BOOKS OF RELATED INTEREST BY SERGE LANG
Linear Algebra, Third Edition 1987, ISBN 96412-6 Undergraduate Algebra, Second Edition 1990, ISBN 97279-X Complex Analysis, Third Edition 1993, ISBN 97886-0 Real and Functional Analysis, Third Edition 1993, ISBN 94001-4 Algebraic Number Theory, Second Edition 1994, ISBN 94225-4 Introduction to Complex Hyperbolic Spaces 1987, ISBN 96447-9 OTHER BOOKS BY LANG PUBLISHED BY SPRINGER-VERLAG
Introduction to Arakelov Theory * Riemann-Roch Algebra (with William Fulton) e
Complex Multiplication * Introduction to Modular Forms ® Modular Units (with Daniel Kubert) * Fundamentals of Diophantine Geometry © Elliptic Functions  Number
Theory III â Cyclotomic Fields I and II SL,(R) Â Abelian Varieties đ Differential and
Riemannian Manifolds « Undergraduate Analysis ¢ Elliptic Curves: Diophantine Analysis ° Introduction to Linear Algebra ¢ Calculus of Several Variables ¢ First Course in Calculus * Basic Mathematics « Geometry: A High School Course (with Gene Murrow) © Math! Encounters with High School Students © The Beauty of Doing Mathematics * THE FILE Serge Lang Introduction to Algebraic and Abelian Functions Second Edition Springer-Verlag
Trang 3Serge Lang Department of Mathematics Yale University New Haven, Connecticut 06520 USA Editorial Board J.H Ewing Department of Mathematics Indiana University Bloomington, IN 47405 USA F.W Gehring Department of Mathematics University of Michigan Ann Arbor, MI 48109 USA P.R Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA AMS Classifications: 14HOJ, 14K25 With 9 illustrations Library of Congress Cataloging in Publication Data Lang, Serge, 1927-
Introduction to algebraic and abelian functions
(Graduate texts in mathematics; 89) Bibliography: p 165 Includes index
1 Functions, Algebraic 2 Functions, Abelian , I Title Il Series QA341.L32 1982 515.9'83 82-5733 AACR2
The first edition of Introduction to Algi
in 1972 by Addison-Wesley Publishing Co., Inc © 1972, Printed and bound by 1982 by Springer-Verlag Ne All rights reserved This worl Teen rica 987
6 5 4 3 2 (Second Corrected printing, 1995)
ebraic and Abelian Functions was published
rporation, Pleasant Hill i a Braun-Brumfi PA Printed in the United States of Mees Ato Berlin Heidelberg Verlag Berlin Heidelberg New York Introduction
This short book gives an introduction to algebraic and abelian functions, with
emphasis on the complex analytic point of view It could be used for a course
or seminar addressed to second year graduate students
The goal is the same as that of the first edition, although I have made a number of additions I have used the Weil proof of the Riemann-Roch the- orem since it is efficient and acquaints the reader with adeles, which are a very useful tool pervading number theory
The proof of the Abel-Jacobi theorem is that given by Artin in a seminar in 1948 As far as I know, the very simple proof for the Jacobi inversion theorem is due to him The Riemann-Roch theorem and the Abel-Jacobi theorem could form a one semester course
The Riemann relations which come at the end of the treatment of Jacobi’s theorem form a bridge with the second part which deals with abelian functions
and theta functions In May 1949, Weil gave a boost to the basic theory of
theta functions in a famous Bourbaki seminar talk I have followed his exposition of a proof of Poincaré that to each divisor on a complex torus there corresponds a theta function on the universal covering space However, the correspondence between divisors and theta functions is not needed for the linear theory of theta functions and the projective embedding of the torus
when there exists a positive non-degenerate Riemann form Therefore I have
given the proof of existence of a theta function corresponding to a divisor only
in the last chapter, so that it does not interfere with the self-contained treat-
ment of the linear theory aie
The linear theory gives a good introduction to abelian varieties, in an
analytic setting Algebraic treatments become more accessible to the reader who has gone through the easier proofs over the complex numbers This
Trang 4; : Introduction
[have included enough material to give all the basic analytic facts neces-
sary in the theory of complex multiplication in Shimura-Taniyama, or my more recent book on the subject, and have thus tried to make this topic accessible at a more elementary level, provided the reader is willing to assume some algebraic results
Ihave also given the example of the Fermat curve, drawing on some recent
results of Rohrlich This curve is both of intrinsic interest, and gives a typical setting for the general theorems proved in the book This example illustrates
both the theory of periods and the theory of divisor classes Again this
example should make it easier for the reader to read more advanced books and
papers listed in the bibliography
New Haven, Connecticut SERGE LANG
Contents
Chapter I
The Riemann-Roch Theorem
§1 Lemmas on Valuations
Bo Thacker ent ot cece ser ale
§3 Remarks on Differential Forms $4 Residues in Power Series Fields §5 The Sum of the Residues §6 The Genus Formula of Hurwitz
§7 Examples
§8 Differentials of Second Kind
§9 Function Fields and Curves
§10 Divisor Classes scivonitusunsastiseieet cd eed cha an Chapter II The Fermat Curve §1 The Genus_ §2 Differentigls -—ccccccccceccecncu
§3 Rational Images of the Fermat Curve
§4 Decomposition of the Divisor Class€s .-‹++ + ‡+*+cs++t
Chapter III
The Riemann Surface
§1 Topology and Analytic Structure -+-++++>strss+s §2, Integration on the Riemann Surface
Trang 5Contents viii Chapter IV ~JacoD1 bi The Theorem of Ta 54 §1, Abelian Integrals 3g §2 Abel's Theorem -: 63 §3 Jacobi’s Theorem 66 §4 Riemann's Ơn eceteeeteetenes 67 §5 Duality ‹.-. - nh ee Chapter V Periods on the Fermat Curve 73 arithm Symbol
e owes the Universal Covering Space 7s
§3 Periods on the Fermat Curve ”
§4 Periods on the Related Curves
Chapter VI -
Linear Theory of Theta Functions
§1 Associated Linear Forms -. - ¡ch nen nh nhe key 83
§2 Degenerate Theta Functions 89
§3 Dimension of the Space of Theta Functions 90 §4 Abelian Functions and Riemann-Roch Theorem on the Torus 97
§5 Translations of Theta Functions 101
§6 Projective Embedding 104
Chapter VII
Homomorphisms and Duality
§1 The Complex and Rational Representations
§2 Rational and p-adic Representations
Š3 Homomorphisms
Š4 Complete Reducibility of Poincaré
§5 The Dual Abelian Manifold
§6 Relations with Theta Functions
§7 The Kummer Pairing §8 Periods and Homology Chapter VIII Riemann Matrices and Classical Theta Functions $1 Riemann Matrices —
$2 The Siegel Upper Half Space
Trang 6CHAPTER I The Riemann-Roch Theorem §1 Lemmas on Valuations
We recall that a discrete valuation ring 0 is a principal ideal ring (and there- fore a unique factorization ring) having only one prime If t is a generator
of this prime, we call ¢ a local parameter Every element x + 0 of such a
ring can be expressed as a product
x=t'y,
where r is an integer = 0, and y is a unit An element of the quotient field
K has therefore a similar expression, where r may be an arbitrary integer,
which is called the order or value of the element If r > 0, we say that x
has a zero at the valuation, and if r < 0, we say that x has a pole We write
r= v(x), or v(x), or ord,(x)
Let p be the maximal ideal of 0 The map of K which is the canonical map
0 — vo/p on 0, and sends an element x € 0 to oo, is called the place of the
valuation ; :
Trang 7I Riemann-Roch Theorem 2 o (or of $8 over P)- If Pg and I, are the value groups over ro: Ty) =6 š la gS, Di above (0,P), or more briefly that B lies i a hat 3$ ¡s „3#) is unramifed above (0,)), or tt at is fication index is equal to 1, that ise = 1
Leta Ek Leto
tion index of © ove
of these valuation ring th We say that the pair (0
above p We say that (OD ;
unramified above P, if the rami
Example Let k be a field and ¢ transcendental over k be the set of rational functions
/(0Jg(), with f(D, g( E Hla] such that g(a) # 0
Then 0 is a discrete valuation ring, whose maximal ideal consists of au such
quotients such that f(a) = 0 This is a typical situation ]n a na ‹ algebraically closed (for simplicity), and consider the extension (a) obtaine with one transcendental element x over k Let 0 be a discrete valuation ring
in k(x) containing k Changing x to I/x if necessary, we may assume that xo Then pf1k{x] # 0, and p M A[x] is therefore generated by an irre-
ducible polynomial p(x), which must be of degree 1 since we assumed k
algebraically closed Thus p(x) = x — 4 for some a © k Then it is clear that the canonical map
o> o/p
induces the map
f(x) > fla)
on polynomials, and it is then immediate that 0 consists of all quotients f(x)/g (x) such that g(a) # 0; in other words, we are back in the situation
described at the beginning of the example
Similarly, let o = &[[1]] be the ring of formal power series in one variable Then 0 is a discrete valuation ring, and its maximal ideal is generated by ¿ Every element of the quotient field has a formal series expansion
FG ™ + tat! +a tattar+ -,
ne a; © k The place maps x on the value ao if x does not have
In the applications, we shall stud nh , y a field K which is a ICh 1s a finite extension of ich is a fi fini i
§1 Lemmas on Valuations
3
valuation ring in E above (0, the root of a polynomial f (Y) cient 1, such that
p) in K Suppose that E = K(y) where yis = 0 having coefficients in 0, leading coeffi- f(y) =0 but f'(y) 3# 0 mod 9%
Then 3Š is unramified over p
z9
Proof There exists a constant Yo © k such that y = yo mod PB By hypothesis, f'(yo) #0 mod $8 Let {yn} be the sequence defined recur- sively by
Yast = Yn —f'Cyn) Fn)
Then we leave to the reader the verification that this sequence converges in
the completion K, of K, and it is also easy to verify that it converges to the root y since y = yo mod ÄŠ but y is not congruent to any other root of f and
38 Hence y lies in this completion, so that the completion Ey is embedded
in Ky, and therefore ‘8 is unramified
We also recall some elementary approximation theorems
Chinese Remainder Theorem Lert R be a ring, and let p\, , Pn be
distinct maximal ideals in that ring Given positive integers r, 5 Tn
and elements a,, , 4, © R, there exists x € R satisfying the con-
gruences
x =a;mod p}' for all i
For the proof, cf Algebra, Chapter II, §2 This theorem is applied to the
integral closure of k[x] in a finite extension
We shall also deal with similar approximations in a slightly different
context, namely a field K and a finite set of discrete valuation rings 0), , 0, of K, as follows
Proposition 1.2 [f 0; and 0, are two discrete valuation rings with quotient
field K, such that 0, C 02, then 0, = 03 ` D4 9ER
Proof We shall first prove that if p, and p2 are their mauimÍ ideals, then
P.C py Let y Cÿ; lfy Œ pị, then 1/y € 0), whence T/y € P2, a con-
tradiction Hence p, C p; Every unit of 0, is a fortiori a unit of Đạo An
Trang 8
1 Riemann-Roch Theorem
Fe A s cannot be a unj
dis not in 0), then 1/15 0n and thus cí oui
ete q unit in 02, and IS _
if wis aunitin 02, position
in > This proves our pro}
valuation rings 9i Wi = ths ca a Deore
» assume that our Ị
on, we assu lations
OW ˆ * re
From n se no inclusion re
distinct, and hence hav
Jement y of K having a zero at 0, and
iti pre exists an e
sition 1.3 There ©
ii = 2) cớ š y tỦ-
a pole at 0; (j = 2+
vill be proved by induction Suppose "= 2 Since there is and 0s, we can find y € 02 and y € o, dz €& ds Then z/y has a zero at 0, and a Proof This w
no inclusion relation between Dị Similarly, we can find z € 0, an pole at 02 as desired
Now suppose we hi
pole at 0ạ, - „ Ủa~I- Letz
ave found an element of K having a zero at 0; anda
be such that z has a zero at 0; and a pole at o,, Then for sufficiently large r, y + 2" satisfies our eQuineeOTE, because we
have schematically zero plus zero = zero, zero plus pole = pote, an the
sum of two elements of K having poles of different order again as a pole
A high power of the element y of Proposition 1.3 has a high zero ave and
a high pole at 9, (j = 2, ,1) Adding 1 to this high power, and
considering 1/(1 + y’) we get
Corollary There exists an element z of K such thatz — | has a high zero
at 0;, and such that z has a high zero at 0; (j = 2, ,M)
Denote by ord; the order of an element of K under the discrete valuation
associated with o, We then have the following approximation theorem Theorem 1.4 Given elements a, , a, of K, and an integer N, there exists an element y © K such that ord;(y — a;) > N
Proof For each i, use the corollary to get z; close to | at 0; and close to
0 at 0; (j # i), or rather at the valuations associated with these valuation Tings Then z,a, + - + - + z,a, has the required property
in particular, we can find an element y having given orders at the valua-
tions arising from the 0; This is used to prove the following inequality
§2 The Riemann-Roch Theorem
Proof Select elements
Mls wm w Plage + © 3 Date vm « 5 Dw
of E such that y, (v= 1, ,¢ have zeroes of high order at the ot the above elements are linearly i
relation of linear dependence
i) Tepresent distinct cosets of in I, and
her valuations vj (j # i) We contend that ndependent over K Suppose we have a
3) c„y„ = 0
Say cj, has maximal value in I’, that is, v(cy) = v(c;,) all i, v Divide the
equation by cy Then we may assume that cy, = 1, and that v(c;,) < 1 Consider the value of our sum taken at vy Allterms yy, C212, + Cle, Yiey
have distinct values because the y’s represent distinct cosets Hence Oily ++ + Chey Yes) = viCyn)
On the other hand, the other terms in our sum have a very small value at
v, by hypothesis Hence again by that property, we have a contradiction,
which proves the corollary
§2 The Riemann-Roch Theorem
Let k be an algebraically closed field, and let K be a function field in one
variable over k (briefly a function field) By this we mean that K is a finite
extension of a purely transcendental extension k(x) of k, of transcendence degree 1 We call k the constant field Elements of K are sometimes called
functions
By a prime, or point, of K over k, we shall mean a discrete valuation ring
of K containing k (or over k) As we saw in the example of §1, the residue class field of this ring is then k itself The set of all such discrete valuation
rings (i.e., the set of all points of K) will be called a curve, whose function
field is K We use the letters P, Q for points of the curve, to suggest geometric terminology
By a divisor (on the curve, or of K over k) we mean an element of the free
abelian group generated by the points Thus a divisor is a formal sum
a= » nịP, = > npP
Trang 9I, Riemann-Roch Theorem a =i np > nj 2 er of cat Pi 8 nà : 0, the then S ite number of points P sucht # hen there is onl ya finite 1 lfx:€ Kandx ÿ y †
) = 0fOor all TỶ x is not
Tê as stant, then ordp (x) i
: d, if x is constan’, ¡ch x has a zero, and one poini
oder er is one point of k(x) at "¬ = nly a finite nuniber
constant, then! sol Each of these points extends to only a ber
at which ae NH is a finite extension of k(x) Hence we can associate
of points of 4, ` a divisor with x, namely
(x) = >) mP
= ordp(x) Divisors a and b are said to be Le dine if nh isthe divisor of a function Ifa = 3P and 6 = % mpP are divisors,
we write
a2b_ ifandonly if np 2 mp for all P
This clearly defines a (partial) ordering among divisors We call a positive
i >
" To ies divisor, we denote by L (a) the set of all elements xEK such that (x) = —-a If a is a positive divisor, then L(a) consists of all the functions
in K which have poles only in a, with multiplicities at most those of a It is
clear that L(a) is a vector space over the constant field & for any divisor a
We let /(a) be its dimension
Our main purpose is to investigate more deeply the dimension /(q) of the
vector space L (a) associated with a divisor a of the curve (we could say of
the function field)
Let P be a point of V, and 0 its local ring in K Let p be its maximal ideal Since k is algebraically closed, o/p is canonically isomorphic to k We know
that o is a valuation ring, belonging to a discrete valuation Let t be a
generator of the maximal ideal Let x be an element of 0 Then for some constant do in k, we can write x = ay mod p The function x — dp is in p,
and has a zero at 0 We can therefore write x — đọ = tyo, where yo is in 0
Again by a similar argument we get yo = a, + ty, with yị €0, and
X= dy + at + y,t?
Continuing this procedure, we obtain an expansion of x into a power series,
XS O+at+aryz ,
It is trivial that if each c efficient a; is =
The quotient field K o} f 0 can be embedded in the power series field k(t) frie Caual to 0, then x = 0
§2 The Riemann-Roch Theorem
as follows If x is in K, then for some power f°, the function t*x lies i
and hence x can be written , a
x es Sai
x= pot tT Fata tees,
If u is another generator of , then clearly k((t)) = k((u)), and our power series field depends only on P We denote it by Kp Anelement & of Kp can
be writen p = Xổ „ dt" with a, #0 Ifm < O, we say that & has a pole
of order —m If m > 0 we say that & has a zero of order m, and we let m = ordp &» Lemma For any divisor a and any point P, we have
a+ P)sl(a) +1,
and I(q) is finite
Proof If a = 0 then /(a) = | and L(q) is the constant field because a function without poles is constant Hence if we prove the stated inequality,
it follows that /(q) is finite for all a Let m be the multiplicity of P in a
Suppose there exists a function z € L(a + P) but z € L(a) Then ordp x = — (m + 1)
Let w € L(a + P) Looking at the leading term of the power series ex-
pansion at P for w, we see that there exists a constant c such that w — cz
has order = —m at P, and hence w € L(a) This proves the inequality,
and also the lemma
Let A* be the cartesian product of all Kp, taken over all points P An element of A* can be viewed as an infinite vector € = ( , &, ) where & is an element of Kp The selection of such an element in A* means that a random power series has been selected at each point P Under component- wise addition and multiplication, A* is a ring It is too big for our purposes,
and we shall work with the subring A consisting of all vectors such that ếp has
no pole at P for all but a finite number of P This ring A will be called the
ring of adeles Note that our function field K is embedded in A under the
mapping
FOG «ody Becae
i.e., at the P-component we take x viewed as a power series in Kp In
particular, the constant field k is also embedded in A, which can be viewed
Trang 10I, Riemann-Roch Theorem
§2 The Riemann-Roch Theorem
‘ 9
3 curve We shall denote by A(a) the Subset of A " h
Letabea divisor on bole that ordy & = —ordp a Then A(a) is imme- indices of the discrete value group in k(y) associated with the point Q, and consisting of all adeles & such k3 h The set of all such A(a) can be taken the extensions of this value group to K These extensions correspond to the diately seen to be a Ko nhborhoods of 0 in A, and đefine a topology points P; We shall now prove that the degree Š e, of € is equal to [K : k(y)]
asa fundamental eo nes a topological ring We senote [A : £0) by n
in A which thereby becon i aha (x) = ais our old vector space L(a), and Let Zips ca Zn be a linear basis of K over k(y) After multiplying each The set of functions + sục a tổ A(q) AK z/ with a suitable polynomial in k[y] we may assume that they are integral is immediately seen to be ae P and let Sn, be its degree The purpose over k[y], i.e., that no place of K which is finite on k[y] is a pole of any
Let a be a divisor, a = hat tan lnÌ and /(a) have the same order of z; All the poles of the z; are therefore among the P; above appearing in ¢ of this chapter is to show eoaeano on [(a) — deg(a) We shall even Hence there is an integer jo such that z, © L(jioc) Let be a large positive
u integer For any integer s satisfying 0 < 5 S p — po we get therefore ‘i ecise ớ NA PB ane a constatl g depending on our field K alone such that tually prove ° ` Ha) = deg(a) + 1 — 8 + 8), a) SLi), and so [(jc) = (ww — po + In
hich is 0 if deg(a) is sufficiently large 8 Let N,, be the integer (A(uc) + K : A(0) + K), oN, = 0 Putting b = 0 :
where 6(a) is a non-negative integer, W
2g — 2) ` ‘ = pei ¿ h
or nies few trivial formulas on which we base further computations and a = wc in the fundamental formula (1), we get
later If B and C are two k-subspaces of A, and B 9 C, then we denote by
(B : C) the dimension of the factor space B mod C over k o(S «) = Nụ +,l(wo) — Ì
Proposition 2.1 Let a and b be two divisors Then A(a) D A(b) if and (2) >=Ny¿ + (w — tạ + l)n — | only if a = b If this is the case, then
1 (A(a) : A(b)) = deg(a) — deg(b), and
2 (A(a) : A(O)) = ((A(a) + K): (A(D) + K))
Dividing (2) by wand letting ps tend to infinity, we get 2 e; = n Taking into account the corollary to Theorem 1.4 we get
+ (A@ NK): (A(O) 1 K)), Theorem 2.2 Let K be the function field of a curve, and y © K a noncon-
ạ stant function If ¢ is the divisor of poles of y, then deg(c) = [K : k()
Proof The first assertion is trivial Formula Ị IS easy to prove as follows Hence the degree of a divisor of a function is equal to 0 (a function has
Ifa point P appears in a with multiplicity d and in 6 with multiplicity e, then as many zeros as poles)
d 2 e If tis an element of order | at P in Kp, then the index (Kp : t °K,) ,
puny equal tod — e, The index in formula | is clearly the sum of the Proof If we let ¢’ be the divisor of zeros of y then c' is the divisor of poles
finite number of local indices of the above type, as P ranges over all points of I/y, and [K : k(1/y)] = n also
in a or b This proves formula 1 As to formula 2, it is an immediate
coi nsequence of the elementary homomorphism theorems for vector Spaces, l ; - Corollary deg(a) is a function of the linear equivalence class of a
and its formal proof will be left as an exercise to the reader
i i i i i lied a class
From Propositi position 2.1 we get a fundamental formula: function We see that the degree is a class function A function depending only on linear equivalence will be ca
(1) deg(a) ~ deg(6) = (Aa) + K = A(b) + K) 4 Ha) — 106) Returning to (2), we can now write
pn =N, + pn — pon tn—1
know t a Ae finite This will be = "¬ : nto two functions of q and ECE we do not : whence
Le unction j mi ss
wnite¢ = Š e,ƑP,, The p ints P ion in K Let ¢ be the divisor of its poles, and - Nụ Sạn —n +]
and this proves that N, is uniformly bounded Hence for large ,
Trang 111 Riemann-Roch Theorem 10 Ny, = (Ain) + KAO) +) sitive integer always a positive 1n
ma of divisors, r(a) = deg() — /(q) Both
functions, the former by Theorem 2.2 and the latter E L(a)isa k-isomorphism between L (a) and
is constant, because iti Now define a new
deg(a) and /(a) are class Ẹ
because the map Z +> Y2 for z L(a - ())- _ mm fundamental formula (1) can be rewritten @) 0 < (A(a) + K: A() + K) = r(8) ~ r() for two divisors a and b such that a 2 b Put 6 = 0 and a = pe, so (A(w© + K: A(0) + K) = r(0 — r()-
This and the result of the preceding paragraph show that r(uc) is uniformly
bounded for all large , / / ;
Let 6 now be any divisor Take a function z € k[ y] having high zeros at
all points of & except at those in common with ¢ (i.¢., poles of y) Then for
some 1, (z) + wc = 6 Putting a = yc in (3) above, and using the fact that
r(a) is a class function, we get
r(b) S r(pe)
and this proves that for an arbitrary divisor 6 the integer r(6) is bounded
(The whole thing is of course pure magic.) This already shows that deg(0)
and /(6) have the same order of magnitude We return to this question later
For the moment, note that if we now keep 0 fixed, and let a vary in (3), then
A(a) can be increased so as to include any element of A On the other hand
the index in that formula is bounded because we have just seen that r(a) is ae Hence for some divisor a it reaches its maximum, and for this visor a we must have A = A(a) + K We State this as a theorem
a riba exists a divisor a such that A = A(a) + K This
e elements of K can be viewed as a lattice in A, and that there is a neighborhood A(q i
ae eo
(a) which when translated along all points of this
or in other words
§2 The Riemann-Roch Theorem 1"
(5) E(a) — deg(a) — 6(a) = 1() — deg(b) — ô(0)
This holds for a = 6 However, since two divisors have a sup, (5) holds for
any two divisors a and 6 The genus of K is defined to be that integer g such
that
l(a) — deg(a) — (a) = 1 — g
It is an invariant of K Putting a = 0 in this definition, we see that g = 8(0),
and hence that g is an integer = 0, g = (A: A() + K) Summarizing, we have
Theorem 2.4 There exists an integer g = 0 depending only on K such that for any divisor a we have
I(a) = deg(a) + 1 — g + 8(a),
where 6(a) = 0
By a differential A of K we shall mean a k-linear functional of A which vanishes on some A(q), and also vanishes on K (considered to be embedded
in A) The first condition means that A is required to be continuous, when
we take the discrete topology on k Having proved that (A : A(a) + K) is
finite, we see that a differential vanishing on A(a) can be viewed as a
functional on the factor space
A mod ‘A(a) + K3)
and that the set of such differentials is the dual space of our factor space, its dimension over k being therefore 6(a)
Note in addition that the differentials form a vector space over K Indeed, if A is a differential vanishing on A(q), if & is an element of A, and y an
Trang 12
I Riemann-Roch Theorem
12
corem it will suffice to
ence to prove our theorem 1 Prove
A atl Pe L(a), so (y) = ~, then YA vanishes
is ; ot because a + (¥) = i" )=0 lfy Jive sy Yn are
ns are yA, ++» Jn A Hence we get
then A vanishes on / that the degree of a
i tai
A(a +fy)) which confalS 2C
a independent over k, then so 5(0) 2 Ha) = deg + 1 8 + PC), Since 5(a) = 0, it follows that dega = (0) + 8 — 1, which proves the desired bound ¬ Vy
Theorem 2.6 The differentials form a 1-dimensional K-space
Proof Suppose we have two differentials A and 4 which are linearly
independent over K Suppose m1, »%, and Yi , Yn Ate tWo sets of
elements of K which are linearly independent over k Then the differentials MA, 2, Xn, Villy +» Ynfe are linearly independent over k, for other- wise we would have a relation
» ayx;A + » by, =0
Letting x = Ề a¡x; and y = 3 b,y,, we get xA + yu = 0, contradicting the
independence of A, yz over K
Both A and /vanish on some parallelotope A(q), for if A vanishes on A(a)
and vanishes on A(q;), we put a = inf (a), a2), and
A(a) = A(a) N A(ay)
a Ú be a arbitrary divisor If y E L(6), so that (y) = —6, then yA van-
es on A(a + (y)) which contains A(a — 6) because a +()>a—Ú
Similarly YP Vanishes on A( ularly, & — 6) and by definition :
beginning of our Proof, we conclude that SG ð(q - b) > 2/460) Using Theorem 2.4, we get Ha — 6) ~ deg(ay 4 deg(b) — 1 + 9 = 21(6) > = 2 (deg(b) + 1 — ø + ô(0)) = 2 deg(b) + 2 — 2ø,
§2 The Riemann-Roch Theorem 13
If we take 6 to be a positive divisor of very large degree, then L(a — 6)
consists of 0 alone, because a function cannot have more zeros than poles Since deg(q@) is constant in the above inequality, we get a contradiction, and
thereby prove the theorem
If A is a non-zero differential, then all differentials are of type yA If A(a) is the maximal parallelotope on which A vanishes, then clearly A(a + (y)) is the maximal parallelotope on which yA vanishes We get therefore a linear equivalence class of divisors: if we define the divisor (A) associated with A to be a, then the divisor associated with yA is a + (y) This divisor class is called the canonical class of K, and a divisor in it is called a canonical divisor
Theorem 2.6 allows us to complete Theorem 2.4 by giving more informa-
tion on 6(a): we can now state the complete Riemann-Roch theorem
Theorem 2.7 Let a be an arbitrary divisor of K Then
l(a) = deg(a) + 1 — gtl(c—a)
where ¢ is any divisor of the canonical class In other words,
6(a) = l(c — a)
Proof Let ¢ be the divisor which is such that A(c) is the maximal paral- lelotope on which a non-zero differential A vanishes If 6 is an arbitrary
divisor and y € L(b), then we know that yA vanishes on A(c — 6) Con-
versely, by Theorem 2.6, any differential vanishing on A(c — 6) is of type
zA for some z € K, and the maximal parallelotope on which zA vanishes is
(z) + ¢, which must therefore contain A(c — 6) This implies that
(Œ)> -—U, ie, z €L(0)
We have therefore proved that 6(c — 6) is equal to /(6) The divisor b was arbitrary, and hence we can replace it by ¢ — a, thereby proving our theorem
Corollary 1 If ¢ is a canonical divisor, then I(c) = g
Proof Put a = 0 in the Riemann-Roch theorem Then L (a) consists of the
constants alone, and so /(a) = 1 Since deg(0) = 0, we get what we want Corollary 2 The degree of the canonical class is 2g — 2
Trang 13
" corem 1
Corollar: 3 / deg(q y If g( ) > 24 2g — 2, th then ô(a) = 0 (a)
8(a) is equal to /(¢ — a) Since a function cannot have more zeros
6(a) i : p—2
fap L(c — a) = 0 if deg(a) > 28 ~ ?
§3 Remarks on Differential Forms
A derivation D of a ring R is a mapping D: R > R ok into itself which is
linear and satisfies the ordinary rule for derivatives, 1.¢.,
D(x + y) = Dx + Dy, and Dixy) = xDy + yDx,
As an example of derivations, consider the polynomial ring &[X ] over a field
k For each variable X, the derivative 4/0X taken in the usual manner is a
derivation of k[X] We also get a derivation of the quotient field in the
obvious manner, i.e., by defining D(u/v) = (vDu — uDv)/v?
We shall work with derivations of a field K A derivation of K is trivial
if Dx = 0 for all x €Ấ It is trivial over a subfield k of K if Dx = 0 for
all x € k A derivation is always trivial over the prime field: one sees that
D(1) = D(1+1) = 2D(1), whence O()=0
We now consider the problem of extending a derivation D on K Let
E= K(x) be generated by one element Iff € K[X], we denote by df /dx the
polynomial df /aX evaluated at x Givena derivation D on K, does there exist a derivation D* on K(x) coinciding with D on K? If f i i "on K(, ? (X) € K[X] is polynomial vanishing on x, then any such D* must satisfy Ta () 0 = D*(f(x)) = f(x) + > (af /ax)D*x, lerivations, The above ne ( Cessary ¢ sufficient Ty condition for the existence of a D* turns out to be b an extension field of K, by x i r
YX in K[X], Then, ifu is an dlenean of ae deal determina
%) satisfying the equation O= (x) +f" (ou,
§3 Remarks on Differential Forms 15
there is one and only one derivation D* of K(x) coinciding with D on K,
and such that D*x = u
Proof The necessity has been shown above Conversely, if g(x), A(x) are
in K[x], and A(x) # 0, one verifies immediately that the mapping D* defined
by the formulas
D*g(x) = g(x) + g(x)u
hD*g — gD*h
D*(g/h) = hề
is well defined and is a derivation of K(x)
Consider the following special cases Let D be a given derivation on K
Case 1 x is separable algebraic over K Let f(X) be the irreducible
polynomial satisfied by x over K Then f'(x) #0 We have 0 = f(x) +f'(u, whence u = —f?(x)/f'(x) Hence D extends to K(x) uniquely If D is trivial on K, then D is trivial on K(x) Case 2 x is transcendental over K Then D extends, and u can be selected arbitrarily in K(x)
Case 3 x is purely inseparable over K, so x? —a =0, witha EK
Then D extends to K(x) if and only if Da = 0 In particular if D is trivial on K, then u can be selected arbitrarily
From these three cases, we see that x is separable algebraic over K if and
only if every derivation D of K(x) which is trivial on K is trivial on K(x)
Indeed, if x is transcendental, we can always define a derivation trivial on K
but not on x, and if x is not separable, but algebraic, then K(x”) # K(x), whence we can find a derivation trivial on K(x?) but not on K(x)
The derivations of a field K form a vector space over K if we define zD for
Zz € K by (zD)(x) = zDx
Let K be a function field over the algebraically closed constant field k
(function field means, as before, function field in one variable) It is an
elementary matter to prove that there exists an element x € K such that K is
separable algebraic over k(x) (cf Algebra) In particular, a derivation on K
is then uniquely determined by its effect on k(x)
We denote by @ the K-vector space of derivations D of K over k, (deri-
Trang 14I Riemann-Roch Theorem 16 ves therefore a K-linear functional : ‘ f K gi
of (9, K) into K Each element z of & 8 We hiive
of B This functional is denoted by dz
d(yz) = ydz + 2dy
diy +2= dy + dz
These linear functionals form a subspace § of the dual space of %, if we
define ydz by
(D, ydz) > yDz
Lemma 3.2 /f K is a function field (in one variable) over the algebraic- ally closed field k, then D has dimension | over K An element tE Kis
such that K over k(t) is separable if and only if dt is a basis of the dual
space of D over K
Proof If K is separable over k(t), then any derivation on K is determined
by its effect ont If Dt = u, then D = uD,, where D, is the derivation such that D)t = 1 Thus & has dimension | over K, and df is a basis of the dual space On the other hand, using cases 2 and 3 of the extension theorem, we
see at once that if K is not separable over k(s), then dr = 0, and hence cannot
be such a basis
The dual space ¥ of D will be called the space of differential forms of K over k Any differential form of K can therefore be written as ydx, where
K is separable over k(x)
§4 Residues in Power Series Fields
The results of this section will be used as lemmas to prove that the sum of the residues of a differential form in a function field of dimension lis O Let k((1)) be a power series field, the
ait
(not necessarily algebraically closed) If | u is an element of that field whi
can be written w = ayt + qạf2 + * with a, # 0, then it is clear that wana
k(Ww) = k(n)
Our i
i
pe ee ee oa admits a derivation D, defined in the obvious
TP 4, dây x aut is an element of k((t)) one verifies immedi-
Oe te hgh ay ví” Í§a derivation We sometimes denote D b
© a derivation D.y defined in the same manner and the
§4 Residues in Power Series Fields 17
classical chain rule D, y-D,u = D,y (or better dy/du+ dujdt = dy/dt) holds
here because it is a formal result
If y = 2 a,t", then a_, (the coefficient of t~') is called the residue of y
with respect to ¢, and denoted by res,(y)
Proposition 4.1 Let x and y be two elements of k((1)), and let u be another
parameter of k((t) Then
resbp\ ares xế "\> du) ~ "P3 ae)
Proof \t clearly suffices to show that for any element y of k((t)) we have
res,(y) = res,(y dt/du) Since the residue is k-linear as a function of power
series, and vanishes on power series which have a zero of high order, it suf-
fices to prove our proposition for y = ¢" (n being an integer) Furthermore,
our result is obviously true under the trivial change of parameter f = au,
where a is a non-zero constant Hence we may assume f = u + aqu?> + + - *,
and dt/du = 1 + 2a,u + +++ We have to show that res,(t"dt/du) = 1
when n = —1, and 0 otherwise
When n = 0, the proposition is obvious, because t” dt/du contains no negative powers of f
When n = —1, we have
1Π1+ 2am+-' 1
tLdu + du Ẻ + u ,
and hence the residue is equal to 1, as desired nee
When n < —1, we consider first the case in which the characteristic is 0 In this case, we have
sauuø = rent (2 m1)
res, (t" dt/du) = res, đư \n +1
and this is 0 forn # —]
For arbitrary characteristic, and fixed n < —1, we have for m > 1, 1 dt l+ 24;u + + + thầu w”"(1 + đau + + *) Nào 7c u
i ial with i fficients It is
where F (a2, a3, ) is a formal polynomial with integer coe :
Trang 15
I Riemann-Roch Theorem
18
nsequence of the result
position is a formal co’ ul
sa eaneeo hh at case, our polynomial
i nce th \ i
on ic 0, because we have just seen that in th
i cterist j “as
a _ ) is identically 0 This proves the proposition
| an expression of type ydx (with
In view of Proposition 4 f that field and the
x and y in the power series / i
residue res(ydx) of that differential form 1s
i d arame f our field
dx /dt) taken with respect to any parameter 1 0 Hd " me er need a more general formula than that of Proposition 4 Given
a power series field k((w)), let r be a non-zero element of that feld.al order m = 1 After multiplying ¢ by a constant if necessary, we can write
t=u”+ bu”! 2 bạu t2 poet
= uM(1 + bie + byw? +77):
Then the power series field k(Œ)) is contained in k((w)) In fact, one sees
immediately that the degree of k((u)) over k(Œ)) is exactly equal to m Indeed, by recursion, one can express any element y of k((u)) in the following
manner
yah) +Adu too + fn(Qun
with f;(t) € k(()) Furthermore, the elements 1,u, , u™~' are linearly
independent over k((t)), because our power series field k((u)) has a discrete
valuation where u is an element of order 1, and ¢ has order m If we had a
relation as above with y = 0, then two terms f;(u' and f;(f)u’ would neces-
sarily have the same absolute value with i # j This obviously cannot be the
case Hence the degree of k((u)) over k((1)) is equal to m, and is equal to
the ramification index of the valuation in k((#)) having f as an element of order 1 with respect to the valuation in k((u)) having u as element of order 1
The following proposition gives the relations between the residues taken
in k((u)) or in k((#)) By Tr we shall denote the trace from k((u)) to k((1))
Proposition 4,2 Let k((u)) be a power series field, and let t be a non-zero
element of that field, of order m = 1 Let y be an element of k((u)) Then
dt
res, b 3u au) = res, (Tr(y) df)
f We have seen that the powers 1, u 1 rer un! fe orm a basis i for
‘over k((1)), and the trace of an element y of k((u)) can be computed
matrix representing y on this basis Multiplying / by a non-zero Not change the validity of the proposition Hence we may
§4 Residues in Power Series Fields
19
t= u" + by trees yn 7Œ + bua + bạu? + 3)
with by € & One can solye recursively
USA) + A(Qu a + frum! where f;(t) are elements of k
polynomials in bị, b;, „„
be written (()), and the Coefficients of f,(2) are universal - » With integer Coefficients, that is cach ƒ,(?) can #0) = 5 Pu(b)r", where each P;,(b) is a polynomial with finite number of b’s The matrix representing an arbitrary element integer coefficients, involving only a Bol) + (Nu t+ + gu of k((u)) is therefore of type Gm=10(1) + + + G—tym—1(0)
where G,,,(t) € k((1)), and where the coefficients of the G,,.(0) are universal
polynomials with integer coefficients in the b’s and in the coefficients of the 8,0) This means that our formula, if it is true, is a formal identity having nothing to do with characteristic p, and that our verification can be carried out
in characteristic 0
This being the case, we can write f = v", where vo =u + cu? +:°°
is another parameter of the field k((u)) This can be done by taking the
binomial expansion for (1 + bju + + + -)!" In view of Proposition 4.1, it
will suffice to prove that
res, (» 2 de) = res,(Tr(y) dt)
By linearity, it suffices to prove this for y = vw, -a <j < too (If y has
a very high order, then both sides are obviously equal to 0, and y can be
written as a sum involving a finite number of terms a;v/, and an element of
very high order.) If we write
Trang 161 Riemann-Roch Theorem 20 J)= pTr(v’)-We have trivially then 0í = 0" and Tr(v / io if j = ms Tr(v’) fr tre r2 lm ifr # whence Tr(e’) =) 9 otherwise Consequently, we get m ifj = — res,(Tr(v’) dt) = if otherwise
ion in terms > is equal to
On the other hand, our first expression in terms of v is equ
res, (vimv™ ' dv),
which is obviously equal to what we just obtained for the right-hand side
This proves our proposition
e started with a power series field k (()) and his section by showing that this situation is
In the preceding discussion, w a subfield k((1)) We conclude t ì
ical of power series field extensions
an F = k(0) be a given power series field over an algebraically closed
field k We have a canonical k-valued place of F, mapping ton 0 Let E be
a finite algebraic extension of F Then the discrete valuation of F extends in at least one way to E, and so does our place Let w be an element of E of order
1 at the extended valuation, which is discrete If e is the ramification index,
then we know by the corollary of Theorem | thate = [E : F] We shall show that e = [E : F] and hence that the extension of our place is unique
An element y of E which is finite under the place has an expansion
Ysataut-e++ +a jue! + my,
where y, is in E and is also finite This comes from the fact that u¢ and t
have the same order in the extended valuation Similarly, y; has also such
an expansion, yị = bọ + blu + +++ + beyut! + py», Substituting this
expression for y; above, and continuing the procedure, we see that we can write
y =fŒ) +fŒ)M + + + + +, i(0w°—! >
: no Ữ a pee series in k(()) Since the powers 1, u y linearly independent over k((1)), this el
ite =[E:
4nd that the extension of the place is unique Proves that = [E : F]
§5 The Sum of the Residues
21
and we can therefore solve recursive ere i inati
recursively for a linear combination of the powers
1,u, , 4°"! with coefficients in k() Summarizing, we get
Proposition 4.3 Let k(()) be a power Series field over an algebraicall
closed field k Then the natural k-valued place of k((1) h TẠI HIỆP
extension to any finite algebraic extension of Ky) If Eis kg
extension, and u is an element of order \ in the extended waluatan ie
E may be identified with the power series field k((u)) and [E : Fl = =
§5 The Sum of the Residues
We return to global considerations, and consider a function field K of dimen-
sion | over an algebraically closed field k The points P of K over k are
identified with the k-valued places of K over k For each such point, we have
an embedding K — Kp» of K into a power series field k(()) = Kp as in §2
Our first task will be to compare the derivations in K with the derivations : ae tions i
k((d) discussed in §3 "
Theorem 5.1 Let y be an element of K Lett © K bea local parameter at the point P, and let z be the element of K which is such that dy = zdt If dy /dt is the derivative of y with respect to t taken formally from the power series expansion of y, then z = dy/dt
Proof The statement of our theorem depends on the fact that every dif- ferential form of K can be written z dt for some z, by §3 We know that K
is separable algebraic over k(1), and the irreducible polynomial equation
S(t, y) = 0 of y over k(t) is such that f, (t, y) # 0 On the one hand, we have
O=fi(t, y) dt + f(t y) dy,
whence z = —f,(t, y)/f,(t, y) (cf Lemma 2 of §3); and on the other hand, if we differentiate with respect to f the relation f(t, y) = 0 in the power series field, we get
đ
0 =/#ứ, y) + 0Œ, y) = š
This proves our theorem
Let w be a differential form of K Let P be a point of K, and fa local
parameter, selected in K Then we can write w = y at for some y € K Referring to Proposition 4.1 of §4, we can define the residue of w at P to be
the residue of y dt at , that is
Trang 17I Riemann-Roch Theorem
22
idue is als' itten resp (x dz) We
i dz forx,z EK, this residue 1s also written res; (
wis written x dz for x, 7 t sid
= state the main theorem of this section
curve over an algebraically
K be the function field of a form of K Then
Theorem 5.2 Let lắt lai bên differenti closed constant fie S| resp (w) = 0 R F ver all points P, but is actually a finite sum since the is taken 0 :
(The sum is ly a finite number of poles.)
differential form has on
Proof The proof is carried out in two steps, first in a rational function field,
i i ition 4.2
i bitrary function field using Proposition
pea caiae tac cane where K = k(x), where xis a single transcendental quantity over k The points Parein ] — Ì correspondence itl the maps of xin k, and with the map l/x > 0 (i.e., the place sending x2 >) te is
not the point sending x to infinity, but, say the point x=a ack, i en
x — acan be selected as parameter at P, and the residue ofa differential orm
ydx is the residue of y in its expansion in terms of x — 4 The situation is the
same as in complex variables
We expand y into partial fractions,
y =D culx — by) + £0)
where f(x) is a polynomial in k[x] To get resp (ydx) we need consider only the coefficient of (x — a)~! and hence the sum of the residues taken over all P finite on x is equal to 2, Cyr
Now suppose P is the point at infinity Then ¢ = 1/x is a local parameter,
and dx = —1/t? dt We must find the coefficient of 1/ in the expression
—ylJr? It is clear that the residue at ¢ of (— 1//?)ƒ(1/0) is equal to 0 The
other expression can be expanded as follows: ` + thy +o var is) thí “|~ t Hi
and from this we get a contribution to th
Which gives precisely — Cui
.aố field
_ Next, suppose we have a finite separable algebraic extension K of a purel
transcendental field F = k(x) of dimension 1 over the algebraically closed
nt field k Let Q be a point of F, and t a local parameter at Q in F
-P be a point of K lying above Q, and let u be a local parameter at P in residue only from the first term, This proves our theorem in the case of a purely
§5 The Sum of the Residues
23
K Under the discrete valuation at P i
ordp f = e'ordp u The power series
degree ¢ of k(n) Thus for each P we get an embedding of K in a finit
algebraic extension of k((1)), and the place on K at P is induced b’ is
canonical place of the power series field k((w) es
Let PG = 1, , 5) be the points of K | i PG = nhở | S
ying above Q Let A
algebraic closure of k(n) The discrete valuation of k(@)) Trai nhức
toa valuation of A, which is discrete on every subfield of A finite over Ko)
(Proposition 4.3 of $4) Suppose K = F(y) is generated by one element y satisfying the irreducible polynomial 8(Y) with leading coefficient 1 over F It splits into irreducible factors over k((1)), say
,
K extending that of Q in F, we have
field k((u)) is a finite extension of
(1) 8Œ) = 81) - - - g.(Y)
of degrees dj(j=1, ,r) Let y; be a root of gj(Y) Then the mapping y > y; induces an isomorphism of K into A Two roots of the same 8; are
conjugate over k((2)), and give rise to conjugate fields By the uniqueness
of the extension of the valuation ring, the induced valuation on K is therefore
the same for two such conjugate embeddings The ramification index relative to this embedding is d;, and we see from (1) that Š dj =n By Theorem 2.2
of §2 we now conclude that two distinct polynomials 8; give rise to two
distinct valuations on K, and thats = r We can therefore identify the fields
k(())(y;) with the fields Kp,
For each i = 1, , r denote by Tr; the trace from the field Kp, to Fo Proposition 5.3 The notation being as above, let Tr be the trace from K
to F Then for any y € K, we have
Tr(y) = ¥ Tri(y)
i=1
Proof Suppose y is a generator of K over F If [K : F] = n, then Tr(y)
is the coefficient of Y"~! in the irreducible polynomial g(Y) as above A
similar remark applies to the local traces, and our formula is then obvious
from (1) If y is not a generator, let z be a generator For some constant
c Ek, w=y + cz isa generator The formula being true for cz and for w,
and both sides of our equation being linear in y, it follows that the equation holds for y, as desired
The next proposition reduces the theorem for an arbitrary function field K
to a rational field k(x)
Proposition 5.4 Let k be algebraically closed Let F = k(x) be a purely
transcendental extension of dimension 1, and Ka Sinite algebraic ate
rable extension of F Let Q be a point øƒF, and Pị ( = Ì,- - + › r) the
Trang 18I Riemann-Roch Theorem
24
: ot Tr denote the ment Of K, and let a
i Lety bean elemer lying above Q i K points of KioF Then trace from resg (Tr() dx) = Fesn (9 40 i=l be a local parameter at Q in k(x), and let hi We have local parame i ị
prod Lette” : ocal trace from Kp, to tọ in K at P; Let Tri denote the |
dx
SY, resp, ( dx) = 5) resp, (5 a a)
dx dt
= > resp, (> a dia au)
g Proposition 4.2 of §4, we see that this is equal to Le
Ss Tes (m ( ) a)
Since dx/dt is an element of k(x), the trace is homogeneous with respect to this element, and the above expression is equal to and usin » TESo (mo “ a) = > res (Tr;(y) dx) II reso (S\ Tr,(y) dx) = reSg (Tr(y) dx) thereby proving our proposition
Theorem 5.2 now follows immediately, because a differential form can be
written ydx, where K is separable algebraic over k(x)
Our theorem will allow us to identify differential forms of a function field K with the differentials introduced in §2, as k-linear functionals on the ring
A of adeles which vanish on some A(q) and on K This is done in the following manner Let £ = (
: : =( , 6 ) b „
differential form of K Then the a ) be an adele Let ydx be a
-linear map of A into k Here, of cour:
e $y and x as elements of Kp
erms Of our sum are 0 course, in the expression resp (&ydx),
- Itis also clear that all but a finite number
35, The Sum of the Residues
25 Our k-linear map vanishes on some A(«), because the differential f h
only a finite number of poles Theorem 5.2 shows that it vai ishes mK I is therefore a differential, and in this way we obtain an embeddi xã s£ - K-vector space of differential forms into the K-vector space oF diffe earatl Since both spaces have dimension 1 over K (the lat
this embedding is surjective (the latter by Theorem 2.6 of §2),
Let ydx be a differential form of K If P isa point of K, we can define the
order of ydx at P easily Indeed, let ¢ be an element of
power series field k(()), the element of order 1.at P TIn the dx
a
is a power series, with a certain order mp independent of the chosen element t
We define mp to be the order of ydx at P, and we let the divisor of ydx be
(ydx) = > mpP
Suppose ordp( ydx) = mp If ordp(é) = —mp, then
ordp(épydx) = 0,
and the residue res p(& ydx) is 0 Hence the differential A vanishes on A(q),
where a = (ydx) On the other hand, if A(b) is the maximal parallelotope on which A vanishes, then A(b) D A(a), and b = a If > a, then for some
P, the coefficient of P in b is > mp, and hence the adele
( 0,0, 1/r*!,0,0, )
lies in A(b) One sees immediately from the definitions that resp(t~”~' ydx) # 0,
and hence A cannot vanish on A(6) Summarizing we have
Theorem 5.3 Let K be a function field of dimension | over the algebra-
Trang 191 Riemann-Roch Theorem 26 iffe tials Furthermore, the divisors (ydx) and (A) of §2 < differen ‘i the K-space of ¢ are equal | ¬ Sóc i r ơ(q) is the dimension of the space of differential y iege l Corollary 1 The in forms w such that (w) 2 a
fi ind if it has es, that is if differential form « is said to be of first kind if it has no
poles, that is if
A differentia
(w) 2 0
forms of first kind is denoted by dfk For any divisor
The caper of differential forms @ such that (w) 2 —a Then
a, let Diff (a) be the spa
dim Diff(a) = 6(-a)
If a = 0 it is clear that Diff (a) contains the space of differentials of first
kind
Corollary 2 For any divisisor a > 0 we have
ô(—a) = dega — l+ 8
and
dim Diff(a)/dfk = deg a — 1
Proof Since (—a) = 0 because a function which has no poles and at least
one zero is identically 0, the formulas are special cases of the Riemann-Roch
theorem
§6 The Genus Formula of Hurwitz
The formula compares the genus of a finite extension, in terms of the ramifica-
tion indices
Theorem 6.1 Let k be algebraically closed, and let K be a function field
with k as constant field Let E be a finite separable extension of K of degree
n Let ge and gx be the genera of E and K respectively For each point
P of K, and each point Q of E above P, assume that the ramification index €g 'S prime to the characteristic of k Then
Bt — 2 = nex ~ 2) +S (eg ~ 1)
Q
§7 Examples
27 Proof If wis any non-zero different
degree is 2g — 2 Such a form can b can also view x, y as elements of E;
compare it with that in K to get the formula, as follows Let P be a point of
K, and let t be a local parameter at P, that is an element of order | at P in
K If «is a local parameter at Q, then
ial form of K, then we know that its
¢ written as ydx, with x, y © K We
we can compute the degree in E, and
t=utv,
where v is a unit at Q Furthermore, dt = ué dv + eu®-'v du Hence ordg ( ydx) = eg ordy (ydx) + (eg — 1)
Summing over all Q over P, and then over all P yields the formula §7 Examples
Fields of genus 0 We leave to the reader as an exercise to prove that k(x)
itself has genus 0 Conversely, let K be a function field of genus 0 and let
P be a point By the Riemann-Roch theorem, there exists a non-constant
function x in L(P), because
(P)=1+1-04+0=2,
and the constants form a I-dimensional subspace of L(P) We contend that K = k(x) Indeed, x has a pole of order | at P, and we know that [K : k(x)] is equal to the degree of the divisor of poles, which is 1 Hence we see that K is the field of rational functions in x
Fields of genus 1 Next let K be a function field of genus 1, and let P again be a point We have 2g — 2 = 0, so the Riemann-Roch theorem shows that
the constant functions are the only elements of L(P) However, since deg(2P) = 2, we have
I2P)=2+1—1=2,
So there exists a function x in K which has a pole of order 2 at P, and no other
pole Also
IGP) =3+1-1=3,
So there exists a function y in K which has a pole of order 3 at P The seven
functions 1, x, x?, x°, xy, y, y? must be linearly dependent because they all
Trang 201 Riemann-Roch Theorem I(6P) = 6 + = ance, the coefficient of y? cannot be 0, for en depen ossible as one sees from the parity of the is is impOoSSIĐ't: ) and this 1s in k(x) at P- x koe ` cS đivisor of poles of x is 2 we have h Ề In a relation h otherwise y © KC Jes of functions Since the degree © [K: k(0Ì =2: Similarly, |K: k()l = 3-
eld between K and k(x), and since y is not
There is no strictly intermediate fi
in k(x), it follows that
K = k(x, y)-
je relation inear È between the above seve ermore ation of lineal dependence
Furthermore, th latioi
n
functions can be written
yỀ= c¡y + GA + x? + GÀ” + 1 + Coe
In characteristic # 2 or 3, we can then make simple transformations of variables, and select x, y so that they satisfy the equation
y? = 4x7 -— cx — 63,
familiar from the theory of elliptic functions
Hyperelliptic fields Let K = k(x, y) where y satisfies the equation
y =f),
and f(x) is a polynomial of degree n, which we may assume has distinct roots Let us assume that the characteristic of k is # 2 Then the genus of
Kis
n-1
_
; Proof Let f(x
is unramified over k(x) at all
x = 4, and also Possibly at
Tamification index is 2, Supp
) = IN (x ~ aj) where the elements a; are distinct Then K points except the points P; corresponding to
those points lying above x = oo At P; the ose first that n is odd Let ¢ = 1/x so that ¢ has
§8 Differentials of Second Kind 29
order | at co in k(x) We write
fo) =" T] C= ta) '
Each power series | — ta, has a square root in k[[1]], while for n odd, the
square root of ¢~" shows that k(x, y) is ramified of order 2 at infinity The Hurwitz genus formula yields
28x —2= 220-2 4+ (2—l)+(2—l)=—4+n +
Solving for 8k yields &k = (n — 1)/2 If n is even, then the ramification index at infinity is | and the Hurwitz formula yields gx = (n — 2)/2 This proves what we wanted
§8 Differentials of Second Kind
In this section all fields are assumed of characteristic 0 A differential form w is called of the second kind if it has no residues, that is if
respw =0 forall P
It is called of the third kind if its poles have order = 1 The spaces dsk and dtk of such forms contain the differentials of first kind
The Riemann-Roch theorem immediately shows that the differentials of
first kind have dimension g, namely 6(0) = g, equal to the genus
A differential form is called exact if it is equal to dz for some function z It is clear that an exact form is of the second kind We shall be interested in the factor space
dsk/exact
Theorem 8.1 Assume that K has characteristic 0 Then
dim dsk/exact = 2g
Proof We define dsk(a) and dtk(a) just as we defined Diff (a), that is, forms of the prescribed kind whose divisor is = —a
Let Pi, , P, be distinct points, and let N be a positive integer such
that (V — 1)r > 2g — 2 Ifa differential form is exact, say equal to dz, and lies in
Trang 211 Riemann-Roch Theorem 0 h 3 e points P, of orders at Most N, then 2 that in other words, if „ and converselY- Note tha ;eL(w=U3? > pydL (UN — 1) = Pi), = U wk( 3 P04 dsk/exact =
vor all Nas above, and all choices of points aken over a it will suffice to prove that each factor the Lo Ung if that is the case, then inereas
er any further contribution to dsk/exact, where the union is ¢
Py a ge Pe To prove ee
space on the right has dimens a
ing N or the set of points cannot y
This will then also prove:
P, be distinct points, and let N be a positive
Theorem 8.2 Let Pies pos De — 3 Then
integer such that (N — Ir > <8
dsk/exact = dsk(N S P)jaL(W = 1) 3ñ)
Note that
dim đ(@ D Š®) =@M= 3B) — 1,
because the only functions z such that dz = 0 are the constants On the other
hand, also note that
dfk M exact = 0,
because a non-constant function z has a pole, and so dz also has a pole
By Riemann-Roch (cf Corollary 5.7) the dimension of the space of dtk having poles at most at the points P; modulo the differentials of first kind has dimension (-EP)—g=r-1 Putting all this together, we find, dim dsk(N © P)L((N — 1) 5 P) = dim Diff(W > P)JdL((N — 1) =P) — dim dtk(> P) = 6(-N > Pp) - (IN )P)— I— @— 1) §9 Function Fields and Curves 31 E3 —~2+Mr+l~g=(@W~Dr—1+g+1—~0-Ð) = 2g
This proves the theorem,
§9 Function Fields and Curves
For technical simplicity, we assume again that k has characteristic zero
Let K be a function field in one variable over a field k This means that K is of transcendence degree 1, and finitely generated If we can write
K = k(x, y), with two generators x, y, then we may call (x, y) the generic
point of a plane curve, defined by the equation f(X, Y) = 0, if f is the
irreducible polynomial vanishing on (x, y), determined up to a constant factor A point (a, b) lies on the curve if and only if f(a, b) = 0 We shall
say that the point is simple if D.f(a, b) #0
If o is a discrete valuation ring of K over k (i.e., containing k) and m its
maximal ideal, then 1 is principal, and any generator of 1m is called a local
parameter of 0 or m1 Assume that the residue class field 0/11 is equal to k
Let @: ð + v/m be the canonical map If K = k(x, y) and x, y € 0, then we let a = g(x), b = ¢(y) We see that (a, b) is a point on the curve
determined by (x, y) Ifz © K,z & 0, we can extend ¢ to all of K by letting (2) = 0 We call ga place of K (over k) We say that the point (a, b) is induced by the place on the curve Local Uniformization Theorem Let K be a function field in one variable over k (1) Let K = k(x, y) where (x, y) satisfy an irreducible polynomial S(X, Y) = 0 over k
Leta, b € k be such that f(a, b) = 0 but Df (a, b) # 0 Then there
exists a unique place y of K over k such that g(x) = a, @(y) = b, and if 0 is the corresponding discrete valuation ring with maximal
ideal m, then x — a is a generator of m
(2) Conversely, let v be a discrete valuation ring of K containing k, with
maximal ideal, m, such that o/m = k Let x be a generator of m
Then there exists y € 0 such that K = k(x, y), and such that the
point induced by the place on the curve is simple
Proof To prove (1), we shall prove that any non-zero element
Trang 221 Riemann-Roch † heorem 32
‘ten in the form
can be written pile A(, 9) gr =O BEY)
This proves that the ring y
¡als, and Z4 “Tạng thất
re polynomials, 3 als gilts Y9/B2(%» 9) Will 656, b) ý 0
oe “i hat y and tha ais generator 8 of its maximal ideal, e, so assume 6 where A, B a consisting of all que is a discrete valuation If g(a, b) $9, nein (a, b) = 0 Write œ are d0! #0) € k[Y] )=~ b4) 0) E k[Y] g(a, Y ar - DA) f(a, Y)
shen f(b) # 0 since Dafta, b) #0 Hence
eta, YAY) = Lea V8)
It follows that
eX, YAW) — FA Ve) = X — OA, ¥)
for some polynomial A) Hence
g(a, y) = (4 - ayAy(x, /fiQ)
e If not, we continue in the same way We
If A\(a, b) #0, we are don \
on for otherwise, we know that there exists some
cannot continue indefinitely, j
place of K over & inducing the given point, and g(x, y) would have a zero of infinite order at the discrete valuation ring belonging to that place, which is
impossible
Conversely, to prove (2), let K = k(x, 2) where z is integral over k[x]
Let z = 2), , 2, (# = 2) be the conjugates of z over k(x), and extend 0 to a valuation ring © of k(x, 21, ., Zn) Let
ZSMt ax tor tax perce
be the power series expansion of z, with a; € k, and let P(x) = ap +++ +a4,x" Fori=1, ,n let 2 = Pr) yet x
§9 Function Fields and Curves 33
If we take r large, then y, has no pole at ©, but Vay ce, yn have poles at ©
The clements vị, , y„ are conjugate over k(x) Let ƒ(X, Y) be the
irreducible polynomial of (x, y) over k Then
LOGY) = XY" $+ Ul
Furthermore, (0) # 0 for some i, otherwise we could factor out some power of X from f(X, Y) We rewrite f(x, Y) in the form
Œ, Y) = iO y2 oe (Y= v(t Y_— )): ‘ (4 y— ))
* Yn
In the valuation determined by ©, we see that the coefficient
= YX) 2 Vn
cannot have a pole (otherwise divide the two expressions for f(x, Y) by this coefficient and read the polynomial modulo the maximal ideal m of 9 to get a contradiction) If we denote by a bar the residue class of an element of 0 mod m, then OFF Y) =(-1)"' WY — 5,) By definition, ¥ = a = 0 We let y = y, and} = b Then Đ;ƒ(a, b) = (11h #0, as was to be shown
Corollary Let K be finite over k(x) There is only a finite number of valuations of k(x) over k which are ramified in K
Proof There exist only a finite number of points (a, b) such that f(a, b)=0 and D,f(a, b) = 0,
and there exist only a finite number of valuation rings of K such that x does
not lie in the valuation ring (i.e., such that x is at infinity)
If K = k(x, y) and f(x, y) = 0 is the irreducible equation for x, y over k,
then one calls the set of solutions (a, b) of the equation f(a, b) = 0 an affine plane curve, which is a model of the function field If all its points are simple, the curve is called non-singular The totality of all places of K (which are
Trang 231 Riemann-Roch Theorem
34 ‘There may of course not be a single
fine +L always excludes
3 “ 10 >, an affine model a y udes
associated with the neat - vhịch is non Further ing to places ‘which map x to 00 This could - of the function
fie 5
model a corresponding“ t models of the Tunction ficld For
oe ae of by considering Oe all k-valued places the complete
be taken ses, it suffices + suffices t0 C8 a) with : this @ sas the curve Ee
"ae
our PUP model and to deal with e is defined to be the genus of its
i jar T enus of a
non-singu genus
For our pu! rhe elements 0! s of R are in bijecti are 1n bi 10N with i function field Let R be the set of poin 5K containing the constant fie f points 0 fK The lemen : » constant field & We speak a ak of
the discrete valuation rings 0! the above complete non-singular model We ts of R as the 4 ference to K once to K -
clemen v) if we WIS > Then the inclusion F C K gives write R(K) if Ki sẽ sa t Supposc that & f rise to a mapping yn field K tof poses, the Git points on | sis h to specify the i finite extension O} œ: R(K) RF)
K associates the ring 0M F of F This
he points of an affine model Indeed, if
ath ., y) = 0, and (a, d) is the
© with irreducible equation f(x, y) ›
rae i es, Kệ Đ and iEF = k(u, v), where g 10 0; DU u, 0 are 1n D and the point
le II (c, d) on (u, v), then we may W rite
which to each valuation ring 0 of
mapping ¢ can be representing on ¢
u = gil, Y)s v= (X,Y)
where g, @ are rational functions whose denominators do not vanish at (a, 6) Then
c = 9,(a, b) and d = oa, b)
§10 Divisor Classes
Let Gp be the group of divisors of degree 0, and %, the subgroup of divisors of functions The factor group
6 = Do/D
is called the group of divisor classes
Suppose that K is a finite extension of F, and let gy: R(K) > R(F)
be the associated map on the curves Then g induces a homomorphism
Px: CK) > GF),
§10 Divisor Classes 35
Indeed, y, is defined to be y on points, and is extended by Z-linearity to
divisors Itis an elementary fact of algebra that ¢, maps divisors of functions
to divisors of functions In fact, if z © K then gy A(z) = (Nez),
in other words, the image of the divisor of (z) under ¢ is the divisor of the
norm For a proof, cf Proposition 22 of Chapter I [La 2]
The map ¢ on divisors also induces a contravariant map
œ@*: €(F) > €(K)
as follows Given a point Q of F, let Pj, , P, be the distinct points of
K lying above Q, and let e, be the ramification index of P, over Q Then we define
r
e*(Q) = > e(P)
isl
Then ¢* also maps the divisor of a function z in F to the divisor of that same
function, viewed as element of K This is obvious from the definition of the
divisor of a function
It is also immediate that
g„œ@* =[K: F],
because if z © F then Ny(z) = 2" where n =[K:F] Or, alternatively, because in the above notation,
r
Trang 24
this chapter is to give a significant example for the notions and
d in the first chapter
The purpose of
theorems prove f Dae
The reader interested in reaching the
course omit this chapter
Abel-Jacobi as fast as possible can of
$1 The Genus
We consider the curve defined by the equation
xt yY =]
over an algebraically closed field k, and assume that N is prime to the
characteristic of k We denote this curve by F (NV) and call it the Fermat curve
of level N We suppose that N = 3 and again let K be its function field,
K = k(x, y)
We observe that the equation defining the curve is non-singular, so the
discrete valuation rings in K are precisely the local rings of points on the
Curve, including the points with x = oo, arising from the projective equation vX N y xe + x" ce 2 with z = 0, We have the expression ¬¬ §2 Differentials 37
where the product is taken over
equal to an N-th root of unity, th
y = 0, and K is ramified of ord
equation Hence
all £¿ € Hy (N-th roots of unity) If x is set
en we obtain a point on the Fermat curve with
er N over this point, as is clear from the above
[K:kQ@)] = N,
and the ramification index of K over the point x = fis N On the other hand, let: = I/x Then có
a 1
Nv N
Y =ngyữứ`—
3 tt ( 1),
and —1 + ris a unit in k[[r]
K, and there exist N distinct
the points at infinity in this se coordinates
1 Hence x = © (ort = 0) is not ramified in
Trang 25
IL The Fermat Curve
38 f wes lie among the points with » < oo oF Ye?) is finite # 0 then the €XPT€ssion int such that - If P is a point such that XP) =9
in shows that dv/y*~' has no Pole at p
is for the differentials of first kind is given by w,, With m 2.1 A basis J? = he Theor’ n thatr +8 = ¡<r, s such "_ tr = Tp Proof Suppose that x(P) = % Pu 1 đi = ~ã dt, and ordp y = ordp x = 1 Then — dt we-Lyscl = Ey? Ons =* 3 r y’ — I then ordp w,,, 2 0 This proves ‘chit atifr +5 =N ae
from which oe a also clear that the differential forms as stated are
> say independent over the constants, and there are precisely g of them, inearly i
where g is the genus of F(N)
We observe that these forms have an additional structure / The group by X py acts as a group of automorphisms of F (N) by the action
(yy? (Gx Gy)
where #w is a fixed primitive N-th root of unity Over the complex numbers,
we usually take ấy = e?”*, Then the form w,,, (without any restriction on
the integers r, s) is an eigenform for the character X,,, such that
X„(, 8) = 89,
The linear independence of the differentials of first kind in Theorem 2.1 can therefore also be seen from the fact that they are eigenforms for this Galois group, with distinct characters
When we view fly X py as a group of automorphisms of F (N) we shall
fin wnite it as G = G(N), and call it simply the group of natural automor-
Phisms of the Fermat curve, or also the Galois group of F(N) over F(1)
Theorem 2,2, The forms w,,, with
1S" SSN ~Jandr +5 # 0 mod N
Constitute a basis for dskJexact
§3 Rational Images of the Fermat Curve 39
Proof Let
N
2 = >) (~,)
ml
be the divisor of points above x = o on F(N), taken with multiplicity |
First we note that the space
dtk(c0)/dfk
has dimension N — | by the Riemann-Roch theorem (Corollary 2 of Theo-
rem 5.3, Chapter I) Checking the order of pole at infinity shows that the
forms
w,; with r+s5 =N and lsr,s
are of the third kind, and obviously linearly independent from the differentials
of first kind in Theorem 2.1 Since they have the right dimension, they form
a basis of dtk(oo)/dfk
Given any differential w, it follows that there exists a homogeneous polynomial h(x, y) of degree N — 2 such that
dx
w — h(x, y) =a y yy
is of the second kind We apply this remark to the forms œ,„ Withr + s #0
mod N We operate with (¢, £) on the above difference, and note that the
automorphisms of F (N) preserve the spaces of dsk Subtracting, we then find that
(= {*)o,, with r+s ¢0modN
is of second kind, whence w,., is of second kind
Finally, we note that the forms w,, with | < r,s <N — landr +5 #0 mod N, taken modulo the exact forms are eigenforms for the Galois group
Hw X py with distinct characters, and hence are linearly independent in
dsk/exact Since the number of such forms is precisely the dimension of dsk/exact, it follows that they form a basis for this factor space, thus proving
the theorem
§3 Rational Images of the Fermat Curve
Throughout this section, we let 1 Sr, sandr+ss N = 1 Such a pair
Trang 2640 - is function feld) ve (subfields of its Fermat curve (su Faddeev [Fa 2]
I The Fermat Curve
following Rohrlich [Ro] after We put preg wax” and vary We let p= g.c.d.(r, s, N) and M=N/D We write per'D,s =s'D so that ứ,s, M) =1 Then u, v are related by the equation 0Ÿ = w/(1 — 85
which, in irreducible form, amounts to
mau (I- uy’
We let F(r, s) be the “non-singular curve” whose function field is k(u, v), so
that we have a map
F(N) > F(r, 8),
given in terms of coordinates by
(x, y) > (u,v) = (XW, x’y’)
If 1 © Z(M) = Z/MZ, we let (t)y be the integer such that
OS(t)y SM — 1l and () = mod MM
If M = N we omit the subscript M from the notation If we let K (N) and
K(r, s) be the function fields of F(N ) and F(r, s) respectively, then K(N) is
Galois over K(r, s) Let G(r, s) be the Galois group K(N) = k(x, y) 6= wy x My | KỨ, 5) = k(u, v) K(1) = ku) G(r, s) §3 Rational Images of the Fermat Curve 41
We note that G(r, s) is the kernel of the character Y, : rs) and that K(r, s) over K(1) is cyclic, of degree M It is in fact a Kumm er extension be
Special case Consider the case when N = P is prime = 3 and r=s=l, so the intermediate curve is defined by 0P = (Ì — n, which is therefore hyperelliptic The change of variables t=2u-—1 changes this equation to P= 1 — 4p",
which is often easier to work with
Let m € Z(M) We say that m is (r, s)-admissible if (mr) and (ms) form an admissible pair, that is
1 S (mr), (ms) and (mr) + (ms) SN — 1
Theorem 3.1 A basis of dfk on F(r, S) is given by the forms
mr) dms)
for all (r, s)-admissible elements m
Proof It is clear that x” y lies in the function field of F(r, s), and
hence that the forms listed above are of the first kind on F(r, s) Conversely,
suppose w is a dfk on F(r, s) Write
o= > Cat qu
where the sum is over all admissible pairs (q, 1) Since @,, is an eigenform OŸ uy x my with eigencharacter Xạ„, and since the factor group
(Hy X My)/ Ker X,5
Trang 27I The Fermat Curve Xạ =ẤP = and f= (ms), as y), as desired ¢ i intege! But then 4g (nr
for some integer /7!-
that m Z(M)*, : ds of F(r, 5) and F
and that is (r, s)-admissible Then the )) are equal, for instance becaus
Suppose (mr), (ms)) are ed a because
function fiel
Ker X,5 = Ker X me) dns)
between the curves in terms of coordinates is given as ce be The corresponden follows M — 1 be prime to M and such that its residue class mod y Letl <m <M — !" is (r, s)-admissible Write ((mr), (ms)) = (rs sy +NG, J) with some ith some pal ir of integers / f integi J hen we have a commutative diagram: F(N) ⁄ N Fữ, s9 — F((mr), (ms))
where the bottom arrow is given by
(u, v) > (u, o™ul(L — 4’),
realizing the automorphism of the function field corresponding to the two
models F(r, s) and F((mr), (ms))
Two admissible pairs (r, s) and (q, f) are called equivalent if there exists
m € Z(M)* such that g = (mr) and t = (ms) It is clear that inequivalent
pairs correspond to distinct subfields K(r, s) and K(q, f) On the other hand, if (r, s) and (q, ?) are equivalent, then
K(r, s) = K(q, 0)
Given the admissible pair (r, s), and m € Z(M)*, we observe that there i Precisely one value of m or —m such that ((mr), (ms)) is admissible This
ollows at once from the fact that for any integer a # 0 mod N we have
(a) + (-a) = N,
We shall now appl
is prime = 3, y this to the most interesting special case when N = P
§4 Decomposition of the Divisor Classes
Theorem 3.2 If (r, 8) the curve F(r, s) has genus (p — 1)/2, and K(r, s) = K(L *) or N = p is prime = 3, then for every admissible pai
a uniquely determined integer s* such that the pair a, s*) is aided Proof The genus can cither be computed directly as we did for the Fermat curve, or one can use Theorem 3.1 The number of m such that ((mr), (ms)) is admissible is trivially computed to be (p — 1)/2, using the remark sreced- ing the theorem The statement that K(r, s) = K(1, s*) is clear *
Of course, instead of (1, s*) we could also have Picked a representative in
the equivalence class of (r, s) to be (r*, 1)
Remark If we define F(p — 1, 1) by
0771 = PT (1 — g),
then F(p — 1, 1) has genus 0, since it is also defined by
where w = v/u, and (1 — w)/u is a fractional linear transform of u, so a
generator of k(w) The function field of F(p — 1, 1) is therefore equal to
k(w)
§4 Decomposition of the Divisor Classes
Let F, = F(1, k) fork =1, , p — 2 and let Si F(p) > Fy
be the associated rational map As we have seen in Chapter I, §10 there is an associated map f;,,, on divisor classes, as well as an inverse map f* Let
Trang 28Il The Fermat Cur ve pre DS he next theorem only over k = C 11 give the proof of We have f*° f
the automorphisms of FIN) induced by (x
We use the same letters for the p For any divisor a of degree 0m ung
d inverse image give " § We sha = pid Theorem 4 1 Let A and B be respectively
he divisor class grou,
definitions of the direct image an Proof (tx, y) and @ cy) automorphisms of t F(p) the elementary the formula St Seg) = >, (A7* BY (a) F pol Hence p=2 pa} ; srof= 5, 2a NB k=1 j=0
We wish to show this is equal to ish ¢ p* id on divisor visor classes W
that this is equivalent to showing this same relation when the * need
" fat
ier applied to differentials of first kind Any general ene ee
is fact The reader can deduce it for instance from the duality ath ee €Orems
5.5 and 5.6 in Chapter IV We as d 5 ssume this fact i Then it s
ae : : en it suffices
h : = ete then is applied to the differential a ° en
<p — 1 Since such forms are eigenforms G ea wil
c \ genforms for the Galois
(p) over F(1), we see that the above relation is Eilinflnt HH relation of s pol i j(s—rk) — k=l h $ Es Sineer+sSp-—1w in ; P 1 we see th at for each pair r D s th ere exists a unique k i ks = r (mod p)
For this value of k, the s i
values of k, th € sum over um over j is equal to j is equal to 0 ts weve làm hà King: s the theorem
The the rem gives us 0) ves as equen ence of maps f ip: ƒ p-2 g — ® %,— %$ P whose composition j en Pid Sor classes is isomorphi In Chapter IV, we shall i to a complex torus, and fi be tna fa :
; it is clear that f, f* &
§4 Decomposition of S f the Divis: or Classes
4
complex analytic ho
rem 4.1 we EiiliiBIiM 2m To over the compl
dimension of the group ot ai o f has finite kenel ex numbers
From Theo-
ivisor classes is equal Nà Ha O the genus the fact that the » we see that
; p>?
dim € = > dim €,
Thus f must be surjecti ° surjective, and =3
Trang 29
chapter is to show how to give a structure of analytic
f points on a curve In the complex numbers, but our
general fields like p-adic numbers We are
plex case, in order to derive the Abel-Jacobj
The purpose of this manifold to the set 0
treatment also applies to more
principally interested in the com theorem in the next chapter
§1 Topology and Analytic Structure
Assume now that k is locally compact It can be shown that & is the real field,
complex field, or a p-adic field A point P of K is called k-rational if the
residue class field o/p of its valuation ring is equal to k itself The local
uniformization theorem shows in fact how to interpret primes as non-singular
points on plane curves We let R be the set of all k-rational points of K over
k, and call it the Riemann surface of K over k We can view the elements of K as functions on R If P ER, we denote by 0; the valuation ring
associated with P, and by mp its maximal ideal If z-E op, then we define
2(P) to be the residue class of z mod Mp Thus
2(P) Ek
ee then we define 2(P) = oo The elements of k are constant func-
Am ah the only constant functions If z(P) = 0, we say that
Tay Sê ue if'2(P) = co, we say that z has a pole at P
the compact space, we let I, = {k, oo} be the Gauss sphere over k, that 1s,
*pace obtained by adjoining to k a point at infinity We let T= I] I, x€K §1, Topology and Analytic Structure § 47 We embed R in in the obvious way: An element P goes on the product
H ae We peice R as a subspace of I’, which amounts to saying that
the topology is the one having the least amount of oj functions x € K are continuous pen sets such that all the 5 2
The product TP is compact, and we contend that R is closed in This
implies that R is compact ‘
Proof Let (a,),ex be in the closure Let 0 be the set of elements x in K
such that a, # oo Then 0 is a valuation ring, whose corresponding point 9
is such that x(Q) = a, for all x This is easily proved We first note that k C 0 because a(P) = a for all a © k and P ER Letx, y Eo Then
a,, @y # 0 By assumption, there exists P ER such that x(P), y(P), and (x + y)(P) are arbitrarily close to a,, dy, Ax+y Tespectively For such P, we
see that x(P) and y(P) + oo, whence (x + y\(P) # 0 Hence x + y€o
Similarly, x — y and xy lie in 0, which is therefore aring Furthermore, the
map x +> a, is a homormophism of ø into k, and is the identity on k This
follows from a continuity argument as above Finally, o is a valuation ring,
for suppose x € 0 Then a, = co Lety =x7! There exists P € R such
that x(P) is close to a, and y(P) is close to a, Since x(P) is close to infinity, it follows that y(P) is close to 0 Hence ay = 0, soy € 0 This proves our
assertion
Let P be a point Lett bea generator of the maximal ideal mp, i.e a lo-
cal uniformizing parameter at P We shall now prove that the map
Q >:(0)
gives a topological isomorphism of a neighborhood of P onto a neighborhood of Oink
According to the local uniformizing theorem, we can find generators t, y
for K such that the point P is represented by a simple point with coordinates (a, 6) in k, and in facta = 0 Split the polynomial f(0, Y) in the algebraic
closure k* of k Then b is a root of multiplicity 1, so we have
£O, Y) = (¥ — bY — by)? ++ + (Y — b,)”
The roots of a polynomial are continuous functions of the coefficients There exists a neighborhood U of 0 in k such that for any element 7 € U, the
polynomiahƒ(7, Y) has exactly one root in k*, with multiplicity 1, and this root is close to b (in the algebraic closure of k) However, using, for instance, the Newton approximation method, starting with the approximate root b, we can
refine ở to a root of f(7, Y) in k itself, if we took U sufficiently small Hence
the map Q +> 1(Q) is injective on the set of Q such that (t(Q), y(Q)) lies in
a suitably small neighborhood of (0, b) Since the topology on R is deter-
Trang 30IH The Riemann Surface 48 gives a topological isomorphism 1, the map er @Q)
for U sufficiently : small, f 0 in k na vohbeo
of U onto a neighborhood hạt 1(U) isa disc on the t-plane We observe that
We can choose U such i ving the local uniformization theorem shows that
apply!
if Q is close to P, then Vormizing parameter at Q, because the condition 1—1(Q) is a local a partial derivative is satisfied by continuity Indeed, if ï ni
concerning the seco Dif (a 7 b) # i det 0 for all (a’, b’) sufficiently close to D;ƒ(a, b) # 0, then De (a, 5) tị chart Theorem 1.1 Taking as ¢ the maps given by local parame! to R
5 discs on t-planes as above, together with ters, gives an analytic manifold structure
arameters at P, then f has a power series PER,3 are two parameters al ", kg Proof If R, andt, u expansion in terms of u, say t= > ayu', ¡=0
and since fis algebraic over k(u), simple estimates show that this power series
is convergent in some neighborhood of the origin Hence
1(Q) = > aiu(Qy
for Q close to 0 in the t-plane, and we see that the functions giving changes
of charts are holomorphic
Theorem | is valid for any field k which is real, complex, or p-adic From now on, we shall assume that k = C is the field of complex numbers
We have the notion of a meromorphic function on R, that is a quotient of holomorphic functions locally at each point If fis such a function, we can
write it locally around a point P as a power series ƒ(0) = Ÿ by where / is a parameter at P, and a finite n occur In this wa Kp ~ C(()), and number of poles
| umber of negative powers of £ may
y, ƒ may be viewed as embedded in the power series field thus can be viewed as an adele because it has only a finite
on R since R is compact
Theorem 1,2, Every meromorphic function on R is in K
Proof roof Let L be the field of meromorphic functions on R If L # K, then
§1 Topology and Analytic Structure
49
the degree (L : K)c¢ of the factor space of L mod K ov
infinite We have: er the complex is
(Li Kye = (L + AO): K + AM)e + (LA AW): K NAO))e
The first term on the right is finite as was shown in Weil’s proof of the
Riemann-Roch theorem The second is 0 because a function having no pole
is a constant (by the maximum modulus principle) Contradiction Theorem 1.3 The Riemann surface is connected
Proof Let S be a connected component Let P € $ and letz € K be a function having a pole only at P (such a z exists by the Riemann-Roch theorem) Then z is holomorphic on any other component, without pole,
hence constant, equal to c on such a component But z — c has infinitely
many zeros, which is impossible since R and S are compact
Theorem 1.4 Let z € K be a non-constant function The points of K induce points of C(z), and thereby induce a mapping of R onto the z-sphere S., which is a ramified topological covering The algebraic ramification index ep at a point P of R is the same as the topological index, and the number of sheets of the covering is
n =([K: C(z)]
Proof Let P be a point of R and ¢ a parameter at P Then P induces a
point z = a on the z-sphere, and f° is equal to z — a times a unit in the power series ring C[[#] Since one can extract nth roots in C, we can find a local parameter u at P on R such that wu‘ = z — a, where a = z(P) The
map
u(Q) +> u(Q) = 2(Q) — a
gives an e to | map of a disc Vp around P onto a disc V, in the z-plane We
shall call such a disc regular for P We have shown that the topological
ramification index is equal to e As to the number of sheets, all but a finite
number of primes of C (z) are unramified in K and, therefore, split completely into n primes of K (by the formula % e; =n) Hence n is the number of
sheets
Theorem 1.5 The Riemann surface is triangulable and orientable
Proof We shall triangulate it in a special way, used later in another
Trang 31
Hl The Riemann Surface
0
5 | ven triangulation) A men high subdivision of the
_ Poin achieves the role ae vertices is ramified, then A jg
PA is a transl eighboriood of a point, with ramification index
aoe " ™ A is a triangle with a ramifie «fed vertex Q, then AC Vo hai, each vertex lifts t0 a certain
i iangulation ntex lifts to
a cert
We can to nf sae A none of whose
vertices is ramified lifts ber of vertices on / * triangles in R tofR ramified above 4 in S:, t Q on R We get a map num uniquely to 7 If Q isa poin' and ¢ a parameter a we let m be its ramification index Vo> Vy by pomeaz—a if we choose f suitably Each pointz — 4 = re’ has m inverse images if we c : ¡8 ) )p= l, , 1 Um = = XỊ + part exp b m
Hence each triangle A with a vertex at ¢ lifts in m ways
As to orientability, we can assume that two triangles of our triangulation,
none of whose vertices are ramified and having an edge in common, are
contained in some regular disc Secondly, if A has a ramified vertex q, and
A, has an edge in common with A, then A, A, are both contained in V,
Now we can orient the triangles on S; such that an edge receives opposite
orientation from the two triangles adjacent to it If A on R covers A, we give
& and its edges the same orientation as A In view of our strengthened
conditions, we can lift the orientation, so R is orientable
Let V, E, T be the number of vertices, edges, triangles on the z-sphere
Then denoting by a prime the same objects on R, we get
F=nE— T'=¡T
On the other hand, let r be the number of primes p of C (z) which are ramified,
ee co Let ny be the number of primes P of K such that P lies above p-
§2 Integration on the Riemann Surface S1 VỀ =nV —mr + ` mụ p = nÝ + 3) ứ„ — n) p p =nV — 3 ( — 1) p PÌp =nv — > (ep - 1) P
where this last sum is taken over all primes P of K
Let X, be the Euler characteristic of S, and Xp that of R We shall prove:
Theorem 1.6 Let g be the topological genus of R Then
2g -2=-2n+D (ep - 1 P
Proof We have X; = By — B, + Bz where B, is the i-th Betti number We have Bo = B = | But also X; = V —E +T, so X, =2
Now Xx = V’ — E' + T' and by our previous result, this is II nV - > Ce — l)— nE + nT nX, — 3) (ep — ]) But Bj = BS = 1 and Bj = 2g Hence Xạ = 2 — 2g, whence 2g —2 = —2n + D (ep — 1) as desired
Corollary The algebraic genus is equal to the topological genus
Proof Both satisfy the same formula |
§2 Integration on the Riemann Surface
As before, K is a function field over C and R its Riemann surface
Trang 32IH The Riemann Surface 52 at fdx ~ gdyif f/g = dy/dx, We /e say th We sa - mp ie
in K y tP if, whenever fis a parameter at
U and x lies! holomorphic al jon on wT te funct ential 1s say that the differ P, and ax fdr =⁄n dt as no pole at P We say that the differentia} ion f dx/dt has no po
at P, then the function f : fit is so at every Po! : every point of U yg
is holomorphic on U if iti Vis a meromorphic function on U such that wo fia dx on eon “ee
A primitive § of ƒ‹ is connected, then two P rimitives differ by a constant, as vas
dg dx = f If U is conne
a a holomorphic differential on a disc V, then w has a primitive g and
' VN DI contained in a disc V, let @ =fdx on V, let
ay=P-Q,
and let g be a primitive of œ on V Then we define
[eo = g(P) - ø()
y
This number is independent of the choice of V and g Indeed, if supp (y) is
contained in another disc V;, and g, is a primitive of fdx on V,, then VO V, contains a connected open set W containing y, and g — g, is constant on W,
If subd, y = y, + ¥2 has one more point A, then
ôy =P —A,ô=Á — Q_ and |“=j w+ [ w
Ỹ hái 2
Hence if y is contained in a disc V, then
[ o =| w
Y subd
¥ = Xn, 9; where oj is a 1-simplex contained in some open disc, we
define the integral of w over ineari
i
y by linearity We get again thé ral
does not change by a subdivision, , sae eee ee
If yis any 1-chain, for r lar
§2 Integration on the Riemann Surface 53
Cauchy’s Theorem Let w be holomorphic on an
be a \-cycle on U, homologous to 0 on U Then |s=®
Y
Proof We have y = 07 where 7 is a 2-chain For some r, Sd’y has
each one of its simplices contained in a disc We have
open set U of R Let y
Sử = Sd'an = aSd'n,
whence it suffices to prove Cauchy’s theorem under the assumption that T¡ and
yare contained in a disc But then w has a primitive, and the result is trivial Corollary Jƒ U is simply connected, then w has a primitive on U The pairing (y, w) > [ @ Y gives a bilinear map A(R, Z) x O(R) > C,
where 1), is the space of differentials of first kind
Theorem 2.1 The kernels of this pairing on both sides are 0 In other words, if w is a differential of first kind whose integral along every cycle is 0, then w = 0 Conversely, if y is a cycle such that the integral along
y of every dfk is 0, then y is homologous to 0
Proof As to the first statement, fix a point O on R Under the hypothesis
that w is orthogonal to every cycle, it follows that the association
*
Pr [ wo
0
is a holomorphic function on R, whence constant, and therefore that w = 0
Trang 33CHAPTER IV
The Theorem of Abel-Jacobl
§1 Abelian Integrals
A differential will be said to be of the first kind if it is holomorphic every-
where on the Riemann surface Such differentials form a vector space over the complex, and by the Riemann-Roch theorem, one sees that the dimension of this space is equal to the genus g of R -
From topology, we now take for granted that our Riemann surface can be
represented as a polygon with identified sides (Fig 1) d= ity G4, %§=1, ,8 : Figure 1
We select a point O inside the polygon, and use it as an origin Let g bea
differential of first kind (writte d i i insi
(which is simply connected), we eae ii P toside the pala fP)=] © dp §1, Abelian Integrals 2 where Ap is any path from O to P lying entirely inside the pol is single valued and holomorphic inside ®, Th PS NI
If A, is an arbitrary path on R, not necessaril:
from O to P, then ly lying inside the polygon, A ~ Ap + > na; with suitable integers (and ~ means homologous) Consequently e={ c+ Sm | Q AY Ap lai The numbers ale) =a; = I g
generate an abelian group which will be called the group of periods of ¢
An integral taken from O to P along any path is well defined modulo periods
The integral over the path A as shown in Fig 2
is also a period, and we write
J?»= 8) |
Consider one of the cycles a; We can find a simply connected open set
U containing a; minus the vertex v We can now define a holomorphic
function f} on U as follows Given any point P on a; — v, we take a path
A> lying entirely inside the polygon except for its end point For any point
Q in U we then define
Trang 34
IV The Theorem 9; f Abe}
“Jacoby
56 gi The Abelian Integrals 57
5 ing taken on a path lying entirely inside U
ral from Pto hae mm - : Oụr »
the iNI€ETf) hoJomOFP II Dán co holomorphic on a, (including th - li ƒ
function» siterential which ish Let @ © Vertex), am jp, - 9 ; + ánh CP) Wi [re > B(o)w where is a period of w roach the vertex as indicated , Qi APP ás P„ 9 Theorem 1.1 Let w be a differential holomorphic on P Then đị 2g
a i Đan va I ,/e =— w & = 247V—-1 Nư¿ế,=2mVET > resp(fw)
where the sum over P is taken over poles of w
as the similar object on the side —a, We gey Proof Our function f is uniquely defined inside the polygon To get the
integral over the polygon itself, where f has been defined separately for the I fio=- [ fi @ two representations of the same side, we approximate the polygon # by a
-a, 4 polygon ®' as indicated (Fig 3)
Similarly, we define fi
Let @i(~) = I ợ 9
Then for P on a; we get
: fi(P) — fi(P) = -ai(¢)
We can now define the symbol
Figure 3
fe
for any differential holomophic on the polygon by such that all the poles of w lie inside P’ Then
amy asl a [fe
the limit being taken as the vertices of ' approach the vertex of 9y
Trang 35_ _—_— — —SMMMNE.MNM
IV The Theorem of Abel-Jacobi 58
and the integral around #'” is, therefore,
the right-hand side of our equality Since
orientation around the poles of @, expression on al to it, thereby proving our
constant, equal to the exp PY, it is equi theorem ae fi K elm rei a ccs ee integral Lo-(, | TA a, a đi Similarly, and call it a vector period, or simply period, from now on @=(G1, - I,
Note that Ay, » Azy and Ai, - - - , Arg differ by a permutation
For P inside the polygon, we let
rợ) =Í_®= 0Œ), - :409 ‘Ap
the f, corresponding to integrals of ¢, We define F * and F; inasimilar way,
so that for P on a; (and # vertex) we have F†(P) - Fï(P) = ~Âi We can now define the vector integral Jo Fw, and Theorem 1 can be formu- lated vectorially Theorem 1.2 If w is holomorphic on P, then 2 — W wiÃ, =2mV—I 3) res(F@), P i=l the residue being the vector of residues, and the sum over P is taken over all poles of w §2 Abel’s Theorem We have fixed a point O on the Riemann surface The vector integral ƒ's=({» .s) = ne §2 Abel's Theorem 59
as P varies over the Riemann surface, taken
is well-defined modulo periods In this way
Riemann surface R into the factor group C* "
can be extended by linearity to the free abelia
group being called the group of cycle: sum
along some path from O to P, we obtain a mapping from the modulo periods This mapping Tan n group generated by points, this Son R, or divisors A divisor is a formal
a= » npP
with integers np, almost all of which are 0 Its
f Its degree is the sum of
sc egree é j a s the np
Those of degree 0 forma group #o, and for these it is clear that the reuiriction
of the above homomorphism is then independent of the origin chosen W thus obtain a homomorphism g sen We
Ẹ
are S(a) = S Np [ ® mod periods,
Ø
from Qo into C*/(periods) This factor group will be denoted by J and is called the Jacobian of R The main theorem is the following statement
Abel-Jacobi Theorem The preceding homomorphism from Do into J is surjective, and its kernel consists of the divisors of functions It establishes an isomorphism between the divisor classes (for linear equivalence) of degree 0, and the Jacobian group C* modulo periods
The statement concerning the kernel of our homomorphism is called Abel’s theorem and will be proved in this section The surjectivity is
postponed to the next seciton
We first prove that divisors of functions are contained in the kernel
other words, if z is a function with divisor In (z) = > npP, we have to show that > np F(P) = 0 mod periods entation of R such that z has no pole on = npF(P) Indeed, tP Then
We can always find a polygon repres
9, and we let w = dz/z in Theorem 1.2 Then resp(F0)
for any holomorphic function f consider a local parameter f at a poin
Trang 36IV The Theorem of Abel-Jacob; ` I dz W se if z has order kat P Thus 1 #2) = f(Oyk wes (fe if) =fO By Theorem 1.2, 2nV=T S mF(P) = ~ D wid
-1 dz, But it is easily shown that we = 27V—1 my, for
Hence we cancel IaV—1 and we get z where Ww; = Sa, some integer i 2e ` 'F(P)= — Ÿ mÃi 3 mF) = = È as was to be shown
For the convenience of the reader, we
which is holomorphic on a cycle a, then
recall the proof that if z is a function
I me integral multiple of 27 V —1
We may assume that a is a closed path The integral is defined by analytic
continuation along the path, over successive discs, say Do, , Dy Starting from a point Py and returning to Pp Let L; be a primitive of dz/z over the
disc D, Let P, be a point in Dy+, M D, so that
đ
Em nẽ na
All terms cancel except Ly(Py) — Lo(Po) But Ly and Lo are two primitives po dele over a disc around Pp, and so they differ by an integral multiple of
2mV—1 Since Py = Po, this proves what we wanted
In order to prove the converse, we need some lemmas
Le mma 2.1, Let x; (i = 1, , 2g) be complex numbers Then a ide
> x4; =0 ifand only if x; = B-A;
§2 Abel's Theorem
61
for some complex vector B = (b,, cre: xà cà b,) The vectors A by), P V€ SA) A span a g-dimensional space over the complex numbers : ee
Proof We prove first that the relation X ` `
X = 0 (for ä complex vector X) Let ø = X-, ae a Semple tat
[,s=Í x:e=x:[ ®=X-A.=0
by hypothesis Hence all the periods of w are 0 Hence the integral {) w is
a ie aes function on R without poles because w is of first kind It is therefore, constant; hence w = 0 and X = 0 becaus independent over C ecause the g, are linearly và i
This shows that A), , Az, generate C£,
Let w = bg, + +++ + be, be adfk Then Fw hd §
tim siy w has no poles and hence -1 z3 0=— resp (Fw) = ———— 5, ti ng Danae iA where § w= > b | © = >, bai = BA v= ay
This proves half the first assertion
Conversely, we want to find the space of relations
HA) +14 xu Ấ =0:
We know from the fact that Ai, , Are generate C# that the relations
X =(B'A, , Br Arg)
form a g-dimensional space, which must therefore be the whole space of all
relations This proves the lemma
A differential w is said to be of second kind if its residue is 0 at all points
It is said to be of third kind if it has at most poles of order | at all points
The next lemma concerns such differentials, abbreviated by dtk
Lemma 2.2 Let a = & npP be a divisor of degree 0 Then there exists
Trang 37
IV The Theorem of Abel Jacob Jacob)
for the case Where all Coef fig;
ssertuion for c ` all icients
suffices ve oi eet, py induction, suppose it proveg cs Proof Its points aw point unequal to the given points le
and Q such that TeSp, w, = Hy and
‘ i 5 only at P; :
points Pu: ˆ ` r) have poles aly W obviously has the required Property
of the space of differentials w such that
where d(a) is the dimension
(w)2 —d
=ø + 1 and hence there exists a dịự We get tO aut the sum of the residues is 0, oat
ˆ 4 r Pa wo
die? LH HH opposite residues Multiplying w by a le a ’
ves what we want
the kernel of our map Leta = = np P be a divisor
ermine „ số me
We can now det ) = 0 We have to show that is the divisor of
of degree 0 such that S(a
a that there exists a dtk wsuch that has poles at P with residues
np, and such that
I yo 2mV—Tn,
for suitable integers n, By the lemma, there exists a dtk w having residue np at P for all P For any P, we have resp(Fw) = npF(P) We are going to change w by a dfk, so as to change the periods but not the residues By Theorem 1.2 we get -% M¡Ã, =2mV—T > resp (Fw) =2mV~I 9 mF(P) =2mV~I 9 mụÃ, for some integers m;, by hypothesis Hence §3 Jacobi's Theorem 63 > (w, — 2mV—T mA, = 0 By Lemma 2.1, we have w¿ — 2mV—T m, = B-A; for some vector B We let w = w — B-® Then whas the same poles and residues as w Furthermore, Í w=[ e-[ B'®=wi—B'A, =2V—T mụ
thereby proving our contention
To construct the desired function z we take essentially
2(P) = exp [ Ử,
0
O being as usual a suitable origin inside our polygon For any point P which
is not a pole of y, and any path from O to P, the exponential of the integral gives a function which is independent of the path and is thus a well-defined meromorphic function For a pole of W, expanding w in terms of a local
parameter around this point shows at once that we get a meromorphic function around the point, whose singularity can only be a pole Abel’s theorem is
proved
§3 Jacobi’s Theorem
We must now prove that our map from divisors of degree 0 into the factor
group C* modulo periods is surjective
A divisor a is said to be non-special if d(—a) = 0, that is if there exists no differential w such that (w) 2 a
Lemma 3.1 There exists g distinct points M,, , Mg such that the
divisor M, + + + Mg is non-special
Proof Let w, # 0 be a dfk, and M, a point which is not zero of w, The space of dfk having a zero at M, has dimension g — 1 by Riemann-Roch
Let ø; # 0 be in it and let M2 be a point which is not a zero of w Continue
g times to get g points, as desired
Theorem 3.2 Let M,, , M, be g distinct points such that the divisor
Trang 38IV The Theorem of Abel-Jacobi P, vy rod [ œ M,
uct of small discs VW.x xX Vy
7 i a produ : Son ZÄE
lytic isomorP ism of @ P rhood of zero in CF
gives an ana M, onto a neighbo: around the points M,. ›
y iti ies Jacobi’s theorem , we show how it implies h
i ote that it suffices to prove Jacobi's theorem fora neighborhood
soroin Ct that is, to prove local surjectivity Indeed,
let X be any vector
"thời Sĩ lýn:X is ina small neighborhood of zero so, by the local result,
Sees fds divisor of degree zero such that F(a) = I/n-X modulo
periods Then F(na) = X modulo periods
Using the notation of Theorem Before proving our theo!
3.2, let a range over divisors
+ Me
with P, ranging over V; Then clearly
P,
Fa) => ƒ ®© _ (mod periods) lM,
and the theorem implies what we want
We now prove Theorem 3.2 Let 1; be a local parameter at M, We contend that the determinant + du) - ‹ - Tt (My) dt, dt, fey) ft dt (i) dt, (M,) is not zero To prove this, consider the homomorphism £ gv er lý (MỊ), tấn |
_ Of differentials of first kind into complex g-space Any dfk in the kernel would be special, so the kernel is trivial, and our map is a linear isomorphism
Hence our determinant is non-zero We write Øi = hụ(h)đn, , øi = hiy( ái, HP h,(n)đn, pees Pe = Ng (t,)dt, sạ, Jacobi's Theorem 65
where (tj) is holomorphic in V; Let hF(t,) be the inteers
series /1y(t)), normalized to vanish at 1; SG The of ihe power
of our map: get a representation
(Pir Pe) ACA) 9
FA ted Hylty st)
where each H,(11, , ty) is holomorphic in the g variables, on Vịt vi Vụ,
Now
oH, = hi(t)), at; Ỷ
and hence the Jacobian determinant evaluated at (0, , 0) is precisely our
preceding determinant, and is non-zero By the implicit function theorem, we
get our local analytic isomorphism, thereby proving Jacobi’s theorem
The theorem can be complemented by an important remark We consider complex g-space as a real 2g-dimensional space We have:
Theorem 3.3 The periods Aj, , Ary are linearly independent over the reals Hence the factor space of C* by the periods is a 2g-real-
dimensional torus
Proof It will suffice to prove that any complex vector X is congruent to
a vector of bounded length modulo periods By the Riemann-Roch theorem, any divisor of degree 0 is linearly equivalent to a divisor of type
(Py) t+ ++ + Pp) -8°0,
for suitable points P;, , Pg, which may be viewed as lying inside or on the polygon For P ranging over the Riemann surface, the integrals
L5
are bounded in the Euclidean norm, if we take
entirely inside the polygon, except if P lies on tl the sum
the path of integration to be
he boundary Consequently,
Trang 39
IV The Theorem of
& heo
an (g times the other bound) Combining uc
has also pounded n0) ctivity theorem s hows that any vector is congruent to one With
Jacobi's SUjÊ h modulo periods, an d concludes the proof of Theater i bounded lengt
i
+5 Relations g4, Riemann
and consider the polygon representation of R, with
i vn .„ ø) following each other (Fig, 4), We lt’ " =| g and Bi = 7 ẹ riods of ø, and œ‡, Ø¡ those oŸ @ -b, = be the canonical pe! a Figure 4 Riemann’s first relations state that & 3 (aiBi - a} B)) = 0 i=!
Proof Let f(P) be the function defined before (integral from O to P, suitably defined on the boundary of the polygon), and consider the integral
[fe
Since ¢' has no poles, we get from Theorem 1.2
0= ([ [+ v ƒ 3
Which gives what we want
: Riemann’s second relations state that ee ee ae Cee: hea \ §5 Duality 67 1 £
Proof This time, we consider the integral
i 7 = left hand side of inequality, and we prove that it is strictly positive Write f=u + iv, in terms of its real and imaginary part Then f = w — iv, and df= @ = du + ¡ do Then fe = $d(u? + 09) + it do — v du)
The first term on the right is exact, and the second gives
d(u dv — v du) = 2du / dv
By Green’s theorem, our integral can be replaced by the double integral
2ï [[ du A do
In the neighborhood of each point, we may take a chart, and express the
functions u, v in terms of a complex variable z, with real parts x, y, that is
z =x + iy By the Cauchy-Riemann equations, we get
2 2
du /\ dv = (2) + B ) dx /\ dy,
which shows that the integral is positive, as desired
§5 Duality
i id the polygon as at the beginning,
Lemma 5.1 Numbering the sides of the p Agen a ee đị, , dạy, and given an index i, there €
Trang 40IV The Theorem of Abe} lM Jacgy: i Re [ g=0 for Ji ay g ioe uy + trị
al and imaginary parts of the periods, and’;
ectors of the above matrix are our biting
complex numbers Ods v are the re Then the column V We wish to find g g) and a dfk where the ¥, therefore, real Ayes? đạp : =O ts 4; net k= 1 sở PI + 249, i iods is amounts to solving, sa’ + having the required periods This am B, Say for i = 1, system of equations xi —yiDl tet Xy tt = y tỷ =1 ago yeti + Xgl — y 03 = 0
This is solvable if the row vectors of the coefficient matrix are linearly
independent over the reals This is indeed the case, because a linear relation
is immediately seen to imply a linear relation between Aj, , A2g over the
reals
Lemma 5.2, A dfk cannot have all its periods pure imaginary
Proof The preceding system of linear equations cannot have a solution when made homogeneous
Theorem 5.3 Given a divisor a = % npP of degree 0, there exits unique dtk w, such that
(1) resp wy = np for all P
(2) The periods of w, are all pure imaginary
ou sede adtk with resp w = np It suffices to find a dfk g having Parts for the canonical periods (integrals around ai) Ths?
$5 Duality 69
immediate from " n Lemma 5.1 The uniqueness is clear, because the difference Le % ~ #
between two differentials sati sfying ying n th € condit 01 id ons of the theo €Orem woul Id be ©
Theorem 5.4 If o is a cycle such that J @ = 0 for all dfk then o ~ 0
Proof Let o@ ~ ma; +++ - 7
that 1 + ñayd„; Find a dfk œ having periods such
Re Í g=1, dị and ä Re [le =0, j#1
Then
0= I y =n + pure imaginary,
and son, = 0 Similarly, n; = 0 all j Theorem 5.5 The pairing
(a, a) > (a, a) = exp Í Wa a
induces a bilinear pairing between the first homology group H\(R) and the
group of divisor classes of degree 0 Its kernels on both sides are 0, and the pairing induces the Pontrjagin duality between the discrete group
H\(R) and the compact torus
Proof Let Bo be the group of divisors of degree 0 If a € Do and a = (2)
is the divisor of a function, then we may take w_ = dz/z All the periods of
dz/z are pure imaginary, of the form 2V—1 m with some integer m Hence (ca, a) = 1 for such divisors, and all cycles o having no point in common with a Furthermore, if 7 ~ 0, then [ Oe F 2V—I > resp 6a =2mV—I times an integer, so (a, a) = 1 Our pairing is therefore well defined as a bilinear map of H,(R) x J into the circle
ition, all periods of w, are of the