Reconstruction of genus hyperelliptic curve

Description

Reconstruction of genus 3 hyperelliptic curve R. Lercier, C. Ritzenthaler GEOCRYPT 2011 Bastia June 20-24

• reconstruction problem


• hyperelliptic curves


• igusa invariants


• x3 ?

Sujets

• Reconstruction conjecture

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 22
• Langue: English
• Poids de l'ouvrage: 1 Mo

Extrait

Reconstruction of genus 3 hyperelliptic curve

R. Lercier, C. Ritzenthaler

GEOCRYPT 2011 Bastia June 20-24

Motivation

Elliptic curvesEk

:y

2 =x

3 +a x+bare classified by

j(E) =1872 4a 3 4 + a 3 27

b 2 

Conversely, for anyj∈k {1728}, we can reconstruct a curveE defined overks.t.j(E) =j, for instance

Ek:y 2 =x 3 − j−1 72j 728 x+ j− 54 1 j 728 

Similarly, genus 2 curves are classified by Igusa invariants, and thanks to [Mestre91], the reconstruction problem is solved .

In this talk, we mostly consider genus 3 hyperelliptic curves.

Motivation

Elliptic curvesEk

:y

2 =x

3 +a x+bare classified by

3 j(E) = 4 2817 a 3 4 + a 27

b 2 

d C e o fi nversely,foranyj j(E ∈ ) k {1728}, we can reconstruct a curveE ned overks.t.=j, for instance

Ek:y 2 =x 3 − j− 72 1 j 728 x+ j−71 45j 28 

Similarly, genus 2 curves are classified by Igusa invariants, and thanks to [Mestre91], the reconstruction problem is solved .

In this talk, we mostly consider genus 3 hyperelliptic curves.

Motivation

Elliptic curvesEk

:y

2 =x

3 +a x+bare classified by

3 j(E) = 4 2817 a 3 4 + a 27

b 2 

d C e o fi n n v e er d se o ly,foranyj (E ∈ ) k =  j {827 ,f 1}, we can reconstruct a curveE verks.t.jor instance

3 27j5 Ek:y 2 =x− −1728 x+ j−71 4j 28  j

Similarly,genus 2curves are classified by Igusa invariants, and thanks to [Mestre91], the reconstruction problem is solved .

In this talk, we mostly consider genus 3 hyperelliptic curves.

Motivation

Elliptic curvesEk

:y

2 =x

3 +a x+bare classified by

3 j(E) =2817 4a 3 4 + a 27

b 2 

Conversely,foran t. yj (E ∈ ) k =  j {172 o 8 r,fi }ev,wecanreconstructacur stn E defined overks.jance

j x 54  Ek:y 2 =x 3 − j−7281 72 + j−17 j 28

Similarly, genus 2 curves are classified by Igusa invariants, and thanks to [Mestre91], the reconstruction problem is solved .

In this talk, we mostly considergenus 3hyperelliptic curves.

Concretely...

> _<x> := PolynomialRing(GF(11)); > H1 := HyperellipticCurve(7*x^8 + 5*x^6 + 5*x^4 + 3*x^2 + x + 9);

> ShiodaInvariants(H1); [ 8, 2, 4, 10, 2, 7, 8, 9, 3 ]

> H2, G := HyperellipticCurveFromShiodaInvariants($1); H2; Hyperelliptic Curve defined by y^2 = x^8 + x^7 + x^6 + 4*x^5 + x^4 + x^2 + 5*x + 8 over GF

> IsQuadraticTwist(H1, H2); true 4 > IdentifyGroup(G); <8, 5>

(11)

Concretely...

> _<x> := PolynomialRing(GF(11)); > H1 := HyperellipticCurve(7*x^8 + 5*x^6 + 5*x^4 + 3*x^2 + x + 9);

> ShiodaInvariants(H1); [ 8, 2, 4, 10, 2, 7, 8, 9, 3 ]

> H2, G := HyperellipticCurveFromShiodaInvariants($1); H2; Hyperelliptic Curve defined by y^2 = x^8 + x^7 + ^6 + 4*x^5 + x^4 + x^2 + 5*x + 8 over GF x

> IsQuadraticTwist(H1, H2); true 4 > IdentifyGroup(G); <8, 5>

(11)

Concretely...

> _<x> := PolynomialRing(GF(11)); > H1 := HyperellipticCurve(7*x^8 + 5*x^6 + 5*x^4 + 3*x^2 + x + 9);

> ShiodaInvariants(H1); [ 8, 2, 4, 10, 2, 7, 8, 9, 3 ]

> H2, G := HyperellipticCurveFromShiodaInvariants($1); H2; Hyperelliptic Curve defined by y^2 = x^8 + x^7 + x^6 + 4*x^5 + x^4 + x^2 + 5*x + 8 over GF

> IsQuadraticTwist(H1, H2); true 4 > IdentifyGroup(G); <8, 5>

(11)