Elliptische Kurve/Differentialform/Translation/Fakt/Beweis2

Wir betrachten die affine Version, also das Differential ${\displaystyle {}{\frac {dx}{y}}}$ auf ${\displaystyle {}R=K[x,y]/{\left(y^{2}-x^{3}-ax-b\right)}}$. Es sei ${\displaystyle {}Q=(u,v)}$. Die Translation mit ${\displaystyle {}Q}$ ist explizit durch

${\displaystyle (x,y)\longmapsto \left(\alpha ^{2}-u-x,\,-\alpha ^{3}+\alpha (u+x)-\beta \right)}$

mit

${\displaystyle {}\alpha =\alpha (x,y)={\frac {y-v}{x-u}}\,}$

und

${\displaystyle {}\beta =\beta (x,y)={\frac {xv-uy}{x-u}}\,}$

gegeben. Diese Formeln definieren eine Abbildung

${\displaystyle \theta \colon R\longrightarrow R_{g}}$

(${\displaystyle {}g}$ ist ein Nenner, der die auftretenden Brüche berücksichtigt) und wir müssen das Bild von ${\displaystyle {}{\frac {dx}{y}}}$ gemäß Fakt  (5) berechnen. Dabei ist

${\displaystyle {}{\begin{aligned}\theta (dx)&=d\theta (x)\\&=d{\left(\alpha ^{2}-u-x\right)}\\&=2\alpha d\alpha -dx\\&=2{\frac {y-v}{x-u}}\cdot {\frac {(x-u)dy-(y-v)dx}{(x-u)^{2}}}-dx\\&=2{\frac {y-v}{(x-u)^{2}}}dy+2{\frac {-(y-v)^{2}}{(x-u)^{3}}}dx-dx.\end{aligned}}}$

Es ist

${\displaystyle {}{\begin{aligned}yd\varphi (x)&=2y{\frac {y-v}{(x-u)^{2}}}dy-2y{\frac {(y-v)^{2}}{(x-u)^{3}}}dx-ydx\\&={\frac {(y-v)(3x^{2}+a)}{(x-u)^{2}}}dx-2y{\frac {(y-v)^{2}}{(x-u)^{3}}}dx-ydx.\end{aligned}}}$

Der gemeinsame Zähler davon ohne ${\displaystyle {}dx}$ zum Nenner ${\displaystyle {}(x-u)^{3}}$ ist

${\displaystyle {}{\begin{aligned}&\,\,\,\,\,\,\,(x-u)(y-v)(3x^{2}+a)-2y(y-v)^{2}-y(x-u)^{3}\\&=-v(x-u)(3x^{2}+a)+y(x-u)(3x^{2}+a)-2y^{3}+4y^{2}v-2yv^{2}-y(x-u)^{3}\\&=v{\left(-(x-u)(3x^{2}+a)+4(x^{3}+ax+b)\right)}+y{\left(-2v^{2}+(x-u)(3x^{2}+a)-(x-u)^{3}-2(x^{3}+ax+b)\right)}\\&=v{\left(-(x-u)(3x^{2}+a)+4(x^{3}+ax+b)\right)}+y{\left(-2(u^{3}+au+b)+(x-u)(3x^{2}+a)-(x-u)^{3}-2(x^{3}+ax+b)\right)}\\&=v{\left(x^{3}+3ux^{2}+3ax+au+4b\right)}+y{\left(-2u^{3}-2au-2b+3x^{3}-3ux^{2}+ax-au-x^{3}+3ux^{2}-3u^{2}x+u^{3}-2x^{3}-2ax-2b\right)}\\&=v{\left(x^{3}+3ux^{2}+3ax+au+4b\right)}+y{\left((-3u^{2}-a)x-u^{3}-3au-4b\right)}\,\end{aligned}}}$

Andererseits ist der Zähler von ${\displaystyle {}-\alpha ^{3}+\alpha (u+x)-\beta }$ zum Nenner ${\displaystyle {}(x-u)^{3}}$ gleich

${\displaystyle {}{\begin{aligned}&\,\,\,\,\,\,\,-(y-v)^{3}+(y-v)(x-u)^{2}(x+u)-(x-u)^{2}(xv-uy)\\&=-y^{3}+3vy^{2}-3v^{2}y+v^{3}-v(x-u)^{2}(x+u)+y(x-u)^{2}(x+u)-(x-u)^{2}xv+(x-u)^{2}uy\\&=3v(x^{3}+ax+b)+v(u^{3}+au+b)-v(x-u)^{2}(x+u)-(x-u)^{2}xv+y{\left(-x^{3}-ax-b-3v^{2}+(x-u)^{2}(x+u)+(x-u)^{2}u\right)}\\&=v{\left(3(x^{3}+ax+b)+u^{3}+au+b-(x-u)^{2}(x+u)-(x-u)^{2}x\right)}+y{\left(-x^{3}-ax-b-3(u^{3}+au+b)+(x-u)^{2}(x+u)+(x-u)^{2}u\right)}\\&=v{\left(3x^{3}+3ax+3b+u^{3}+au+b-(x^{2}-2ux+u^{2})(2x+u)\right)}+y{\left(-x^{3}-ax-b-3u^{3}-3au-3b+(x^{2}-2ux+u^{2})(x+2u)\right)}\\&=v{\left(x^{3}+3ux^{2}+3ax+au+4b\right)}+y{\left(-x^{3}-ax-3u^{3}-3au-4b+x^{3}-3u^{2}x+2u^{3}\right)}\\&=v{\left(x^{3}+3ux^{2}+3ax+au+4b\right)}+y{\left((-3u^{2}-a)x-u^{3}-3au-4b\right)},\,\end{aligned}}}$

die Ergebnisse stimmen also überein. Es ist also

${\displaystyle {}{\frac {yd\theta (x)}{(x-u)^{3}}}={\frac {\theta (y)dx}{(x-u)^{3}}}\,}$

und damit

${\displaystyle {}{\frac {d\theta (x)}{\theta (y)}}={\frac {dx}{y}}\,.}$