87 Pages
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# Reconstruction of genus hyperelliptic curve

87 Pages
English Description

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

• reconstruction problem

• hyperelliptic curves

• igusa invariants

• x3 ?

Subjects

##### Reconstruction conjecture

Informations

Exrait Reconstruction of genus 3 hyperelliptic curve
R. Lercier, C. Ritzenthaler
GEOCRYPT 2011 Bastia June 20-24 Motivation
Elliptic curvesEk
:y
2=x
3+a x+bare classiﬁed by
j(E) =18724a34+a327
b2
Conversely, for anyjk {1728}, we can reconstruct a curveE deﬁned overks.t.j(E) =j, for instance
Ek:y2=x3j172j728x+j541j728
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 curvesEk
:y
2=x
3+a x+bare classiﬁed by
3 j(E) =42817a34+a27
b2
dCeonversely,foranyjj(E)k {1728}, we can reconstruct a curveE ned overks.t.=j, for instance
Ek:y2=x3j721j728x+j7145j28
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 curvesEk
:y
2=x
3+a x+bare classiﬁed by
3 j(E) =42817a34+a27
b2
dCeonnveerdseoly,foranyj(E)k=j{827,f1}, we can reconstruct a curveE verks.t.jor instance
327j5 Ek:y2=x1728x+j714j28j
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 curvesEk
:y
2=x
3+a x+bare classiﬁed by
3 j(E) =28174a34+a27
b2
Conversely,forant.yj(E)k=j{172o8r,fi}ev,wecanreconstructacurstnE deﬁned overks.jance
jx54Ek:y2=x3j728172+j17j28
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)