Bad reduction of genus 3 curves with Complex Multiplication

Bad reduction of genus 3 curves with Complex Multiplication Elisa Lorenzo García Universiteit Leiden Joint work with Bouw, Cooley, Lauter, Manes, Newton, Ozman. October 1, 2015 Elisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 1 / 17

Index 1 Motivation Gross-Zagier Formula Gross-Zagier g=2 Gross-Zagier g=3 2 Set up Abelian Varieties with CM Bad reduction 3 Main Theorem The statement The Proof 4 Removing the assumptions lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 2 / 17

1 Motivation Gross-Zagier Formula Gross-Zagier g=2 Gross-Zagier g=3 2 Set up Abelian Varieties with CM Bad reduction 3 Main Theorem The statement The Proof 4 Removing the assumptions lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 3 / 17

Gross-Zagier g=1 Let be O 1, O 2 be two dierent orders of discriminant d i in Q( d i ). We wonder whenever there are two elliptic curves E i /C with CM by O i and a prime p such that E 1 E 2 mod p. In order to answer it, Gross and Zagier dened the number: lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 4 / 17

Gross-Zagier g=1 Let be O 1, O 2 be two dierent orders of discriminant d i in Q( d i ). We wonder whenever there are two elliptic curves E i /C with CM by O i and a prime p such that E 1 E 2 mod p. In order to answer it, Gross and Zagier dened the number: 8 J(d 1, d 2 ) = (j(τ 1 ) j(τ 2 )) [τ 1 ], [τ 2 ] disc(τ i ) = d i where the τ i run over equivalence classes, and w i is the number of units in O i. w 1 w 2, lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 4 / 17

Gross-Zagier g=1 Let be O 1, O 2 be two dierent orders of discriminant d i in Q( d i ). We wonder whenever there are two elliptic curves E i /C with CM by O i and a prime p such that E 1 E 2 mod p. In order to answer it, Gross and Zagier dened the number: 8 J(d 1, d 2 ) = (j(τ 1 ) j(τ 2 )) [τ 1 ], [τ 2 ] disc(τ i ) = d i where the τ i run over equivalence classes, and w i is the number of units in O i. Under some assumptions, GZ show that J(d 1, d 2 ) Z, and their main result gives a formula for its factorization. Lauter and Viray generalize the result for other disc. [LV14]. w 1 w 2, lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 4 / 17

Gross-Zagier g=1 Let be O 1, O 2 be two dierent orders of discriminant d i in Q( d i ). We wonder whenever there are two elliptic curves E i /C with CM by O i and a prime p such that E 1 E 2 mod p. In order to answer it, Gross and Zagier dened the number: 8 J(d 1, d 2 ) = (j(τ 1 ) j(τ 2 )) [τ 1 ], [τ 2 ] disc(τ i ) = d i where the τ i run over equivalence classes, and w i is the number of units in O i. Remark Under some assumptions, GZ show that J(d 1, d 2 ) Z, and their main result gives a formula for its factorization. Lauter and Viray generalize the result for other disc. [LV14]. J(d 1, d 2 ) = Res(H d1, H d2 ) (Hilbert polynomials CM-method) w 1 w 2, lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 4 / 17

Gross-Zagier g=1 Roughly speaking, { factorization J(d 1, d 2 ) } { Counting Isom n (E 1, E 2 ) } Counting embeddings ι : End(E 2 ) B p, lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 5 / 17

Gross-Zagier g=1 Roughly speaking, { factorization J(d 1, d 2 ) They prove } { Counting Isom n (E 1, E 2 ) } v l (j 1 j 2 ) = 1 #Isom n (E 1, E 2 ). 2 n Counting embeddings ι : End(E 2 ) B p, lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 5 / 17

Gross-Zagier g=1 Roughly speaking, { factorization J(d 1, d 2 ) They prove } { Counting Isom n (E 1, E 2 ) } v l (j 1 j 2 ) = 1 #Isom n (E 1, E 2 ). 2 n Counting embeddings ι : End(E 2 ) B p, Idea If Q( d 1 ) Q( d 2 ), we need End 0 (Ē i ) B p,. ι : End(E 2 ) End(Ē 2 ) End(Ē 1 ) B p, lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 5 / 17

Gross-Zagier g=2 Generalizations g = 2: Igusa invariants: i i : M 2 A 2 C, i {1, 2, 3} lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 6 / 17

Gross-Zagier g=2 Generalizations g = 2: Igusa invariants: i i : M 2 A 2 C, i {1, 2, 3} For C 1, C 2 curves with CM by orders O 1, O 2 in quartic CM-elds, we want to dene J(O 1, O 2 ) = g.c.d(i 1 i 1, i 2 i 2, i 3 i 3) lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 6 / 17

Gross-Zagier g=2 Generalizations g = 2: Igusa invariants: i i : M 2 A 2 C, i {1, 2, 3} For C 1, C 2 curves with CM by orders O 1, O 2 in quartic CM-elds, we want to dene J(O 1, O 2 ) = g.c.d(i 1 i 1, i 2 i 2, i 3 i 3) PROBLEM!! the invariants are not integer numbers any more!! but, still rational... Numerators: (Gross-Zagier formula) Goren and Lauter. {factorization} { Counting Isomorphisms } { Counting embeddings ι : End(C 2 ) B p, L } lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 6 / 17

Gross-Zagier g=2 Generalizations g = 2: Igusa invariants: i i : M 2 A 2 C, i {1, 2, 3} For C 1, C 2 curves with CM by orders O 1, O 2 in quartic CM-elds, we want to dene J(O 1, O 2 ) = g.c.d(i 1 i 1, i 2 i 2, i 3 i 3) PROBLEM!! the invariants are not integer numbers any more!! but, still rational... Numerators: (Gross-Zagier formula) Goren and Lauter. {factorization} { Counting Isomorphisms } { Counting embeddings ι : End(C 2 ) B p, L Denominators: (Cryptographic purposes) Goren & Lauter, Bruinier & Yang, Lauter & Viray, Streng. { } { } Bad Counting embeddings {factorization} Reduction ι : End(C 2 ) M 2 (B p, ) lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 6 / 17 }

Gross-Zagier g=2 Theorem (Lauter-Viray) p divides denominators (χ 10 ) Bad Reduction Solution to emb. prob. Igusa invariants: i i = some modular form χ 10, p χ 10 J decomposable. Bad reduction: C = C 1 C 2 lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 7 / 17

Gross-Zagier g=2 Theorem (Lauter-Viray) p divides denominators (χ 10 ) Bad Reduction Solution to emb. prob. Igusa invariants: i i = some modular form χ 10, p χ 10 J decomposable. Bad reduction: C = C 1 C 2 Idea If p is a prime of bad reduction, then C = C 1 C 2 and J = J(C 1 ) J(C 2 ) J E F J E 2 ι : O M 2 (B p, ) We will look for a bound, so we don't care about the other direction of the implication. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 7 / 17

Gross-Zagier g=3 PROBLEM: there are not invariants!! lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 8 / 17

Gross-Zagier g=3 PROBLEM: there are not invariants!! but..., we can still study the embedding problem! ι : O M 3 (B p, ) Bad reduction J(C)) E 3 with E supersingular we have a solution to the embedding problem. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 8 / 17

Gross-Zagier g=3 PROBLEM: there are not invariants!! but..., we can still study the embedding problem! ι : O M 3 (B p, ) Bad reduction J(C)) E 3 with E supersingular we have a solution to the embedding problem. Conditions: Optimal Rosati involution given by complex conjugation γ γ = λ 1 γ λ lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 8 / 17

Abelian Varieties with CM Proposition If A is an abelian variety dened over a number eld M and with CM, then it has potentially good reduction everywhere. Proposition (Lang) Let A be an abelian variety with CM by K and dened over a eld of characteristic zero. The CM-type (K, ϕ) is primitive if and only if the abelian variety A is simple. If g = 2: (K, ϕ) primitive i K does not contain any imaginary quadratic subeld K 1. This is not true any more if g = 3. (R1) Restriction 1: we assume that K does not contain any K 1. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 10 / 17

Abelian Varieties with CM Proposition If A is an abelian variety dened over a number eld M and with CM, then it has potentially good reduction everywhere. Proposition (Lang) Let A be an abelian variety with CM by K and dened over a eld of characteristic zero. The CM-type (K, ϕ) is primitive if and only if the abelian variety A is simple. If g = 2: (K, ϕ) primitive i K does not contain any imaginary quadratic subeld K 1. This is not true any more if g = 3. (R1) Restriction 1: we assume that K does not contain any K 1. If g = 2: in the previous setting, we have that λ is given by a diagonal matrix. (R2) Restriction 2: we still assume that λ is diagonal. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 10 / 17

Curves with CM Let C be a genus 3 curve with CM by K. We denote by J its Jacobian. Proposition One of the following three possibilities holds for the irreducible components of C of positive genus: (i) (good reduction) C is a smooth curve of genus 3, (ii) C has three irreducible components of genus 1, (iii) C has an irreducible component of genus 1 and one of genus 2. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 11 / 17

Curves with CM Let C be a genus 3 curve with CM by K. We denote by J its Jacobian. Proposition One of the following three possibilities holds for the irreducible components of C of positive genus: (i) (good reduction) C is a smooth curve of genus 3, (ii) C has three irreducible components of genus 1, (iii) C has an irreducible component of genus 1 and one of genus 2. Theorem If J is not simple, then J is isogenous to E 3. In particular, if C has bad reduction, then J E 3. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 11 / 17

The main Theorem We are ready to state the main theorem in our paper: Theorem Let C be a genus 3 curves with CM by a CM-eld K (with a primitive CM-type). Write K = Q( α) for some totally negative element α K + /Z with α O = End(J(C)). Assume further that we are under restrictions (R1) and (R2). Then any prime p p for which there is a solution to the embedding problem, in particular, for primes of bad reduction, is bounded by p 4Tr K + /Q(α) 6 /3 6. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 13 / 17

Sketch of the proof Proof (Sketch). If p is a prime of bad reduction, then there exists an embedding ι : K = End 0 (J) End 0 (J) = M 3 (B p, ) such that complex conjugation on the LHS corresponds to the Rosati involution on the RHS. Recall that K = Q( α) and K + = Q(α) with α a totally negative element. Let us call γ = ι(α). Then (using R2) 0 > Tr(α) = Tr(γγ ) = Tr(γλγ λ 1 ) = Nr(γ ij ) Goren & Lauter's Lemma about small norm elements implies that we have an embedding ι : K M 3 (K 1 ) with K 1 an at most degree 2 CM-eld. (R1) K 1 = Q. This gives us a contradiction with [Q( α) : Q] = 6. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 14 / 17

Restrictions How can we remove the restrictions?? (R1) Restriction 1: we need to use more ne arguments, not only dimension ones. We still didn't use the primitivity of the CM-type. The reduction of the CM-type is not well-dened, we have to work with the Lie type, and prove that if Q( d) K, then ι( d) is a matrix equivalent to ± d 1 0 0 0 1 0 0 0 1. lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 16 / 17

Restrictions How can we remove the restrictions?? (R1) Restriction 1: we need to use more ne arguments, not only dimension ones. We still didn't use the primitivity of the CM-type. The reduction of the CM-type is not well-dened, we have to work with the Lie type, and prove that if Q( d) K, then ι( d) is a matrix equivalent to ± d 1 0 0 0 1 0 0 0 1 (R2) Restriction 2: We need to study the case in which C has an irreducible component of genus 1 and one of genus 2. That is, consider the case in which the polarization may be given by λ = ( A 0 0 1. ), A M 2 (B p, ) lisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 16 / 17

Thank you! Elisa Lorenzo García Universiteit Leiden (Joint Bad reduction work withofbouw, genus Cooley, 3 curveslauter, with CManes, Newton, October Ozman.) 1, 2015 17 / 17