bad reduction of genus 3 curves with complex multiplication · bad reduction of genus 3 curves with...

Bad reduction of genus 3 curves with Complex

Multiplication

Elisa Lorenzo GarcíaUniversiteit Leiden

Joint work with Bouw, Cooley, Lauter, Manes, Newton, Ozman.

October 1, 2015

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

1 MotivationGross-Zagier FormulaGross-Zagier g=2Gross-Zagier g=3

2 Set upAbelian Varieties with CMBad reduction

3 Main TheoremThe statementThe Proof

4 Removing the assumptions

Gross-Zagier g=1

Let be O1, O2 be two di�erent orders of discriminant di in Q(√di ). We

wonder whenever there are two elliptic curves Ei/C with CM by Oi and aprime p such that E1 ' E2mod p. In order to answer it, Gross and Zagierde�ned the number:

J(d1, d2) =

∏[τ1], [τ2]

disc(τi ) = di

(j(τ1)− j(τ2))

where the τi run over equivalence classes, and wi is the number of units inOi .

Under some assumptions, GZ show that J(d1, d2) ∈ Z, and their mainresult gives a formula for its factorization.

Lauter and Viray generalize the result for other disc. [LV14].

Remark

J(d1, d2) = Res(Hd1 ,Hd2) (Hilbert polynomials ⇒ CM-method)

Gross-Zagier g=1

J(d1, d2) =

∏[τ1], [τ2]

disc(τi ) = di

(j(τ1)− j(τ2))

Remark

Gross-Zagier g=1

J(d1, d2) =

∏[τ1], [τ2]

disc(τi ) = di

(j(τ1)− j(τ2))

Remark

Gross-Zagier g=1

J(d1, d2) =

∏[τ1], [τ2]

disc(τi ) = di

(j(τ1)− j(τ2))

Remark

Gross-Zagier g=1

Roughly speaking,

{factorizationJ(d1, d2)

CountingIsomn(E1,E2)

Counting

embeddingsι : End(E2) ↪→ Bp,∞

They prove

vl(j1 − j2) =1

#Isomn(E1,E2).

If Q(√d1) 6= Q(

√d2), we need End0(Ei ) ' Bp,∞.

ι : End(E2) ↪→ End(E2)→ End(E1) ' Bp,∞

Gross-Zagier g=1

Roughly speaking,

Counting

They prove

vl(j1 − j2) =1

#Isomn(E1,E2).

If Q(√d1) 6= Q(

Gross-Zagier g=1

Roughly speaking,

Counting

They prove

vl(j1 − j2) =1

#Isomn(E1,E2).

If Q(√d1) 6= Q(

Gross-Zagier g=2

Generalizations g = 2:

Igusa invariants: ii : M2 ↪→ A2 → C, i ∈ {1, 2, 3}

For C1, C2 curves with CM by orders O1, O2 in quartic CM-�elds, wewant to de�ne

J(O1,O2) = g.c.d(i1 − i ′1, i2 − i ′2, i3 − i ′3)

PROBLEM!! the invariants are not integer numbers any more!! but, stillrational ....Numerators: (Gross-Zagier formula) Goren and Lauter.

{factorization} ↔{

Counting

Isomorphisms

Counting embeddings

ι : End(C2) ↪→ Bp,∞ ⊗ L

Denominators: (Cryptographic purposes) Goren & Lauter, Bruinier & Yang,Lauter & Viray, Streng.

Reduction

Counting embeddings

ι : End(C2) ↪→M2(Bp,∞)

Gross-Zagier g=2

Igusa invariants: ii : M2 ↪→ A2 → C, i ∈ {1, 2, 3}For C1, C2 curves with CM by orders O1, O2 in quartic CM-�elds, wewant to de�ne

J(O1,O2) = g.c.d(i1 − i ′1, i2 − i ′2, i3 − i ′3)

Counting

Isomorphisms

Counting embeddings

ι : End(C2) ↪→ Bp,∞ ⊗ L

Reduction

Counting embeddings

ι : End(C2) ↪→M2(Bp,∞)

Gross-Zagier g=2

J(O1,O2) = g.c.d(i1 − i ′1, i2 − i ′2, i3 − i ′3)

PROBLEM!! the invariants are not integer numbers any more!! but, stillrational ....

Numerators: (Gross-Zagier formula) Goren and Lauter.

Counting

Isomorphisms

Counting embeddings

ι : End(C2) ↪→ Bp,∞ ⊗ L

Reduction

Counting embeddings

ι : End(C2) ↪→M2(Bp,∞)

Gross-Zagier g=2

J(O1,O2) = g.c.d(i1 − i ′1, i2 − i ′2, i3 − i ′3)

Counting

Isomorphisms

Counting embeddings

ι : End(C2) ↪→ Bp,∞ ⊗ L

Reduction

Counting embeddings

ι : End(C2) ↪→M2(Bp,∞)

Gross-Zagier g=2

J(O1,O2) = g.c.d(i1 − i ′1, i2 − i ′2, i3 − i ′3)

Counting

Isomorphisms

Counting embeddings

ι : End(C2) ↪→ Bp,∞ ⊗ L

Reduction

Counting embeddings

ι : End(C2) ↪→M2(Bp,∞)

}Elisa Lorenzo García Universiteit Leiden (Joint work with Bouw, Cooley, Lauter, Manes, Newton, Ozman.)Bad reduction of genus 3 curves with CM October 1, 2015 6 / 17

Gross-Zagier g=2

Theorem (Lauter-Viray)

p divides denominators (χ10) ⇔ Bad Reduction ⇔ Solution to emb. prob.

Igusa invariants: ii = some modular formχ10

, p | χ10 ⇔ J decomposable.

Bad reduction: ⇔ C = C1 ∪ C2

If p is a prime of bad reduction, then C = C1 ∪ C2 and J = J(C1)× J(C2)⇒ J ' E × F ⇒ J ∼ E 2 ⇒ ι : O ↪→M2(Bp,∞)

We will look for a bound, so we don't care about the other direction of theimplication.

Gross-Zagier g=2

Theorem (Lauter-Viray)

p divides denominators (χ10) ⇔ Bad Reduction ⇔ Solution to emb. prob.

Igusa invariants: ii = some modular formχ10

, p | χ10 ⇔ J decomposable.

Bad reduction: ⇔ C = C1 ∪ C2

If p is a prime of bad reduction, then C = C1 ∪ C2 and J = J(C1)× J(C2)⇒ J ' E × F ⇒ J ∼ E 2 ⇒ ι : O ↪→M2(Bp,∞)

We will look for a bound, so we don't care about the other direction of theimplication.

Gross-Zagier g=3

PROBLEM: there are not invariants!!

but ..., we can still study the embedding problem!

ι : O ↪→M3(Bp,∞)

Bad reduction ⇒ J(C )) ∼ E 3 with E supersingular ⇒ we have a solutionto the embedding problem.

Conditions:

Optimal

Rosati involution given by complex conjugation

γ → γ† = λ−1 · γ∨ · λ

Gross-Zagier g=3

ι : O ↪→M3(Bp,∞)

Conditions:

Optimal

γ → γ† = λ−1 · γ∨ · λ

Gross-Zagier g=3

ι : O ↪→M3(Bp,∞)

Conditions:

Optimal

γ → γ† = λ−1 · γ∨ · λ

Abelian Varieties with CM

Proposition

If A is an abelian variety de�ned 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 de�ned 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 quadraticsub�eld K1. This is not true any more if g = 3.

(R1) Restriction 1: we assume that K does not contain any K1.

If g = 2: in the previous setting, we have that λ is given by a diagonalmatrix.

(R2) Restriction 2: we still assume that λ is diagonal.

Proposition

Proposition (Lang)

Proposition

Proposition (Lang)

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.

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.

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 ≤ 4TrK+/Q(α)6/36.

Sketch of the proof

Proof (Sketch).

If p is a prime of bad reduction, then there exists an embedding

ι : K = End0(J) ↪→ End0(J) =M3(Bp,∞)

such that complex conjugation on the LHS corresponds to the Rosatiinvolution 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(γij)

Goren & Lauter's Lemma about small norm elements implies that we havean embedding ι : K ↪→M3(K1) with K1 an at most degree 2 CM-�eld.(R1) ⇒ K1 = Q. This gives us a contradiction with [Q(

√α) : Q] = 6.

Restrictions

How can we remove the restrictions??

(R1) Restriction 1: we need to use more �ne arguments, not onlydimension ones. We still didn't use the primitivity of the CM-type. Thereduction of the CM-type is not well-de�ned, we have to work with the Lietype, and prove that if Q(

√−d) ⊆ K , then ι(−

√d) is a matrix equivalent

±√−d

0 0 −1

(R2) Restriction 2: We need to study the case in which C has anirreducible component of genus 1 and one of genus 2. That is, consider thecase in which the polarization may be given by

(A 00 1

), A ∈M2(Bp,∞)

Restrictions

How can we remove the restrictions??

(R1) Restriction 1: we need to use more �ne arguments, not onlydimension ones. We still didn't use the primitivity of the CM-type. Thereduction of the CM-type is not well-de�ned, we have to work with the Lietype, and prove that if Q(

√−d) ⊆ K , then ι(−

√d) is a matrix equivalent

±√−d

0 0 −1

(R2) Restriction 2: We need to study the case in which C has anirreducible component of genus 1 and one of genus 2. That is, consider thecase in which the polarization may be given by

(A 00 1

), A ∈M2(Bp,∞)

Thank you!

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