Weierstrass normal form

Question

How can I show that the Weierstrass normal form $u^3 + v^3 = \alpha$,with $x=12\alpha/(u+v)$ and $y=36\alpha (u-v)/(u+v)$, satisfy $y^2=x^3-432α^2$ ?

Answer 1

Since $u^3 + v^3 = (u+v)(u^2 - uv + v^2)$, we can clear the fractions in the expressions for $x$ and $y$ by multiplying the numerator and denominator by $(u^2 - uv + v^2)$: $$ x = 12 (u^2 - uv + v^2)$$ $$ y= 36(u-v)(u^2 - uv + v^2)$$ Then substitute these expressions for $x, y, \alpha$ in terms of $u$ and $v$ into $y^2 = x^3 - 432\alpha^2$. Simplify, and everything cancels out, as desired.

This proof is not particularly enlightening, but it works.

Answer 2

Oh, this is an ooold question here. Note that the given substitution is birational from the $(u,v)$-world to the $(x,y)$-world, in formulas: $$ \begin{aligned} x &= 12a\cdot\frac 1{u+v}\ , & y &= 36a\cdot \frac{u-v}{u+v}\ ,\\ u &= \frac 1{6x}(36a-y)\ , & v &= \frac 1{6x}(36a+y)\ . \end{aligned} $$ I am writing $a$ instead of $\alpha$. The above substitutions are functions $f:(u,v)\to (x,y)$ given by the first line, and $g:(x,y)\to(u,v)$ given by the second line. It is easy to see that in a generic point $(u,v)$ resp. $(x,y)$ we have $gf(u,v)=(u,v)$, resp. $fg(x,y)=(x,y)$. For instance: $$ \begin{aligned} fg(x,y) &=f\left(\frac 1{6x}(36a-y)\ , \ \frac 1{6x}(36a+y)\right) \\ &=\left(\ 12a\cdot \frac 1{\frac 1{6x}(36a-y)+ \frac 1{6x}(36a+y)} \ ,\ 36a\cdot \frac {\frac 1{6x}(36a-y)- \frac 1{6x}(36a+y)} {\frac 1{6x}(36a-y)+ \frac 1{6x}(36a+y)}\ \right) \\ &=(x,y)\ . \end{aligned} $$

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Let us show that using these substitutions the two equations $$ \begin{aligned} &u^3 + v^3 = a\ ,\\ &y^2 = x^3 - 432a^2\ , \end{aligned} $$ are mapped in each other. Assume $u^3+v^3=a$. Then: $$ \begin{aligned} x^3 - y^2 &= 432a^2\frac{4a}{(u+v)^2} - 432a^2\frac{3(u-v)^2}{(u+v)^2} \\ &= 432a^2\cdot \frac{4a\ - \ (3u^3-3u^2v-3uv^3+3v^3)}{u^3+3u^2v+3uv^3+v^3} \\ &= 432a^2\cdot \frac{4a\ - \ (3a-3u^2v-3uv^3)}{a+3u^2v+3uv^3} \\ &= 432a^2\ . \end{aligned} $$ Conversely, assume $y^2 = x^3 - 432a^2$. Then: $$ \begin{aligned} u^3 + v^2 &= \frac 1{(6x)^3}\Big(\ (36a-y)^3 + (36a+y)^3\ \Big) \\ &= \frac 2{(6x)^3}\Big(\ (36a)^3 + 3\cdot(36a)\;y^2\ \Big) \\ &= \frac {2\cdot3\cdot(36a)}{(6x)^3}\Big(\ 432a^2 + y^2\ \Big) \\ &= \frac {2\cdot3\cdot(36a)}{(6x)^3}\cdot x^3 \\ &=a\ . \end{aligned} $$