Proving the condition for two elliptic curves given in Weierstrass form to be isomorphic

Question

I'm taking a course on elliptic curves and trying to understand the proof of

Proposition 3.2. Let $E$, $E'$ be elliptic curves over $K$ in Weierstrass form: $$E:y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ $$E':y'^2+a_1'x'y'+a_3'y'=x'^3+a_2'x'^2+a'_4x'+a_6'.$$ Then $E\cong E'$ over $K$ iff the Weierstrass equations are related by a substitution of the form $$\begin{array}{ll}x&= u^2x'+r \\ y &=u^3y'+u^2sx'+t \end{array}$$ for some $u,r,s,t \in K$, $u \not=0$.

Now, the proof proceeds by examining the degrees of freedom we had when we found a Weierstrass form in Theorem 3.1:

Theorem 3.1 Every elliptic curve $E$ over $K$ is isomorphic to a curve in Weierstrass form, via an isomorphism taking $\mathcal{O}_E$ to $(0:1:0)$.

(We defined an elliptic curve $E$ over $K$ to be a smooth projective curve of genus 1, defined over $K$, with a specified $K$-rational point $\mathcal{O}_E$.)

The proof goes like this: we pick a basis $1, x$ for $\mathcal{L}(2.\mathcal{O}_E)$ and extend it to a basis $1, x, y$ of $\mathcal{L}(3.\mathcal{O}_E)$. Then we see that the seven elements $1,x,y,x^2,xy,x^3,y^2$ of $\mathcal{L}(6.\mathcal{O}_E)$ must satisfy some dependence relation with the coefficients of $x^3$ and $y^2$ non-zero. Rescaling $x$ and $y$ we get $$E':y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.$$ Then we show that the rational map $$\begin{array}{llll} \phi: & E & \rightarrow &E' \subset \mathbb{P}^2 \\ &P & \mapsto & (x(P):y(P):1)\end{array}$$ is an isomorphism.

I don't see why all Weierstrass equations for $E$ can be found this way, which is what the proof of 3.2 relies on. I think this is because I don't really understand what is going on in the proof of 3.1. It seems like we do a bunch of stuff which will definitely give us the right kind of equation (i.e. a thing in Weierstrass form), but I don't see why this really ought to be isomorphic to our $E$. (Apart from believing the proof, which shows that $[K(E):K(x)]=2$ and $[K(E):K(y)]=3$ so by the Tower Law $[K(E):K(x,y)]=1$.)

Answer 1

I am not sure I understand correctly your question. If $$ y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6 $$ is a Weierstrass equation of $E$, define $O_E$ be the point with homogeneous coordinates $x=0, y=1, z=0$. Consider $x,y $ as rational functions on $E$. Then the computation of the divisors of rationals functions shows that $\{1, x\}$ is a basis of $L(2O_E)$ and $\{1, x, y\}$ is a basis of $L(3O_E)$. For instance the pole divisor of $x$ is $-2O_E$ and that of $y$ is $-3O_E$.