3. Clifford's Theorem The elementary proof

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1 3. Clifford's Theorem The elementary proof Objekttyp: Chapter Zeitschrift: L'Enseignement Mathématique Band (Jahr): 30 (1984) Heft 1-2: L'ENSEIGNEMENT MATHÉMATIQUE PDF erstellt am: Nutzungsbedingungen Die ETH-Bibliothek ist Anbieterin der digitalisierten Zeitschriften. Sie besitzt keine Urheberrechte an den Inhalten der Zeitschriften. Die Rechte liegen in der Regel bei den Herausgebern. Die auf der Plattform e-periodica veröffentlichten Dokumente stehen für nicht-kommerzielle Zwecke in Lehre und Forschung sowie für die private Nutzung frei zur Verfügung. Einzelne Dateien oder Ausdrucke aus diesem Angebot können zusammen mit diesen Nutzungsbedingungen und den korrekten Herkunftsbezeichnungen weitergegeben werden. Das Veröffentlichen von Bildern in Print- und Online-Publikationen ist nur mit vorheriger Genehmigung der Rechteinhaber erlaubt. Die systematische Speicherung von Teilen des elektronischen Angebots auf anderen Servern bedarf ebenfalls des schriftlichen Einverständnisses der Rechteinhaber. Haftungsausschluss Alle Angaben erfolgen ohne Gewähr für Vollständigkeit oder Richtigkeit. Es wird keine Haftung übernommen für Schäden durch die Verwendung von Informationen aus diesem Online-Angebot oder durch das Fehlen von Informationen. Dies gilt auch für Inhalte Dritter, die über dieses Angebot zugänglich sind. Ein Dienst der ETH-Bibliothek ETH Zürich, Rämistrasse 101, 8092 Zürich, Schweiz,

2 Theorem (Riemann-Roch). Let C be a projective nonsingular algebraic curve. The genus of C is a nonnegative integer g. For ail divisors D on C, // the strict inequality holds, D is spécial. For ail spécial divisors D Corollary. deg jf = 2g -2; dim Jf = g- 1; an d ail divisors D of degree >2g?2 are nonspecial. 3. Clifford's Theorem? The elementary proof Clifford's Theorem compléments Riemann-Roch by providing information about spécial divisors, which of necessity are of small degree. The theorem also gives a sufficient condition that the curve C is hyperelliptic. (The theorem owes its name to the appearance of its first part in [I].) The proof I give hère is elementary ; more typical modem proofs [e.g. 4, Ch. IV, section 5 and 3, Ch. 2, section 3] involve considering whether the canonical morphism C -? P^-i defined by the canonical divisor X is an embedding. Définition. C is a hyperelliptic curve if its genus g is at least 2, and if C admits a. g\. Remarks 1. C is hyperelliptic if and only if there is a rational map C->Px of degree 2. y2y 2. This happens if and only if C has an (affine) équation of the form 2 = /M- 3. Part (3) of Clifford's Theorem shows that a hyperelliptic curve has a unique g\. Contrast this to the case of an elliptic curve, where g = 1. Hère any divisor of degree 2 defines a g\. Yet choosing distinct points P, Q one sees easily that the divisors 2? and P + Q are not linearly équivalent, and so define distinct g \ 's. Theorem (CHfford). Let C be a curve of genus g, and let D be an effective spécial divisor on C. Then

3 (1) dimd <idegd. (2) Equality holds in only 3 cases : (a) D = 0;or (3) // (b) D = JÏT;or (c) C is a hyperelliptic curve. Case 2c holds then C admits a unique g\, deg D=2r for some integer r 1, and D~ r-g\. Proof of (1 ). Since D is effective spécial, the vector spaces L(D) and L(Jf?D) are both of positive dimension. Define a map u: L(D) x L(Jf?D) -> L(Jf) by \x{f g) = f-g. (Since (/) + D > 0 and fo) + X - Z) > 0, (fg) + JT = (/) + te) + JT = [(/) + /)] + [te) + jf-d] > Oso fgelijt).) This map is bi-injective, so dim L(Jf) dim L(D) + dim L(Jf?D)?lby Clifford's Lemma. Since l(d) = dim D \ \? 1, one has (1) (2) On the other hand, Riemann-Roch guarantees that Adding thèse, and recalling that dim C/f = g? 1, one gets deg D \ 2 dim D.? Implicit in the proof is a resuit I will need later. Lemma 1. For the effective spécial divisor D, dim D =? deg D if and only if dim Jf= dim D+ dim Jf-D. T/iïs fcoms i/ and only if g-1 dim D+ dim Jf-D. Further, equality holds for D if and only if it holds for (any effective divisor linearly équivalent to) Jf?D.? Proof ôf (2). Assume that equality holds, and that Dis neither 0 nor C/f. Notice that if deg D = 2, or deg Jf - D = 2, then D, or Jf - D, defines a #2 and C is hyperelliptic. Thus, I may assume that deg D and deg Jf? D are both at least 4, so dim D and dim Jf? D are both at least 2. Fix a point PinC. Since dim Jf?D 2I can choose a divisor E=P + Z^R in Jf? D. Now fix a point Q on C but not in the support of E (i.e. e Q = 0). Because dim D 2I can choose a divisor (sloppily I call it D) in D whose support contains both P and g,

4 Set / = inf (D, E) Since P is that dim / and in /, and Q is - deg /, = /< (/) + are ord R (f) > -e R, D and an d / < (D, E). Then not, we hâve 0 < deg /< deg D. Once I show by descent I will hâve shown that C is hyperelliptic. Notice that L(I) = L(D) because = sup S c L( ). The inclusion L(I) f) E. On the other hand, if / L(D) f] Ufl) holds f) L( ), (/) + D ovd R (f) -d R and g L(D) both effective. Then, for ail points R, /g 0 and L(I). Similarly, one so ord*(/) + min(^,^) that L(D) + L(E) c L(S). Since D < S and E < S both L(D) and L( ) are subspaces of L(S). If 8g UD) and sg L( ), then for ail R, ord K (ô +s) min = -max (d R, e R ). This shows > min (ordn(8), ord^(s)) 8 that + s g US). As subspaces of L(S), we see that sees (-^, -^) Rewriting this in terms of linear Systems gives ~ Jf? Since E / + Yet S D, Lemma = D + shows that dim E ~ Jf, applied to D gives 1 so S ~ JT? /. Lemma 1, now applied to / / = - deg /. \Z To prove the third part of the theorem I need some technical lemmas. We may assume that the curve C is hyperelliptic and so cornes equipped with a given g\. On any such curve I can define a function tu :C-*C, by defining n(p) to be the unique point Q such that P + Q is a divisor in the given g\. To verify that n(p) is well defined, notice that if P + Q and P + # both belong to the given g\, then Q ~ R. Since g > 0, Q must.r [4, IL ]; this shows that k(p) is well-defmed. Notice that since kp + P is in the gj, n(np) = P. equal Lemma 2. = /(Jf) - Proof. 1+ dim For any point P, L(jf-P) = L(jT -P-tcP) am* /(JT-P) 1. P+ n{p) is a Jf-P-nP\= g\ so dim dim Jf. P+nP\ Since = 1 and by Lemma 1, jr-p-tcp<jt-p

5 < Jf, one sees that L(jf -P-ttP) ci L(jT-P) cz L(X). To prove L(jf -P) = L(jf-P-nP) it suffices to show that L(Jf -P) L(Jf). Yet if thèse were equal, the diviser P would be an effective spécial divisor of degree 1 with dim \jf?p\= dim Jf. By Lemma 1, then dim P would equal - deg P, which is absurd!? Définition. The points Pi,..., P k on C form a disjoint set of points if for each i, P t # tt(p;) and if the divisors P + np t are pairwise disjoint. Lemma 3. Let {Px,...,P x,..., P n } fre a disjoint set of points, with n g Then Proofi Since l(jf? P t ) = /(JT)? 1, the intersection has dimension /(Jf)? n. Choose points P n+ 1,_...,1,_..., P such that g {P1,...,P 1,..., Pj is a disjoint set. Then If dim f] L(Jf-Pt ) > \X) - n, then i This shows that there is an effective divisor E ~ JT -? E(P t + 7iP f ) ; but this is impossible since deg (Jf- Z(P + 7cP f )) <0. D Corollary. Le {P x, P3,...,P 3,..., P n } be disjoint. Then Proofi Since L(Jf -2PJ a L(Jf'-PJ, by the lemma L(jf-2P I)f p L(jf -P^ is contained in the vector space Lpf -PJ f fj L(^T-P^ of 3 3 dimension g? n + 1. If thèse vector spaces were equal, then they would both equal

6 P19...,P Choosing more points Pn+1,...,Pn+l,..., P g as in the proof of the lemma would give; similarly, Again, we get a contradiction since this shows that the divisor Jf - 2P 1 g 3 _ Kp i _ of négative degree is linearly équivalent to an effective divisor. D Now I can finally prove (3). Proof of (3). Given an effective spécial divisor Dof degree 2r and with P1,...,P dim \D\ = r, choose points 1,..., P r forming a disjoint set. Notice that since 2< deg D and 2< deg (X'-D), then 1 r g-2. Then there is a divisor, call it D, in D of the form I claim A = np np r. This could fail in two ways Case 1: If A contains some point Q which is not equal to any of Pt-)P l9...,p r or np 19...,nP r, then L(Jf-D) c -) f) L(Jf- t f] Q). L(X- Yet i /(Jf?D)= dim Jf?D+l =g?r while, by Lemma 3, the intersection has dimension g? (r -h 1). This shows that Case 1 cannot occur. Case 2: If A contains some P t, or contains some np t twice, (after interchanging P t and np { if necessary and renumbering) we can write where B is effective, of degree r - 1. Hère, L(Jf -D)-c L(jf -2PJ f) r P L(Jf? P f ). Again, /(Jf?D)=g?r, and by the corollary the dimension 2 of the intersection is g? (r-f 1). Case 2 cannot occur either. Thus, D - P P r + TrPi cP r so D ~ r? o. In particular, if D is any divisor on C of degree 2 with dim D = 1, D is linearly équivalent to a divisor in the given g\. Thus a hyperelliptic curve has a unique g \. It is interesting to compare the results of Clifford's theorem with those of the Riemann-Roch theorem, for hyperelliptic curves. Clifford's theorem shows that any spécial effective divisor D with dim D \ = - deg D is linearly n

7 équivalent to a multiple of the unique g\. In particular, for the canonical Conversely, the Riemann-Roch theorem divisor Jf we hâve Jf ~ (0? 1)?? g i shows that any divisor D~r? g\, -where 1 r<g?l, satisfies dim D\ = - deg D. To see this, note that the proof of part (3) shows that if D ~ r? gj I can write for a disjoint set of points {Pi,..., P r }. Then By lemma 3 this set has dimension g? r; in other words, dim Jf = 0-r-l=-deg (Jf-D). By lemma 1, dim D \ - - deg D.? D \ REFERENCES [1] Clifford, William Kingdon. On the Classification of Loci. XXXIII in Collected Papers, London, Macmillan & Co., [2] Fulton, William. Algebraic Curves. Reading, MA, Addison-Wesley, [3] Griffiths, Philipp and Joseph Harris. Principles of Algebraic Geometry. New York, John Wiley & Sons, [4] Hartshorne, Robin. Algebraic Geometry. New York, Springer-Verlag, [5] Walker, Robert J. Algebraic Curves. New York, Dover, (Reçu le 2 juin 1983) William J. Gordon Department of Mathematics State University of New York at Buffalo Buffalo, New York 14214