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Let

$$E:y^2 = 4x^3-g_2 x - g_3$$

be an elliptic curve and

$$j=\frac{g_2^3}{g_2^3-27 g_3^2}$$

denote to its $j$-invariant. I want to transform $E$ to find $f$ and $g$ s.t.

$$E:y^2=4x^3-f(j)x-g(j).$$

I have no clue what ansatz I should take for $f$ and $g$, nor how to transform $E$. Who can help me?

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1 Answer 1

up vote 4 down vote accepted

The formula is well known to the experts, which number I do not count myself among. Since none of them has answered, let me lay it out here: $$ y^2= 4x^3 + \frac{3j}{1-j}x +\frac j{1-j}\,. $$ And here’s how you derive it: from $$ j = \frac{g_2^3}{g_2^3-27g_3^2} = \frac1{1-\frac{27g_3^2}{g_2^3}} $$ you conclude $$ \frac{27g_3^2}{g_2^3} = \frac{j-1}j\,. $$ So you just adjust $g_2$ and $g_3$ so that they satisfy this requirement. In particular, $g_2=3j/(j-1)$ and $g_3=j/(j-1)$ work.

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Similarly, I considered $E:y^2=x^3-cx-c$ and $j=\frac{c}{c-27}$. This leads to $c=\frac{27j}{j-1}$. –  robertmiles Jun 1 '13 at 22:07
    
And probably with less work than I expended trying to rediscover the formula. –  Lubin Jun 1 '13 at 23:52
    
Sorry, I just found this way before checking back to the thread. I will therefore mark your answer as the solution. Thanks! ;) –  robertmiles Jun 2 '13 at 4:19

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