the $j$-invariant classifies elliptic curves up to isomorphismJInvariantClassifiesEllipticCurvesUpToIsomorphism2005-03-01 16:00:112005-03-01 16:01:35Theoremj-invariant% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Rats}{\mathbb{Q}}In this entry, an isomorphism over $K$ should be understood in the sense of the entry isomorphism of varieties.
\begin{thm}
Let $K$ be a field, and let $\overline{K}$ be a fixed algebraic closure of $K$.
\begin{enumerate}
\item Two elliptic curves $E_1$ and $E_2$ are \PMlinkname{isomorphic}{IsomorphismOfVarieties} (over $\overline{K}$) if and only if they have the same $j$-invariant, i.e. $j(E_1)=j(E_2)$.\\
\item Let $j_0\in \overline{K}$ be fixed. There exists an elliptic curve $E$ defined over the field $K(j_0)$ such that $j(E)=j_0$.
\end{enumerate}
\end{thm}
\begin{proof}
For part $2$:
\begin{itemize}
\item For $j_0=0$, the curve $E_0\colon y^2+y=x^3$ satisfies $j(E)=0$;
\item For $j_0=1728$, the curve $E_{1728} \colon y^2=x^3+x$ satisfies $j(E_{1728})=1728$;
\item If $j_0\neq 0, 1728$ consider the elliptic curve:
$$E=E_{j_0} \colon y^2+xy=x^3-\frac{36}{j_0-1728}x-\frac{1}{j_0-1728}.$$
It satisfies $j(E)=j_0$ and it is defined over $K(j_0)$.
\end{itemize}
\end{proof}