < δ ⇒ |z| < ε.

Moreover, for L ∈ U1,

(5.2) W ⊂ z ∈ Cm : |z| ≤ CL(1 + |L(z)|)for some CL > 0.

LetU = L ∈ LR(m,n) : L ∈ U1.

Obviously, U is a dense and Zariski open subset of LR(m,n). Take L ∈ U and ε > 0.Assume to the contrary that there exists a sequence xν ∈ X such that |xν | → ∞,|L(f(xν))| → 0 and |f(xν)| ≥ ε. By (5.2) we may assume that f(xν)→ y0 for somey0 ∈ W . Since F |X\F−1(V ) : X \ F−1(V ) → W \ V is a proper mapping, we havey0 ∈ V . So, |y0| ≥ ε and L(y0) = 0. This contradicts (5.1) and ends the proof inthe case n = k.

Let now, k > n and let

U = L = (L1, . . . , Lk) ∈ LR(m, k) : (L1, . . . , Ln) ∈ U1.Then for any L = (L1, . . . , Lk) ∈ U and x ∈ Rn we have

|(L1, . . . , Ln) F (x)| ≤ |L F (x)|,so, the assertion immediately follows from the previous case.

Proof of Theorem 2.3 (cf. proof of Theorem 2.1 in [21]). Since for non-zero L ∈LR(m, k) we have |L F (x)| ≤ ||L|||F (x)| and ||L|| > 0, then by the definition ofthe Łojasiewicz exponent at infinity we obtain the first part of the assertion. Wewill prove the second part of the assertion.



Since F−1(0) is a compact set, by Proposition 4.1, there exists a dense Zariskiopen subset U of LR(m, k) such that

U ⊂ L ∈ LR(m, k) : (L F )−1(0) is a compact set.

So, for the generic L ∈ LR(m, k) the set (L F )−1(0) is compact.If LR

∞(F |X) < 0, the assertion (2.7) follows from Lemmas 3.1, 3.2 and 5.1.Assume that LR

∞(F |X) = 0. Then there exist C, R > 0 such that |F (x)| ≥ C as|x| ≥ R. Moreover, there exists a sequence xν ∈ X such that |xν | → ∞ as ν →∞and |F (xν)| is a bounded sequence. So, by Lemma 5.1 for the generic L ∈ U andε = C there exist r, δ > 0 such that |L F (x)| ≥ δ as |x| > r, so LR

∞(L F |X) ≥ 0.Since |L F (xν)| is a bounded sequence, we have LR

∞(L F |X) ≤ 0. Summing upLR∞(L F |X) = LR

∞(F |X) in the considered case.In the case LR

∞(F |X) > 0, we obtain the assertion analogously as in the proofof Theorem 2.1 in [21].

References

[1] R. Achilles, P. Tworzewski and T. Winiarski, On improper isolated intersection in complexanalytic geometry. Ann. Polon. Math. 51 (1990), 21–36.

[2] J. Bochnak, M. Coste and M-F. Roy, Real algebraic geometry. Springer-Verlag, Berlin, 1998.[3] J. Bochnak and J.J. Risler, Sur les exposants de Łojasiewicz. Comment. Math. Helv. 50

(1975), 493–507.[4] J. Chadzyński, On proper polynomial mappings. Bull. Polish Acad. Sci. Math. 31 (1983),

115–120.[5] E. Cygan, A note on separation of algebraic sets and the Łojasiewicz exponent for polynomial

mappings. Bull. Sci. Math. 129 (2005), no. 2, 139–147.[6] E. Cygan, T. Krasiński and P. Tworzewski, Separation of algebraic sets and the Łojasiewicz

exponent of polynomial mappings. Invent. Math. 136 (1999), no. 1, 75–87.[7] Z. Jelonek, On the effective Nullstellensatz. Invent. Math. 162 (2005), no. 1, 1–17.[8] Z. Jelonek, On the Łojasiewicz exponent. Hokkaido Math. J. 35 (2006), no. 2, 471–485.[9] S. Ji, J. Kollár and B. Shiffman, A global Łojasiewicz inequality for algebraic varieties. Trans.

Amer. Math. Soc. 329 (1992), 813–818.[10] J. Kollár, Sharp effective Nullstellensatz. J. Amer. Math. Soc. 1 (1988), 963–975.[11] J. Kollár, An effective Łojasiewicz inequality for real polynomials. Period. Math. Hungar. 38

(1999), no. 3, 213–221.[12] K. Kurdyka, T. Mostowski and A. Parusiński, Proof of the gradient conjecture of R. Thom.

Ann. of Math. (2) 152 (2000), no. 3, 763–792.[13] K. Kurdyka and S. Spodzieja, Convexifying Positive Polynomials and Sums of Squares Ap-

proximation. SIAM J. Optim. 25 (2015), no. 4, 2512–2536.[14] M. Lejeune-Jalabert and B. Teissier, Clôture intégrale des idéaux et équisingularité. Centre

de Mathématiques Ecole Polytechnique Palaiseau, 1974.[15] A. Płoski,Multiplicity and the Łojasiewicz exponent. Banach Center Publications 20, Warsaw

(1988), 353–364.[16] T. Rodak and S. Spodzieja, Effective formulas for the Łojasiewicz exponent at infinity.

J. Pure Appl. Algebra 213 (2009), 1816–1822.[17] T. Rodak and S. Spodzieja, Effective formulas for the local Łojasiewicz exponent. Math. Z.

268 (2011), 37–44.[18] T. Rodak and S. Spodzieja, Łojasiewicz exponent near the fibre of a mapping. Proc. Amer.

Math. Soc. 139 (2011), 1201–1213.



[19] T. Rodak and S. Spodzieja, Equivalence of mappings at infinity. Bull. Sci. Math. 136 (2012),no. 6, 679–686.

[20] S. Spodzieja, O włóknach odwzorowań wielomianowych. Materials of the XIX Conferenceof Complex Analytic and Algebraic Geometry, Publisher University of Lodz, Lodz, (2001),23–37 (in polish).

[21] S. Spodzieja, The Łojasiewicz exponent at infinity for overdetermined polynomial mappings.Ann. Polon. Math. 78 (2002), 1–10.

[22] S. Spodzieja, Multiplicity and the Łojasiewicz exponent. Ann. Polon. Math. 73 (2000), 257–267

[23] S. Spodzieja, The Łojasiewicz exponent of subanalytic sets. Ann. Polon. Math. 87 (2005),247–263.

[24] S. Spodzieja and A. Szlachcińska, Łojasiewicz exponent of overdetermined mappings. Bull.Pol. Acad. Sci. Math. 61 (2013), no. 1, 27–34.

[25] B. Teissier, Variétés polaires. I. Invariants polaires des singularités d’hypersurfaces. Invent.Math. 40 (1977), no. 3, 267–292.

[26] H. Whitney, Tangents to an analytic variety. Ann. of Math. 81 (1965), 496–549.

Faculty of Mathematics and Computer Science, University of Łódźul. S. Banacha 22, 90-238 Łódź, Poland

E-mail address, Stanisław Spodzieja: [email protected]

E-mail address, Anna Szlachcińska: [email protected]

wydawnictwo uniwersytetu Łódzkiego...poem: shreeram s. abhyankar , wielomiany i szeregi potęgowe...

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