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It seems that in whatever proof of the theorem that an elliptic curve can be put in Weierstrass form that you look at, the next step after getting an equation:

$$\alpha Y^2Z + a_1XY Z + a_3Y Z^2= \beta X^3+ a_2X^2Z + a_4XZ^2+ a_6Z^3$$

is to multiply through by $\beta^2/\alpha^3$ in order to make the linear change of variables $$\left(X\mapsto \frac\alpha\beta X, Y\mapsto \frac\alpha\beta Y\right)$$ to get rid of the leading coefficients. My question is: why are we restricted to linear change of variables that rescale $X$ and $Y$ in the same way (but still can send $Y\mapsto Y + mZ+nX$).

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  • $\begingroup$ Your question is not very clear. Are you asking if changes of variables of the form $X'=aX$, $Y'=bY$ are allowed? What would that accomplish? $\endgroup$ Feb 19, 2014 at 2:08
  • $\begingroup$ @ÁlvaroLozano-Robledo yeah, I certainly expected that the change of variables $X'=\beta^{1/3}X, Y'=\sqrt{\alpha}Y$ would be allowed. I'd use that to get rid of the leading coefficients without touching on the $a_6$. But instead my reference insists that first one should multiply by $\beta^2/\alpha^3$ and then make the proportional change of variables. $\endgroup$
    – Rodrigo
    Feb 19, 2014 at 17:21

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It is not that we can't rescale $X,Y$ independently, but rather that we don't want to rescale them by a square, or cubic root (as I suggested in my comment above $X'=\beta^{1/3}, Y'=\alpha^{1/2}Y$), which might not exist in the field $K$ over which the curve is defined.

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