Elliptische Kurve/Differentialform/Rückzug auf Produkt/Fakt/Beweis

Beweis

Die Addition ${}+\colon E\times E\rightarrow E$ wird gemäß Fakt durch

$(x_{1},y_{1};x_{2},y_{2})\mapsto \left(\alpha ^{2}-x_{1}-x_{2},\,-\alpha ^{3}+\alpha (2x_{1}+x_{2})-y_{1}\right)=(x,y)$

mit ${}\alpha ={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$ beschrieben. Diese Terme kann man als funktionale Ausdrücke auf der affinen offenen Menge ${}D_{+}(z)$ bzw. als Ringelemente in

${}R=K[x,y]/(y^{2}-x^{3}-ax-b)\,$

bzw. im Tensorprodukt

${}R\otimes _{K}R=K[x_{1},y_{1},x_{2},y_{2}]/(y_{1}^{2}-x_{1}^{3}-ax_{1}-b,y_{2}^{2}-x_{2}^{3}-ax_{2}-b)\,$

auffassen, bei der letzten Interpretation geht es um den Ringhomomorphismus

$R\longrightarrow R\otimes _{K}R,$

der durch die Einsetzungen ${}x\mapsto \alpha ^{2}-x_{1}-x_{2}$ und ${}y\mapsto -\alpha ^{3}+\alpha (2x_{1}+x_{2})-y_{1}$ festgelegt ist. Unter diesem Ringhomomorphismus wird ${}dx$ nach Fakt  (5) auf

${}{\begin{aligned}&\,\,\,\,\,\,\,d{\left(\alpha ^{2}-x_{1}-x_{2}\right)}\\&=2\alpha d\alpha -dx_{1}-dx_{2}\\&=2{\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}d\left({\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\right)-dx_{1}-dx_{2}\\&=2{\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\cdot {\frac {(x_{2}-x_{1})d(y_{2}-y_{1})-(y_{2}-y_{1})d(x_{2}-x_{1})}{(x_{2}-x_{1})^{2}}}-dx_{1}-dx_{2}\\&=2{\frac {y_{2}-y_{1}}{(x_{2}-x_{1})^{2}}}d(y_{2}-y_{1})-2{\frac {(y_{2}-y_{1})^{2}}{(x_{2}-x_{1})^{3}}}d(x_{2}-x_{1})-dx_{1}-dx_{2}\\&={\frac {y_{2}-y_{1}}{(x_{2}-x_{1})^{2}}}{\left({\frac {3x_{2}^{2}+a}{y_{2}}}dx_{2}-{\frac {3x_{1}^{2}+a}{y_{1}}}dx_{1}\right)}-2{\frac {y_{2}^{2}-2y_{1}y_{2}+y_{1}^{2}}{(x_{2}-x_{1})^{3}}}d(x_{2}-x_{1})-dx_{1}-dx_{2}\\&={\left(-{\frac {y_{2}-y_{1}}{(x_{2}-x_{1})^{2}}}\cdot {\frac {3x_{1}^{2}+a}{y_{1}}}+2{\frac {y_{2}^{2}-2y_{1}y_{2}+y_{1}^{2}}{(x_{2}-x_{1})^{3}}}-1\right)}dx_{1}+*dx_{2}\,\end{aligned}}$

abgebildet, wobei der nicht angeführte Term vor ${}dx_{2}$ symmetrisch gebildet ist.

Der Term vor ${}dx_{1}$ ist

${}{\begin{aligned}&\,\,\,\,\,\,\,-{\frac {y_{2}-y_{1}}{(x_{2}-x_{1})^{2}}}\cdot {\frac {3x_{1}^{2}+a}{y_{1}}}+2{\frac {y_{2}^{2}-2y_{1}y_{2}+y_{1}^{2}}{(x_{2}-x_{1})^{3}}}-1\\&=-{\frac {y_{1}y_{2}-y_{1}^{2}}{(x_{2}-x_{1})^{2}}}\cdot {\frac {3x_{1}^{2}+a}{x_{1}^{3}+ax_{1}+b}}+2{\frac {{\left(x_{2}^{3}+ax_{2}+b\right)}-2y_{1}y_{2}+{\left(x_{1}^{3}+ax_{1}+b\right)}}{(x_{2}-x_{1})^{3}}}-1\\&={\frac {-y_{1}y_{2}+{\left(x_{1}^{3}+ax_{1}+b\right)}}{(x_{2}-x_{1})^{2}}}\cdot {\frac {3x_{1}^{2}+a}{x_{1}^{3}+ax_{1}+b}}+2{\frac {x_{1}^{3}+x_{2}^{3}+a{\left(x_{1}+x_{2}\right)}+2b-2y_{1}y_{2}}{(x_{2}-x_{1})^{3}}}-1\\&={\frac {x_{1}^{3}+ax_{1}+b}{(x_{2}-x_{1})^{2}}}\cdot {\frac {3x_{1}^{2}+a}{x_{1}^{3}+ax_{1}+b}}+2{\frac {x_{1}^{3}+x_{2}^{3}+a{\left(x_{1}+x_{2}\right)}+2b}{(x_{2}-x_{1})^{3}}}-1+y_{1}y_{2}{\left(-{\frac {3x_{1}^{2}+a}{{\left(x_{1}^{3}+ax_{1}+b\right)}{\left(x_{2}-x_{1}\right)}^{2}}}-{\frac {4}{{\left(x_{2}-x_{1}\right)}^{3}}}\right)}\\&={\frac {\left(3x_{1}^{2}+a\right)}{(x_{2}-x_{1})^{2}}}+2{\frac {x_{1}^{3}+x_{2}^{3}+a(x_{1}+x_{2})+2b}{(x_{2}-x_{1})^{3}}}-1+y_{1}y_{2}{\left(-{\frac {3x_{1}^{2}+a}{{\left(x_{1}^{3}+ax_{1}+b\right)}{\left(x_{2}-x_{1}\right)}^{2}}}-{\frac {4}{{\left(x_{2}-x_{1}\right)}^{3}}}\right)}\\&={\frac {{\left(3x_{1}^{2}+a\right)}{\left(x_{2}-x_{1}\right)}+2{\left(x_{1}^{3}+x_{2}^{3}+a(x_{1}+x_{2})+2b\right)}-(x_{2}-x_{1})^{3}}{(x_{2}-x_{1})^{3}}}+y_{1}y_{2}{\frac {-{\left(3x_{1}^{2}+a\right)}{\left(x_{2}-x_{1}\right)}-4x_{1}^{3}-4ax_{1}-4b}{{\left(x_{1}^{3}+ax_{1}+b\right)}{\left(x_{2}-x_{1}\right)}^{3}}}\,\end{aligned}}$

Der Zähler links ist

${}{\begin{aligned}r&={\left(3x_{1}^{2}+a\right)}{\left(x_{2}-x_{1}\right)}+2{\left(x_{1}^{3}+x_{2}^{3}+a{\left(x_{1}+x_{2}\right)}+2b\right)}-(x_{2}-x_{1})^{3}\\&=-3x_{1}^{3}+3x_{1}^{2}x_{2}+a{\left(x_{2}-x_{1}\right)}+2x_{1}^{3}+2x_{2}^{3}+2a{\left(x_{1}+x_{2}\right)}+4b-x_{2}^{3}+3x_{1}x_{2}^{2}-3x_{1}^{2}x_{2}+x_{1}^{3}\\&=x_{2}^{3}+3x_{1}x_{2}^{2}+ax_{1}+3ax_{2}+4b.\end{aligned}}$

Der Zähler rechts ist

${}{\begin{aligned}s&=-{\left(3x_{1}^{2}+a\right)}{\left(x_{2}-x_{1}\right)}-4x_{1}^{3}-4ax_{1}-4b\\&=3x_{1}^{3}-3x_{1}^{2}x_{2}-a{\left(x_{2}-x_{1}\right)}-4x_{1}^{3}-4ax_{1}-4b\\&=-x_{1}^{3}-3x_{1}^{2}x_{2}-a{\left(x_{2}+3x_{1}\right)}-4b.\end{aligned}}$

Wir haben also eine Beschreibung der zurückgezogenen Differentialform ${}+^{*}dx$ der Gestalt

${\left({\frac {r}{{\left(x_{2}-x_{1}\right)}^{3}}}+{\frac {s}{{\left(x_{2}-x_{1}\right)}^{3}{\left(x_{1}^{3}+ax_{1}+b\right)}}}y_{1}y_{2}\right)}dx_{1}+*dx_{2}$

mit explizit bestimmten Polynomen ${}r$ und ${}s$ in ${}x_{1},x_{2}$. Für die Differentialform ${}{\frac {dx}{y}}$ gilt

${}+^{*}{\frac {dx}{y}}={\frac {+^{*}dx}{+^{*}y}}\,,$

wobei ${}+^{*}y$ einfach den Rückzug der Funktion ${}y$ bezeichnet, also einfach ${}-\alpha ^{3}+\alpha (2x_{1}+x_{2})-y_{1}$. In Aufgabe haben wir dies berechnet, es ist

${}-\alpha ^{3}+\alpha {\left(2x_{1}+x_{2}\right)}-y_{1}={\frac {fy_{1}+gy_{2}}{{\left(x_{2}-x_{1}\right)}^{3}}}\,$

mit

${}f=x_{2}^{3}+3x_{1}x_{2}^{2}+a{\left(x_{1}+3x_{2}\right)}+4b\,$

und

${}g=-x_{1}^{3}-3x_{1}^{2}x_{2}-a{\left(3x_{1}+x_{2}\right)}-4b\,.$

Die behauptete Gleichheit

${}+^{*}{\frac {dx}{y}}={\frac {{\left({\frac {r}{{\left(x_{2}-x_{1}\right)}^{3}}}+{\frac {s}{{\left(x_{2}-x_{1}\right)}^{3}{\left(x_{1}^{3}+ax_{1}+b\right)}}}y_{1}y_{2}\right)}dx_{1}+*dx_{2}}{\frac {fy_{1}+gy_{2}}{{\left(x_{2}-x_{1}\right)}^{3}}}}={\frac {dx_{1}}{y_{1}}}+{\frac {dx_{2}}{y_{2}}}\,$

ergibt sich somit (für den ${}dx_{1}$-Anteil) aus

${}r{\left(x_{1}^{3}+ax_{1}+b\right)}y_{1}+sy_{1}^{2}y_{2}={\left(fy_{1}+gy_{2}\right)}{\left(x_{1}^{3}+ax_{1}+b\right)}\,.$

Dies folgt nun direkt aus ${}r=f$ und ${}s=g$.

Zur bewiesenen Aussage