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A binary quartic form

$aX^4+bX^3Y+cX^2Y^2+dXY^3+Y^4$

decomposes as a product of linear factors $Y-t_jX$, $j=1,...,4$. I would like to have an explicit formula for symmetrization of the crossratio of $t_j$.

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Do you mean the j-invariant of the elliptic curve $y^2=ax^4+bx^3+cx^2+dx+1$? –  Robin Chapman Jul 7 '10 at 13:54
    
Yes, that is exactly what I am looking for. –  David Marín Jul 7 '10 at 14:00
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If you have access to a computer algebra system, you can do the following. Let $\xi$ denote a solution to $f(1,\xi)=0$ where $f$ is your quartic. Then $f(X,Y+\xi X)=b'X^3Y+\cdots+Y^4$. The elliptic curve is now isomorphic to $y^2=b'x^3+c'x^2+d'x+1$. Transform it to the usual Weierstrass form and take the $j$-invariant. Note that $b'$ etc. will have $\xi$s in them, but they should all cancel out via the equation $f(1,\xi)=0$ in the final result. –  Robin Chapman Jul 7 '10 at 14:23
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1 Answer 1

up vote 7 down vote accepted

The $j$ invariant is

$j=\frac{S^3}{S^3-27T^2}$

where

$S=a-\frac{bd}{4}+\frac{c^2}{12}$

and

$T=\frac{ac}{6}+\frac{bcd}{48}-\frac{c^3}{216}-\frac{ad^2}{16}-\frac{b^2}{16}$

for more details see my article J. Algebra 303 (2006) 771-788.

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