Ergebnisse der Mathematik und ihrer Grenzgebiete 3. Folge
A Series of Modern Surveys in Mathematics
Editorial Board M. Gromov, Bures-sur-Yvette J. Jost, Leipzig J. Kollár, Princeton H. W. Lenstra, Jr., Leiden J. Tits, Paris D. B. Zagier, Bonn G. Ziegler, Berlin Managing Editor R. Remmert, Münster
Volume 11
Michael D. Fried • Moshe Jarden
Field Arithmetic Third Edition Revised by Moshe Jarden
Michael D. Fried Department of Mathematics Montana State University – Billings Billings MT 59101 USA
[email protected]
ISBN 978-3-540-77269-9
Moshe Jarden School of Mathematics Tel Aviv University Ramat Aviv, Tel Aviv 69978 Israel
[email protected]
e-ISBN 978-3-540-77270-5
DOI 10.1007/978-3-540-77270-5 Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics ISSN 0071-1136 Library of Congress Control Number: 2008924174 Mathematics Subject Classification (2000): 12E30 c 2008 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: WMXDesign GmbH, Heidelberg Printed on acid-free paper 9 8 7 6 5 4 3 2 1 springer.com
To those precious colleagues who can appreciate the goals of and connections to other areas. To those who acknowledge the depth of what we already know from the absorbed contribution of previous generations before we address our papers. To those who can transcend the hubris of today’s mathematical community.
Table of Contents Chapter 1. Infinite Galois Theory and Profinite Groups 1.1 Inverse Limits . . . . . . . . . . . . . . . 1.2 Profinite Groups . . . . . . . . . . . . . . 1.3 Infinite Galois Theory . . . . . . . . . . . . 1.4 The p-adic Integers and the Pr¨ ufer Group . . . 1.5 The Absolute Galois Group of a Finite Field . . Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .
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Chapter 2. Valuations and Linear Disjointness . 2.1 Valuations, Places, and Valuation Rings . . 2.2 Discrete Valuations . . . . . . . . . . . 2.3 Extensions of Valuations and Places. . . . 2.4 Integral Extensions and Dedekind Domains 2.5 Linear Disjointness of Fields . . . . . . . 2.6 Separable, Regular, and Primary Extensions 2.7 The Imperfect Degree of a Field . . . . . 2.8 Derivatives . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .
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19 19 21 24 30 34 38 44 48 50 51
Chapter 3. Algebraic Function Fields of One Variable . . 3.1 Function Fields of One Variable . . . . . . . . . 3.2 The Riemann-Roch Theorem . . . . . . . . . . 3.3 Holomorphy Rings . . . . . . . . . . . . . . . 3.4 Extensions of Function Fields . . . . . . . . . . 3.5 Completions . . . . . . . . . . . . . . . . . . 3.6 The Different . . . . . . . . . . . . . . . . . 3.7 Hyperelliptic Fields . . . . . . . . . . . . . . . 3.8 Hyperelliptic Fields with a Rational quadratic Subfield Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .
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Chapter 4. The Riemann Hypothesis for Function Fields . 4.1 Class Numbers . . . . . . . . . . . . . . . . . 4.2 Zeta Functions . . . . . . . . . . . . . . . . . 4.3 Zeta Functions under Constant Field Extensions . . 4.4 The Functional Equation . . . . . . . . . . . . 4.5 The Riemann Hypothesis and Degree 1 Prime Divisors 4.6 Reduction Steps . . . . . . . . . . . . . . . . 4.7 An Upper Bound . . . . . . . . . . . . . . . . 4.8 A Lower Bound . . . . . . . . . . . . . . . .
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Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5. Plane Curves . . . . . . . 5.1 Affine and Projective Plane Curves 5.2 Points and prime divisors . . . . 5.3 The Genus of a Plane Curve . . . 5.4 Points on a Curve over a Finite Field Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .
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Chapter 6. The Chebotarev Density Theorem 6.1 Decomposition Groups . . . . . . . 6.2 The Artin Symbol over Global Fields . 6.3 Dirichlet Density . . . . . . . . . . 6.4 Function Fields . . . . . . . . . . 6.5 Number Fields . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . .
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107 107 111 113 115 121 129 130
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132 132 134 135 137 138 139 141 145 147 147 148
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149 149 152 154 156 157 159 160 161 162
Chapter 7. Ultraproducts . . . . . . . 7.1 First Order Predicate Calculus . . 7.2 Structures . . . . . . . . . . . 7.3 Models . . . . . . . . . . . . 7.4 Elementary Substructures . . . . 7.5 Ultrafilters . . . . . . . . . . 7.6 Regular Ultrafilters . . . . . . . 7.7 Ultraproducts . . . . . . . . . 7.8 Regular Ultraproducts . . . . . 7.9 Nonprincipal Ultraproducts of Finite Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .
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Chapter 8. Decision Procedures . . . . . . . . 8.1 Deduction Theory . . . . . . . . . . . 8.2 G¨odel’s Completeness Theorem . . . . . 8.3 Primitive Recursive Functions . . . . . . 8.4 Primitive Recursive Relations . . . . . . 8.5 Recursive Functions . . . . . . . . . . 8.6 Recursive and Primitive Recursive Procedures 8.7 A Reduction Step in Decidability Procedures Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .
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Chapter 9. Algebraically Closed Fields . 9.1 Elimination of Quantifiers . . . . 9.2 A Quantifiers Elimination Procedure 9.3 Effectiveness . . . . . . . . . . 9.4 Applications . . . . . . . . . . Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .
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163 163 165 168 169 170 170
Chapter 10. Elements of Algebraic Geometry . . 10.1 Algebraic Sets . . . . . . . . . . . . 10.2 Varieties . . . . . . . . . . . . . . . 10.3 Substitutions in Irreducible Polynomials . 10.4 Rational Maps . . . . . . . . . . . . 10.5 Hyperplane Sections . . . . . . . . . . 10.6 Descent . . . . . . . . . . . . . . . 10.7 Projective Varieties . . . . . . . . . . 10.8 About the Language of Algebraic Geometry Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .
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172 172 175 176 178 180 182 185 187 190 191
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192 192 193 199 201 203 207 211 217 218
Chapter 11. Pseudo Algebraically Closed Fields . . 11.1 PAC Fields . . . . . . . . . . . . . . . 11.2 Reduction to Plane Curves . . . . . . . . 11.3 The PAC Property is an Elementary Statement 11.4 PAC Fields of Positive Characteristic . . . 11.5 PAC Fields with Valuations . . . . . . . . 11.6 The Absolute Galois Group of a PAC Field . . . . . 11.7 A non-PAC Field K with Kins PAC Exercises . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . .
Chapter 12. Hilbertian Fields . . . . . . . . . . . . 12.1 Hilbert Sets and Reduction Lemmas . . . . . . 12.2 Hilbert Sets under Separable Algebraic Extensions 12.3 Purely Inseparable Extensions . . . . . . . . 12.4 Imperfect fields . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .
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219 219 223 224 228 229 230
Chapter 13. The Classical Hilbertian Fields 13.1 Further Reduction . . . . . . . . 13.2 Function Fields over Infinite Fields 13.3 Global Fields . . . . . . . . . . 13.4 Hilbertian Rings . . . . . . . . 13.5 Hilbertianity via Coverings . . . .
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13.6 Non-Hilbertian g-Hilbertian Fields 13.7 Twisted Wreath Products . . . 13.8 The Diamond Theorem . . . . 13.9 Weissauer’s Theorem . . . . . Exercises . . . . . . . . . . . . . Notes . . . . . . . . . . . . . .
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248 252 258 262 264 266
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267 267 268 270 272 274 275 276
Chapter 15. Nonstandard Approach to Hilbert’s Irreducibility Theorem 15.1 Criteria for Hilbertianity . . . . . . . . . . . . 15.2 Arithmetical Primes Versus Functional Primes . . 15.3 Fields with the Product Formula . . . . . . . . 15.4 Generalized Krull Domains . . . . . . . . . . . 15.5 Examples . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .
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277 277 279 281 283 286 289 290
Chapter 16. Galois Groups over Hilbertian Fields . . . . . . 16.1 Galois Groups of Polynomials . . . . . . . . . . . . 16.2 Stable Polynomials . . . . . . . . . . . . . . . . 16.3 Regular Realization of Finite Abelian Groups . . . . . 16.4 Split Embedding Problems with Abelian Kernels . . . 16.5 Embedding Quadratic Extensions in Z/2n Z-extensions . 16.6 Zp -Extensions of Hilbertian Fields . . . . . . . . . . 16.7 Symmetric and Alternating Groups over Hilbertian Fields 16.8 GAR-Realizations . . . . . . . . . . . . . . . . . 16.9 Embedding Problems over Hilbertian Fields . . . . . 16.10 Finitely Generated Profinite Groups . . . . . . . . 16.11 Abelian Extensions of Hilbertian Fields . . . . . . . 16.12 Regularity of Finite Groups over Complete Discrete Valued Fields . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .
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291 291 294 298 302 306 308 315 321 325 328 332
Chapter 14. Nonstandard Structures . 14.1 Higher Order Predicate Calculus 14.2 Enlargements . . . . . . . . 14.3 Concurrent Relations . . . . 14.4 The Existence of Enlargements 14.5 Examples . . . . . . . . . . Exercises . . . . . . . . . . . . Notes . . . . . . . . . . . . .
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Chapter 17. Free Profinite Groups . . . . . . . . . . . . . . . 338 17.1 The Rank of a Profinite Group . . . . . . . . . . . . . 338
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17.2 Profinite Completions of Groups . . . 17.3 Formations of Finite Groups . . . . . 17.4 Free pro-C Groups . . . . . . . . . . 17.5 Subgroups of Free Discrete Groups . . 17.6 Open Subgroups of Free Profinite Groups 17.7 An Embedding Property . . . . . . . Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .
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340 344 346 350 358 360 361 362
Chapter 18. The Haar Measure . . . . . . . . . . 18.1 The Haar Measure of a Profinite Group . . . 18.2 Existence of the Haar Measure . . . . . . . 18.3 Independence . . . . . . . . . . . . . . . 18.4 Cartesian Product of Haar Measures . . . . . 18.5 The Haar Measure of the Absolute Galois Group 18.6 The PAC Nullstellensatz . . . . . . . . . . 18.7 The Bottom Theorem . . . . . . . . . . . 18.8 PAC Fields over Uncountable Hilbertian Fields 18.9 On the Stability of Fields . . . . . . . . . . 18.10 PAC Galois Extensions of Hilbertian Fields . 18.11 Algebraic Groups . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .
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363 363 366 370 376 378 380 382 386 390 394 397 400 401
Chapter 19. Effective Field Theory and Algebraic Geometry . 19.1 Presented Rings and Fields . . . . . . . . . . . . . 19.2 Extensions of Presented Fields . . . . . . . . . . . 19.3 Galois Extensions of Presented Fields . . . . . . . . 19.4 The Algebraic and Separable Closures of Presented Fields 19.5 Constructive Algebraic Geometry . . . . . . . . . . 19.6 Presented Rings and Constructible Sets . . . . . . . 19.7 Basic Normal Stratification . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .
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403 403 406 411 412 413 422 425 427 428
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Chapter 20. The Elementary Theory of e-Free PAC Fields 20.1 ℵ1 -Saturated PAC Fields . . . . . . . . . . . . 20.2 The Elementary Equivalence Theorem of ℵ1 -Saturated PAC Fields . . . 20.3 Elementary Equivalence of PAC Fields . . . . . . 20.4 On e-Free PAC Fields . . . . . . . . . . . . . 20.5 The Elementary Theory of Perfect e-Free PAC Fields 20.6 The Probable Truth of a Sentence . . . . . . . . 20.7 Change of Base Field . . . . . . . . . . . . . . . . . . . . . . . . 20.8 The Fields Ks (σ1 , . . . , σe )
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20.9 The Transfer Theorem . . . . . . . . 20.10 The Elementary Theory of Finite Fields Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .
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446 448 451 453
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454 454 455 460 462 463 467 472 479 489 493 495
Chapter 22. Projective Groups and Frattini Covers 22.1 The Frattini Groups of a Profinite Group . 22.2 Cartesian Squares . . . . . . . . . . . 22.3 On C-Projective Groups . . . . . . . . 22.4 Projective Groups . . . . . . . . . . . 22.5 Frattini Covers . . . . . . . . . . . . 22.6 The Universal Frattini Cover . . . . . . 22.7 Projective Pro-p-Groups . . . . . . . . 22.8 Supernatural Numbers . . . . . . . . . 22.9 The Sylow Theorems . . . . . . . . . 22.10 On Complements of Normal Subgroups . 22.11 The Universal Frattini p-Cover . . . . . 22.12 Examples of Universal Frattini p-Covers . . . 22.13 The Special Linear Group SL(2, Zp ) 22.14 The General Linear Group GL(2, Zp ) . . Exercises . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . .
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497 497 499 502 506 508 513 515 520 522 524 528 532 534 537 539 542
Chapter 23. PAC Fields and Projective Absolute Galois Groups 23.1 Projective Groups as Absolute Galois Groups . . . . . 23.2 Countably Generated Projective Groups . . . . . . . 23.3 Perfect PAC Fields of Bounded Corank . . . . . . . 23.4 Basic Elementary Statements . . . . . . . . . . . . 23.5 Reduction Steps . . . . . . . . . . . . . . . . . . 23.6 Application of Ultraproducts . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .
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544 544 546 549 550 554 558 561 561
Chapter 21. Problems of Arithmetical Geometry 21.1 The Decomposition-Intersection Procedure 21.2 Ci -Fields and Weakly Ci -Fields . . . . . . . 21.3 Perfect PAC Fields which are Ci 21.4 The Existential Theory of PAC Fields . 21.5 Kronecker Classes of Number Fields . . 21.6 Davenport’s Problem . . . . . . . . 21.7 On permutation Groups . . . . . . . 21.8 Schur’s Conjecture . . . . . . . . . . 21.9 Generalized Carlitz’s Conjecture . . . . Exercises . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . .
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Chapter 24. Frobenius Fields . . . . . . . . . . . . . . 24.1 The Field Crossing Argument . . . . . . . . . . . 24.2 The Beckmann-Black Problem . . . . . . . . . . 24.3 The Embedding Property and Maximal Frattini Covers 24.4 The Smallest Embedding Cover of a Profinite Group . 24.5 A Decision Procedure . . . . . . . . . . . . . . 24.6 Examples . . . . . . . . . . . . . . . . . . . . 24.7 Non-projective Smallest Embedding Cover . . . . . 24.8 A Theorem of Iwasawa . . . . . . . . . . . . . . 24.9 Free Profinite Groups of at most Countable Rank . . 24.10 Application of the Nielsen-Schreier Formula . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . .
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Chapter 25. Free Profinite Groups of Infinite Rank . . 25.1 Characterization of Free Profinite Groups by Embedding Problems 25.2 Applications of Theorem 25.1.7 . . . . . . . . 25.3 The Pro-C Completion of a Free Discrete Group . 25.4 The Group Theoretic Diamond Theorem . . . . 25.5 The Melnikov Group of a Profinite Group . . . 25.6 Homogeneous Pro-C Groups . . . . . . . . . 25.7 The S-rank of Closed Normal Subgroups . . . . 25.8 Closed Normal Subgroups with a Basis Element . 25.9 Accessible Subgroups . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .
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595 601 604 606 613 615 620 623 625 633
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Chapter 26. Random Elements in Free Profinite Groups . 26.1 Random Elements in a Free Profinite Group . . . 26.2 Random Elements in Free pro-p Groups . . . . . ˆn . . . . . . . . . . . . . 26.3 Random e-tuples in Z 26.4 On the Index of Normal Subgroups Generated by Random Elements . 26.5 Freeness of Normal Subgroups Generated by Random Elements Notes . . . . . . . . . . . . . . . . . . . . . . Chapter 27. Omega-Free PAC Fields . . . . . . . . . 27.1 Model Companions . . . . . . . . . . . . . 27.2 The Model Companion in an Augmented Theory of 27.3 New Non-Classical Hilbertian Fields . . . . . . 27.4 An abundance of ω-Free PAC Fields . . . . . . Notes . . . . . . . . . . . . . . . . . . . . .
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Chapter 28. Undecidability . . . . . . . . . . . . . . 28.1 Turing Machines . . . . . . . . . . . . . . . 28.2 Computation of Functions by Turing Machines . . 28.3 Recursive Inseparability of Sets of Turing Machines 28.4 The Predicate Calculus . . . . . . . . . . . . 28.5 Undecidability in the Theory of Graphs . . . . . 28.6 Assigning Graphs to Profinite Groups . . . . . . 28.7 The Graph Conditions . . . . . . . . . . . . . 28.8 Assigning Profinite Groups to Graphs . . . . . . 28.9 Assigning Fields to Graphs . . . . . . . . . . . 28.10 Interpretation of the Theory of Graphs in the Theory of Fields Exercises . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .
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671 671 672 676 679 682 687 688 690 694
. . . 694 . . . . 697 . . . . 697
Chapter 29. Algebraically Closed Fields with Distinguished Automorphisms 29.1 The Base Field K . . . . . . . . . . . . . . . . . 29.2 Coding in PAC Fields with Monadic Quantifiers . . . . ˜ σ1 , . . . , σe i’s . . . . . . 29.3 The Theory of Almost all hK, 29.4 The Probability of Truth Sentences . . . . . . . . .
. . . . .
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. 698 698 700 704 706
Chapter 30. Galois Stratification . . . . 30.1 The Artin Symbol . . . . . . . . 30.2 Conjugacy Domains under Projection 30.3 Normal Stratification . . . . . . 30.4 Elimination of One Variable . . . 30.5 The Complete Elimination Procedure 30.6 Model-Theoretic Applications . . . 30.7 A Limit of Theories . . . . . . . Exercises . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . .
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708 708 710 715 717 720 722 725 726 729
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Chapter 31. Galois Stratification over Finite Fields . 31.1 The Elementary Theory of Frobenius Fields . 31.2 The Elementary Theory of Finite Fields . . . 31.3 Near Rationality of the Zeta Function of a Galois Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . Formula . . . . . . . .
. . 730 . . 730 . . 735 . 739 . . 748 . . 750
Chapter 32. Problems of Field Arithmetic . . . . 32.1 Open Problems of the First Edition . . . . 32.2 Open Problems of the Second Edition . . . 32.3 Open problems . . . . . . . . . . . . .
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751 751 754 758
Table of Contents
References Index
xv
. . . . . . . . . . . . . . . . . . . . . . . . . 761
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Introduction to the Third Edition The third edition of “Field Arithmetic” improves the second edition in two ways. First it removes many typos and mathematical inaccuracies that occur in the second edition. In particular, it fills out a big gap in the References of the second edition, where unfortunately all references between “Gilmore and Robinson” and “Kantor and Lubotzky” are missing. Secondly, the third edition reports on five open problems of the second edition that were solved since that edition appeared in 2005. J´anos Koll´ar solved Problem 2 by proving that if each projective plane curve defined over a field K has a K-rational point, then K is PAC. J´anos Koll´ar also solved Problem 3 and proved that if K is a PAC field, ˜ and V is a variety defined over K, then V (K) is w is a valuation of K, ˜ w-dense in V (K). J´anos Koll´ar partially settled Problem 21. He proved that every PAC field of characteristic 0 is C1 . Problem 31 was affirmatively solved by Lior Bary-Soroker by establishing an analog of the diamond theorem for the finitely generated non-Abelian free profinite groups. Finally, Eric Rosen suggested to reorganize Corollary 28.5.3 of the second edition that led to an affirmative solution of Problem 33. Unfortunately, a full account of the first four solutions is out of the scope of the present volume. Much of the improvment made in the present edition is due to Arno Fehm and Dan Haran. I am really indebted to them for their contribution. Tel Aviv, Autumn 2007
Moshe Jarden
Introduction to the Second Edition The first edition of “Field Arithmetic” appeared in 1986. At the end of that edition we gave a list of twenty-two open problems. It is remarkable that since then fifteen of them were partially or fully solved. Parallel to this, Field Arithmetic has developed in many directions establishing itself as an independent branch of Algebra and Number Theory. Some of these developments have been documented in books. We mention here “Groups as Galois groups” [V¨ olklein] on consequences of the Riemann existence theorem, “Inverse Galois Groups” [Malle-Matzat] with a comprehensive report on finite Galois groups over number fields, “Profinite groups” [Ribes-Zalesskii] including the cohomology of profinite groups, “Analytic prop Groups” [Dixon-du.Sautoy-Mann-Segal] on closed subgroups of GL(n, Zp ), “Subgroup Growth” [Lubotzky-Segal] on counting the number of subgroups of finitely generating groups, and “Multi-Valued Fields” [Ershov7] on the model theory of fields with several valuations. This led to an official recognition of Field Arithmetic by the Mathematical Reviews in the form of MSC number 12E30.
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The extent which Field Arithmetic has reached makes it impossible for us to report in one extended volume about all exciting results which have been achieved. We have therefore made several choices which best suit the spirit of this book but do not extend beyond the scope of one volume. The new results and additional topics have made it necessary to reorganize and to enlarge the sections dealing with background material. Of course, we took the opportunity afforded by editing a second edition to correct flaws and mistakes which occurred in the first edition and to add more details to proofs wherever it seemed useful. We list the major changes and additions we made in the book: Chapter 2 has been reorganized. Sections 2.5–2.9 of the first edition, which survey the theory of algebraic function fields of one variable, were moved to Chapter 3. Sections 2.5–2.8 dealing with linear disjointness, regular extensions, and separability appeared in the first edition as sections 9.1– 9.3. A nice application of linear disjointness is Leptin’s construction (which preceded that of Warehouse) of a Galois group isomorphic to a given profinite group (Proposition 2.6.12). In addition to the introductory material about the theory of algebraic function fields of one variable, Chapter 3 now includes a proof of the RiemannHurwitz formula and a discussion of hyperelliptic curves. The proof of Theorem 4.9 of the first edition, estimating the number of zeros of an absolutely irreducible polynomial over a finite field, had a flaw. This has been fixed in the proof of Theorem 5.4.1. Likewise, the inequality given by [Fried-Jarden3, Prop. 5.16] is inaccurate. This inaccuracy is fixed in Proposition 6.4.8. We find it more convenient to use the language of algebraic sets as introduced in [Weil5] for model theoretic applications. Section 10.8 translates the basic concepts of that language to the now more commonly used language of schemes. Theorem 10.14 of the first edition (due to Frey-Prestel) says that the Henselian hull of a PAC field K is Ks . Proposition 11.5.3 (due to Prestel) ˜ for every valuation strengthens this theorem. It says that K is w-dense in K ˜ w of K. What we called “a separably Hilbertian field” in the first edition, is now called “a Hilbertian field” (Section 12.1). This agrees with the common usage and seems more appropriate for applications. Section 13.5 gives an alternative definition of Hilbertianity via coverings leading to the notion of “g-Hilbertianity”. This sets the stage for a generalization of a theorem of Zannier: Every global field has an infinite normal extension N which is g-Hilbertian but not Hilbertian (Theorem 13.6.2). Moreover, there is a unique factorization subring R of N with infinitely many irreducible elements (Example 15.5.8). This answers negatively Problems 14.20 and 14.21 of the first edition. Chapter 13 includes now one of the major results of Field Arithmetic which we call “Haran’s diamond theorem”: Let M1 and M2 be Galois ex-
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Introduction to the Second Edition
tensions of a Hilbertian field K and M a field between K and M1 M2 not contained in M1 nor in M2 . Then M is Hilbertian (Theorem 13.8.3). In particular, if N is a Galois extension of K, then N is not the compositum of two Galois extensions of K neither of which is contained in the other. This settles Problems 12.18 and 12.19 of the first edition. The immediate goal of Hilbert’s irreducibility theorem was to realize the groups Sn and An as Galois groups over Q. Chapter 16 is dedicated to realizations of Galois groups over arbitrary Hilbertian fields. One of the most important of these results is due to Harbater (Proposition 16.12.1): Let K be a complete valued field, t an indeterminate, and G a finite group. Then G is regular over K, that is, K(t) has a finite Galois extension F , regular over K, with Gal(F/K(t)) ∼ = G. Unfortunately, none of the three proofs of this theorem fits into the scope of this book. Section 16.6 proves a theorem of Whaples: Let K be a field and p a prime number. Suppose Z/pZ (resp. Z/4Z if p = 2) occurs as a Galois group over K. Then Zp is realizable over K. Section 16.7 generalizes a theorem of Hilbert: Let K be a field and n ≥ 2 an integer with char(K) - (n − 1)n. Then An is regular over K. One of the most far-reaching attempts to realize arbitrary finite Galois groups over Hilbertian fields uses Matzat’s notion of GAR realization of simple finite groups: Let K be a Hilbertian field and α: G → Gal(L/K) a finite embedding problem over K. Suppose every composition factor of Ker(α) has a GAR realization over K. Then the embedding problem is solvable. This leads in particular to the realization of many finite groups over Q (Remark 16.9.5). Chapter 17 deals mainly with Melnikov’s formations C (i.e. sets consisting of all finite groups whose composition factors belong to a given set of finite simple groups). We prove that every free abstract group F is residuallyC. Thus, if the free pro-C group with a given rank m exists, then the canonical injection of F into Fˆm (C) is injective (Proposition 17.5.11 – Ribes-Zalesskii). Konrad Neumann improved former results of Fried-Geyer-Jarden and proved that every field is stable (Theorem 18.9.3). This allows the construction of PAC Hilbertian Galois extensions of arbitrary countable Hilbertian fields (Theorems 18.10.2 and 18.10.3). We survey Neumann’s proof in Section 18.9. The full proof unfortunately falls outside the scope of this book. It seemed to be well known that the concept of absolute irreducibility of a variety is elementary. Unfortunately, we could find no solid proof for it in the literature. Proposition 19.5.9 fills in the gap by proving that result. Section 21.2 includes now the classical results about Ci -fields and not only the corresponding results about weakly Ci -fields as was the case in Section 19.2 of the first edition. Sections 21.8 gives a complete proof of Schur’s Conjecture: If f (X) is a polynomial with coefficients in a global field K with char(K) - deg(f ) and f permutes OK /p for infinitely many primes p of K, then each composition factor of f is linearly related over K to a Dickson polynomial of a prime
Introduction to the Second Edition
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degree. Section 21.7 proves all lemmas about permutation groups which are used in the proof of Schur’s Conjecture (Theorem 21.8.13). This includes the classification of subgroups of AGL(1, Fl ) (Lemma 21.7.2), and the theorems of Schur (Proposition 21.7.7) and Burnside (Proposition 21.7.8) about doubly transitive permutation groups. Section 21.9 contains the Fried-Cohen version of Lenstra’s proof of the generalized Carlitz’s Conjecture: Let p be a prime number, q a power of p, and f ∈ Fq [X] a polynomial of degree n > 1 which is not a power of p. Suppose f permutes infinitely many finite extensions of Fq . Then gcd(n, q − 1) = 1. The universal Frattini p-cover of a finite group plays a central role in Fried’s theory of modular towers. Section 22.11 introduces the former concept and proves its basic properties. Corollary 22.13.4 shows then that PSL(2, Zp ) is a p-Frattini cover of PSL(2, Fp ) although it is not the universal p-Frattini cover. Chapter 23 puts together material on PAC fields which appeared in Section 20.5 and Chapter 21 of the first edition. The Beckmann-Black Problem is a refinement of the inverse problem of Galois Theory. D´ebes proved that the problem has an affirmative solution over PAC fields (Theorem 24.2.2). Chapter 25 substantially extends the study of free profinite groups F of infinite rank which appeared in Section 24.4 of the first edition. Most of the material goes back to Melnikov. We characterize closed normal subgroups of F by their S-ranks, and prove that a closed subgroup of F is accessible if and only if it is homogeneous. The first part of Chapter 25 reproduces the group theoretic version of Haran’s diamond theorem. Chapter 26 is completely new. It describes the properties of the closed subgroup hxi and the closed normal subgroup [x] generated by a random etuple x = (x1 , . . . , xe ) of elements of a finitely generated free profinite group F of finite rank n ≥ 2. For example, with probability 1, hxi ∼ = Fˆe (Proposition 26.1.7). This solves Problem 16.16 of the first edition. In addition, with a positive probability, [x] has infinite rank and is isomorphic to Fˆω (Theorem 26.4.5 and Corollary 26.5.7). The latter result is based on the Golod-Shafarevich Inequality. Chapter 28 considers an infinite field K which is finitely generated over its base field. It proves that for e ≥ 2 the theory of all sentences θ which hold ˜ σ1 , . . . , σe i with (σ1 , . . . , σe ) ∈ Gal(K)e is undein almost all structures hK, ˜ σ 1 , . . . , σe i cidable. Moreover, the probability that a sentence θ hold in hK, is in general a nonrational number. Perhaps the most significant achievement of Field Arithmetic since the first edition appeared is the solution of Problem 24.41 of that edition: The absolute Galois group of a countable PAC Hilbertian field is free of rank ℵ0 . It was originally proved in characteristic 0 with complex analysis by Fried-V¨olklein. Then it was proved in the general case by Pop using rigid geometry and by Haran-Jarden-V¨ olklein using “algebraic patching”. The two
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Introduction to the First Edition
latter methods also lead to the proof that Gal(C(t)) is a free profinite group if C is an arbitrary algebraically closed field (Harbater, Pop, Haran-Jarden). The method of Fried-V¨ olklein led to the theory of modular towers of Fried. A remote goal in Galois theory is the classification of absolute Galois groups among all profinite groups. In this framework, one tries to construct new absolute Galois groups out of existing ones. For example, for all fields K1 , . . . , Kn there exists a field K with Gal(K) isomorphic to the free product of Gal(K1 ), . . . , Gal(Kn ) (Pop, Melnikov, Ershov, Koenigsmann). Generalization of this result to infinite families of closed subgroups generalize the concepts “projective groups” and “PRC fields” or “PpC fields” to “relatively projective groups” and “pseudo closed fields” (Haran-Jarden-Pop). They generalize the classification of projective groups as those profinite groups appearing as absolute Galois groups of PAC fields. All of the exciting material mentioned in the preceding two paragraphs lie unfortunately outside the scope of this volume. It is my pleasure to thank colleagues and friends who critically read parts of the manuscript of the present edition of “Field Arithmetic”: Michael Bensimhoun, David Brink, Gregory Cherlin, Michael Fried, Wulf-Dieter Geyer, Peter M¨ uller, Dan Haran, Wolfgang Herfort, Alexander Lubotzky, Nikolay Nikolov, Dan Segal, Aharon Razon, and Irene Zimmermann. Tel Aviv, Spring 2004
Moshe Jarden
Introduction to the First Edition Our topic is the use of algebraic tools — coming mainly from algebraic geometry, number theory, and the theory of profinite groups — in the study of the elementary properties of classes of fields, and related algorithmic problems. (We take the precise definition of “elementary” from first order logic.) This subject has its more distant roots in Tarski’s observation that, as a consequence of elimination theory, the full elementary theory of the class of all algebraically closed fields is decidable; this relies on the Euclid algorithm of finding the greatest common divisor of two polynomials in one variable over a field. In its first phase this line of thought led to similar results on real closed fields and p-adic fields. The subject took a new turn with the work of James Ax [Ax2] on the elementary theory of the class of finite fields, which represents a radical departure in terms of the algebraic methods used. The analysis is based entirely on three properties of a finite field K: (1a) K is perfect. (1b) K has a unique extension of each degree. (1c) There is an explicitly computable function q(d, m) such that any absolutely irreducible variety V defined over K will have a K-rational point if |K| > q(dim(V ), deg(V )).
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The validity of the third condition for finite fields is a consequence of Riemann’s hypothesis for curves over finite fields. Methods of logic, specifically ultraproducts, led Ax to consider this condition for infinite fields as well, in which case the lower bound afforded by the function q is vacuous, and the condition becomes: (2) Every absolutely irreducible variety over K has a K-rational point. Fields satisfying (2) are said to be pseudo algebraically closed, or PAC. The second condition may be interpreted as a description of the absolute Galois group Gal(K) as a profinite group: Gal(K) is the free profinite group on one generator. In Ax’ approach it was convenient to have an Abelian absolute Galois group, but a strong trend in later work has been the systematic analysis of situations involving progressively more general Galois groups. One of our central goals here is the presentation of the general theory of PAC fields in its modern form, and its connections with other branches of algebra. From what we have said so far, some connections with algebraic geometry and profinite groups are visible; a number theoretic connection will appear shortly. One important feature of PAC fields is that they occur in profusion in nature and are in fact typical in the following sense. Since the absolute Galois group Gal(Q) of the rationals is a compact topological group, it carries a canonical invariant probability measure, the Haar measure. We can therefore ˜ ask for the probability that the fixed field Q(σ) of a sequence σ = (σ1 , ..., σe ) ˜ will be PAC; and we find that this occurs with probof automorphisms of Q ˜ ability 1. In addition, the absolute Galois group of Q(σ) is free on the e generators σ1 , ..., σe , again with probability 1. These facts are consequences of Hilbert’s irreducibility theorem for Q (Chapter 13), at least in the context of countable fields. We will develop other connections between the PAC property and Hilbertianity. There are also remarkable connections with number theory via the Chebotarev density theorem (Chapters 6, 13, 16, 20, 21, 31). For example, the ˜ probability that a given elementary statement ψ holds for the field Q(σ) coincides with the Dirichlet density of the set of primes for which it holds for the field Fp , and this density is rational. Thus, the “probability 1” theory of ˜ the fixed fields Q(σ) coincides with the theory of “all sufficiently large” finite fields, which by Ax’ work is an algorithmically decidable theory. ˜ Ax’ results extend to the “probability 1” theory of the fields Q(σ) for σ of length e > 1, by somewhat different methods (Chapter 20), although the connection with finite fields is lost. The elementary theories of such fields are largely determined by three properties: PAC, characteristic zero, and having an absolute Galois group which is free on e generators. To determine the full elementary theory of one such field K, it is also necessary to describe the ˜ intersection K ∩ Q. Although the absolute Galois group of a PAC field need not be free, it can be shown to be projective in a natural sense, and conversely any
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Introduction to the First Edition
projective profinite group occurs as the Galois group of some PAC field. In extending the theory from PAC fields with free Galois group to the general (projective) case, certain obstacles arise: for example, the algorithmic results do not extend. There is nonetheless a quite general theory, which enables us to identify some broad classes of projective profinite groups for which the associated classes of profinite groups behave well, and also to pinpoint unruly behavior in other case. One approach to the algorithmic problems associated with PAC fields leads to the study of profinite groups G with the embedding property (the terminology reflects a preoccupation with the corresponding fields): for each pair of continuous epimorphisms ϕ: G → A, α: B → A, where B is a finite quotient of G, we require that ϕ should factor through α. A perfect PAC field whose absolute Galois group is a group with the embedding property is called a Frobenius field. The elementary theory of all Frobenius fields can be computed quite explicitly. The algorithm has some relationship with elimination theory as used by Tarski. We associate to each elementary statement in the language of PAC fields a stratification of affine space into basic normal locally closed algebraic sets, each equipped with a Galois extension of its function field, and the given statement is reinterpreted as a statement about conjugacy classes of subgroups of the specified Galois groups. When the initial statement has no quantifiers this is a fairly trivial procedure, but addition of quantifiers corresponds to a special kind of “projection” of these Galois stratifications. This procedure has not yet been closely examined from the point of view of computational complexity. Like most procedures which operate by tracing through a series of projections, it is effective but hopelessly inefficient in its present form. It is not yet clear whether it is substantially less efficient than Tarski’s procedure for algebraically closed fields, nor whether, like that procedure, it can significantly reorganized and sped up. The Galois stratification algorithm relies on techniques of effective algebraic geometry, and also involves substantial algorithmic problems of a new type connected with the theory of profinite groups. Specifically, it is necessary to determine, given two collections A1 , ..., Am and B1 , ..., Bn of finite groups, whether or not there is a projective group with the embedding property which has each Ai as (continuous) image, but none of the groups Bj . The solution to this problem depends on recent work on projective covers (Chapter 22) and embedding covers (Chapter 24). Ultimately our decision problem reduces to the determination of the finite quotients of the projective cover of the embedding cover of a single finite group. The theory of projective covers leads also to the undecidability results alluded to earlier. A fairly natural encoding of graphs into profinite groups is lifted by this theory into the class of projective profinite groups, and then by looking at the corresponding PAC fields we see that their elementary theories encode algorithmically undecidable problems (the analogous results for graphs are well known).
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In the final chapter we return to our point of departure, the theory of finite fields. The zeta function of a Galois formula over a finite field is defined, and using a result of Dwork and Bombieri we show that some integral power of each such function is a product of an exponential and a rational function over Q. One of the goals of this book is to serve as a bridge between algebraists and logicians. For the algebraist there is a self contained introduction to the logic and model theory background for PAC fields (Chapter 7). Chapter 14 gives the “nonstandard” framework that suffices for Weissauer’s proof of Hilbert’s irreducibility theorem (Chapter 15), and Chapters 8 and 28 include basic recursion theory. On the other hand, for logicians with basic algebraic background (e.g. Lang’s book “Algebra”) Chapter 4 has the Stepanov-Bombieri elementary proof of the Riemann hypotheses for curves, and Chapter 6 gives an elementary proof of the Chebotarev density theorem. Both groups of readers may find the extensive treatment of profinite groups (Chapters 1, 17, 18, 22, 24, 25 and 26) and of Hilbertian fields (Chapters 12, 13, 15, and 16) valuable. Although PAC fields arise over arithmetically rich fields, they themselves lack properties that we associate with the arithmetic, say, of the rationals. For example, a PAC field F admits no orderings and all Henselizations of F are separably closed (Section 11.5). Many PAC field results generalize to pseudo real closed (PRC) fields. A field F is PRC if each absolutely irreducible variety defined over F has an F -rational point provided it has a nonsingular Fˆ -rational point in each real closure Fˆ of F . Thus, a PRC field without orderings is PAC. This, and the development of the theory of pseudo p-adically closed PpC fields are outside the scope of this book. We refer to [Prestel1], to [Jarden12], [Haran-Jarden2], and to [Haran-Jarden3] for literature about PRC fields and to [Haran-Jarden4] for PpC fields. Similarly, we give no account of the theories of real closed fields and p-adically closed fields that preceded the development of the theory of PAC fields. In particular, for Hilbert’s 17th problem and the Ax-Kochen-Ershov p-adic theory, we refer the reader to [Prestel2], [Ax-Kochen1, Ax-Kochen2, and Ax-Kochen3], and [Prestel-Roquette]. Acknowledgement: We are indebted to several colleagues who corrected errors in the process of critically reading the manuscript. In particular, WulfDieter Geyer, Gregory L. Cherlin, and Dan Haran made crucial contributions. Michael D. Fried, Gainesville, Florida Moshe Jarden, Tel Aviv, Israel Summer 1986
Notation and Convention Z = the ring of rational integers. Zp = the ring of p-adic integers. Q = the field of rational numbers. R = the field of real numbers. C = the field of complex numbers. Fq = the field with q elements. Ks = the separable closure of a field K. Kins = the maximal purely inseparable extension of a field K. ˜ = the algebraic closure of a field K. K Gal(L/K) = the Galois group of a Galois extension L/K. ˜ We call a polynomial f ∈ K[X] separable if f has no multiple root in K. Gal(f, K) = the Galois group of a separable polynomial f ∈ K[X] over a field K viewed as a permutation group of the roots of f . Gal(K) = Gal(Ks /K) = the absolute Galois group of a field K. irr(x, K) = the monic irreducible polynomial of an algebraic element x over a field K. Whenever we form the compositum EF of field extensions of a field K we tacitly assume that E and F are contained in a common field. |A| = #A = the cardinality of a set A. R× = the group of invertible elements of a ring R. Quot(R) = the quotient field of an integral domain R. A ⊂ B means “the set A is properly contained in the set B”. ax = x−1 ax, for elements a and x of a group G. H x = {hx | h ∈ H}, for a subgroup H of G. Given subgroups A, B of a group G, we use “A ≤ B” for “A is a subgroup of B” and “A < B” for “A is a proper subgroup of B”. Given an Abelian (additive) group A and a positive integer n, we write An for the Ssubgroup {a ∈ A | na = 0}. For a prime number p we let ∞ Ap∞ = i=1 Api . For a group B that acts on a group A from the right, we use B n A to denote the semidirect product of A and B. Bold face letters stand for n-tuples, e.g. x = (x1 , . . . , xn ). ord(x) is the order of an element x in a group G. For a positive integer n and an integer a with gcd(a, n) = 1, we use ordn a to denote the order of a modulo n. Thus, ordn a is the minimal positive integer d with ad ≡ 1 mod n. In the context of groups, Sn (resp. An ) stands for the full permutation group (resp. alternative group) of {1, . . . , n}. In the context of groups, Cn stands for the cyclic multiplicative group of order n. Likewise we use Z/nZ for the additive multiplicative group of order n. In S the context of fields, ζn stands for a primitive root of unity of order n. · i∈I Bi is the disjoint union of sets Bi , i ∈ I.
Chapter 1. Infinite Galois Theory and Profinite Groups The usual Galois correspondence between subgroups of Galois groups of finite Galois extensions and intermediate fields is not valid for infinite Galois extensions. The Krull topology restores this correspondence for closed subgroups (Proposition 1.3.1). Since Galois groups are inverse limits of finite groups, they are profinite. Conversely, we define profinite groups, independently of Galois theoretic properties. Each profinite group actually appears as a Galois group ˆ and (Corollary 1.3.4). In particular, we study the procyclic groups Zp and Z prove that every finite field has the latter as its absolute Galois group.
1.1 Inverse Limits Our interest in inverse limits comes from infinite Galois theory: Infinite Galois groups are inverse limits of finite Galois groups. As a preparation to the study of “profinite groups” we define in this section inverse limits of topological spaces and characterize inverse limits of finite topological spaces. Let I be a set with a partial ordering ≤; that is, ≤ is a binary relation which is reflexive, transitive, and a ≤ b and b ≤ a imply a = b. We call (I, ≤) a directed partially ordered set if in addition (1) for all i, j ∈ I there exists k ∈ I with i ≤ k and j ≤ k. An inverse system (also called a projective system) over a directed partially ordered set (I, ≤) is a data (Si , πji )i,j∈I where Si is a set and πji : Sj → Si is a map for all i, j ∈ I with i ≤ j satisfying the following rules: (2a) πii the identity map for each i ∈ I. (2b) πki = πji ◦ πkj if i ≤ j ≤ k. Q Let S be the subset of the cartesian product i∈I Si consisting of all elements Q s = (si )i∈I with πji (sj ) = si for all i ≤ j. Note: S may be empty. Let pri : j∈I Sj → Si be the projection on the ith coordinate. Denote the restriction of pri to S by πi . Then πi = πji ◦ πj for every i ≤ j. We say (S, πi )i∈I is the inverse (or projective) limit of the family (Si )i∈I with respect to the maps πji . Denote S by lim Si . ←− 0 )i,j∈I be another inverse system over I. Suppose for each Let (Si0 , πji 0 ◦ θj = θi ◦ πji for all i ≤ j. (We i ∈ I we are given a map θi : Si → Si0 with πji say that the maps θi , i ∈ I, are compatible.) Then there exists a unique map θ: lim Si → lim Si0 satisfying πi0 ◦ θ = θi ◦ πi for each i ∈ I: θ maps ←− ←− s = (si ) ∈ lim Si onto θ(s) with θ(s)i = θi (si ). Denote θ by lim θi . ←− ←− Similarly, let X be a set and for each i ∈ I let θi : X → Si be a map satisfying πji ◦ θj = θi whenever i ≤ j (Again, we say that the maps θi ,
2
Chapter 1. Infinite Galois Theory and Profinite Groups
i ∈ I, are compatible.) Then there exists a unique map θ: X → lim Si with ←− πi ◦ θ = θi for each i ∈ I. When Si are topological spaces, we assume πji are continuous. Then we equip lim Si with the topology induced from the product topology of ←− Q Q i∈I Si . Recall Q that the product topology on i∈I Si has a basis consisting of the sets i∈I Ui , with Ui open in Si for each i ∈ I, and Ui = Si for all but finitely many i ∈ I. Since pri is continuous, so is πi , i ∈ I. If θi : Si → Si0 are continuous, then θ: lim Si → lim Si0 is also continuous. ←− ←− Lemma 1.1.1: The collection of all subsets of S = lim Si of the form πi−1 (Ui ) ←− with Ui open in Si is a basis for the topology of S. Proof: Let s ∈ S. each basic open neighborhood of s has the Q Q By definition, V × form V = S ∩ j∈J j i∈I r J Sj , where J is a finite subset of I and subset of S , j ∈ J. Take k ∈ I with k ≥ j for all j ∈ J. Vj is an open j T −1 (Vj ) is an open subset of Sk and πk−1 (Uk ) is an open Then Uk = j∈J πkj neighborhood of s in V . Therefore, the collection πi−1 (Ui ), i ∈ I, is a basis for the topology of S. Lemma 1.1.2: In the notation above, if each Si , i ∈ I, is a Hausdorff space, Q then lim Si is a closed subset of Si . ←− Q Proof: Suppose s = (si ) ∈ Si does not belong to lim Si . Then there are ←− Ui i, j ∈ I with i ≤ j and πji (sj ) 6= si . Take open disjoint neighborhoods Q −1 0 and Ui0 of si and πji (sj ), respectively. Then U × π (U ) × S is an i k i ji k6 = i,j Q open neighborhood of s in Si that does not intersect lim Si . ←− Q
If, in addition, each Si is compact, then Tychonoff’s theorem implies Si is also compact. Thus, lim Si with the induced topology is compact. ←−
Lemma 1.1.3: The inverse limit S of an inverse system of nonempty compact Hausdorff spaces Si , i ∈ I, is a nonempty compact Hausdorff space. T Proof: We only need Q to prove that S is nonempty. Indeed, S = k≥j Q Rkj , Si → where Rkj = {s ∈ Si | πkj (sk ) = sj }. The natural map prk × prj : Sk × Sj is continuous. The Hausdorff property of Sj implies T = {(sk , sj ) ∈ Sk × Sj | πkj (sk ) = sj } is a closed Q subset of QSk × Sj . Hence, Rkj = (prk × prj )−1 (T ) is a closed subset of Si . Since Si is compact, we only need to show that the intersection of finitely many of the Rkj is nonempty. Indeed, let j1 ≤ k1 , . . . , jn ≤ kn be n pairs in I. Choose l ∈ I with ki ≤ l, i = 1, . . . , n, and choose sl ∈ Sl . Define sji = πl,ji (sl ) and ski = πl,ki (sl ), for i = 1, . . . , n. For each r ∈ I r 1 , . . . , jn , k1 , . . . , kn } let sr an arbitrary element of Sr . T{j n Then s = (si ) ∈ i=1 Rki ,ji .
1.1 Inverse Limits
3
Corollary 1.1.4: The inverse limit of an inverse system of nonempty finite sets is nonempty. Proof: Equip each of the finite sets with the discrete topology.
0 )i,j∈I be inverse systems of Corollary 1.1.5: Let (Si , πji )i,j∈I and (Si0 , πji 0 compact Hausdorff spaces. Let θi : Si → Si be a compatible system of surjective continuous maps. Put S = lim Si , S 0 = lim Si0 , and θ = lim θi . Then ←− ←− ←− θ: S → S 0 is surjective.
Proof: Let s0 = (s0i ) be an element of S 0 . Then (θi−1 (s0i ), πji )i,j∈I is an inverse system of nonempty compact Hausdorff spaces. By Lemma 1.1.3, the inverse limit of θi−1 (s0i ) is nonempty. Each element in the inverse limit is mapped by θ onto s0 . Corollary 1.1.6: Let X be a compact space, (Si , πji )i,j∈I an inverse system of Hausdorff spaces, and θi : X → Si a compatible system of continuous surjective maps. Put θ = lim θi . Then θ: X → lim Si is surjective. ←− ←− Consider s = (si )i∈I ∈ lim Si . Then θi−1 (si ) is a closed nonempty ←− subset of X. For i1 , . . . , in ∈ I there is a j ∈ I with i1 , . . . , in ≤ j. Then (si1 ) ∩ · · · ∩ θi−1 (sin ), so θi−1 (si1 ) ∩ · · · ∩ θi−1 (sin ) is nonempty. θj−1 (sj ) ⊆ θi−1 1 n 1 n Since X is compact, there is an x ∈ X which belongs to θi−1 (si ) for every i ∈ I. It satisfies, θ(x) = s. Thus, θ is surjective. Proof:
A profinite space is an inverse limit of an inverse system of finite discrete spaces. Lemma 1.1.7: A compact Hausdorff space S is profinite if and only if its topology has a basis consisting of open-closed sets. Proof: Suppose first S = lim Si is an inverse limit of finite discrete spaces ←− Si . Let πi : S → Si be the projection on the ith coordinate, i ∈ I. By Lemma 1.1.1, the sets πi−1 (Ui ), where i ∈ I and Ui is a subset of Si , form a basis of the topology of S. Since Si is discrete, Ui is open-closed. Hence, so is πi−1 (Ui ). Now suppose the topology of S has a basis consisting of open-closed sets. An open-closed partition of S is a finite set A of nonempty openclosed disjoint subsets of S whose union is S. Denote the set of all openclosed partitions of S by A. Let A, A0 , B ∈ A. Write A ≤ B if for each B ∈ B there is an A ∈ A with B ⊆ A. Then A is unique. Next note that C = {A ∩ A0 | A ∈ A, A0 ∈ A0 } belongs to A and satisfies A, A0 ≤ C. Thus, (A, ≤) is a directed partially ordered set. When A ≤ B, define a map πB,A from B to A by πB,A (B) = A, where A is the unique element of A containing B. Equip each A ∈ A with the discrete topology. Then (A, πB,A )A,B∈A is an inverse system of finite discrete spaces.
4
Chapter 1. Infinite Galois Theory and Profinite Groups
Its limit S 0 = lim A is a profinite space. We construct a homeomorphism of ←− S onto S 0 . Let s ∈ S and A ∈ A. Define θA (s) to be the unique A in A which contains s. Then θA : S → A is a continuous surjective map. If A ≤ B, then πB,A ◦ θB = θA . Hence, by Corollary 1.1.6, there is a continuous surjective map θ: S → S 0 satisfying πA ◦ θ = θA for each A ∈ A. Suppose s, s0 are distinct elements of S. Since S is Hausdorff, there are disjoint open-closed subsets A and A0 of S with s ∈ A and s0 ∈ A0 . Put A00 = S r(A ∪ A0 ). Then A = {A, A0 , A00 } is an open-closed partition of S, θA (s) = A, and θA (s0 ) = A0 . Thus, θA (s) 6= θA (s0 ), so θ(s) 6= θ(s0 ). Therefore, θ is bijective. Since S is compact and S 0 is Hausdorff, θ is a homeomorphism. Consequently, S is a profinite space. Remark 1.1.8: Totally disconnected spaces. Let S be a compact Hausdorff space. Suppose S has a basis for its topology consisting of open-closed sets. It is not difficult to see that S is totally disconnected; that is each s in S is its own connected component. Conversely, if S is totally disconnected, then the topology of S has a basis consisting of open-closed subsets [Ribes-Zalesski, Thm. 1.1.12].
1.2 Profinite Groups We survey here the basic properties of compact groups. In particular, this will apply to profinite groups. Topological Groups. A topological group is a group G equipped with a topology in which the product (x, y) 7→ xy and the inverse map x 7→ x−1 are continuous. It follows that for each a ∈ G the maps x 7→ ax, x 7→ xa, and x 7→ x−1 are homeomorphisms. We always assume {1} is a closed subset of G. Consequently, {a} is a closed subset of G for each a ∈ G. It follows that G is a Hausdorff space. Indeed, let a, b be distinct elements of G. The identity 1 · 1−1 = 1 and the continuity of the group operations give open neighborhoods U and W of 1 with U W −1 ⊆ G r{a−1 b}. Thus, aU and bW are disjoint open neighborhoods of a and b, respectively, as needed. In addition, each closed subgroup H of GSof finite index is open. Indeed, there are ai ∈ G, i ∈ I, with I finite and G = · i∈I ai H. Each of theSsets ai H is closed and there is a j ∈ I with aj H = H. Therefore, H = G r · i6=j ai H is open. Conversely, if G is compact, then every open subgroup H of G is of finite index. Otherwise, G would be a disjoint union of infinitely many cosets of H, each of which is open. Let N be a closed normal subgroup of G and π: G → G/N the quotient ¯ of G/N map. Equip G/N with the quotient topology. Thus, a subset U −1 ¯ is open if and only if π (U ) is open. It follows that the group operations of G/N are continuous. In addition, a subset C¯ of G/N is closed if and
1.2 Profinite Groups
5
¯ is closed. In particular, since N = π −1 (1) is closed, {1} only if π −1 (C) is a closed subset of G/N . Consequently, G/N is a topological group and π: G → G/N is a continuous map. Moreover, if U is an open subset of G, S then π −1 (π(U )) = n∈N nU , so π(U ) is open in G/N . Therefore, π is an open map. Suppose again G is compact. Let θ be a continuous homomorphism of G into a topological group H. Since H is Hausdorff, θ is a closed map. Suppose in addition, θ is surjective. Put N = Ker(θ). Let π: G → G/N ¯ G/N → H the map induced by θ. Then θ¯ is be the quotient map and θ: an isomorphism of abstract groups. In addition, θ¯ is a continuous bijective map of the compact group G/N onto the Hausdorff group H. Hence, θ¯ is an isomorphism of topological groups and θ is an open map. This is the first isomorphism theorem for compact groups. Let H be a closed subgroup and N a closed normal subgroup of G. Then HN = {hn | h ∈ H, n ∈ N } is the image of the compact group H × N under the continuous map (h, n) 7→ hn. Hence, HN is a closed subgroup of G. As in the preceding paragraph, the group theoretic isomorphism θ: HN/N → H/H ∩ N defined by θ(hN ) = h(H ∩ N ), h ∈ H, is a homeomorphism, so θ is an isomorphism of topological groups. Similarly, if M and N are closed normal subgroups of G with N ≤ M , then M/N is a closed normal subgroup of G/N and the map G/M → (G/N )/(M/N ) given by gM 7→ (gN )M/N , g ∈ G, is an isomorphism of topological groups. Here is one way to equip an abstract group G with a topology. Let N be a family of normal subgroups of G, closed under finite intersections, such that the intersection of all N ∈ N is 1. Take N to be a basis for the open neighborhoods of 1. A basis for the open neighborhoods of a ∈ G is the family Na = {aN | N ∈ N }. The union of the Na is then a basis for a group topology of G. For example, the identity xN · yN = xyN for normal subgroups N implies multiplication is continuous. Let (Gi , πji )i,j∈I be an inverse system of topological groups and continuous homomorphisms πji : Gj → Gi , for each i, j ∈ I with j ≥ i. Then G = lim Gi is a topological group and the projections πi : G → Gi are con←− 0 ii,j∈I be another system of topological tinuous homomorphisms. Let hG0i , πji 0 0 groups with G = lim Gi . Suppose θi : Gi → G0i , i ∈ I, is a compatible system ←− of continuous homomorphisms. Then the corresponding map θ: G → G0 is a continuous homomorphism. Profinite Groups. In this book, we primarily consider an inverse system of finite groups (Gi , πji )i,j∈I , each equipped with the discrete topology. We call the inverse limit G = lim Gi a profinite group. By Lemma 1.1.3, G is ←− a compact group. By Lemma 1.1.1, the open-closed sets πi−1 (gi ), gi ∈ Gi , i ∈ I, form a basis for the topology on G. In particular, the open normal subgroups of G form a basis for the open neighborhoods of 1.
6
Chapter 1. Infinite Galois Theory and Profinite Groups
Remark 1.2.1: Basic rules. Here are some basic rules for profinite groups G and H which we use in this book without explicit reference: (a) A subgroup H of G is open if and only if H is closed of a finite index. The intersection of all open normal subgroups of G is 1. Every open subset of G is a union of cosets gi Ni with Ni open normal and gi ∈ G. (b) Every profinite group is compact, Hausdorff, and has a basis for its topology consisting of open-closed sets (Lemmas 1.1.3 and 1.1.7). (c) A subset C of a profinite group is closed if and only if C is compact (use (b)). (d) A subset B of a profinite group is open-closed if it is a union of finitely many cosets gi N with N open normal and gi ∈ G (use (a) and the compactness of B). (e) Every homomorphism ϕ: G → H is tacitly assumed to be continuous. In particular, ϕ maps compact subsets of G onto compact subsets of H. Hence, ϕ maps closed subsets of G onto closed subsets of H (use (c)). (f) By the first isomorphism theorem for compact groups, every epimorphism ϕ: G → H of profinite groups is an open map. In particular, ϕ maps open subgroups of G onto open subgroups of H. We list below special properties of profinite groups not shared by all compact groups: Lemma 1.2.2: Let {Hi | i ∈ I} be a directed family of closed subset of a profinite group G; that is: T (1) for every finite subset J of I there is an i ∈ I with Hi ≤ j∈J Hj . T Put H = i∈I Hi . Then: (a) For every open subgroup U of G containing H, there is an i ∈ I with H Ti ≤ U . T (b) i∈I KHi = KH and i∈I Hi K = HK for T every closedTsubset K of G. ¯ be an epimorphism. Then ϕ( (c) Let ϕ: G → G i∈I Hi ) = i∈I ϕ(Hi ). (d) Let K and K 0 be closed subgroups of G. Then the set S = {σ ∈ G | K σ = K 0 } is closed. (e) Let K and K 0 be closed subgroups of G which contain H. Suppose each Hi is normal and KHi and K 0 Hi are conjugate. Then K and K 0 are conjugate. (f) Let G = lim Gi be an inverse limit of finite groups, πi : G → Gi the ←− quotient maps, and K, K 0 closed subgroups of G. Suppose πi (K) and πi (K 0 ) are conjugate in Gi for each i. Then K and K 0 are conjugates. T Proof of (a): The set Hi ∩ (G r U ) is closed for every i ∈ I and i∈I Hi ∩ r (G r U ) = H ∩ (G T U ) = ∅. Since G is compact, there exists a finite subset J of ITsuch that j∈J Hj ∩ (G r U ) = ∅. By (1) there exists i ∈ I such that Hi ≤ j∈J Hj . Hence, Hi ≤ U . T Proof of (b): Let g ∈ i∈I KHi . Then, for each i ∈ I there are ki ∈ K and hi ∈ Hi with g = ki hi . Hence, the closed subset Hi ∩ g −1 K of G is nonempty.
1.2 Profinite Groups
7
−1 By (1), the intersection of finitely many T of the sets Hi ∩ g K is nonempty. Since G is compact, there exists h ∈ i∈I Hi ∩ g −1 K. It satisfies h ∈ H and h = g −1 k for some k ∈ K. Therefore, g = kh−1 ∈ KH.
Proof of (c): Let K = Ker(ϕ). Then ϕ induces a bijection between the set of closed subgroups lying between K and G and the set of closed subgroups ¯ Hence, by (b), ϕ(H) = ϕ(KH) = ϕ(T KHi ) = T ϕ(KHi ) = of G. i∈I i∈I T i∈I ϕ(Hi ). Proof of (d): Denote the set of all open normal subgroups of G by N . For each N ∈ N , let SN be the inverse image under the quotient map G → G/N of the finite set {s ∈ G/N | (KN/N )s = K 0 N/N }. If s ∈ SN for each N ∈ T N , then, by (b), K s = Then, SN is closed. T T s 0 0 N ∈N K N = N ∈N K N = K . Hence, S = N ∈N SN . Therefore, S is closed. Proof of (e): For each i ∈ I, Si = {σ ∈ G | K σ Hi = K 0 Hi } is a nonempty T subset of G. By (d), Si is closed. In theTnotation of (1) we have Si ⊆ j∈J Sj . Hence, by compactness, there is a σ ∈ i∈I Si . Thus, K σ Hi = K 0 Hi for each i ∈ I. We conclude from (b) that K σ = K 0 . Proof of (f): Apply (e) to Hi = Ker(πi ).
Lemma 1.2.3: Each closed subgroup H of a profinite group G is the intersection of open subgroups. Proof: Let T N be the set of all open normal subgroups of G. By Lemma 1.2.2(b), N ∈N N H = H. The following result gives a sufficient condition for a compact group to be profinite. We use it below to prove that the category of profinite groups is closed under various natural operations. Lemma 1.2.4: Let G be a compact group and {Ni | i ∈ I} a directed family of closed T normal subgroups of G of finite index satisfying (2) i∈I Ni = 1. Then G = lim G/Ni and G is a profinite group. ←− Proof: By definition, lim G/Ni is a profinite group. We only need to prove ←− the isomorphism. The quotient maps G → G/Ni define a continuous embedding θ of G into lim G/Ni . If (gi Ni )i∈I is an element of the latter group, ←− then the closed subsets gi Ni ofT G have the finite intersection property. Since G is compact, there exists g ∈ i∈I gi Ni . Then, θ(g) = (gi Ni )i∈I . Thus, θ is bijective. Compactness of G and the Hausdorff property of lim G/Ni imply ←− that θ is an isomorphism of topological groups.
8
Chapter 1. Infinite Galois Theory and Profinite Groups
Lemma 1.2.5: The following statements hold for each closed subgroup H of a profinite group G: (a) The group H is profinite. Moreover, for each open normal subgroup M of H there exists an open normal subgroup N of G with H ∩ N ≤ M . If M is normal in G, then N can be chosen such that H ∩ N = M . (b) Every open subgroup H0 of H is the intersection of an open subgroup of G with H. (c) Every homomorphism ϕ0 of H into a finite group A extends to a homomorphism ϕ: H 0 → A, where H 0 is an open subgroup of G containing H. If H is normal in G, then H 0 can be chosen to be normal. Proof of (a): If N is an open normal subgroup of G, then H ∩ N is an open normal subgroup of H. The family of all groups H ∩ N is directed and its intersection is trivial. Hence, H = lim H/H ∩ N is a profinite group. ←− Let M be an open normal subgroup of H. By Lemma 1.2.2(a), H ∩ N ≤ M for some open normal subgroup N of G. If M / G, then M N / G and H ∩ MN = M. Proof of (b): The intersection of all conjugates of H0 in H is an open normal subgroup of H. Hence, by the first statement of the lemma, there is an open normal subgroup N of G such that M = H ∩ N ≤ H0 . Then G0 = H0 N is an open subgroup of G and H ∩ G0 = H0 . Proof of (c): By (a), G has an open normal subgroup N with H ∩ N ≤ Ker(ϕ0 ). Put H 0 = HN and define ϕ: H 0 → A by ϕ(hn) = ϕ0 (h). Then ϕ is a well defined homomorphism which extends ϕ0 . Lemma 1.2.6: These statements hold: (a) If N is a closed normal subgroup of a profinite group G, then G/N is profinite. Q (b) The cartesian product, G = i∈I Gi , of profinite groups is profinite. (c) Every inverse limit, G = lim Gi , of profinite groups is profinite. ←− Proof of (a): A closed normal group N is an intersection of open subgroups (Lemma 1.2.3) and therefore of open normal subgroups Ni , i ∈ I, of G. Now apply Lemma 1.2.4 to G/N and {Ni /N | i ∈ I}. Proof of (b): Consider a finite subset J of Q I. For each Q j ∈ J let Nj be an open normal subgroup of Gj . Then N = j∈J Nj × i∈I r J Gi is an open normal subgroup of G. The family of all these normal subgroups is directed and its intersection is 1. By Lemma 1.2.4, G = lim G/N is a profinite group. ←− Proof of (c): Observe that G is a closed subgroup of the cartesian product Q i∈I Gi (Lemma 1.1.2). By (b) this product is profinite. Now apply Lemma 1.2.5.
1.3 Infinite Galois Theory
9
Lemma 1.2.7: Each epimorphism ϕ: G → A of profinite groups has a continuous set theoretic section ϕ0 : A → G. That is, ϕ0 is a continuous map satisfying ϕ ◦ ϕ0 = idA . Proof: Let K = Ker(ϕ). We split the proof into two parts. Part A: K is finite. Lemma 1.2.5 gives an open subgroup H of G with K ∩ H = 1. Thus, ϕ maps H bijectively onto B = ϕ(H). As both H and B are Hausdorff and compact, ϕ|H is a homeomorphism. Set β = (ϕ|H )−1 . Next S n let (A : B) = n, choose a1 , . . . , an ∈ A and g1 , . . . , gn ∈ G with A = · i=1 ai B and ϕ(gi ) = ai , i = 1, . . . , n. Define ϕ0 : A → G by ϕ0 (ai b) = gi β(b), b ∈ B. Then ϕ0 is a continuous set theoretic section of ϕ. Part B: K is arbitrary. Denote the set of all closed normal subgroups of G which are contained in K by L. For each L ∈ L, let ϕL : G/L → A be the epimorphism induced by ϕ. For each L0 ∈ L with L0 ≤ L let ϕL0 ,L : G/L0 → G/L be the quotient map. Let Φ0 be the set of all pairs (L, ϕ0L ) where L ∈ L and ϕ0L : A → G/L is a set theoretic section of ϕL . Since ϕK : G/K → A is an isomorphism, it has an inverse ϕ0K . Hence, (K, ϕ0K ) is in Φ0 . Define a partial ordering on Φ0 as follows: (L0 , ϕ0L0 ) ≤ (L, ϕ0L ) if L0 ≤ L and ϕL0 ,L ◦ ϕ0L0 = ϕ0L . Suppose Φ00 = {(Li , ϕ0i ) | i ∈ I} is a descending chain in Φ0 ; that is, every two elements of Φ00 are comparable. Put L00 = T 00 lim G/Li . Thus, i∈I Li . Then, by Lemmas 1.2.6(a) and 1.2.4, G/L = ←− 0 0 the compatible maps ϕi : A → G/Li give a section ϕL00 : A → G/L00 to ϕL00 . Thus, (L00 , ϕ0L00 ) is a lower bound for Φ00 . Zorn’s lemma gives a minimal element (L, ϕ0L ) of Φ0 . Assume L 6= 1. Then L has a proper open subgroup L0 which is normal in G. Consider the epimorphism ϕL0 ,L : G/L0 → G/L with the finite kernel L/L0 . Part A gives a set theoretic section ϕ0L,L0 to ϕL0 ,L . Set ϕ0L0 = ϕ0L,L0 ◦ ϕ0L . Then (L0 , ϕ0L0 ) is an element of Φ0 which is smaller than (L, ϕ0L ). This contradiction to the minimality of (L, ϕ0L ) proves that L = 1. Put ϕ0 = ϕ0L . Then ϕ0 is a continuous set theoretic section of ϕ.
1.3 Infinite Galois Theory Let N be a Galois extension of a field K. The Galois group Gal(N/K) associated with N/K consists of all automorphisms of N that fix each element of K. If N/K is a finite extension and H1 , H2 are subgroups of Gal(N/K) with the same fixed fields in N , then H1 = H2 . This is not the case any more if N/K is infinite. Consider S∞ for example the case where K = Fp for some prime number p and N = i=1 Fpi . Let ϕ be the Frobenius automorphism of N/Fp . It is defined by the rule ϕx = xp for each x ∈ N . Let G0 be the discrete subgroup of Gal(N/Fp ) generated by ϕ. It is a countable group and Fp is its fixed field in N . On the other hand, each element of Gal(Fp2i /Fp ) has exactly
10
Chapter 1. Infinite Galois Theory and Profinite Groups
two extensions to Fp2i+1 . Hence, there are 2ℵ0 sequences (σ1 , σ2 , σ3 , . . .) with σi ∈ Gal(Fp2i /Fp ) such that the restriction of σi+1 to Fp2i is σi for i = 1, 2, 3, . . . . Each such sequence defines a unique σ ∈ Gal(N/Fp ) whose restriction to Fp2i is σi , i = 1, 2, 3, . . . . It follows that the cardinality of Gal(N/Fp ) is 2ℵ0 . In particular, Gal(N/Fp ) is different from G0 but has the same fixed field, namely Fp . The Galois correspondence is restored for closed subgroups of Gal(N/K) in the “Krull topology” which we now introduce. Denote the set of all intermediate fields K ⊆ L ⊆ N , with L/K finite and Galois, by L. To each L ∈ L associate the (finite) Galois group Gal(L/K). If L0 ∈ L and L ⊆ L0 , then resL : Gal(L0 /K) → Gal(L/K) is an epimorphism. Consider the inverse limit lim Gal(L/K), with L ranging over L. Every σ ∈ ←− Gal(N/K) defines a unique element (resL σ)L∈L of lim Gal(L/K). Con←− versely, every (σL )L∈L ∈ lim Gal(L/K) defines a unique σ ∈ Gal(N/K) ←− with resL σ = σL for each L ∈ L. Thus, σ 7→ (resL σ)L∈L is an isomorphism Gal(N/K) ∼ = lim Gal(L/K). This isomorphism induces a topology ←− on Gal(N/K) through the topology on lim Gal(L/K): the Krull topology. ←− Thus, under the Krull topology, Gal(N/K) becomes a profinite group and the family N = {Gal(N/L) | L ∈ L} is a basis for the open neighborhoods of 1. If N/K is a finite extension, then the Krull topology is discrete. Suppose L is a finite extension of K contained in N . Then its Galois ˆ is the smallest Galois extension of K that contains L. It is finite closure L over K and is contained in N . Write Gal(N/L) as a union of right cosets of ˆ to see that Gal(N/L) is an open closed subgroup of Gal(N/K). Gal(N/L) Suppose L is an arbitrary extension of K in N . Then L is the union of a T family {Li | i ∈ I} of finite extensions of K. Hence, Gal(N/L) = i∈I Gal(N/Li ). Therefore, Gal(N/L) is a closed subgroup of Gal(N/K). If S is a set of automorphisms of N , then N (S) = {x ∈ N | σx = x for every σ ∈ S} is the fixed field of S in N . N (S) is also the fixed field in N of the closed subgroup hSi of Gal(N/K) generated by S. If S = {σ1 , . . . , σe } is a finite set, replace N (S) by N (σ1 , . . . , σe ). Let M be a Galois extension of K in N . Denote by resM (or res) the homomorphism from Gal(N/K) into Gal(M/K) that maps σ ∈ Gal(N/K) onto its restriction resM σ to M . It is a continuous surjective map. Proposition 1.3.1: Let N be a Galois extension of a field K. Then L 7→ Gal(N/L) is a bijection from the family of fields L lying between K and N onto the family of closed subgroups of G = Gal(N/K). The inverse map is H 7→ N (H). Proof: Consider a field extension L of K in N . Put H = Gal(N/L). Then L ⊆ N (H). Each x ∈ N (H) is contained in a finite Galois extension M ⊆ N
1.3 Infinite Galois Theory
11
of L. Since the map res: Gal(N/L) → Gal(M/L) is surjective, σx = x for every σ ∈ Gal(M/L). By finite Galois theory, x ∈ L. Hence, N (Gal(N/L)) = L. Conversely, let H be a closed subgroup of G and put L = N (H). Then H ≤ Gal(N/L). Consider σ ∈ Gal(N/L). In order to prove σ ∈ H it suffices to show that σ is in the closure of H. Indeed, let M ⊆ N be a finite Galois extension of K. Then M ∩ L = M (resM H). Finite Galois theory shows resM σ ∈ Gal(M/M ∩ L) = resM H. Hence, H ∩ σGal(N/M ) is nonempty. It follows that Gal(N/N (H)) = H. As in finite Galois theory [Lang7, pp. 192–199], Proposition 1.3.1 gives the following rules for the Galois correspondence: (1a) L1 ⊆ L2 ⇐⇒ Gal(N/L2 ) ≤ Gal(N/L1 ). (1b) H1 ≤ H2 ⇐⇒ N (H2 ) ⊆ N (H1 ). (1c) N (H1 ) ∩ N (H2 ) = N (hH1 , H2 i), where hH1 , H2 i is the closed subgroup of G generated by the closed subgroups H1 and H2 . (1d) Gal(N/L1 ∩ L2 ) = hGal(N/L1 ), Gal(N/L2 )i. (1e) Gal(N/L1 L2 ) = Gal(N/L1 ) ∩ Gal(N/L2 ). (1f) N (H1 ∩ H2 ) = N (H1 )N (H2 ). Since Galois groups are compact, their images under restriction are closed. As in the finite case this produces other theorems of infinite Galois theory. (2a) N (σHσ −1 ) = σN (H) and (2b) Gal(N/σL) = σGal(N/L)σ −1 , for every σ ∈ G. (2c) A closed subgroup H of G is normal if and only if L = N (H) is a Galois extension of K. (2d) If L is a Galois extension of K and L ⊆ N , then res: Gal(N/K) → Gal(L/K) is a continuous open epimorphism with kernel Gal(N/L) and Gal(L/K) ∼ = Gal(N/K)/Gal(N/L). (2e) res: Gal(LM/M ) → Gal(L/L ∩ M ) is an isomorphism for every Galois extension L of K and every extension M of K. (2f) If, in (2e), M is also a Galois extension of K, then σ 7→ (resL σ, resM σ) is an isomorphism Gal(LM/L ∩ M ) ∼ = Gal(L/L ∩ M ) × Gal(M/L ∩ M ) and Gal(LM/K) ∼ = {(σ, τ ) ∈ Gal(L/K) × Gal(M/K) | resL∩M σ=resL∩M τ }. In both cases we use the product topology on products of groups. The first isomorphism of (2f) is a special case of the second one. To prove the second isomorphism, note first that the map σ 7→ (resL σ, resM σ) is a continuous injective map of the left hand side onto the right hand side. Hence, it suffices to prove surjectivity. Thus, consider ρ ∈ Gal(L/K) and τ ∈ Gal(M/K) with resL∩M ρ = resL∩M τ . Extend ρ to an automorphism ρ1
12
Chapter 1. Infinite Galois Theory and Profinite Groups
of LM and let ρ0 = resM ρ1 . Then ρ−1 0 τ ∈ Gal(M/L ∩ M ). By (2e) there is τ . The element σ = ρ1 λ of Gal(LM/K) a λ ∈ Gal(LM/L) with resM λ = ρ−1 0 satisfies resL σ = ρ and resM σ = τ , as desired. These statements are useful when N is the separable closure Ks of K. Denote Gal(Ks /K) by Gal(K), the absolute Galois group of K. Next we show that profinite groups are Galois groups. Lemma 1.3.2: Suppose a profinite group G acts faithfully as automorphisms of a field F . Suppose for each x ∈ F , the stabilizer S(x) = {σ ∈ G | σx = x}, is an open subgroup of G. Then F is a Galois extension of the fixed field K = F (G) and G = Gal(F/K). Proof: If G is a finite group, this is a result of Artin [Lang7, p. 264]. In general the group H = S(x1 ) ∩ · · · ∩ S(xn ) is open in G for every x1 , . . . , xn ∈ F . Therefore, so is the intersection N of all the conjugates of H (Exercise 4). The finite quotient group G/N acts faithfully on the field L = K(Gx1 , . . . , Gxn ). It has K as its fixed field. Thus, L is a finite Galois extension of K and G/N ∼ = Gal(L/K). The field F is the union of all above L and 1 is the intersection of all the above N . Hence, F is a Galois extension of K and Gal(F/K) ∼ = lim Gal(L/K) ∼ = lim G/N ∼ = G. ←− ←− Proposition 1.3.3: Let L/K be a Galois extension and α: G → Gal(L/K) an epimorphism of profinite groups. Then there is a Galois extension F/E and an isomorphism ϕ: Gal(F/E) → G such that F is a purely transcendental extension of L, L ∩ E = K, and α ◦ ϕ = resL . Proof: Let X be the disjoint union of all quotient groups G/N , where N ranges over all open normal subgroups of G. Consider the elements of X as independent over L and set F = L(X). Define an action of each σ of G on F by σ(τ N ) = στ N , for τ N ∈ X, and σ(a) = α(σ)(a) for a ∈ L. This action of G is faithful. We have S(τ N ) = N and S(a) = α−1 (Gal(L/K(a))). Any u ∈ F is a rational function with integral coefficients in a1 , . . . , am ∈ L and x1 , . . . , xn ∈ X. The stabilizer S(u) of u contains the open subgroup S(a1 ) ∩ · · · ∩ S(am ) ∩ S(x1 ) ∩ · · · ∩ S(xn ). Hence, S(u) is an open subgroup. Let E be the fixed field of G in F . By Lemma 1.3.2, G = Gal(F/E) and the conclusion of the proposition follows from the definitions. Corollary 1.3.4 (Leptin): Every profinite group is isomorphic to a Galois group of some Galois extension.
1.4 The p-adic Integers and the Pr¨ ufer Group The first examples of profinite groups are the group Zp of p-adic numbers ˆ = lim Z/nZ. and the Pr¨ ufer group Z ←−
1.4 The p-adic Integers and the Pr¨ ufer Group
13
The p-adic group Zp . Let p be a rational prime. Consider the quotient rings Z/pi Z with their canonical homomorphisms Z/pj Z → Z/pi Z given for j ≥ i by x + pj Z 7→ x + pi Z. The inverse limit Zp = lim Z/pi Z is the ring ←− of p-adic integers. It is a profinite ring. Each x ∈ Zp is a sequence (xi + pi Z)i∈N where xi ∈ Z and xj ≡ xi mod pi Z for j ≥ i. Each integer m ≥ 0 corresponds to a basic neighborhood of x consisting of all elements y = (yi + pi Z)i∈N with ym ≡ xm mod pm Z. The map a 7→ (a + pi Z)i∈N is an embedding of Z into Zp . Identify Z with its image in Zp . The sequence (xi )i∈N converges to x = (xi + pi Z)i∈N in the p-adic topology. Hence, Z is dense in Zp . Yet, Z is not equal to Zp . For Pn−1 example, if p 6= 2, then ( i=0 pi + pn Z)n∈N belongs to Zp but not to Z. For Pn−1 p = 2, ( i=0 4i + 2n Z)n∈N belongs to Z2 but not to Z (see also Exercise 15). Lemma 1.4.1: The ring Zp has the following properties: (a) An element x = (xi + pi Z)i∈N is invertible if and only if p - x1 . (b) Zp is an integral domain. Proof of (a): Suppose x0 = (x0i + pi Z)i∈N is an inverse of x. Then x01 x1 ≡ 1 mod p, hence p - x1 . Conversely, suppose p - x1 . Then for each i, p - xi . Hence, there exists x0i ∈ Z which is unique modulo pi with x0i xi ≡ 1 mod pi . Thus, x0 = (x0i + pi Z)i∈N is in Zp and x0 x = 1. Proof of (b): Let x = (xi + pi )i∈N and y = (yi + pi )i∈N be nonzero elements of Zp . Then there are m, n ∈ N with xm 6≡ 0 mod pm and yn 6≡ 0 mod pn . Hence, xm+n ym+n 6≡ 0 mod pm+n . Therefore, xy 6= 0. Consequently, Zp is an integral domain. Lemma 1.4.2: (a) For each i, pi Zp is the kernel of the projection πi : Zp → Z/pi Z. Thus, pi Zp is an open subgroup of Zp of index pi . (b) If H is a subgroup of Zp of a finite index, then H = pi Zp for some i ∈ N. (c) 0 is the only closed subgroup of Zp of infinite index. (d) pZp is the unique closed maximal subgroup of Zp . (e) All nonzero closed subgroups of Zp are isomorphic to Zp . Proof of (a): Suppose x = (xj + pj Z)j∈N belongs to Ker(πi ). Then xi ≡ 0 mod pi . Hence, xj ≡ 0 mod pi for each j ≥ i. Write xj = pi yj for j ≥ i and yj = yi for j < i. Let z = (yj+i + pj Z)j∈N . Then z ∈ Zp and pi z = x. Indeed, pi zj = pi yj+i = xj+i ≡ xj mod pj for every positive integer j. Proof of (b): Conversely, let H be a subgroup of Zp of index n. Suppose n = kpi where p - k. By Lemma 1.4.1(a), k is invertible in the ring Zp , so nZp = pi Zp . Thus, pi Zp = nZp ≤ H. It follows, pi = (Zp : pi Zp ) ≥ (Zp : H) = kpi . Therefore, k = 1 and H = pi Zp . Proof of (c): A closed subgroup J of Zp of infinite index is the intersection of infinitely many open groups (Lemma 1.2.3). Hence, by (a), all subgroups pi Zp contain J. Consequently J = 0.
14
Chapter 1. Infinite Galois Theory and Profinite Groups
Proof of (d): By (a), (b), and (c), pZp is the unique closed maximal subgroup of Zp . Proof of (e): By Lemma 1.4.1(b), the map x 7→ pi x is an isomorphism of Zp onto pi Zp . Every element xP= (xi + pi Z)i∈N has a unique representation as a ∞ i formal power series i=0 ai p , with 0 ≤ ai < p for all i. Indeed, xn ≡ Pn−1 i n i=0 ai p mod p , for every n ∈ N. Lemma 1.4.3: Let α: Zp → Z/pn Z be an epimorphism with n ≥ 1 and H a closed subgroup of Zp . Suppose α(H) = Z/pn Z. Then H = Zp . Proof: By Lemma 1.4.2(a), Ker(α) = pn Zp . Thus, by assumption, H + pn Zp = Zp . Assume H 6= Zp . Then, by Lemma 1.4.2(d), H ≤ pZp . Therefore, Zp = H + pn Zp ≤ pZp < Zp . It follows from this contradiction that H = Zp . In the terminology of Section 22.5, Lemma 1.4.3 says that α: Zp → Z/pn Z is a Frattini cover. ¨ fer group. For each n ∈ N consider the quotient group Z/nZ The Pru and the canonical homomorphisms Z/nZ → Z/mZ defined for m|n by x + ˆ = lim Z/nZ is the Pr¨ nZ 7→ x + mZ. The inverse limit Z ufer group. Like ←− ˆ by x 7→ (x + nZ)n∈N . Thus, Z ˆ with Zp , embed Z as a dense subgroup of Z ˆ is the closure of the subgroup generated by 1. Write Z = h1i and say that 1 ˆ Also, the subgroups nZ of Z form a basis for the neighborhoods generates Z. of 0 in the induced topology. ˆ is an open subgroup of Z ˆ of index n and Lemma 1.4.4: For each n ∈ N, nZ ∼ ˆ ˆ ˆ ˆ nZ = Z. If H is a subgroup of Z of index n, then H = nZ. Proof of: Suppose x = (xk + kZ)k∈Z lies in the kernel Zn of the projection ˆ → Z/nZ. Then xn ≡ 0 mod n. Hence, for each r ∈ N we have xrn ≡ Z xn ≡ 0 mod n, so xrn = nyrn for some yrn ∈ Z. Let z = (yrn + rZ)r∈Z . If r0 is a multiple of r, then nyr0 n ≡ nyrn mod rn, so yr0 n ≡ yrn mod r. ˆ Moreover, xr ≡ xrn ≡ nyrn = nzr mod r. Hence, x = nz. Therefore, z ∈ Z. ˆ is an open subgroup of Z ˆ of index n. Consequently, nZ ˆ onto nZ. ˆ Indeed, Next note that the map x 7→ nx is an isomorphism of Z if nx = 0, then nxrn ≡ 0 mod rn, so xr ≡ xrn ≡ 0 mod r for each r ∈ N. Hence, x = 0. ˆ of index n, then nZ ˆ is contained in H Finally, if H is a subgroup of Z ˆ and has the same index. Therefore, H = nZ. ˆ to the groups Zp . We conclude by relating Z
1.5 The Absolute Galois Group of a Finite Field
15
ˆ is topologically isomorphic to the cartesian prodLemma1.4.5: The group Z Q uct Zp where p ranges over all primes numbers. Q Proof: Let n = pkp be the decomposition of a positive integer n into a product of prime Q powers. The Chinese remainder theorem gives a canonthe projection ical isomorphism Z/pkp Z → Z/nZ. Combine this with Q Q Q kp Zp → Z/nZ. Zp → Z/p Z to obtain a continuous epimorphism fn : The maps Q fn form a compatible system, so they give a continuous homomorˆ Since Q Zp is compact and Z ˆ is Hausdorff, Im(f ) is a phism f : Zp → Z. ˆ Moreover, Z embeds diagonally in Q Zp and f (m) = m closed subgroup of Z. ˆ ˆ for each m ∈ Z. Thus, Z ⊆ Im(f ). Since T Z is dense in Z, we have Im(f ) = Z, so f is surjective. The kernel of f is Ker(fn ) = 0. Hence, f is also injective. The compactness and Hausdorff properties imply that f is a topological isomorphism. As a consequence of Lemma 1.4.5, Zp is both a closed subgroup and a ˆ for each prime p. quotient of Z,
1.5 The Absolute Galois Group of a Finite Field For every prime power q there exists a field Fq (unique up to isomorphism) with q elements. It is characterized within its algebraic closure F˜q by Fq = {x ∈ F˜q | xq = x}. The field Fq has, for each n ∈ N, exactly one extension, Fqn , of degree n. It is Galois with a cyclic group generated by the Frobenius automorphism a is πq,n defined by πq,n (x) = xq for x ∈ Fqn . The map a + nZ 7→ πq,n an isomorphism of Z/nZ onto Gal(Fqn /Fq ). If m|n, there is a canonical commutative diagram / Z/mZ
Z/nZ Gal(Fqn /Fq )
res
/ Gal(Fqm /Fq ).
ˆ∼ Take the inverse limits to obtain an isomorphism Z = Gal(Fq ) mapping the ˆ to the Frobenius automorphism πq , defined on all identity element 1 of Z ˜ q by πq (x) = xq . of F S∞ (l) Let l be a prime number and Fq = i=1 Fqli . Then the projection ˆ 7→ Zl corresponds to res: Gal(Fq ) → Gal(F(l) Z q /Fq ). By Lemma 1.4.5, Q (l) ∼ Gal(Fq ) = Gal(Fq /Fq ). On the other hand, let Nl be the fixed field of Zl (l) ˜ q . Then Gal(Nl ) = Zl , F(l) ˜ in F q Nl = Fq , and Fq ∩ Nl = Fq . It follows, Y ∼ Zl0 × Zl . Gal(Fq ) = Gal(F(l) q ) × Gal(Nl ) = l0 6=l
16
Chapter 1. Infinite Galois Theory and Profinite Groups
Exercises 1. Let (Si , πji ) be an inverse system of finite sets with all πji surjective and let S = lim Si . Use Lemma 1.1.3 to prove that all maps πi : S → Si ←− determined by the πji ’s are surjective. 2.
Let H1 , . . . , Hr be closed subgroups of a profinite group G. Prove that \ H1 ∩ · · · ∩ Hr = (H1 N ∩ · · · ∩ Hr N ), N
where N ranges over all open normal subgroups of G. Hint: Use Lemma 1.2.2(b). 3. Suppose H is a closed subgroup of a profinite group G. Prove: If HN/N = G/N for every open normal subgroup N of G, then H = G. 4. Let H be an open subgroup of index n of a profinite group G. Denote the intersection of all conjugates of H in G by N . Note: Multiplication of G on the left cosets of H induces a homomorphism of G into the symmetric group Sn with kernel N . Conclude that G/N is isomorphic to a subgroup of Sn and (G : N ) ≤ n!. 5. Let G be a compact group and H an open subgroup. Suppose H is profinite. Prove: G is profinite. Hint: Use Lemma 1.2.4. 6. Let S be a set of rational primes. Consider the profinite group ZS = lim Z/nZ, with n running over all positive integers with prime factors in S. ←− (a) Prove: The finite homomorphic images of ZS are exactly the groups Z/nZ, where the prime factors of n belong to S. (b) Embed Z in ZS and determine the topology on Z induced by that of ZS . (c) Prove that ZS is procyclic (i.e. ZS is the closure of a group generated by one element). Q (d) Follow the proof of Lemma 1.4.5 to prove that ZS ∼ = p∈S Zp . 7. Let G be a procyclic group (Exercise 6). Use that G is a homomorphic ˆ to show there exists a set S of primes with G = Q image of Z p∈S Gp , where ip ∼ for each p ∈ S either Gp ∼ Z for some i ∈ N or G . Z/p Z = p p = p In particular, Q Z . if G is torsion free, then G ∼ = p∈S p ˆ Prove that every finite quotient of G is 8. Let G be a closed subgroup of Z. a cyclic group. Conclude that G is procyclic and therefore that there exists Q a set S of prime numbers with G ∼ = l∈S Zl . 9. Let G be a profinite group. Prove that each of the following statements ˆ is equivalent to G ∼ = Z. (a) G has exactly one open subgroup of each index n. ˆ (b) G is procyclic and there is an epimorphism π: G → Z.
Exercises
17
10. Let G be a procyclic group. Use Exercise 8 to prove that each epimorˆ is an isomorphism. phism π: G → Z 11. Let G be a profinite group with at most one open subgroup of every index n. (a) Prove that every open subgroup is normal. ¯ of G. Con(b) Observe that (a) holds for every finite homomorphic image G ¯ clude that G is nilpotent. (c) Let P be a finite p-group with the above properties. Prove that every element x of P of maximal order generates P . (d) Conclude that G is a procyclic group. ˆ in the following 12. Define powers in a profinite group G with exponents in Z ˆ way: Let g ∈ G and ν ∈ Z. Then there exists a sequence {ν1 , ν2 , ν3 , . . .} of elements of Z that converges to ν. By compactness there is a subsequence of {g ν1 , g ν2 , g ν3 , . . .} that converges to an element h of G. (a) Prove h does not depend on the sequence {ν1 , ν2 , ν3 , . . .}. So we may denote h by g ν . Hint: If N is a normal subgroup of G of index n, then xn ∈ N for every x ∈ G. (b) Prove the usual rules for the power operations. For example, g µ g ν = g µ+ν ,
(g µ )ν = g µν ,
and
g ν hν = (gh)ν if gh = hg.
ˆ into G is continuous. (c) Prove that the map (g, ν) 7→ g ν of G × Z 13. Multiplication in the groups Z/pi Z is compatible with the canonical maps Z/pi+1 Z → Z/pi Z. Therefore, it defines a multiplication in the additive group Zp = lim Z/pi Z. ←− (a) Prove: Zp is an integral domain (the quotient field of which, Qp , is the field of p-adic numbers). (b) Show: Every closed subgroup of Zp is an ideal of Zp . (c) Show: pZp is the unique maximal ideal of Zp ; observe that Zp /pZp ∼ = Fp . (d) Deduce: α ∈ Zp is a unit (i.e. invertible in this ring) if and only if α is congruent modulo pZp to one of the numbers 1, 2, . . . , p − 1. Hint: Form the inverse of 1 + βp, β ∈ Zp using the geometric series for 1/(1 + x). ˆ in a manner analogous to 14. Define multiplication in the additive group Z ˆ a commutative the definition of multiplication in Zp . Prove that this makes Z topological ring with zero divisors. Q ˆ∼ (a) Prove that the isomorphism of additive groups Z = Zp established in Lemma 1.4.5 is an isomorphism of rings. ˆ is also an ideal. (b) Prove that every closed subgroup of Z 15. Use the power series representation of the elements of Zp (Section 1.4) to show |Zp | = 2ℵ0 . Conclude that Qp has elements that are transcendental over Q.
18
Chapter 1. Infinite Galois Theory and Profinite Groups
√ 16. For a prime number p, let K be Q(ζ ) if p = 6 2 and Q( −1) if p = 2. p p S∞ Also, let Lp = i=1 Q(ζpi ). (a) Prove that if K 0 is a field such that Kp ⊆ K 0 ⊂ Lp , then Gal(Lp /K 0 ) ∼ = Zp and [K 0 : Q] < ∞. (b) Prove that Gal(Lp /Q) is isomorphic to Zp × Z/(p − 1)Z if p 6= 2 and to Z2 × Z/2Z if p = 2.
Notes More about topological groups can be found in [Pontryagin]. A detailed exposition on Galois theory of finite extensions appears in [Lang7, Chapter VI, Section 1]. For finite fields see [Langl, Chapter V, Section 5]. Leptin’s proof of Corollary 1.3.4 uses linear disjointness of fields [Leptin]. We reproduce it in Proposition 2.6.12. The proofs of Lemma 1.3.2 and Proposition 1.3.3 appear in [Waterhouse]. This chapter overlaps with [Ribes, Chapter 1].
Chapter 2. Valuations and Linear Disjointness Sections 2.1–2.4 introduce the basic elements of the theory of valuations, especially discrete valuations, and of Dedekind domains. These sections are primarily a survey. We prove that an overring of a Dedekind domain is again a Dedekind domain (Proposition 2.4.7). The rest of the chapter centers around the notion of linear disjointness of fields. We use this notion to define separable, regular, and primary extensions of fields. In particular, we prove that an extension F/K with a K-rational place is regular. Section 2.8 gives a useful criterion for separability with derivatives.
2.1 Valuations, Places, and Valuation Rings The literature treats arithmetic theory of fields through three intimately connected classes of objects: valuations, places, and valuation rings. We briefly review the basic definitions. Call an Abelian (additive) group Γ with a binary relation < an ordered group if the following statements hold for all α, β, γ ∈ Γ. (1a) Either α < β, or α = β, or β < α. (1b) If α < β and β < γ, then α < γ. (1c) If α < β, then α + γ < β + γ. Some examples of ordered groups are the additive groups Z, R, and Z⊕Z with the order (m, n) < (m0 , n0 ) if either m < m0 or m = m0 and n < n0 (the lexicographic order). A valuation v of a field F is a map of F into a set Γ ∪ {∞}, where Γ is an ordered group, with these properties: (2a) v(ab) = v(a) + v(b). (2b) v(a + b) ≥ min v(a), v(b) . (2c) v(a) = ∞ if and only if a = 0. (2d) There exists a ∈ F × with v(a) 6= 0. By definition the symbol ∞ satisfies these rules: (3a) ∞ + ∞ = α + ∞ = ∞ + α = ∞; and (3b) α < ∞ for each α ∈ Γ. Condition (2) implies several more properties of v: (4a) v(1) = 0, v(−a) = v(a). (4b) If v(a) < v(b), then v(a + b) = v(a) (Use the identity a = (a + b) − b and P(2b)); n (4c) If i=1 ai = 0, then there exist i 6= j such that v(ai ) = v(aj ) and v(ai ) = min(v(a1 ), . . . , v(an )) (Use (2b) and (4b)). We refer to the pair (F, v) as a valued field.
20
Chapter 2. Valuations and Linear Disjointness
The subgroup Γv = v(F × ) of Γ is the value group of v. The set Ov = {a ∈ F | v(a) ≥ 0} is the valuation ring of v. It has a unique maximal ideal mv = {a ∈ F | v(a) > 0}. Refer to the residue field F¯v = Ov /mv as the residue field of F at v. Likewise, whenever there is no ambiguity, we ¯ and call it the residue of a at v. denote the coset a + mv by a Two valuations v1 , v2 of a field F with value groups Γ1 , Γ2 are equivalent if there exists an isomorphism f : Γ1 → Γ2 with v2 = f ◦ v1 . Starting from Section 2.2, we abuse our language and say that v1 and v2 are distinct if they are inequivalent. A place of a field F is a map ϕ of F into a set M ∪ {∞}, where M is a field, with these properties: (5a) ϕ(a + b) = ϕ(a) + ϕ(b). (5b) ϕ(ab) = ϕ(a)ϕ(b). (5c) There exist a, b ∈ F with ϕ(a) = ∞ and ϕ(b) 6= 0, ∞. By definition the symbol ∞ satisfies the following rules: (6a) x + ∞ = ∞ + x = ∞ for each x ∈ M . (6b) x · ∞ = ∞ · x = ∞ · ∞ = ∞ for each x ∈ M × . (6c) Neither ∞ + ∞, nor 0 · ∞ are defined. It is understood that (5a) and (5b) hold whenever the right hand side is defined. These conditions imply that ϕ(1) = 1, ϕ(0) = 0 and ϕ(x−1 ) = ϕ(x)−1 . In particular, if x 6= 0, then ϕ(x) = 0 if and only if ϕ(x−1 ) = ∞. We call an element x ∈ F with ϕ(x) 6= ∞ finite at ϕ, and say that ϕ is finite at x. The subring of all elements finite at ϕ, Oϕ = {a ∈ F | ϕ(a) 6= ∞}, is the valuation ring of ϕ. It has a unique maximal ideal mϕ = {a ∈ F | ϕ(a) = 0}. The quotient ring Oϕ /mϕ is a field which is canonically isomorphic to the residue field F¯ϕ = {ϕ(a) | a ∈ Oϕ } of F at ϕ. The latter is a subfield of M . Call ϕ a K-place if K is a subfield of F and ϕ(a) = a for each a ∈ K. Two places ϕ1 and ϕ2 of a field F with residue fields M1 and M2 are equivalent if there exists an isomorphism λ: M1 → M2 with ϕ2 = λ ◦ ϕ1 . A valuation ring of a field F is a proper subring O of F such that if / O} is x ∈ F × , then x ∈ O or x−1 ∈ O. The subset m = {x ∈ O | x−1 ∈ the unique maximal ideal of O (Exercise 1). The map ϕ: F → O/m ∪ {∞} which maps x ∈ O onto its residue class modulo m and maps x ∈ F r O onto ∞ is a place of F with valuation ring O. Denote the units of O by U = {x ∈ O | x−1 ∈ O}. Then F × /U is a multiplicative group ordered by the rule xU ≤ yU ⇐⇒ yx−1 ∈ O. The map x 7→ xU defines a valuation of F with O being its valuation ring. These definitions easily give a bijective correspondence between the valuation classes, the place classes and the valuation rings of a field F . An isomorphism σ: F → F 0 of fields induces a bijective map of the valuations and places of F onto those of F 0 according to the following rule: If v is a valuation of F , then σ(v) is defined by σ(v)(x) = v(σ −1 x) for every x ∈ F 0 . If ϕ is a place of F , then σ(ϕ)(x) = ϕ(σ −1 x). In particular, σ
2.2 Discrete Valuations
21
0 induces an isomorphism F¯ϕ ∼ of residue fields. It is also clear that if = F¯σ(ϕ) ϕ corresponds to v, then σ(ϕ) corresponds to σ(v). A valuation v of a field F is real (or of rank 1) if Γv is isomorphic to a subgroup of R. Real valuations satisfy the so called weak approximation theorem, a generalization of the Chinese remainder theorem [CasselsFr¨ohlich, p. 48]:
Proposition 2.1.1: Consider the following objects: inequivalent real valuations v1 , . . . , vn of a field F , elements x1 , . . . , xn of F , and real numbers γ1 , . . . , γn . Then there exists x ∈ F with vi (x − xi ) ≥ γi , i = 1, . . . , n.
2.2 Discrete Valuations ∼ Z. In this case we normalize A valuation v of a field F is discrete if v(F × ) = v by replacing it with an equivalent valuation such that v(F × ) = Z. Each element π ∈ F with v(π) = 1 is a prime element of Ov . Prime elements of a unique factorization domain R produce discrete valuations of F = Quot(R). If p is a prime element of R, then every element x of F × has a unique representation as x = upm , where u is relatively prime to p and m ∈ Z. Define vp (x) to be m. Then vp is a discrete valuation of F . Suppose p0 is another prime element of R. Then vp0 is equivalent to vp if and only if p0 R = pR, that is if p0 = up with u ∈ R× . Example 2.2.1: Basic examples of discrete valuations. (a) The ring of integers Z is a unique factorization domain. For each prime number p the residue field of Q at vp is Fp . When p ranges over all prime numbers, vp ranges over all valuations of Q (Exercise 3). (b) Let R = K[t] be the ring of polynomials in an indeterminate t over a field K. Then R is a unique factorization domain. Then prime elements of R are the irreducible polynomials p over K. Units of R are the elements u of K × , so vp (u) = 0 and we say vp is trivial on K. The residue field of K(t) at vp is isomorphic to the field K(a), where a is a root of p. There is one additional valuation, v∞ , of K(t) which is trivial on K. It is defined for a quotient fg of elements of K[t] by the formula v∞ ( fg ) = deg(g) − deg(f ). The set of vp ’s and v∞ give all valuation of K(t) trivial on K. Thus, all valuations of K(t) which are trivial on K are discrete (Exercise 4). ˜ An arbitrary irreducible polynomial p may have several roots a ∈ K. ˜ ∪{∞} by ϕa (t) = a and ϕa (c) = c Each of them defines a place ϕa : K(t) → K for each c ∈ K. These places are equivalent. If p(t) = t − a, then ϕa is the unique place of K(t) corresponding to vp . Similarly, there is a unique place ϕ∞ corresponding to v∞ . It is defined by ϕ∞ (t) = ∞. We may view each f (t) ∈ K(t) as a function from K ∪ {∞} into itg(t) with g, h ∈ K[X] and self: f (a) = ϕa (f (t)). Explicitly, write f (t) = h(t) gcd(g, h) = 1. Let a ∈ K. Then f (a) =
g(a) h(a)
if h(a) 6= 0 and f (a) = ∞
22
Chapter 2. Valuations and Linear Disjointness
if h(a) = 0. To compute f (∞) let u = t−1 and write f (t) =
g1 (u) h1 (u)
with
g1 , h1 ∈ K[X] and gcd(g1 , h1 ) = 1. Then f (∞) = hg11(0) (0) if h1 (0) 6= 0 and f (∞) = ∞ if h1 (0) = 0. Suppose for example f (t) ∈ K[t] and f 6= 0. Then f maps K into itself and f (∞) = ∞. Now suppose f (t) = at+b ct+d with ad − bc 6= 0 and c 6= 0, then f (∞) = ac . When K is algebraically closed, each irreducible polynomial is linear. Hence, each valuation of K(t) which is trivial over K is either vt−a for some a ∈ K or v∞ . More examples of discrete valuations arise through extensions of the basic examples (Section 2.3). Lemma 2.2.2: Every discrete valuation ring R is a principal ideal domain. Proof: Let v be the valuation of K = Quot(R) with Ov = R and v(K × ) = Z. Choose a prime element π of R. Now consider a nonzero ideal a of R. Then the minimal integer m with π m ∈ a is positive. It satisfies, a = π m R. As a consequence of Lemma 2.2.2, finitely generated modules over R have a simple structure. Proposition 2.2.3: Let R be a discrete valuation ring, p a prime element ¯ = R/pR. of R, K = Quot(R), and M a finitely generated R-module. Put K Let r = dimK M ⊗R K, n = dimK¯ M/pM , and m = n − r. Then there is a unique m-tuple of positive integers (k1 , k2 , . . . , km ) with k1 ≤ k2 ≤ · · · ≤ km and M ∼ = R/pkm R ⊕ · · · ⊕ R/pk1 R ⊕ Rr . Moreover, r is the maximal number of elements of M which are linearly independent over R and n is the minimal number of generators of M . Proof: By Lemma 2.2.2, R is a principal ideal domain, so M = Mtor ⊕ N , where Mtor = {m ∈ M | rm = 0 for some r ∈ R, r 6= 0} and N is a free R-module [Lang7, p. 147, Thm. 7.3]. Both Mtor and N are finitely generated [Lang7, p. 147, Cor. 7.2]. In particular, N ∼ = Rs for some integer s ≥ 0. Suppose m ∈ Mtor and am = 0 with a ∈ R, a 6= 0. Then, m⊗1 = am⊗ a1 = 0. Hence, Mtor ⊗R K = 0 and M ⊗R K ∼ = K s . Therefore, s = r. By [Lang7, p. 151, Thm. 7.7], Mtor ∼ = R/qm0 R ⊕ · · · ⊕ R/q1 R where q1 , . . . , qm0 are elements of R which are neither zero nor units and qi |qi+1 , i = 1, . . . , m0 − 1. Multiplying each qi by a unit, we may assume qi = pki with ki an integer and 1 ≤ k1 ≤ k2 ≤ · · · ≤ km0 . Moreover, the above cited theorem assures Rq1 , . . . , Rqm0 are uniquely determined by the above conditions. Hence, k1 , . . . , km0 are also uniquely determined. Combining the first two paragraphs gives: M∼ = R/pkm0 R ⊕ · · · ⊕ R/pk1 R ⊕ Rr . 0 ¯ m0 +r , so n = m0 + r and m0 = m. Hence, M/pM = (R/pR)m +r ∼ =K
2.2 Discrete Valuations
23
Now recall Ps that elements v1 , . . . , vs of M are linearly independent over R if i=1 ai vi = 0 with a1 , . . . , as ∈ R implies a1 = · · · = as = 0. Alternatively, v1 ⊗ 1, . . . , vs ⊗ 1 are linearly independent over K. Thus, r is the maximal number of R-linearly independent elements of M . Finally, by Nakayama’s lemma [Lang7, p. 425, Lemma 4.3], n is the minimal number of generators of M . Definition 2.2.4: Let R be an integral domain with quotient field F . An overring of R is a ring R ⊆ R0 ⊂ F . It is said to be proper if R 6= R0 . Lemma 2.2.5: A discrete valuation ring O has no proper overrings. Proof: Let R be an overring of O. Assume there exists x ∈ R r O. Then x−1 is a nonunit of O. Choose a prime element π for O. Then x = uπ −m for some u ∈ O× and a positive integer m. Hence, π −1 = u−1 π m−1 x ∈ R. Therefore, u0 π k ∈ R for all u0 ∈ O× and k ∈ Z. We conclude that R = Quot(O). Composita of places attached to discrete valuations of rational function fields of one variable give rise to useful places of rational function fields of several variables. Construction 2.2.6: Composition of places. Suppose ψ is a place of a field K with residue field L and ϕ is a place of L with residue field M . Then ψ −1 (Oϕ ) is a valuation ring of K with maximal ideal ψ −1 (mϕ ) and residue field ψ −1 (Oϕ )/ψ −1 (mϕ ) ∼ = Oϕ /mϕ ∼ = M . Define a map ϕ ◦ ψ: K → M ∪ {∞} as follows: ϕ ◦ ψ(x) = ϕ(ψ(x)) if ψ(x) 6= ∞ and ϕ ◦ ψ(x) = ∞ if ψ(x) = ∞. Then ϕ ◦ ψ is a homomorphism on ψ −1 (Oϕ ) and {x ∈ K | ϕ ◦ ψ(x) = ∞} = K r ψ −1 (Oϕ ). Therefore, ϕ ◦ ψ is a place of K, called the compositum of ψ and ϕ, Oϕ◦ψ = ψ −1 (Oϕ ), and mϕ◦ψ = ψ −1 (mϕ ). K Oψ Oϕ◦ψ mϕ◦ψ mψ
/ L ∪ {∞}
ϕ
/ M ∪ {∞}
ψ
/L
ϕ
/ M ∪ {∞}
ψ
/ Oϕ
ϕ
/M
ψ
/ mϕ
ϕ
/0
ψ
/0
ϕ
/0
ψ
¯ ψ and M = L ¯ϕ = K ¯ ϕ◦ψ . In addition, L = K
˜ t1 , . . . , tr indeLemma 2.2.7: Let K be a field, a1 , . . . , ar elements of K, terminates, and L a finite extension of K. Then there exists a K-place ϕ: K(t) → K(a) ∪ {∞} such that ϕ(ti ) = ai , i = 1, . . . , r. Moreover, every extension of ϕ to an L-place of L(t) maps L(t) onto L(a) ∪ {∞}. Proof: For each i there is a K(a1 , . . . , ai−1 , ti+1 , . . . , tr )-place ϕi : K(a1 , . . . , ai−1 , ti , ti+1 . . . , tr ) → K(a1 , . . . , ai−1 , ai , ti+1 , . . . , tr )
24
Chapter 2. Valuations and Linear Disjointness
with ϕi (ti ) = ai (Example 2.2.1). The compositum ϕ = ϕr ◦ · · · ◦ ϕ1 is a K-place of K(t1 , . . . , tr ) with residue field K(a1 , . . . , ar ) and ϕ(ti ) = ai , i = 1, . . . , r. Let now ϕ be an extension of ϕ to an L-place of L(t). Choose a basis also a basis for L(t)/K(t). Hence, b1 , . . . , bn for L/K. Then b1 , . . . , bn is P n each f ∈ L(t) has a presentation f = i=1 bi fi with fi ∈ K(t). Assume Pn without loss that ff1i is finite under ϕ for i = 1, . . . , n. Then f = f1 i=1 ff1i bi and ϕ(f ) ∈ L(a1 , . . . , ar ) ∪ {∞}. Thus, ϕ(L(t)) = L(a) ∪ {∞}.
2.3 Extensions of Valuations and Places The examples of Section 2.2 and the following extension results give a handle on describing valuations of function fields in one variable. Proposition 2.3.1 (Chevalley [Lang4, p. 8, Thm. 1]): Let ϕ0 be a homomorphism of an integral domain R into an algebraically closed field M and let F be a field containing R. Then ϕ0 extends either to an embedding ϕ of F into M or to a place ϕ of F into M ∪ {∞}. When F is algebraic over R, the proposition has a more precise form: Let f ∈ R[X] be an irreducible polynomial over E = Quot(R) and f¯ ∈ M [X] the result of applying ϕ0 to the coefficients of f . Suppose f¯ is ˜ and M , not identically zero. Assume x and x ¯ are roots of f and f¯ in E respectively. Then ϕ0 extends to a place ϕ of E(x) into M ∪ {∞} with ϕ(x) = x ¯ [Lang4, p. 10, Thm. 2]. Moreover, if ϕ0 is injective, so is ϕ [Lang4, p. 8, Prop. 2]. In particular, suppose v is a valuation of a field E and F is an extension of E. Then v extends to a valuation w0 of F . Each valuation w of F which is equivalent to w0 lies over v. Thus, w lies over v if and only if Ov ⊆ Ow and mv = mw ∩ Ov . The number ew/v = (w(F × ) : w(E × )) is the ramification index of w over v (and also over E). The field degree [F : E] ¯v embeds in F¯w to give the inequality bounds ew/v (Exercise 5). Similarly, E ¯v ] ≤ [F : E] (Exercise 7). Both the ramification index and fw/v = [F¯w : E the residue field degree are multiplicative. Thus, if (F 0 , w0 ) is an extension of (F, w), then ew0 /v = ew0 /w ew/v and fw0 /v = fw0 /w fw/v . If [F : E] < ∞, then the number of valuations of F that lie over v is finite (a consequence of Proposition 2.3.2). Proposition 2.3.2: Let F/E be a finite extension of fields and v a valuation of E. Let w1 , . . . , wg be all inequivalent extensions of v to F . Then
(1)
g X
ewi /v fwi /v ≤ [F : E]
i=1
[Bourbaki2, p. 420, Thm. 1]. If, in addition, v is discrete and F/E is separa-
2.3 Extensions of Valuations and Places
25
ble, then each wi is discrete and (see [Bourbaki2, p. 425, Cor. 1]) g X
(2)
ewi /v fwi /v = [F : E].
i=1
Suppose (F, w)/(E, v) is an extension of discrete valued fields. In particular, w(a) = v(a) for each a ∈ E. By definition, ew/v = (w(F × ) : v(E × )). However, as in Section 2.2, it is customary to replace v and w by equivalent valuations with v(E × ) = w(F × ) = Z. The new valuations satisfy w(a) = ew/v v(a)
for each a ∈ E.
Whenever we speak about an extension of discrete valuations, we mean they are normalized and satisfy the latter relation. Suppose F is a finite Galois extension of E with a Galois group G. Let w be a discrete valuation of F and let σ ∈ G. Then, σ(w) is a valuation of F (Section 2.1), both w and σ(w) lie over the same valuation v of E, and ew/v = eσ(w)/v
and fw/v = fσ(w)/v .
Conversely, suppose w and w0 are two discrete valuations of F over the same valuation v of E. Then there exists σ ∈ G such that σ(w) = w0 (Exercise 9). Thus, if w1 , . . . , wg are all distinct valuations of F that lie over v, then they all have the same residue degree f and ramification index e over v. In this case formula (2) simplifies to (3)
ef g = [F : E]. The subgroups Dw = Dw/v = {σ ∈ G | σOw = Ow } Iw = Iw/v = {σ ∈ G | w(x − σx) > 0 for all x ∈ Ow }
are the decomposition group and the inertia group, respectively, of w ¯v is separable, then [Serre3, p. 33] over E. Obviously Iw / Dw . If F¯w /E (4)
|Iw | = ew/v
and |Dw | = ew/v fw/v .
Section 2.6 generalizes the notion of separable algebraic extension of fields to arbitrary extensions of fields. In particular, purely transcendental extensions of fields are separable. We use this notion in the following definition. Suppose (F, w)/(E, v) is an arbitrary extension of valued fields. We say ¯v w is unramified (resp. tamely ramified) over v (or also over E) if F¯w /E ¯ is a separable extension and ew/v = 1 (resp. char(Ev ) - ew/v ). We say v is unramified (resp. tamely ramified) in F if each extension of v to F is unramified (resp. tamely ramified) over v.
26
Chapter 2. Valuations and Linear Disjointness
Example 2.3.3: Purely transcendental extensions. Let (E, v) be a valued field. Consider a transcendental element t over E. Extend v to a valuation v 0 of E(t) as follows. First define v 0 on E[t] by the following rule: v0
(5)
m X
ai ti = min v(a0 ), . . . , v(am )
i=0
for a0 , . . . , am ∈ E. The same argument used to prove Gauss’ Lemma proves 0 (g) for all f, g ∈ E[t].P that v 0 (f g) = v 0 (f ) + vP m n Indeed, let f (t) = i=0 ai ti and g(t) = j=0 bj tj . Let r be the minimal integer with v(ar ) = min v(a0 ), . . . , v(am ) and let s be the minimal integer with v(bs ) = min v(b0 ), . . . , v(bn ) . If i + j = r + s and (i, j) 6= (r, s), then either i < r or j < s. In both cases v(ar ) + v(bs ) < v(ai ) + v(bj ). Hence v0
m X
n X ai ti + v 0 bj tj = v(ar ) + v(bs )
i=0
j=0
X
= min
v(ai bj ) | k = 0, . . . , m + n
i+j=k
= v0
m X
ai ti ·
i=0
n X
b j tj ,
j=0
as claimed. We extend v 0 to E(t) by the rule v 0 ( fg ) = v 0 (f ) − v 0 (g). Then we prove v 0 (u1 + u2 ) ≥ min v 0 (u1 ), v 0 (u2 ) first for u1 , u2 ∈ E[t] and then for u1 , u2 ∈ Note that the residue of t at v 0 is E(t). Thus, v 0 is a valuation of E(t). P n ¯ ¯i t¯i = 0 for some a0 , . . . , an ∈ transcendental over Ev . Indeed, suppose i=0 a P n 0 i > 0. Hence, a ¯i = 0, Ov . Then min v(a0 ), . . . , v(an ) = v i=0 ai t i = 0, . . . , n. ¯v (t¯) is a rational function field over E ¯v . By It follows that, E(t)v0 = E 0 definition, Γv0 = Γv . In particular, if v is discrete, then so is v and ev0 /v = 1. Suppose v 00 is another extension of v to E(t) with the residue of t at v 00 ¯v . We show that v 00 = v 0 . Indeed, for a0 , . . . , an ∈ E, transcendental over E not all zero, choose j between 0 and n with v(aj ) = min v(a0 ), . . . , v(an ) . Pn Then i=0 ai /aj t¯i 6= 0. Therefore, v
00
n X i=0
ai t
i
= v(aj ) + v
00
n X
(ai /aj )ti
i=0 n X = min v(a0 ), . . . , v(an ) = v 0 ai ti , i=0
as claimed.
2.3 Extensions of Valuations and Places
27
Lemma 2.3.4: Let v be a discrete valuation of a field E, h ∈ Ov [X] a monic ˜ and F = E(x). irreducible polynomial of degree n, x a root of h(X) in E, ¯ Suppose the residue polynomial h(X) is separable. Then v is unramified in F. Qr ¯ ¯v [X] are distinct Proof: By assumption, h(X) = i=1 hi (X), where hi ∈ E monic irreducible polynomials. For each i between 1 and r choose a root ai ¯v )s . Use Proposition 2.3.1 to extend the residue map Ov → E ¯v of hi (X) in (E to a place ϕi of F with ϕi (x) = ai . Denote the corresponding valuation by ¯v (ai ) ⊆ F¯w . Since hi (X) and hj (X) have no common root for wi . Then E i i 6= j, the valuations w1 , . . . , wr are mutually inequivalent extensions of v. Label any further extensions of v to valuations of F as wr+1 , . . . , wg . By (1) n=
r X
deg(hi ) =
i=1
r X
¯v (ai ) : E ¯v ] ≤ [E
i=1
g X
ewi /v fwi /v ≤ n.
i=1
¯v (ai ) = F¯w for i = 1, . . . , r. Moreover, w1 , . . . , wr are Hence, ewi /v = 1 and E i all extensions of v to F and each of them is unramified over E. Therefore, v is unramified in F . ¯v to be infinite. The converse of Lemma 2.3.4 requires E Lemma 2.3.5: Let v be a discrete valuation of a field E. Let F be a separable ¯v is an infinite extension of E of degree n. Suppose v is unramified in F and E field. Then F/E has a primitive element x with irr(x, E) ∈ Ov [X] and the residue of irr(x, E) at v is a separable polynomial. Proof: Let w1 , . . . , wg be all extensions of v to F . By (2), [F : E] = P g ¯ ¯ ¯ ¯ i=1 [Fwi : Ev ]. Moreover, for each i the extension Fwi /Ev is finite and separable. Hence, we may choose ci in F with wi (ci ) = 0 and the residue ¯v . Let hi = irr(¯ ¯v ). Since ci , E c¯i of ci at wi is a primitive element of F¯wi /E ¯v is infinite, we may choose c1 , . . . , cg such that c¯1 , . . . , c¯g are mutually E ¯v . Thus, h1 , . . . , hg are relatively prime. nonconjugate over E Use Proposition 2.1.1 to find x ∈ F with wi (x − ci ) > 0, i = 1, . . . , g. Then, wi (x) = 0, i = 1, . . . , g. Extend each wi to the Galois closure of F/E. Then all E-conjugates of x have nonnegative values under each extended valuation. Hence, the elementary symmetric polynomials in the E-conjugates of x belong to Ov . Therefore, f (X) = irr(x, E) ∈ Ov [X]. Let f¯ be the residue of f at v. By construction, Q f¯(¯ ci ) = 0, therefore g ¯ hi |f , i = 1, . . . , g. Since h1 , . . . , hg are relatively prime, i=1 hi |f¯. Hence, [F : E] =
g X i=1
¯v ] = [F¯wi : E
g X
deg(hi )
i=1
≤ deg(f¯) = deg(f ) = [E(x) : E] ≤ [F : E]. Consequently, E(x) = F , as desired.
28
Chapter 2. Valuations and Linear Disjointness
¯v to be infinite is necessary Example 3.5.4 shows the assumption on E for Lemma 2.3.5 to hold. The next lemma says that arbitrary change of the base field preserves unramified discrete valuations. Lemma 2.3.6: Let (E, v) be a discrete valued field. Consider a separable algebraic extension F of E and a discrete valued field (E1 , v1 ) which extends (E, v). Suppose v is unramified in F . Then v1 is unramified in F E1 . Proof: Suppose without loss that [F : E] < ∞. Let F1 = F E1 . Suppose ¯v is infinite. Choose x as in Lemma 2.3.5 and let f (X) = irr(x, E). first that E Then F = E(x) and f¯(X) is separable. Hence, F1 = E1 (x) and f¯(X) is still separable. By Lemma 2.3.4, v1 is unramified in F1 . In the general case we consider an extension w1 of v1 to a valuation of F1 . Denote the restriction of w1 to F by w. Let t be transcendental over F1 . Example 2.3.3 extends v (resp. w, v1 , w1 ) in a canonical way to a discrete valuation v 0 (resp. w0 , v10 , w10 ) of E(t) (resp. F (t), E1 (t), F1 (t)). Further, ¯v (t¯) ev0 /v = 1 (resp. ew0 /w = 1, ev10 /v1 = 1, ew10 /w1 = 1) and E(t)v0 = E ¯1,v (t¯), F1 (t) 0 = F¯1,w (t¯)), where t¯ is (resp. F (t)w0 = F¯w (t¯), E1 (t)v0 = E 1 1 w1 1 transcendental over F¯1,w1 . Moreover, w10 extends w0 and v10 extends v 0 giving this diagram: (F (t), w0 ) p p ppp
(F1 (t), w10 ) n n nnn (F1 , w1 )
(E(t), v 0 ) p p ppp
(E1 (t), v10 ) n n nnn
(F, w)
(E, v)
(E1 , v1 )
¯v (t¯) is a separable We claim v 0 is unramified in F (t). Indeed, F (t)w0 = F¯w · E extension of E(t)v0 . Also, ew0 /v0 = ew0 /v0 ev0 /v = ew0 /v = ew0 /w ew/v = 1. Hence, w0 is unramified over v 0 . If u∗ is an arbitrary extension of v 0 to F (t) and u is its restriction to F , then the residue of t at u∗ is t¯, which is transcendental over F¯u . Thus, by uniqueness of the construction in Example 2.3.3, u∗ = u0 , where u0 is the canonical extension of u to F (t). By the above, u∗ is unramified over v 0 . Since E(t)v0 is infinite, the first paragraph of the proof implies v10 is ¯1,v (t¯) is a separable extension and unramified in F1 (t). Thus, F¯1,w1 (t¯)/E 1 ¯1,v is a separable extension and ew10 /v10 = 1. Therefore, F¯1,w1 /E 1 ew1 /v1 = ew1 /v1 ew10 /w1 = ew10 /v1 = ew10 /v10 ev10 /v1 = 1. Consequently, v1 is unramified in F1 .
Combine the multiplicativity of the ramification index and the residue field degree with Lemma 2.3.6 to prove:
2.3 Extensions of Valuations and Places
29
Corollary 2.3.7: Let (E, v) ⊆ (E 0 , v 0 ) ⊆ (E 00 , v 00 ) be a tower of discrete valued fields. The following hold: (a) v 00 /v is unramified if and only if v 00 /v 0 and v 0 /v are unramified. (b) v is unramified in E 00 if and only if v is unramified in E 0 and each extension of v to E 0 is unramified in E 00 . (c) Let F1 and F2 be field extensions of E which are contained in a common field. Suppose F1 /E is separable algebraic and v is unramified in F1 and in F2 . Then v is unramified in F1 F2 . Example 2.3.8: Radical extensions. Let (E, v) be a discrete valued field and ¯v ) - n. Consider an extension F = E(x) of n a positive integer with char(E degree n of E where xn = a is in E. Let w be an extension of v to a valuation of F and let e = ew/v . Assume both v and w are normalized. Then (6)
nw(x) = ev(a)
and e ≤ n.
There are three cases to consider: Case A: gcd n, v(a) = 1. By (6), n|e, so n = e. By (2), w is the unique extension of v to F . Therefore, v totally ramifies in F . Case B: n - v(a). By (6), e 6= 1. Hence, w ramifies over E. Case C: n|v(a). Choose π ∈ E with v(π) = 1. Write a = bπ kn with k ∈ Z and b ∈ E such that v(b) = 0. Then y = xπ −k satisfies y n = b ¯v )s into distinct linear and F = E(y). Moreover, Y n − ¯b decomposes over (E factors. Therefore, by Lemma 2.3.4, v is unramified in F . Example 2.3.9: Artin-Schreier Extensions. Let (E, v) be a discrete valued field of positive characteristic p. An Artin-Schreier extension F of degree p has the form E(x) where xp − x = a with a ∈ E. We consider two cases: Case A: v(a) < 0 and p - v(a). Let w be an extension of v to F . Then w(x) must be negative and w(xp ) < w(x). Hence, pw(x) = ev(a), where e = ew/v . Hence, p = e and w(x) = v(a). Thus, v totally ramifies in F . Case B: v(a) ≥ 0. Then X p − X − a ¯ is a separable polynomial. By Lemma 2.3.4, v is unramified in F . Qp−1 ¯v . Hence, by In particular, if v(a) > 0, then X p − X = i=0 (X − i) in E Proposition 2.3.2, v has exactly p extensions to F . Label them v0 , . . . , vp−1 with vi (x − i) > 0, i = 0, . . . , p − 1. Since vi (x − i) < vi (x − i)p , we conclude from (x − i)p − (x − i) = a that vi (x − i) = v(a). Lemma 2.3.10 (Eisenstein’s Criterion): Let R be a unique factorization domain, p a prime element of R, and f (X) = an X n + an−1 X n−1 + · · · + a0 a polynomial with coefficients ai ∈ R. Then each of the following conditions suffices for f to be irreducible over Quot(R): (a) p - an , p divides a0 , . . . , an−1 , and p2 - a0 .
30
Chapter 2. Valuations and Linear Disjointness
(b) p - a0 , p divides a1 , . . . , an , and p2 - an . Proof of (a): See [Lang7, p. 183]. Proof of (b): By (a), the polynomial X n f (X −1 ) = an + an−1 X + · · · + a0 X n is irreducible over K. Therefore, f (X) is irreducible. Example 2.3.11: Ramification at infinity. Let K be a field, t an indeterminate, and f (X) = an X n + · · · + a0 ∈ K[X] with an 6= 0. By Eisenstein crite] ˜ rion, f (X) − t is irreducible over K(t). Choose a root x of f (X) = t in K(t). Let v = v∞ be the valuation of K(t) with v(t) = −1 which is trivial on K and let w be a valuation of K(x) lying over v. The relation an xn + · · · + a0 = t implies w(x) < 0. Hence, −ew/v = w(t) = w(f (x)) = nw(x). Since ew/v ≤ [K(x) : K(t)] ≤ n, this implies ew/v = [K(x) : K(t)] = n and w(x) = −1. Hence, v is totally ramified in K(x). In particular, w is the unique valuation of K(x) lying over K(t).
2.4 Integral Extensions and Dedekind Domains Integral extensions of Z in number fields are Dedekind domains. Although they are in general not unique factorization domain, their ideals uniquely factor as products of prime ideals. In this section we survey the concepts of integral extensions of rings and of Dedekind domains and prove that every overring of a Dedekind domain is again a Dedekind domain. Let F be a field containing an integral domain R. An element x ∈ F is integral over R if it satisfies an equation of the form xn + an−1 xn−1 + · · · + a0 = 0 with a1 , . . . , an ∈ R. The set of all elements of F which are integral over R form a ring (e.g. by Proposition 2.4.1 below), the integral closure of R in F . Call R integrally closed if R coincides with its integral closure in Quot(R). For example, every valuation ring O of F is integrally closed. Indeed, assume x ∈ F r O and x is integral over O. Then xn + an−1 xn−1 + · · · + a0 = 0 for some a0 , . . . , an−1 ∈ O. Then x−1 is in the maximal ideal m of O and 1 + an−1 x−1 + · · · + a0 x−n = 0. Thus, 1 ∈ m, a contradiction. Proposition 2.4.1 ([Lang4, p. 12]): An element x of F is integral over R if and only if every place of F finite on R is finite at x. Thus, the integral closure of R in F is the intersection of all valuation rings of F which contain R. In particular, every valuation ring of F is integrally closed. Suppose ϕ is a place of a field F and K is a subfield of F . We say that ϕ is trivial on K, or also that ϕ is a place of F/K, if ϕ(x) 6= ∞ for all x ∈ K. Then ϕ(y) 6= 0 for all y ∈ K × . Thus, ϕ maps K isomorphically onto ϕ(K). Lemma 2.4.2: Let K ⊆ L ⊆ F be a tower of fields and ϕ a place of F . Suppose ϕ is trivial on K and L is algebraic over K. Then ϕ is trivial on L. Proof: Each x ∈ L is integral over K, so by Proposition 2.4.1, ϕ(x) 6= ∞. Thus, ϕ is also trivial on L.
2.4 Integral Extensions and Dedekind Domains
31
Let S be a subring of F containing R. Call S integral over R if every element of S is integral over R. If S = R[x1 , . . . , xm ] and S is integral over R, then S is a finitely generated R-module. Indeed, every element of S is a linear αm 1 α2 combination with coefficients in R of the set of monomials xα 1 x2 · · · xm , where 0 ≤ αi < deg irr(xi , Quot(R)) . Propositions 2.3.1 and 2.4.1 give the following: Proposition 2.4.3: Let R ⊆ S be integral domains with S finitely generated as an R-algebra. Suppose S is integral over R. Then the following hold: (a) S is finitely generated as an R-module. (b) Let ϕ: R → M be a homomorphism into an algebraically closed field M . Then the set of all homomorphisms ψ: S → M that extend ϕ is finite and nonempty. Suppose R1 ⊆ R2 ⊆ R3 are integral domains. Proposition 2.4.1 implies that R3 is integral over R1 if and only if R2 is integral over R1 and R3 is integral over R2 . Call an integral domain R Noetherian if every ideal of R is finitely generated. For example, since a discrete valuation ring O is a principal ideal domain, it is integrally closed and Noetherian. If R is an integral domain and p is a prime ideal of R, then Rp =
a | a∈R b
and
b ∈ Rrp
is the local ring of R at p. It has a unique maximal ideal, pRp . If R is a Noetherian domain, then Rp is also Noetherian. If R is integrally closed, then so is Rp . T Lemma 2.4.4: Suppose R is an integral domain. Then R = T Rm , where m ranges over all maximal ideals of R. More generally, a = aRm for each ideal a of R. Proof: Suppose x belongs to each aRm . For each m, x = am /bm , with am ∈ a and bm ∈ R r m. Denote the ideal generated by all the bm ’s by b. If b 6= R, then b is contained in a maximal ideal Pm. Hence, bm ∈ m, a contradiction. Hence, b = R. In particular, 1 = m∈M bm cm where M is a P finite set of maximal P ideals, and cm ∈ R for each m ∈ M . Therefore x = m∈M xbm cm = m∈M am cm ∈ a. Let R be an integral domain with the quotient field F . A nonzero Rsubmodule a of F is said to be a fractional ideal of R if there exists a nonzero x ∈ R with xa ⊆ R. In particular, every ideal of R is a fractional ideal. Define the product, ab, of two fractional ideals a and b to be the R-submodule generated by the products ab, with a ∈ a and b ∈ b. Define the inverse of a fractional ideal a as a−1 = {x ∈ F | xa ⊆ R}. If a ∈ a, then aa−1 ⊆ R. Therefore, both ab and a−1 are fractional ideals.
32
Chapter 2. Valuations and Linear Disjointness
Proposition 2.4.5 ([Cassels-Fr¨ ohlich, p. 6]): The following conditions on an integral domain R are equivalent: (a) R is Noetherian, integrally closed, and its nonzero prime ideals are maximal. (b) R is Noetherian and the local ring, Rp , of every nonzero prime ideal p is a discrete valuation ring. (c) Every fractional ideal a is invertible (i.e. aa−1 = R). When these conditions hold, R is called a Dedekind domain. By Proposition 2.4.5, the set of all fractional ideals of a Dedekind domain R forms an Abelian group, with R being the unit. One proves that this group is free and the maximal ideals of R are free generators of this group. mr 1 m2 Thus, every ideal a of R has a unique presentation a = pm 1 p2 · · · pr , as the product of powers of maximal ideals with positive exponents [CasselsFr¨ohlich, p. 8]. Every principal ideal domain is a Dedekind domain. Thus, Z and K[x], where x is a transcendental element over a field K, are Dedekind domains. By the same reason, every discrete valuation ring is a Dedekind domain. In the notation of Proposition 2.4.5(b), Rp is the valuation ring of a discrete valuation vp of K = Quot(R). The corresponding place ϕp is finite on R. Conversely, if ϕ is such a place, then p = {x ∈ R | ϕ(x) = 0} is a nonzero prime ideal of R. Since Rp ⊆ Oϕ , Lemma 2.2.5 implies that Rp = Oϕ . This establishes a bijection between the nonzero prime ideals of R and the equivalence classes of places of K finite on R. Proposition 2.4.6 ([Cassels-Fr¨ ohlich, p. 13]): Let S be the integral closure of a Dedekind domain R in a finite algebraic extension of Quot(R). Then S is also a Dedekind domain. Let p be a prime ideal of R. Then pS = Pe11 Pe22 · · · Perr , where P1 , P2 , . . . , Pr are the distinct prime ideals of S that lie over p; that is, Pi ∩ R = p, i = 1, . . . , r. For each i we have pSPi = Pei i SPi . Hence, ei is the ramification index of vPi over vp . We say Pi is unramified over K if vPi /vp is unramified; that is, ei = 1 and S/Pi is a separable extension of R/p. The prime ideal p is unramified in L if each Pi is unramified over K. By Proposition 2.4.6, the integral closure of Z in a finite extension L of Q is a Dedekind domain, OL , called the ring of integers of L. Proposition 2.4.7 (Noether-Grell): Every overring R0 of a Dedekind domain R is a Dedekind domain. Proof: We show that R0 satisfies Condition (b) of Proposition 2.4.5. Part A: An injective map. If p0 is a nonzero prime ideal of R0 , then p = R ∩ p0 is a nonzero prime ideal of R. Indeed, for 0 6= x ∈ p0 , write x = ab , where a, b ∈ R. Thus, 0 6= a = bx ∈ R ∩ p0 = p. Since Rp ⊆ Rp0 0 and Rp is a discrete valuation ring, Lemma 2.2.5 implies that Rp = Rp0 0 . Hence, (1)
pRp = p0 Rp0 0 .
2.4 Integral Extensions and Dedekind Domains
33
In addition, p0 Rp0 0 ∩ R0 = p0 . Therefore, the map p0 7→ R ∩ p0 from the set of nonzero prime ideals of R0 into the set of nonzero prime ideals of R is injective. Part B: A finiteness condition. Let x be a nonzero element of R0 , p0 a prime ideal of R0 which contains x, and p = R ∩ p0 . Then Rp = Rp0 0 . Hence, vp (x) > 0, where vp is the valuation of Quot(R) corresponding to p. But this relation holds only for the finitely many prime ideals of R that appear with positive exponents in the factorization of the fractional ideal xR. Hence, by Part A, x belongs to only finitely many prime ideals of R0 . Part C: The ring R0 is Noetherian. Let a be a nonzero ideal of R0 . Choose a nonzero element x ∈ a and denote the finite set of prime ideals of R0 that contain x by P . For each p ∈ P the local ring Rp0 is a discrete valuation domain. Hence, there exists ap ∈ a such that aRp0 = ap Rp0 . Denote the ideal of R0 generated by x and by all ap , for p ∈ P , by a0 . It is contained in a. To show that a is finitely generated, we need only prove that a ⊆ a0 . Indeed, consider a prime ideal q of R0 not in P . Then x ∈ / q,Tso a0 6⊆ q. Rq0 . It followsT from Lemma T2.4.4 that a0 = p∈P a0 Rp0 . Hence, a0 Rq0 = T Therefore, a ⊆ p∈P aRp0 = p∈P ap Rp0 ⊆ p∈P a0 Rp0 = a0 , as desired. Lemma 2.4.8: Let (E, v) be a discrete valued field, F1 , F2 , F finite separable extensions of E with F = F1 F2 , and w an extension of v to F . Suppose v is unramified in F1 . Then the residue fields with respect to w satisfy F¯ = F¯1 F¯2 . Proof: Choose a finite Galois extension N of E which contains F and an extension w0 of w to N . Denote the decomposition groups of w0 over E, F1 , F2 , F by DE , DF1 , DF2 , DF , respectively. Let E 0 , F10 , F20 , F 0 be the fixed fields in N of DE , DF1 , DF2 , DF , respectively. Let v 0 = w0 |E 0 . Since all valuations of N lying over v 0 are conjugate over E 0 , the definition of E 0 as the fixed field of DE implies that w0 is the unique extension of v 0 to N . Also, DF1 = Gal(N/F1 ) ∩ DE , so F1 E 0 = F10 . By Lemma 2.3.6, v 0 is unramified in F10 . Finally, by [Serre3, p. 32, Prop. 21(c)], the residue fields of E, F1 , F2 , F at w coincide with the residue fields of E 0 , F10 , F20 , F 0 at w0 , respectively. We may therefore replace E, F1 , F2 , F , respectively, by E 0 , F10 , F20 , F 0 , if necessary, to assume that w|F1 is the unique extension of v to F1 . Now put wi = w|Fi , i = 1, 2. By Proposition 2.4.1, Ow1 is the integral closure of Ov in ¯ F1 . Since v is unramified in F1 , Proposition 2.3.2 implies [F1 : E] = [F¯1 : E], where the bar denotes reduction modulo w. Choose x ∈ Ow1 such that x ¯ is a primitive element for the separable ¯ Let f = irr(x, E) and p = irr(¯ ¯ Then f ∈ Ov [X] and extension F¯1 /E. x, E). f (x) = 0. Hence, f¯(¯ x) = 0 and p|f¯. Therefore, ¯ = [F1 : E]. [F1 : E] ≥ deg(f ) ≥ deg(¯ p) = [F¯1 : E] ¯ x). Consequently, p = f¯, F1 = E(x), and F¯1 = E(¯
34
Chapter 2. Valuations and Linear Disjointness
By Lemma 2.3.6, w2 is unramified in F . Thus, we may apply the result x). Conseof the preceding paragraph to F/F2 and conclude that F¯ = F¯2 (¯ ¯ x)F¯2 = F¯2 (¯ x) = F¯ . quently, F¯1 F¯2 = E(¯
2.5 Linear Disjointness of Fields Central to field theory is the concept “linear disjointness of fields”, an analog of linear independence of vectors. We repeat the convention made in “Notation and Convention” that whenever we form the compositum of fields, we tacitly assume they are contained in a common field. Lemma 2.5.1: Let E and F be extensions of a field K. The following conditions are equivalent: (a) Each m-tuple (x1 , . . . , xm ) of elements of E which is linearly independent over K is also linearly independent over F . (b) Each n-tuple (y1 , . . . , yn ) of elements of F which is linearly independent over K is also linearly independent over E. Proof: It suffices to prove that (a) implies (b). Let y1 , . . . , yn be elements of F for which there exist a1 , . . . , an ∈ E with a1 y1 + · · · +Pan yn = 0. Let {xj | j ∈ J} be a linear basis for E over K and write ai = j∈J aij xj with aij elements of K, only finitely many different from 0. Then n X X j∈J
aij yi xj = 0.
i=1
P aij yi = 0 for By (a), {xj | j ∈ J} is linearly independent over F . Hence, every j. If y1 , . . . , ym are linearly independent over K, then aij = 0 for every i and j, so ai = 0, i = 1, . . . , m. Thus, y1 , . . . , ym are linearly independent over E. This proves (b). Definition: With E and F field extensions of a field K, refer to E and F as linearly disjoint over K if (a) (or (b)) of Lemma 2.5.1 holds. Corollary 2.5.2: Let E and F be extensions of a field K such that [E : K] < ∞. Then E and F are linearly disjoint over K if and only if [E : K] = [EF : F ]. If in addition [F : K] < ∞, then this is equivalent to [EF : K] = [E : K][F : K]. Proof: If E and F are linearly disjoint over K and w1 , . . . , wn is a basis for E/K, then w1 , . . . , wn is also a basis for EF over F . Hence, [EF : F ] = n = [E : K]. Conversely, suppose [E : K] = [EF : F ] and let x1 , . . . , xm ∈ E be linearly independent over K. Extend {x1 , . . . , xm } to a basis {x1 , . . . , xn } of E/K. Since {x1 , . . . , xn } generates EF over F and n = [EF : F ], {x1 , . . . , xn } is a basis of EF/F . In particular, x1 , . . . , xm are linearly independent over F .
2.5 Linear Disjointness of Fields
35
Let E/K be a finite Galois extension. If E ∩ F = K, then, by Corollary 2.5.2, E and F are linearly disjoint over K. The condition, E ∩ F = K is equivalent to “res: Gal(EF/F ) → Gal(E/K) is an isomorphism” and also to “res: Gal(F ) → Gal(E/K) is surjective.” For arbitrary extensions this condition is clearly necessary, but not sufficient. Let L be a degree n > 1 extension of K for which L0 is conjugate to L over K and L0 ∩ L = K. Then 2.5.2, L and L0 are not [LL0 : K] ≤ n(n − 1). Thus, according to Corollary √ 3 linearly √ disjoint over K. For example, Q( 2) is not linearly disjoint from Q(ζ3 3 2) over Q although their intersection is Q. Lemma 2.5.3 (Tower Property): Let K ⊆ E and K ⊆ L ⊆ F be four fields. Then E is linearly disjoint from F over K if and only if E is linearly disjoint from L over K and EL is linearly disjoint from F over L. Proof: The only nontrivial part is to show that if E and F are linearly disjoint over K, then EL and F are linearly disjoint over L. Apply Lemma 2.5.1. Suppose that y1 , . . . , ym are elements of F which are linearly independent over L, but a1 , . . . , am are elements of EL such Pm that Pi=1 ai yi = 0. Clear denominators to assume that ai ∈ L[E], so that xj with aij ∈ L, where {xj | j ∈ J} is a linear basis for E over K. ai = PaijP yi )xj = 0. By assumption, the xj are linearly independent Then j ( i aijP over F . Hence, j aij yi = 0, so aij = 0 for all i and j. Consequently, ai = 0, i = 1, . . . , m. Lemma 2.5.4: Let L be a separable algebraic extension of a field K and let M be a purely inseparable extension of K. Then L and M are linearly disjoint over K. ˆ be the Galois closure of L/K. Then L ˆ ∩ M = K. Hence, L ˆ Proof: Let L and M are linearly disjoint over K. Therefore, by Lemma 2.5.3, L and M are linearly disjoint over K. Let E1 , . . . , En be n extensions of a field K. We say that E1 , . . . , En are linearly disjoint over K if E1 · · · Em−1 and Em are linearly disjoint over K for m = 2, . . . , n. Induction on n shows that this is the case if of and only if the following condition holds: If wi,ji , ji ∈ Ji , are elements Qn Ei which are linearly independent over K, i = 1, . . . , n, then i=1 wi,ji , (j1 , . . . , jn ) ∈ J1 × · · · × Jn , are linearly independent over K. It follows that E1 , . . . , En are linearly disjoint over K if and only if the canonical homomorphism of E1 ⊗K · · · ⊗K En into E1 · · · En that maps x1 ⊗ · · · ⊗ xn onto x1 · · · xn is injective. It also follows that if E1 , . . . , En are linearly disjoint over K, then Eπ(1) , . . . , Eπ(n) are linearly disjoint over K for every permutation π of {1, . . . , n}. The application of tensor products makes the following lemma an easy observation. Lemma 2.5.5: Let E1 , . . . , En (resp. F1 , . . . , Fn ) be linearly disjoint field extensions of K (resp. L). For each i between 1 and n let ϕi : Ei → Fi ∪ {∞},
36
Chapter 2. Valuations and Linear Disjointness
be either a place or an embedding. Suppose ϕ1 , . . . , ϕn coincide on K and ϕi (K) = L, i = 1, . . . , n. Let E = E1 · · · En and F = F1 · · · Fn . Then there exists a place ϕ: E → F˜ ∪ {∞} that extends each of the ϕi ’s. If each ϕi is an isomorphism of Ei onto Fi , then ϕ is an isomorphism of E onto F . Proof: Let Oi be the valuation ring of ϕi if ϕi is a place and Ei if ϕi is an isomorphism. By assumption, the map x1 · · · xn → x1 ⊗ · · · ⊗ xn is an isomorphism O1 · · · On ∼ = O1 ⊗K · · · ⊗K On of rings. Hence, there exists a ring homomorphism ϕ0 : O1 · · · On → F such that ϕ0 (x) = ϕi (x) for each x ∈ Oi , i = 1, . . . , n. Extend ϕ0 to a place ϕ: E → F˜ ∪ {∞} (Proposition 2.3.1). If x ∈ Ei r Oi , then ϕ(x−1 ) = ϕi (x−1 ) = 0, so ϕ(x) = ϕi (x) = ∞. We conclude that ϕ coincides with ϕi on Ei . Finally, define a family {Ei | i ∈ I} of field extensions of K to be linearly disjoint over K if every finite subfamily is linearly disjoint over K. It follows from the discussion preceding Lemma 2.5.5 that a sequence (E1 , E2 , E3 , . . .) of fields extensions of K is linearly disjoint over K if En is linearly disjoint from E1 · · · En−1 for 2, 3, 4, . . . . Then, Eπ(1) , Eπ(2) , Eπ(3) , . . . are linearly disjoint for every permutation π of N. disjoint family of Galois extenLemma 2.5.6: Let {Li | i ∈ I} Q be a linearlyQ sions of a field K. Then Gal( i∈I Li /K) ∼ = i∈I Gal(Li /K). Q Q Proof: Since i∈I Gal(Li /K) ∼ = lim i∈I0 Gal(Li /K), we may assume I ←− Q Q is finite. In this case, the embedding Gal( i∈I Li /K) → i∈I Gal(Li /K) given by σ 7→ (σ|Li )i∈I is surjective (Lemma 2.5.5). Therefore, it is an isomorphism. Lemma 2.5.7: Let K be a field, K1 , K2 , K3 , . . . a linearly disjoint sequence of extensions of K, and L a finite separable extension of K. Then there exists a positive integer n such that L, Kn , Kn+1 , Kn+2 , . . . are linearly disjoint over K. Proof: Replace L by its Galois closure over K, if necessary, to assume L is Galois over K. Assume for each positive integer n the field L is not linearly disjoint from Kn Kn+1 Kn+2 · · · over K. Then Ln = L ∩ Kn Kn+1 Kn+2 · · · is a proper extension of K. Since L has only finitely many extensions that contain K and since Ln ⊇ Ln+1 ⊇ Ln+2 ⊇ · · ·, there is an m such that Ln = Lm for all n ≥ m. Since Lm is a finite extension of K, there is an n > m with Lm ⊆ Km · · · Kn−1 . Similarly, there exists r > n with Lm ⊆ Kn · · · Kr−1 . By assumption, Km · · · Kn−1 and Kn · · · Kr−1 are linearly disjoint over K. In particular, their intersection is K. Therefore, Lm = K. This contradiction proves there exists n such that L, Kn , Kn+1 , Kn+2 , . . . are linearly disjoint over K. Lemma 2.5.8: Let v be a discrete valuation of a field K and L, M finite extensions of K. Suppose v is unramified in L but totally ramified in M . Then L and M are linearly disjoint over K.
2.5 Linear Disjointness of Fields
37
Proof: Let L0 be the maximal separable extension of K in L and v0 an extension of v to L0 . Then L/L0 is purely inseparable. Hence, v0 is ramified in L. Therefore, L = L0 and L/K is separable. Since v is unramified in each of the conjugates of L over K, it is unramified in their compositum (Corollary 2.3.7). We may therefore replace L by the Galois closure of L/K, if necessary, to assume L/K is Galois. Let m = [L ∩ M : K]. Choose an extension w of v to L ∩ M . Then e(w/v) = 1 on one hand and e(w/v) = m on the other hand. Thus, L ∩ M = K. Therefore, L is linearly disjoint from M over K. Example 2.5.9: Roots of unity. For each n consider the Galois extension Q(ζn ) of Q obtained by adjoining a primitive root of unity of order n. It is well known that ϕ(n) = [Q(ζn ) : Q] is the number of integers between 1 and n which are relatively prime to n [Lang7, p. 278, Thm. 3.1]. If m is relatively prime to n, then ϕ(mn) = ϕ(m)ϕ(n) [LeVeque, p. 28, Thm. 3-7]. In addition, Q(ζm , ζn ) = Q(ζmn ). Hence, [Q(ζm , ζn ) : Q] = [Q(ζmn ) : Q] = ϕ(mn) = ϕ(m)ϕ(n) = [Q(ζm ) : Q][Q(ζn ) : Q]. It follows from Corollary 2.5.2 that Q(ζm ) and Q(ζn ) are linearly disjoint over Q. Here is an application of linear disjointness to integral closures of domains. Lemma 2.5.10: Let K be a field, L a separable algebraic extension of K, and R an integrally closed integral domain containing K. Let E = Quot(R), F = EL, and S the integral closure of R in F . Suppose E and L are linearly disjoint over K. Then S = RL ∼ = R ⊗K L. Proof: Assume without loss L/K is finite. Choose a basis w1 , . . . , wn for L/K. Let σ1 , . . . , σn be the distinct K-embeddings of L into Ks . Then det(σi wj ) 6= 0. Each element of L is integral over K, hence over R, so RL ⊆ S. Conversely, let xP ∈ S. By the linear disjointness, w1 , . . . , wn form a basis for F/E. n Hence, x = j=1 ej wj with ej ∈ E, j = 1, . . . , n. Also, each σi extends to Pn an E-embedding of F into Es (Lemma 2.5.5). Thus, σi x = j=1 ej σi wj , i = 1, . . . , n. Apply Kramer’s law to present each ek as a polynomial in σi x, σi wj , with i, j = 1, . . . , n, divided by det(σi wj ). Thus, ek is an element of E which is integral over R. Since R is integrally closed, ek ∈ R, k = 1, . . . , n. Consequently, x ∈ RL, as needed. We generalize the tower property to families of field extensions: Lemma 2.5.11: Let K be a field and I a set. For each i ∈ I let Fi /Ei be a field extension with K ⊆ Ei . Suppose {Fi | i ∈ I} is linearly disjoint over K. Denote the compositum of all Ei ’s by E. Then the set {Fi E | i ∈ I} is linearly disjoint over E. Moreover, for each i ∈ I, the field Fi is linearly disjoint from E over Ei . Proof: It suffices to consider the case where I = {1, 2, . . . , n}. By induction suppose Fi E1 · · · En−1 , i = 1, . . . , n − 1, are linearly disjoint over E1 · · · En .
38
Chapter 2. Valuations and Linear Disjointness
By assumption, F1 · · · Fn−1 is linearly disjoint from Fn over K. Hence, by the tower property, F1 · · · Fn−1 is linearly disjoint from E over E1 , . . . , En−1 , so Fi E, i = 1, . . . , n − 1, are linearly disjoint over E. F1 · · · Fn−1
F1 · · · Fn−1 E
Fi E1 · · · En−1
Fi E
E1 · · · En−1
E
EFn
K
En
Fn
Moreover, F1 · · · Fn−1 E is linearly disjoint from EFn over E. Consequently, E is linearly disjoint from Fn over En and Fi E, i = 1, . . . , n are linearly disjoint over E, as claimed.
2.6 Separable, Regular, and Primary Extensions Based on the notion of linear disjointness we define here three type of field extensions. We say that a field extension F/K is separable (resp. regular, ˜ Ks ) over K. primary) if F is linearly disjoint from Kins (resp. K, Separable Extensions. We generalize the notion of “separable algebraic extension” to arbitrary field extensions. Let K be a field of positive characteristic p. The field generated over K by the pth roots of all elements of K is denoted K 1/p . We denote the ∞ maximal purely inseparable extension of K by Kins (or K 1/p ). Let F be a finitely generated extension of K. A collection t1 , . . . , tr ∈ F of elements algebraically independent over K is a separating transcendence basis if F/K(t1 , . . . , tr ) is a finite separable extension. Lemma 2.6.1: An extension F of a field K is separable if it satisfies one of the following equivalent conditions: (a) F is linearly disjoint from Kins over K. (b) F is linearly disjoint from K 1/p over K. (c) Every finitely generated extension E of K which is contained in F has a separating transcendence basis. Moreover, a separating transcendence basis can be selected from a given set of generators for F/K. Proof: The implications “(a) => (b)” and “(c) => (a)” are immediate consequences of the tower property (Lemma 2.5.3). For “(b) => (c)” see [Lang 4, p. 54]. Lemma 19.2.4 gives a constructive proof. In particular, every separable algebraic extension satisfies conditions (a), (b), and (c) of Lemma 2.6.1. Now apply the rules of linear disjointness.
2.6 Separable, Regular, and Primary Extensions
39
Corollary 2.6.2: (a) If E/K and F/E are separable extensions, then F/K is also separable. (b) If F/K is a separable extension, then E/K is separable for every field K ⊆ E ⊆ F. (c) Every extension of a perfect field is separable. (d) If E/K is a purely inseparable extension and F/K is a separable extension, then E and F are linearly disjoint over K. Example 2.6.3: A separable tower does not imply separable steps. Consider the tower of fields Fp ⊂ Fp (tp ) ⊂ Fp (t), where t is transcendental over Fp . The extension Fp (t)/Fp is separable, but Fp (t)/Fp (tp ) is not. Regular Extensions. Finitely generated regular extensions characterize absolutely irreducible varieties (Section 10.2) Lemma 2.6.4: A field extension F/K is regular if it satisfies one of the following equivalent conditions: (a) F/K is separable and K is algebraically closed in F . ˜ over K. (b) F is linearly disjoint from K Proof: The implication “(b) => (a)” is immediate. To prove “(a) => (b)”, it suffices to assume that F/K is finitely generated. Then F/K has a separating transcendence basis, t1 , . . . , tr , which is also a separating transcendence basis for the extension F Ks /Ks . Since ˜ over ˜ = (Ks )ins , Lemma 2.6.1 implies that F Ks is linearly disjoint from K K Ks . Also, Ks /K is a Galois extension and F ∩ Ks = K. Hence, F is linearly disjoint from Ks over K. Therefore, by Lemma 2.5.3, F is linearly disjoint ˜ over K. from K Corollary 2.6.5: (a) If E/K and F/E are regular extensions, then F/K is regular. (b) If F/K is a regular extension, then E/K is regular for every field E lying between K and F . (c) Every extension of an algebraically closed field is regular. (d) Let m be a cardinal number andSKα , α ≤ m, an ascending transfinite sequence of fields such that Kγ = α<γ Kα for each limit ordinal number γ ≤ m. Suppose Kγ+1 is a regular extension of Kγ for all γ < m. Then Km is a regular extension of each Kβ with β < m. Proof of (d): Let δ ≤ m be a transfinite number. By transfinite induction assume Kγ is regular extension of Kβ for all β ≤ γ < δ. Now distinguish between two cases: Case A: δ is a limit number. Consider β < δ, elements a1 , . . . , ar ∈ ˜ K Pβr linearly independent over Kβ , and elements u1 , . . . , ur ∈ Kδ satisfying i=1 ai ui = 0. Then there exists an ordinal number γ with β ≤ γ < δ and ˜ β is linearly disjoint from Kγ u1 , . . . , ur ∈ Kγ . Since Kγ /Kβ is regular, K
40
Chapter 2. Valuations and Linear Disjointness
˜β over Kβ , so a1 = · · · = ar = 0. Therefore, Kδ is linearly disjoint from K over Kβ . Case B: δ = γ + 1 is a successor number. By assumption, both Kγ /Kβ and Kγ+1 /Kγ are regular extensions. Hence, by (a), Kδ /Kβ is a regular extension. The next lemma gives a criterion for a regular extension F/K to be linearly disjoint from another extension of K in terms of “algebraic independence”. To define this notion consider an arbitrary field extension F/K and a subset T of F . We say that T is algebraically independent over K if f (t1 , . . . , tn ) 6= 0 for all t1 , . . . , tn ∈ T and for each nonzero f ∈ K[X1 , . . . , Xn ]. If in addition F/K(T ) is an algebraic extension, then T is a transcendence base of F/K. The cardinality of T depends only on F/K. It is the transcendence degree of F/K. We denote it by trans.deg(F/K). For example, trans.deg(F/K) = 0 if and only if F/K is an algebraic extension. If S is a subset of F such that F/K(S) is algebraic, then S contains a transcendence base for F/K [Lang7, p. 356, Thm. 1.1]. In particular, if F/K is finitely generated, then trans.deg(F/K) < ∞. The converse is false. For ˜ ˜ example, trans.deg(Q/Q) = 0 although Q/Q is not finitely generated. If T0 is a subset of F which is algebraically independent over K, choose a transcendence base T1 for F/K(T0 ). Then T0 ∩ T1 = ∅ and T0 ∪ T1 is a transcendence base for F/K. This argument also gives the additivity of the transcendence degree for a tower K ⊆ E ⊆ F of fields: (1)
trans.deg(F/K) = trans.deg(E/K) + trans.deg(F/E).
Now consider two extensions E and F of a field K. We say that E and F are algebraically independent over K if (2) every m-tuple (t1 , . . . , tm ) of elements of E which is algebraically independent over K is also algebraically independent over F . It follows that E and F are algebraically independent over K if and only if E0 and F are algebraically independent over K for every subfield E0 of E which is finitely generated over K. Hence, in order to prove that algebraic independence is a symmetric relation, we may consider finitely generated extensions E and F of K, assume that (2) holds, and prove condition (2) with the roles of E and F exchanged. Indeed, let u1 , . . . , un be elements of F which are algebraically independent over K. Enlarge n, if necessary, to assume that u1 , . . . , un form a transcendence base of F/K. Then F/K(u) is algebraic and therefore so is EF/E(u). After reordering the ui , we may assume u1 , . . . , um form a transcendence base for EF/E. Assumption (2) implies that trans.deg(E/K) = trans.deg(EF/F ). Hence, by (1), m = trans.deg(EF/E) = trans.deg(EF/K) − trans.deg(E/K) = trans.deg(EF/K) − trans.deg(EF/F ) = trans.deg(F/K) = n. Therefore, u1 , . . . , un are algebraically independent over E, as desired. Like linear disjointness, algebraic independence has the tower property: Let K ⊆ L ⊆ M and K 0 ⊆ L0 ⊆ M 0 be fields with K ⊆ K 0 , L0 = K 0 L and
2.6 Separable, Regular, and Primary Extensions
41
M 0 = L0 M . Then trans.deg(M/K) = trans.deg(L/K)+trans.deg(M/L) and trans.deg(M 0 /K 0 ) = trans.deg(L0 /K 0 ) + trans.deg(M 0 /L0 ). Also, trans.deg(L0 /K 0 ) ≤ trans.deg(L/K), trans.deg(M 0 /L0 ) ≤ trans.deg(M/L). Hence, M is algebraically independent from K 0 over K if and only if L is algebraically independent from K 0 over K and M is algebraically independent from L0 over L. By considering monomials in elements x1 , . . . , xn of E, it is clear that if E and F are linearly disjoint over K, then they are also algebraically independent over K. The converse, however, is false: Any two extensions of K one of which is algebraic are algebraically independent over K. Lemma 2.6.7 below gives a partial converse. Lemma 2.6.6: Let F and F¯ be fields, T (resp. T¯) an algebraically independent set over F (resp. over F¯ ), ϕ0 : F → F¯ ∪ {∞} a place, and ϕ1 : T → T¯ a bijective map. Then there exists a place ϕ: F (T ) → F¯ (T¯) ∪ {∞} extending both ϕ0 and ϕ1 . Proof: The case where T consists of one element t is covered by Example 2.3.3. In the general case well order T and apply transfinite induction. Lemma 2.6.7: Let E be a regular extension of a field K and let F be an extension of K. If E and F are algebraically independent over K, then E and F are linearly disjoint over K. Proof (Artin): Let x1 , . . . , xn be elements of E for which there exist P ai xi = 0. Use Proposition 2.3.1 a1 , . . . , an ∈ F , not all zero, such that ˜ ∪ {∞}. Let T be a transcendence base to choose a K-place ϕ of F into K for E over K. Then the elements of T are algebraically independent over F . Hence, by Lemma 2.6.6, ϕ extends to a K(T )-place of F (T ). Since E is an ˜ ∪ {∞} algebraic extension of K(T ), ϕ extends to an E-place of EF into E (Lemma 2.4.2). say a1 , to assume that a1 = 1 With no loss we may divide a1 , . . . , an by, P ϕ(ai )xi = 0 is a nontrivial and that all the ai are finite under ϕ. Thus, ˜ But E is linearly disjoint from K ˜ over linear combination of the xi over K. K. Hence, x1 , . . . , xn are also linearly dependent over K. Corollary 2.6.8: (a) Let E be a regular extension of a field K, algebraically independent from an extension F of K. Then EF is a regular extension of F . (b) If two regular extensions E and F of K are algebraically independent, then EF/K is regular. Proof: For (a) note that E is also algebraically independent from F˜ over K. By Lemma 2.6.7, E is linearly disjoint from F˜ over K. Hence, by Lemma 2.5.3, EF is linearly disjoint from F˜ over F . Therefore, EF/F is regular. For (b) use (a) and Corollary 2.6.5(a).
42
Chapter 2. Valuations and Linear Disjointness
Lemma 2.6.9: Each of the following conditions on a field extension F/K implies that F/K is regular: ˜ ∪ {∞} with (a) For all u1 , . . . , un ∈ F × , there exists a K-place ϕ: F → K ϕ(u1 ), . . . , ϕ(un ) ∈ K × . (b) There exists a K-place ϕ: F → K ∪ {∞}. Proof: We prove that F/K satisfies Condition (b) of Lemma 2.6.4. Consider ˜ which are linearly independent over K. Assume there exist w1 , . . . , wn in K Pn u1 , . . . , un in F , not all zero, such that i=1 ui wi = 0. Omitting the terms with ui = 0, we may assume ui 6= 0 for all i. In Case (a) choose a K-place ˜ ∪ {∞} with ϕ(ui ) ∈ K × for each i such that ui 6= 0. In Case (b) ϕ: F → K divide u1 , . . . , un with one of them, say with u1 , to assume that u1 = 1 and that ϕ(ui ) ∈ K for i = 1, . . . , n. Now apply Proposition 2.3.1 and extend ϕ ˜ →K ˜ ∪ {∞}. By Lemma 2.4.2, ϕ˜ is trivial on K. ˜ In other to a place ϕ: ˜ FK ˜ words, the restriction of ϕ˜ to K is an automorphism. ˜ n ) are linearly independent over K. Since ˜ 1 ), . . . , ϕ(w Pn In particular, ϕ(w ˜ i ) = 0, this implies ϕ(ui ) = 0 for i = 1, . . . , n. This contrai=1 ϕ(ui )ϕ(w diction proves that w1 , . . . , wn are linearly independent over F . We conclude ˜ that F is linearly disjoint from K over K. Example 2.6.10: Purely transcendental extensions. Let t1 , . . . , tn be algebraically independent elements over a field K. For each i between 1 and n the map ti → 0 extends to a K(t1 , . . . , ti−1 )-place ϕi of K(t1 , . . . , ti ) onto K(t1 , . . . , ti−1 ) ∪ {∞}. Hence, by Lemma 2.6.9(b), K(t1 , . . . , ti )/K(t1 , . . . , ti−1 ) is a regular extension. Therefore, by Corollary 2.6.9(a), K(t)/K is a regular extension. Of course, we can also prove the latter result directly: Let f1 , . . . , fm ˜ be elements of K(t) which are linearly Pmdependent over K. Thus, there ˜ ˜i fi = 0. Clearing denominaare c˜1 , . . . , c˜m ∈ K not all zero with i=0 c P j1 jn tors, we may assume all fi ∈ K[t]. Write fi (t) = j aij t1 · · · tn . Then j1 P P P Pm m m ˜i aij t1 · · · tjnn = i=1 c˜i fi (t) = 0. Hence, i=1 c˜i aij = 0 for all j i=1 c Pm j. Thus, the homogeneous linear system of equations i=1 Xi aij = 0 with ˜ n . Therefore, it has a nonzero coefficients aij ∈ K has a nonzero solution in K n solution in K . In other words, thereP are c1 , . . . , cn ∈ K not all zero with Pm m i=1 ci aij = 0 for all j. They satisfy i=1 ci fi = 0. Hence, f1 , . . . , fm are linearly dependent over K. This completes the direct proof that K(t)/K is a regular extension. We have not defined composition of places. But if we had, we could compose the places ϕi : K(t1 , . . . , ti ) → K(t1 , . . . , ti−1 ), i = n, n − 1, . . . , 1, of the first paragraph to a K-place ϕ: K(t) → K satisfying ϕ(ti ) = 0, i = 1, . . . , n. Again, by Lemma 2.6.9(b), this would prove that K(t)/K is regular. Also, given a1 , . . . , an ∈ K, we can replace ti by ti − ai to produce a K-place ψ: K(t) → K with ψ(ti ) = ai , i = 1, . . . , n.
2.6 Separable, Regular, and Primary Extensions
43
Let now T be an arbitrary set of algebraically independent elements over K and ϕ0 : T → K a map. Then every finitely generated subextension of K(T )/K is regular. Therefore, K(T )/K is regular. Moreover, using transfinite induction, it is possible to construct a K-place ϕ: K(T ) → K which extends ϕ0 . Example 2.6.11: Absolutely irreducible polynomials. Now consider a polynomial f ∈ K[T1 , . . . , Tn , X]. Suppose f is absolutely irreducible; that ] ˜ 1 , . . . , Tn , X]. Let x be a root of f (t, X) in K(t). is, f is irreducible in K[T ˜ ˜ Then [K(t, x) : K(t)] = degX f = [K(t, x) : K(t)]. By Corollary 2.5.2, ˜ K(t, x) is linearly disjoint from K(t) over K(t). By Example 2.6.10, K(t) ˜ is linearly disjoint from K over K. Hence, by the tower property (Lemma ˜ over K. Therefore, K(t, x)/K is a 2.5.3), K(t, x) is linearly disjoint from K regular extension. Conversely, suppose f is irreducible in K[T, X] and K(t, x)/K is a regular extension. Reversing the above arguments shows that f is absolutely irreducible. As an application we rephrase Corollary 1.3.4 and supply a new proof. It is a simplified version of [Leptin]. Proposition 2.6.12: Let G be a profinite group and K a field. Then there is a Galois extension F/E with K ⊆ E and Gal(F/E) ∼ = G. Proof: Write G as a projective limit lim Gi of finite groups Gi with i ranging ←− Q over a directed set I. By definition, G is a closed subgroup of i∈I Gi . Suppose we have Q constructed an algebraic extension F/E with K ⊆ E and Gal(F/E) ∼ = = i∈I Gi . Let E 0 be the fixed field of G in F . Then Gal(F/E 0 ) ∼ G. Q In order to construct F/E with Gal(F/E) ∼ = i∈I Gi , we choose a family (xσi )i∈I, σ∈Gi of algebraically independent elements over K. For each i ∈ I let Fi = K(xσi | σ ∈ Gi ). The group Gi acts on Fi by the rule (xσi )τ = xστ i and aτ = a for a ∈ K. Let Ei be the fixed field. Then K ⊆ Ei and Fi /Ei is a Galois extension with Galois group Gi [Lang7, p. 264]. Denote the compositum of all Ei ’s by E and the compositum of all Fi ’s by F . By Example 2.6.10, each Fi is a regular extension of K. By construction, the set {Fi | i ∈ I} is algebraically independent over K. Hence, by Lemma 2.6.7, the set {Fi | i ∈ I} is linearly disjoint over K. It follows from Lemma 2.5.11, that the set {EFi | i ∈ I} is linearly disjoint over E. Moreover, E is ∼ ∼ linearly disjoint from Fi over Ei . Therefore, Gal(EF Q i /E) = Gal(Fi /Ei ) = Gi . It follows from Lemma 2.5.6 that Gal(F/E) ∼ = i∈I Gi , as desired. Primary Extensions. (Section 21.2).
We use primary extensions in the study of Ci -fields
Lemma 2.6.13: A field extension F/K is primary if it satisfies one of the following equivalent conditions:
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Chapter 2. Valuations and Linear Disjointness
˜ (a) F ∩ K/K is a purely inseparable extension. (b) The field F is linearly disjoint from Ks over K. Proof: Clearly “(b) => (a).” The implication “(a) => (b)” holds since F ∩ Ks = K and Ks /K is a Galois extension. Corollary 2.6.14: (a) If E/K and F/E are primary extensions, then so is F/K. (b) If F/K is a primary extension, then E/K is primary, for every field K ⊆ E ⊆ F. (c) Every extension of a separably closed field is primary. (d) An extension F/K is regular if and only if it is separable and primary. Lemma 2.6.15: (a) Let E be a primary extension of a field K which is algebraically independent from an extension F of K. Then EF is a primary extension of F. (b) If two primary extensions E and F of K are algebraically independent, then EF/K is primary. Proof: Assertion (b) follows from (a) and from Corollary 2.6.14(a). To prove (a), choose a transcendence base T for E/K and let M be the maximal separable extension of K(T ) in E. Then M is a separable and primary extension of K. Hence, by Lemma 2.6.14(d), it is regular. Also, M is algebraically independent from Fs over K. By Lemma 2.6.7, M F is linearly disjoint from Fs over F . Since EF is a purely inseparable extension of M F , it is linearly disjoint from M Fs . It follows that EF is linearly disjoint from Fs over F ; that is, EF is a primary extension of F .
2.7 The Imperfect Degree of a Field We classify fields of positive characteristic by their imperfect degree and characterize those fields for which every finite extension has a primitive element as fields of imperfect degree 1. Let F be a field of positive characteristic p. Consider a subfield F0 of F that contains the field F p of all pth powers in F . Observe that for x1 , . . . , xn ∈ F , the set of monomials (1)
xi11 · · · xinn ,
0 ≤ i1 , . . . , in ≤ p − 1,
generates F0 (x) over F0 . Hence, [F0 (x) : F0 ] ≤ pn . If [F0 (x) : F0 ] = pn , then x1 , . . . , xn are said to be p-independent over F0 . Equivalently, each of the fields F0 (x1 ), . . . , F0 (xn ) has degree p over F0 and they are linearly disjoint over F0 . This means that the set of monomials (1) is linearly independent over F0 . A subset B of F is p-independent over F0 , if every finite subset of B is p-independent over F0 . If in addition F0 (B) = F , then B is said to be a p-basis for F over F0 . As in the theory of vector spaces, each maximal p-independent subset of F over F0 is a p-basis for F over F0 .
2.7 The Imperfect Degree of a Field
45
If x1 , . . . , xn ∈ F are p-independent over F p , we call them p-independent elements of F . The p-power pn = [F : F p ] is the imperfect degree of F and n is the imperfect exponent of F . We say that F is n-imperfect. Thus, a perfect field has imperfect exponent 0. Both quantities are infinite if [F : F p ] = ∞. In this case F is ∞-imperfect. Lemma 2.7.1 (Exchange Principle): Let F0 be a subfield of F which contains F p. (a) Let x1 , . . . , xm , y1 , . . . , yn ∈ F be such that x1 , . . . , xm are p-independent over F0 and x1 , . . . , xm ∈ F0 (y1 , . . . , yn ). Then m ≤ n, and there is a reordering of y1 , . . . , yn so that y1 , . . . , ym ∈ F0 (x1 , . . . , xm , ym+1 , . . . , yn ). (b) Every subset of F which is p-independent over F0 extends to a p-basis for F over F0 . Proof: We use induction on m. Assume the lemma is true for m = k. Thus, for m = k + 1 we may assume that xk+1 ∈ F0 (x1 , . . . , xk , yk+1 , . . . , yn ) = F1 . Then [F1 : F0 ] ≤ pn and there exists l between k + 1 and n such that yl ∈ F0 (x1 , . . . , xk+1 , yk+1 , . . . , yl−1 ), since otherwise [F1 : F0 ] ≥ pn+1 , a contradiction. Thus, yl can be exchanged for xk+1 . This proves the first part of the lemma for m = k + 1. For the last part start from a subset A of K which is p-independent over F0 . Use Zorn’s lemma to prove the existence of a maximal subset B of F which contains A and which is p-independent over F0 . Then B is a p-basis of F over F0 . Lemma 2.7.2: Suppose F is a finitely generated extension of transcendence degree n of a perfect field K of positive characteristic p. Then the imperfect exponent of F is n. Proof: Choose a separating transcendence basis t1 , . . . , tn for F/K. Then K(t)p = K(tp ) and t1 , . . . , tn is a p-basis for K(t)/K(tp ); that is, [K(t) : K(tp )] = pn . Since K(t) is a purely inseparable extension of K(tp ) and F p is a separable extension of K(tp ), these extensions of K(tp ) are linearly disjoint. Also, F is both a separable extension and a purely inseparable extension of K(t)F p . Hence, F = K(t)F p . Consequently, [F : F p ] = [K(t) : K(tp )] = pn , as claimed. Lemma 2.7.3: Let B a subset of F which is p-independent over F p and F 0 a separable extension of F . Then B is p-independent over (F 0 )p . If, in addition, F 0 is separable algebraic over F , then the imperfect degree of F 0 is equal to that of F . Proof: Assume without loss that B consists of n elements. Then [(F 0 )p (B) : (F 0 )p ] = [F p (B) : F p ] = pn . Hence, B is p-independent over (F 0 )p .
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Chapter 2. Valuations and Linear Disjointness
Suppose now F 0 /F is separably algebraic. Then F 0 is both separably and purely inseparable over F (F 0 )p , so, F 0 = F (F 0 )p . Hence, [F 0 : (F 0 )p ] = [F : F p ]. Therefore, the imperfect degree of F 0 is equal to that of F . Lemma 2.7.4: Let K be a field of positive characteristic p, let a, b1 , . . . , bm be p-independent elements of K, and let x1 , . . . , xm be algebraically independent over K. Suppose y1 , . . . , ym satisfy (2)
axpi + bi yip = 1,
i = 1, . . . , m.
Then K is algebraically closed in K(x, y) = Km . Proof: We use induction on m. Part A: m = 1. Let x = x1 , y = y1 , and b = b1 and assume that u is a nonzero element of K1 which is algebraic over K. Then u is also algebraic over K(a1/p , b1/p ). But K(x, y, a1/p , b1/p ) = K(x, a1/p , b1/p ) is a purely transcendental extension of K(a1/p , b1/p ). Hence, u ∈ K(a1/p , b1/p ) and therefore up ∈ K. Write (3)
u=
h0 (x) h1 (x) hk (x) k + y + ··· + y h(x) h(x) h(x)
with k ≤ p − 1, h(x), h0 (x), . . . , hk (x) ∈ K[x] and h(x), hk (x) 6= 0. With no loss we may assume that x does not divide the greatest common divisor of h(x), h0 (x), . . . , hk (x). Raise (3) to the pth power, multiply it by h(x)p and substitute y p = (1 − axp )b−1 to obtain: (4) (h(x)u)p = h0 (x)p + h1 (x)p (1 − axp )b−1 + · · · + hk (x)p (1 − axp )k b−k . If h(0) = 0, then the substitution x = 0 in (4) gives 0 = h0 (0)p + h1 (0)p b−1 + · · · + hk (0)p b−k , Therefore, h0 (0) = h1 (0) = · · · = hk (0) = 0, contrary to assumption. Thus, we may assume h(0) 6= 0. Then the substitution x = 0 in (4) shows that u ∈ K(b1/p ). Similarly, u ∈ K(a1/p ). Since a and b are p-independent in K, u ∈ K(a1/p ) ∩ K(b1/p ) = K. Thus, K is algebraically closed in K(x, y). Part B: Induction. Assume the Lemma is true for m − 1. Then K is algebraically closed in Km−1 = K(x1 , . . . , xm−1 , y1 , . . . , ym−1 ). If we prove that a and bm are p-independent in Km−1 , then with Km−1 replacing K in Part A, Km−1 is algebraically closed in Km , so K is algebraically closed in Km . Since x1 , . . . , xm are algebraically independent over K, the field 1/p 1/p 1/p K(a , b1 , . . . , bm ) is linearly disjoint from Em−1 = K(x1 , . . . , xm−1 ) over K. Thus, (5)
1/p
m+1 . [Em−1 (a1/p , b1 , . . . , b1/p m ) : Em−1 ] = p
2.7 The Imperfect Degree of a Field
47
Also, from (2) Km−1 = Em−1 (y1 , . . . , ym−1 )
and 1/p
1/p Km−1 (a1/p , b1/p , b1 , . . . , b1/p m ) = Em−1 (a m ).
Thus, (6)
[Km−1 : Em−1 ] ≤ pm−1
and
2 [Km−1 (a1/p , b1/p m ) : Km−1 ] ≤ p .
Combine (5) and (6) to conclude that (6) consists of equalities. In particular, a and bm are p-independent in Km−1 . Lemma 2.7.5: The following conditions on a field K of positive characteristic p are equivalent: (a) The imperfect exponent of K is at most 1. (b) Every finite extension of K has a primitive element. (c) If K is algebraically closed in a field extension F , then F is regular over K. Proof: If K is perfect, then (a), (b), and (c) are true. Therefore, we may assume char(K) = p > 0 and K is imperfect. Proof of “(a) =⇒ (b)”: By assumption, [K 1/p : K] = [K : K p ] = p. Hence, K1 = K 1/p is the unique purely inseparable extension of K of degree p. n Moreover, K1 = K(a1/p ) for some a ∈ K, so Kn = K(a1/p ) is a purely inseparable extension of K of degree pn . Assume that for each m ≤ n, Km is the unique purely inseparable extension of K of degree pm . Let L be a purely inseparable extension of K of degree pn+1 . If we prove that L = Kn+1 , then we may conclude by induction that each finite purely inseparable extension of K has a primitive element. To this end choose x ∈ L r Kn . Let m be the smallest positive integer m with xp ∈ K. Then K(x) is a purely inseparable extension of K of degree pm . If m ≤ n, then by the induction hypothesis K(x) = Km ⊆ Kn , so x ∈ Kn . This contradiction proves that m = n + 1 and L = K(x). The same argument implies that xp ∈ Kn . Hence, with q = pn , we have Pq−1 n p x = i=0 ci ai/p for some c0 , . . . , cq−1 ∈ K. Therefore, x=
q−1 X
1/p
ci ai/p
n+1
∈ K1 (a1/p
n+1
) = Kn+1 .
i=0
It follows that L ⊆ Kn+1 . As both fields have degree pn+1 over K, they coincide, as desired. Now let E be a finite extension of K. Denote the maximal separable extension of K in E by E0 . By the primitive element theorem, E0 = K(x). Since E0 is both separable and purely inseparable over KE0p we have E0 = KE0p . Therefore [E0 : E0p ] = [K : K p ] = p. Apply the first part of the proof
48
Chapter 2. Valuations and Linear Disjointness
to E0 and conclude that E = E0 (y), for some element y. Thus, E = K(x, y) with x separable over K. By [Waerden3, §6.10], E/K has a primitive element Proof of “(b) =⇒ (c)”: Let K(x) be a finite extension of K and let f = irr(x, K). If K is algebraically closed in F , then f remains irreducible over F . Otherwise, its factors would have coefficients algebraic over K and in F , and therefore in K. Thus, F is linearly disjoint from K(x) over K. Hence, (b) implies that F is regular over K. Proof of “(c) =⇒ (a)”: Assume a and b are p-independent elements of K. Then [K(a1/p , b1/p ) : K] = p2 . Let x and y be transcendental elements over K with axp + by p = 1. Put F = K(x, y). By Lemma 2.7.4, K is algebraically closed in F . Hence, by (c), F is regular over K. Therefore, [F (a1/p , b1/p ) : F ] = [K(a1/p , b1/p ) : K] = p2 . On the other hand, , F (a1/p ) = F (b1/p ), so [F (a1/p , b1/p ) : F ] ≤ p. This contradiction proves that the imperfect exponent of K is at most 1. Remark 2.7.6: Relative algebraic closedness does not imply regularity. Let K be a field of positive characteristic p. Suppose K has p-independent elements a, b (e.g. K = Fp (t, u) where t, u are algebraically independent over Fp ). Let x, y be transcendental elements over K with axp + by p = 1. Put F = K(x, y). The proof of “(c) =⇒ (a)” then shows that K is algebraically closed in F but F is not linearly disjoint from K 1/p over K. Thus, F is not a separable extension of K. A fortiori, F/K is not regular.
2.8 Derivatives We develop a criterion for a finitely generated field extension of positive characteristic p to be separable in terms of derivatives.. Definition 2.8.1: A map D: F → F is called a derivation of the field F if D(x + y) = D(x) + D(y) and D(xy) = D(x)y + xD(y) for all x, y ∈ F . If D vanishes on a subfield K of F , then D is a derivation of F over K (or a K-derivation). Let F (x) be a field extension of F and f ∈ F [X]. Suppose D extends to F (x). Then D satisfies the classical chain rule: (1)
D(f (x)) = f D (x) + f 0 (x)D(x),
where f D is the polynomial obtained by applying D to the coefficients of f and f 0 is the usual derivative of f . There are three cases: Case 1: x is separably algebraic over F . Then, with f = irr(x, F ), f 0 (x) 6= 0. By (1), 0 = f D (x) + f 0 (x)D(x). Thus, D extends uniquely to F (x). Case 2: x is transcendental. may be chosen arbitrarily.
Then D extends to F (x) by rule (1) and D(x)
2.8 Derivatives
49 m
Case 3: x satisfies xp = a ∈ F , for some m. Then D extends to F (x) if and only if D(a) = 0. In this case D(x) may be chosen arbitrarily. Lemma 2.8.2: A necessary and sufficient condition for a finitely generated extension F/K to be separably algebraic is that 0 is the only K-derivation of F . Proof: Necessity follows from Case 1. Now suppose F/K is not separably algebraic. Then we may write F = K(x1 , . . . , xn ) such that xi is transcendental over K(x1 , . . . , xi−1 ) for i = 1, . . . , k, xi is separably algebraic over K(x1 , . . . , xi−1 ) for i = k+1, . . . , l, and xi is purely inseparable over K(x1 , . . . , xi−1 ) for i = l + 1, . . . , n. Moreover, either n > l or n = l and k > 0. If n > l, then Case 1 allows us to extend the zero derivation of K(x1 , . . . , xn−1 ) to a nonzero derivation of F . If n = l and k > 0, then by Case 2, the zero derivation of K(x1 , . . . , xk−1 ) extends to a nonzero derivation D of K(x1 , . . . , xk ). Applying Case 3 several times, we may then extend D to a derivation of F . Lemma 2.8.3: Let F/K be a finitely generated extension of positive characteristic p and transcendence degree n. Then F/K is separable if and only if [F : KF p ] = pn . In this case t1 , . . . , tn form a p-basis for F over KF p if and only they form a separating transcendence basis for F/K. Proof: Suppose first [F : KF p ] = pn . Let t1 , . . . , tn be a p-basis for F/KF p . Every derivation D of F vanishes on F p . If D vanishes on K(t), it vanishes on F = K(t) · F p . By Lemma 2.8.2, F/K(t) is separably algebraic and t1 , . . . , tn is a separating transcendence basis for F/K. Conversely, suppose F/K is separable. Let t1 , . . . , tn be a separating transcendence basis for F/K. The extension F/K(t) · KF p is both separable and purely inseparable. Hence, F = K(t) · KF p . Since F p /K(t)p is separably algebraic and since K(tp )F p = KF p , we conclude that KF p /K(tp ) is separably algebraic. K(t) F
K(tp )
KF p
K(t)p
Fp
Therefore, KF p is linearly disjoint from K(t) over K(tp ), and [F : KF p ] = [K(t) : K(tp )] = pn . Moreover, t is a p-basis for F/KF p . Corollary 2.8.4: Let F/K be a finitely generated separable extension of positive characteristic p and let t ∈ F . (a) If there exists a derivation D of F/K such that D(t) 6= 0, then F is a separable extension of K(t).
50
Chapter 2. Valuations and Linear Disjointness
(b) If t is transcendental over K and F/K(t) is separable, then there exists a derivation D of F/K such that D(t) 6= 0. Proof of (a): By assumption, t ∈ / KF p . Let n = trans.deg(F/K). By p n Lemma 2.8.3, [F : KF ] = p . Hence, t can be extended to a p-basis t, t2 , . . . , tn for F/KF p . Again, by Lemma 2.8.3, t, t2 , . . . , tn is a separating transcendence basis for F/K. Therefore, F is a separable extension of K(t). Proof of (b): Let t2 , . . . , tn be a separating transcendence basis for F/K(t). By Case 2, there exists a derivation D0 of K(t, t2 , . . . , tn )/K such that D0 (t) = 1, D0 (t2 ) = 0, . . . , D0 (tn ) = 0. By Case 1, D0 extends to a derivation D of F/K.
Exercises 1. Let O be a valuation ring of a field F and consider the subset m = {x ∈ / O}. Show that if x ∈ m and a ∈ O, then ax ∈ m. Prove that m is O | x−1 ∈ closed under addition. Hint: Use the identity x + y = (1 + xy −1 )y for y 6= 0. Show that m is the unique maximal ideal of O. 2.
Use Exercise 1 to prove that every valuation ring is integrally closed.
3. Let v be a valuation of Q. Observe that v(n) ≥ v(1) = 0, for each n ∈ N. Hence, there exists a smallest p ∈ N such that v(p) > 0. Prove that p is a prime element of Ov and v is equivalent to vp . Hint: If a positive integer m is relatively prime to p, then there exist x, y ∈ Z such that xp + ym = 1. 4. Let v be a valuation of the rational function field F = K(t) which is trivial on K. Suppose there exists p ∈ K[t] with v(p) > 0. Now suppose p has smallest degree with this property. Show that v is equivalent to vp . Otherwise, there exists f ∈ K[t] such that v(f (t)) < 0. Conclude that v(t) < 0, and that v is equivalent to v∞ . 5. Let F/E be a field extension, w a valuation of F , and x1 , . . . , xe elements of F such that w(x1 ), . . . , w(xe ) represent distinct classes of w(F × ) modulo w(E × ). Show that x1 , . . . , xe are linearly independent over E. Thus, (w(F × ) : w(E × )) ≤ [F : E]. Hint: Use (4b) of Section 2.1. 6. Let ∆ be an ordered group containing Z as a subgroup of index e. Show there exists no positive element δ ∈ ∆ such that eδ < 1. Conclude that ∆ contains a smallest positive element and hence that ∆ ∼ = Z. Combine this with Exercise 5 to prove that if the restriction of w to E is discrete, then w is discrete. 7. In the notation of Exercise 5, let v be the restriction of w to E. Let y1 , . . . , yf be elements of F with w(y1 ), . . . , w(yf ) ≥ 0 with residue classes ¯v . Show that y1 , . . . , yf are linearly y¯1 , . . . , y¯f linearly independent over E ¯v ] ≤ [F : E]. Hint: If a1 , . . . , af ∈ ¯ independent over E. Conclude that [Fw : E
Notes
F are not all zero, then there exists j, 1 ≤ j ≤ f such that v 0.
51 a1 aj
,...,v
af aj
≥
8. Let v be a discrete valuation of a field K and let w be an extension of v to a finite Galois extension L of K. Assume that w0 is also an extension of v to L such that w0 6= σ(w) for all σ ∈ Gal(L/K). Combine Exercise 7 with Proposition 2.1.1 to produce x ∈ L such that w0 (x) > 0 and w(σx−1) > 0 for all σ ∈ Gal(L/K). With y = NL/K (x), conclude that the former condition gives v(y) > 0, while the latter implies v(y − 1) > 0. Use this contradiction to prove that Gal(L/K) acts transitively on the extensions of v to L. 9. Let L, K1 , . . . , Kn be extensions of a field K. Let Li = Ki L, i = 1, . . . , n. Suppose Ki is linearly disjoint from L over K for i = 1, . . . , n and L1 , . . . , Ln are linearly disjoint over L. Prove that K1 , . . . , Kn are linearly disjoint over K. 10. Let v be a discrete valuation of a field K and let L and M be two finite extensions of K such that v is unramified in L and totally ramified in M . Prove that L and M are linearly disjoint over K. Hint: Consider the Galois ˆ of L/K. hull L 11. Let E be a regular extension of a perfect field K and let F be a purely inseparable extension of E. Prove that F/K is a regular extension. 12. Let K be a field algebraically closed in an extension F . Prove that K(x) ˜ Hint: Check the irreducibility of is linearly disjoint from F for every x ∈ K. irr(x, K) over F . ˜ = 13. Prove that a field extension F/K is primary if and only if F Kins ∩ K Kins . Use this criterion to give another proof to Lemma 2.6.14(a). 14. Let F/K be a finitely generated field extension of characteristic p > 0 n and of transcendence degree 1. Prove that for each positive integer n, KF p is the unique subfield E of F which contains K such that F/E is a purely inseparable extension of degree pn . 15. (Geyer) The following example shows that Lemma 2.4.8 is false for arbitrary real valuations. Consider the field Q2 of 2-adic numbers. Show that √ the field K = Q2 ( n 2 | n ∈ N) is a totally ramified extension of Q2 with value group Q. Hence, of K is unramified. Prove that the residue √ each extension √ 3) and K( −1) is F2 . However, their compositum contains field of both K( √ K( −3) and therefore has F4 as its residue field.
Notes The terminology “algebraic independence” for field extensions replaces “freeness” which we used in [Fried-Jarden3]. Corollary 4 of [Lang4, p. 61] proves Lemma 2.6.15(a) only under the condition (our notation) that E is a separable extension of K.
Chapter 3. Algebraic Function Fields of One Variable Sections 3.1–3.4 survey the theory of functions of one variable; the RiemannRoch Theorem; properties of holomorphy rings of function fields; and extensions of the field of constants. Sections 3.5–3.6 include a proof of the Riemann-Hurwitz formula. The rest of the chapter applies these concepts and results to hyperelliptic curves.
3.1 Function Fields of One Variable Call a field extension F/K an algebraic function field of one variable (briefly a function field) if these conditions hold: (1a) The transcendence degree of F/K is 1. (1b) F/K is finitely generated and regular. In this case there exists t ∈ F , transcendental over K, with F/K(t) a finite separable extension. All valuations of K(t) trivial on K are discrete (Example 2.2.1), so their extensions to F are also discrete (Proposition 2.3.2). Also, since the residue fields of the valuations of K(t) are finite extensions of K, so are the residue fields of the valuations of F . We define a prime divisor of F/K as an equivalence class of K-places of F . For p a prime divisor of F/K, choose a place ϕp in p. Then ϕp fixes ˜ ∪ {∞}. Denote its residue field by F¯p . the elements of K and maps F into K ¯ As mentioned above, Fp is a finite extension of K of degree deg(p) = [F¯p : K] which we call the degree of p. Also, choose a valuation vp corresponding to p and normalize it so that vp (F × ) = Z. Each element π of F with vp (π) = 1 is a local parameter of F at p. Denote the free Abelian group that the prime divisors of F/K generate by Div(F/K). P Each element a of Div(F/K) is a divisor of F/K. It has the form a = αp p, where p runs over the prime divisors of F/K, the αp are integers and all but finitely many of them are zero. Define a homomorphism vp : Div(F/K) → Z by vp (a) = αp . The symbol vp appears for two distinct functions. We show these uses are compatibleP as follows. Introduce the divisor of a nonzero element x of F as div(x) = vp (x)p. Since vp (x) = 0 for all but finitely many p the right hand side is well defined. If x is a constant (i.e. x ∈ K), then div(x) = 0. If x ∈ F r K, then the place of K(x) taking x to 0 (resp. to ∞) has finitely many extensions to the field F (Proposition 2.3.1). Equivalence classes of these extensions are the zeros (resp. poles) of x. Thus, div(x) is not zero if x is not a constant. Define the divisor of zeros and the divisor of poles
3.2 The Riemann-Roch Theorem
53
of x as follows: div0 (x) =
X
vp (x)p,
vp (x)>0
div∞ (x) = −
X
vp (x)p.
vp (x)<0
Then div(x) = div0 (x) − div∞ (x). In particular, vp (x) (using vp as a valuation) is the same as vp applied to the divisor div(x). P Define the degree deg(a) of a divisor a to be vp (a) deg(p). Then deg : Div(F/K) → Z is a homomorphism. Since deg(div0 (x)) = deg(div∞ (x)) = [F : K(x)] for x ∈ F r K ([Deuring3, p. 27] or Proposition 2.3.2), deg(div(x)) = 0. To each divisor a attach a vector space L(a) over K: L(a) = {x ∈ F × | div(x) + a ≥ 0}. The phrase ‘b ≥ 0’ for a divisor b means that vp (b) ≥ 0 for every prime divisor p. The K-vector space L(a) is finite dimensional [Deuring3, p. 23] (Note: Deuring uses the notation L(a) for L(−a)). Denote the nonnegative integer dimK L(a) by dim(a). Extend the notation 0 ≤ b to a partial ordering on Div(F/K) by writing a ≤ b if vp (a) ≤ vp (b) for every prime divisor p of F/K. This implies L(a) ⊆ L(b). In this case, L(a) ⊂ L(b) is equivalent to dim(a) < dim(b). For divisors a1 , . . . , an of F/K, write min(a1 , . . . , an ) =
X
min vp (a1 ), . . . , vp (an ) p.
p
This is the maximal divisor of F/K less than or equal to ai for i = 1, . . . , n . Similarly, define max(a1 , . . . , an ). Call a divisor of the form div(x), where x ∈ F × , a principal divisor. Since div(xy) = div(x) + div(y), the set of principal divisors of F/K is a subgroup of Div(F/K). The quotient group is called the group of divisor classes of F/K. Denote it by C. Every element of C is a class of divisors. Call two divisors linearly equivalent if they differ by a principal divisor. Two linearly equivalent divisors have the same degree and the same dimension. This defines the degree and the dimension of a class of divisors. × Lemma Pm 3.1.1: Let F/K be a function field andSx∞1 , . . . , xm ∈ F . Put a = i=1 div∞ (xi ) and R = K[x1 , . . . , xm ]. Then k=1 L(ka) is the integral closure of R in F .
Proof: Denote the integral closure of R in F by S and the set of poles of x1 , . . . , xm by P . First consider a nonzero z ∈ L(ka). Then div(z) + ka ≥ 0. Hence, vp (z) ≥ 0 for each prime divisor p of F/K which is not in P ; that is for each p with ϕp finite on R. By Proposition 2.4.1, z ∈ S.
54
Chapter 3. Algebraic Function Fields of One Variable
Conversely, let z ∈ S. By Proposition 2.4.1, vp (z) ≥ 0 for each p ∈ / P. m | v (x ) < 0} is not empty. Choose For each p ∈ P the set Ip = {1 ≤ i ≤ p i P a positive integer k with vp (z) − k i∈Ip vp (xi ) ≥ 0 for all p ∈ P . Then z ∈ L(ka).
3.2 The Riemann-Roch Theorem The degree of a divisor a of a function field F/K can be often read out of its definition. It is more difficult to compute the dimension of a directly from the definition. The Riemann-Roch theorem allows us in many cases to compute dim(a) from deg(a) and a constant of F/K called the “genus”. In this section we present the Riemann-Roch theorem in three forms and explain how to apply it in the computation of dim(a). Let F/K be a function field. A repartition of F/K is a function α from prime divisors of F/K into F (denote the image of p by αp ) with vp (αp ) ≥ 0 for all but finitely many p. Denote the set of all repartitions of F/K by A. The following definitions turn A into an F -algebra: For α, β ∈ A and x ∈ F , (α + β)p = αp + βp ,
(αβ)p = αp βp
and (xα)p = x · αp .
Extend vp to A by vp (α) = vp (αp ). If a is a divisor of F/K, then Λ(a) = {α ∈ A | vp (α) + vp (a) ≥ 0 for every p} is a vector space over K. Identify F as a subset of A by the diagonal mapping. Then Λ(a) contains L(a) and A/(Λ(a) + F ) is a finite dimensional K-vector space. Moreover, there exists a nonnegative integer g, called the genus of F/K, which is independent of a such that (1)
dim(a) − deg(a) + g − 1 = dimK
A . Λ(a) + F
This is the first form of the Riemann-Roch theorem [Deuring3, p. 34]. The second form of the theorem interprets the right hand side of (1) as the dimension of a space of differentials. A differential (also called Weil differential) of the function field F/K is a K-linear map of A into K that vanishes on some subspace of the form Λ(a) + F , where a is a divisor of F/K. The set of all differentials that vanish on Λ(a) + F , denoted Ω(a), is the dual space of A/(Λ(a) + F ). We denote its dimension by δ(a) and rewrite (1) as (2)
dim(a) = deg(a) − g + 1 + δ(a).
The third form of the theorem identifies δ(a) with the dimension of a divisor of F/K related to a. To each nonzero differential ω of F/K there corresponds a unique divisor of F/K, denoted div(ω), with div(ω) ≥ a if and only if ω ∈ Ω(a) [Deuring3, p. 39].
3.2 The Riemann-Roch Theorem
55
Theorem 3.2.1: Let ω be a nonzero differential of an algebraic function field F/K of one variable of genus g. Put w = div(ω). Then, for every divisor a of F/K we have (3)
dim(a) = deg(a) − g + 1 + dim(w − a).
Proof: For each x ∈ F × define a K-linear map xω: A → K by (xω)(α) = ω(xα). If ω vanishes on Λ(a), then xω vanishes on Λ(div(x) + a). Hence, xω is a differential and div(xω) = div(x) + div(ω). The map x 7→ xω gives an isomorphism of L(div(ω)−a) onto Ω(a). Hence, dim(div(ω)−a) = dim Ω(a) = δ(a). Thus, (3) is a reformulation of (2). If ω 0 is another nonzero differential, then there exists x ∈ F × with ω = xω. Thus, div(ω 0 ) = div(x) + div(ω) [Deuring3, p. 37] and the divisor of a differential is determined up to a principal divisor. The class of these divisors is the canonical class of F/K. Each of its elements is a canonical divisor. A few additional observations help us apply the third form of the Riemann-Roch theorem in (3): 0
Lemma 3.2.2: The following holds for each function field F/K of genus g: (a) dim(0) = 1, and if x ∈ F × , then deg(div(x)) = 0 and dim(div(x)) = 1. (b) If w is a canonical divisor, then dim(w) = g and deg(w) = 2g − 2 Hence, dimK (A/Λ(0) + F ) = dim(Ω(0)) = dim(L(w)) = g. (c) deg(a) < 0 implies dim(a) = 0. (d) deg(a) > 2g − 2 implies δ(a) = 0 and dim(a) = deg(a) − g + 1 . Proof of (a): L(0) = K because each x ∈ F r K has a pole. Similarly, L(div(x)) = Kx−1 . Proof of (b): Take a = 0 and then a = w in (3). Proof of (c): Use that deg(div(x)) = 0. Proof of (d): By (b) deg(w − a) < 0. Hence, by (c), δ(a) = dim(w − a) = 0. Thus, (3) simplifies to dim(a) = deg(a) + 1 − g. Here is our first application of Lemma 3.2.2(d): Lemma 3.2.3: Let F/K be a function field of genus g, p a prime divisor of F/K, and n a positive integer satisfying (n − 1) deg(p) > 2g − 2. Then there exists x ∈ F × with div∞ (x) = np. Proof: By Lemma 3.2.2(d), dim((n − 1)p) = (n − 1) deg(p) + 1 − g and dim(np) = n deg(p)+1−g. Hence, dim((n−1)p) < dim(np), so L((n−1)p) ⊂ L(np). Every element x ∈ L(np) r L((n − 1)p) will satisfy div∞ (x) = np.
56
Chapter 3. Algebraic Function Fields of One Variable
Example 3.2.4: Rational function field. Let F = K(t), where t is a transcendental element over K. Denote the prime divisor corresponding to the valuation v∞ (Section 2.2) by p∞ . We determine the linear space L(np∞ ), where n is a positive integer. If u ∈ L(np∞ ), then vp (u) ≥ 0 for every prime divisor p 6= p∞ . This means u ∈ K[t]. Also vp∞ (u) ≥ −n, so the degree of u as a polynomial in t is bounded by n. Thus, L(np∞ ) = {u ∈ K[t] | deg(u) ≤ n} and dim(np∞ ) = n + 1. For n > 2g − 2, Lemma 3.2.2(d) gives n + 1 = n − g + 1. Hence, g = 0. Conversely, let F/K be a function field of one variable of genus 0. Assume F/K has a prime divisor p of degree 1 (e.g. if K is algebraically closed). Lemma 3.2.3 gives t ∈ F × with div∞ (t) = p. By Section 3.1, [F : K(t)] = deg(div∞ (t)) = 1. Consequently, F = K(t). Pn Consider now a polynomial f (t) = i=0 ai ti with ai ∈ K and an 6= 0. For each i < n we have v∞ (an tn ) < v∞ (ai ti ), so v∞ (f (t)) = −n. For p 6= p∞ we have vp (t) ≥ 0, hence vp (f (t)) ≥ 0. Thus, div∞ (f (t)) = deg(f )p∞ . (t) Suppose g(t) ∈ K[t] is relatively prime to f (t). Put u = fg(t) . Then t is a root of the polynomial h(u, T ) = u · g(T ) − f (T ). Since h(u, T ) is linear in u and gcd(f, g) = 1, h(u, T ) is irreducible over K(u). In addition, degT (h) = max(deg(f ), deg(g)). Therefore,
(4)
[K(t) : K(u)] = max deg(f ), deg(g) .
In particular, suppose K(u) = K(t). Then, f (t) and g(t) are linear and relatively prime. This means, u = at+b ct+d with a, b, c, d ∈ K and ad − bc 6= 0.
3.3 Holomorphy Rings Let F/K be a function field of genus g. Denote the set of prime divisors of F/K by R. For each p ∈ R let Op = {x ∈ F | vp (x) ≥ 0} be the corresponding valuation ring. To every subset S of R we attach the holomorphy ring T OS = p∈S Op . By definition, K ⊆ OS . If S is empty, then, by definition, OS = F . If S = R, then the elements of OS have no poles. They are therefore constants. Thus, OR = K. The case where S is a nonempty proper subset of R requires a strengthening of the weak approximation theorem (Proposition 2.1.1). Proposition 3.3.1 (Strong Approximation Theorem): Let S be a finite subset of R. Consider q ∈ R r S and let S 0 = S ∪ {q}. Suppose for each p ∈ S we have an xp ∈ F and a positive integer mp . Then there exists x ∈ F with (1)
vp (x − xp ) = mp for each p ∈ S and vp (x) ≥ 0 for each p ∈ R r S 0 .
3.3 Holomorphy Rings
57
Moreover, if m is an integer with (2)
m · deg(q) > 2g − 2 +
X
(mp + 1) deg(p),
p∈S
then x can be chosen such that, in addition to (1), it satisfies vq (x) ≥ −m. Proof: Let P m be a positive integer satisfying (2). Consider the divisor a = mq − p∈S mp p. Then deg(a) > 2g − 2. By Lemma 3.2.2(d), δ(a) = 0, so A = F +Λ(a). Define ξ ∈ A by ξp = xp for p ∈ S and ξp = 0 for p ∈ R r S. Then there exists y ∈ F such that y − ξ ∈ Λ(a): (3) vp (y − xp ) ≥ mp for p ∈ S, vq (y) ≥ −m, and vp (y) ≥ 0 for p ∈ R r S 0 . P Now consider the divisor b = mq − p∈S (mp + 1)p. For each p ∈ S we have deg(b + p) > deg(b) > 2g − 2. By Lemma 3.2.2(d), dim(b + p) = deg(b + p) − g + 1 > deg(b) − g + 1 = dim(b). Hence, L(b) ⊂ L(b + p). Choose zp ∈ L(b + p) r L(b). Then vp (zp ) = mp . Also vp0 (zp ) ≥ mp0 + 1 if p0 ∈ S r{p}, vq (zp ) ≥ −m, and vp0 (zp ) ≥ 0 for p0 ∈ R r S 0 . r P Let P = {p ∈ S | vp (y − xp ) > mp } and let Q = S P . Then z = p∈P zp has the following property: (4) vp (z) = mp if p ∈ P , vp (z) ≥ mp + 1 if p ∈ Q, vq (z) ≥ −m, and vp (z) ≥ 0 for p ∈ R r S 0 . Combine (3) and (4) to see that x = z + y satisfies vp (x − xp ) = mp , for p ∈ S, vq (x) ≥ −m, and vp (x) ≥ 0 for p ∈ R r S 0 . If p belongs to a subset S of R, then OS ⊆ Op . Also, P = {x ∈ OS | vp (x) > 0} is a prime ideal of OS , the center of p at OS . Denote the local ring of OS at P by OS,P . Proposition 3.3.2 (Holomorphy Ring Theorem): Let S be a nonempty proper subset of R. Then S has these properties: (a) Quot(OS ) = F . (b) If p ∈ S and P is the center of p at OS , then Op = OS,P . (c) If q ∈ R r S, then OS 6⊆ Oq . (d) Every nonzero prime ideal of OS is the center of a prime p ∈ S. (e) Distinct primes in S have distinct centers at OS , and the center of each p ∈ S is a maximal ideal of OS . (f) OS is a Dedekind domain. Proof of (a): Consider z ∈ F r K. Since S ⊂ R, there is a q ∈ R r S. There are only finitely many p ∈ S with vp (z) < 0. Hence, Proposition 3.3.1 gives y ∈ F such that the following holds for each p ∈ S: If vp (z) < 0, then vp (y − z −1 ) = vp (z −1 ) + 1, so vp (y) = −vp (z); while if vp (z) ≥ 0, then
58
Chapter 3. Algebraic Function Fields of One Variable
vp (y) ≥ 0. Let x = yz. Then both x and y belong to OS . If z ∈ OS , then / OS , then there is a p ∈ S with vp (z) < 0. Hence, z ∈ Quot(OS ). If z ∈ vp (y) = vp (z −1 ) 6= ∞, so y 6= 0. Therefore, z = xy −1 ∈ Quot(OS ). Proof of (b): Let z ∈ Op r K. As in the proof of (a), there exists y ∈ F such that vp (y) = 0, vp0 (y) = −vp0 (z) if p0 ∈ S r{p} and vp0 (z) < 0, while vp0 (y) ≥ 0 if p0 ∈ S r{p} and vp0 (z) ≥ 0. Therefore, x = yz is in OS and z ∈ OS,P . Since the inclusion OS,P ⊆ Op is clear, Op = OS,P . Proof of (c): If S ∪ {q} = R and OS ⊆ Oq , then OS = OR = K, a contradiction to (a). Therefore, assume that S ∪ {q} is a proper subset of R. By Proposition 3.3.1, there exists x ∈ F such that vq (x) = −1 and vp (x) ≥ 0 for each p ∈ S. This element belongs to OS but not to Oq . Proof of (d): Let P be a nonzero prime ideal of OS . Proposition 2.3.1 extends the quotient map OS → OS /P to a place ϕ of F trivial on K. Let p be the prime divisor of F/K which is defined by ϕ. Then OS ⊆ Op and P = {x ∈ OS | vp (x) > 0}. By (c), p ∈ S. Proof of (e): Let p and p0 be two distinct prime divisors in S. By the strong approximation theorem (Proposition 3.3.1), there exists x ∈ OS with vp (x) > 0 and vp0 (x) = 0. This means the center P of p is not contained in the center of p0 . The maximality of P now follows from (d). Proof of (f): Let q ∈ R r S, let S 0 = R r{q} and choose p0 ∈ S 0 . Since OS is an overring of OS 0 it suffices (Proposition 2.4.7) to prove that OS 0 is a Dedekind domain. The strong approximation theorem gives x ∈ OS 0 with vp0 (x) > 0. Since x must have a pole, it must be q. Thus,T K[x] ⊆ Op if and only if p ∈ S 0 . Therefore, by Proposition 2.4.1, OS 0 = p∈S 0 Op is the integral closure of K[x] in F . Since K[x] is a Dedekind domain, OS 0 is also a Dedekind domain (Proposition 2.4.6). Corollary 3.3.3: The following conditions are equivalent for a nonempty subset S of R: (a) S = R; (b) OS = K; and (c) OS is a field. Proof: Every nonconstant element of F has a pole. Therefore, (a) implies (b). The implication “(b) => (c)” is trivial. To prove that (c) implies (a), note that OS 6= F , since S is nonempty. If S 6= R, then, by Proposition 3.3.2(a), the quotient field of OS is F . Hence, OS is not a field. The following converse to Proposition 3.3.2 is useful: Proposition 3.3.4: Let F/K be a function field and R a proper subring of F containing K. Suppose R is integrally closed and Quot(R) = F . Then there is a nonempty subset S of R with R = OS . Thus, R is a Dedekind domain. Proof: Let S be the set of all prime divisors of F/K which are finite on R. By Proposition 2.4.1, OS = R. As R ⊂ F , S is nonempty. By Corollary 3.3.3, S 6= R. Hence, by Proposition 3.3.2(f), R is a Dedekind domain.
3.4 Extensions of Function Fields
59
3.4 Extensions of Function Fields Let E/K and F/L be algebraic function fields of one variable. We say that F/L is an extension of E/K if E ⊆ F , K ⊆ L, and L ∩ E = K. We call F/L a constant field extension of E/K if E is linearly disjoint from L over K and F = EL. Thus, in this case [F : E] = [L : K]. Recall that E/K is a regular extension ((1b) of Section 3.1). Hence, the linear disjointness of E and L over K is automatic if L is an algebraic extension of K. Likewise, E is linearly disjoint from L over K if L = K(x1 , . . . , xn ) and x1 , . . . , xn are algebraically independent over E (Lemma 2.6.7). Let F/L be an extension of E/K, p be a prime divisor of E/K, and P be a prime divisor of F/L that lies over p, that is vP lies over vp . Denote the ramification index of vP over vp by eP/p . Since both vP and vp are discrete and normalized, vP (x) = eP/p vp (x) for each x ∈ E. Refer to P as unramified over E if vP is unramified over E. If F is separable algebraic over E, then only finitely many prime divisors of F/L are ramified over E [Deuring3, p. 111]. Over every prime divisor of E/K there lie only finitely many prime divisors of F/L [Deuring3, p. 96]. Use this result to embed the group of divisors Div(E/K) of E/K into Div(F/L) as follows: For p a prime divisor of E/K and P1 , . . . , Pd the prime divisors of F/L lying over p, map p to the Pd divisor i=1 ePi /p Pi of F/L. Extend this map to Div(E/K) by linearity. The principal divisor of x (in Div(E/K)) maps to the principal divisor of x in Div(F/L), so there is no ambiguity using div(x) for that divisor. In particular, for every divisor a of E/K we have LE (a) = E ∩ LF (a). Suppose [F : E] < ∞. If F/E is separable, apply (2) of Section 2.3 to conclude: (1)
d X
¯p ] = [F : E]. ePi /p [F¯Pi : E
i=1
Even if F/E is not separable, the following argument shows (1) still holds: By Lemma 3.2.3, there are an integer m and x ∈ E with div∞(x) = mp. Apply the rules degF div∞ (x) = [F : L(x)] and degE div∞ (x) = [E : K(x)] to conclude that (1) is true in general [Deuring3, p. 97]. Next consider a function field F/K and let E be a proper extension of ˜ over K, so is E. In K contained in F . Since F is linearly disjoint from K particular, there is a transcendental element x in E. Then [F : K(x)] = degF div∞ (x) < ∞. Thus, [E : K(x)] < ∞, E/K is a function field of one variable, and [F : E] < ∞. Lemma 3.4.1: Let E/K and F/K be algebraic function fields of one variable with E ⊆ F . Then, degF (a) = [F : E] degE (a) for each a ∈ Div(E/K). Proof: Assume by linearity that a = p is a prime divisor. Let P1 , . . . , Pd ¯p ⊆ F¯P , so be the prime divisors of F/K which lie over p. Then K ⊆ E i
60
Chapter 3. Algebraic Function Fields of One Variable
degF Pi = fPi /p degE (p), i = 1, . . . , d. By (1),
degF (p) =
d X
ePi /p degF (Pi )
i=1
=
d X
ePi /p fPi /p degE (p) = [F : E] degE (p).
i=1
Proposition 3.4.2: Let F/L be a constant field extension of an algebraic function field E/K of one variable with L/K separable. Then: (a) For a divisor a of E/K, degE (a) = degF (a), dimE (a) = dimF (a), and LE (a)L = LF (a). (b) genus(E/K) = genus(F/L). (c) If p and P are respective prime divisors of E/K and F/L, with P lying ¯p and P is unramified over p. over p, then F¯P = LE (d) Let R be an integrally closed subring of E containing K. Then RL is the integral closure of R in F . (e) Let p be a prime divisor of E/K of degree 1. Then there is a unique prime divisor P of F/L lying over p and deg(P) = 1. (f) Let x1 , . . . , xm be elements of E. Denote the integral closure of K[x] in E by S. Then SL is the integral closure of L[x] in F . Proof of (a): [Deuring3, p. 126] shows degE (a) = degF (a). We show dimE (a) = dimF (a). Let B be a basis of LE (a) over K. Then B is contained in LF (a) and is linearly independent over L. Therefore, dimE (a) ≤ dimF (a). Conversely, let B 0 be a basis of LF (a). Then B 0 is contained in EL0 for some finitely generated extension L0 /K of L/K. By [Deuring3, p. 132], dimEL0 (a) = dimE (a). Hence, dimF (a) = |B 0 | ≤ dimEL0 (a) = dimE (a). Therefore, dimE (a) = dimF (a). It follows that LE (a)L ⊆ LF (a) have the same dimension over L, so LE (a)L = LF (a). Proof of (b): Let gE = genus(E/K) and gF = genus(F/L). Choose a divisor a of E/K with degE (a) > max(2gE − 2, 2gF − 2). By Lemma 3.2.2(d), dimE (a) = degE (a) + 1 − gE and dimF (a) = degF (a) + 1 − gF . Therefore, by (a), gE = gF . Proof of (c): The general case reduces to the case where L/K is finitely ¯p . generated. In this case [Deuring3, p. 128] proves that F¯P = LE Now choose a primitive element c of L/K and let f = irr(c, K). By ¯ The reduction of f modulo P is f the preceding paragraph, F¯P = E(c). itself. Since L/K is separable, f is separable. Hence, by Lemma 2.3.4, P/p is unramified. Proof of (d): Assume without loss that [L : K] < ∞. Choose a basis w1 , . . . , wn of L/K. Let σ1 , . . . , σn be the K-embeddings of L into Ks . Since
3.5 Completions
61
E is linearly disjoint from L over K each σi extends uniquely to an Eembedding of F into Es . Pn Now consider z ∈ F which is integral over R. Write z = j=1 aj wj with Pn a1 , . . . , an ∈ E. Then σi z = j=1 aj σi wj , i = 1, . . . , n. By Cramer’s rule, b
b ∆
aj = ∆j = ∆j 2 , where bj is in the ring generated over R by the ak ’s and the σk wl , and ∆ = det(σk wl )1≤k,l≤n ∈ K × . By assumption, each bj is integral over R. Since L/K is separable, ∆2 ∈ K × [Lang7, p. 286, Cor. 5.4]. Hence, each aj is integral over R. Since R is integrally closed, aj ∈ R, j = 1, . . . , n. Consequently, z ∈ RL. Proof of (e): To prove (e), it suffices to consider two cases. In one case, L = K(u) with u purely transcendental over K. [Deuring3, pp. 128–129] handles this case. The other case is when L/K is separable and finite. Let in this case P1 , . . . , Pd be the prime divisors of F/L lying over p. By (c), ¯p = K. Hence, by ePi /p = 1, i = 1, . . . , m. By assumption, deg(p) = 1, so E (c), F¯Pi = L, i = 1, . . . , d. Since F/L is a constant field extension of E/K, we have [F : E] = [L : K]. Therefore, by (1), d = 1 and F¯P1 = L. Thus, deg(P1 ) = 1. S∞ Pm Proof of (f): Put S∞(xi ). By Lemma 3.1.1, S = k=1 LE (ka). S∞ a = i=1 div∞,E By (a), SL = k=1 LE (ka)L = k=1 LF (ka). Hence, by Lemma 3.1.1, SL is the integral closure of L[x] in F .
3.5 Completions The completion of a function field F/K at a prime divisor p gives a powerful tool to investigate the behavior of F at p. For example, it allows us to determine the decomposition of p to prime divisors in finite extensions of F (Proposition 3.5.3). We also use completions to define the ‘different’ of an extension. This notion plays a central role in the Riemann-Hurwitz genus formula, to be introduced in the next section. Let v be a rank-1 valuation of a field F . Then v induces a topology on F . Two elements x, y of F are ‘close’ in this topology if v(x − y) is ‘large’. A sequence {xi }∞ i=1 of elements of F is a Cauchy sequence if for every integer m there exists k such that i, j ≥ k implies v(xi − xj ) ≥ m. If every Cauchy sequence converges, F is complete. Each F embeds as a dense subfield in a complete field Fˆv with a valuation v extending the valuation of F [BorevichShafarevich, Chap. 1, Sec. 4.1]. In particular, Fˆv has the same residue field and value group at v as F . We call (Fˆv , v) (or also just Fˆv ) the completion of (F, v). We also say that Fˆv is the completion of F at v. The completion (Fˆv , v) of (F, v) is unique up to an F -automorphism. Example 3.5.1: Qp and K((t)). The completion of Q at the p-adic valuation vp (Section 2.2) is the field Qp of p-adic numbers. Every element x of Q× p has a unique presentation as a convergent (in the vp -topology) power series
62
Chapter 3. Algebraic Function Fields of One Variable
P∞
n n=m an p with m = vp (x), an ∈ Z, 0 ≤ an ≤ p − 1, and am 6= 0. The valuation ring Zp of Qp consists of all x ∈ Qp with m ≥ 0. Next consider a field K and a transcendental element t over K. Let v be the unique valuation of K(t) with v(t) = 1 (Section 2.2). The completion of K(t) at v is the field K((t)) of formal power series in t with coefficients in K. Each nonzero element f of K((t)) has a unique presentation f = P∞ n n=m an t with m = v(f ) and an ∈ K. The valuation ring of K((t)) is the ring K[[t]] of formal power series in t with coefficients in K. The residue field of K((t)) under v is K. By Lemma 2.6.9(b), K((t)) is a regular extension of K. Consider now a finite extension L of K with a basis wP 1 , . . . , wd . Then ∞ w1 , . . . , wd are linearly independent over K((t)). Let x = n=m an tn with Pd an ∈ L be an element of L((t)). For each n write an = i=1 ani wi with Pd P∞ ani ∈ K. Then, x = i=1 ( n=m ani tn )wi . Therefore, L((t)) = K((t))L and w1 , . . . , wd form a basis for L((t))/K((t)).
The next result contains various versions of Hensel’s Lemma: Proposition 3.5.2: Let (F, v) be a complete discrete valued field. (a) Let f ∈ Ov [X] and a ∈ Ov with v(f (a)) > 2v(f 0 (a)). Then there is a unique x ∈ F with f (x) = 0 and v(x − a) ≥ v(f (a)) − v(f 0 (a)) [CasselsFr¨ohlich, p. 83]. (b) Let f ∈ Ov [X] be a monic polynomial. Denote reduction at v by a bar. Suppose f¯(X) = ζ(X)η(X) with ζ, η ∈ F¯v [X] monic and relatively prime. Then there are monic polynomials g, h ∈ Ov [X] with g¯ = ζ, ¯ = η, and f (X) = g(X)h(X) [Zariski-Samuel2, p. 279, Thm. 17]. h (c) Let F 0 be a finite algebraic extension of F . Then v has a unique extension v 0 to F 0 and F 0 is complete under v 0 [Cassels-Fr¨ohlich, p. 56, Thm.]. Here is a global application of completions: Proposition 3.5.3: Let (E, v) be a discrete valued field. Denote its comˆ vˆ). Consider a finite separable extension F of E. Let z be a pletion by (E, primitive element for F/E which is integral over Ov . Put h = irr(z, E). Let h = h1 · · · hr be the decomposition of h into a product of irreducible polyˆs . Denote the unique ˆ For each i let zi be a root of hi in E nomials over E. ˆ ˆ extension of vˆ to Fi = E(zi ) by vˆi . The following holds: (a) The map z 7→ zi extends to a E-embedding of F into Fˆi . (b) Fˆi is the completion of F at the restriction vi of vˆi to F . (c) The valuations v1 , . . . , vr are mutually nonequivalent. Every extension of v to F coincides with one of the vi ’s. ¯ i = he with hi0 ∈ E[X] ¯ (d) evˆi /ˆv = evi /v , fvˆi /ˆv = fvi /v , and h irreducible i0 and e ∈ N. ˆ ˆ (e) The map g(z) 7→ g(z1 ), . . . , g(zr )) for g ∈ E[X] is an E-isomorphism of L r ˆ ˆ E ⊗E F onto i=1 Fi .
3.5 Completions
(f) Each z in F satisfies traceF/E z = Qr ˆ zi . i=1 normFˆi /E
63
Pr
i=1
traceFˆi /Eˆ zi and normF/E z =
Proof: See [Cassels-Fr¨ ohlich, p. 57, Thm.]. The last statement of (d) follows from Proposition 3.5.2(b). Example 3.5.4: Dedekind. We apply Proposition 3.5.3 to show the necessity ¯v is infinite” in Lemma 2.3.5. of the assumption “E Consider the polynomial f (X) = X 3 − X 2 − 2X − 8. Observe that f (X) has no root modulo 3, hence no root in Q. It is therefore irreducible. Let z ˜ and F = Q(z). Then [F : Q] = 3. Next observe that be a root of f (X) in Q 0 v2 (f (0)) = 3, v2 (f (0)) = 1, v2 (f (1)) = 1, v2 (f 0 (1)) = 0, v2 (f (2)) = 3, and v2 (f 0 (2)) = 1. Hence, by Proposition 3.5.2(a), f has three roots x1 , x2 , x3 in Q2 with v2 (x1 ) ≥ 2, v2 (x2 − 1) ≥ 1, and v2 (x3 − 2) ≥ 2. Thus, f (X) = (X − x1 )(X − x2 )(X − x3 ) is a decomposition of f (X) over Q2 into three distinct irreducible polynomials. By Proposition 3.5.3, v2 has three distinct extensions to F : w, w0 , and w00 . By Proposition 2.3.2, F¯w = F2 . Now assume F/Q has a primitive element z with g = irr(z, Q) ∈ Ov2 [X] such that g¯ is separable. Then deg(g) = 3. By the preceding paragraph, F¯w = F2 . Hence, all three distinct roots of g¯ belong to F2 . But F2 has only two elements, so z does not exist. Lemma 3.5.5: Let (F, v) be a discrete valued field and I a nonzero Ov submodule of F which is not F . Then I is a fractional ideal of Ov and there exists m ∈ Z with I = m−m v . If F is a finite extension of a field E, and Ov is the unique valuation ring of F lying over Ov ∩ E (e.g. (E, v) is complete), then traceF/E I is a fractional ideal of Ov ∩ E. Proof: Choose π ∈ F with v(π) = 1. Let x be a nonzero element of I. If 0 x0 ∈ F and v(x0 ) ≥ v(x), then x0 = xx x ∈ I. Since I 6= F , this implies v(I) is bounded from below. Hence, −m = inf(v(x) | x ∈ I) is an integer. Therefore, I = π −m Ov . Now suppose Ov is the unique valuation ring of F over O = Ov ∩ E. By Proposition 2.4.1, Ov is the integral closure of O in F . Let a be an element of A with v(a) sufficiently large. By the preceding paragraph, aI ⊆ Ov . Then a · traceF/E (I) = traceF/E (aI) ⊆ traceF/E (Ov ) ⊆ A. Thus, traceF/E (I) is a fractional ideal of A. Using completions, we generalize the notions of repartition and differential (Section 3.2) of a function field. Let p be a prime divisor of F/K. Suppose the residue field F¯p of F at p is separable over K. By Hensel’s Lemma, F¯p embeds into Fˆp . Indeed, a). Put f = irr(¯ a, K). Then vp (f (a)) > 0 and choose a ∈ F with F¯p = K(¯ vp (f 0 (a)) = 0. By Proposition 3.5.2(a), there is an x0 ∈ Fˆp with f (x0 ) = 0. The map a ¯ 7→ x0 extends to a K-embedding of F¯p into Fˆp . Let π be an element of Fˆp with vp (π) = 1. Then the completion Fˆp is isomorphic to the field F¯p ((π)) of formal power series in π over F¯p . Every
64
Chapter 3. Algebraic Function Fields of One Variable
P∞ element of this field has the form α = i=m ai π i , where m is an integer and ai ∈ F¯p [Chevalley2, p. 46]. If am 6= 0, then vp (α) = m. Q Consider the cartesian product Fˆp with p ranging over all prime diviQ sors of F/K. An α ∈ Fˆp is an adele if vp (αp ) ≥ 0 for all but finitely many p. In particular, each repartition of F/K is an adele. Denote the set of adeles ˆ (or by A ˆ F , if F is not clear from the context). It is an F -subalgebra of by A Q ˆ Fp which contains the algebra A of repartitions. For each prime divisor p ˆ as follows. Identify x ∈ Fˆp with the adele ξ having of F/K embed Fˆp in A 0 ξp = x and ξp = 0 if p0 6= p. For each a ∈ Div(F/K) consider the K-vector space ˆ | vp (α) + vp (a) ≥ 0 for each p}, ˆ Λ(a) = {α ∈ A ˆ = Λ(a) and A + where we have abbreviated vp (αp ) by vp (α). Then A ∩ Λ(a) ˆ Indeed, let α ˆ For each p choose αp ∈ F with vp (αp − α ˆ Λ(a) = A. ˆ ∈ A. ˆp) ≥ ˆ ˆ is in Λ(a), and α ˆ = α − (α − α ˆ) ∈ −vp (a). Then α = (αp ) belongs to A, α − α ˆ A + Λ(a). Therefore (1)
ˆ ˆ ˆ A ∩ (Λ(a) + F ) = Λ(a) + F and A + (Λ(a) + F ) = A.
∼ A/ Λ(a) + F . Thus, in the notation of Section 3.2, ˆ Λ(a) ˆ Hence, A/ +F = (2)
ˆ Λ(a) ˆ + F ) = δ(a) = dim(w − a), dimK A/(
where w is a canonical divisor of F/K. Recall that a differential of F/K is a K-linear map, ω: A → K, which vanishes on a subspace of the form Λ(a)+F . By (1) we can extend ω uniquely ˆ → K which vanishes on Λ(a) ˆ + F . So, from now on, a to a K-linear map ω ˆ: A ˆ ˆ differential of F/K is a K-linear map ω ˆ : A → K which vanishes on Λ(a)+F for some a ∈ Div(F/K). The restriction of ω to A is a differential in the old sense. The divisor of ω ˆ is the maximum of all a ∈ Div(F/K) such that ω ˆ ˆ vanishes on Λ(a). Denote it by div(ˆ ω ). By the above, div(ˆ ω ) = div(ω). ˆ Denote the K-vector space of all differentials that vanish on Λ(a) +F ˆ ˆ by Ω(a). The natural map ω 7→ ω ˆ is an isomorphism of Ω(a) onto Ω(a). In particular, these spaces have the same dimension δ(a). Suppose F/E is a finite separable extension of function fields of one ˆ E . Given a prime divisor P of F/K, denote its variable over K. Let α ∈ A ˆ E as a subalgebra of restriction to E by p and let αP = αp . This identifies A L ˆ F . The isomorphisms E ˆp ⊗E F ∼ A = P|p FˆP of Proposition 3.5.3(e) combine to an isomorphism (3)
ˆ E ⊗E F ∼ ˆF A =A
3.5 Completions
65
[Cassels-Fr¨ohlich, p. 64 or Artin3, p. 244, Thm. 2]. In addition, define a trace ˆF → A ˆE: function traceF/E : A traceF/E (β)p =
X
traceFˆP /Eˆp (βP ).
P|p
Proposition 3.5.6: Let F/E be a finite separable extension of function ˆF : fields of one variable over a field K. The following holds for all α, β ∈ A (a) traceF/E (α + β) = traceF/E (α) + traceF/E (β). ˆE. (b) traceF/E (αβ) = α · traceF/E (β) if α ∈ A (c) The trace of an element x of F coincides with the trace of x as an adele. (d) Let P be a prime divisor of F/K and p = P|E . Then, the trace of an ˆp coincides with the trace of x as an adele. element x of FˆP to E ˆ F such that traceF/E (α) 6= 0. (e) There exists α ∈ A Proof: Statements (a) and (b) follow from the corresponding properties of the trace function on fields. Statement (c) follows from Proposition 3.5.3(f). Statement (d) follows from the definition. Finally, use (c) and the corresponding fact for the trace of fields [Lang7, p. 286, Thm. 5.2] to prove (e). Remark 3.5.7: Complementary modules. (a) Let (E, v) be a complete discrete valued field. Denote the valuation ring of (E, v) by OE . Let F be a finite separable extension of E. Denote the unique extension of v to F by v. Choose a generator πF of the maximal ideal of OF . The complementary module of OF over E is 0 OF/E = {x ∈ F | traceF/E (xOF ) ⊆ OE }.
It is a fractional ideal of OF which contains OF [Lang5, p. 58, Cor.]. By −d
0 Lemma 3.5.5, OF/E = πF F /E OF for some nonnegative integer dF/E . Call dF/E the different exponent of F/E. It is known [Lang5, p. 62, Prop. 8] that dF/E > 0 if and only if F/E is ramified. Moreover, dF/E ≥ eF/E − 1. If the residue field extension is separable, then equality holds if and only if F/E is tamely ramified [Serre4, p. 67, Prop. 13]. (b) Let F/E be a finite separable extension of function fields of one variable over a field K. Consider prime divisors p and P of E/K and F/K, ˆp and FˆP with respectively, with P lying over p. Choose completions E ˆp and FˆP , by O ˆ p and O ˆ P , reˆp ⊆ FˆP . Denote the valuation rings of E E spectively. Occasionally write dFˆP /Eˆp as dP/p or dP/E and call dP/p the different exponent of P over p (or over E). By (a), dP/p > 0 if and only if P/p is ramified. This happens for only finitely many P’s.
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Chapter 3. Algebraic Function Fields of One Variable
Lemma 3.5.8: Let E ⊆ F ⊆ F 0 be function fields of one variable over a field K with F 0 /E separable. Consider prime divisors p, P, P0 of E/K, F/K, F 0 /K, respectively, with P lying over p and P0 lying over P. Then dP0 /p = eP0 /P dP/p + dP0 /P . In particular, if P0 /P is unramified, then dP0 /p = dP/p . If P/p is unramified, then dP0 /p = dP0 /P . Proof: Suppose the formula holds. If P0 /P is unramified, then dP0 /P = 0 and eP0 /P = 1. Hence, dP0 /p = dP/p . If P/p is unramified, then dP/p = 0, so dP0 /p = dP0 /P . To prove the formula, assume without loss, E, F , and F 0 are complete with respective valuation rings OE , OF , and OF 0 . Choose prime elements πE , πF , and πF 0 for the respective maximal ideals. Put d = dF/E , d0 = dF 0 /F , and e0 = eF 0 /F . We have to prove that e0 d + d0 = dF 0 /E . 0
First note that πFe 0 OF 0 = πF OF 0 . Hence, 0
0
0
traceF 0 /E (πF−e0 d−d OF 0 ) = traceF/E (traceF 0 /F (πF−d πF−d0 OF 0 )) 0
= traceF/E (πF−d traceF 0 /F (πF−d0 OF 0 )) ⊆ traceF/E (πF−d OF ) ⊆ OE . Thus, 0
0
πF−e0 d−d OF 0 ⊆ OF0 0 /E .
(4)
0 By definition, πF−d−1 ∈ / OF/E . Hence, traceF/E (πF−d−1 OF ) 6⊆ OE . By the
second part of Lemma 3.5.5, traceF/E (πF−d−1 OF ) is a fractional ideal of OE . −m Hence, by the first part of Lemma 3.5.5, traceF/E (πF−d−1 OF ) = πE OE for −1 −d−1 some positive integer m. Therefore, πE OE ⊆ traceF/E (πF OF ). Simi0
larly, πF−1 OF ⊆ traceF 0 /F (πF−d0 −1 OF 0 ). 0
0
0
0
Assume πF−e0 d−d OF 0 6= OF0 0 /E . Then, by (4), πF−e0 d−d −1 OF 0 ⊆ OF0 0 /E . Therefore, by the preceding paragraph, 0
0
OE ⊇ traceF 0 /E (πF−e0 d−d −1 OF 0 ) 0
= traceF/E (πF−d traceF 0 /F (πF−d0 −1 OF 0 )) −1 ⊇ traceF/E (πF−d−1 OF ) ⊇ πE OE , 0
0
−d
which is a contradiction. Therefore, πF−e0 d−d OF 0 = OF0 0 /E = πF 0 F Consequently, e0 d + d0 = dF 0 /E , as claimed.
0 /E
OF 0 .
3.6 The Different
67
3.6 The Different The Riemann-Hurwitz genus formula enables us to compute the genus of a function field of one variable F over a field K from the genus of a function subfield E in terms of the ‘different’ of the extension F/E. This formula is in particular useful when the genus of E is known, e.g. when E is the field of rational functions over K. Let F/E be a finite separable extension of function fields of one variable over a field K. The different of F/E is a divisor of F/K: Diff(F/E) =
X
dP/E P,
where P ranges over all prime divisors of F/K and dP/E is the different exponents of P/E. By Remark 3.5.7(a), dP/E ≥ 0 for all P. Moreover, dP/E > 0 if and only if P is ramified over E. Since only finitely many P ramify over E, Diff(F/E) is well defined, Diff(F/E) ≥ 0, and deg(Diff(F/E)) ≥ 0. Theorem 3.6.1 (Riemann-Hurwitz Genus Formula): Let F/E be a finite separable extension of function fields of one variable over a field K. Put gE = genus(E/K) and gF = genus(F/K). Then (1)
2gF − 2 = [F : E](2gE − 2) + deg(Diff(F/E)).
Proof: Let pi , i ∈ I, be the prime divisors of E/K. Choose a nonzero P differential ω of E/K. Write div(ω) = i∈I ki pi with ki ∈ Z and ki = 0 for all but finitely many i P ∈ I. By definition, P ˆ ˆE( (2) ω vanishes on Λ i∈I ki pi ) but not on ΛE ((kr + 1)pr + i6=r ki pi ) for any r ∈ I. ˆ F → K: Define a map Ω: A Ω(α) = ω(traceF/E (α)). It is K-linear and vanishes P on F (Proposition 3.5.6). For each i ∈ I let pi = j∈Ji eij pij , where the pij are the distinct prime divisors of F/K lying over pi and the eij are the corresponding ramification ˆp , indices. We simplify notation and for all i ∈ I and j ∈ Ji let Ei = E i ˆ ˆ ˆ Oi = Opi , vi = vpi , Fij = Fpij , Oij = Opij , vij = vpij , and (3)
0 dij = vij (Diff(F/E)) = −vij (Oij ),
0 is the complementary module of Oij with respect to traceF/E where Oij ˆ E and β ∈ A ˆ F let αi = αp and βij = βp . (Remark 3.5.7). Also, for all α ∈ A i ij We prove that ˆ F (P (eij ki + dij )pij ) and (4a) Ω vanishes on Λ i,j (4b) for each r ∈ I and each sP ∈ Jr the differential Ω does not vanish on ˆ F (ers kr + drs + 1)prs + Λ (e k + d )p ij ij . (i,j)6=(r,s) ij i
68
Chapter 3. Algebraic Function Fields of One Variable
This will imply that divF (Ω) =
X
(eij ki + dij )pij = divE (ω) + Diff(F/E).
i,j
The formulas deg(divF (Ω)) = 2gF − 2 and deg(divE (ω)) = [F : E](2gE − 2) (Lemmas 3.2.2(b) and 3.4.1) will give (1). For all i, j choose πi ∈ E and πij ∈ F with vi (πi ) = 1 and vij (πij ) = 1. ˆ F (P (eij ki + dij )pij ). Then vij (α) ≥ Proof of (4a): Consider α ∈ Λ ij −eij ki − dij .
Hence, vij (πiki α) ≥ −dij .
0 By (3), (πiki α)ij ∈ Oij , so
traceFij /Ei (πiki α)ij ∈ Oi . Thus, (traceF/E (πiki α))i =
X
traceFij /Ei (πiki α) ∈ Oi ,
j∈Ji
hence vi (traceF/E (α)) + ki ≥ 0 for each i ∈ I. Therefore, traceF/E (α) ∈ ˆ E (P ki pi ). Consequently, by (2), Ω(α) = ω(traceF/E (α)) = 0. Λ i∈I
Proof of (4b): Assume there exist r ∈ I and s ∈ Jr such that Ω vanishes on X ˆ F (ers kr + drs + 1)prs + (eij ki + dij )pij . V =Λ (i,j)6=(r,s)
By Lemma 3.5.5, there is an m ∈ Z with −drs −1 Ors ) = πr−m Or Irs = traceFrs /Er (πr−kr πrs
We distinguish between two cases: ˆ E (kr + 1)pr + P Case A: m ≥ kr + 1. Consider β ∈ Λ i6=r ki pi . Then ˆ E (P ki pi ), so ω(β) = vr (β) ≥ −kr − 1. If vr (β) ≥ −kr , then β ∈ Λ i∈I 0 (by (2)). Otherwise, vr (β) = −kr − 1. In this case write β = α + γ, ˆ E satisfy αr = (1 − πr )βr , γr = πr βr , αi = 0, and γi = βi where α, γ ∈ A for i 6= r. Then, αr ∈ πr−kr −1 Or ⊆ πr−m Or = Irs , so there is a δrs ∈ −drs −1 Ors with traceFrs /Er δrs = αr . For (i, j) 6= (r, s) let δij = 0. πr−kr πrs Then δ ∈ V , α = traceF/E δ, and ω(α) = ω(trace(δ)) = Ω(δ) = 0. Also, ˆ E (P ki pi ), so ω(γ) = 0 (by (2)). It follows that ω(β) = 0. Thus, ω γ∈Λ i∈I ˆ E (kr + 1)pr + P vanishes on Λ ki pi , in contradiction to (2). i6=r
Case B: m ≤ kr . Then, −drs −1 Ors ) = πr−m Or ⊆ πr−kr Or . traceFrs /Er (πr−kr πrs −drs −1 −drs −1 0 Hence, traceFrs /Er (πrs Ors ) ⊆ Or . Therefore, πrs Ors ⊆ Ors , so 0 vrs (Ors ) ≤ −drs − 1. This contradiction to (3) completes the proof of Case B and the proof of the whole theorem.
3.6 The Different
69
Remark 3.6.2: Applications of the Riemann-Hurwitz formula. Let F/E be a finite separable extension of function fields of one variable of a field K. We say that F/E is unramified (resp. tamely ramified) if each prime divisor of F/K is unramified (resp. tamely ramified) over E. Let gE = genus(E/K) and gF = genus(F/K). Suppose [F : E] ≥ 2. (a) Comparison of genera: We have mentioned at the beginning of this section that deg(Div(F/E)) ≥ 0. Hence, by (1), gF ≥ gE . Both gE and gF have the same value if and only if gE = 1 and F/E is unramified, or gE = 0 and deg(Diff(F/E)) = 2 [F : E] − 1 . In particular, if F = K(t), then gF = 0 (Example 3.2.4). Hence, gE = 0. Each prime divisor of F/K of degree 1 induces a prime divisor of E/K of degree 1. We conclude from Example 3.2.4 that E = K(u) is also a rational function field. This is L¨ uroth’s theorem. This theorem actually holds for arbitrary algebraic extension K(t)/E and not only for separable extensions and may be proved by elementary arguments on polynomials [Waerden1, p. 218]. (b) An analog of a theorem of Minkowski: Suppose F/E is unramified. In this case the Riemann-Hurwitz formula simplifies to gF − 1 = [F : E](gE − 1). Hence, gE > 0. In other words, a function field E/K of genus 0 has no proper finite unramified extension F which is regular over K. In particular, K(t) has no finite proper separable unramified extension F which is regular over K. This is an analog of a theorem of Minkowski saying that Q has no proper unramified extensions [Janusz, p. 57, Cor. 11.11]. (c) The Hurwitz-Riemann formula for tamely ramified extensions: Suppose F/E is tamely ramified. By Remark 3.5.7, the Riemann-Hurwitz formula simplifies to XX eP/p − 1 deg(P). 2gF − 2 = [F : E](2gE − 2) + p
P|p
(d) An analog of Minkowski’s theorem in the tamely ramified case: Suppose K is algebraically closed and gE = 0. Then E = K(t) and the degree of each prime divisor is 1. Suppose F/E is a proper tamely ramified extension. Then E has at least two prime divisors that ramify in F . Indeed, assume E has only one prime divisor p that ramifies in F . Let P1 , . . . , Pr be the prime divisors of p in F . Then −2 ≤ 2gF − 2 = −2[F : E] +
r X
ePi /p − 1
i=1
= −2[F : E] + [F : E] − r = −[F : E] − r Hence, 3 ≤ [F : E] + r ≤ 2, a contradiction.
70
Chapter 3. Algebraic Function Fields of One Variable
In particular, if K is algebraically closed and char(K) = 0, then every proper extension of K(t) is ramified over at least two prime divisors. (e) Generation of Galois groups by inertia groups: Let K be an algebraically closed field, E = K(t), and F a finite Galois extension of E. Denote the prime divisors of F/K which ramify over E by P1 , . . . , Pr . For each i let Di be the decomposition group of Pi over E. Since K is algebraically closed, Di is also the inertia group of Pi over E. Let Ei be the fixed field of Di in F . Then Pi |Ei is unramified over E. Hence, E0 = E1 ∩ · · · ∩ Er is unramified over E. By (b), E0 = E. In other words, the inertia groups of the prime divisors of F/K which ramify over E generate Gal(F/E). (f) Quasi-p groups: Let K be an algebraically closed field of positive characteristic p. Consider the rational field E = K(t), a prime divisor p of E/K, and a finite Galois extension F of E which is ramified only over p. Denote the fixed field in F of all p-Sylow subgroups of Gal(F/E) by Ep . Then Ep /E is a Galois extension of degree relatively prime to p. Hence, Ep /E is tamely ramified. The only prime divisor of E/K which is possibly ramified in Ep is p. It follows from (d) that Ep = E. In other words, Gal(F/E) is generated by its p-Sylow subgroups. One says that Gal(F/E) is quasi-p. (g) Abhyankar’s conjecture: Let K be an algebraically closed field of positive characteristic p. Let E = K(t) and p a prime divisor of E/K. In 1957, Abhyankar conjectured that for each finite quasi-p group G there exists a finite Galois extension F of E which is unramified outside p and Gal(F/E) ∼ = G [Abhyankar]. Serre proved the conjecture for solvable G in 1990 [Serre9]. Raynaud treated all other cases in 1991 [Raynaud]. (h) The generalized Abhyankar’s conjecture: Let again K be an algebraically closed field of positive characteristic p. Put E = K(t). Consider a set S = {p1 , . . . , pr } of prime divisors of E/K and a finite Galois extension F of E which is unramified over E outside S. Let G = Gal(F/E). Denote the subgroup of G generated by all p-Sylow subgroups of G by G(p). Let Ep be the fixed field of G(p) in F . Then Ep is a Galois extension of E which is tamely ramified over E and unramified outside S. By [Grothendieck, XIII, Cor. 2.12], Gal(Ep /E) is generated by r − 1 elements. This led Abhyankar to conjecture that for every finite group G such that G/G(p) is generated by r − 1 elements there is a Galois extension F of E which is unramified outside S and Gal(F/E) ∼ = G. Harbater proved this conjecture in [Harbater2] by reducing it to the special case r = 1 proved by Raynaud.
3.7 Hyperelliptic Fields We demonstrate the concepts and results of this chapter to a study of a special kind of algebraic function fields of one variable which we now introduce. A function field F/K is hyperelliptic if its genus is at least 2 and if it is a quadratic extension of a function field E/K of genus 0. We then say E is a quadratic subfield of F . It turns out that E is uniquely determined.
3.7 Hyperelliptic Fields
71
Moreover, we will be able to identify E from the arithmetic of F . Lemma 3.7.1: Let F/K be a function field of genus g. Let a, b be nonnegative divisors of F/K and let w be a canonical divisor of F/K. (a) If L(b − a) = L(b), and dim(b) ≥ 1, then dim(a) = 1. (b) If g ≥ 1 and a > 0, then dim(w − a) < dim(w). (c) If g ≥ 1 and x1 , . . . , xg is a basis for L(w), then −w = min div(x1 ), . . . , div(xg ) . Proof of (a): Since a ≥ 0, we have K ⊆ L(a). Conversely, let x ∈ L(a). Then div(x) + a ≥ 0. Consider y ∈ L(b). By assumption, y ∈ L(b − a), hence div(y) + b − a ≥ 0. Therefore, div(xy) + b ≥ 0, so xy ∈ L(b). Apply this Pn result to a basis y1 , . . . , yn of L(b). Find aij ∈ K such that xyi = j=1 aij yj , i = 1, . . . , n. Hence, det(xI − A) = 0, where A = (aij )1≤i,j≤n . Therefore, x satisfies a monic equation with coefficients in K. Since K is algebraically closed in F , we have x ∈ K. Consequently, dim(a) = 1. Proof of (b): Assume dim(w − a) = dim(w). Then L(w − a) = L(w). By Lemma 3.2.2(b), dim(w − a) = dim(w) = g. So, by (a), dim(a) = 1. By Theorem 3.2.1, dim(a) = deg(a) + 1 − g + dim(w − a). Hence, deg(a) = 0, which is a contradiction. Therefore, dim(w − a) < dim(w). Proof of (c): Denote min div(x1 ), . . . , div(xg ) by m. Since div(xi ) + w ≥ 0 we have m ≥ −w. If m > −w, then there exists a prime divisor p of F/K with m − p ≥ −w. Hence, div(xi ) + w − p ≥ 0, so xi ∈ L(w − p) for i = 1, . . . , g. Therefore, L(w − p) = L(w), contradicting (b). Consequently, m = −w. Proposition 3.7.2: Let F/K be a function field of genus g ≥ 2. Consider a canonical divisor w of F/K. Let x1 , . . . , xg be a basis of L(w). If E = x K xx12 , . . . , xg1 is a proper subfield of F , then genus(E/K) = 0 and [F : E] = 2. Thus, F/K is a hyperelliptic field. Proof: Let w0 = div(x1 ) + w. Then w0 is also a canonical divisor and x 1, xx21 , . . . , xg1 is a basis of L(w0 ). Replace w by w0 , if necessary, to assume that x1 = 1 and E = K(x2 , . . . , xg ). Since g ≥ 2, the element x2 is transcendental over K. Hence, E is also an algebraic function field of one variable over K and d = [F : E] < ∞ (Section divisor of E/K. 3.4). Denote genus(E/K) by gE and let wE be a canonical By Lemma 3.7.1(c), w = − min 0, div(x2 ), . . . , div(xg ) . In particular, w ≥ 0. By Lemma 3.2.2(b), deg(w) = 2g − 2 ≥ 2. Therefore, (1)
w > 0.
Observe that div(xi ) ∈ Div(E/K), i = 2, . . . , g, so w ∈ Div(E/K). We may therefore apply Riemann-Roch to E/K and w: (2) dim LE (w) = degE (w) + 1 − gE + dim LE (wE − w) .
72
Chapter 3. Algebraic Function Fields of One Variable
Since 1, x1 , . . . , xg are in E and generateLF (w), we have LE (w) = LF (w). Hence, by Lemma 3.2.2(b), dim LE (w) = g. Applying Lemma 3.2.2(b) again and Lemma 3.4.1, we have 2g−2 = degF (w) = d·degE (w). Substituting this in (2) gives (3)
g=
2g − 2 + 1 − gE + dim LE (wE − w) , d
which may be rewritten as (4)
d dim LE (wE − w) − gE ) = (g − 1)(d − 2)
By assumption, both g and d are at least 2. Hence, the right hand side of (4) is at least 0. On the other hand, by Lemma 3.2.2(b) and by (1), (5) dim LE (wE − w) ≤ dim LE (wE ) = gE . Hence, both sides of (4) are 0. Since g ≥ 2, this gives d = 2. Finally, by (3), gE = dim(LE (wE − w)). Hence, by (5), dim(LE (wE − w)) = dim(LE (wE )). We conclude from Lemma 3.7.1(b) that gE = 0. Lemma 3.7.3: Let F/K be a function field of genus g ≥ 1, w a canonical divisor of F/K, and x1P , . . . , xg a basis of LF (w). Let E/K be a subfield of g genus 0 of F/K. Then i=1 Exi 6= F . ˆ E (0)xi ⊆ Λ ˆ F (w). Hence, Proof: For each i we have Λ g X
(6)
ˆ F (w) + F. ˆ E (0) + E xi ⊆ Λ Λ
i=1
Choose a canonical divisor wE of E/K. By (2) of Section 3.5, ˆ E /(Λ ˆ E (0) + E) = dim(wE ) = genus(E/K) = 0. dimK A Hence, ˆE = Λ ˆ E (0) + E. A
(7) On the other hand,
ˆ F /(Λ ˆ F (w) + F ) = dim(0) = 1. dimK A ˆ F , so by (6) and (7) ˆ F (w) + F ⊂ A Hence, Λ (8)
g X
ˆ E xi ⊂ A ˆF . A
i=1
Pg Pg ˆ ˆ If i=1 Exi = F , then Pg by (3) of Section 3.5, AF = i=1 AE xi , contradicting (8). Consequently, i=1 Exi 6= F .
3.8 Hyperelliptic Fields with a Rational Quadratic Subfield
73
Proposition 3.7.4: Let F/K be a hyperelliptic function field, w a canonical x divisor of F/K, and x1 , . . . , xg a basis of LF (w). Then E = K xx21 , . . . , xg1 is the only quadratic subfield of F . Proof: By definition, F has subfields of genus 0. Let E 0 be one Pquadratic g 0 of them. By Lemma 3.7.3, i=1 E xi 6= F . Since [F : E 0 ] = 2, this implies xi ∈ E 0 , i = 1, . . . , g. It follows that E ⊆ E 0 . In particular, E ⊂ F . By Proposition 3.7.2, [F : E] = 2. Comparing degrees implies E 0 = E.
3.8 Hyperelliptic Fields with a Rational Quadratic Subfield A hyperelliptic field with a rational quadratic subfield is generated by two generators which satisfy an equation of a special type. This situation arises, for example, when the hyperelliptic field has a prime divisor of degree 1 (Example 3.2.4). Proposition 3.8.1: Let F/K be a hyperelliptic field of genus g. Suppose the quadratic subfield of F is K(x) with x indeterminate. Then F = K(x, y), where y satisfies a relation y 2 + h1 (x)y + h2 (x) = 0 with h1 , h2 ∈ K[X], deg(h1 ) ≤ g + 1, and deg(h2 ) ≤ 2g + 2. If char(K) 6= 2, we may choose y with y 2 = f (x), where f ∈ K[X] is a polynomial with no multiple root and deg(f (x)) ≤ 2g + 2. Proof: Let p∞ be the pole of x in K(x)/K. By Lemma 3.4.1, degF (np∞ ) = 2n for each positive integer n. If n ≥ g, then 2n > 2g − 2. By Lemma 3.2.2(d), dim(LF (np∞ )) = 2n + 1 − g. In particular, for n = g the elements 1, x, . . . , xg form a basis for LF (gp∞ ). For n = g + 1 we have dim LF ((g + 1)p∞ ) = g + 3. Hence, two more elements are needed to complete 1, x, . . . , xg to a basis of LF (g + 1)p∞ . We take one of them as xg+1 and denote the other one by y. By RiemannRoch, dim(LK(x) ((g + 1)p∞ )) = g + 2. So, 1, x, . . . , xg+1 form a basis of LK(x) (g + 1)p∞ . If y is in K(x), then y is also in LK(x) (g + 1)p∞ . This implies that 1, x, . . . , xg+1 , y are linearly dependent over K. We conclude from this contradiction that y ∈ / K(x), so F = K(x, y). Next note that dim LF ((2g + 2)p∞ ) = 3g + 5. All 3g + 6 elements 1, x, . . . , x2g+2 , y, yx, . . . , yxg+1 , y 2 belong to LF (2g + 2)p∞ . Therefore, P2g+2 Pg+1 there are ai , bj , c ∈ K, not all zero, with i=0 ai xi + j=0 bj yxj + cy 2 = 0. Pg+1 −1 bj X j , h2 (X) = Since y ∈ / K(x), we have c 6= 0. Let h1 (X) = j=0 c P2g+2 −1 ai X i . Then i=0 c (1)
y 2 + h1 (x)y + h2 (x) = 0.
If char(K) 6= 2, replace y by 2y + h1 (x), if necessary, to assume (1) has the form y 2 = f (x). Here f ∈ K[X] has degree at most 2g + 2. Finally, if f has
74
Chapter 3. Algebraic Function Fields of One Variable
multiple roots, then f = g 2 h with h having no multiple roots. Replace y by yg(x)−1 to assume f has no multiple roots. Our first task is to compute the genus of the hyperelliptic field in characteristic 6= 2 from deg(f ). Proposition 3.8.2: Let K be a field of characteristic 6= 2 and f ∈ K[X] a polynomial of degree d ≥ 1 with no multiple roots. Let F = K(x, y), with x transcendental over K and y 2 = f (x). Then F/K is an algebraic function d−2 field of one variable of genus d−1 2 if d is odd and of genus 2 if d is even. In particular, if d ≥ 5, then F/K is hyperelliptic. ˜ Proof: Since f (x) is not a square in K(x), ˜ ˜ [K(x, y) : K(x)] = [K(x, y) : K(x)] = 2. ˜ Hence, K(x, y) is linearly disjoint from K(x) over K(x). Since x is transcen˜ over K. Hence, by the tower dental over K, K(x) is linearly disjoint form K ˜ property of linear disjointness (Lemma 2.5.3), F is linearly disjoint from K over K. It follows that F/K is an algebraic function field of one variable. Since char(K) 6= 2, F/K(x) is tamely ramified. Since the genus of K(x) is 0, the Riemann-Hurwitz formula reduces to XX (eP/p − 1) deg(P) (2) 2g − 2 = −4 + p
P|p
(Remark 3.6.2(c)). If p is a prime divisor of K(x) which ramifies in F , then p has only one extension P to F , the ramification index of P is 2, and its residue degree is 1. Hence, (2) simplifies to (3)
2g = −2 +
X
deg(p),
where p ranges over all prime divisors of K(x)/K which ramify in F . Let f (x) = p1 (x) · · · pr (x) be the decomposition of f (x) into a product of distinct irreducible polynomials in K[x]. To each pi there corresponds a prime divisor pi of K(x)/K of degree deg(pi ) and a valuation vi such that vi (f (x)) = vi (pi (x)) = 1 (Example 2.2.1(b)). By Example 2.3.8, each pi ramifies in F . In addition, let v∞ be the valuation of K(x)/K with v∞ (x) = −1 and let p∞ be the corresponding prime divisor; its degree is 1. Since v∞ (f (x)) = −d, the prime divisor p∞ is ramified in F if d is odd and unramified if d is even (Example 2.3.8). All other prime divisors of K(x) are unramified in F . The Pr sum of the degrees of the ramified prime divisors is δ + i=1 deg(pi ) = δ + d, where δ = 1 if d is odd and δ = 0 if d is even. It follows from (3) that g = d−1 2 if d is odd and g = d−2 2 if d is even. In characteristic 2 we compute the genus of only a special type of a hyperelliptic field.
Exercises
75
Lemma 3.8.3: Let (K, v) be a valued field of characteristic 2 and t an element of K with v(t) = 1. Consider an Artin-Schreier extension L = K(x) with x2 + x = 1t . Denote the unique extension of v to L by w. Let y = tx. Then 0 = y −2 Ow . w(y) = 1, Ow = Ov [y] is the integral closure of Ov in L, and Ow Proof: By Example 2.3.11, v has a unique extension w to L. It totally ramifies over K and satisfies w(x) = −1, w(t) = 2, and w(y) = 1. In particular, Ow is the integral closure of Ov in L (Proposition 2.4.1). Since ¯ w , each z ∈ Ow can be written as z = a + by with a ∈ Ov and b ∈ Ow . ¯v = L K Hence, Ow = Ov [y] [Lang5, p. 26, Prop. 23]. 0 To compute Ow = {z ∈ L | traceL/K (zOw ) ⊆ Ov } observe that g(X) = 2 0 = t−1 Ow = y −2 Ow irr(y, K) = Y + tY + t and g 0 (X) = t. Therefore, Ow 0 [Lang5, p. 59, Cor.]. Of course, one may also compute Ow directly. Proposition 3.8.4: Let K be a field of characteristic 2 and a1 , . . . , ad distinct elements of K. Let F = K(x, y) with x transcendental over K, h ∈ K[X] with deg(h) ≤ d, and y 2 + y = (x−a1h(x) )···(x−ad ) . Then F/K is an algebraic function field of one variable of genus d − 1. In particular, if d ≥ 3, then F/K is hyperelliptic. Proof: For each i between 1 and d let vi be the discrete normalized valuation of K(x)/K with vi (x − ai ) = 1. Let t = (x − a1 ) · · · (x − ad ). Then vi (t) = 1. ˜ ˜ If y is in K(x), ˜ K. then 2˜ vi (y) = −1, Extend vi to a valuation v˜i of K(x)/ ˜ which is a contradiction. Hence, y is not in K(x), so [K(x, y) : K(x)] = ˜ ˜ ˜ over K, so [K(x, y) : K(x)] = 2. It follows that F is linearly disjoint from K F/K is an algebraic function field of one variable. By Example 2.3.9, v1 , . . . , vd are the only valuations of K(x)/K that ramify in F . Hence, only v1 , . . . , vd contribute to the different of F/K(x). [ (y). By Let wi be the unique extension of vi to F . Then Fˆwi = K(x) vi Lemma 3.8.3, the contribution of wi to deg Diff(F/K(x)) is 2. Hence, deg(Diff(F/K(x))) = 2d. Let g = genus(F/K). By Riemann-Hurwitz (Theorem 3.6.1), 2g − 2 = −4 + 2d. Hence, g = d − 1, as claimed.
Exercises In the following exercises F is a function field of one variable over a field K and t is a transcendental element over K. 1. Let a be a divisor of F/K. Note: If a ≥ 0 and deg(a) = 0, then a = 0. (a) Prove that if deg(a) = 0 and a is not a principal divisor, then dim(a) = 0. (b) Let g be the genus of F . Suppose a is a noncanonical divisor with deg(a) = 2g − 2. Show that dim(a) = g − 1. Hint: Use Riemann-Roch and (a). 2. Let a and b be divisors of F/K with b ≥ 0. Use Riemann-Roch to prove that (a) dim(a) ≤ dim(a + b) ≤ dim(a) + deg(b); and
76
Chapter 3. Algebraic Function Fields of One Variable
(b) dim(a) ≤ max(0, deg(a) + 1). Hint: Write a as the difference of its “positive” and “negative” parts. 3. Let p1 , . . . , pn be distinct Pnprime divisors of F/K and let m1 , . . . , mn be If K is infinite, prove positive integers such that i=1 mi deg(pi ) > 2g − P1. n × such that div (x) = that there exists x ∈ F ∞ i=1 Pn mi pi . In particular, Pn [F : K(x)] = i=1 mi deg(pi ). Hint: Consider a = i=1 mi pi and aj = a−pj for j = 1, . . . , n. For each j, 1 ≤ j ≤ n, distinguish between the cases deg(aj ) ≤ 2g − 2 and deg(aj ) > 2g − 2, and prove that L(aj ) ⊂ L(a). 4. Suppose F = K(t). Prove, in the notation of Example 3.2.4, that −2p∞ is a canonical divisor of F/K. 5. Suppose the genus of F/K is 0. Prove that every divisor a with deg(a) = 0 is principal. Hint: Compute dim(a) and apply Exercise 1(a). kr ∈ N, and 6. In the notation of Proposition 3.3.2 let p1 , . . . , pr ∈ S, k1 , . . . , Q n 1, . . . , r. Consider the ideal A = i=1 Piki of Pi be the center of pi at OS , i =P r OS . Prove that the divisor a = i=1 ki pi of F/K satisfies deg(a) = (OS : A). Qr Hint: Prove that OS /A ∼ = i=1 OS /Piki and that OS /Pi ∼ = Pik /Pik+1 , as groups, for each nonnegative integer k. 7. Prove that every algebraic function field F/K of genus 2 is hyperelliptic with a rational quadratic subfield. Hint: Choose a positive canonical divisor w of F/K. Then choose a nonconstant x in L(w). 8. Let K be a field of characteristic that does not divide n. Consider the function field F = K(x, y) over K with x, y satisfying xn +y n = 1. Prove that . Hint: Use Example 2.3.8 and the Riemann-Hurwitz genus(F ) = (n−1)(n−2) 2 formula (Remark 3.6.2(c)).
Notes In addition to [Chevalley2] one may find a proof of the Riemann-Roch theorem in [Lang4, Chapter 10, Section 2], [Deuring3, Section 15], and [Stichtenoth, Section I.5]. The content of Section 3.7 is borrowed from [Artin3]. One may also find it in [Stichtenoth, Section VI.2] along with computations of genera of various function fields using the Riemann-Hurwitz genus formula. However, in contrast to our exposition, [Stichtenoth] assumes the fields of constants to be perfect.
Chapter 4. The Riemann Hypothesis for Function Fields In this chapter K is a finite field of characteristic p with q elements. Let F be an algebraic function field of one variable over K and g the genus of F/K. Denote the group of divisors and the group of divisor classes of F/K by D and C, respectively. P ∞ The series ζ(s) = n=1 n−s defines the classical Riemann zeta function of a complex variable. It converges absolutely for Re(s) > 1. Hence, ζ(s) is an analytic function in this domain. The series diverges for s = 1. The function, however, can be analytically continued to a function meromorphic on the whole s-plane. The production of this requires two stages: an analytic 1 to the half plane Re(s) > 0 by a rearrangement of continuation of ζ(s) + 1−s the series via Abel summation; and then, (the difficult part) a demonstration ζ(s) is the product of the classical Gamma function and an elementary that ζ(1−s) function in the domain 0 < Re(s) < 1. The expression that results from the last stage is called the functional equation for ζ(s). The resulting analytic continuation of ζ(s) yields a function with a simple pole at s = 1 with residue 1 and zeros at the points −2, −4, −6, −8, . . . . There are no other zeros in the domains Re(s) ≥ 1 and Re(s) ≤ 0 [Titchmarsh, p. 30]. The classical Riemann hypothesis is still unproven. It states that the only zeros of ζ(s) in the strip 0 < Re(s) < 1 lie on the line Re(s) = 12 ; its applications are legion. There is an analog for F/K of the Riemann zeta function (Section 4.2). It satisfies a functional equation (Proposition 4.4.1). Our main goal is the proof of an analog to the Riemann hypothesis (Theorem 4.5.1). In Chapter 5 we extract from this an explicit estimate for the number of points on any curve over a finite field.
4.1 Class Numbers The assumption that K is finite results in the finiteness of other sets connected to F/K. For example, F/K has only finitely many ideal classes of degree 0 and only finitely many nonnegative divisors of a given degree n. The main result of this section (Lemma 4.1.4) computes the latter number in terms of the former one. Let p be a prime divisor of F/K. Its residue field F¯p is a finite extension of K of degree deg(p). Thus, F¯p is a finite field whose order we indicate by N p = q deg(p) , the norm of p. Extend the definition of the norm to arbitrary divisors by the formula N a = q deg(a) . Then N (a + b) = N a · N b. Definition 4.1.1: Denote the number of divisor classes of F/K of degree zero by h; the class number of F/K. If g = 0, then every divisor of degree 0 of F/K is principal (Exercise 5
78
Chapter 4. The Riemann Hypothesis for Function Fields
of Chapter 6). Therefore, h = 1 if F is K(x) or any genus 0 function field. Lemma 4.1.2: Only finitely many nonnegative divisors of F/K have degree equal to a given integer m. In addition, the class number of F/K is finite. Proof: Let x ∈ F be transcendental element over K and let E = K(x). Denote the collection of all prime divisors of E/K of degree ≤ m by S0 . Each element of S0 , except possibly p∞ , corresponds to a monic irreducible polynomial in K[x] of degree ≤ m. Hence, S0 is a finite set. Only finitely many prime divisors of F/K lie over a given prime divisor p0 of E/K. Each of them has degree at least as large as deg(p0 ). Thus, there are only finitely many prime divisors of F/K of degree ≤ m. Therefore, the set Am of all nonnegative divisors of F/K of degree m is finite. For the second part of the lemma choose a nonnegative divisor m of degree m ≥ g. Denote the set of divisor classes of degree m by Cm . For each b ∈ Cm Riemann-Roch (Theorem 3.2.1) implies that dim(b) ≥ m − g + 1 ≥ 1. Hence, there is an x ∈ F × with div(x) + b ≥ 0, so the class of b contains a nonnegative divisor. It follows that the map Am → Cm mapping each a ∈ Am onto each class is surjective. Therefore, by the preceding paragraph, Cm is finite. Finally, the map a 7→ a + m induces a bijective map of C0 onto Cm . Consequently, C0 is a finite group. Lemma 4.1.3: The number of nonnegative divisors in a given class of divisors C of F/K is
q dim(C) −1 . q−1
Proof: If a ≥ 0, then div(a) + a ≥ 0 for each a ∈ K × , so K ⊆ L(a). Hence, dim(a) ≥ 1. This gives the formula if dim(C) = 0. Suppose dim(C) = n > 0. Let c be a divisor in C. The number of nonnegative divisors in C is equal to . , xn be a the number of principal divisors div(x) with x ∈ L(c). Let x1 , . .P n basis for L(c) over K. The number of elements of F × of the form i=1 ai xi , with a1 , . . . , an ∈ K is equal to q n − 1. Since div(x) = div(x0 ) if and only if there exists an a ∈ K × such that x0 = ax, the formula follows. Denote the greatest common divisor of the degrees of the divisors of F/K by δ. Eventually we prove that δ = 1. In the meantime notice that a positive integer n is a multiple of δ if and only if there exists a divisor of degree n. Indeed, there are divisors a1 , . . . , ar with δ = gcd(deg(a Pr 1 ), . . . , deg(ar )). For , . . . , a ∈ Z with each multiple n of δ there are a 1 r i=1 ai deg(ai ) = n. The Pr divisor a = i=1 ai ai satisfies deg(a) = n. In particular, since the degree of the canonical divisor is 2g − 2 (Lemma 3.2.2(b)) δ divides 2g − 2. Lemma 4.1.4: Let An be the number of nonnegative divisors of F/K of degree n. If n ≥ 0 is a multiple of δ larger than 2g − 2, then An = h q
n−g+1
q−1
−1
.
Proof: Let n be a multiple of δ with n > 2g − 2. Choose a divisor c of degree n. By Riemann-Roch, dim(c) = n − g + 1. Moreover, the map a 7→ a + c defines a bijection of the set of divisor classes of degree 0 with the set of
4.2 Zeta Functions
79
divisor classes of degree m. Hence, there are h divisor classes of degree n. It n−g+1 follows that, An = h q q−1 −1 .
4.2 Zeta Functions The zeta function of a function field over K is a rational function with coefficients in Q and with simple poles at several points including 1. We define the zeta function of the function field F/K to be the Dirichlet series X (N a)−s (1) ζ(s) = ζF/K (s) = a≥0
where a runs over the nonnegative divisors of F/K. Check the domain of convergence of the series (1) using the substitution t = q −s and the identity N a = q deg(a) . We obtain a power series for ζ(s) in terms of t: (2)
Z(t) =
X
tdeg(a) =
∞ X
An tn ,
n=0
a≥0
where An is the number of nonnegative divisors of degree n. By Lemma 4.1.4 Z(t) =
2g−2 X
An tn + h
n=0
∞ X q mδ−g+1 − 1 mδ t q−1
m=d
where d = 2g−2+δ . The right hand side converges for |t| < q −1 ; (i.e. for δ Re(s) > 1) and (3)
Z(t) = Φ(t) +
hq g−1+δ t2g−2+δ t2g−2+δ h · · − , q−1 1 − (qt)δ q − 1 1 − tδ
where Φ(t) =
2g−2 X
An tn
n=0
is a polynomial of degree ≤ 2g − 2. We summarize: Proposition 4.2.1: The power series Z(t) in (2) converges in the circle |t| < q −1 . Formula (3) continues Z(t) to a meromorphic function on the whole plane. The only poles of Z(t) occur for values of t with tδ = 1 or tδ = q −δ , and they are simple. The Dirichlet series for ζ(s) in (1) converges in the right half plane Re(s) > 1. The substitution t = q −s in (3) continues ζ(s) to a meromorphic function in the whole plane. Like the Riemann zeta function, ζF/K (s) has a multiplicative presentation. Let a1 , a2 , a3 , . . . be a sequence of complex numbers of absolute value
80
Chapter 4. The Riemann Hypothesis for Function Fields
less than 1. We say that the infinite product solutely converges) if the limit n Y
1 n→∞ 1 − ai i=1 lim
Q∞
1 i=1 1−ai
converges (resp. ab-
n Y
1 n→∞ 1 − |ai | i=1
resp. lim
converges to a nonzero complex number. Proposition 4.2.2: If Re(s) > 1 and |t| < q −1 , then Y Y 1 1 = , (4) ζ(s) = −s deg(p) 1 − (N p) 1 − t p p where p runs over the prime divisors of F/K. The product converges absolutely. Therefore, it is independent of the order of the factors. In particular, if Re(s) > 1, then ζ(s) 6= 0. Proof: The prime divisors are free generators of the group of divisors. Thus, for every positive integer m, if Re(s) > 1, then Y N p≤m
∞ Y X X X0 1 = (N p)−sk = (N a)−s + (N a)−s , −s 1 − (N p) a≥0 a≥0 N p≤m k=0
N a>m
N a≤m
where the prime in the second sum means that a runs over all nonnegative divisors with norm exceeding m whose prime divisors p satisfy N p ≤ m. It follows that X Y 1 |(N a)−s | ≤ ζ(s) − 1 − (N p)−s a≥0 N p≤m
N a>m
and the right hand side converges to zero as m tends to infinity. Now we prove that ζ(s) 6= 0 when Re(s) > 1. Indeed, consider a positive integer m. Then, Y Y 1 1 − (N p)−s = ζ(s) 1 − (N p)−s N p>m N p≤m X X00 (N a)−s ≥ 1 − (N a)Re(s) > 0, = 1 + a>0 N a>m
(where the double primes mean summation over all positive divisors a with only prime divisors p satisfying N p > m) if m is large enough. Therefore, ζ(s) 6= 0. Q In particular, p 1−(N p)1 −Re(s) converges. Since, |N p−s | = N p−Re(s) , this Q means that p 1−(N1p)−s absolutely converges. By Exercise 3, the value of the product is independent of the order of the factors.
4.3 Zeta Functions under Constant Field Extensions
81
4.3 Zeta Functions under Constant Field Extensions The analytic properties of the zeta function of a function field F/K proved in Proposition 4.2.2 result in the conclusion that δ = 1 (Corollary 4.3.3). Denote the unique extension of K of degree r by Kr . Then Fr = F Kr is a function field of one variable over Kr . Use r as a subscript to denote the “extension” of objects of F to Fr . For example, if a is a divisor of F/K, then degr a denotes the degree of a as a divisor of Fr /Kr . We have already noted that degr a = deg(a), dimr a = dim(a) and gr = g (Proposition 3.4.2). Lemma 4.3.1: Let p be a prime divisor of F/K. Then p decomposes in Fr as p = P1 + P2 + · · · + Pd , where P1 , P2 , . . . , Pd are distinct prime divisors of Fr /Kr , deg(Pi ) = r−1 · lcm(r, deg(p)) and d = gcd(r, deg(p)). Proof:
Put m = deg(p). Since p is unramified in Fr (Proposition 3.4.2), P1 , . . . , Pd are distinct. Moreover, (Fr )Pi = Kr F¯p . Hence, [(Fr )Pi : K] =
lcm(r, m), and thus deg(Pi ) = [(Fr )Pi : Kr ] = r−1 · lcm(r, m), i = 1, . . . , d. Also, by Propositions 2.6 and 3.4.2, r = [Kr : K] = [Fr : F ] = d · [F¯r,Pi : F¯p ] = d · lcm(r, m)m−1 . Therefore, d = gcd(r, m). Proposition 4.3.2: For every complex number t, Y (1) Zr (tr ) = Z(ξt), ξ r =1
where ξ runs over the rth roots of unity. Proof: Since both sides of (1) are rational functions of t (by (3) of Section 4.2), it suffices to prove (1) for |t| < q −r . First apply the product formula (4) of Section 4.2, then (Lemma 4.3.1): YY Y (1 − tr·deg(P) ) = (1 − tlcm(r,deg(p)) )gcd(r,deg(p)) (2) Zr (tr )−1 = p P|p
(3)
Y ξ r =1
Z(ξt)−1 =
p
Y Y p
(1 − (ξt)deg(p) ).
ξ r =1
Thus, (1) follows if we show equality of the corresponding factors on the right hand sides of (2) and (3). Indeed, for a fixed p let m = deg(p) and d = gcd(r, m). We must show that Y (1 − (ξt)m ). (4) (1 − trm/d )d = ξ r =1
Substitute tm = x−1 in (4) to rewrite it as Y (x − ξ m ) . (5) (xr/d − 1)d = ξ r =1
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Chapter 4. The Riemann Hypothesis for Function Fields
Both monic polynomials in (5) have each (r/d)th root of unity as a zero of multiplicity d. Indeed, if ζr is a primitive root of unity of order r, then ζrm is a primitive root of unity of order r/d and each power of ζrm appears d times (r−1)m among 1, ζrm , ζr2m , . . . , ζr . Therefore, the polynomials are equal. Corollary 4.3.3 (F. K. Schmidt): δ = 1. P∞ Proof: By (2) of Section 4.2, Z(t) = m=0 Amδ tmδ . Hence, if ξ δ = 1, then Z(ξt) = Z(t). From (1), Zδ (tδ ) = Z(t)δ . However, by (3) of Section 4.2 (applied to Zδ (tδ ) instead of to Z(t)), Zδ (tδ ) has a simple pole at t = 1, while Z(t)δ has a pole of order δ at t = 1. Consequently, δ = 1. Corollary 4.3.4: For every integer n there are exactly h divisor classes of F/K of degree n. Proof: By Corollary 4.3.3 there is a divisor c of degree n. The map a → a+c induces a bijection of the set of divisor classes of F/K of degree 0 onto the set of divisor classes of degree m. Hence, the number of elements in the latter set equals the number of elements in the former set, namely h. Example 4.3.5: By definition, A1 is the number of prime divisors of F/K of degree 1. If g = 0, then h = 1 and A1 = q +1 (Lemma 4.1.4). Thus, F/K has prime divisors of degree 1. By Example 3.2.4, F = K(x) is a rational function field. If δ = 1 (Corollary 4.3.3) and g = 1, then A1 = h (Lemma 4.1.4); in other words the class number is equal to the number of prime divisors of degree 1. Since this is the order of a group (Definition 4.1.1 ), it is again 1 . positive. By Lemma 4.1.4 and by (2) of Section 4.1.2, Z(t) = (1−t)(1−qt)
4.4 The Functional Equation Like the Riemann zeta functions, Z(t) satisfies a functional equation relating 1 . The main tool in the proof is the Riemann-Roch its values in t and qt theorem. Proposition 4.4.1: Z(t) satisfies the functional equation 1 √ √ ( qt)1−g Z(t) = ( qt)g−1 Z . qt Proof:
If g = 0, the result follows from the explicit presentation Z(t) = in Example 4.3.5. Therefore, assume g > 0. The basic idea is to split Z(t) into the sum of a polynomial P (t) and an infinite series Q(t), each of which satisfies the same functional equation in the statement of the proposition. Apply Lemmas 4.1.3 and 4.1.4 with δ = 1 (Corollary 4.3.3) to obtain: 1 (1−t)(1−qt)
Z(t) =
X a≥0
tdeg(a) =
2g−2 X deg(C)=0
X a≥0 a∈C
tdeg(a) +
∞ X n=2g−1
An tn
4.4 The Functional Equation
=
2g−2 X deg(C)=0
=
h 1 q−1
83
∞ X q dim(C) − 1 deg(C) q n−g+1 − 1 n t t + h q−1 q−1 n=2g−1 2g−2 X
q dim(C) tdeg(C)
i
deg(C)=0
+
∞ ∞ h h X h X ni q n−g+1 tn − t q − 1 n=2g−1 q − 1 n=0
= P (t) + Q(t), where P (t) (resp. Q(t)) is the expression in the first (resp. second) brackets. We have also used that F/K has exactly h divisor classes of each degree (Corollary 4.3.4). First we analyze P (t). The Riemann-Roch theorem relates dim(C) to dim(W − C), where W is the canonical class. Recall that deg(W ) = 2g − 2 (Lemma 3.2.2(b)). Hence 1 1 deg(C) = deg(C) + 1 − g + dim(W − C) 2 2 1 = dim(W − C) − deg(W − C). 2 As C varies over all divisor classes of degree between 0 and 2g − 2 so does W − C. Hence √ 2g−2 ( qt)2−2g X √ 2−2g ( qt) P (t) = q dim(C) tdeg(C) q−1 dim(C) −
deg(C)=0
=
=
=
1 q−1 1 q−1 1 q−1
2g−2 X
1 √ q dim(C)− 2 deg(C) ( qt)2−2g+deg(C)
deg(C)=0 2g−2 X
1 √ q dim(W −C)− 2 deg(W −C) ( qt)− deg(W −C)
deg(C)=0 2g−2 X
0
0
q dim(C ) (qt)− deg(C ) = P
deg(C 0 )=0
1 . qt
Now evaluate the geometric series involved in the expression for Q(t): ∞ ∞ h h q g t2g−1 h h X n−g+1 n X n i 1 i Q(t) = − q t − t = q − 1 n=2g−1 q − 1 1 − qt 1−t n=0 A direct computation shows that √
( qt)
2−2g
1 . Q(t) = Q qt
84
Chapter 4. The Riemann Hypothesis for Function Fields
This completes the proof of the proposition.
4.5 The Riemann Hypothesis and Degree 1 Prime Divisors We reformulate here the Riemann hypothesis for a function field F of one variable over a finite field and draw an estimate for the number of prime divisors of F of degree 1. Rewrite formula (3) of Section 4.2 with δ = 1 (Corollary 4.3.3) as Z(t) = Φ(t) +
t2g−1 h t2g−1 hq g · − · q − 1 1 − qt q − 1 1 − t
where Φ(t) a polynomial of degree ≤ 2g − 2. Hence Z(t) =
L(t) (1 − t)(1 − qt)
with L(t) = a0 + a1 t + · · · + a2g t2g a polynomial with rational coefficients. We determine some of these coefficients: First: a0 = L(0) = Z(0) = A0 = 1, since the zero divisor is the only nonnegative divisor of degree 0. Second: A1 is equal to the number of prime divisors of F/K of degree 1. Write N = A1 , so that L(t) = (1 − t)(1 − qt)
∞ X
An tn ≡ 1 + (N − (q + 1))t mod t2 .
n=0
Therefore, a1 = N − (q + 1)
(1)
Now let x ∈ F be transcendental over K and write F0 = K(x). The Zeta 1 (Example 4.3.5). By Proposition function of F0 /K is Z0 (t) = (1−t)(1−qt) 4.4.1 1 q g−1 t2g−2 Z qt Z(t) 1 g 2g (2) L(t) = = −1 −2 . t L = q 1 Z0 (t) qt q t Z0 qt This functional equation for L(t), written explicitly, has the form 2g X i=0
ai ti =
2g X
a2g−i q i−g ti .
i=0
Equivalently, ai = q i−g a2g−i . In particular, deduce for i = 0 and i = 1 that (3) a2g = q g and a2g−1 = q g−1 N − (q + 1) .
4.5 The Riemann Hypothesis and Degree 1 Prime Divisors
85
These formulas imply that deg(L(t)) = 2g. Decompose L(t) over C as (4)
L(t) =
2g Y
(1 − ωi t)
i=1
where the ωi−1 ’s are the zeros of L(t). Formulas (1), (3), and (4) give (5)
qg =
2g Y
ωi
and N − (q + 1) = −
i=1
2g X
ωi .
i=1
Moreover, the functional equation (2) for L(t) implies that 1 ωi = 0. L = 0 if and only if L ωi q √ √ Rename the roots ω1 , . . . , ω2g as ω1 , ω10 , . . . , ωf , ωf0 , q, . . . , q, √ √ − q, . . . , − q with f ≤ g such that ωi ωi0 = q, i = 1, . . . , f , and q √ (resp. − q) appear k (resp. l) times. Then 2f + k + l = 2g shows that if k is odd, then l is odd. In this case (5) gives q g = q f q k/2 (−1)l q l/2 = −q g , a contradiction. Hence, both k and l are even and we may take f = g. Thus, √
L(t) =
g Y
(1 − ωi t)(1 − ωi0 t),
i=1
with ωi ωi0 = q for i = 1, . . . , g. Here is a reformulation of the Riemann hypothesis for the function field F/K. Sections 4.6 - 4.8 complete the proof. Theorem 4.5.1: (a) The zeros of the function ζF/K (s) lie on the line Re(s) = 12 . 1
(b) The zeros of the function ZF/K (t) lie on the circle |t| = q − 2 . (c) If the ωi are the inverses of the zeros of the polynomial LF/K (t), then |ωi | =
√ q,
i = 1, 2, . . . , 2g.
Note that (c) is equivalent to (b), since the poles of Z(t) are t = 1 and t = q −1 . Theorem 4.5.1 with (5) provides an estimate on the number of prime divisors of degree 1: Theorem 4.5.2: Let F be a function field of one variable over a finite field K of q elements. Denote the genus of F/K by g and let N be the number of √ prime divisors of F/K of degree 1. Then |N − (q + 1)| ≤ 2g q.
86
Chapter 4. The Riemann Hypothesis for Function Fields
4.6 Reduction Steps Theorem 4.5.2 is a consequence of the Riemann hypothesis. As a first step, this section shows that an appropriate version of Theorem 4.5.2 implies the Riemann hypothesis. As in Section 4.3, denote the unique extension of K of degree r by Kr . Lemma 4.6.1: The Riemann hypothesis holds for the function field F/K if and only if it holds for the function field Fr /Kr . Proof: Use (2) and (4) of Section 4.5 to express L(t). Then apply Proposition 4.3.2 to compute Lr (tr ): Lr (tr ) =
Y Z(ξt) Y Zr (tr ) = = L(ξt) r Zr,0 (t ) Z0 (ξt) r r ξ =1
ξ =1
=
2g Y Y ξ r =1
(1 − ωi ξt) =
i=1
2g Y
(1 − ωir tr ).
i=1
Q2g Hence, Lr (t) = i=1 (1 − ωir t). Thus, r (1) ω1r , . . . , ω2g are the inverses of the zeros of Lr . √ √ Since |ωi | = q if and only if |ωir | = q r , the lemma follows.
Denote the number of prime divisors of Fr /Kr of degree 1 by Nr . Lemma 4.6.2: Let F be a function field of one variable over a field K of q elements. If there exists a constant c such that |Nr − (q r + 1)| ≤ cq r/2 for every positive integer r, then the Riemann hypothesis holds for F/K. log Proof: Apply the differential operator D = −t ddt to both sides of the Q2g formula L(t) = i=1 (1 − ωi t):
(2)
D(L(t)) =
2g X i=1
2g ∞ X X ωi t = ωir tr 1 − ωi t n=1 r=1
P2g Combine (1) with (5) of Section 4.5 to obtain − i=1 ωir = Nr − (q r + 1). P2g r The hypothesis of the lemma thus implies | i=1 ωi | ≤ cq r/2 . Therefore 1 the radius of convergence R of the right hand side of (2) satisfies R ≥ q − 2 . But (2) implies that the ωi−1 are the only singularities of D(L(t)). Hence, √ R = min1≤i≤2g |ωi−1 |. Therefore, |ωi | ≤ q for i = 1, . . . , 2g. This together Q2g √ with the equality q g = i=1 ωi ((5) of Section 4.5), implies that |ωi | = q for i = 1, . . . , 2g.
4.7 An Upper Bound
87
4.7 An Upper Bound Assume, by extension of constants if necessary, that K and F satisfy these conditions: (1a) q = a2 is a square; (1b) q > (g + 1)4 ; and (1c) F has a prime divisor o of degree 1. By Lemma 4.6.1, a proof of the Riemann hypothesis under these conditions suffices for the general case. We prove a result that has, as a special case, the inequality √ N − (q + 1) < (2g + 1) q.
(2)
Let σ be an automorphism of F over K. It induces a permutation of the prime divisors of F/K. If p is a prime divisor of F/K, the pσ is the prime divisor corresponding to the place ϕσp (x) = ϕp (σx) for x ∈ F (where, as in Section 2.1, ϕσp = σ −1 ϕp ). Also, recall that the map x 7→ xq is an automorphism F˜ over K. Hence, the formula ϕqp (x) = ϕp (x)q defines a place ϕqp of F/K. Although ϕqp is equivalent to ϕp , it is convenient to use ϕqp , because ϕp = ϕqp if and only if deg(p) = 1. The remainder of this section investigates the expression N (σ) =
X
deg(p)
q ϕσ p =ϕp
in order to show that (3)
√ N (σ) − (q + 1) < (2g + 1) q.
If σ is the identity automorphism, then N (σ) = N and (3) becomes (2). With the notation m = a − 1, n = a + 2g, and r = m + an, rewrite (3) as (4)
N (σ) − 1 ≤ r.
With o as in (1c), consider the ascending sequence of K-vector spaces L(o) ⊆ L(2o) ⊆ L(3o) ⊆ · · ·. By Exercise 2 of Chapter 3, (5)
dim(L(io)) − dim(L((i − 1))o) ≤ 1.
88
Chapter 4. The Riemann Hypothesis for Function Fields
With k a positive integer, let Ik be the set of i, 1 ≤ i ≤ k for which equality holds in (5). For each i ∈ Ik , choose ui ∈ L(io)−L((i−1)o). Then div∞ (ui ) = io and the system {ui | i ∈ Ik } is a basis for L(ko). In particular, this holds for k = m. Since a2 = q, a is a power of char(K). Thus, the set L(no)a = {y a | y ∈ L(no)}, consisting of elements in the field F a is a Kvector space with basis {uaj | j ∈ In } and the same dimension as L(no). Therefore, o nX ui yia | yi ∈ L(no) L= i∈Im
is a K-vector space generated by the set U = {ui uaj | i ∈ Im and j ∈ In }. Lemma 4.7.1: The set U is linearly independent over K. Proof: It suffices to prove that {ui | i ∈ Im } is linearly independent over a . Indeed, assume that there exist yi ∈ F , i ∈ Im , not all zero, the field FP a such that i∈Im ui yi = 0. Then there exist distinct i, j ∈ Im such that a a vo (ui yi ) = vo (uj yj ), yi 6= 0, yj 6= 0 ((2) of Section 2.1). Thus, −i+avo (yi ) = −j + avo (yj ), so i ∼ = j mod a. Since 1 ≤ i, j ≤ m < a, this is a contradiction. By Lemma 4.7.1, dim(L) = dim(L(mo)) · dim(L(no)). Apply RiemannRoch to the right hand side terms: (6)
dim(L) ≥ (m − g + 1)(n − g + 1) = q +
√ q − g(g + 1).
Now consider the K-vector space L0 =
nX
o (σ −1 ui )a yi | yi ∈ L(no) .
i∈Im
Check that L0 ⊆ L(maoσ + no) and deg(maoσ + no) = q + 2g > 2g − 2. By Riemann-Roch, (7)
dim(L0 ) ≤ dim(maoσ + no) = deg(maoσ + no) − g + 1 = q + g + 1.
√ By (1b), q − g(g + 1) > g + 1. Thus, the right side of (6) is greater than the right side of (7). Therefore, dim(L0 ) < dim(L).
(8)
Finally, define an additive map σ ∗ from L into L0 by σ∗
X i∈Im
X ui yia = (σ −1 ui )a yi i∈Im
4.8 A Lower Bound
89
By (8), the kernel of σ ∗ is nontrivial. Hence, there exist yi ∈ L(no), i ∈ Im , not all zero, with X (9) (σ −1 ui )a yi = 0. i∈Im
In particular, u = i∈Im ui yia is a nonzero element of L(ro). If p is a prime divisor of F/K and p 6= o, then ϕp (yi ) 6= ∞ and ϕp (ui ) 6= ∞. If in addition ϕσp = ϕqp , then (9) implies X X ϕp (u) = ϕp (ui )ϕp (yi )a = ϕp (σ −1 ui )q ϕp (yi )a P
i∈Im
= ϕp
i∈Im
X
(σ −1 ui )a yi
a
=0
i∈Im
Hence, p occurs in the divisor of zeros of u. In other words, X p ≤ div0 (u). p6=o pσ =pq
Thus, N (σ) − 1 ≤ deg(div0 (u)) = deg(div∞ (u)) ≤ r and the proof of (4) is complete.
4.8 A Lower Bound In the notation of Section 4.7, we establish a lower bound inequality of the √ form N (σ) − (q + 1) > c0 q where c0 is an explicit constant depending only on F . This will complete the construction of Lemma 4.6.2 toward the proof of the Riemann hypothesis for F/K. Lemma 4.8.1: Let F be a function field of one variable over a field K with q elements. Let F 0 be a finite Galois extension of F with a Galois group G such that K algebraically closed in F 0 and let σ ∈ G. Then N (σ) (F ) = Pis also (στ 0 −1 ) (F 0 ). [F : F ] τ ∈G N Proof: Let p0 be a prime divisor of F 0 /K and let p be its restriction to F . q Then ϕσp = ϕqp if and only if there exists τ ∈ G such that ϕστ p0 = ϕp0 . The 0 number of such τ is the ramification index ep of p over p (Section 2.3). Put fp0 /p = [F¯p0 0 : F¯p ] and denote the number of prime divisors of F 0 lying over p by gp0 /p . Then: X X X X X N (στ ) (F 0 ) = deg(p0 ) = ep0 /p deg(p0 ) q τ ∈G ϕστ 0 =ϕ 0
τ ∈G
p
=
X
p
q 0 ϕσ p =ϕp p |p
ep0 /p fp0 /p gp0 /p deg(p)
q ϕσ p =ϕp
= [F 0 : F ]
X q ϕσ p =ϕp
deg(p) = [F 0 : F ]N (σ) (F ).
90
Chapter 4. The Riemann Hypothesis for Function Fields
Let F be a function field of one variable over a field K of q elements and let σ be an automorphism of F over K of finite order. Denote the fixed field of σ in F by E. Then F is a finite Galois extension of E. As a finite field, K is perfect. Hence, there exists x ∈ E, transcendental over K, such that E is a finite separable extension of K(x). Let Fˆ be the Galois closure of F/K(x) ˆ be the algebraic closure of K in Fˆ . Then Fˆ , as well as F K, ˆ are and let K ˆ and σ extends to an automorphism of function fields of one variable over K ˆ ˆ (if necessary), assume Fˆ over K(x). After an additional finite extension of K ˆ and therefore for F K/ ˆ K ˆ that Condition (1) of Section 4.7 holds for Fˆ /K, (use Proposition 3.4.2). Starting with a given F we have extended the field of constants so as to assume these conditions: (1a) F/K has a separating transcendence element x; the field F has a finite extension Fˆ , which is Galois over K(x), and K is algebraically closed in Fˆ ; ˆ and (1b) q is a square larger than (ˆ g + 1)4 where gˆ is the genus of Fˆ /K; ˆ has a prime divisor of degree 1. (1c) Fˆ /K Lemma 4.8.2: Under these conditions N (σ) (F ) − (q + 1) ≥ −
(2)
n−m √ (2ˆ g + 1) q, m
where m = [Fˆ : F ] and n = [Fˆ : K(x)]. Proof: Let H = Gal(Fˆ /F ) and G = Gal(Fˆ /K(x)). From Lemma 4.8.1 (3)
N (σ) (F ) =
1 X (στ ) ˆ 1 X (θ) ˆ N (F ) and q + 1 = N (K(x)) = N (F ). m n τ ∈H
θ∈G
Apply inequality (3) of Section 4.7: X
N (θ) (Fˆ ) =
X
N (στ ) (Fˆ ) +
τ ∈H
θ∈G
≤
X
=
N (θ) (Fˆ )
θ∈G r σH
N (στ ) (Fˆ ) +
τ ∈H
X
X X
√ (q + 1 + (2ˆ g + 1) q)
θ∈G r σH
N
(στ )
√ (Fˆ ) + (n − m)(q + 1 + (2ˆ g + 1) q).
τ ∈H
From the second half of (3), X τ ∈H
√ N (στ ) (Fˆ ) ≥ n(q + 1) − (n − m)(q + 1 + (2ˆ g + 1) q) √ = m(q + 1) − (n − m)(2ˆ g + 1) q.
Exercises
91
Thus, the first half of (3) implies that N (σ) (F ) ≥ (q + 1) −
n−m √ (2ˆ g + 1) q. m
Note that the condition (1), as well as the numbers gˆ, m, n, are independent of extension of the field of constants. We may therefore combine Lemma 4.8.2 with the results of Section 4.7 to conclude: Proposition 4.8.3: Let F be a function field of one variable over a finite field K and let σ be an automorphism of F over K of finite order. Then K has a finite extension K 0 with q 0 elements and there exists a positive constant c such that for every positive integer r we have |N (σ) (Fr0 ) − ((q 0 )r − 1)| ≤ c(q 0 )r/2 , where Kr0 is the unique extension of K 0 of degree r, Fr0 = F 0 Kr0 , and σ extends to an automorphism, also denoted by σ, of Fr0 over Kr0 . In particular, Proposition 4.8.3 is valid in the case σ = 1. By Lemma 4.6.2, the Riemann hypothesis is true for the function field F 0 /K 0 . It follows from Lemma 4.6.1 that it is also true for F/K.
Exercises 1.
For real valued functions f, g write f (x) = O(g(x)) as x → a
if there exists a positive constant c such that |f (x)| ≤ c|g(x)| for all values of x in a neighborhood of a. Let F be a function field of one variable over a field K of q elements. Denote the set of prime divisors of F/K of degree r by Pr (F/K). Follow these instructions to prove that |Pr (F/K)| =
1 r q + O(q r/2 ) r
(a) Use Theorem 4.5.2 to prove that |P1 (Fr /Kr )| = q r + O(q r/2 ). (b) Observe that if P ∈ P1 (Fr /Kr ) and if p is the prime divisor of F/K that lies below P, then deg(p)|r. (c) Deduce from Lemma 4.3.1 that if d|r, then over each p ∈ Pd (F/K) there lie exactly d elements of P1 (Fr /Kr ). Thus, X |P1 (Fr /Kr )| = d|Pd (F/K)|. d|r
(d) Use the estimates |Pd (F/K)| = O(q d ) and d ≤ 2r for proper divisors d of r r r and the inequality q + q 2 + · · · + q 2 ≤ 2q 2 to complete the proof.
92
Chapter 4. The Riemann Hypothesis for Function Fields
2. Let F be a function field of one variable over a field K of q elements and genus g. Let σ be an automorphism of F over K of finite order and let N (σ) √ be as in Section 4.7. Prove that |N (σ) − (q + 1)| ≤ 2g q. Hint: Extend σ to ˜ K ˜ by σ ˜ Then let E be an automorphism σ ˜ of F K/ ˜ x = xq for each x ∈ K. ˜ the fixed field of σ ˜ in F K. Show that E is a function field of one variable ˜ = F K. ˜ over K and that E K 3. This exercise establishes basic facts about infinite products which lie behind the multiplicative presentation of the zeta functions. Consider a se∞ quence {zi }Q i=1 of nonzero complex numbers. If the sequence of the partial n to a nonzero complex number z, we say that products i=1 zi converges Q∞ z converges and the infinite product i=1 i Q∞that z is its value. We say that Qn (1 + z ) absolutely converges if i i=1 i=1 (1 + |zi |) converges. The logarithm function makes a connection between the theory of infinite products and the theory of infinite series: Proposition ([Knopp, p. 434]): Q∞ Let ai 6= −1, i = 1, 2, 3, . . . be complex numbers. Then the product i=1 (1 + ai ) converges if and only if, the series P∞ i=1 log(1+ai ) whose terms are the Q∞principal values of log(1+ai ) converges. If l is the sum of this series, then i=1 (1 + ai ) = el . (a) Prove that if |z| ≤ 12 , then |z| ≤ 2| log(1 + z)| ≤ 2|z|. (b) Suppose 0 < |ai | ≤ 12 for i = 1, 2, 3, . . . and consider the following series: ∞ ∞ ∞ X X X | log(1 + ai )|, |ai |, log(1 + |ai |). i=1
i=1
i=1
Use (a) to prove that the convergence of each these series implies the convergence of the two others. Q∞ − |ai |)−1 con(c) Suppose 0 < |ai | ≤ 12 for i = 1, 2, 3, . . . and i=1 Q(1 ∞ verges. Then for each permutation π of N, the product i=1 (1 − aπ(i) )−1 converges and its value is independent of π. 4. Let F be a function field of one variable over a field K with q elements. Consider σ ∈ Aut(F/K). As in Section 4.7, define N (σ) to be the sum of all deg(p), where p ranges over all prime divisors of F/K for which ϕσp = ϕqp . ˜ by the formula σ ˜ Extend σ to an automorphism of F K ˜ x = xq for each x ∈ K. Use Lemma 4.3.1 for r a multiple of deg(p) to prove that N (σ) is also the ˜ K ˜ for which ϕσ˜ = ϕq . number of prime divisors P of F K/ P P 5. Let F be a function field of one variable over a field with q elements and of genus g. Let σ be an element of Aut(F/K) of order n. Define N (σ) as in Exercise 4. Prove that √ |N (σ) − (q + 1)| ≤ 2g q. ˜ by F 0 . Prove Hint: Let σ ˜ be as in Exercise 4. Denote the fixed field of σ ˜ in F K 0 ˜ in F Kr for each multiple r of n. Deduce that F is also the fixed field of σ
Notes
93
˜ = F 0 K. ˜ Use that F 0 is a function field of one variable over K such that F K Exercise 4 to prove that N (σ) is the number of prime divisors of F 0 /K of degree 1. Then apply Theorem 4.5.2. 6. Let q be a power of an odd prime. Prove that ax2 + by 2 = c has a solution (x, y) ∈ F2q if a, b, c ∈ F× q . Let (x0 , y0 ) be one of these solutions. Use the substitution x = x0 + tz, y = y0 + t and solve for t in terms of z to show that the function field of aX 2 + bY 2 − c = 0 is isomorphic to Fq (z). Conclude that the function field has exactly q + 1 degree 1 places. Hint: Assume a = 1 and observe that c − by 2 takes on (q + 1)/2 values in Fq . On the other hand there are q+1 2 squares in Fq . 7. Let q be a prime power ≡ 1 mod 3 and let α be a generator of F× q . Follow these instructions to prove that αX 3 + α2 Y 3 = 1 has a solution in Fq : Let Ai = {αi x3 | x ∈ F× q }, i = 0, 1, 2 and note that αA0 = A1 , αA1 = A2 and αA2 = A0 . Let Ai + Aj = {x + y | x ∈ Ai ∧ y ∈ Aj }. Now assume that A1 + A2 ⊆ A1 ∪ A2 to obtain a contradiction. For this use the notation A1 + A2 = m1 A1 ∪ m2 A2 to indicate that for each ai ∈ Ai there are mi pairs (x, y) ∈ A1 × A2 such that x + y = ai (independent of ai ∈ Ai ), i = 1, 2. Multiply this “equality” by α (resp. α2 ) to compute A2 + A0 (resp. A0 + A1 ). For a0 ∈ A0 , use this to compute the number of triples (x, y, z) ∈ A0 ×A1 ×A2 such that x + y + z = a0 in two ways: first compute (A0 + A1 ) + A2 and then A0 + (A1 + A2 ). The resulting expressions for m1 and m2 will lead to the contradiction m1 = m2 = 0. The attractiveness of the Riemann hypothesis for curves over finite fields has resulted in extensive lists of problems in a number of books treating the combinatorics of finite fields. For the sake of completeness we reference two such problem sources: [Ireland-Rosen, Chap. 8, pp. 105–107; Chap. 11, pp. 169–171] [Lidl-Niederreiter, see notes, Chap. 6, pp. 339–346]. In those chapters that use the Riemann hypothesis, our material and problems will tend to concentrate on the connections between the Riemann hypothesis and arithmetic properties of fields that are special to this book (e.g. an explicit form of Hilbert’s irreducibility theorem for global fields that follows from the Riemann hypothesis - Theorem 13.3.4).
Notes An extensive survey of the literature giving estimates on the number of points on an affine variety V appears in [Lidl-Niederreiter, pp. 317–339]. Although considerable literature on the Riemann hypothesis for curves over finite fields (Theorem 4.5.1) existed long before the two proofs given by Weil [Weil2], subsequent concerns included two sophisticated — and interrelated — developments. Both of Weil’s proofs employed elements of the theory of algebraic geometry outside the domain of algebraic curves. Indeed, Weil’s far reaching generalization of Theorem 4.5.1 was suggested to him
94
Chapter 4. The Riemann Hypothesis for Function Fields
by the latter proof. This generalization, now called the Riemann hypothesis for nonsingular projective varieties over finite fields (proved by Deligne in [Deligne]). Since, however, Weil’s theorem had so many applications to apparently elementary results about finite fields, many practitioners were anxious for a more accessible proof. Stepanov [Stepanov1] was the first to make serious progress on an elementary proof (e.g. not applying the theory of algebraic surfaces) of Weil’s result. He introduced elements of diophantine approximation to the problem in the case of hyperelliptic curves, in a style suggested by the Thue-SiegelRoth theorem. To do this he constructed an auxiliary nonvanishing function on the hyperelliptic curve with a “lot” of prescribed zeros of “high” multiplicity. Eventually [Stepanov2] realized Theorem 4.5.1 for all hyperelliptic curves over finite fields. Continuing with the purely diophantine approximation approach, Stepanov [Stepanov3], for prime fields (and some extra conditions on the equations) and W. M. Schmidt [W.M.Schmidt1 and W.M.Schmidt2] were able to prove Theorem 4.5.1 for all curves. Indeed, Schmidt’s method were even applicable to prove results like those of Deligne even for some complete intersections. Our proof, however, of Theorem 4.5.1 follows [Bombieri1]. In this proof the Riemann hypotheses replaces diophantine approximation to give the construction of the auxiliary functions that appear in Stepanov’s proof (the function u at the end of Section 4.7). Voloch [Voloch] has an elementary proof of the Riemann hypothesis for function fields that sometimes gives a better bound than Weil’s estimate (see also [St¨ohr-Voloch]).
Chapter 5. Plane Curves The estimate on the number of prime divisors of degree 1 of a function field F over Fq (Theorem 4.5.2) leads in this chapter to an estimate on the number N of K-rational zeros of an absolutely irreducible polynomial f ∈ Fq [X, Y ]. √ We prove (Theorem 5.4.1) that |N + (q + 1)| ≤ (d − 1)(d − 2) q + d, where d = deg(f ).
5.1 Affine and Projective Plane Curves Let K be a field and let Ω be an algebraically closed extension of K of infinite transcendence degree over K. We denote by A2 the affine plane: all pairs (x, y) ∈ Ω2 . We denote by P2 the projective plane: all nonzero triples x = (x0 , x1 , x2 ) ∈ Ω3 modulo the equivalence relation x ∼ x0 if and only if x0 = cx for some c ∈ Ω. We denote the equivalence class of (x0 , x1 , x2 ) by (x0 :x1 :x2 ). Embed the affine plane A2 in P2 by the map (x, y) → (1:x:y). With this understood, the points of A2 are then referred to as the finite points on P2 , whereas the points of the form (0:x1 :x2 ) are the points at infinity on P2 . An affine plane curve defined over K is a set (1)
Γ = {(x, y) ∈ A2 | f (x, y) = 0},
where f ∈ K[X, Y ] is a nonconstant absolutely irreducible polynomial. Write it in the form (2)
f (X, Y ) = fd (X, Y ) + fd−1 (X, Y ) + · · · + f0 (X, Y )
where fk (X, Y ) is a homogeneous polynomial of degree k, for k = 0, . . . , d, and fd (X, Y ) 6= 0. Then d is the degree of Γ. Attach to f the homogeneous polynomial f ∗ (X0 , X1 , X2 ) of degree d: (3) f ∗ (X0 , X1 , X2 ) = fd (X1 , X2 ) + X0 fd−1 (X1 , X2 ) + · · · + X0d f0 (X1 , X2 ) and let Γ∗ = {(x0 :x1 :x2 ) ∈ P2 | f ∗ (x0 , x1 , x2 ) = 0} Then Γ∗ is the projective plane curve corresponding to Γ. It is also called the projective completion of Γ. We have Γ = Γ∗ ∩ A2 , and Γ∗ r Γ is a finite set corresponding to the points of fd (X1 , X2 ) = 0 in P1 . The infinite points of Γ∗ are sometimes referred to as the points at infinity on Γ. If f (X, Y ) = a+bX+cY with a, b, c ∈ K and b 6= 0 or c 6= 0 (i.e. deg(f ) = 1), then Γ is a line. The points on the corresponding projective line satisfy aX0 + bX1 + cX2 = 0. The line at infinity is given by X0 = 0.
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Chapter 5. Plane Curves
Call a point (x, y) of an affine curve Γ generic if trans.degK K(x, y) = 1. Because f is absolutely irreducible, the field F = K(x, y) is a regular extension of K. Thus, F is a function field of one variable which we call the function field of Γ over K. Up to K-isomorphism it is independent of the choice of the generic point. The map (X, Y ) 7→ (x, y) extends to a K-epimorphism of rings K[X, Y ] → K[x, y] with f (X, Y )K[X, Y ] as the kernel (Gauss’ lemma). Thus, a polynomial g ∈ K[X, Y ] vanishes on Γ if and only if g(x, y) = 0, or, equivalently, g is a multiple of f . Define the genus of Γ (and of Γ∗ ) to be the genus of F . The coordinate ring of Γ over K is R = K[x, y]. There is a bijective correspondence between the set of all maximal ideals m of R with quotient field R/m = K and the set Γ(K) of all K-rational points of Γ. If (a, b) ∈ Γ(K), then the corresponding maximal ideal of R is m = {g(x, y) ∈ R | g(a, b) = 0}. If p is a prime ideal of R and p ⊆ m, then the transcendence degree of the quotient field of R/p is either 0 or 1. In the latter case p = 0, and in the former case R/p is already a field so that p = m. The local ring of Γ at a = (a, b) over K is the local ring of R at m: OΓ,(a,b),K = Rm =
o n g(x, y) g(x, y), h(x, y) ∈ R and h(a, b) 6= 0 . h(x, y)
The unique nonzero prime ideal of Rm is generated by the elements of m. As a local ring of a Noetherian domain, Rm is itself a Noetherian domain. Similarly, a point (x0 :x1 :x2 ) of Γ∗ with x0 6= 0 is said to be generic over K, if xx10 , xx20 is a generic point of Γ. Define the local ring of a point a = (a0 :a1 :a2 ) of Γ∗ over K as OΓ∗ ,a,K =
n g(x , x , x ) g, h are homogeneous polynomials of o 0 1 2 h(x0 , x1 , x2 ) the same degree and h(a0 , a1 , a2 ) 6= 0
If a0 6= 0, then this ring coincides with the local ring of the corresponding point aa10 , aa20 of Γ. If a1 6= 0, then regard a as a point on the affine curve Γ1 , defined by the equation f ∗ (X0 , 1, X2 ) = 0. Then the projective completion of Γ1 is also Γ∗ , but the line X1 = 0 is taken as “the line at infinity”. This shows that the local ring of a projective plane curve at each point is equal to the local ring of some affine curve at a point. It is therefore a Noetherian domain. In any case, call a K-rational point of Γ (or of Γ∗ ) K-normal if its local ring over K is integrally closed. Two affine plane curves Γ1 and Γ2 are K- isomorphic if their coordinate rings are K-isomorphic. In this case generators of the coordinate ring of Γ2 (resp. Γ1 ) can be expressed as polynomials in the generators of the coordinate ring of Γ1 (resp. Γ2 ), and the composition of these two polynomial maps is the identity when applied to generators of the coordinate ring of Γ1 (resp. Γ2 ).
5.2 Points and Prime Divisors
97
Two projective plane curves Γ∗1 and Γ∗2 are K-isomorphic if (4a) for each x ∈ Γ∗1 there exist homogeneous polynomials g0 , g1 , g2 in K[X0 , X1 , X2 ] of the same degree such that yi = gi (x) 6= 0 for at least one i, 0 ≤ i ≤ 2, y ∈ Γ∗2 , and there exist homogeneous polynomials h0 , h1 , h2 in K[X0 , X1 , X2 ] of the same degree such that hi (y) = xi , i = 0, 1, 2; and (4b) the same condition with the roles of Γ∗1 and Γ∗2 exchanged. For example, if g0 , g1 , g2 are linear polynomials with a nonsingular coefficient matrix, then (g0 , g1 , g2 ) is called nonsingular homogeneous linear transformation. The function fields of two isomorphic plane curves are K-isomorphic. So are the local rings of corresponding points. It follows that the genera of isomorphic plane curves are the same; and if p1 and p2 are corresponding points of the curves, then p1 is K-normal if and only if p2 is K-normal. In particular, both curves have the same number of nonnormal points. If Γ is a plane curve defined over a field K and L is an algebraic extension of K, then Γ is also defined over L. The function field FL of Γ over L is the extension of F by the field of constants L. By Proposition 3.4.2, the genus of Γ remains unchanged if L is separable over K.
5.2 Points and Prime Divisors Let Γ be an affine plane curve of degree d defined by an absolutely irreducible equation f (X, Y ) = 0 over a field K. We establish a bijective correspondence between the K-rational points of f and the prime divisors of the function field of f of degree 1 (Lemma 5.2.2). We prove that a K-rational of Γ is normal if and only if it is simple (Lemma 5.2.3). Let (x, y) be a generic point of Γ over K. Denote the coordinate ring and the function field, respectively, of Γ over K by R = K[x, y] and F = K(x, y). Consider a K-rational point (a, b) of Γ. Then the map (x, y) 7→ (a, b) uniquely extends to a K-homomorphism ϕ of the local ring O(a,b) into K. The ˜ homomorphism ϕ extends further (not necessarily uniquely) to a K-valued 0 place ϕ of F . Call (a, b) the center of the corresponding prime divisor p of F/K. Then p lies over the unique prime divisor p0 of K(x)/K determined by the map x 7→ a. Since [F : K(x)] ≤ d, there exist at most d prime divisors p of F/K with the point (a, b) as a center on Γ. Since each point of Γ∗ is a finite point of some affine representative of Γ∗ (Section 5.1) this holds for each K-rational point of the projective completion Γ∗ of Γ. If a point p of Γ∗ (K) is K-normal, then its local ring Op is a Noetherian integrally closed domain. By Section 5.1 , each nonzero prime ideal of Op is maximal. Hence, by Proposition 2.4.5, Op is a discrete valuation domain. Therefore, there exists a unique prime divisor p of F/K with p as a center on Γ. The degree of p is 1. Conversely, consider a prime divisor p of F/K of degree 1. Suppose that ϕp is finite on R. Then (a, b) = ϕp (x, y) is a K-rational point of Γ and it is
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Chapter 5. Plane Curves
the center of p on Γ. The following result shows that the number of prime divisors p of F/K which are not finite on R does not exceed d. Proposition 5.2.1 (Noether’s Normalization Theorem): Let K[x1 , . . . , xn ] be a finitely generated integral domain over a field K with a quotient field F . If the transcendence degree of F over K is r, then there exist elements t1 , . . . , tr in K[x] such that K[x] = K[t1 , . . . , tr , x1 , . . . , xn−r ] (after rearranging the xi ’s) and K[x] is integral over K[t] [Lang4, p. 22]. If K is an infinite field, then t1 , . . . , tr can be chosen to be linear combinations of x1 , . . . , xr with coefficients in K [Zariski-Samuel1, p. 266]. Lemma 19.5.1 gives a constructive version of Noether’s normalization theorem. Return now to the plane curve Γ. Noether’s normalization theorem, in its linear form, allows us to replace x by a linear combination t of x and y such that R is integral over K[t] and [F : K(t)] ≤ d. If K is finite, replace K by a suitable finite extension to achieve the linear dependence of t on x and y. If a place ϕp of F is not finite on R, then it is infinite at t (Lemma 2.4.1). Hence, p lies over the infinite prime divisor p∞ of K(t)/K. There are at most d prime divisors of F/K that lie over p∞ . This proves our contention. Now we summarize. Lemma 5.2.2: Let Γ be an affine plane curve of degree d defined over a field K. Denote the coordinate ring and the function field, respectively, of Γ over K by R and F . Then: (a) For each K-normal point p ∈ Γ(K) there exists exactly one prime divisor p of F/K with center p on Γ; the degree of p is 1. (b) There are at most d prime divisors of F/K whose centers on Γ are a given point p. (c) If a prime divisor p of F/K is of degree 1 and if ϕp is finite on R, then its center is in Γ(K). (d) There are at most d prime divisors of F/K which are not finite on R. Finally, we point out that the K-normal points and simple points on ∂f (a, b) 6= 0 or Γ are the same. Here, a point (a, b) on Γ is simple if ∂X ∂f ∂Y (a, b) 6= 0. Lemma 5.2.3: A K-rational point (a, b) of Γ is normal if and only if it is ∂f (a, b) 6= 0, then F has a discrete normalized valuation simple. Moreover, if ∂X v with v(y − b) = 1. Proof: Suppose first that (a, b) is normal. Then, its local ring R = O(a,b) over K is a discrete valuation domain. Hence, R(x − a) ⊆ R(y − b) or R(y − b) ⊆ R(x − a). Suppose for example that R(x − a) ⊆ R(y − b). Then there exist g, h, q ∈ K[X, Y ] such that h(a, b) 6= 0 and h(X, Y )(X − a) = g(X, Y )(Y − b) + q(X, Y )f (X, Y ).
5.3 The Genus of a Plane Curve
Apply
∂ ∂X
99
to both sides and substitute (a, b) for (X, Y ): h(a, b) = q(a, b)
∂f (a, b). ∂X
∂f (a, b) 6= 0 and therefore (a, b) is simple. Hence, ∂X ∂f (a, b) 6= 0. We show that y − b generates Conversely, suppose that ∂X the maximal ideal m of R. Indeed,
f (X, Y ) =
∂f ∂f (a, b)(X − a) + (a, b)(Y − b) + higher terms, ∂X ∂Y
and 0 = f (x, y) =
∂f ∂f (a, b)(x − a) + (a, b)(y − b) + (x − a)u + (y − b)v, ∂X ∂Y
with u, v ∈ m. Hence, 0=
∂f (a, b) + u (x − a) + (a, b) + v (y − b). ∂X ∂Y
∂f
∂f (a, b) is a nonzero constant and u ∈ m, the coefficient of x − a But, since ∂X is a unit of R. Therefore, x − a ∈ R(y − b),Tso m = R(x − a, y − b) = R(y − b). ∞ Since R is Noetherian, the ideal a = n=1 mn is a finitely generated Rmodule. Since ma = a, Nakayama’s Lemma [Lang4, p. 195, Prop. 1] implies that a = 0 (alternatively, use Krull’s intersection theorem [Eisenbud, p. 152]). Hence, each z ∈ R has a unique representation z = w(y − b)n , with w a unit of R and n ≥ 0. Thus, R is a discrete valuation domain. Therefore, R is integrally closed (Exercise 3 of Chapter 2). Let v be the normalized discrete valuation of F corresponding to R. Since y − b generates m, we have v(y − b) = 1.
For a projective curve Γ∗ defined by f ∗ (X0 , X1 , X2 ) = 0 as in (3) of ∂f (a0 , a1 , a2 ) 6= 0, for some i, Section 5.1, the point (a0 :a1 :a2 ) is simple if ∂X i i = 0, 1 or 2. Otherwise, (a0 :a1 :a2 ) is singular.
5.3 The Genus of a Plane Curve Here we bound the genus and the number of nonnormal points of a plane curve by a function of its degree. We first prove a finiteness result for the coordinate ring of a plane curve. Lemma 5.3.1: Let R = K[x, y] be an integral domain with quotient field F of transcendence degree 1 over K. Let S be the integral closure of R in F . Then S/R is a finitely generated K-vector space. Proof: It is well known that S is a finitely generated R-module ([Lang4, p. 120] or [Zariski-Samuel1, p. 267]). Hence, there exists a nonzero element
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Chapter 5. Plane Curves
z ∈ R such that zS ⊆ R. Thus, S/R ∼ = zS/zR ⊆ R/zR. It suffices to prove that dimK R/zR < ∞. By Noether’s normalization theorem (Proposition 5.2.1), we may assume without loss that R is integral over K[x]. The element z satisfies an equation of the form z m + a(x)z m−1 + · · · + g(x) = 0, with 0 6= g(x) ∈ K[x] of degree, say, k. Since g(x) belongs to zR, every power of x is a linear combination modulo zR of 1, x, . . . , xk−1 with coefficients in K. If d is the degree of a monic equation for y over K[x], then each element of R can be written as h(x, y), with h ∈ K[X, Y ] is of degree at most d − 1 in Y . Our lemma follows. In the notation of the proof of Lemma 5.3.1, S is the integral closure in F of the Dedekind domain K[x]. Hence, S is a Dedekind domain and therefore every nonzero prime ideal of S is maximal. By Section 5.2, every nonzero prime ideal of R is maximal. Consider a nonzero prime ideal p of R. Let Rp be the local ring of R at p. Then the integral closure of Rp in F is Sp = sb | s ∈ S and b ∈ R r p . Lemma 5.3.2: P (a) The quotient S/R is isomorphic to the direct sum Sp /Rp , where p runs over the nonzero prime ideals of R. (b) Let p1 , . . . , pk be the prime ideals of S that lie over a prime ideal p of R. Then dimK Sp /Rp ≥ k − 1. Proof of (a): Denote the conductor of S over R by c = {c ∈ R | cS ⊆ R}. It is a nonzero ideal of both R and S. The local ring Rp is integrally closed, and therefore equal to Sp , if and only if c 6⊆ p [Zariski-Samuel1, p. 269]. Let Qk Ql(i) e c = i=1 j=1 piji,j be a factorization of c into the product of prime ideal powers of S where pi1 , . . . , pi,l(i) all lie over the same prime ideal pi of R, i = 1, . . . , k. Then Spij ⊆ Rpi , i = 1, . . . , k. If p is a nonzero prime ideal of R not in the set {p1 , . . . , pk }, then Rp = Sp . Hence, with Ri T = Rpi and Si = Spi , i = 1, . . . , k, Lemma 2.4.4 gives S ∩ R1 ∩ · · · ∩ Rk = Rp = R, where p runs over the nonzero prime ideals of R. Therefore, the map Pk s+R 7→ (s+R1 , . . . , s+Rk ) for s ∈ S, is an injective homomorphism into i=1 Si /Ri . The proof of the lemma is complete if we show that the map is surjective. / pi . Let s1 , . . . , sk be elements of S and let a1 , . . . , ak ∈ R such that ai ∈ Ql(i) ei,j The ideals qi = j=1 pij of S are pairwise relatively prime, i = 1, . . . , k. Hence, by the weak approximation theorem (Proposition 2.1.1), there exist s ∈ S, qi ∈ qi , and a0i ∈ R r pi such that s−
qi si = 0 ai ai
i = 1, . . . , k.
Q For each i choose bi ∈ (R r pi ) ∩ j6=i qj . Then ci = bi qi ∈ c ⊆ R and s = asii + bicai 0 . Thus, s + Ri = asii + Ri for i = 1, . . . , k. i
Proof of (b): Now let p1 , . . . , pk be the prime ideals of S that lie over p. For i = 1, . . . , k − 1 use the weak approximation theorem to find si ∈ S
5.3 The Genus of a Plane Curve
101
such that si ≡ 1 mod pi and si ≡ 0 mod pj , for j = 1, . . . , k and j 6= i. We need only show that s1 , . . . , sk−1 are K-linearly independent modulo Rp . Pk−1 Indeed, if r = i=1 ai si ∈ Rp with ai ∈ K (and si ∈ pk ), then, r belongs to Pk−1 pRp . Since aj ≡ i=1 ai si ≡ 0 mod pk , this gives aj = 0, j = 1, . . . , k − 1. Consequently, dimK Sp /Rp ≥ k − 1. Lemma 5.3.3: Let Γ be an affine plane curve defined over a field K with a generic point x = (x1 , x2 ) and let a ∈ Γ(K). Consider a separable algebraic 0 ) be the local ring of Γ at a over extension L of K and let Oa,K (resp. Oa,K K (resp. its integral closure). Then (a) Oa,L ∩ K(x) = Oa,K ; and 0 0 (b) dimK Oa,K /Oa,K ≤ dimL Oa,L /Oa,L . Proof: Without loss assume that L/K is finite. Let w1 , . . . , wn be a linear basis for L/K. Since Γ is defined by an absolutely irreducible polynomial, K(x)/K is a regular extension of K. Hence, K(x) and L are linearly disjoint over K. Therefore, w1 , . . . , wn is also a linear basis for L(x)/K(x). Consider u ∈ Oa,L ∩ K(x). Then P f (x), g(x) ∈ P u = f (x)/g(x) with = gi (x)wi , with L[x] and g(a) 6= 0. Write f (x) = fP i (x)wi and g(x) P ugi (x)wi = fi (x)wi deduce that fi (x), gi (x) ∈ K[x], i = 1, . . . , n. From ugi (x) = fi (x), i = 1, . . . , n. Since there is an i with gi (a) 6= 0, we have u ∈ Oa,K . This proves (a). 0 To prove (b), consider u1 , . . . , um ∈ Oa,K linearly dependent over L modulo Oa,L . We show that they are also linearly dependent over K modulo that Oa,K . Indeed, there existPb1 , . . . , bm ∈ L not all zero and v ∈ Oa,L Psuch P n m = j=1 aij wj with aij ∈ K and let vj = i=1 aij ui . bi ui = v. Write bi P Then vj ∈ K(x) and vj wj = v. For each σ of L(x) into P K(x)-embedding the algebraic closure of K(x) we have vj wjσ = v σ . Since L(x)/K(x) is ) 6= 0 [Lang7, p. 286, Cor. 5.4]. Apply Cramer’s rule to separable, det(wjσP solve the system vj wjσ = v σ , all σ, and write vj as a linear combination σ ˆ of L/K. Since a is of the v ’s with coefficients in the Galois closure L K-rational, each v σ is in Oa,Lˆ . Thus, vj ∈ Oa,Lˆ ∩ K(x). Hence, by (a), vj ∈ Oa,K . Therefore, the ui ’s are linearly dependent over K modulo Oa,K . Proposition 5.3.4: Let K be an algebraically closed field and Γ∗ a projective plane curve of degree d and genus g defined over K. Then (1)
g=
X 1 (d − 1)(d − 2) − dimK Op0 /Op , 2
where p runs over the points of Γ∗ (K), Op denotes the local ring of Γ∗ at p and Op0 is the integral closure of Op in the function field of Γ∗ over K. Proof: The five parts of the proof draw information from the form of Γ∗ after a change of variables that puts d distinct points at infinity.
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Chapter 5. Plane Curves
Part A: A linear transformation. Choose a line L over K passing through no singular point of Γ∗ and is not tangent to Γ∗ . Then L cuts Γ∗ in d distinct simple points p1 , . . . , pd of Γ∗ [Seidenberg, p. 37]. By Lemma 5.2.3, they are normal. After a suitable linear homogeneous transformation [Seidenberg, Chapter 5] we may assume that L is the line at infinity and that the infinite points on the X-axis and the Y -axis, (0:1:0) and (0:0:1) do not belong to the set {p1 , . . . , pd }. Such a transformation does not change the genus or the degree of the curve. Moreover, corresponding points have the same local rings. View Γ∗ as the projective completion of an affine curve Γ defined by an equation f (X, Y ) = 0, as given by (1) of Section 5.1. Let (x, y) be a generic point of Γ over K, let R = K[x, y] be the coordinate ring of Γ, and let F be its function field. Part B: The divisor at ∞. By Part A, pi = (0:ai :1), with ai ∈ K × for i = 1, . . . , d, and a1 , . . . , ad distinct. In the notation of (2) of Section 5.1, obtain the factorization (2)
fd (X, Y ) = c
d Y
(X − ai Y ), with 0 6= c ∈ K.
i=1
Therefore, f (x, Y ) is an irreducible polynomial of degree d over K[x]. Thus, R is integral over K[x] and each element of R can be uniquely expressed as a polynomial h(x, y) with coefficients in K[x] such that degY (h) ≤ d − 1. Similarly R is integral over K[y]. Over the infinite prime divisor p∞ of K(x) there lie exactly d distinct prime divisors p1 , . . . , pd of F/K, with pi being the unique prime divisor with center pi . In particular, p∞ is unramified in F (Proposition 2.3.2) and we may therefore normalize vpi such that vpi(x) = −1. Since R is integral over K[x], this implies vpi (y) < 0, so vpi xy ≥ 0. by (2) of Section 5.1, Qd c i=1 xy − ai + fd−1 xy , 1 y −1 + · · · + f0 xy , 1 y −d = 0. Hence, xy has residue ai at pi , so vpi (y) = −1. Let o = p1 + · · · + pd . Then the pole divisors div∞ (x) and div∞ (y) of x and y in F are both o. Part C: The integral closure of R in F . Let S be the integral closure of R S∞ in F . By Lemma 3.1.1, S = n=1 Sn , where Sn = L(no). Let Rn = R ∩ Sn , n = 1, 2, . . . . By Lemma 5.3.1, dimK S/R < ∞. Hence, for n sufficiently large R + Sn = R + Sn+1 = R + Sn+2 = . . . . Therefore, S = R + Sn , so Sn /Rn ∼ = S/R. For n > 2g − 2, Lemma 3.2.2(d) implies that dimK Sn = nd − g + 1. It follows that (3) Part D: (4)
nd − g + 1 = dimK Rn + dimK S/R. For n > d, we have Rn = {h(x, y) ∈ R | deg(h) ≤ n and degY (h) ≤ d − 1}.
5.3 The Genus of a Plane Curve
103
Indeed, let k ≥ 0 be an integer and hk ∈ K[X, Y ] a homogeneous polynomial of degree k with degY (hk ) ≤ d − 1. If i + j = k, then div∞ (xi y j ) = ko (Part B). Hence, hk (x, y) ∈ Sk . / Sk−1 : Since K is algebraically closed, we may We prove that hk (x, y) ∈ factor hk (x, y) as hk (x, y) =
d−1 Y
(x − bj y) · axk−d+1
j=1
with bj ∈ K and a ∈ K × . Then vpi (hk (x, y)) =
d−1 X
vpi (x − bj y) − (k − d + 1).
j=1
Pd−1 If hk (x, y) ∈ Sk−1 , then for each i we have j=1 vpi (x − bj y) − (k − d + 1) + k − 1 ≥ 0. Hence, there exists 1 ≤ j ≤ d − 1 with vpi (x − bj y) ≥ 0. Therefore, bj = ai . But this implies that two of the ai ’s are equal, a contradiction. To complete Pm Part D, write each h(x, y) in K[x, y] with degY (h) ≤ d − 1 as h(x, y) = k=1 hk (x, y) with hk as above and hm (x, y) 6= 0. If m ≤ n, then hk (x, y) ∈ Sk ⊆ Sn , k = 0, . . . , m, and therefore h(x, y) ∈ Rn . If m > n, / Rn . then h(x, y) ∈ Sm r Sm−1 and therefore h(x, y) ∈ Sd−1 Part E: Computation of the genus. By Part D, the set j=0 {xi y j | i = 0, . . . , n − j} is a basis of Rn . Hence dimK Rn =
d−1 X
1 (n − j + 1) = nd + 1 − (d − 1)(d − 2). 2 j=0
Substitute this in (3) to conclude that 1 (d − 1)(d − 2) − dimK S/R. 2 L By Lemma 5.3.2(a), S/R = SP /RP , where P ranges over all nonzero prime ideals of R. Since K is algebraically closed, there is a bijective correspondence between the nonzero prime ideals P ofL R and the finite points p of Γ∗ and 0 we have RP = Op . Hence, S/R = p∈Γ∗ (K) Op /Op . Substituting this expression for S/R in (5), we get (1). (5)
g=
Corollary 5.3.5: Let Γ∗ be a projective plane curve of degree d defined over a perfect field K. Then genus(Γ∗ ) ≤ 12 (d − 1)(d − 2). Proof: The genus of Γ∗ does not change by going from K to its algebraic closure (Proposition 3.4.2). Now apply Lemma 5.3.3 and Proposition 5.3.4.
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Chapter 5. Plane Curves
Corollary 5.3.6: Let Γ∗ be a projective plane curve of degree d defined over an algebraically closed field K. Then genus(Γ∗ ) = 12 (d − 1)(d − 2) if and only if Γ∗ is smooth; that is, all K-rational points of Γ∗ are simple. Proof: The condition “Γ∗ is smooth” is equivalent to Op = Op0 for all K rational points p of Γ∗ . Now apply Proposition 5.3.4.
5.4 Points on a Curve over a Finite Field Using the formula for the genus of an absolutely irreducible curve Γ over Fq (Proposition 5.3.4), we translate the estimate of the number of prime divisors of degree 1 of the function field of Γ to an estimate on |Γ(Fq )| which involves only q and the degree of Γ. Theorem 5.4.1: Let f ∈ Fq [X, Y ] be an absolutely irreducible polynomial of degree d. Denote the affine curve defined by the equation f (X, Y ) = 0 by Γ. Then √ √ q + 1 − (d − 1)(d − 2) q − d ≤ |Γ(Fq )| ≤ q + 1 + (d − 1)(d − 2) q. Proof: Put K = Fq . Denote the function field of Γ over K by F , and denote the number of prime divisors of degree 1 of F/K by N . By Theorem 4.5.2, √ √ (1) q + 1 − 2g q ≤ N ≤ q + 1 + 2g q , where g = genus(F/K). Let Γ∗ be the projective completion of Γ. For each p ∈ Γ∗ (K) let k(p) (resp. k1 (p)) be the number of prime divisors (resp. prime divisors of degree 1) of F/K with center at p. Then X X (2) N − |Γ∗ (K)| = (k1 (p) − 1) ≤ (k(p) − 1). p∈Γ∗ (K)
p∈Γ∗ (K)
Let (3)
δ=
X
0 dimK Op,K /Op,K and δ˜ =
p∈Γ∗ (K)
X
0 dimK˜ Op, ˜. ˜ /Op,K K
p∈Γ∗ (K)
0 /Op,K = 0 and Consider p ∈ Γ∗ (K). If p is K-normal, then dimK Op,K k1 (p) = k(p) = 1 (Lemma 5.2.2(a)). If p is not K-normal, then 0 dimK Op,K /Op,K ≥ 1. Hence, by (3), and by Lemma 5.3.3,
X
˜ (1 − k1 (p)) ≤ δ ≤ δ.
p∈Γ∗ (K)
We conclude from (2), (1), and Proposition 5.3.4 that X √ |Γ(K)| ≤ |Γ∗ (K)| = N + (1 − k1 (p)) ≤ q + 1 + 2g q + δ˜ p∈Γ∗ (K)
√ √ √ = q + 1 + (d − 1)(d − 2) q − 2δ˜ q + δ˜ ≤ q + 1 + (d − 1)(d − 2) q.
Exercises
105
Next we give a lower bound for |Γ(K)|. Note first that Γ∗ has at most d points at infinity [Seidenberg, p. 37]. Hence, |Γ(K)| ≥ |Γ∗ (K)|−d. Therefore, by (2), Lemma 5.3.2(b), (1), and Proposition 5.3.4 |Γ(K)| ≥ |Γ∗ (K)| − d ≥ N −
X
(k(p) − 1) − d
p∈Γ∗ (K)
≥N−
X
√ 0 dimK OK,p /OK,p − d ≥ q + 1 − 2g q − δ˜ − d
p∈Γ∗ (K)
√ √ = q + 1 − (d − 1)(d − 2) q + 2δ˜ q − δ˜ − d √ ≥ q + 1 − (d − 1)(d − 2) q − d. This completes the proof of the theorem.
Corollary 5.4.2: The curve Γ of Theorem 5.4.1 satisfies the following conditions: (a) For each m there exists q0 such that |Γ(Fq )| ≥ m for all q ≥ q0 .
