the $j$-invariant classifies elliptic curves up to isomorphism

In this entry, an isomorphism over $K$ should be understood in the sense of the entry isomorphism of varieties.

Theorem 1.

Let $K$ be a field, and let $\overline{K}$ be a fixed algebraic closure of $K$.

1. Two elliptic curves $E_1$ and $E_2$ are isomorphic (http://planetmath.org/IsomorphismOfVarieties) (over $\overline{K}$) if and only if they have the same $j$-invariant, i.e. $j(E_1)=j(E_2)$.
   

2. Let $j_0\in \overline{K}$ be fixed. There exists an elliptic curve $E$ defined over the field $K(j_0)$ such that $j(E)=j_0$.

Proof.

For part $2$:

• For $j_0=0$, the curve $E_0:y^2+y=x^3$ satisfies $j(E)=0$;

• For $j_0=1728$, the curve $E_{1728}:y^2=x^3+x$ satisfies $j(E_{1728})=1728$;

• If $j_0\ne 0,1728$ consider the elliptic curve:

  $$E=E_{j_0}:y^2+xy=x^3-\frac{36}{j_0-1728}x-\frac{1}{j_0-1728}.$$

  It satisfies $j(E)=j_0$ and it is defined over $K(j_0)$.

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