Let $K$ be a field. Consider two elliptic curves $E,E^{\prime}$ over $K$ isomorphic over $K^{\textrm{sep}}$ or equivalently having the same $j$-invariant. Suppose there exists a non-zero $K$-isogeny $E^{\prime}\to E$. Is it always true that $E$ is isomorphic to $E^{\prime}$ over $K$? I am particularly interested in the case when $K$ is a number field.
The statement turned out to be false for curves with complex multiplication though it is true for non CM-curves.
Best Answer
If $E'$ doesn't have CM then yes. Let $f:E'\to E$ be the isogeny, since $j(E)=j(E')$ there is an isomorphism $g:E\to E'$ defined over $K^{sep}$ so that $g\circ f$ is an endomorphism of $E'$, which must be $[n]$. The $Gal(\overline{K}/K)$ Galois action commutes with $f$ and $[n]$ so it must commute with $g$ which is defined over $K$.
If $E'$ has CM then it can fail. Let $$E'/\Bbb{Q}:y^2=x^3+x$$ then $$[1+i](x,y)=(x,y)+(-x,iy)$$ is an endomorphism defined over $\Bbb{Q}(i)$ but whose kernel $$\{O,(0,0)\}$$ belongs to $E'(\Bbb{Q})$, and all the endomorphisms $[\pm 1\pm i]$ with the same kernel are defined over $\Bbb{Q}(i)$.
So $E=E'/\ker(1+i)$ is an elliptic curve defined over $\Bbb{Q}$ and we have an isogeny $E'\to E$ defined over $\Bbb{Q}$,
$j(E')=j(E)$ since $[1+i]$ is an endomorphism,
but $E$ can't be isomorphic to $E'$ over $\Bbb{Q}$,
as composing that isomorphism $E\to E'$ with the isogeny $E'\to E$ would give an endomorphism of $E'$ defined over $\Bbb{Q}$ and with kernel $\ker(1+i)$.
Edit: $E$ has the equation $Y^2=X^3-4X$, this is because $(x,y)+(0,0)=(1/x,-y/x^2)$
so $\Bbb{Q}(E)=Tr_{\Bbb{Q}(E')/\Bbb{Q}(E)}(\Bbb{Q}(E'))=\Bbb{Q}( x+1/x, y-y/x^2)$
and the isogeny $E'\to E$ is $(x,y)\to (x+1/x,y-y/x^2)$