Field extension of isomorphism of elliptic curves "If $A,B$ are elliptic curves over a field $k$ of characteristic not 2 or 3 and if $A,$ isomorphic over an extension of $k$, then they become isomorphic over an extension of $k$ of degree $\leq 6$. "
$\textbf{Q:}$ Since $A,B$ are exactly described by $g_2$ and $g_3$ or equivalently their $j$ invariants, suppose $g_2^A=c^4g_2^B$ and $g_3^A=c^6g_3^B$ with $c^4,c^6\in k$. The statement is saying there is an isomorphism over an extension of $k$. The only possibility is degree 2 extension or degree 6 extension. What does the book mean isomorphic over an extension of $k$? I can only see application of such conclusion for degree 2 or degree 6 extension and application of such conclusion for extension beyond 6 is redundant. Should not $j$ invariant automatically conclude isomorphism over arbitrary extension?
Ref. Lang, Elliptic Functions, Chpt 1, Sec 4. pg 18


Should not $j$-invariant automatically conclude isomorphism over arbitrary extension?

No, the $j$-invariant only classifies elliptic curves over an algebraically closed field. Curves with the same $j$-invariant may not be isomorphic over $k$ as they could be twists. That is, a curve and its twist might not be isomorphic over $k$, but there is an isomorphism defined over some extension of $k$.
For instance, consider the elliptic curves $E_1 : y^2 = x^3 - 1$ and $E_2: y^2 = x^3 + 1$. One can show that these elliptic curves are not isomorphic over $\mathbb{Q}$. However, there is an easy isomorphism over $\mathbb{Q}(i)$:
\begin{align*}
E_1 &\to E_2\\
(x,y) &\mapsto (-x,iy) \, .
\end{align*}
(Indeed, I cooked up this example by forming the quadratic twist of $E_1$ by $-1$.)
