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Suppose I have the following Weierstrass equation:

$y^2-y=x^3-x$. I want to make a change of variable to get the form $y^2=x^3+ax+b$. In Milne's book, p.50, we have the following statement:

"Let $E$ be an elliptic curve over $k$. Any equation of the form

$Y^2 Z+a_1 XYZ+ a_3Y Z^2 +a_3YZ^2 =X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3$

is called a Weierstrass equation for the elliptic curve. When k has characteristic $\neq 2,3$ a change of variables $X' = X +a_2/3; Y'= Y + a_3/2 ; Z'=Z$

will then eliminate the $X^2$ and $Y^2$ terms. Thus we get the form $y^2=x^3+ax+b$"

I applied directly this technique. I wrote $x'=x-1/3, y'=y-1/2$.

i.e. $x=x'+1/3$ and $y=y'+1/2$. Replacing this in $y^2-y=x^3+x$, I get $y^2=x^3-x/3+19/108$

However, I thought the coefficients had to be in $\mathbb{Z}$. I need them in $\mathbb{Z}$ because I want to use Lutz Nagell to find the torsion points.

3 Answers 3

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If an elliptic curve is defined over $\Bbb{Z}$ so is its minimal Weierstrass model. However, that minimal equation need not be in the standard simple form.

If you want an equation of the form $y^2=x^3+ax+b$ with $a,b\in\Bbb{Z}$ you need to accept a non-minimal version. So multiply your equation by $6^6$ and rewrite it to read $$ (216y)^2=(36x)^3-\frac{6^4}3(36x)+19\frac{6^6}{108}. $$ In terms of new variables $X=36x$ and $Y=216y$ this reads $$ Y^2=X^3-2^4\cdot3^3X+2^4\cdot3^3\cdot19. $$ Hope you can work with that instead.

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    See [this MO question](http://mathoverflow.net/q/10607/15503) or some lecture notes on minimal Weierstrass models for more. – 2017-02-26
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For a general field $k$, it makes no sense to ask that the coefficients be in $\mathbb{Z}$. With your equation, a very simple change of variables will get rid of the denominators.

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    the problem is that if I multiply everything by 108 the coefficient of $y^2$ will not be 1.. – 2017-02-25
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Must you use Lutz-Nagell?
The curve as given has good supersingular reduction at both $2$ and $3$, and there must be a theorem in Silverman saying that if a curve over $\Bbb Z_p$ has good supersingular reduction, the torsion injects into the group of $\Bbb F_p$-points. Since there are five points over $\Bbb F_2$ and seven over $\Bbb F_3$, the curve is supersingular at both, and this “theorem” applies to show that the $\Bbb Q$-torsion is trivial.