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Rationality of intersection points of a line and a quartic Christophe Ritzenthaler Joint work with Roger Oyono Institut de Mathématiques de Luminy, CNRS Istanbul, 27-30 June 2010 Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 1 / 12 Initial problem How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k ? Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 2 / 12 Initial problem How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k ? This problem was addressed and solved geometrically in Flon-Oyono-R. 08. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 2 / 12 Initial problem How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k ? This problem was addressed and solved geometrically in Flon-Oyono-R. 08. Summary: non-hyperelliptic genus 3 curves = smooth plane quartics. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 2 / 12 Initial problem How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k ? This problem was addressed and solved geometrically in Flon-Oyono-R. 08. Summary: non-hyperelliptic genus 3 curves = smooth plane quartics. points on the Jacobian are represented by the sum D of 3 points on the curve C . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 2 / 12 Initial problem How to add points on the Jacobian of non-hyperelliptic genus 3 curves C /k ? This problem was addressed and solved geometrically in Flon-Oyono-R. 08. Summary: non-hyperelliptic genus 3 curves = smooth plane quartics. points on the Jacobian are represented by the sum D of 3 points on the curve C . Choice of a good divisor at infinity condition (∗): There is a rational line `∞ which crosses the quartic C in four k-points P1∞ , P2∞ , P3∞ , P4∞ . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 2 / 12 A geometric addition algorithm Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 3 / 12 A geometric addition algorithm Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 . 1 Take a cubic E which goes (with multiplicity) through the support of D1 , D2 and P1∞ , P2∞ , P4∞ . This cubic also crosses C in the residual effective divisor D3 . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 3 / 12 A geometric addition algorithm Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 . 1 2 Take a cubic E which goes (with multiplicity) through the support of D1 , D2 and P1∞ , P2∞ , P4∞ . This cubic also crosses C in the residual effective divisor D3 . Take a conic Q which goes through the support of D3 and P1∞ , P2∞ . This conic also crosses C in the residual effective divisor D + . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 3 / 12 A geometric addition algorithm Aim: given D1 , D2 find D + such that D + + D ∞ ∼ D1 + D2 . 1 2 Take a cubic E which goes (with multiplicity) through the support of D1 , D2 and P1∞ , P2∞ , P4∞ . This cubic also crosses C in the residual effective divisor D3 . Take a conic Q which goes through the support of D3 and P1∞ , P2∞ . This conic also crosses C in the residual effective divisor D + . Why ? Because (`∞ · C ) ∼ κ, (Q · C ) ∼ 2κ and (E · C ) ∼ 3κ where κ is the canonical divisor on C . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 3 / 12 A chord construction Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 4 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), deg(f4 ) ≤ 4, Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), 1 deg(f4 ) ≤ 4, deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case); Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), deg(f4 ) ≤ 4, 1 deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case); 2 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case). Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), deg(f4 ) ≤ 4, 1 deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case); 2 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case). 3 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex case). These curves are the C3,4 -curves. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), deg(f4 ) ≤ 4, 1 deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case); 2 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case). 3 4 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex case). These curves are the C3,4 -curves. If char(k) 6= 3, C : y 3 = f4 (x) (Picard curves) iif P1∞ is a rational Galois point. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), deg(f4 ) ≤ 4, 1 deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case); 2 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case). 3 4 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex case). These curves are the C3,4 -curves. If char(k) 6= 3, C : y 3 = f4 (x) (Picard curves) iif P1∞ is a rational Galois point. : the more special, the faster. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Special forms of the curve C admits an equation of the form C : y 3 + h1 (x)y 2 + h2 (x)y = f4 (x), deg(f4 ) ≤ 4, 1 deg(h1 ) ≤ 2 and deg(h2 ) ≤ 3 if P1∞ = P2∞ (tangent case); 2 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 3 if P1∞ = P2∞ = P4∞ (flex case). 3 4 deg(h1 ) ≤ 1 and deg(h2 ) ≤ 2 if P1∞ = P2∞ = P3∞ = P4∞ (hyperflex case). These curves are the C3,4 -curves. If char(k) 6= 3, C : y 3 = f4 (x) (Picard curves) iif P1∞ is a rational Galois point. : the more special, the faster. Remark: Quartics with a hyperflex form a sub-variety of codimension 1. So generically the two last cases do not occur. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 5 / 12 Study of the condition (∗) over Fq (Oyono-R.) Theorem Let C be a smooth plane quartic over the finite field Fq with q = p n elements. If q ≥ 127, then there exists a line ` which intersects C at rational points only. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 6 / 12 Study of the condition (∗) over Fq (Oyono-R.) Theorem Let C be a smooth plane quartic over the finite field Fq with q = p n elements. If q ≥ 127, then there exists a line ` which intersects C at rational points only. Theorem Let C be a smooth plane quartic over Fq . If q ≥ 662 + 1, then there exists a tangent to C which intersects C at rational points only. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 6 / 12 Study of the condition (∗) over Fq (Oyono-R.) Theorem Let C be a smooth plane quartic over the finite field Fq with q = p n elements. If q ≥ 127, then there exists a line ` which intersects C at rational points only. Theorem Let C be a smooth plane quartic over Fq . If q ≥ 662 + 1, then there exists a tangent to C which intersects C at rational points only. Moreover, under a certain conjecture, the probability for a plane smooth quartic over a finite field to have a rational flex is about 0.63. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 6 / 12 Proof of the generic case Follow an idea of (Diem-Thomé 08). Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 7 / 12 Proof of the generic case Follow an idea of (Diem-Thomé 08). 1 Let P ∈ C (k). Consider the separable geometric cover φ : C → |κ − P| = P1 of degree 3 induced by the linear system |κ − P|. This parametrizes lines through the point P. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 7 / 12 Proof of the generic case Follow an idea of (Diem-Thomé 08). 1 Let P ∈ C (k). Consider the separable geometric cover φ : C → |κ − P| = P1 of degree 3 induced by the linear system |κ − P|. This parametrizes lines through the point P. 2 Using effective Chebotarev’s density theorem for function fields, one gets estimation on the number of completely split divisors in |κ − P|. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 7 / 12 Proof for the tangent case Let T : C → Sym2 (C ), P 7→ TP (C ) · C − 2P be the tangential correspondence. We associate to it its correspondence curve XC = {(P, Q) ∈ C × C : Q ∈ T (P)} which is defined over k. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 8 / 12 Proof for the tangent case Let T : C → Sym2 (C ), P 7→ TP (C ) · C − 2P be the tangential correspondence. We associate to it its correspondence curve XC = {(P, Q) ∈ C × C : Q ∈ T (P)} which is defined over k. We want to prove that there is a rational point on XC when q is big enough. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 8 / 12 Proof for the tangent case Let T : C → Sym2 (C ), P 7→ TP (C ) · C − 2P be the tangential correspondence. We associate to it its correspondence curve XC = {(P, Q) ∈ C × C : Q ∈ T (P)} which is defined over k. We want to prove that there is a rational point on XC when q is big enough. Proposition (Aubry, Perret 95) Let X /Fq be a geometrically irreducible curve of arithmetic genus πX . Then √ |#X (Fq ) − (q + 1)| ≤ 2πX q. In particular if q ≥ (2πX )2 then X has a rational point. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 8 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. The morphism π1 a degree 2-cover which ramification points are the bitangency points. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. The morphism π1 a degree 2-cover which ramification points are the bitangency points. Moreover if a bitangency point P is not a hyperflex, then the points in the fiber of π1−1 (P) are smooth points on XC . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. The morphism π1 a degree 2-cover which ramification points are the bitangency points. Moreover if a bitangency point P is not a hyperflex, then the points in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is absolutely irreducible. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. The morphism π1 a degree 2-cover which ramification points are the bitangency points. Moreover if a bitangency point P is not a hyperflex, then the points in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is absolutely irreducible. • The case p = 2 (not in the article, see Arxiv). We prove that if XC is not absolutely irreducible, then XC is reducible and each component is isomorphic to C . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. The morphism π1 a degree 2-cover which ramification points are the bitangency points. Moreover if a bitangency point P is not a hyperflex, then the points in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is absolutely irreducible. • The case p = 2 (not in the article, see Arxiv). We prove that if XC is not absolutely irreducible, then XC is reducible and each component is isomorphic to C .Apply the previous bound to a component. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 The question of the absolute irreducibility of XC • The case of p 6= 2. Let πi : XC → C be the projections on the first and second factors. The morphism π1 a degree 2-cover which ramification points are the bitangency points. Moreover if a bitangency point P is not a hyperflex, then the points in the fiber of π1−1 (P) are smooth points on XC .This implies that XC is absolutely irreducible. • The case p = 2 (not in the article, see Arxiv). We prove that if XC is not absolutely irreducible, then XC is reducible and each component is isomorphic to C .Apply the previous bound to a component. Remark: the only case where XC is reducible is when C is geometrically isomorphic to the Klein quartic x 3 y + y 3 z + z 3 x = 0 (in particular Conjecture 1 in the paper is false). Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a line and Istanbul, a quartic 27-30 June 2010 9 / 12 Computation of the arithmetic genus of XC If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz formula with the degree 2-cover π1 : XC → C to get gXC . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 10 / 12 Computation of the arithmetic genus of XC If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a flat family argument to get πXC in general. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 10 / 12 Computation of the arithmetic genus of XC If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a flat family argument to get πXC in general. When p = 2, wild ramification prevents to use Riemann-Hurwitz formula so easily. We use instead intersection theory in C × C . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 10 / 12 Computation of the arithmetic genus of XC If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a flat family argument to get πXC in general. When p = 2, wild ramification prevents to use Riemann-Hurwitz formula so easily. We use instead intersection theory in C × C . In both cases, we find that πXC = 33. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 10 / 12 Computation of the arithmetic genus of XC If p 6= 2 and there is no hyperflex on C , one can use Riemann-Hurwitz formula with the degree 2-cover π1 : XC → C to get gXC . Then one uses a flat family argument to get πXC in general. When p = 2, wild ramification prevents to use Riemann-Hurwitz formula so easily. We use instead intersection theory in C × C . In both cases, we find that πXC = 33. Remark : unfortunately, there is a mistake in a remark of our paper (corrected in the Arxiv version). We claimed that πXC = 9 when p = 2. This is due to a confusion between the degree of the dual curve and the degree of the dual map π2 (which is inseparable in this case). Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 10 / 12 The case of flexes Let P14 be the linear system of all plane quartics over a field K and I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 11 / 12 The case of flexes Let P14 be the linear system of all plane quartics over a field K and I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Let I4 ⊂ P14 × I0 be the locus I4 = {(C , (P, `)), C is smooth and P is a flex of C with tangent line `}. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 11 / 12 The case of flexes Let P14 be the linear system of all plane quartics over a field K and I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Let I4 ⊂ P14 × I0 be the locus I4 = {(C , (P, `)), C is smooth and P is a flex of C with tangent line `}. Theorem (Harris 79) The Galois group of the cover I4 → P14 over C is the full symmetric group S24 . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 11 / 12 The case of flexes Let P14 be the linear system of all plane quartics over a field K and I0 = {(P, `), P ∈ `} ⊂ P2 × (P2 )∗ . Let I4 ⊂ P14 × I0 be the locus I4 = {(C , (P, `)), C is smooth and P is a flex of C with tangent line `}. Theorem (Harris 79) The Galois group of the cover I4 → P14 over C is the full symmetric group S24 . If this result is still true over F̄p , using Chebotarev density theorem, one gets that the probability to have a plane quartic with a rational flex tends to 1− 1 1 1 + − ... − ≈ 1 − exp(−1) ≈ 0.63 2! 3! 24! when q tends to infinity. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 11 / 12 Harris’ result over finite fields General reduction arguments show that the Galois group remains S24 for almost all p. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 12 / 12 Harris’ result over finite fields General reduction arguments show that the Galois group remains S24 for almost all p.However Proposition The Galois group over F̄3 is S8 . Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 12 / 12 Harris’ result over finite fields General reduction arguments show that the Galois group remains S24 for almost all p.However Proposition The Galois group over F̄3 is S8 . This is due to the fact that the flexes of C are on a conic, which is not the case when p 6= 3. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 12 / 12 Harris’ result over finite fields General reduction arguments show that the Galois group remains S24 for almost all p.However Proposition The Galois group over F̄3 is S8 . This is due to the fact that the flexes of C are on a conic, which is not the case when p 6= 3. We conjecture that p = 3 is the only exception. Christophe Ritzenthaler Joint work with Rationality Roger Oyono of intersection (IML) points of a lineIstanbul, and a quartic 27-30 June 2010 12 / 12