Proof of Theorem 3.1. $L(2.0_E)\subset L(3.0_E)$. Pick basis $1,x$ for $L(2.0_E)$ and $1,x,y$ for $L(3.0_E)$.

Note: ${\mathrm{ord}}_{0_E}(x)=-2$, ${\mathrm{ord}}_{0_E}(y)=-3$.

The 7 elements $1,x,y,x^2,xy,x^3,y^2$ in the 6-dimensional space $L(6.0_E)$ must satisfy a dependence relation.

Leaving out $x^3$ or $y^2$ gives a basis for $L(6.0_E)$ since each term has a different order pole at $0_E$.

Therefore the coefficients of $x^3$ and $y^2$ are non-zero. Rescaling $x$ and $y$ (if necessary) we get

$$y^2+a_{1}xy+a_{3}y=x^3+a_2{x}^2+a_{4}x+a_6$$

for some $a_i\in K$.

Let $E^{\prime }$ be the curve defined by this equation (or rather its projective closure). There is a morphism

$$\begin{array}{llll}\hfill \varphi :E& \to E^{\prime }\subset {\mathbb{P}}^2& \hfill & \\ \hfill P& \mapsto (x(P):y(P):1)=\left(\frac{x}{y}(P):1:\frac{1}{y}(P)\right)& \hfill & \\ \hfill 0_E& \mapsto (0:1:0)& \hfill & \end{array}$$

Then

$$\begin{array}{llllll}\hfill [K(E):K(x)]& =\mathrm{deg}(E\stackrel{x}{\to }{\mathbb{P}}^1)& \hfill & & \hfill & \\ \hfill & ={\mathrm{ord}}_{0_E}\left(\frac{1}{x}\right)& \hfill & & \hfill & \\ \hfill & =2[K(E):K(y)]& \hfill =\mathrm{deg}(E\stackrel{y}{\to }{\mathbb{P}}^1)& & \hfill & & \hfill \\ \hfill & ={\mathrm{ord}}_{0_E}\left(\frac{1}{y}\right)& \hfill & & \hfill & \\ \hfill & =3& \hfill & & \hfill & \end{array}$$

This gives us a diagram of field extensions:

By the Tower Law (since $2,3$ are coprime), we get that $[K(E):K(x,y)]=1$. Hence $K(E)=K(x,y)={\varphi }^{\ast }K(E^{\prime })$, so $\mathrm{deg}\varphi =1$, so $\varphi$ is birational.

If $E^{\prime }$ is singular then $E$ and $E^{\prime }$ are rational, contradiction.

So $E^{\prime }$ is smooth. Then Remark 2.9 implies that ${\varphi }^{-1}$ is a morphism. So $\varphi$ is an isomorphism. □

Proof.

• $\Leftarrow$ Obvious.
• $\Rightarrow$ $⟨1,x⟩=L(2.0_E)=⟨1,x^{\prime }⟩$ hence $x=\lambda x^{\prime }+r$ for some $\lambda ,r\in K$ with $\lambda \ne 0$.

  $⟨1,x,y⟩=L(3.0_E)=⟨1,x^{\prime },y^{\prime }⟩$ implies $y=\mu y^{\prime }+\sigma x^{\prime }+t$ for some $\mu ,\sigma ,t\in K$ with $\mu \ne 0$.

  Looking at coefficients of $x^3$ and $y^2$, we get ${\lambda }^3={\mu }^2$. So $\lambda =u^2$, $\mu =u^3$ for some $0\ne u\in K$.

  Put $s=\frac{\sigma }{u^2}$. □

Proof. $E$ and $E^{\prime }$ are related by a substitution as in Proposition 3.2 with $r=s=t=0$. □

Proof.

$$\begin{array}{llll}\hfill E\cong E^{\prime }& ⟺\{\begin{array}{c}{a}^{\prime }=u^{4}a\hfill \\ b^{\prime }=u^{6}b\hfill \end{array}\text{ for some }u\in K^{\ast }& \hfill & \\ \hfill & ⟹(a^3:b^2)=({(a^{\prime })}^3:{(b^{\prime })}^2)& \hfill & \\ \hfill & ⟺j(E)=j(E^{\prime })& \hfill & \end{array}$$

and the converse holds if $K=\overline{K}$ (to go backwards on the $⟹$ step, we only need to take some kind of $n$-th root). □