Curba eliptica y^2=x^3-372x+2761: punctele de ordin finit
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Cezar Lupu
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Curba eliptica y^2=x^3-372x+2761: punctele de ordin finit
Sa se determine punctele de ordin finit (de torsiune) de pe curba eliptica \( \mathcal{C}:\ y^2=x^3-372x+2671 \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Cristi Popa
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\( C(\mathbb{Q})\ :\ y^2=x^3-372x+2761=f(x) \)
Daca \( f(x)=x^3+ax^2+bx+c,\ a,b,c\in\mathbb{Z} \), atunci \( \Delta_f=a^2b^2-4a^3c-4b^3+18abc-27c^2 \) este discriminantul lui \( f \).
In cazul nostru, \( \Delta_f=-4\cdot(-372)^3-27\cdot2761^2=205915392-205824267=91125=3^6\cdot5^3 \)
Vom folosi urmatoarea:
Propozitie. Fie curba eliptica \( C(\mathbb{Q})\ :\ y^2=x^3+ax^2+bx+c=f(x),\ a,b,c\in\mathbb{Z},\ \Delta_f\neq0 \), \( p \) un numar prim a.i. \( p \) nu divide \( 2\Delta_f. \)
Fie \( C(\mathbb{Z}_p)\ :\ y^2=x^3+\overline{a}x^2+\overline{b}x+\overline{c},\ \overline{a},\ \overline{b},\ \overline{c}\in\mathbb{Z}_p \) si \( \mathcal{M} \) grupul punctelor de ordin finit ale lui \( C(\mathbb{Q}). \)
Atunci \( |\mathcal{M}|\ |\ |C(\mathbb{Z}_p)|. \)
Fie \( p=7\Rightarrow C(\mathbb{Z}_7)\ :\ y^2=x^3-x+\overline{3} \).
\( y^2\equiv0,1,2,4\ (mod\ 7) \)
Daca \( x=\overline{0}\Rightarrow y^2=\overline{3} \) (nu are solutii modulo 7).
\( x=\overline{1}\Rightarrow y^2=\overline{3} \)
\( x=\overline{2}\Rightarrow y^2=\overline{2}\Rightarrow(\overline{2},\overline{\pm3})\in C(\mathbb{Z}_7) \)
\( x=\overline{3}\Rightarrow y^2=\overline{-1} \)
\( x=\overline{4}\Rightarrow y^2=\overline{0}\Rightarrow(\overline{4},\overline{0})\in C(\mathbb{Z}_7) \)
\( x=\overline{5}\Rightarrow y^2=\overline{4}\Rightarrow(\overline{5},\overline{\pm2})\in C(\mathbb{Z}_7) \)
\( x=\overline{6}\Rightarrow y^2=\overline{3} \)
Adaugand si punctul de la infinit \( \mathcal{O} \), obtinem ca \( |C(\mathbb{Z}_7)|=6. \)
Fie \( p=11\Rightarrow C(\mathbb{Z}_{11})\ :\ y^2=x^3+\overline{2}x. \)
Putem arata ca mai sus ca \( |C(\mathbb{Z}_{11})|=12 \) sau folosind urmatoarea:
Observatie. Fie curba eliptica \( C(\mathbb{Q})\ :\ y^2=x^3+bx,\ p \) un numar prim a.i. \( p\equiv3\ (mod\ 4) \) si \( p \) nu divide \( 2\Delta_f. \)
Atunci \( |C(\mathbb{Z}_p)|=p+1 \).
In cazul nostru \( b=2,\ p=11\equiv3\ (mod\ 4) \) si nu divide \( 2\Delta_f=-4b^3=-32. \) Rezulta ca \( |C(\mathbb{Z}_{11})|=12 \).
Deci, \( |\mathcal{M}| \) divide atat pe 6, cat si pe 12 rezulta ca \( |\mathcal{M}|\ |\ (6,12)=6 \).
Fie punctul \( P(2,45) \). Calculand \( 5P \) obtinem \( 5P=-P\Rightarrow 6P=\mathcal{O}\Rightarrow ord(P)=6 \).
Deci, \( \mathcal{M}\simeq\mathbb{Z}_6. \)
Daca \( f(x)=x^3+ax^2+bx+c,\ a,b,c\in\mathbb{Z} \), atunci \( \Delta_f=a^2b^2-4a^3c-4b^3+18abc-27c^2 \) este discriminantul lui \( f \).
In cazul nostru, \( \Delta_f=-4\cdot(-372)^3-27\cdot2761^2=205915392-205824267=91125=3^6\cdot5^3 \)
Vom folosi urmatoarea:
Propozitie. Fie curba eliptica \( C(\mathbb{Q})\ :\ y^2=x^3+ax^2+bx+c=f(x),\ a,b,c\in\mathbb{Z},\ \Delta_f\neq0 \), \( p \) un numar prim a.i. \( p \) nu divide \( 2\Delta_f. \)
Fie \( C(\mathbb{Z}_p)\ :\ y^2=x^3+\overline{a}x^2+\overline{b}x+\overline{c},\ \overline{a},\ \overline{b},\ \overline{c}\in\mathbb{Z}_p \) si \( \mathcal{M} \) grupul punctelor de ordin finit ale lui \( C(\mathbb{Q}). \)
Atunci \( |\mathcal{M}|\ |\ |C(\mathbb{Z}_p)|. \)
Fie \( p=7\Rightarrow C(\mathbb{Z}_7)\ :\ y^2=x^3-x+\overline{3} \).
\( y^2\equiv0,1,2,4\ (mod\ 7) \)
Daca \( x=\overline{0}\Rightarrow y^2=\overline{3} \) (nu are solutii modulo 7).
\( x=\overline{1}\Rightarrow y^2=\overline{3} \)
\( x=\overline{2}\Rightarrow y^2=\overline{2}\Rightarrow(\overline{2},\overline{\pm3})\in C(\mathbb{Z}_7) \)
\( x=\overline{3}\Rightarrow y^2=\overline{-1} \)
\( x=\overline{4}\Rightarrow y^2=\overline{0}\Rightarrow(\overline{4},\overline{0})\in C(\mathbb{Z}_7) \)
\( x=\overline{5}\Rightarrow y^2=\overline{4}\Rightarrow(\overline{5},\overline{\pm2})\in C(\mathbb{Z}_7) \)
\( x=\overline{6}\Rightarrow y^2=\overline{3} \)
Adaugand si punctul de la infinit \( \mathcal{O} \), obtinem ca \( |C(\mathbb{Z}_7)|=6. \)
Fie \( p=11\Rightarrow C(\mathbb{Z}_{11})\ :\ y^2=x^3+\overline{2}x. \)
Putem arata ca mai sus ca \( |C(\mathbb{Z}_{11})|=12 \) sau folosind urmatoarea:
Observatie. Fie curba eliptica \( C(\mathbb{Q})\ :\ y^2=x^3+bx,\ p \) un numar prim a.i. \( p\equiv3\ (mod\ 4) \) si \( p \) nu divide \( 2\Delta_f. \)
Atunci \( |C(\mathbb{Z}_p)|=p+1 \).
In cazul nostru \( b=2,\ p=11\equiv3\ (mod\ 4) \) si nu divide \( 2\Delta_f=-4b^3=-32. \) Rezulta ca \( |C(\mathbb{Z}_{11})|=12 \).
Deci, \( |\mathcal{M}| \) divide atat pe 6, cat si pe 12 rezulta ca \( |\mathcal{M}|\ |\ (6,12)=6 \).
Fie punctul \( P(2,45) \). Calculand \( 5P \) obtinem \( 5P=-P\Rightarrow 6P=\mathcal{O}\Rightarrow ord(P)=6 \).
Deci, \( \mathcal{M}\simeq\mathbb{Z}_6. \)