One can put every cubic curve into Weierstrass form, how unique is this form?

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    $\begingroup$ The substitution $y\rightarrow u^3y$, $x\rightarrow u^2x$ is there, when $p>3$. This will turn a curve in the Weierstrass form $y^2=x^3+Ax+B$ to another $y^2=x^3+Au^{-4}x+Bu^{-6}$. Probably something similar is there in the cases $p=$ and $p=3$, but I don't remember the details. $\endgroup$ Commented Jun 10, 2011 at 15:27

1 Answer 1


As Jyrki points out in the comments to your question, there is not a unique Weierstrass form of an elliptic curve, but a bit more can be said. Given an elliptic curve E/K with Weierstrass form

$y^2 +a_1 xy +a_3 y = x^3 + a_2 x^2 + a_6 $ , $a_i \in$ K

we can make the change of variables $(x,y) \to (x',y')$ where $x=u^2 x' + r$ and $y=u^3 y' + su^2 x' + t$ where $r,s,t \in K$ and $u \in K^*$. Any such change of variables will yield another Weierstrass equation, and it will also induce a change in the discrimant $\Delta = u^{12} \Delta '$ .

When K=$\mathbf{Q}$, there exists a global minimal Weierstrass model for $E/\mathbf{Q}$, that is, a Weierstrass equation for $E$ such that its discriminant $\Delta$ is minimal with respect to all primes $p \mid \Delta$. (I believe I recall reading somewhere that the minimal Weierstrass equation with integer coefficients and such that $a_1, a_2=$0 or 1 and $a_3=$ -1,0, or 1 is unique, but I don't remember the source. Sorry! Maybe someone else can chime in?) This theorem is not true over general number fields.

Over any number field K, however, and for a fixed valuation $\nu$, you can find a Weierstrass equation for $E/K$ which is minimal with respect to $\nu$, and which is unique up to isomorphism. See for example Joseph Silverman, Arithmetic of Elliptic Curves, VII.1.


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