Some consequences of the Riemann-Roch theorem

Transcription

1 Some consequences of the Riemann-Roch theorem

2 Proposition Let g 0 Z and W 0 D F be such that for all A D F, dim A = deg A + 1 g 0 + dim(w 0 A). Then g 0 = g and W 0 is a canonical divisor. Proof We have already seen (where?) that the prerequisites of the theorem imply dim W 0 = g 0 and deg W 0 = 2g 2. Let A D F be such that deg A max{2g 2, 2g 0 2}. Then dim A = deg A + 1 g and dim A = deg A + 1 g 0, implying g 0 = g. Finally, let W be a canonical divisor. Then dim W = g = (2g 2) + 1 g + dim(w 0 W ). Therefore, dim(w 0 W ) = 1, which, together with deg(w 0 W ) = 0, implies that W 0 W is principal. Therefore, W 0 W, i.e., W 0 id also canonical. 1

3 Proposition dim B g. A divisor B is canonical if and only if deg B = 2g 2 and Proof Let deg B = 2g 2 and dim B g and let W be a canonical divisor. Then g dim B = deg B + 1 g + dim(w B) = g 1 + dim(w B) (why?). Therefore, dim(w B) 1 and, since deg(w B) = 0, W B is principal... Proposition An algebraic function field F/K is rational if and only if g = 0 and there is a divisor A D F such that deg A = 1. Proof We have already seen the only if direction of the proposition. 2

4 For the proof of the if direction, let A D F be a divisor of degree 1. Since deg A = 1 2g 1 = 1, dim A = deg A + 1 g = 2. Therefore, there is an integral divisor A [A]. Since dim A = 2, there is a non-zero element x L(A ) \ K such that (x) + A 0, which is possible if and only if A = (x), because deg A 0 and deg A = 1. Therefore, That is, F = K(x). [F : K(x)] = deg(x) = deg A = 1. 3

5 Theorem (Strong Approximation Theorem) Let S be a proper subset of P F and let P 1,..., P r S. Let x 1,..., x r F and let n 1,..., n r Z. Then there exists an element x of F such that v Pi (x x i ) = n i, for all i = 1,..., r, and v P (x) 0, for all P S \ {P 1,..., P r }. Proof Let the adele α = (α P ) P PF A F be defined by α P = { xi if P = P i, i = 1,..., r 0 otherwise. Let Q P F \ S. For sufficiently large m N, A F = A F (mq ) r (n i + 1)P i + F i=1 (why?). 4

6 Therefore, there is an element z of F such that z α A F (mq In particular, ) r (n i + 1)P i. i=1 v Pi (z x i ) > n i, for all i = 1,..., r, and v P (z) 0, for all P S \ {P 1,..., P r }. Let y 1,..., y r F be such that v Pi (y i ) = n i, i = 1,..., r, and let y F be such that v Pi (y y i ) > n i, for all i = 1,..., r, and v P (y) 0, for all P S \ {P 1,..., P r }. (Why there is such a y?) 5

7 Then (why?) and, for x = y + z, v Pi (y) = v Pi ((y y i ) + y i ) = n i, i = 1,..., r v Pi (x x i ) = v Pi (y + (z x i )) = n i, i = 1,..., r (why?) and, for P S \ {P 1,..., P r }, v P (x) = v P (y + z) 0 (why?). 6

8 Proposition Let P P F. Then, for any n 2g, there is an element x of F such that (x) = np. Proof Since (why?), and dim(n 1)P = (n 1) deg P + 1 g dim np = n deg P + 1 g (why?), L((n 1)P ) is a proper subspace of L(nP ). Therefore, for all x L(nP ) \ L((n 1)P ), (x) = np (why?). Definition Let P P F. A non-negative integer n is called a pole number of P if there is an element x of F such that (x) = np. Otherwise, n is called a gap number of P. 7

9 Remark An non-negative integer n is a pole number of P if and only if dim(np ) > dim((n 1)P ). An non-negative integer n is a gap number of P if and only if L(nP ) = L((n 1)P ). If n 1 and n 2 are pole numbers of P, then n 1 + n 2 is also a pole number of P. Theorem (Weierstrass Gap Theorem) Let F/K be of positive genus g and let P be a place of degree one. Then there are exactly g gap numbers 1 = i 1 < < i g 2g 1 of P. 8

10 Proof Consider the following sequence of vector spaces K = L(0) L(P ) L(2P ) L((2g 1)P ). Since dim L(0) = 1, dim L((2g 1)P ) = g (why?), and for all i, dim L(iP ) dim L((i 1)P ) + 1 (why?), there are exactly g 1 numbers between 1 and 2g 1 such that L(iP ) is a proper subspace of L((i 1)P ). The remaining g numbers are gap numbers of P. It remains to show that 1 is a gap number of P. Were 1 a pole number of P, there would not be gap numbers of P at all (why?), which is impossible, because g > 0 (why?). 9

11 Definition A divisor A D F is called non-special if i(a) = 0. Otherwise A is called special. Remarks (a) A is non-special if and only if dim A = deg A + 1 g. (b) If deg A > 2g 2, then A is non-special. (c) The specialty of A depends only on the class [A] of A in the divisor class group C F. (d) Canonical divisors are special. (e) Any divisor A with dim A > 0 and deg A < g is special. (f) If A is non-special and B A, then B is non-special. 10

12 Lemma Let P 1,..., P g P F be pairwise different places of degree one and let A D F, A 0, be such that dim A = 1 and deg A g 1. Then for some j = 1,..., g, dim(a + P j ) = 1. Proof Assume to the contrary that dim(a + P j ) > 1 for all j = 1,..., g, and let z j L(A + P j ) \ L(A), j = 1,..., g. Then v Pj (z j ) = v Pj (A) 1 and v Pi (z j ) v Pi (A) for i j, and, by Strict Triangle Inequality, the g + 1 elements 1, z 1,..., z g of F are linearly independent over K. Let D A + P P g be of degree 2g 1. Then 1, z 1,..., z g L(D), implying dim D g + 1. However, by the Riemann-Roch theorem, dim D = deg D + 1 g g. Therefore, our assumption was wrong. 11

13 Proposition If there are g pairwise different places P 1,..., P g P F of degree one, then there exists a non-special divisor B 0 such that deg B = g and supp B {P 1,..., P g }. Proof By the lemma, there is a sequence of divisors 0 < P i1 < P i1 + P i2 < < P i1 + P i2 + + P ig = B, {i 1,..., i g } {1,..., g}, such that dim(p i1 + P i2 + + P ij ) = 1, j = 1,..., g. In particular, dim B = 1, implying deg B + 1 g = g + 1 g = 1 = dim B. That is, B is non-special. 12

14 Lemma Let A, B D F be such that dim A, dim B > 0. Then dim A + dim B 1 + dim(a + B). In this course we make the additional assumption that K is infinite. Proof of the lemma Since dim A, dim B > 0, there are positive divisors A 0 and B 0 such that A 0 A and B 0 B. Let X = {D D F : D A 0 and L(D) = L(A 0 )}. Since, by definition, A 0 X, X. Let D 0 be an element of X of minimal degree (why there is such an element?). 13

15 Let supp B 0 = {P 1,..., P r }. Since L(D 0 P i ) is a proper subspace of L(D 0 ) (why?), i = 1,..., r, and a vector space over an infinite field is not a union of finitely many proper subspaces (why?), for some z F, z L(D 0 ) \ r i=1 L(D 0 P i ). Let ϕ : L(B 0 ) L(D 0 + B 0 )/L(D 0 ) be the (K-linear) map defined by ϕ(x) = zx + L(D 0 ). We contend that Ker φ = K. The inclusion K Ker φ is obvious (why?), and, for the converse inclusion, let x L(B 0 )\K. Then each pole of x is in supp B 0 and x has a pole. That is, for some i = 1,..., r, v Pi (x) < 0. Therefore, since v Pi (z) = v Pi (D 0 ) (why?), v Pi (zx) = v Pi (z) + v Pi (x) = v Pi (D 0 ) + v Pi (x) < v Pi (D 0 ), implying ϕ(x) L(D 0 )). That is, ϕ(x) 0. 14

16 Therefore, implying Finally, dim B 0 1 dim(d 0 + B 0 ) dim D 0, dim D 0 + dim B dim(d 0 + B 0 ). dim A + dim B = dim A 0 + dim B 0 = dim D 0 + dim B dim(d 0 + B 0 ) 1 + dim(a 0 + B 0 ) = 1 + dim(a + B). 15

17 Theorem (Clifford s Theorem) Let A D F be such that 0 deg A 2g 2. Then dim A deg A. 2 Proof The case of dim A = 0 is trivial, and if dim(w A) = 0, where W is canonical, then dim A = deg A + 1 g = deg A (deg A 2g) deg A. If both dim A and dim(w A) are positive, then, by the lemma, dim A + dim(w A) 1 + dim W = 1 + g and, by the Riemann-Roch theorem, dim A dim(w A) = deg A + 1 g Adding the last two (in)equalities and dividing by 2 yields the desired result. 16