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It is true that some quartics can be changed to Weierstrass form birationally. but is it possible to do the inverse always? is it possible to transform a weierstrass form of genus 1 curve to quartic? i mean transforming this equation: $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ to :$$y^2=ax^4+bx^3+cx^2+dx+e$$ if the answer would be positive, how? and if negative, why? it is very important for me to write the curve in the form that decrease the coefficients.(one way is to write it in short weierstrass form , then put $1/x$ instead of $x$ and then multiply it by x^4, then you have a quartic, but this is useless cause doesn't decrease the coefficients.)

thanks

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    $\begingroup$ Over which field are you working? $\endgroup$ – abx Apr 5 '17 at 13:49
  • $\begingroup$ rational field , in fact $\endgroup$ – user371596 Apr 5 '17 at 13:50
  • $\begingroup$ Over a not-algebraically-closed field, if none of the four zeros of the quartic is rational, it's hard to see how you'd accomplish this. At least the easy argument over an algebraically closed field does not succeed... $\endgroup$ – paul garrett Apr 5 '17 at 13:53
  • $\begingroup$ like exactly the inverse operations of what is in I.connell's "elliptic curve handbook" $\endgroup$ – user371596 Apr 5 '17 at 14:31
  • $\begingroup$ even it is okay if you work with extension field and get a rational quartic for this elliptic curve, even there is no problem that the quartic have higher genus , it's ok $\endgroup$ – user371596 Apr 5 '17 at 17:42
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Why do you want to decrease the coefficients? If it is to search for points, then possibly what you're looking for is the theory of homogeneous spaces. Thus an equation of the form $C:y^2=ax^4+\cdots+e$, even if it does not have a rational point, represents an element of order $2$ in the Weil-Chatelet group of its Jacobean, which is an elliptic curve $E$, and there is a map $C\to E$ defined over your base field. Conversely, starting with an elliptic curve $E$, there exist homogeneous spaces of various orders $m$, and, to some extent, the coefficients of these homogeneous spaces will be smaller than the coefficients of the original curve. If $m$ is not too large, say $2\le m\le 6$, it's probably feasible to write down (some of) the associated homogeneous spaces. Note that rational points on the homogeneous spaces will then map to rational points on $E$.

This is closely related to performing an $m$-descent on $E$. The first place I'd suggest looking is Cremona's book [1], which in addition to a lot of tables (since expanded and transferred online) includes a very nice description of practically constructing homogeneous spaces and indicates why their coefficients are often smaller than those on the original curve. See in particular Chapter 3.

[1] MR1628193 Cremona, J. E. Algorithms for modular elliptic curves. Second edition. Cambridge University Press, Cambridge, 1997. Freely available online at http://homepages.warwick.ac.uk/~masgaj/book/fulltext/index.html

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  • $\begingroup$ extraordinary as always -:). thanks. yes i'd like to search for points $\endgroup$ – user371596 Apr 5 '17 at 16:22

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