The Riemann-Roch Theorem § 1 Lemmas on Valuations
A discrete valuation ring is a principal ideal ring with a unique prime, making it a unique factorization ring In this context, if t serves as a generator for the prime, it is referred to as a local parameter Every element x in the ring can be represented as a product of the form x = t^r * y, where r is a non-negative integer and y is a unit This structure allows for a clear understanding of the elements within the quotient field of the ring.
In the context of K, an element's order or value is represented by an arbitrary integer r When r is greater than zero, we indicate that x has a zero at the valuation, while if r is less than zero, x is said to have a pole This relationship is expressed as r = vo(x), v(x), or simply v(x).
Let tJ be the maximal ideal of o The map of K which is the canonical map o ~ o/tJ on 0, and sends an element x f/= 0 to 00, is called the place of the valuation
We shall take for granted a few basic facts concerning valuations, all of which can be found in my Algebra Especially, if E is a finite extension of
K and 0 is a discrete valuation ring in K with maximal ideal tJ, then there exists a discrete valuation ring () in E, with prime ~, such that and tJ = ~ n K
If u is a prime element of (), then t () = u e (), and e is called the ramifica-
2 1 Riemann-Roch Theorem tion index of () over 0 (or of ~ over lJ) If r D and ro are the value groups of these valuation rings, then (fD : ro) = e
The pair ô(),~) is considered to lie above (o,lJ), indicating that ~ is positioned above lJ Additionally, we define ô(),~) as unramified above (o,lJ) or state that ~ is unramified above lJ when the ramification index is equal to 1, meaning e = 1.
Example Let k be a field and t transcendental over k Let a E k Let 0 be the set of rational functions f(t)/g(t), with f(t), g(t) E k[t] such that g(a) =1= O
The discrete valuation ring 0 consists of all quotients f(a) = O, which is a common scenario Considering an algebraically closed field k and the extension k(x) with a transcendental element x, we can assume x is in 0 by adjusting it to 1/x if necessary The intersection of 0 with k[x] is non-empty and generated by an irreducible polynomial p(x) of degree 1, expressed as p(x) = x - a for some a in k Consequently, the canonical map induces a transformation f(x) to f(a) on polynomials, leading to the conclusion that 0 includes all quotients f(x)/g(x) where g(a) ≠ 0, returning us to the initial scenario described.
Similarly, let 0 = k[[t]] be the ring of formal power series in one variable Then 0 is a discrete valuation ring, and its maximal ideal is generated by t
Every element of the quotient field has a formal series expansion with coefficients ai E k The place maps x on the value ao if x does not have a pole
In this article, we explore a field K that serves as a finite extension of a transcendental extension k(x), where k is an algebraically closed field and x is transcendental over k This type of field is referred to as a function field in one variable.
If that is the case, then the residue class field of any discrete valuation ring o containing k is equal to k itself, since we assumed k algebraically closed
Proposition 1.1 states that if E is a finite extension of K and ô(•, •) is a discrete valuation ring in E that lies above (o, lJ) in K, then E can be expressed as K(y), where y is the root of a polynomial f(Y) with coefficients in 0 This polynomial has a leading coefficient of 1, and it satisfies f(y) = 0 while ensuring that the derivative f'(y) is not congruent to 0 modulo ~.
Then ~ is unramified over lJ
Proof There exists a constant Yo E k such that y == Yo mod ~ By hypothesis, f'(yo) =F 0 mod~ Let {yn} be the sequence defined recur- sively by
The reader is invited to verify that this sequence converges in the completion Kp of K It can also be easily established that it converges to the root y, as y is congruent to Yo modulo ~ However, y is distinct from any other root of the function f.
~ Hence y lies in this completion, so that the completion E'13 is embedded in K p, and therefore ~ is unramified
We also recall some elementary approximation theorems
The Chinese Remainder Theorem is a fundamental concept in algebra, which states that for a given ring R and distinct maximal ideals I1, I2, , In, there exists a solution x in R that satisfies the congruences x ≡ ai mod Ii for all i, where ai are elements in R and ri are positive integers This theorem provides a unique solution for the system of congruences, enabling the simultaneous solution of multiple modular equations.
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 01,
Proposition 1.2 If 01 and 02 are two discrete valuation rings with quotient field K, such that 01 Co2, then 01 = 02
Proof We shall first prove that if lJ I and lJ2 are their maximal ideals, then
Let \( Y \) be an element in \( lJ2 \) If \( Y \) is not equal to \( lJt \), then \( 1/Y \) belongs to \( Ot \), leading to a contradiction since \( 1/Y \) should also be in \( lJ2 \) Therefore, we conclude that \( lJ2 \) is a subset of \( lJl \) Every unit in \( O1 \) is also a unit in \( O2 \) An element \( y \) from \( lJ2 \) can be expressed as \( y = \tau{Iu} \), where \( u \) is a unit from \( O1 \) and \( \tau \) is an element of order 1 in \( lJl \) If \( \tau \) is not in \( lJ2 \), it would imply that it is a unit in \( O2 \), which is a contradiction Thus, \( \tau \) must be in \( lJ2 \), leading to the conclusion that \( lJ1 = O1 \tau \) is also in \( lJ2 \) This establishes that \( lJ2 \) equals \( lJ1 \).
4 1 Riemann-Roch Theorem if u is a unit in 02, and is not in OI then llu is PI and thus cannot be a unit in 02 This proves our proposition
From now on, we assume that our valuation rings OJ (i = 1, , n) are distinct, and hence have no inclusion relations
Proposition 1.3 There exists an element y of K having a zero at 01 and a pole at OJ (j = 2, , n)
Proof This will be proved by induction Suppose n = 2 Since there is no inclusion relation between 01 and 02, we can find y E 02 and y $ 01
Similarly, we can find Z E 01 and Z $ 02 Then zly has a zero at 01 and a pole at 02 as desired
Now suppose we have found an element y of K having a zero at 01 and a pole at 02, , On-I Let z be such that z has a zero at 01 and a pole at on
For sufficiently large values of r, the expression y + zr meets our criteria, as it demonstrates that the sum of zero and zero results in zero, the combination of zero and a pole yields a pole, and the addition of two elements from K with poles of differing orders also produces a pole.
A high power of the element y of Proposition 1.3 has a high zero at 0 I and a high pole at OJ (j = 2, ,n) Adding 1 to this high power, and considering 11(1 + y r) we get
Corollary There exists an element z of K such that z - 1 has a high zero at 010 and such that z has a high zero at OJ (j = 2, , n)
Denote by ordj the order of an element of K under the discrete valuation associated with OJ We then have the following approximation theorem
Theorem 1.4 Given elements ai, , an of K, and an integer N, there exists an element y E K such that ordj(y - aj) > N
To demonstrate the proof, for each index i, we apply the corollary to ensure that Zj approaches 1 at OJ and approaches 0 at OJ for j ≠ i, specifically at the valuations linked to these valuation rings Consequently, the expression Zl a1 + + Znan exhibits the necessary properties.
In particular, we can find an element y having given orders at the valua- tions arising from the OJ This is used to prove the following inequality
In the context of a finite algebraic extension E of a field K, consider the value group f associated with a discrete valuation of K The value groups f_j correspond to a finite number of inequivalent discrete valuations of E that extend the valuation of K Furthermore, let e_j denote the index of f within the group C.
2: ej :;§i [E : K] §2 The Riemann-Roch Theorem 5
In this article, we explore the distinct cosets YII, Ylep, Yrh, and Yre of E, where Yiv (for v = 1, , e) exhibit high-order zeroes at the valuations Vj (for j ≠ 0) We assert that these elements are linearly independent over the field K To support this claim, we examine the implications of a potential linear dependence relation among them.
In the context of CII having a maximal value in f, we can express this relationship as V(CII) approximately equal to V(Civ) for all i and v By dividing the equation by CII, we can simplify our assumptions to CII being equal to 1, leading to V(Civ) also being approximately 1 When evaluating our sum at VI, it is important to note that all terms, including Yll, C12Y12, and Clel Ylel, possess distinct values due to the unique representation of the y's as distinct cosets.
On the other hand, the other terms in our sum have a very small value at
VI 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 \( K \) represent a function field in one variable over \( k \), defined as a finite extension of the purely transcendental extension \( k(x) \) with a transcendence degree of 1 In this context, \( k \) is referred to as the constant field, and the elements of \( K \) are often called functions.
The purpose of this chapter is to give a significant example for the notions and theorems proved in the first chapter
The reader interested in reaching the Abel-Jacobi as fast as possible can of course omit this chapter §t The Genus
The Fermat curve of level N, denoted as F(N), is defined by an equation over an algebraically closed field k, with N being prime to the characteristic of k For our discussion, we assume N is approximately 3 and let K represent its function field.
The equation that defines the curve is non-singular, indicating that the discrete valuation rings in K correspond directly to the local rings of points on the curve This includes points where x = 0, which originate from the projective equation where z = 0.
The expression §2 Differentials 37 involves the product over all E/LN (N-th roots of unity) By setting x as an N-th root of unity, we derive a point on the Fermat curve where y = 0, indicating that K is ramified of order N at this point, as demonstrated by the preceding equation.
[K : k(x)] = N, and the ramification index of K over the point x = lis N
On the other hand, let t = l/x Then
Y = N (t - 1), t and -1 + t N is a unit in k[[t]] Hence x = 00 (or t = 0) is not ramified in
K, and there exist N distinct points of F(N) (or K) lying over x = 00, called the points at infinity in this section If we put z = ty, then theseN points have coordinates z = l for l E /LN'
By the Hurwitz genus formula, if we let g be the genus of F(N), we get:
Theorem 1.1 The genus of F(N) lS g = 2 §2 Differentials
For integers r, s such that 1 ~ r, s we let
We can also write r-\ _\ dx
38 II The Fermat Curve we see that the only possible poles of W r•s lie among the points with x = 00
If point P has a finite value for x(P) equal to F0, then the expression dy/X N-1 indicates that W r•s does not have a pole at P Conversely, if x(P) equals 0 and y(P) is non-zero, the expression dx/yN-1 demonstrates that there is also no pole at P.
Theorem 2.1 A basis for the differentials of first kind is given by W r•s with
Proof Suppose that x(P) = 00 Put t = l/x Then dx = -"2 t 1 dt, and ord p y = ord p x = 1 Then r-I s-I (-1) dt
The theorem is proven by demonstrating that if \( r + s \) is approximately equal to \( N - 1 \), then the order of the differential form \( W_{r,s} \) approaches zero Additionally, it is evident that the differential forms in question are linearly independent over the constants, and there are exactly \( g \) such forms, where \( g \) represents the genus of \( F(N) \).
We observe that these forms have an additional structure The group
The group /LN X /oLN operates as a set of automorphisms on F(N) through a specific action, utilizing Ctv as a designated primitive N-th root of unity, typically represented as Ctv = e^(2πi/N) in the realm of complex numbers The form Wr•s, which imposes no limitations on the integers r and s, serves as an eigenform corresponding to the character X r•s.
The linear independence of the first kind differentials in Theorem 2.1 is evident as they serve as eigenforms for the Galois group, each possessing distinct characters.
The group of automorphisms of the Fermat curve F(N), denoted as G = G(N), is referred to as the group of natural automorphisms or the Galois group of F(N) over F(1).
Theorem 2.2 The forms wr,s with
1 ~ r, s ~ N - 1 and r + s i= 0 mod N constitute a basis for dsk/exact §3 Rational Images of the Fermat Curve 39
According to the Riemann-Roch theorem, the dimension of the space dtk(oo)/dfk is N - 1 Additionally, an analysis of the pole's order at infinity reveals significant insights into the forms involved.
The functions \( r \) and \( s \) belong to the third kind and are linearly independent from the first kind differentials as established in Theorem 2.1 Given their appropriate dimensionality, they serve as a basis for the space \( dtk(\infty)/dfk \).
Given any differential w, it follows that there exists a homogeneous polynomial h(x, y) of degree N - 2 such that
Y belongs to the second kind, and we apply this observation to the forms W r.s, where r + s * 0 mod N By operating with ô(, () on the previously mentioned difference, we observe that the automorphisms of F(N) maintain the integrity of the dsk spaces Upon subtracting, we arrive at a significant finding.
(1 - r+S)w r•s with r + s * 0 mod N is of second kind, whence W r•s is of second kind
Finally, we note that the forms W r•s with 1 ~ r, s ~ N - 1 and r + s * 0 mod N, taken modulo the exact forms are eigenforms for the Galois group
The JLN x JLN forms exhibit unique characteristics, establishing their linear independence within the dsk/exact framework Given that the quantity of these forms matches the dimension of dsk/exact, they effectively constitute a basis for this factor space, thereby validating the theorem In Section 3, we explore the rational images of the Fermat Curve.
Throughout this section, we let 1 ~ r, sand r + s ~ N - 1 Such a pair
(r, s) will be called admissible We shall consider rational images of the
Fennat curve (subfields of its function field) following Rohrlich [Ro] after Faddeev [Fa 2]
Then u, v are related by the equation v N = u r (1 - uy, which, in irreducible fonn, amounts to
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, s), given in tenns of coordinates by
If t E Z(M) = Z/MZ, we let (t)M be the integer such that o ~ (t)M ~ M , 1 and (t)M == t mod M
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(1) I = k(u) §3 Rational Images of the Fermat Curve 41
We note that G(r, s) is the kernel of the character X r and that K(r, s) over
K(1) is cyclic, of degree M It is in fact a Kummer extension
In a special case where N equals the prime number P, specifically 3, and both r and s are set to 1, the intermediate curve is defined by the equation vP = u(1 - u), characterizing it as hyperelliptic By applying the variable transformation t = 2u - 1, this equation can be simplified, making it more manageable for further analysis.
Let m E Z(M) We say that m is (r, s)-admissible if (mr) and (ms) form an admissible pair, that is
1 ~ (mr), (ms) and (mr) + (ms) ~ N - 1
Theorem 3.1 A basis of dfk on F(r, s) is given by the forms
W(mr),(ms) for all (r, s)-admissible elements m
It is evident that x (mr) y (ms) exists within the function field of F (r, s), indicating that the aforementioned forms are categorized as the first kind on F(r, s) Conversely, if W is a differential form on F(r, s), then it can be expressed as follows.
The expression W = 2: cq,l wq,l represents a summation over all admissible pairs (q, t) Given that wq,l is an eigenform of JLN x JLN with the eigencharacter X q,,, and considering that the factor group is cyclic, it can be concluded that if cq,l is not equal to zero, then certain properties hold true.
42 n The Fermat Curve for some integer m But then q = (mr) and t = (ms), as desired
Suppose that m E Z(M)*, and that m is (r, s)-admissible Then the function fields of F(r, s) and Fômr), (msằ are equal, for instance because
The correspondence between the curves in terms of coordinates is given as follows
Let 1 ~ m ~ M - 1 be prime to M and such that its residue class mod M is (r, s )-admissible Write ômr), (msằ = m(r, s) + N(i, j) with some pair of integers i, j Then we have a commutative diagram:
F(r, s) -+ Fômr), (msằ where the bottom arrow is given by realizing the automorphism of the function field corresponding to the two models F(r, s) and Fômr), (msằ