Elliptische Kurven/Morphismus/Differentialform/Rechnung/Fakt/Beweis

Beweis

Es wird ${}{\frac {du}{v}}$ auf ${}{\frac {d\varphi }{\psi }}$ . Wir wissen

{}{\frac {d\varphi }{\psi }}=\gamma {\frac {dx}{y}}\,

und bestimmen ${}\gamma$ . Es ist ( ${}f'$ bedeutet die formale Ableitung nach ${}x$ )

{}{\begin{aligned}2y{\frac {d\varphi }{\psi }}&=2y{\frac {d(f+gy)}{p+qy}}\\&={\frac {2y}{p+qy}}{\left((f'+g'y)dx+gdy\right)}\\&={\frac {2y(f'+g'y)}{p+qy}}dx+{\frac {g}{p+qy}}2ydy\\&={\frac {2y(f'+g'y)}{p+qy}}dx+{\frac {g}{p+qy}}{\left(3x^{2}+a\right)}dx\\&={\frac {2yf'+2g'{\left(x^{3}+ax+b\right)}+g{\left(3x^{2}+a\right)}}{p+qy}}dx\\&={\frac {{\left(2yf'+2g'{\left(x^{3}+ax+b\right)}+g{\left(3x^{2}+a\right)}\right)}{\left(p-qy\right)}}{(p+qy)(p-qy)}}dx\\&={\frac {{\left(2g'{\left(x^{3}+ax+b\right)}+g{\left(3x^{2}+a\right)}\right)}p-2f'q{\left(x^{3}+ax+b\right)}+{\left(2f'p-q{\left(2g'{\left(x^{3}+ax+b\right)}+g{\left(3x^{2}+a\right)}\right)}\right)}y}{p^{2}-q^{2}{\left(x^{3}+ax+b\right)}}}dx.\end{aligned}}